Physical Chemistry Quantum Chemistry and Molecular Interactions 1st Edition Andrew Cooksy Solutions by Ace Test Banks

Physical Chemistry Quantum Mechanics, Spectroscopy, and Molecular Interactions Solutions Manual by Andrew Cooksy

February 4, 2014

Contents Contents

Objectives Review Questions

Chapter Problems Notes on Maple and Mathematica commands . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11 11 11 23 39 63 95 134 154 184 205 229 249 263 274 282

Sources and Acknowledgments

293

Bibliography

297

Objectives Review Questions Chapter 1 1.1 We use Eq. 1.1 to convert wavelength to frequency (taking advantage of the constant speed of light c) and we use Eq. 1.2 to convert frequency to photon energy: ν=

c 2.998 · 108 m s−1 = = 3.0 · 1011 s−1 λ 1.0 · 10−3 m

by Eq. 1.1

Ephoton = hν = (6.626 · 10−34 J s)(3.0 · 1011 s−1 ) = 2.0 · 10−22 J.

Eq. 1.2

1.2 The de Broglie wavelength (Eq. 1.3) is our measure of the degree of quantum character in our system. We calculate λdB and compare it to the domain to determine if we need quantum mechanics to describe the physics. In this case, to find the de Broglie wavelength we need to calculate the momentum p from the kinetic energy, but we can do that: p2 mv 2 = 2mp √2 p = 2mK = 2(1.008 amu)(1.661 · 10−27 kg amu−1 )(4.0 · 10−21 J) = 3.66 · 10−24 kg m s−1

λdB =

h 6.626 · 10−34 J s = 1.8 · 10−10 m = 1.8 Å. = p 3.66 · 10−24 kg m s−1

Because 1.8 · 10−10 m ≪ 1.0µm = 1.0 · 10−6 m, it is unlikely that quantum effects arising from this motion will be significant. 1.3 The atom is in an n = 2 state, and we can use the Bohr model of the atom to calculate the correct values of the energies. From Eq. 1.15 we can calculate the total energy, and from Eq. 3.7 we can calculate the potential energy. The question does not specify units, and the most convenient units for the total energy are Eh : Z2 22 En = − 2 Eh = − Eh = −0.5 Eh. 2n 2(22 ) The potential energy depends on the radius of the electron orbit in the Bohr model, rn =

22 n2 a0 = a0 = 2a0 , Z 2

which gives us U =−

Ze2 2e2 e2 =− =− = −1.00 Eh. 4πǫ0 r (4πǫ0 )(2a0 ) (4πǫ0 )a0 )

Chapter 2 1

2.1 We apply the operator to the function, and see if we can find the original function again afterward: 1 1 d 3x e2x = 3e2x + 3x(2e2x ) α̂f (x) = x dx x 1 2x 2 1 2x = 3e + 2 · 3e = f (x). + x x2 x The result is equal to f (x) times another function of x, so this is not an eigenvalue equation. 2 1 d 2x2 1 = 3e 3(4x)e2x x dx x 2 = 4 · 3e2x = 4f (x).

α̂g(x) =

But g(x) is an eigenfunction of α̂, because the result is the original function g(x) times the eigenvalue 4. 2.2 We use the average value theorem, Eq. 2.10, integrating between 0 and a. The integral over x4 sin(cx) can be found using a symbolic math program. Setting c = 2π/a for now to simplify the notation, we have: Z Z a 2 a 2 sin (cx) x4 dx = 0.176a4. ψ ∗ (x4 ) ψ dx = a 0 o 2.3 To write the Schrödinger equation we need the Hamiltonian, which consists of the kinetic energy operator −(h̄2 /2m)∂ 2 /∂x2 , and the potential energy function described in the problem. In this case, the potential energy is given by the formula for a line, to which we assign a slope U0 . We can also add a constant, but it will have no effect on the relative energies or the wavefunctions, so we may as well set it equal to zero. Our potential energy function therefore is U0 x, and the Schrödinger equation becomes h̄2 ∂ 2 + U x ψ = Eψ. − 0 2m ∂x2 2.4 We use Eq. 2.41 to calculate the energy, with a mass mp and the volume given: h2 (n2 + n2y + n2z ) 8mV 2/3 x (6.626 · 10−34 J s)2 E100,1,1 = (1002 + 12 + 12 ) 8(1.673 · 10−27 kg)(1.0 · 10−18 m3 )2/3

Enx ,ny ,nz =

Eq. 2.41

= 3.28 · 10−25 J.

Chapter 3 3.1 We combine the radial and angular parts of the wavefunction as dictated by the quantum numbers, and also substitute Z = 3 for lithium: ψ3,1,−1 (r, θ, φ) = R3,1 (r) Y1−1 (θ, φ) r √ 3/2 r 3r 3 3 4 2 −r/a0 1− e sin θ e−iφ . = √ a a 2a 8π 27 3 0 0 0 3.2 We are using an integral to find an average value, so we use the average value theorem (Eq. 2.10), where the operator is r (the distance from the nucleus) and the wavefunction is given by ψ3,1,−1 (r, θφ) with Z = 3 for lithium: 3 Z ∞ 2 2 32 3 r 3r 1 − e−2r/a0 r3 dr = 25a0 /3. 3 7 a0 a 2a 0 0 0 2

3.3 The number of angular nodes is given by l, which is 1 for a p orbital, and the number of radial nodes is equal to n − l − 1 = 3 − 1 − 1 = 1: 1 angular node, 1 radial node. Chapter 4 4.1 This ion has three electrons and an atomic number Z = 4. We need one kinetic energy term for each electron, three terms for the attraction of each electron for the nucleus, and then three terms for electron–electron repulsions, one for each distinct pairing of the electrons: 1 and 2, 2 and 3, and 1 and 3:

h̄2 4e2 4e2 e2 4e2 − − − + ∇(1)2 + ∇(2)2 + ∇(3)2 − 2me 4πǫ0 r1 4πǫ0 r2 4πǫ0 r3 4πǫ0

1 1 1 + + r12 r23 r13

4.2 Neutral beryllium has 4 electrons, so Be+ has 3 electrons, which we place in the lowest energy subshells 1s and 2s for an electron configuration 1s2 2s1 . The zero-order energy is then the sum of the energies we would calculate if each electron were alone in that subshell. That one-electron energy is −Z 2 Eh /(2n2 ). We have two n = 1 electrons and one n = 2 electron in the configuration, and Z = 4 for Be, so we arrive at 42 1 1 1 E0 = − Eh + + 2 12 12 22 16 9 = −18 Eh. = 2 4 4.3 We use Eq. 4.30, which calculates the effective atomic number by treating the electron as though it were a single electron in an atom with a variable atomic number: 1/2 1/2 2ǫ(i)n2 2(0.182) Eh(32 ) Zeff = − = = 1.81. Eh Eh 4.4 We reverse the labels 1 and 2 in the function and then check to see whether the function has changed sign: Then we find that P̂21 ψ(1, 2) = cos(−x2 ) cos(y1 ) − cos(−x1 ) cos(y2 ) = − cos(−x1 ) cos(y2 ) + cos(−x2 ) cos(y1 ) = −ψ(1, 2). Therefore, the function is antisymmetric. 4.5 According to the arrow diagrams, we have 1s1

ml = 0 ↑ ↑ ↑ ↑ ↑ ↑ ↓ ↓ ↓ ↓ ↓ ↓

ml 0 0 0 0 0 0 0 0 0 0 0 0

ms +1/2 2p1 +1/2 +1/2 +1/2 +1/2 +1/2 −1/2 −1/2 −1/2 −1/2 −1/2 −1/2

ml = −1 ↑ ↓ ↑ ↓

0 ↑ ↓ ↑ ↓

↑ ↓ ↑ ↓

ml −1 0 +1 −1 0 +1 −1 0 +1 −1 0 +1

ms +1/2 +1/2 +1/2 −1/2 −1/2 −1/2 +1/2 +1/2 +1/2 −1/2 −1/2 −1/2

ML −1 0 +1 −1 0 +1 −1 0 +1 −1 0 +1

MS +1 +1 +1 0 0 0 0 0 0 −1 −1 −1

P P 3 P 3 P 3 P 3 P 1 P 1 P 1 P 3 P 3 P 3 P 3

where we found maximum values of L = 1 and S = 1 initially (based on the largest values of ML and MS ), and then after assigning the 9 3 P states, we were left with three MS = 0 states, which gave L = 1, S = 0 for the 1 P term. Breaking the 3 P into its component J values from L−S = 0 to L+S = 2, and ordering according to Hund’s rules, the final list of states is 3 P0 , 3 P1 , 3 P2 , 1 P1 . Chapter 5 5.1 This molecular ion has 2 nuclei with ZLi = 3 and ZH = 1 and 3 electrons, for a total of 5 particles. That means we should have 5 kinetic energy terms (one for each particle), 6 electron–nucleus attraction terms, 1 nucleus–nucleus repulsion term, and 3 electron–electron repulsion terms. (The total number of potential energy terms for N particles is always N (N − 1)/2, in this case 5 · 4/2 = 10, which gives the number of distinct pairs of particles. There is a potential energy term for each pair of interacting particles.) Using the standard form of the kinetic energy operator for each particle and the Coulomb potential for each pair of particles, we end up with the following: h̄2 1 3 1 3 1 e2 3 Ĥ = − − − − − − ∇(1)2 + ∇(2)2 + ∇(3)2 + − 2me 4πǫ0 rLi1 rH1 rLi2 rH2 rLi3 rH3 h̄2 1 1 3 h̄2 1 − + + + ∇(Li)2 − ∇(H)2 . + r12 r23 r13 RAB 2mLi 2mH 5.2 The orbital we’re constructing combines an s orbital (spherical) with a p orbital lying along the bond axis. If we keep the same orientation of nuclei A and B with respect to the z axis direction that is used elsewhere in the chapter, then the s and p orbitals have the same phase where they overlap, so we will get constructive interference between the two nuclei. However, we expect a node (where the new wavefunction will change sign) somewhere to the +z side of nucleus B, where the negative phase of the p orbital cancels the positive phase of the exponentially decaying s orbital.

5.3 The problem describes a curve such as Fig. 5.14, but a with minimum at R = 1.5 Å where the potential energy reaches a value U = −200 kJ mol−1.

E (kJ mol −1)

1.5A 0

−200

5.4 We can deduce from the orientation of the orbitals that (i) only the s, px , and py atomic orbitals are involved (because the orbitals lie in the xy plane) and (ii) orbital 1 consists of only s and px character (because it points along the x axis). All of the original px orbital density must be distributed somewhere among all three hybrid orbitals, so if we increase the amount of px in orbital 1, then the px character of orbitals 2 and 3 must decrease. The px orbital character tends to elongate the hybrid orbital along the x axis. By removing that character from orbitals 2 and 3, we elongate them more along the y axis instead, which will increase the angle between orbitals 2 and 3. (That angle approaches 180◦ in the limit that only s and py character remains, because then you have an sp hybrid, rather than an sp2 4

hybrid.) As the angle between orbitals 2 and 3 diverges, the angles between 1 and 2 and between 1 and 3 decrease. 5.5 For the proton NMR, we have two chemical shifts, one at about δ = 3.0 for the protons adjacent to the Br and another at about δ = 3.5 for the protons adjacent to the Cl. Each of these is then split into a triplet by the interaction with the two protons at the neighboring carbon atom. The actual chemical shifts turn out to be shifted downfield (to higher δ) because we have two electronegative atoms relatively close together: 1 H: 2 triplets, equal intensity, δ ≈ 3.55, 3.75. For the 13 C, we get two chemical shifts (one for each carbon atom). The J-coupling to the protons is usually very large and potentially a source of confusion, so 13 C spectra are usually purposely gathered under conditions that eliminate the J-coupling, resulting in singlets: 13 C: 2 singlets, equal intensity, δ ≈ 25, 35. Chapter 6 6.1 For this one, we can use the mathematical approach: ˆ Iˆσ̂xy ψ(x, y, z) = Iψ(x, y, −z) = ψ(−x, −y, z) = Ĉ2 (z)ψ(x, y, z). Therefore, Iˆσ̂xy = Ĉ2 (z). 6.2

H C

C H

This molecule has a Ĉ2 symmetry axis along the C C bond and two vertical mirror planes that contain that bond. Those and the identity are the only symmetry elements, so the point group is C2v and the symmetry elements are Ê, Ĉ2 , σ̂xz , σ̂yz . 6.3 1,1-Dibromoethene is in the point group C2v . We evaluate the results of the direct product Γi ⊗ Γµ , where Γµ may be any of A1 , B1 , and B2 for electric dipole transitions (because these correspond to the functions x, y, and z), or A1 , A2 , B1 , and B2 for Raman transitions (because each of these corresponds to some quadratic function such as x2 or xy). Therefore, the possible upper states for an electric dipole transition would be 1 A1 , 1 B1 , 1 B2 , and for a Raman transition would be 1 A1 , 1 B1 , 1 B2 , 1 A2 . 6.4 The molecule is in the point group C2h . The π bond lies perpendicular to the plane of the nuclei— the xy plane. That π orbital must be symmetric under the Ĉ2 rotation of the molecule (which doesn’t change what lies above or below the xy plane) but changes sign under inversion and under reflection through the xy plane. The representation is therefore au . Chapter 7 7.1 The molecule has 7 electrons, but 4 of them are in the 1s core orbitals and are not expected to contribute to the bonding. In the ground state, we would put 2 of the remaining 3 electrons into the 2σg bonding orbital and the last electron in the 2σu antibonding orbital, predicting a bond order of (2 − 1)/2 = 1/2. 7.2 The molecule has 7 electrons, and we would predict the MO configuration 1σg2 1σu2 2σg2 2σu1 . 5

7.3 Because the superscripts are not the same, the spins of the two states are not the same and the transition is forbidden by the spin selection rule ∆S = 0. 7.4 The transition would be an emission transition, forbidden by the spin selection rule (because ∆S 6= 0). Therefore, the transition would occur by the process of phosphorescence. Chapter 8 8.1 According to Eq. 8.17, 1 Ev = v + ωe = (1450 cm−1)(2.5) = 3625 cm−1. 2 The wavefunction for the v = 2 state we can assemble from the normalization constant A2 , the Hermite 2 polynomial H2 (y), and the exponential function e−y /2 (Eq. 8.16):

kµ h̄2

1/8

1 √ 8 π

1/2

(4y 2 − 2)e−y /2

where y = (R − Re )(kµ/h̄2 )1/4 . 8.2 The spacing increases with E (because the walls are steeper than in the harmonic oscillator potential), but more slowly than particle in box (because the walls are not infinitely steep). The wavefunctions are similar to those in the harmonic oscillator, but with less variation in amplitude from the center to the walls (because the bottom of the well is flatter). And because the walls are steeper than in the harmonic oscillator, the tunneling does not extend as far.

8.3 The reduced mass µ of a diatomic is given by mA mB /(mA + mB ), and if mA = mB (i.e., any homonuclear diatomic), then µ = mA /2. For 39 K2 , where each atom has a mass of 39.098 amu, the reduced mass is µ=19.55 amu. To estimate the vibrational constant, we need a guess of the force constant. Choosing k ≈ 12 N m−1 , halfway between the values of 17 N m−1 for Na2 and 7 N m−1 for Cs2 in Table 8.2, we predict s r k ( N m−1 ) 12 −1 ωe ( cm ) = 130.28 ≈ (130.28) = 102 cm−1. µ ( amu) 19.55 8.4 The point group is D∞h . There are two equivalent C H bonds. In group theory, we consider all the equivalent bonds at the same time, so we can either have the two bonds move in phase to get the symmetric stretch, which has σg symmetry, or they can move exactly out of phase to get the antisymmetric stretch, which has symmetry σu . Checking the functions for whether these correspond to functions for an IR active mode (x, y, or z) or a Raman active mode (any quadratic function of x, y, and z), we obtain the following results, in summary: σg (Raman active), σu (IR active). Chapter 9 6

9.1 We use Eq. 9.5, combining the reduced mass of 7.55 amu and the equilibrium bond length to get Be ( cm−1 ) =

16.858 16.858 = 1.755 cm−1 . = 2 (7.55)(1.128)2 µ ( amu) Re ( Å)

9.2 From Table 9.2 we obtain the values A = B = 9.94 cm−1 and C = 6.30 cm−1, which corresponds to an oblate top with energy levels given by Eq. 9.22: Erot = Kc2 (C − B) + BJ(J + 1) = (12 )(6.30 − 9.94) + (9.94)(2)(3) cm−1 = 56.0 cm−1 .

9.3 The rotational constant will increase, according to Eq. 9.16, as the moment of inertia decreases. Of the three molecules, 12 C16 O has the lowest moment of inertia, because it has only two atoms (of the same masses as the atoms that make up the CO2 molecules). Between the two CO2 isotopologues, the 18 O-substituted molecule has the greater isotopic mass, so the greater moment of inertia. The ordering is therefore 12 C18 O2 , 12 C16 O2 , 12 C16 O. 9.4 We fold the values of the rotational constants from Table 9.1 into Eqs. 9.8 and 9.9, setting v = 1, to find the rotational energies for J = 2 and J = 3. The difference between those energies is the transition energy ∆E that we’re looking for: 1 αe = [20.9557 − (1.5) (0.798)] cm−1 = 19.759 cm−1 Bv=1 = Be − v + 2 Erot = Bv J(J + 1) − Dv [J(J + 1)]

Erot (J = 2) = (19.759 cm−1)(2)(3) − (2.15 · 10−3 cm−1 ) [(2)(3)]2

Erot (J = 3) = (19.759 cm−1)(3)(4) − (2.15 · 10−3 cm−1 ) [(3)(4)]

∆E = Erot (J = 3) − Erot (J = 2) = 118.32 cm−1 . Chapter 10 10.1 Repulsion applies to all examples, and is expected to be roughly proportional to e−aR , although other forms are used to model this (such as the R−12 repulsive term in the Lennard–Jones potential). (a) dispersion: U (R) ∝ αA αB /R6 (by Eq. 10.38). (b) H-bonding: U (R) ∝ µ2A µ2B /R6 (by Eq. 10.19), although a case can be made that hydrogen bonding is strong enough that the structures cannot be treated as freely rotating at typical temperatures, in which case the µA µB /R3 of Eq. 10.16 would be more appropriate. In at least one popular model potential used to predict biochemical structure, the distance-dependence of hydrogen bonding is given using either a Lennard–Jones 6-12 potential (as in our Eq. 10.44) or a 10-12 potential (in which the distance dependence is a much more quickly decaying R−10 [1]). (c) dipole–dipole, dipole–induced dipole: U (R) ∝ µ2A µ2B /R6 , µA αB /R6 (Eqs. 10.19, 10.23). The dipole–induced dipole is important here because CO is only very weakly polar, but has a relatively high polarizability. Note the important distinction between these two terms: polar means the positive and negative charges in the molecule are already well separated to create a permanent dipole moment, while polarizable means that an external electric field can easily separate the charges, whether or not they are already well separated. (d) dipole–dipole (non-rotating): U (R) ∝ µ2A αB /R3 (by Eq. 10.16). You could make a good case that dipole–induced dipole is important here as well. I have only left it off because the interaction between non-rotating dipoles, varying as R3 , will tend to be more important if only because it decays so much more slowly than the dipole–induced dipole interaction. ′

10.2 From Eq. 10.17, the interaction between a multipole of 2k charges and another of 2k charges has ′ a potential energy proportional to R−k−k −1 . A quadrupole is formed by an arrangement of 4 charges, for which k = 2, so the interaction between two quadrupoles will vary in proportion to R−2−2−1 = R−5 . 7

10.3 The dipole moment of HI is 0.45 D and the polarizability of N2 is 0.20 Å . Plugging these values and the distance of 3.90 Å into Eq. 10.23 yields the following: 4µ2A α (4πǫ0 )R6 4(0.45 D)2 (3.3356 · 10−30 C m/D)2 (0.20 · 10−30 m3 ) =− (1.113 · 10−10 C2 J−1 m−1 )(3.90 · 10−10 m)6

u2−2∗ (R) = −

= −4.60 · 10−24 J = −0.00277 kJ mol−1.

This interaction energy is quite weak, partly because the N2 polarizability is so small. 10.4 We can expect dispersion to be most important, because the bromine atoms are so large (and consequently have high polarizabilities). The molecule should also be quite polar, given the electronegativity of the Br atom, and so dipole–dipole interactions will also run high. The dipole–induced dipole attraction will also be present, but is usually weaker than the direct dipole–dipole interaction: b<a<c. 10.5 The well should extend from 3.61 Å to 5.42 Å and be 190 K (132 cm−1 ) deep, as in the following graph.

10.6 The degrees of freedom that couple most strongly are usually those that have the most similar energy spacing. In this case, there is a transfer of energy away from vibrations in molecule A. Rotational energy spacings tend to be at least two orders of magnitude smaller than the vibrational spacings, and electronic transitions are typically an order of magnitude greater. Therefore, the likeliest place for vibrational energy to go is into other vibrations. In this case, if A has lost vibrational energy, it is most likely to have gone into vibrational excitation of B. Chapter 11 11.1 The dispersion force binds CO2 molecules together in a cluster, because there is no monopole (ionic charge) or dipole moment to bind them. As the cluster adds more units, the average binding energy per molecule will generally decrease when the cluster is very small. For small clusters, removing one unit reduces the overall binding significantly. But the effect depends on the cluster and the cluster size. For larger clusters, the loss of one unit has little effect on the remaining cluster. We expect the molecules at the surface of the cluster to be the most weakly bound, because they are not completely surrounded by the stabilizing neighbors. As the cluster size increases, the ratio of the volume (proportional to the cluster size N ) to the surface area (proportional to N 2/3 ) steadily increases, assuming the cluster shape remains roughly spherical, so on average for large clusters the binding energy may increase slightly with N . 11.2 The weak bonding interactions will be modeled by the van der Waals term (which includes dispersion forces) and the electrostatic potential energy term (which accounts for charge-charge interactions across distances longer than the typical chemical bond): UvdW and Uelectrostatic . 8

Chapter 12 12.1 The pair correlation function for this system should resemble the curves shown for water in Figs. 12.3g–i. The oscillations should become smoother as we approach the boiling point. low temp

near boiling point

12.2 Yes. Molecules of the two substances will attract one another, principally through dispersion. Chapter 13 13.1 The unit cell has two square faces opposite one another, and the rest of the faces are equivalent rectangles. There is one Ĉ4 principal rotation axis, a horizontal mirror plane, and four vertical mirror planes. The crystallographic point group is therefore D4h . 13.2 No. The reason is that no charge separation can exist in a regular monatomic crystal, and polarity is one of the requirements for piezoelectricity.

Chapter Problems Notes on Maple and Mathematica commands Several solutions in this manual have added notes on how to set up expressions under the symbolic mathematics programs Maple (Waterloo Maple, Inc.) and Mathematica (Wolfram Research) to assist in solving the problem. Please note, however, that a symbolic math program is useful only after we have overcome the conceptual challenges of the problem. Where these programs have their greatest use is when we need to solve simultaneous or transcendental equations, or obtain numerical values or algebraic expressions for integrals. We still have to know how to set up an integral, and we have to decide where and how to apply approximations (which turns out to be more important than one might think). Once we have taken these steps, the program may be able to take over. Even then, however, the math in the majority of our problems is limited to relatively straightforward algebraic manipulations, however complex the concepts may be. For most of the cases where the Maple and Mathematica syntax is not shown, little would differ from the written mathematics already appearing in the solution. This is largely the goal of symbolic math programs in the first place: to accept input in the form that one would write the problem on paper. The following problems in this volume give specific commands for use with Maple or Mathematica: A.10 A.18 3.29 3.31

1.9 1.20 2.3 2.4 2.10 2.35 3.47 4.26 9.19

Chapter A A.1 This problem uses a common manipulation, one of the features of logarithms that makes them so useful: pKa = − log10 Ka = − log10 e−∆G/(RT ) ∆G =− − log10 e RT ∆G = (0.434). RT

log xa = a log x

This shows, if you don’t mind us getting ahead of ourselves a little, that the pKa is directly proportional to the free energy of dissociation, ∆G, and inversely proportional to the temperature, T . A.2 The idea here is that, even if we think at first we have no idea what the number ought to be, a closer look at the available choices makes it clear that we can spot some potentially ridiculous answers: a. 2 · 1010 m s−1 is faster than the speed of light. 11

b. 2 · 105 m s−1 has no obvious objections. c. 2 m s−1 is the speed of a slow walk, and would imply, for example, that you could send an e-mail message over a cable connection to a friend half a mile away, and then run the half-mile to arrive and deliver the message in person before the e-mail finishes traveling through the wires. When we have calculations that toss around factors of 10−34 , for one example, this is a significant skill. The correct answer is 2 · 105 m s−1 . 3

A.3 The volume is roughly 125 Å , which we can show is not big enough to hold more than about 15 atoms. Chemical bonds, formed between overlapping atoms, are roughly 1 Å long, and so typical atomic 3 diameters are roughly 2 Å or more, and occupy a volume on the order of (2 Å)3 = 8 Å . A volume of 3 125 Å , therefore, cannot hold more than about 125/8 = 15.6 atoms. Among the choices, the only reasonable value is 8. A.4

a. Chemical bond lengths in molecules are always in the range 0.6–4.0 Å, or 0.6 · 10−10 to 4.0 · 10−10 m. 25 · 10−8 m is much too large for a bond length. no b. Six carbon atoms have a mass of 6 · 12 = 72 amu. With the added mass of a few hydrogen atoms at 1 amu each, 78 amu is a reasonable value. yes

A.5

a. The derivatives d[A] and dt have the same units as the parameters [A] and t, respectively. Both sides of the equation should therefore have units of mol L−1 s−1 . That means that k needs to provide the units of s−1 and cancel one factor of concentration units on the righthand side. k has units of L s−1 mol−1 . b. The argument of the exponential function must be unitless, so kB must cancel units of energy (J) in the numerator and temperature (K) in the denominator. The correct units are J K−1 . c. The units all cancel, and Keq is unitless. d. Squaring both sides of the equation, we can solve for k: µω 2 = k. k must therefore have units of kg s−2 . 2

A.6 There are two factors on the lefthand side, (2x+ 1)2 and e−ax . For the product to be zero, at least one of these factors must be zero. If (2x + 1) = 0, then All three are valid solutions.

x = − 12 . If e−ax = 0, then

x→ ± ∞.

A.7 In general, for any complex number (a + ib), the complex conjugate is (a + b)∗ = a − ib. We look for the imaginary component and and invert its sign: a. x − iy : a = x b = −y, x + iy. b. ix2 y 2 : a = 0 b = x2 y 2 ,

−ix2 y 2 .

c. xy(x + iy + z) : a = x2 y + xyz d. a = x/z

b = y/z,

b = xy 2 , x2 y + xyz − ixy 2 ,

xy(x − iy + z).

(x − iy)/z .

e. eix = 1 + ix − x2 − ix3 + x4 + ix5 − . . . a = 1 − x2 + x4 − . . .

b = x − x3 + x5 − . . .

a − ib = 1 − ix − x2 + ix3 + x4 − ix5 − . . . = e−ix . 12

f. 54.3: a = 54.3 b = 0, 54.3. A.8 This problem tests a few algebraic operations involving vectors, particularly useful to know when we look at angular momentum and (often related) magnetic field effects. √ ~ = 02 + 22 + 12 = a. The length of a vector is calculated using the Pythagorean theorem: |C| √ 5. ~ +B ~ = (1 + 1, 0 + 0, 0 + 1) = (2, 0, 1). b. We add vectors one coordinate at a time: A c. The dot product of two vectors multiplies the values for each coordinate of the two vectors and ~ · B ~ = (1 · 1) + (0 · 0) + (0 · 1) = 1. sums the results: A ~ · C ~ = (1 · 0) + (0 · 2) + (0 · 1) = 0. d. In the case of perpendicular vectors, this gives us zero: A e. The cross product involves a little more work, and yields a new vector, perpendicular to the two ~ × B ~ = (0 · 1 − 0 · 0, 0 · 1 − 1 · 1, 1 · 0 − 0 · 1) = (0, −1, 0). original vectors: A A.9 If we accept that the Taylor series expansion is exact if we take it to infinite order, then the Euler formula can be proven by the expansions of ex (Eq. A.25), sin x (Eq. A.26), and cos x (Eq. A.27): eix =

∞ X 1 (ix)n n! n=0

1 i 5 i x − ... = 1 + ix − 12 x2 − x3 + x4 + 6 24 120 1 1 5 1 = (1 − 12 x2 + x4 − . . .) + i(x − x3 + x − . . .) 24 6 120 = cos x + i sin x . This equation is of practical importance to us, and is famous among mathematicians for tying together three fundamental mathematical values—π, i, and e—in one equation: eiπ = 1. A.10

• Maple: We can have Maple solve the equation P − Va2 (Vm − b) m

directly using the solve command. After checking that all of the units are indeed consistent, enter the Maple command solve((1.000-(3.716/Vˆ2))*(V-0.0408)/(0.083145*298.15)=1,V); The resulting solution, 24.98, is in the same units as b, so our final value to three significant digits is 25.0 L mol−1 . • Mathematica: In Mathematica we use the Solve command to find the value of a variable in an algebraic equation: Solve[(1.000 - (3.716/Vˆ2))*(V - 0.0408)/(0.083145*298.15) == 1, V] Note that we need to use a double equals sign in the command. This command gives 24.979, which rounds to 25.0 L mol−1 . 13

• Successive approximation: There are several ways to solve this, corresponding to different forms of the equation that leaves Vm on one side. One way to set up the equation quickly is to recognize that (Vm − b) will vary rapidly compared to P − (a/Vm2 ), so we can isolate Vm as follows: P − Va2 (Vm − b) m =1 RT a P − 2 (Vm − b) = RT Vm RT Vm − b = P − Va2 m

Vm =

P − Va2

+ b.

Substituting in the values for P , a, b, R, and T (making sure that the units are all compatible), we can reduce the equation to the following: Vm ( L mol−1 ) =

24.790 (0.083145)(298.15) + 0.0408 = + 0.0408. 1 − 3.716 1 − 3.716 V2 V2 m

Guessing an initial value of 1 L mol

−1

yields the following series of approximations: 24.790 + 0.0408 = −9.0864 Vm = 1 + 3.716 12 24.790 Vm = 3.716 + 0.0408 = 25.999 1 + 5.330 2 24.790 Vm = 3.716 + 0.0408 = 24.967 1 + 22.099 2 24.790 Vm = 3.716 + 0.0408 = 24.979 1 + 24.796 2 24.790 Vm = 3.716 + 0.0408 = 24.979. 1 + 24.834 2

The series has converged to the three significant digits requested. The final value for Vm is 25.0 L mol−1 . A.11 Here we apply the rules of differentiation summarized in Table A.3. a. f (x) = (x + 1)1/2 df = 12 (x + 1)−1/2 . dx b. f (x) = [x/(x + 1)]1/2 −1/2 x df 1 dx 1 d 1 = 2 +x dx x+1 x + 1 dx dx x + 1 −1/2 x 1 x . − = 21 x+1 x + 1 (x + 1)2 14

c. i d h df x1/2 = exp x1/2 dx hdx i = 21 x−1/2 exp x1/2 .

d df = exp cos x2 (cos x2 ) dx dx d = exp cos x2 (− sin x2 ) (x2 ) dx = −2x sin x2 exp cos x2 .

A.12 This problem tests our ability to use a few of the analytic integration results given in Table A.5. a.

R∞

1 e−ax dx = − a1 e−ax |∞ 0 = − a (0 − 1) =

x2 dx = 13 x3 |51 = 31 (125 − 1) =

x−3/2 dx = −2x−1/2 |51 =

1 . a

124 . 3

1 −2( √ − 1). 5

R 2π Rπ 2π π d. r2 0 dφ 0 sin θdθ = r2 (φ)|0 (− cos θ)|0 = r2 (2π − 0)[−(−1) − (−1)] = 4πr2 .

A.13 We use the Coulomb force law, Eq. A.41, using the charge of the electron −e for both charges and r12 set to 1.00 Å: FCoulomb = =

e2 4πǫ0 r2 (1.602 · 10−19 C)2

= −1 (1.113 · 10−10 C2 J−1 m−1 )(1.00 Å)2 (10−10 m Å )2

2.31 · 10−8 N .

A.14 This problem relies on the definitions of the linear momentum p and the kinetic energy K (Eq. A.36): p = mv p2 . 2m

K = 21 mv 2 =

A.15 We’re calling the altitude r. Because the acceleration is downward but r increases in the upward direction, the acceleration is negative: −9.80 m s−2 . We invoke the relationship between force and the potential energy, and find that we have to solve an integral: Z r Z r U (r) = − F (r′ ) dr′ = − (−mg) dr′ = mgr. 0

A.16 |FCoulomb | =

(1.602 · 10−19 C)2 e2 = −1 2 4πǫ0 r (1.113 · 10−10 C2 J−1 m−1 )(0.529 Å)2 (10−10 m Å )2 = 8.23 · 10−8 N

|Fgravity | = mH g

= (1.008 amu)(1.661 · 10−27 kg amu−1 ) 9.80 m s2

= 1.64 · 10−26 N.

Sure enough, the gravitational force is smaller than the Coulomb force by orders of magnitude, and the motions of these particles will be dictated—as well as we can measure them—exclusively by the Coulomb force. A.17 We are proving an equation that depends on L and x and t and vx , which may look like too many variables. If we use the definition of L to put this equation in terms of K and U , then we can at least put K in terms of speed. Then, because speed itself is a function of position and time, the number of variables is quite manageable. Nonetheless, keeping things in terms of K and U is useful, because of their straightforward dependence on only v and x, respectively. To prove the equation, we could try working from both sides and seeing if the results meet in the middle. First the lefthand side: ∂K ∂U ∂L − = ∂x ∂x ∂x |{z}

K not a function of x

Fx = −dU/dx

= Fx = ma 2

d x dt2

acceleration = d2 x/dt2

Next the righthand side: d ∂L d 1 ∂vx2 ∂U = − m dt ∂vx dt 2 ∂vx ∂vx   =

 d  mvx − ∂U   dt ∂vx  |{z}

U not a function of vx

dvx d2 x =m =m 2. dt dt

And there we are. One of the useful features of the Lagrangian is that the equation proved here can be made to hold for different choices of coordinates. This enables the mechanics problems to be written in coordinates that take advantage of symmetry (for example, if the only force is a radial one, attracting or repelling particles from a single point), and the Lagrangian then provides a starting point to develop relationships between the positions and velocities of the particles. A.18 The overall energy before the collision is the sum of the two kinetic energies: K = 21 m1 v12 + 21 m2 v22 , 16

and this must equal the energy after the collision: 2

K = 21 m1 v1′ + 21 m2 v2′ . Similarly, we may set the expressions for the linear momentum before and after the collision equal to each other: p = m1 v1 + m2 v2 = m1 v1′ + m2 v2′ . So there are two equations and two unknowns. At this point, the problem is ready to solve with a symbolic math program. Maple. The problem can be solved in a single step by asking Maple to solve the conservation of energy and conservation of momentum equations simultaneously to get the final speeds (here vf [1] and vf [2] in terms of the masses and initial speeds: solve({m[1]*v[1]+m[2]*v[2] = m[1]*vf[1]+m[2]*vf[2], (1/2) * m[1] * v[1]ˆ2+(1/2) * m[2] * v[2]ˆ2 = (1/2) * m[1] * vf[1]ˆ2+(1/2) * m[2]*vf[2]ˆ2}, [vf[1], vf[2]]); Mathematica. We are looking for an algebraic expression, rather than a numerical value, in the solution to this problem. We can use the Solve command for either case: Solve[m1*v1 + m2*v2 == m1*vf1 + m2*vf2 && (1/2)*m1*v1ˆ2 + (1/2)*m2*v2ˆ2 == (1/2)*m1*vf1ˆ2 + (1/2)*m2*vf2ˆ2, vf1, vf2] On paper. This last equation lets us eliminate one variable by writing, for example, the final speed v2′ in terms of v1′ : m1 v1 + m2 v2 − m1 v1′ . v2′ = m2 Now we can put this value into the equation for K, and solve for v1′ : K = 12 m1 v12 + 21 m2 v22

(a)

2 2 = 12 m1 v1′ + 21 m2 v2′ 2 = 12 m1 v1′ + 21 m2

m1 v1 + m2 v2 − m1 v1′ m2

This is going to be an equation that depends on v1′ and v1′ , so we can solve it using the quadratic formula. In that case, it’s easiest to put all the quantities on one side of the equation:

2 m1 v1 + m2 v2 − m1 v1′ − 12 m1 v12 + 21 m2 v22 m2   2 m21 v12 + m22 v22 + m21 v1′    +2m1 m2 v1 v2 − 2m21 v1 v1′ − 2m1 m2 v1′ v2  2  = 21 m1 v1′ + 21 m2    m22   2

0 = 12 m1 v1′ + 21 m2

subtract (a) above

expand the square

− 12 m1 v12 − 21 m2 v22 2

= m1 v1′ +

m21 2 m2 2 v1 + m2 v22 + 1 v1′ + 2m1 v1 v2 m2 m2

divide by 1/2

m21 v1 v1′ − 2m1 v1′ v2 − m1 v12 − m2 v22 m2 m21 m21 ′ ′2 + v1 −2 v1 − 2m1 v2 = v1 m1 + m2 m2

−2

group by power of v1′ 17

m21 2 v1 + m2 v22 + 2m1 v1 v2 − m1 v12 − m2 v22 m2 m2 m2 2 m1 + 1 + v1′ −2 1 v1 − 2m1 v2 = v1′ m2 m2 2 m1 + − m1 v12 + 2m1 v1 v2 m2 −1 m21 m21 ′ quadratic formula 2 v1 + 2m1 v2 v1 = 2m1 + 2 m2 m2  " #1/2  2 2 m21 m1 m21 2 ± 2 . v1 + 2m1 v2 − 4 m1 + − m1 v1 + 2m1 v1 v2  m2 m2 m2 +

To deal with this equation, we can expand the multiplication inside the square brackets: 2 m2 m3 m4 2 1 v1 + 2m1 v2 = 4 21 v12 + 8 1 v1 v2 + 4m21 v22 m2 m2 m2 2 3 2 m m m4 m3 m1 −4 m1 + 1 − m1 v12 + 2m1 v1 v2 = −4 1 v12 − 4 21 v12 + 4m21 v12 + 4 1 v12 m2 m2 m2 m2 m2 3 m − 8m21 v1 v2 − 8 1 v1 v2 . m2

Nearly all of these terms cancel when we add these two expressions together, leaving: 4m21 v22 + 4m21 v12 − 8m21 v1 v2 . In the quadratic equation, we have to take the square root of this, but that turns out to be easy: 1/2 1/2 2 2 = 2m1 v22 + v12 − 2v1 v2 4m1 v2 + 4m21 v12 − 8m21 v1 v2 = 2m1 (v2 − v1 ) .

Finally, putting this back into our equation for v1′ , we get −1 m2 m2 2 1 v1 + 2m1 v2 ± 2m1 (v2 − v1 ) 2m1 + 2 1 m2 m2 −1 m1 m1 = 1+ v1 + v2 ± (v2 − v1 ) . m2 m2

v1′ =

divide out 2m1

This is correct as far as it goes, but we have two solutions, corresponding to either the + or − sign. If we use the − sign, then we get v1′ =

−1 m1 m1 v1 + v2 − v2 + v1 = v1 . 1+ m2 m2

This is the solution if the collision doesn’t occur; particle 1 just keeps moving at the same speed as 18

before. The + sign gives us the correct solution: −1 m1 m1 1+ v1 + v2 + v2 − v1 m2 m2 −1 m1 m1 − 1 v1 + 2v2 = 1+ m2 m2 1 = [(m1 − m2 ) v1 + 2m2 v2 ] . m1 + m2

v1′ =

We can now use the conservation of momentum to solve for v2′ . I’m going to factor out a 1/(m1 + m2 ) to get an equation similar to the one for v1′ : m1 v1 + m2 v2 − m1 v1′ m2 m1 1 m1 v1 + m2 v2 − [(m1 − m2 ) v1 + 2m2 v2 ] = m2 m1 + m2 m1 m1 − m2 m1 m1 = v1 + v2 − v1 − 2 v2 m2 m1 + m2 m2 m1 + m2 m1 (m1 + m2 ) m1 (m1 − m2 ) 1 v1 − 2m1 v2 v1 + (m1 + m2 )v2 − = m1 + m2 m2 m2 1 = [(m2 − m1 ) v2 + 2m1 v1 ] . m1 + m2

v2′ =

Because there is nothing in the problem that determines which particle is labeled 1 and which is labeled 2, the equations for v1′ and v2′ must be exactly the same, with all the labels 1 and 2 switched. If you haven’t seen this result or simply don’t remember it, it’s worthwhile to check a few values. For example, if the two particles have equal mass (m1 = m2 ), then the final speeds are v1′ = v2 and v2′ = v1 ; i.e., the particles simply exchange speeds. Another example: if particle 1 is initially at rest (v1 = 0), then it picks up a speed 2m2 v2 /(m1 + m2 ) from the collision. In that case, if particle 2 dominates the mass (m2 ≫ m1 ), then particle 1 will find itself with a final speed equal to 2v2 . In contrast, if particle 1 is much more massive than 2, then the collision will hardly affect it (v1′ ≈ 0) and particle 2 will simply reverse direction (v1′ ≈ −v2 ). Note that the two particles don’t have to be moving in opposite directions. If particle 1 is behind 2 but moving faster and in the same direction, then they will strike each other, and particle 2 will acquire particle 1’s higher speed. A.19 K =U =−

(1.602 · 10−19 C)2 e2 −18 = J −1 = 2.31 · 10 4πǫ0 r (1.113 · 10−10 C2 J−1 m−1 )(1.0 Å)(10−10 m Å )

L = |~r × p~| = rp, since ~r ⊥ ~ p. p 1/2 p = 2me K = 2(9.109 · 10−31 kg) · (2.31 · 10−18 J) = 2.05 · 10−24 kg m s−1

L = (1.0 Å)(10−10 m Å A.20

−1

)(2.05 · 10−24 kg m s−1 ) = 2.05 · 10−34 kg m2 s−1 . (0)

a. Find the center of mass positions ~ri at collision. Let’s call the center of mass of the entire system the origin. The particles have equal mass, so the origin will always lie exactly in 19

between the two particles. At the time of the collision, we may draw a right triangle for each particle, connecting the particle’s center of mass, the origin, and with the right angle resting on the z axis. The hypotenuse of the triangle connects the center of mass to the point of contact between the two particles, and must be of length d/2 (the radius of the particle). The other two sides are of length (d/2) cos θ (along the z axis) and (d/2) sin θ (along the x axis), based on the definitions of the sine and cosine functions in Eqs. A.5. These correspond to the magnitudes of the z and x coordinates, respectively, of the particle centers of mass at the collision. The signs of the values may be determined by inspection of the figure: at the time of the collision, x1 and z2 are positive while x2 and z1 are negative, so the position vectors are: (0)

~r1 = ((d/2) sin θ, 0, −(d/2) cos θ) (0)

~r2 = (−(d/2) sin θ, 0, (d/2) cos θ) . b. Find the velocities ~vi′ after collision. Simple collisions obey a simple reflection law: the angle of incidence is equal to the angle of reflection. These are the angles between the velocity vectors and the normal vector—the line at angle θ from the z axis. (This is the normal vector because it lies perpendicular to the plane that lies between the two spheres at the point of collision; this plane is effectively the surface of reflection for the collision.) Therefore, the velocity vector after the collision is at an angle 2θ from the z axis, and the velocity vectors after the collision are ~v1′ = v0 (sin 2θ, 0, − cos 2θ)

~v2′ = v0 (− sin 2θ, 0, cos 2θ). Notice that the speed after the collision is still v0 for each particle. Because they each began with the same magnitude of linear momentum, the momentum transfer that takes place only affects the trajectories. ~ is conserved before and after the collision. We now have position and velocity c. Show that L vectors before and after the collision: ~r1 = ((d/2) sin θ, 0, −(d/2) cos θ) + ~v1 t

~v1′′ = v0 (0, 0, 1) ~v1′ = v0 (sin 2θ, 0, − cos 2θ)

~r2 = (−(d/2) sin θ, 0, (d/2) cos θ) + ~v1 t ~v2′′ = v0 (0, 0, −1)

~v2′ = v0 (− sin 2θ, 0, cos 2θ).

~ for each particle, and we add these We take the cross products of these for each particle to get L together to get the total angular momentum for the system. Before the collision, ~r1′′ = ((d/2) sin θ, 0, −(d/2) cos θ) + v0 t(0, 0, 1)

~ ′′ = m~r′′ × ~v ′′ L 1 1 1

′′ ′′ ′′ = m y1′′ vz1 − z1′′ vy1 , z1′′ vx1 − x′′1 vz1 , x′′1 vy1 − y1′′ vx1

= m (0, −(dv0 /2) sin θ, 0)

~ ′′ : and similarly for L 2 ~ ′′ = m (0, −(dv0 /2) sin θ, 0) L 2 and combining these yields: ~ ′′ = L ~ ′′ + L ~ ′′ = −mdv0 (0, sin θ, 0). L 1 2 20

All of the position or velocity vectors have only zero y components, and therefore only the y component of the cross product survives. After the collision, ~r1′ = ((d/2) sin θ, 0, −(d/2) cos θ) + v0 t(sin 2θ, 0, − cos 2θ)

~ ′ = m~r′ × ~v ′ L 1 1 1

which has a y component ~ ′ = m {−(d/2) cos θ sin 2θ − v0 t cos 2θ sin 2θ − [(d/2) sin θ(− cos 2θ) + v0 t sin 2θ(− cos 2θ)]} L y1 ~′ : and similarly for L y2 ~ ′ = m {(d/2) cos θ(− sin 2θ) + v0 t cos 2θ(− sin 2θ) − [−(d/2) sin θ cos 2θ − (−v0 t sin 2θ) cos 2θ]} . L y1 Adding the two components together we find that all the t-dependent terms cancel, and trigonometric identities from Table A.2 simplify the rest: ~ ′2y ~ ′y = L ~ ′1y + L L mdv0 = [−2 cos θ sin 2θ + 2 sin θ cos 2θ] 2 sin 2θ = 2 sin θ cos θ cos 2θ = 2 cos2 θ − 1 ~ ′ = 2mdv0 − cos θ (2 sin θ cos θ) + sin θ 2 cos2 θ − 1 L y 2 = mdv0 −2 cos2 θ sin θ + 2 cos2 θ sin θ − sin θ = −mdv0 sin θ.

~ ′ , and the x and z components are again zero in the cross products, This is the y component of L ~ ′ and L ~ ′′ are equal to so we have shown that both L ~ = mdv0 (0, sin θ, 0). L If the particles hit head-on, then θ = 0 and the angular momentum is zero. As θ increases, L increases to a maximum value of mdv0 when the two particles just barely touch each other in passing. If we had used the conservation of L at the outset, we could have found this solution quickly. Because the angular momentum does not depend on the size of the particles, we can replace our two objects here with point masses. It won’t matter that they now won’t collide, because if L is conserved we have to get the same answer before the collision takes place anyway. In fact, because L is conserved, we can pick any point in time that’s convenient for us to calculate L, so I would pick the time when the two particles reach z = 0. At this time, both particles are traveling on trajectories that are exactly perpendicular to their position vectors (~vi is perpendicular to ~ri ). This makes the cross product for each particle easy to evaluate: ~ i = m~ri × ~vi = m(±(d/2) sin θ, 0, 0) × (0, 0, ±v0 ) = mdv0 /2(0, sin θ, 0), L where the minus sign applies to particle 2. There are two particles, so we multiply this vector by 2, ~ as above. arriving at the same L A.21

a. Write E~1 in vector form. The magnitude of the electric field generated by particle 1 is given by F = q2 E1 , and this force must be equal to the Coulomb force F = −q1 q2 /(4πǫ0 r2 ). The 21

force vector points along the axis separating the two particles, and we can include this directiondependence by multiplying the magnitude of the vector by ~r/r. The Cartesian form of the vector ~r from particle 1 to 2, just working off part (b) of the figure, may be written (rv1 /c, y2 , 0) and has length 1/2 rv1 2 2 r= + y2 . c Therefore, the force vector is

F~ =

q1 q2 4πǫ0 r2

~r q1 q2 (rv1 /c, y2 , 0) = r 4πǫ0 r3

and the electric field vector is F~ q1 E~1 = = (rv1 /c, y2 , 0). q2 4πǫ0 r3 ~ in vector form. Here we just have to be careful to correctly evaluate the cross product. b. Write B ~ = 12 E~1 × ~v1 , and we have an equation for E~2 already. The velocity We are using the equation B c vector consists only of an x-velocity component: ~v1 = (v1 , 0, 0). Notice that because these two vectors lie in the xy plane, their cross product—which is perpendicular to both vectors—will lie along the z axis. The z component of the cross product ~a × ~b is equal to ax by − ay bx , so we have q1 ~ = 1 E~1 × ~v1 = (0, 0, v1 y2 ) . B c2 4πǫ0 c2 r3 c. Find the magnetic force vector. Again, we take a cross product with the velocity. This time, ~ vector lies along z, and ~v1 lies along x, so the cross product lies along y: the B ~ = q1 q2 (0, v12 y2 , 0) . F~mag = q2~v1 × B 4πǫ0 c2 r3 d. Calculate the difference between the actual and classical values of the Coulomb force. To compute the actual Coulomb force, we use the distance r, so F~ has a magnitude q1 q2 . F = 4πǫ0 r2 The classical Coulomb force would be F′ =

q1 q2 , 4πǫ0 y22

and the difference between the two forces is q1 q2 1 1 F − F′ = . − 4πǫ0 r2 y22 We can simplify this by relating r2 and y 2 : rv 2 1 r2 = + y22 c v 2 −1 1 . r2 = y22 1 − c

So, finally, we have v 2 1 q1 q2 1 1 − 1 − 4πǫ0 y 2 c y22 q1 q2 v12 q1 q2 v1 2 =− = − . 2 4πǫ0 y2 c 4πǫ0 c2 y22

F − F′ =

In comparison, the magnitude of the magnetic force we calculated from the standard equations is Fmag = q1 q2 v12 y2 /(4πǫ0 c2 r3 ), and for v ≪ c, we can allow r ≈ y2 , so that Fmag = q1 q2 v12 /(4πǫ0 c2 y22 ). Magnetic forces are a natural result of the motion of electrical charge when special relativity is taken into account. It was this relationship between electric and magnetic forces that was the basis of Einstein’s original paper on special relativity.

Chapter 1 1.1 Radiation behaves more classically at large wavelengths, because the smaller frequency allows smaller energy increments to be absorbed or emitted by matter. Therefore, the energy of longwavelength radiation appears more like a continuous, rather than quantized, variable. Radiation can be expected to behave like a classical wave provided that its wavelength is long compared to the system of interest. For example, radiofrequency radiation can be directed by reflection and refraction more easily than the relatively particle-like x-ray and γ radiation. Therefore, a correspondence principle would suggest that radiation must be treated by quantum mechanics as the wavelength becomes small compared to the system of interest. (Furthermore, it turns out that photons do carry a momentum, described exactly by de Broglie’s equation p = h/λ, and this decidedly non-classical property is most apparent at short wavelengths.) 1.2 The quantized picture, while accurate in some aspects, is the misleading one. Radiofrequency photons, in particular, carry so little energy that single-photon detection at those wavelengths is quite challenging. Related to this is the fact that, in order to maintain a detectable signal, our transmitter must emit photons at an enormous rate. A tiny 100 W transmitter emits more than 1027 photons per second at a frequency of 100 MHz. And these are big photons, with wavelengths of a few meters. Even 10 km away from the source, nearly 1018 photons fly through each square meter of space every second. The numbers are so big that as we move away from the transmitter, the signal appears to drop continuously. Furthermore, as we get to very great distances, photons will continue to strike the source, but at a reduced rate. Although the photon energy itself is a discrete quantity, the rate at which the photons impinge on our antenna remains a continuous variable, and the signal deteriorates smoothly until we can no longer detect it. That said, experiments that detect single photons are not rare, but they are normally carried out with higher energy photons, such as in the visible or UV. 1.3 Although the electron beam leaving the source was incoherent (with random phases), the reflection off the surface sets a boundary condition where the phase goes to zero, in the same way that the wavefunction of our particle in a box goes to zero at the walls of the box. From that point on, the electrons reflected towards the detector are coherent, and the interference pattern becomes observable. 23

1.4 Mass and energy and charge must each be conserved, at least in chemistry, so the electrons cannot just disappear. Because they have wave-like character, the electrons are not constrained to a single trajectory: they can be detected at angles other than the normal reflection angle for particles. If some particular reflection angle corresponds to destructive interference—where no electron signal is detected – then there will be other angles at which an excess signal be detected. 1.5 An exact value of λdB requires an exact value of the momentum, which the uncertainty principle tells me is only possible for a particle with infinite uncertainty in position. So the answer is yes: if particle 1 has a momentum that is known precisely, then it effectively has an infinite size (it is possible for the particle to be detected anywhere). But this is true of both my measurement λdB in the laboratory and the student’s measurement in the reference frame of particle 2. In practice, if I have any idea where the particle is (so the position uncertainty is finite), then I can’t know exactly what the momentum is; in other words, I will not always get the same value for p when I measure it. The student on particle 2 will find the same thing—the measured momentum will cover a range of different values. The mean p values (and therefore the apparent de Broglie wavelengths) will be different in the two sets of measurements, but the student will always find that the mean value of λdB is finite. 1.6

a. The spectrum of the neutral helium atom. b. The wavefunction of the electron in the neutral hydrogen atom.

c. The second ionization energy of the helium atom. The Bohr model works only for one-electron atoms, and does not correctly predict the distribution of the electron. 1.7 This should be more energy than for the same transition in the H atom, because the greater nuclear charge on He+ makes the electron more tightly bound and harder to pull out to an excited state orbital. The He atom has two electrons, and the He+ ion has one electron. He+ therefore obeys the equations for a one-electron atom, with the atomic number Z = 2. Z2 (1 Eh ) 2n2 3 1 Z2 1 − 2 Eh = Z 2 Eh E2 − E1 = − 2 2 2 1 8 3 2 E2 − E1 = − 2 Eh = 1.5 Eh = 6.54 · 10−18 J. 8 En = −

He+ : Z = 2

1.8 This will be a small transition energy, because the energy levels get very close together at high n. What does that imply for the wavelength λ? That it will be long, because λ ∝ 1/Ephoton . The wavelength, frequency, and energy of radiation are all related through Planck’s law E = hν and the constancy of the speed of light ν = c/λ. For helium, Z = 2. 1 Z2 1 Eh − ∆E = 2 n′′ 2 n′ 2 1 1 Eh = (2) − 1002 1012 = 3.94 · 10−6 Eh = 0.865 cm−1 1 λ= = 1.16 cm = 0.0116 m, ∆E( cm−1 )

which is in the microwave. 24

1.9 Energy has been added to the atom by the first photon (to reach state n1 ) and removed with the second photon (landing us in the final state n2 ), so the final energy change in the atom ∆En2 ,n1 is equal to the difference in energy of the two photons: 1 Z2 1 Eh − ∆En2 ,n1 = 2 n21 n22 9 1 = 1 − 2 Eh Z = 3, n1 = 1 2 n2 hc hc ∆Ephoton = − λ1 λ2 1 1 −34 8 −1 − = (6.626 · 10 J s)(2.998 · 10 m s ) 10.4 · 10−9 m 828 · 10−9 m 9 1 = 1.89 · 10−17 J = 4.33 Eh = 1 − 2 Eh 2 n2 −1/2 9 2 n2 = = 5. − 4.32 2 9 Maple. By the time you’ve entered the constants and worked out the relationships, a symbolic math program may not save you much effort on a problem like this, but the same setup can be applied to numerous problems: • First, declare the constants in the problem: h := 0.66260755e-33: c := 299792468.: Eh:=4.35980e-18: Z:=3; n1:=1; lambda1 := 10.4e-9; lambda2 := 828e-9; • Define any equations that relate the variables and parameter values: DeltaE :=(n1,n2)->((Zˆ2)/2)*((1/n1ˆ2)-(1/n2ˆ2))*Eh; • And use the solve function to extract the value you’re looking for: solve(DeltaE(n1,n2)=h*c*(1/lambda1-1/lambda2),n2); In this case, as with many others, there is more than one solution and you have to choose the one that is valid. Here, the solutions are ±5, and you have to recognize that n2 must be positive. Mathematica. • First, declare the constants in the problem. Note Mathematica’s method of indicating scientific notation: h=6.6261*ˆ-34; c=299792468.; Eh=4.35980*ˆ-18; Z=3; lam2=828.*ˆ-9; lam1=10.4*ˆ9; • Define any equations that relate the variables and parameter values: DeltaE:=((Zˆ2)/2)*((1/n1ˆ2)-(1/n2ˆ2))*Eh • And use the Solve function to extract the value you’re looking for: Solve[DeltaE==h*c*(1/lam1-1/lam2),n2] The numeric result is n2=±5.08589. We eliminate the negative solution and round the positive value to the nearest integer to get n2=5. 1.10 The longest wavelength corresponds to the lowest transition energy: λ=

hc , ∆E

and n = 2 → 3 is the lowest-energy absorption from n = 2. Z2 1 1 ∆E = − − 2 Eh 2 n22 n 1 1 1 = −2 Eh − 9 4 5 Eh = 0.278 Eh. = 18 1.11 [Thinking Ahead: Why are these energy values so close together? Based on the pattern of energy levels for the one electron atom, this means that the upper state energies are approaching the ionization energy. This means the upper state n values almost don’t matter – it’s enough to know that they are large.] Z2 ∆E = En − E1 = − 2 (1 har) − 2n

−Z 2 2

(1 har) 1 Eh = 27.2 eV

The absorption energies differ by small fractions; therefore, n is a large number. Z2 n large, ∆E ≈ −E1 = (1 har) ≈ 1626 eV 2 s 2 · 1626 eV Z≈ −1 = 10.93 (27.2 eV (1 har) The atom is

Z = 11

Na10 .

1.12 Coulomb forces control almost all molecular interactions and structure. In total, these forces wield sufficient strength, for example, to prevent the atoms of a car from sliding between the atoms of the Brooklyn Bridge to fall into the East River. But these forces work individually over tiny, molecule-sized areas. In order to add up to a macroscopic force over a large area, they need to be fairly strong individually. Quantitatively, with charges on the order of 10−19 C and distances of 10−10 m, the forces should be of magnitude (don’t forget the 4πǫ0 , which is about 10−10 in SI units) (10−19 )2 /(10−10 )(10−10 )2 = 10−8 N, larger than you might have expected. r1 = a0 = 5.292 · 10−11 m

Z =1

FCoulomb = − =−

Ze2 4πǫ0 r2

(1)(1.602 · 10−19 C)2 (1.113 · 10−10 C2 J−1 m−1 )(5.292 · 10−11 m)2

= −8.24 · 10−8 N .

1.13 [Thinking Ahead: Is this a positive or negative number, and how does it compare to the 2 Eh ionization energy of He+ ? We’ve defined U = 0 for the ionized electron, and the n = 1 electron must be at a lower potential energy in order to be bound to the nucleus, so U is a negative number. To ionize the atom, we have to overcome the stabilization represented by this negative potential energy. The kinetic energy of the n = 1 state is a positive number, so some of the negative potential energy is already canceled before we ionize the He+ . Therefore, U must be lower than −2 Eh to begin with. We expect to find that the potential energy is less than −2 Eh .] This question takes advantage of the fact that the 26

distance r between the nucleus and the electron is a constant for each state n in the Bohr model. Since the potential energy only depends on r, the potential energy of the electron is also constant: Un = −

Ze2 Z 2 me e 4 =− 4πǫ0 rn (4πǫ0 )2 n2 h̄2

because 4πǫ0 n2 h̄2 Zme e2 n = 1.

rn =

Un=1 = −

Z=2 for He+

(2)2 (9.109 · 10−31 kg)(1.602 · 10−19 C)4 (1.113 · 10−10 C2 J−1 m−1 )2 (1)2 (1.055 · 10−34 J s)2

= −1.745 · 10−17 J

or, more quickly by using atomic units, Z2 Un=1 = − 2 n

me e 4 (4πǫ0 )2 h̄2

= −4 Eh .

1.14 [Thinking Ahead: Does kinetic energy decrease or increase with Z and with n? It increases with Z (the electron has to move faster when the nucleus pulls on it more strongly) and decreases with n (the electron requires less centripetal force to balance the weaker nuclear attraction at large distances).] The kinetic energy is always mv 2 /2, which is a handy form in this case because we know the mass (me ) and the speed is obtainable from the Bohr model: vn =

Ze2 4πǫ0 nh̄

2 Ze2 me me vn2 = 2 2 4πǫ0 nh̄ 2 4 Z me e = = −En . 2(4πǫ0 )2 n2 h̄2

Kn =

The kinetic energy is equal to the total energy times −1. This result is predicted by the virial theorem: for any stable, dynamic system involving a central force law (i.e., the force depends only on r), the average kinetic and potential energies are related as follows: U = −2K. Since the total energy is the sum of these two contributions, E = K + U = −K. 1.15 [Thinking Ahead: Do the changes in Z and n push these numbers in the same direction? No. Increasing Z from 2 to 3 increases the attraction between electron and nucleus, decreasing r and U (making it more negative) while increasing v. Increasing n from 2 to 3 pulls the electron further away, having all the opposite effects. ] Given the values for He+ n = 2, we only need to know how Z and n affect the equations for these parameters, and then substitute Z and n = 3 for Z and n = 2. The radius rn varies as n2 /Z (Eq. 1.13), vn varies as Z/n (Eq. 1.14), and the potential energy Un as Z/rn or Z 2 /n2 . He+ n = 2 Li2+ n = 3 rn 1.06 Å 1.59 Å vn 2.19 · 106 m s−1 2.19 · 106 m s−1 Un −4.36 · 10−18 J −4.36 · 10−18 J 27

1.16 −

AµA µB mv 2 = − r4 r AµA µB = mv 2 r3 r r = m2 v 2 r 2 = L2 m m AµA µB =

rn =

r m

Fd−d = Fcentripetal multiply through by r4 L = mvr for circular orbit

(nh̄)2

L = nh̄

AµA µB m . n2 h̄2

This is a weird result if you look at it, because the orbit actually gets smaller as you increase n. Going further and evaluating the total energy shows that the system is not stable under these assumptions. 1.17 4πǫ0 n2 h̄2 Zme e2 Ze2 vn = 4πǫ0 nh̄ me vn2 Fn = rn 2 Zme e2 Ze2 = me 4πǫ0 nh̄ 4πǫ0 n2 h̄2 rn =

Z 3 m2e e6 . (4πǫ0 )3 n4 h̄4

The centripetal force should decrease with n, because it takes less force to keep a particle in a circular orbit when the circle is bigger. (You could also point out that the the Coulomb force, which is equal to the centripetal force, is weaker at higher n because the orbital radius rn is bigger. Or simply that r increases and v decreases with n, and therefore Fn = me rvn2 /rn must decrease.) 1.18 parameter n=1 momentum (kg m s−1 ) 1.99 · 10−24 de Broglie wavelength (nm) 0.333 kinetic energy ( Eh ) 0.500 transition energy to n = 3 ( Eh ) 0.444 For the transition energy in the last row, ∆E = E3 − E2 = −

Z2 Eh 2

n=2 9.95 · 10−25 (p = mv, v ∝ 1/n) 0.666 (λdB ∝ 1/p ∝ n) 0.125 (K ∝ v 2 ∝ 1/n2 ) 0.0694

1 1 − 2 32 2

= 0.0694 Eh.

1.19 We set the speed greater than 0.1 c and solve for Z: vn =

Ze2 > 0.1 c 4πǫ0 nh̄ 0.1(1.113 · 10−10 C2 J−1 m−1 )(1.055 · 10−34 J s)(2.998 · 108 m s−1 ) 0.1 (4πǫ0 )h̄c = Z> e2 (1.602 · 10−19 C)2

n=1

= 13.7.

The smallest value of Z (rounding up to the nearest integer) is 14. So relativistic corrections are likely to start becoming important for atoms beyond aluminum in the periodic table. 1.20 [Thinking Ahead: How does this transition energy compare to the ionization energy? Given the ionization energy of Z 2 /2 = 2 Eh for He+ , this transition energy is a small fraction of the energy range of the He+ quantum states. The transition must therefore be among the closely grouped, higher energy n values.] n = n′′ → n′′ + 1 Z =2 2 1 1 Z − ′′ 2 Eh ∆E = − 2 (n′′ + 1)2 n 1 1 − ′′ =2 Eh (n + 1)2 n′′ 2 = 0.045 Eh. This can be solved numerically. • Maple: A command to solve this equation for n′′ is solve(2*((1/nˆ2)-(1/((n+1)ˆ2)))=0.045,n); • Mathematica: Solve[2*((1/nˆ2) - (1/((n + 1)ˆ2))) == 0.045, n] • Iteration:

′′

n =

1 0.045 + ′′ 2 (n + 1)2

−1/2

Start with n′′ = 1 on the lefthand side; this predicts n′′ = 1.92. Plug this into the lefthand side, and this predicts n′′ = 2.67, then 3.22, 3.56, 3.77, 3.88, converging to n′′ = 4. The transition is

n = 4 → 5.

1.21 The linear momentum p is mass times speed, and we have an equation for the speed (Eq. 1.14): p = me vn vn =

me = 9.109 · 10−28 g

Ze2 2(1.602 · 10−19 C)2 = = 2.187 · 106 m s−1 4πǫ0 nh̄ (1.113 · 10−10 C2 J−1 m−1 )2(1.055 · 10−34 J s)

p = (9.109 · 10−31 kg)(2.187 · 106 m s−1 ) = 1.99 · 10−24 kg m s−1

1.22 [Thinking Ahead: Is this a high or low frequency, compared for example to radiofrequency? High. The speed of the electron in a typical Bohr atom is a significant fraction of the speed of light, and the distance to cover is microscopic. Therefore, the oscillation of the dipole moment would be quite fast.] Radiation is emitted by an oscillating dipole because the electric field generated by the dipole must also be oscillating, and that oscillation propagates through the surroundings at the speed of light. 29

Therefore, the frequency of the radiation – the frequency at which the electric field of the radiation oscillates back and forth—is the same as the frequency at which the dipole oscillates back and forth. For the Bohr atom, with a dipole arising from the separation between the positively charged nucleus and the negatively charged electron, this frequency is the frequency at which the electron orbits the nucleus. This we can obtain from the speed vn of the Bohr atom electron and the circumference 2πrn , which gives the distance the electron travels in one orbit. vn =

Ze2 4πǫ0 nh̄

rn =

4πǫ0 n2 h̄2 . Zme e2

The time for one revolution of the electron around the nucleus is τn =

2π(4πǫ0 )2 n3 h̄3 2πrn = , vn Z 2 me e 4

so the frequency of oscillation, and the frequency of any emitted radiation, is νn =

1 Z 2 me e 4 . = τn 2π(4πǫ0 )2 n3 h̄3

For Li+2 , Z = 3. With n = 4, νn=4 =

(3)2 (9.109 · 10−31 kg)(1.602 · 10−19 C)4 2π(1.113 · 10−10 C2 J−1 m−1 )2 (4)3 (1.055 · 10−34 J s)3

= 9.247 · 1014 s−1 ,

which corresponds to near-ultraviolet radiation. 1.23 This is an application of Bohr’s expression for energy levels of one-electron atoms. The lower quantum state is known (ground state, so n1 = 1) and the energy is known, so the upper quantum Z2 states can be found: En = − 2n 2 Eh , where Z = 4 for Be. Z2 ∆E(n → n′ ) = En′ − En = − 2 2 1 Z =− −1 2 n′ 2 − 1 2∆E 2 ′ n = 1− Z2 − 1 2∆E 2 = 1− 16 ∆E ∆E

1 1 − 2 n n′ 2

Z=4

= 7.50 Eh = 7.68 Eh

n′ = 4 n′ = 5

1.24 The n → ∞ transition energy gets smaller as n climbs, and the photon wavelength increases. Therefore, we are looking for the lowest value of n′′ such that the photon wavelength is more than 30

1.0 mm, where Z = 2 for He+ : Z2 1 1 1 Z2 Z2 ∆E = − − Eh − E = − E = h h 2 ∞2 n2 2 n2 2n2 hc hc hcn2 λ= ≥ 1.0 · 10−3 m = 2 = ∆E Z Eh /(2n2 ) 2 Eh 2(1.0 · 10−3 m) Eh n2 ≥ hc 1/2 1/2 2(1.0 · 10−3 m) Eh 2(1.0 · 10−3 m)(4.360 · 10−18 J) n≥ = = 209.5. hc (6.626 · 10−34 J s)(2.998 · 108 m s−1 ) So n is the smallest integer greater than 209.5, so n = 210. 1.25 [Thinking Ahead: What about the energy level distribution allows us to solve for two unknown n values with a single equation? It’s the converging value of the energy as n increases. It is straightforward to prove that no two values of n have the same energy spacing as any other two values.] This problem approaches atomic spectroscopy from a more realistic perspective: when the transition energy is measured, how does the spectroscopist determine the quantum states involved? The Li atom has atomic number Z = 3, so Li+2 is a one-electron atom. This time we have the energy and need to obtain n. En = −

me e 4 Z2 n2 2(4πǫ0 )2 h̄2

(9.109 · 10−31 kg)(1.602 · 10−19 C)4 (3)2 n2 2(1.113 · 10−10 C2 J−1 m−1 )2 (1.055 · 10−34 J s)2 1.96 · 10−17 J =− n2 1 1 −17 −17 − 2 1.74 · 10 J = 1.96 · 10 n21 n2 1 1 1.74 − 2 = = 0.888 n21 n2 1.96 =−

We have one equation and two unknowns. However, the pattern of energy levels is such that all the energy levels except n = 1 are crowded together within En=1 /4 of zero. Therefore, transitions between any two states n 6= 1 must have transition energy less than En=1 /4. Since 1.74 · 10−17 J > − then

n1 = 1

En=1 1.96 · 10−17 = J, 4 4

, so n2 = [−(0.888 − 1)]−1/2 = 3.

1.26 These parameters are all related to the speed at which the electron travels the circumference of its orbit, which can be calculated from Eqs. 1.13 and 1.14. a. Time to complete one orbit τ = distance/speed: τn =

(2π(4πǫ0 )n2 h̄2 /Zme e2 ) 2π(4πǫ0 )2 n3 h̄3 2πrn = = vn Ze2 /(4πǫ0 )nh̄) Z 2 me e 4

Number of orbits per second = τ1n ≡ fn :

fn =

Z 2 me e 4 2π(4πǫ0 )2 n3 h̄3

b. Z 2 me e 4 1 1 − 43 2π(4πǫ0 )2 h̄3 53 (1)2 (9.109 · 10−31 kg)(1.602 · 10−19 C)4 (−7.625 · 10−3 ) = 2π(1.113 · 10−10 C2 J−1 m−1 )2 (1.055 · 10−34 J s)3

∆fn = f5 − f4 =

= −5.01 · 1013 s−1

c. ν=

Z 2 (4.360 · 10−18 J Eh−1 ) ∆E E5 − E4 = =− h h 2h

=−

(1)2 (4.360 · 10−18 J Eh−1 ) (−0.0225) 2(6.626 · 10−34 J s)

1 1 − 2 2 5 4

= 7.40 · 1013 s−1

This value and the magnitude of the answer in part (b) become more similar as n′′ and n′ increase. 1.27 [Thinking Ahead: Is this going to be high Z, given that the wavelength for the H atom ionization is 91.1 nm? No. Increasing Z increases the energy as Z 2 and shortens the wavelength for the transition as 1/Z 2 . Here the wavelength is only about a factor of 4 lower.] In units of Eh : Z2 Z2 + 2n ′ 2 2n′′ 2 2 1 1 Z − + = 2 9 1 4Z 2 = 9

∆E = En′ − En′′ = −

hc (4.36 · 10−18 J Eh−1 ) 1/2 9(6.626 · 10−34 J s)(2.998 · 108 m s−1 ) Z= = 2. 4(4.36 · 10−18 )(25.6 · 10−9 ) λ = 25.6 · 10−9 m =

hc = ∆E

4Z 2 9

The atom is He+ . 1.28 This expression for Fgrav has the same dependence on distance as FCoulomb . The difference is that the constants Ze2 /(4πǫ0 ) have been replaced by m1 m2 G, where (since the masses are unchanged) m1 m2 = me mp . We could follow the same steps used to obtain the Bohr energy expression, but since all we have done is change one set of constants for another, we can just take the result for the Bohr energy and change out those constants: Z 2 me e 4 2(4πǫ0 )n2 h̄2 (me mp G)2 me En = − 2n2 h̄2 3 2 2 me mp G =− . 2n2 h̄2 En = −

real atom, Eq. 1.20 gravity atom

For the ground state, we set n = 1: E1 = −

(9.109 · 10−31 kg)3 (1.673 · 10−27 kg)2 (6.67 · 10−11 m3 kg−1 s−2 )2 = −4.23 · 10−97 J. 2(1.055 · 10−34 J s)2

Comparing this to the ground state energy using the charged particles (E1 = 2.18 · 10−18 J), you can see why we never worry about gravity when solving the quantum mechanics of atoms and molecules. Note: one topic of ongoing speculation among physicists is whether the gravitational force law stays the same at all distance scales. Although gravitational forces have been measured for individual neutrons, that was over distances of several micrometers—much larger than the distances separating subatomic particles in an atom. For a further exercise, try calculating the radius of the orbit of the particles in this gravity-atom. 1.29 If there’s only one electron, we need to know only the atomic number Z to identify the ion. From Eqs. 1.13 and 1.14: rn =

n2 a0 = 1.286a0 Z

vn =

n2 = 1.286 Z Z = (2.3332)(1.286) = 7.

Ze2 e2 = 2.333 4πǫ0 nh̄ 4πǫ0 h̄

Z = 2.333 n

The ion is N+6 . 1.30 [Thinking Ahead: Is this longer or shorter than λdB for the electron in hydrogen? Shorter. The heavier mass of the muon compared to the electron makes the system more classical, with wave properties harder to measure. That implies a smaller λdB .] The solution will be identical to the Bohr atom, except with the muon mass mµ substituted for me .

-e mµ +e mp

λdB =

Ze2 4πǫ0 nh̄

Z=1

n=1

(1.113 · 10−10 C2 J−1 m−1 )(6.626 · 10−34 J s)(1.055 · 10−34 J s) 4πǫ0 hh̄ h = = 2 mµ v mµ e (1.884 · 10−28 kg)(1.602 · 10−19 C)2 = 1.61 · 10−12 m.

1.31 The total power is the sum of the contribution from each photon emitted, the product of the number of photons per second and the energy per photon. Setting the average photon wavelength equal to λ̄, the energy per photon is approximately given by (6.626 · 10−34 J s)(2.998 · 108 m s−1 ) hc = 590 · 10−9 m λ̄ −19 = 3.37 · 10 J/photon.

Ephoton ≈

The number of photons per second is therefore the total power divided by the energy per photon: 10 J s−1 = 2.97 · 1019 photon s−1 . 3.37 · 10−19 J/photon 1.32 Call the unknown atomic number Z and set the two transition energies equal: 1 1 22 Eh = 1.98 Eh − ∆EHe+ = − 2 102 12 1 Z2 1 Eh = 0.006111 Z 2 Eh − =− 2 62 52 1.98 Z2 = = 324 0.006111 Z = 18 The one-electron atom is Ar17+ . 1.33 c λi νf = 3νi νi =

3hc λi −34 3(6.626 · 10 J s)(2.998 · 108 m s−1 ) = 1064 · 10−9 m

Ef = hνf = 3hνi =

= 5.601 · 10−19 J.

1.34 c c c = = ν3 |ν2 − ν1 | |(c/λ2 ) − (c/λ1 )| −1 1 1 = 2136 nm = 2.14 µm. − = 551 438

λ3 =

1.35 If the speed is well-defined, then wavefunction of the neutrinos will be sin(2πx/λdB ), which is a sine wave with wavelength λdB = h/(mv). Using the speed to rewrite this as a function of time, we replace x with vt and obtain sin(2πvt/λdB ) = sin(2πmv 2 t/h). This goes through one cycle of the wavefunction in time τ = h/(mv 2 ) = 7.4 · 10−10 s for v1 and τ = 1.8 · 10−10 s for v2 . The signal for v2 oscillates at 4 times the rate, because the speed is two times greater and the λdB is cut in half. The detector sees only the probability density, which is proportional to the square of the wavefunction. (a) For v1 :

(b) For v2 : 34

The whole experiment is speculative in the extreme. We don’t actually have any idea what the neutrino mass is, although recent work indicates that neutrinos have masses less than 0.28 eV or 5 · 10−37 kg. Neutrinos are also notoriously difficult to detect, let alone control to an extent were we can determine their speed. Still, they are intriguing because they are by far the lightest components of matter known to exist as independent particles, and λdB may be really large. 1.36 Each Ar+ ion that undergoes this electronic transition releases one photon with energy E = hν =

hc λ

where λ = 514 · 10−9 m. This energy per photon is (6.626 · 10−34 J s)(2.998 · 108 m s−1 ) 514 · 10−9 m −19 = 3.86 · 10 J.

To provide 8 W = 8 J s−1 of power, the number of transitions per second must be: 8 J s−1 = 2 · 1019 s−1 . 3.86 · 10−19 J 1.37 For the two particles (the electron and the 56 Fe atom): p2 2m h h λdB = = √ p 2mK K = 12 mv 2 =

For the visible light: λ=

c 2.998 · 108 m s−1 = 5.71 · 10−7 m = ν 5.25 · 1014 s−1

We set λ equal to λdB and solve for K: K=

6.73 · 10−55 J kg h2 = 2 2mλ m

a. m = me = 9.109 · 10−31 kg K=

6.73 · 10−55 J kg = 7.39 · 10−25 J = 4.61 · 10−6 eV 9.109 · 10−31 kg

b. m = 5.9 amu = 9.28 · 10−26 kg K=

6.73 · 10−55 J kg = 7.25 · 10−30 J = 4.53 · 10−11 eV 9.28 · 10−26 kg 35

These particles must be cooled to very low energies to give them wavelengths comparable to that of visible light. 1.38 The relationship between wavelength and momentum for matter is given by de Broglie’s equation λdB =

h h = . mv p

Assuming the same relationship for photons, we have p =

6.626 · 10−34 J s h = = 1.472 · 10−27 kg m s−1 . λ 450 · 10−9 m

We’re hoping to get a 0.0010 kg marble to pick up an additional 0.010 m s−1 of speed; in other words, to transfer 1.0 · 10−5 kg m s−1 of momentum. We will need a lot of photons: 1.0 · 10−5 = 6.8 · 1021 photons. 1.472 · 10−27 Each of these photons has energy hν = hc/λ, so the total energy of this beam is Etotal =

(6.8 · 1021 )(6.626 · 10−34 J s)(2.998 · 108 m s−1 ) 450 · 10−9 m

= 3.0 · 103 J.

Isn’t it surprising how inefficient this is? If the marble is initially at rest, we only need to add a kinetic energy of mv 2 /2 = 5 · 10−8 J to bring it to the desired speed. If you work through this for a general case, it works out that the ratio of the total photon energy to the kinetic energy needed is 2c/v, so the efficiency is always low. But for pushing small things around, photons have many desirable characteristics, as discussed later in Section 11.3.

1.39

λB =

h mv

In our model, the proton is the center of mass of the atom, so its average speed

is the speed of the atom: λB (proton) =

6.626 · 10−34 J s = 2 · 10−10 m. (1.673 · 10−27 kg)(2 · 103 m s−1 )

The electron is always moving much faster (classically) because it is simultaneously orbiting the proton at a speed vn=1 =

(1.602 · 10−19 C)2 e2 = = 2.19 · 106 m s−1 , 4πǫ0 h̄ (1.113 · 10−10 C2 J−1 m−1 )(1.055 · 10−34 J s)

compared to which the overall speed of the atom is insignificant. This yields a de Broglie wavelength of λB (electron) =

6.626 · 10−34 J s = 3 · 10−10 m. (9.109 · 10−31 kg)(2.19 · 106 m s−1 )

1.40 [Thinking Ahead: How will this wavelength compare to the roughly 1 Å value we found for an electron in an atom? It should be of comparable magnitude, because the speed is similar, near 1% the speed of light. ] This is a direct application of de Broglie’s law. λdB =

6.626 · 10−27 erg s h = = 1.212 · 10−9 m = 12.12 Å me v (9.109 · 10−28 g)(6 · 105 m s−1 ) 36

1.41 The de Broglie wavelength is straightforward enough: λdB = =

h mv 6.626 · 10−34 J s (16 amu)(1.661 · 10−27 kg amu−1 )(8 · 103 m s−1 )

= 3 · 10−12 m.

This wavelength is equal to 0.03 Å, which is much smaller than the travel distance of the atoms, so we expect that a quantum analysis is not critical to the analysis of this problem. 1.42

a. The physical limit on λdB = h/(mv) is that no object’s speed can exceed the speed of light, so v ≤ c. Therefore, according to our equation, λdB ≥

6.626 · 10−34 J s h = = 2.43 · 10−12 m s−1 . mc (9.109 · 10−31 kg)(2.998 · 108 m s−1 )

b. The relativistic correction increases the momentum over the classical value mv, and therefore reduces the de Broglie wavelength for given mass and speed. Therefore, quantum effects should be less noticeable as the motion becomes more relativistic. It may seem odd, but this does not always mean that the higher overall energy states are the most relativistic. The particle in a one-dimensional box becomes more relativistic as the quantum number n increases, but the reverse is true for the electron in a one-electron atom, where lower states have higher kinetic energy. As a bound state electron is promoted to a higher energy state, it trades some kinetic energy for even more potential energy, which slows it down. Therefore, it is the lowest energy electrons in an atom which move at relativistic speeds. 1.43 [Thinking Ahead: Based on the mass alone, do you expect λdB to be bigger or smaller for the nucleus than for the electron? The nuclear mass is much greater than me , no matter what nucleus we have, so that alone would tend to make λdB smaller. Next, how will the speed of the nucleus compare to that of the electron, and how will this effect the final λdB ? Without doing the math, one can reasonably expect that the nucleus, being more massive, is not moving as fast as the electron in the same atom. Lower speed will tend to drive λdB back up.] This problem raises an interesting point regarding our treatment of the nucleus as a point charge. If the center of mass of the atom is fixed in space, the nucleus (which is not quite at the center of mass) must be moving slightly as the electron moves. Its de Broglie wavelength can be determined from its mass and speed. rn + re = r +

Ze rn q s mnuc COM

−e s me

rn =

me mnuc + me

re =

mnuc mnuc + me

2πre 2πrn vn = t t t is the time for one orbit of the electron or the nucleus about the center of the mass. ve =

h/(mnuc vn ) me ve λdB (nucleus) = = λdB (electron) h/(me ve ) mnuc vn me mnuc /(mnuc + me ) me re me (2πre /t) = = = mnuc (2πrn /t) mnuc rn mnuc me /(mnuc + me ) = 1 37

Surprisingly, the nucleus has exactly the same de Broglie wavelength as the electron. Why, then, can we treat the nucleus as a point charge when our result suggests that it’s as broadly smeared out as the electron? It is true that atoms are never at rest, but this doesn’t help, because typical kinetic energies for translational motion of atoms at room temperature, or for vibrations in chemical bonds, yield similar de Broglie wavelengths. A partial answer is provided by our analysis of the Bohr model using the de Broglie wavelength, to show that the electron is not localized at a particular point in space, but instead is evenly distributed about the nucleus. In this picture, the nucleus is the center of mass of the system, and the analysis just given is simply invalid. But that’s not a great answer. Rigorous solutions to the quantum mechanics of atoms do find small differences in the electronic energies between different nuclear isotopes of the same atom, arising from the shift in the center of mass as described in the problem. In fact, our concern about the large nuclear λdB is only worse if we pin the nucleus at rest in our coordinate system, because at zero velocity its de Broglie wavelength becomes infinite. One way out of this digs into the subatomic physics of the nucleus. Even if you could keep the center of mass of a single proton at rest, there is still considerable kinetic energy bound up in orbital motion of the quarks that make up the proton. The uncertainty in the momentum of these sub-nuclear particles is sufficient to allow the position of the nucleus to be well-determined and the size of the nucleus to be much smaller than that of the electron density. This obviously gets into awkward territory for a chemistry text, but you can see how difficult it is to draw a line between the physics we need and the physics we can ignore if we want to understand atoms and molecules. 1.44 For the minimum uncertainty, use the “=” sign: h̄ 2 δx 6.626 · 10−34 J s h = 2.4 · 10−10 m λdB = = p (9.109 · 10−31 kg)(3.0 · 106 m s−1 ) h dλdB =− 2 dp p δλdB h = 2 δp p h h h̄ δλdB = 2 δp = 2 p p 2 δx (2.4 · 10−10 m)2 λ2 = 4.6 · 10−11 m. = dB = 4π δx 4π(1.0 · 10−10 m) δp =

1.45 If we change λdB to λ, then we can use Planck’s law to put things in terms of Ephoton : δx δ

h̄ h Ephoton = δx δ ≥ λ c 2 x h̄ δEphoton ≥ δ c 2 h̄ δt δEphoton ≥ 2

Ephoton =

hc λ

The uncertainty in distance, δx, divided by the speed of the radiation, c, is the uncertainty in x/c = time t for the photon to pass some fixed point. This is not a rigorous derivation, but the result is correct. As a result, lasers that put out very short pulses of light (small δt) have a poorly defined photon energy (large δE). 38

Chapter 2 2.1 Because eax is not finite at all points, it cannot be a valid wavefunction over all space. The free particle can travel anywhere, and the exponential function does not converge at large x. However, the exponential function is indeed a valid solution over regions of the system where the potential energy becomes larger than the energy of the system (as considered in Chapter 3), but the particles in those cases are no longer free. 2.2

a. The mass of the particle. b. The spacing between the walls. c. The de Broglie wavelength of the particle. d. The speed of the particle. Increasing λdB relative to the domain increases quantum behavior.

2.3 To normalize the wavefunction ψ, we need to find the normalization constant A such that Z (Aψ)∗ (Aψ)dτ = 1. allspace

2 2 2 √ We begin with ψ(x, y, z) = Ae−(x +y +z ) xyz for x, y, z positive, and plug into the above equation: Z ∞ Z ∞ Z ∞ 1= dx dy dz |ψ|2 0 0 0 Z ∞ Z ∞ Z ∞ 2 2 2 = A2 dz e−2(x +y +z ) xyz dy dx 0 Z ∞0 Z 0∞ Z ∞ 2 2 −2x2 = e xdx e−2y ydy e−2z zdz A2 0 0 0 1 −2y2 ∞ 1 −2z2 ∞ 2 1 −2x2 ∞ |0 − e |0 − e |0 A = − e 4 4 4 2 1 1 1 −∞ = A2 e−2x |∞ − e0 = −1 0 = e 4 4 4 A2 = 64 A= 8

The normalized wavefunction is 2 2 2 √ ψ(x, y, z) = 8e−(x +y +z ) xyz

Maple. This problem we can solve with a single command that combines the int and solve functions: solve(int(int(int(Aˆ2*x*y*z*exp(-2*(xˆ2+yˆ2+zˆ2)), x = 0 .. infinity), y = 0 .. infinity), z = 0 .. infinity) = 1, A) Note that both the positive and negative solutions are valid, because the absolute phase of the wavefunction is not experimentally meaningful. Mathematica. This problem we can solve with a single command that combines the Integrate and Solve functions: Solve[Integrate[Aˆ2*x*y*z * Exp[-2*(xˆ2 + yˆ2 + zˆ2)], {x, 0, Infinity}, {y, 0, Infinity}, {z, 0, Infinity}==1,A] Note that both the positive and negative solutions are valid, because the absolute phase of the wavefunction is not experimentally meaningful. 39

2.4 [Thinking Ahead: Will the normalization constant depend on r, θ, or φ? Also, the wavefunction is complex. Does that mean the normalization constant is complex too? No, no, no, and no. The normalization constant is a constant, after all, and never depends on the variables in the function. Its purpose is to ensure that the probability density (not the wavefunction) has the right magnitude, but the phase of the normalization constant (which doesn’t affect the magnitude of |ψ|2 ) is up to us. Because the probability density is calculated so as to guarantee that it is pure real, all information about whether the original wavefunction was real or complex is lost during normalization. We can choose a real, imaginary, or complex normalization constant, but that choice has nothing to do with the unnormalized wavefunction. To keep things simple, we’ll choose pure real normalization constants.] If you are already familiar with some symbolic math program, you might be able to work this quickly with one of those, but the triple integral and the complex phase factors make it non-trivial. This solution works everything out on paper, and one example of how to apply a more technologically advanced method is given at the end of the solution. The wavefunction is r 1 r 1 ψ(r, θ, φ) = e−r/(2a0 ) cos θ + √ e−iφ sin θ − √ eiφ sin θ + 2 1 − a0 2a0 2 2 ∗

ψ (r, θ, φ) = e

−r/(2a0 )

r a0

1 −iφ r 1 iφ sin θ + 2 1 − cos θ + √ e sin θ − √ e 2a0 2 2

To normalize ψ, we want to multiply it by a number A such that Z ∞ 0

r dr

Z 2π 0

dφ

Z π

sin θdθ(Aψ)(Aψ)∗ = 1.

Before evaluating the integral, we should simplify (Aψ)(Aψ)∗ = A2 |ψ|2 as much as possible. Let’s deal with the imaginary terms first: ψ = f + e−iφ g − eiθ g where r r r e−r/(2a0 ) sin θ cos θ + 2 1 − g= √ f = e−r/(2a0 ) a0 2a0 2a0 |ψ|2 = ψψ ∗ = (f + e−iφ g − eiφ g)(f + eiφ g − e−iφ g)

= f 2 + f g(eiφ − e−iφ + e−iφ − eiφ ) + g 2 (1 − e−2iφ − e2iφ + 1) = f 2 + g 2 (2 − e2iφ − e−2iφ ).

This still appears to have an imaginary component until we look more closely at −(e2iφ + e−2iφ ) using power series or the Euler formula (Problem A.9): e

2iφ

−2iφ

"∞ X 1 n=0

(2iφ) +

"∞ X 1

n=0 3

(−2iφ)

8iφ 16φ4 4φ2 − + + ... = 1 + 2iφ − 2 6 24 2 3 4φ 8iφ 16φ4 = 1 − 2iφ − + + − ... 2 6 24 4 2 16φ 4φ + + . . .) = 2 cos 2φ = 4 cos2 φ − 2 = 2(1 − 2 24 |ψ|2 = f 2 + g 2 (2 − 2 cos 2φ) = f 2 + g 2 (4 − 4 cos2 φ) = f 2 + 4g 2 sin2 φ 40

all

space

|Aψ|2 dτ = A2

Z ∞

r2 dr

Z 2π

dφ

Z π

h sin θdθ e−2r/(2a0 )

# 2 r r 4r2 −2r/(2a0 ) 2 2 × cos θ + 2 1 − sin θ sin φ + 2e a0 2a0 2a0 Z ∞ Z 2π Z π = A2 r2 dr dφ sin θdθe−r/a0 0 0 0 2 2r2 4r r2 4r 2r2 r + 2+ cos θ + 2 sin2 θ sin2 θ sin2 φ × 2 cos2 θ − 2 cos θ + 4 − a0 a0 a0 a0 a0 a0

At this point, a table of integrals is useful. We need the following integrals from Table A.5: R ∞ 2 −r/a 0 dr = 2a30 r e R0∞ 3 −r/a 0 dr = 6a40 r e R0∞ 4 −r/a 0 dr = 24a50 0 r e Z

allspace

|Aψ| dτ = A

Rπ sin θdθ = 2 R0π sin θ cos θdθ = 0 R0π sin θ cos2 θdθ = 23 R0π 3 sin θdθ = 43 0

(24a30 )

R 2π dφ = 2π R02π 2 sin φdφ = π 0

2 (2π) − 2(6a30 )(0)(2π) + 4(2a30 )(2)(2π) 3

−4(6a30 )(2)(2π) + (24a30 )(2)(2π) + 4(6a30 )(0)(2π) 4 +2(24a30 ) (π) 3 3 2 = A 32πa0 − 0 + 32a30 − 96πa30 + 96πa3o + 0 + 64πa30 = 128πa30 A2

= 1 if Aψ is normalized. The normalization constant is A=

1 −3/2 = 0.0499a0 . 128πa30

Maple. This is a little tougher than you might hope. The conjugate function does not formulate a useful form of ψ ∗ , so perhaps the simplest approach is to define the complex conjugate yourself. In the commands below, I have broken up ψ(r, θ, φ) into f and g and the φ-dependent phase factors as in the math outlined above. The triple integral does not pose much trouble, but (as with other integrals over exponentials) we need to specify that the constant a0 in the exponent is positive. This time we use the assume function in the first command to establish that a0 > 0: • assume(a0>0); • f := (r,theta)-> exp(-r/(2*a0))*(r*cos(theta)/a0+2-r/a0) ; • g := (r,theta)->r/(sqrt(2)*a0)*exp(-r/(2*a0))*sin(theta); • psi := (r,theta,phi)-> f(r,theta)+g(r,theta)*(exp(-I*phi)-exp(I*phi)); • psiconj := (r,theta,phi)->f(r,theta)+g(r,theta)*(exp(I*phi)-exp(-I*phi)); • int(int(int(psi(r, theta, phi)*psiconj(r, theta, phi)*rˆ2*sin(theta), phi = 0 .. 2*Pi), theta = 0 .. Pi), r = 0 .. infinity); • solve(%*Aˆ2 = 1, A); 41

• evalf(%); Don’t forget to add the r2 sin θ components of the volume element before evaluating the integral. Mathematica. These steps follow the same procedure as summarized above for Maple: • $Assumptions = a0>0 • f:=Exp[-r/(2*a0)]*(r*Cos[theta]/a0+2-r/a0) • g:=r/(Sqrt[2]*a0)*Exp[-r/(2*a0)]*Sin[theta] • psi:=f+g*(Exp[-I*phi]-Exp[I*phi]) • psiconj:=f+g*(Exp[I*phi]-Exp[-I*phi]) • Integrate[psi*psiconj*rˆ2*Sin[theta], {phi,0,2*Pi}, {theta,0,Pi}, {r,0,Infinity}] • Solve[%*Aˆ2==1,A] • And finally, to get the floating point value: N[%,4] Note that a single Integrate command can be applied to all three coordinates. 2.5 Because the wavefunction depends only on x and x must be positive, we will integrate the probability density from 0 to ∞ along x: Z ∞ Z ∞ 2 2 |ψ(x)| dx = A (x + 1)−a ln(x + 1) dx. 0

This must be equal to 1 if the function is normalized. Our job is to find A. So we integrate. We can simplify the look of the integrand by setting y = x + 1, which changes the lower limit from x = 0 to y = 1, but leaves dy = d(x + 1) = dx the same: Z ∞ 1 = A2 y −a ln(y) dy. 1

The integral isn’t bad, because we can set u = ln(y) and v = y 1−a /(1 − a), so that Z Z u dv = uv − v du Z y 1−a ln(y) 1 = − y 1−a−1 dy 1−a 1−a 1 1 y 1−a ln(y) − = 1−a 1−a and evaluate this from 1 to ∞ Z ∞ y −a ln(y) dy = 1

∞ 1 1 1−a ln(y) − y 1−a 1−a 1 2 1 = . 1−a

Adding the factor of A2 for normalization gives 1=

A 1−a

A = ±(a − 1). 42

Keep in mind that we only determine the magnitude of a normalization constant. The phase of a normalization constant is arbitrary, unless we are about to combine one wavefunction with another to obtain a specified result. We normally assume that the normalization constant is a pure real number, so here we show only two solutions (determined by the choice of + or − sign), but in general the phase may be any value of eiφ , which is in general complex. 2.6 This is a combination of a linear term r and a very rapidly decaying exponential function. Because the decay in the exponential proceeds at such a high rate, we expect that almost all of the probability density appears at small values of r. We’d better normalize the wavefunction first. Although r is measured in three dimensions (because it is the distance in any direction from the origin), the volume element for integrating over r is r2 dr. In this case, because we are examining a wavefunction ψ(r) that depends only on r, we do not need to also integrate over the angles θ and φ. As long as we are calculating properties that depend only on r (such as the probability in this problem)—so that we will integrate ψ(r) only over r—we don’t need to include the angular contribution to the normalization constant. To simplify the notation, let’s replace r/a0 by s. To get the normalized form, we multiply ψ(s) by A such that Z ∞ Z ∞ Z 5 A2 ∞ −4s5 1= |Aψ(s)|2 s2 ds = A2 (s4 e−4s )ds = − e d(−4s5 ) 20 0 0 0 A2 −4s5 ∞ A2 A2 =− e . = − (0 − 1) = 20 20 20 0 √ So A = 20. We’ve done most of the work now, and we have to plug in the limits to get the probability density in that region. The region we’re interested in is now from s = 0 to s = a0 /a0 = 1: P(s < 1) =

2.7 The integral

Z 1 0

A2 −4s5 1 e 20 0 20 −4(15 ) − 1) = 0.98. = − (e 20

|Aψ(s)|2 s2 ds = −

Z x0

A sin(ax)B cos(bx)dx = 0

−x0

because sin(ax) is an odd function and cos(bx) is an even function. This means that sin(ax) = − sin(−ax), whereas cos(bx) = cos(−bx). Therefore, for any value of x, A sin(ax)B cos(bx) = −A sin(−ax)B cos(−bx). The integral Z 0

A sin(ax)B cos(bx)dx =

−x0

A sin(ax)B cos(bx)dx,

−x0

and hence Z x0

Z x0

A sin(ax)B cos(bx)dx = −

Z 0

A sin(ax)B cos(bx)dx +

Z x0

A sin(ax)B cos(bx)dx

−x0

=0.

The functions are always orthogonal over this interval. 43

2.8 This range of θ values is centered around 90◦ or π/2 radians, whereas the function lies along the z axis and its magnitude drops quickly as θ varies from 0 or π towards π/2. The 2p orbital is symmetric about θ = 90◦ , and the probability density |ψ|2 = 3 cos2 θ/2 is equal measuring in either direction from θ = 90◦ . Therefore, the probability density between 0◦ and 90◦ must be 0.5. Likewise, if the density between 60◦ and 120◦ equals 0.125, the density between 60◦ and 90◦ must be half that, or 0.0625. The difference is the density between 0◦ and 60◦ : Z 60◦ 0

|ψ2p |2 sin θ dθ = 0.5 − 0.0625 = 0.4375.

The total charge (call it q) measured in this region is the fraction of the total charge times this density: q = 0.4375 e = 7.01 · 10−20 C. 2.9 The probability density P over a particular range is just given by the integral of the square modulus of the wavefunction in that interval: Z 2 Z 2 1 2 (x − 2i)∗ (x − 2i) dx P= |ψ(x)| dx = 21 0 0 3 2 Z 2 Z 2 x 1 1 1 + 4x (x + 2i)(x − 2i) dx = (x2 + 4) dx = = 21 0 21 0 21 3 0 1 8 = + 8 = 0.51. 21 3 2.10 Because the function is not an eigenfunction of K̂, we need to use the average value theorem. This means solving the integral over ψ ∗ K̂ψ over the interval −a/2 to a/2. Outside that interval, the wavefunction is zero and so is the value of the integral. If we slog forward, we find that we can solve the integral analytically: hKi =

Z a/2

−a/2

ψ ∗ K̂ψ dx

h̄2 =− 2me

30 a5

Z a/2

−a/2

(a/2) − x

∂2 2 2 dx (a/2) − x ∂x2

Z a/2 h̄2 30 (a/2)2 − x2 [−2] dx 5 2me a −a/2 Z a/2 2 h̄ 30 = (a/2)2 − x2 dx 5 me a −a/2 a/2 2 x3 h̄ 30 2 (a/2) x − = me a 5 3 −a/2 3 3 2 a −a3 −a3 30 a h̄ − − − = me a 5 8 8 24 24 3 2 2 h̄ 5h̄ a 30 = = . me a 5 6 me a 2 =−

We can compare that to π 2 h̄2 /(2me a2 ), which is the ground state (n = 1) kinetic energy of an electron in a one-dimensional box of length a. The system in our problem has a kinetic energy that differs by a factor of 5/(π 2 /2) = 1.013. The parabolic wavefunction in this problem is similar in shape to the half sine wave that is the correct n = 1 particle-in-a-box wavefunction, so it gives a similar average kinetic 44

energy. The energy in this problem is a little higher, because the particle in a box gives the lowest possible kinetic energy for a system of length a, as per the variational principle described in Section 4.2. Maple. The only trick in putting the kinetic energy operator into Maple format is the second derivative, which I carry out in the example below just by using the diff function twice in a row: • psi := x -> sqrt(30/aˆ5)*((1/4)*aˆ2-xˆ2); • avgK := int(-psi(x)*hbarˆ2*(diff(diff(psi(x), x), x))/(2*me), x = -(1/2)*a .. (1/2)*a); Mathematica. We use D[psi,{x,2}] below to obtain the second derivative of psi with respect to x, and we have to constrain the constant a to be a real number for Mathematica to evaluate the integral: • $Assumptions = a \[Element] Reals • psi:=Sqrt[30/aˆ5]*((1/4)*aˆ2-xˆ2) • avgK=Integrate[-psi*hbarˆ2*D[psi,{x,2}]/(2*me),{x,-a/2,a/2}] 2.11 Carry out the operation with m and k undetermined and see what you get. The factor of (0.805) is just a normalization constant that will be unaffected by the operation, so I call it A and remember that it is part of the wavefunction at the end: 2 d A exp(−mxk ) x5 dx 2A = 5 −mkxk−1 exp(−mxk ) x = −2mkxk−6 A exp(−mxk ) = −2mkxk−6 ψ(x).

α̂ψ(x) =

In order for this to be an eigenvalue equation, the factor multiplying ψ(x) must not depend on x, which means that k = 6 (so that we get a factor of xk−6 = x0 = 1 multiplying ψ). Now we have an eigenvalue of −2 km = −12 m multiplying ψ(x), and since the eigenvalue is equal to −6, m = 1/2. 2

d 2.12 Looking at the operators, we see that the dx 2 part of α̂ will reduce the x part of the function to zero if we operate with α̂ first, and will reduce the 1/x2 part of the function to zero if we operate with α̂ last. So it makes sense to operate with α̂ in the middle. We can also tell that β̂ and γ̂ are essentially the same operator, except for a factor of 2, which can be factored out of all the operations, so the ordering

of β̂ and γ̂ doesn’t matter. Either β̂ α̂γ̂ or γ̂ α̂β̂ will work, and in either case the eigenvalue is 2 : 2 1 1 d 1 2 β̂ α̂γ̂ x + 2 = 2x x + − 2 x dx2 x x 2 1 1 1 = 2x 2 + 3 − 1 − 3 = 2x 1 + 3 = 2 x + 2 . x x x x 2.13 ∂ ∂ (x4 y 2 /3) + 2y Â(x4 y 2 /3) = 2x ∂x ∂y ∂ ∂ = 2x (x4 y 2 /3) + 2y (x4 y 2 /3) ∂x ∂y = 2x(4x3 y 2 /3) + 2y(2x4 y/3) = 8x4 y 2 /3 + 4x4 y 2 /3 = 12(x4 y 2 /3). The constant multiplying the original function at the end is our eigenvalue; hence, the eigenvalue is 12. 45

2.14 One convenient thing here is that no matter how many derivatives of eax you take, eax is still part of the solution. So the derivative dψ/dφ will give 2iψ, for example, which only needs to be multiplied by −2i to get 4ψ. More than one solution exists, but two of the simplest are Â = −2i 2.15

d dφ

B̂ = −

d2 . dφ2

a. e−ax e−bx

d dx

e−ax e−bx = e−(a+b)x d −(a+b)x d −(a+b)x = −(a + b) e e dx dx = (a + b)2 e−(a+b)x

eigenvalue = (a + b)2 b. b sin(ax) cos(ax) d d d (b sin(ax) cos(ax)) = b [sin(ax)(−a sin(ax)) + a cos(ax) cos(ax)] dx dx dx d = b [−a sin2 (ax) + a cos2 (ax)] dx = ab[−2a sin(ax) cos(ax) + 2a cos(ax)(− sin(ax))] = −4a2 b sin(ax) cos(ax) eigenvalue = −4a2 c. cos(ax)e−ax d cos(ax)e−ax = [−a sin(ax) − a cos(ax)] e−ax dx d2 cos(ax)e−ax = −a2 cos(ax) + a2 sin(ax) − a [−a sin(ax) − a cos(ax)] e−ax 2 dx = 2a2 sin(ax) e−ax Not an eigenfunction of the operator. 2.16 Use the Euler equation to rewrite eikx in terms of the trigonometric functions, drop the imaginary part, and then use one of our trig identities in Table A.2 to write what’s left in terms of the sine function: ! ! r r 3 3 i i ikx e = (cos(kx) + i sin(kx)) + + 4 2 4 2 r 3 i2 cos(kx) + sin(kx) = 4 2 ! r 1 3 sin(kx) + cos(kx) +i 4 2 # r ! " r 3 3 i 1 eikx = + cos(kx) − sin(kx) ℜ 4 2 4 2 = sin(kx + φ) 46

= sin φ cos(kx) + cos φ sin(kx) q 3 √ sin φ 4 = tan φ = 1 = − 3 cos φ −2 √ φ = arctan(− 3) = −60◦ . 2.17 We’ve learned only one way to get the momentum of a particle from its wavefunction, and that’s to apply the momentum operator to that function: h̄ d [2 cos(3kx) + 2i sin(3kx)] i dx h̄ [2(−3k) sin(3kx) + 2i(3k) cos(3kx)] = i h̄ = (3ki) [2i sin(3kx) + 2 cos(3kx)] i = 3kh̄ψ(x)

p̂ψ(x) =

The momentum is the eigenvalue in this equation, 3kh̄. 2.18 For this problem, it is necessary to express the momentum operator p̂ in terms of x using the ∂ expression ih̄ ∂x . Then we leave f (x) unknown and see what happens: ∂ ∂ + ih̄ x f (x) (x̂p̂x − p̂x x̂)f (x) = −ih̄x ∂x ∂x ∂f (x) ∂x ∂f (x) + ih̄ x + f (x) = −ih̄x ∂x ∂x ∂x ∂f (x) ∂f (x) = ih̄x − + ih̄f (x) + ∂x ∂x = ih̄f (x) 6= 0, so x̂ and p̂x do not commute. x̂p̂x − p̂x x̂ =

ih̄

Interestingly, most of the wavefunctions we will work with are not eigenfunctions of either position x or momentum p̂x . This problem demonstrates that our wavefunctions are nevertheless eigenfunctions of the commutator [x̂, p̂x ] of the two operators, and that the eigenvalue will always be −ih̄. 2.19 This is a test of our ability to assemble the integral and carry out the operations in the correct sequence. Abbreviating the volume element of the integral by dτ , we have h i d3 d d2 −x 3 Â, B̂ = 2 x dx dx dx h Dh i iE Z Â, B̂ = ψ ∗ Â, B̂ ψdτ 2 Z d ′ ′′′ dτ xψ − xψ = ψ∗ dx2 Z d = ψ∗ (ψ ′ + xψ ′′ ) − xψ ′′′ dτ dx Z = ψ ∗ [ψ ′′ + xψ ′′′ + ψ ′′ − xψ ′′′ ] dτ Z = 2 ψ ∗ ψ ′′ dτ 47

2.20 Here we have a solution to a legitimate quantum mechanics problem, one which we take up in Chapter 8. The hard work has been done, though, and all we have to do is show that the wavefunction is an eigenfunction of the Hamiltonian. The first thing to look at is the result of the differentiation in the kinetic energy operator: −

h̄2 d2 h −(kµ)1/2 x2 /(2h̄) i Ah̄2 d (kµ)1/2 −(kµ)1/2 x2 /(2h̄) = − − Ae (2x)e 2µ dx2 2µ dx 2h̄ 1/2 1/2 2 1/2 2 Ah̄(kµ) (kµ)1/2 = e−(kµ) x /(2h̄) + x − (2x)e−(kµ) x /(2h̄) 2µ 2h̄ s (kµ)1/2 2 −(kµ)1/2 x2 /(2h̄) Ah̄ k 1− x e = 2 µ h̄ " s # h̄ k k 2 = − x ψ(x). 2 µ 2

Now combine that result with the potential energy term 21 kx2 , and look for the original wavefunction in the result. Whatever isn’t the wavefunction, that’s the eigenvalue, which is the energy when our operator is the Hamiltonian:

s # " s k k h̄ h̄2 d2 h̄ k k + 21 kx2 ψ(x) = − x2 + x2 ψ(x) = ψ(x) − 2 2µ dx 2 µ 2 2 2 µ

h̄ 2

k . µ

2.21 Set 2π/λ0 = k0 to simplify the notation. If we take the second derivative of the wavefunction, we get d2 ψ(x) d2 = [f (x) cos(k0 x)] dx2 dx2 d ′ [f (x) cos(k0 x) − f (x) sin(k0 x)] = dx = f ′′ (x) cos(k0 x) − 2k0 f ′ (x) sin(k0 x) − k02 f (x) cos(k0 x). If we only look at values of xn = 2nλ0 /π, we can set the sine term to zero and the cosine terms to one: d2 ψ(x) = f ′′ (xn ) − k02 f (xn ). dx2 x=xn To evaluate the local value p(x)2 , we use the operator p̂2 , where p̂(x)2 ψ(x) = −h̄2

∂ 2 ψ(x) . ∂x2 2

h̄ Therefore, the value of p(x)2 can be found by multiplying our second derivative by − ψ(x) , keeping in

mind that the cosine term in ψ(xn ) is still one: h̄2 ′′ f (xn ) − k02 f (xn ) f (xn ) f ′′ (xn ) = h̄2 k02 1 − 2 k0 f (xn ) 2 f ′′ (xn ) 2π 2 1 − 2 = h̄ λ0 k0 f (xn ) ′′ f (xn ) 2 = p0 1 − 2 . k0 f (xn )

p(xn )2 = −

So what exactly does this show? Well, first notice again that if f (x) is a constant, then f ′′ (x) is zero and the squared momentum is equal to p20 at all xn . In other words, the wavepacket that extends over all x is allowed to have an exactly defined momentum. What’s interesting now is that as the function f (x), which defines an envelope for the wavepacket, becomes more localized, it must develop some curvature. For f (x) to be a well-behaved function, it must be single-valued and have continuous derivatives at all x. For f (x) to be localized—non-zero over a finite range of x values – it must also be composed of both concave up and concave down parts (having positive and negative second derivatives, respectively). For 2 example, if f (x) is a Gaussian function e−ax , it is concave down near the middle, which pushes the function towards zero to keep it localized, but concave up near the edges, so that the first derivative stays continuous as the function approaches zero. That means that f ′′ (x)/f (x) is positive at some places and negative at others; in particular, we would expect the local squared momentum to be higher near the middle of the envelope (f ′′ (x)/f (x) negative) and to diminish as the envelope approaches zero at the ends (where f ′′ (x)/f (x) is positive). The more narrowly localized the wavepacket, the faster the curvature has to change and the larger the magnitude of f ′′ (x). From our solution, that means that a smaller δx requires a larger spread in possible values for p, meaning a larger δp. 2.22 This combines three basic points: (i) λdB depends on the momentum, (ii) kinetic energy can be written in terms of the momentum, and (iii) the energy of a particle in a box is all kinetic energy, because U = 0 inside the box: mv 2 p2 E=K = = 2 2m √ p = 2mE √ h λdB = = h/( 2mE). p 2.23 The purpose of this problem is to illustrate that even a small transition energy such as 1 cm−1 for a light particle such as an electron corresponds to a microscopically small box. The energy gap, 1 cm−1 , is small compared to many of the transition energies we’ll see in the next chapter, but it’s still measurable. Therefore, we don’t expect the box to be very big, because the correspondence principle dictates that the energy will approach being a continuous variable as the size of the box increases. π 2 h̄2 n2 2ma2 π 2 h̄2 2 ∆E(n = 2 → 3) = 3 − 22 2ma2 5π 2 h̄2 = 1 cm−1 = 2ma2 E=

1/2 5π 2 h̄2 a= 2me (1 cm−1 ) 1/2 5π 2 (1.055 · 10−34 J s)2 = 2(9.109 · 10−31 kg)(1 cm−1 )(1.986 · 10−23 J/ cm−1 )

= 1.23 · 10−7 m

= 123 nm

2.24 [Thinking Ahead: Based on the correspondence principle, is the new state at high n or low n? Higher energies correspond to shorter de Broglie wavelengths. If λdB is still half the size of the box, it’s comparable to the domain of the particle, and therefore we’re still far from the classical limit, at low energy.] The wavefunctions of the particle in a one-dimensional box are pure sine waves that extend over a distance nλdB /2, where n is the same quantum number that appears in the energy expression En = n2 h̄2 /(2ma2 ). The state with two full wavelengths in the box has four λdB /2’s, and therefore is the n = 4 state. The n = 4 energy is 42 = 16 times greater than the energy of the lowest state n = 1. Therefore, the energy is 16E1 = 2.08 · 10−18 J. Here’s an alternative way to look at it. The particle in a box has the useful feature that, within the box, the energy is all kinetic energy. Kinetic energy is mv 2 /2 = p2 /(2m) so perhaps a more rigorous way to approach the problem is to see that h h =√ p 2mEn √ h 2ma2 = q 2m(π 2 n2 h̄2 )

λdB =

2πa 2a ha = = πnh̄ πn n a = 2 2a = 4, n= a/2 =

which gives the same answer as above. 2.25 The lowest energy transition for a particle in a one-dimensional box is the n = 1 → 2. The spacing between the energy levels increases with n, so the n = 1 and 2 states are closer together than any other pair of energy levels. So we just need to find an equation for the transition energy and solve for the mass: ∆E = E2 − E1 =

π 2 h̄2 2 (2 − 12 ) 2ma2

3π 2 h̄2 2ma2 3π 2 h̄2 m= 2 2a ∆E 3π 2 (1.055 · 10−34 J s)2 = 2(3.00 · 10−10 m)2 (7.60 · 10−21 J) =

= 2.41 · 10−28 kg.

2.26 This is the solution to the particle in a one-dimensional box after we have applied the first boundary condition ψ(0) = 0, 50

but before we apply the second boundary condition ψ(a) = 0, because the first boundary condition is the result of there being a wall at x = 0, and the second condition is the result of the wall at x = a. Take away the second wall, and we have the free particle wavefunctions under the constraint that ψ(0) = 0, which means all the cosine terms are zero. The unnormalized wavefunction is ψ(x) = c2 sin

! 2mE x . h̄2

2.27 This is a particle in a one-dimensional box problem, and therefore may take advantage of the energy expression obtained for that potential energy function. The problem is asking for a transition frequency, and energies and frequencies are related by Planck’s law. En =

n2 π 2 h̄2 2ma2

mNa = 3.82 · 10−26 kg a = 2 cm

En = n2

−34

π (1.055 · 10 J s) 2(3.82 · 10−26 kg)(0.020 m)2

= n2 (3.59 · 10−35 J) E2 − E1 νn=1→2 = h (3.59 · 10−35 J)(4 − 1) = 6.626 · 10−34 J s = 0.163 s−1

This is a very small energy difference. Because this box is large by atomic standards, the energies are close together, essentially continuous, as expected from the correspondence principle. 2.28 The energy of the particle in a one-dimensional box is all kinetic energy, which we can set equal 1/2 to mv 2 /2. From this we can solve for v 2 and take the square root to get v 2 : n2 π 2 h̄2 mv 2 = 2ma2 2 2 2 2 n π h̄ 2E = v2 = m m2 a 2 E=

2.29

1/2

nπh̄ . ma

a. Write the kinetic energy operator for a particle of mass m anywhere along x. Solution: h̄2 ∂ 2 The same as for any x-dependent system, − 2m ∂x2 . b. Write the boundary condition for the wavefunction at ψ(x = 0) and ψ(x = a) and ψ(x = a/2) (in other words, what is the value of ψ at those points). Solution: Because we have U = ∞ at each of these three points, the wavefunction must be zero at all three points: ψ(x = 0) = ψ(x = a) = ψ(x = a/2) = 0. 51

c. Write an equation for the normalized wavefunctions that satisfy all of these boundary conditions. Solution: We need wavefunctions that satisfy the same conditions as the one-dimensional box wavefunctions and the additional condition that ψ(x = a/2) = 0. That additional condition is satisfied by all the wavefunctions qwith even n (because they go to zero in the middle of the box). The solutions we need are ψ =

2 2nπx a sin a , with n any integer greater than 0.

2.30 The quick way to answer this starts from the solution to the energy of the particle in a box, which is all kinetic energy, mv 2 /2 = p2 /(2m). (Because the wavefunction is an eigenfunction of the Hamiltonian, where Ĥ = −p̂2 /(2m), ψ(x) must also be an eigenfunction of p̂2 .) p2 n2 π 2 h̄2 = 2 2ma 2m 2 2 2 n π h̄ p2 = 2mEn = a2 2 2 2 π (1.055 · 10−34 J s)2 = 1.1 · 10−47 kg2 m2 s−2 . = (2.0 · 10−10 m)2

En =

Operating on ψ(x) with p̂2 = −h̄2 ∂ 2 /∂x2 will yield the same result. 2.31 The n = 21 state should have bigger density in this region, because the n = 1 state rises from the walls and peaks in the middle of the box, whereas higher n states oscillate up and down across the length of the box and are on average more evenly distributed. r nπx 2 sin ψ(x) = a a The probability density between 0 and a/4 is Z a/4 Z 2 a/4 2 nπx a ∗ = ψ(x) ψ(x)dx = dx P 0≤x< sin 4 a 0 a 0 " #a/4 2 x sin 2nπx a = − 4nπ a 2 a 0 nπ i 2 ha a = − sin a 8 4nπ 2 1 1 nπ = − sin 4 2nπ 2 π 1 1 sin = 0.091 n=1 P= − 4 2π 2 1 21π 1 = 0.242 . sin n = 21 P= − 4 42π 2 Classically, the particle spends equal time in each quarter of the box, so P=

1 = 0.250. 4

The higher energy state is much more classical in distribution than the ground state, in concordance with the correspondence principle. 2.32 We are asked to evaluate the integral of the probability density across these five regions. If you can recognize it, it often pays off to consider any symmetry that may simplify the problem: 52

ψ (x)

f3 f2

n=1

f4 f5

x 0

ψn=1 (x) =

πx 2 sin a a

The second half of the curve is a mirror image of the first half, and therefore the fraction of time in the region 0 to a/5 is the same as in the region 4a/5 to a, etc.: f1 ≡ f2 ≡

Z a/5 0

Z 2a/5 a/5

|ψ|2 dx = |ψ|2 dx =

Z a

4a/5

|ψ|2 dx ≡ f5

Z 4a/5 3a/5

|ψ|2 dx ≡ f4

The results 2 a

Z x2 x1

sin

πx a

# x2 x sin 2πx a − 4π 2 a x1 x2 (x2 − x1 ) 2πx1 2πx2 1 sin − sin = + a 2x a a x1

2 dx = a

1 [sin(0) − sin(2π · 0.2)] = 0.04863 2π 1 f2 = f4 = 0.2 + [sin(2π · 0.2) − sin(2π · 0.4)] = 0.25782 2π 1 f3 = 0.2 + [sin(2π · 0.4) − sin(2π · 0.6)] = 0.38709 2π f1 = f5 = 0.2 +

The five terms add up to 1.0 as they should. Note how different this is from a classical particle, for which these values would all be 0.20. The quantum particle in this state is about eight times more likely to be in the middle fifth than at either of the ends. 2.33 [Thinking Ahead: If you are a particle in this box, what properties of the box can you tell have changed? None. The value of x at the beginning of the box is a choice made by the person writing the problem, and if you’re trapped in the box, it’s possible they haven’t bothered to tell you what value they’ve chosen. The length of the box and the mass of the particle—things that you would be aware of—are still the same.] This is almost exactly the same as the particle in a box problem solved in the text. All that has changed is the location of the coordinate system: the origin of the x axis has been shifted by a/2. Since the physics of the system cannot depend on our choice of coordinates, the energies must be unchanged and the wavefunctions are different only by a shift in the x value by a/2: 53

En =

n2 π 2 h̄2 2ma2 " # nπ x + a2 2 sin a a

ψ(x) =

h πx π i 2 sin n + a a 2 nπx 2 . cos a a

R 2.34 Orthogonality is the characteristic shared by two functions ψa and ψb for which all ψa∗ ψb dτ = 0. space For this problem, we need to establish that this relationship is true for the lowest energy wavefunctions of the particle in a one-dimensional box: q q 2 2 πx 2πx ψn=1 (x) = ψn=2 (x) = a sin a a sin a R∞ If orthogonal, −∞ ψ1 (x)ψ2 (x)dx = 0. Z ∞

Z πx 2 a 2πx sin dx sin a 0 a a Z a πx πx 2 2 sin2 cos dx = a 0 a a h i πx a 4 a sin3 = a 3π a 0 4 (0 − 0) = 0 = 3π

ψ1 (x)ψ2 (x)dx =

−∞

sin(2x) = 2 sin x cos x

The n = 1 and n = 2 states are orthogonal. 2.35 [Thinking Ahead: What is the answer for the classical system? Classically, the average position of the particle would be in the middle of the box, no matter how much energy we gave it, because the particle moves at a constant speed whether in the first half of the box or the second half. One can see from the wavefunctions in Fig. 2.6 that the probability distributions will be symmetric about the middle of the box in the quantum case, so the average will still be at a/2.] We use the average value theorem to calculate the average of x for the general wavefunction ψn (x): hxi =

Z a 0

ψ ∗ xψdx = A2

Z a 0

sin2

nπx a

x dx.

The indefinite integral we need is Z

sin2 (cx)x dx =

2 x sin(2cx) 1 x − − 2 cos(2cx) . 4 4c 8c 54

In our case, c = nπ/a, so: 2 a x x sin(2cx) 1 hxi = A − − 2 cos(2cx) 4 4c 8c 0 2 a cos(0) 2 a sin(2ca) cos(2ca) = − 0 − 0 − − − a 4 4c 8c2 8c2   1 0 2 1  a sin(2nπ) cos(2nπ) 2 a + 2 − =  − a 4 4c 8c2 8c 2

a . 2

Maple. Particularly if you already know what the answer will be, you should be forgiven for not wanting to grind through the integration by hand. In this case, we need to let Maple know that n can only have integer values; otherwise it won’t be able to simplify functions such as cos(2nπ): • assume(n::integer); • psi := x -> sqrt(2/a)*sin(n*Pi*x/a) ; avgx := int(psi(x)ˆ2*x, x = 0 .. a); Mathematica. We need to let Mathematica know that n is an integer, so that it can simplify the integrals over functions such as cos(2nπ). In this set of commands, I enforce that constraint in the last step, specifying that n is an element of the set of Integers: • psi:=Sqrt[2/a]*Sin[n*Pi*x/a] • avgx=Integrate[psiˆ2*x,{x,0,a}] • Simplify[avgx,{n \[Element] Integers}] 1/2

instead of hpx i is that the particle has equal momentum in either 2.36 The reason we use p2x direction when averaged over time, so the average is always zero: Z Z h̄ a nπx nπh̄ a nπx nπx ∂ nπx hpx i = sin dx = − cos dx = 0. sin sin i 0 a ∂x a ia 0 a a 1/2

Two ways to solve for p2x are shown below. Which one is preferable depends on how much is already known about the system. The short way to find this answer is to recall that all the energy in the system is kinetic energy, and therefore p2 , 2m 2 2 2 n π h̄ n2 π 2 h̄2 p2 = 2mE = 2m = 2ma2 a2 E=

1/2

nπh̄ . a

Please realize, however, that such a straightforward answer is available only because p2 happens to be directly proportional to the total energy in this example. Our wavefunctions have been selected to satisfy the Schrödinger equation, which means they are eigenfunctions of the energy and, in this case, also of p2 . As soon as we go to examples for which the energy includes some contribution from a potential energy that changes with position, p2 stops being proportional to E, and we can no longer use the simple relation we used here. In such cases, the rms momentum can be evaluated by taking the square root of the expectation value of p2x , which can be obtained from the average value theorem. r nπx 2 ψn (x) = sin a a Z ∞ 1/2 ψn (x < 0) = ψn (x > a) = 0 2 1/2 ∗ 2 px = ψn (x) p̂x ψn (x)dx ψn∗ (x) = ψn (x) −∞ 1/2 Z a ∂2 1/2 = p2x ψn (x) −h̄2 2 ψn (x)dx ∂x 0 1/2 2 Z a 2h̄ nπx ∂ 2 h nπx i dx = − sin sin a 0 a ∂x2 a Z Z 2 1/2 a 2h̄ −n2 π 2 x sin(2cx) 2 nπx = − dx +C sin sin2 (cx)dx = − 2 a a a 2 4c 0 " #a #1/2 " x sin 2nπx 2n2 π 2 h̄2 a = − a3 2 4 nπ a 0 1/2 2 2 2 a 2n π h̄ −0 = a3 2 =

n2 π 2 h̄2 a2

1/2

nπh̄ a

2.37 This is a problem typical of those appearing in subsequent chapters on quantum mechanics. We have computers capable of solving difficult integrals, but we must be able to spell out the integrals we want to evaluate, including the integrand, the volume element, and the limits: r 3πx h̄ ∂ 2 sin ψ= p̂ = a a i ∂x Z ∞ xp2 = ψ ∗ (x)(xp̂2 ψ(x)dx −∞ # # "r Z a "r 2 3πx 3πx 2 2 2 ∂ = −xh̄ dx sin sin a a ∂x2 a a 0 2 Z a 3πx 2h̄2 3π 2 dx. x sin =+ a a a 0 2.38 Region A has zero potential energy, and therefore only the kinetic term appears in the Hamiltonian. ∂2 For motion along x, that kinetic energy term is always −h̄2 /(2m) ∂x 2 . Region B has the kinetic term plus a potential energy which increases linearly with x at a rate of 2 Eh /4 Å. 2

∂ h̄ • ĤA = − 2m ∂x2

∂ h̄ • ĤB = − 2m ∂x2 + (0.5 Eh / Å)x

2.39 Normalizing any function ψ means evaluating the integral the integral is exactly equal to one.

all

space

A2 ψ ∗ ψdτ and adjusting A so that

ny πy nz πz nx πx sin sin ψ(x, y, z) = A sin a Z a Z aa Z a Z aZ aZ a a ny πy nz πz 2 nx πx ∗ 2 ψ(x, y, z) ψ(x, y, z)dxdydz = A dx dy dz sin sin2 sin2 a a a 0 0 0 0 0 0 a" " #a #a n πy sin2 nxaπx sin2 nzaπz y sin2 ya  z 2 x  =A − 4nx π − − 4nz π 4ny π 2 2 2 a a 0 0 a 0 hai hai hai = A2 2 2 2 3 a = A2 3 = 1 2 23 2 A= a 2.40 We can check the energies for each of those states using Eq. 2.41: E=

π 2 h̄2 (n2 + n2y + n2z ) 2mV 2/3 x g

72 + 52 + 52 72 + 62 + 22 82 + 42 + 32 92 + 22 + 22

= 99 = 89 = 89 = 89

6 6 3 g(E) = 15

The (7,5,5) state does not belong; the correct degeneracy is 15. 2.41

a. What is the value of |ψ|2 in the center of the box for the state (1,1,1)? Solution: The center of the box is at a/2 along each axis, where sin(πx/a) = sin(π/2) = 1. So the value of p ψ is 8/a3 and the value of |ψ|2 is 8/a3 = 0.125 Å−3 .

b. What is the value of |ψ|2 at the point x, y, z = 0,a,a for the state (1,1,1)? Solution: The wavefunction must go to zero at the walls (when any coordinate = 0, or when any coordinate = a) and so does |ψ|2 .

c. What is the value of the probability density integrated from the origin to x, y, z = a/2, a/2, a/2 for the state (1,1,1)? Solution: The sine waves have equal density in either half of the box along each coordinate, so integrating |ψ|2 along x from 0 to a/2 gives 1/2. You get the same factor from integrating halfway along y and halfway along z, for a total of 1/8, or 0.125. d. What are the x, y, z coordinates at each position where the state (2,1,1) reaches its maximum probability density? Solution: With ny = nz = 1, the y and z components of the wavefunction reach their maximum values in the middle of the box at y = z = a/2. For nx = 2, the x component of the wavefunction is a sine wave reaching a maximum at a/4 and a minimum of equal magnitude at 3a/4. The probability density peaks at these two points: (a/4, a/2, a/2) and (3a/4, a/2, a/2). 57

2.42 Why—mathematically and physically—shouldn’t I be able to normalize this equation? A free particle can travel anywhere, and our wavefunctions only describe the time-independent properties of the particle, and over all time our particle can travel all the way from −∞ to ∞ along every axis. So, physically, the lack of normalization results from the fact that this wavefunction tells you absolutely nothing about where the particle could be. On the other hand, conserved quantities, such as the energy and momentum, can be obtained from these wavefunctions. Mathematically, the problem is that the R integral |ψ|2 dx dy dz approaches ∞ when integrated over all space, so the normalization constant approaches zero. The Hamiltonian for the three-dimensional free particle is Ĥ =

p̂2y p̂2 p̂2 p̂2 h̄2 = x + + z =− 2m 2m 2m 2m 2m

∂2 ∂2 ∂2 + 2+ 2 2 ∂x ∂y ∂z

Because the Hamiltonian is divided into three independent terms, the coordinates are separable, meaning that the wavefunction is of the form ψ(x, y, z) = ψ(x)ψ(y)ψ(z) and the energy is of the form E = Ex + Ey + Ez

where Ĥx ψ(x) = Ex ψ(x) , and so forth.

These are the wavefunctions of the one-dimensional free particle: r 2mEx ikx ψ(x) = cx e kx = . h̄2 Keeping in mind that Ex , Ey , and Ez are independent parameters, the solution is: ψ(x, y, z) = cx cy cz eikx eiky y eikz z , where kx =

2mEx h̄2

ky =

2mEy h̄2

kz =

2mEz . h̄2

2.43 We set the ground state energies equal, using the two different masses, and cancel all the common factors on both sides before solving for c:

2 2

π h̄ 2mp

2.44

E1,1,1 (e) = E1,1,1 (p) π 2 h̄2 1 1 1 1 1 1 = + + + + a2 a2 c2 2me a2 a2 a2 1 1 1 2 3 + 2 = mp a 2 c me a 2 2 mp 3 mp 3 1 − = ≈ c2 me a 2 a2 me a 2 s r me 9.109 · 10−31 kg a= (10.0 nm) = 0.135 nm. c= 3mp 3(1.673 · 10−27 kg)

a. Motion along this axis makes the greatest contribution to the total kinetic energy. x The kinetic energy along each direction is proportional to n2 /d2 , where d is the length of the box along that axis. The x axis has the largest n value and smallest length, so the greatest kinetic energy. 58

b. Along this axis the wavefunction has the greatest number of nodes. x The number of nodes along any axis is equal to n − 1. The x axis has the greatest n value, and therefore the most nodes. c. Along this axis the wavefunction has the greatest de Broglie wavelength. z De Broglie wavelength is inversely proportional to momentum, so the coordinate with the lowest kinetic energy (and hence lowest momentum) must have the greatest λdB . That’s the z axis. d. Along this axis the probability density has exactly two maxima. y The probability density is proportional to the square of the wavefunction. Since the ny = 2 component of the wavefunction has one positive maximum and one negative minimum, the probability density will have two maxima. 2.45

a. The zero-point energy (in J). Solution: For a cubical box, the energy of the (2,2,2) state is at 22 + 22 + 22 = 12 times ε0 = π 2 h̄2 /(2ma2 ), while the ground state is at 3ε0 . The energy of the ground state is equal to the zero-point energy in this case (because the potential energy is a constant), 3/12 of the (2,2,2) energy is equal to 1.5 · 10−20 J. b. λdB (in Å) of the particle along any axis. Solution: Each side of the box has length 2.0 Å. With n = 2 along each axis, the wavefunction undergoes one complete cycle from one end of the box to the other, so λdB is the length of the side: 2.0 Å. c. The new energy in the (2,2,2) state if we double the length of each side of the box. Solution: Doubling the length a will decrease the value of ε0 = π 2 h̄2 /(2ma2 ) by a factor of 22 = 4, so the new energy is again 1.5 · 10−20 J. d. The mass (in kg) of the particle. Solution: For this one, we actually need a calculator. Solve the energy E2,2,2 = 12ε0 for m: m=

12π 2 (1.055 · 10−34 J s)2 12π 2 h̄2 = = 2.7 · 10−28 kg. 2 2a E2,2,2 2(2.0 · 10−10 m)2 (6.0 · 10−20 J)

2.46 The degeneracy is 2 because (1,2,1) and (2,1,1) both have the same energy, whereas (1,1,2) will be higher energy. 2.47 To start off, try writing the general expression for this energy, and then set it equal to the value in the problem and see what constraints appear: ! ! n2y n2y π 2 h̄2 n2x n2x n2z n2z = + 2 + + 2 + 2 a2 a (2a)2 2m a2 a 4a 2 2 2 2 π h̄ 1 π h̄ 4n2x + 4n2y + n2z = 4n2x + 4n2y + n2z = 2 2 2m 4a 8ma

π 2 h̄2 E= 2m

7π 2 h̄2 ma2 56 = 4n2x + 4n2y + n2z =

So now we need to find all combinations of nx , ny , nz such that the sum of the squares is 56, keeping in mind that the quantum numbers must all be integers. One way to do it is to recognize that nz < 8 59

and to start checking the possible values of nz and seeing if any values of nx , ny work out: nz 7 6 5 4 3 2 1

4(n2x + n2z ) n2x + n2y nx , ny 7 7/4 not integer 20 5 1, 2 2, 1 31 31/4 not integer 40 10 1, 3 3, 1 47 47/4 not integer 52 13 2, 3 3, 2 55 55/4 not integer

The solutions are (1,2,6), (2,1,6) (1,3,4), (3,1,4), (2,3,2), (3,2,2). 2.48 The number of nodes is n − 1 for each coordinate. The state is (2, 4, 1). 2.49 The energy is proportional to n2y n2x n2z + + a2 b2 c2

where in this case a and b are equal to c/2: E∝

n2 4n2x 4n2y + 2 + 2z 2 c c c

1 4n2x + 4n2y + n2z . 2 c

We want the lowest value for this sum where more than one combination of values (nx , ny , nz ) give the same energy. The lowest energy values will have nx and ny equal to 1, but at least one of these must change to 2 in the lowest energy set of degenerate states (because all the states with nx = ny = 1 have different energies, and are not degenerate). The minimum energy for any set of degenerate states is therefore at least 4(22 ) + 4(12 ) + (12 ) = 21, where we’ve set nx = 2 and left ny and nz equal to 1. Because the lengths of the box are the same along the x and y axes, the state (2,1,1) must have the same energy as the state (1,2,1), so these are the lowest pair of degenerate states. The next lowest energy degenerate states are (1,1,4), (2,1,2), (1,2,2). 2.50 How many quantum numbers will be needed? Two: one quantum number for each coordinate in the problem. A quantum number tells us about the amount of energy stored in motion along some coordinate of the system, so the number of quantum numbers goes hand-in-hand with the number of coordinates in which the system is free to move. Call the coordinates in the plane of the crystal x and y. These are separable coordinates, as for the three-dimensional box, with each coordinate q contributing an energy h2 2 Eq = n . 8ma2 q The total energy is E = Ex + Ey =

h2 (n2 + n2y ) 8ma2 x

where h is Planck’s constant, m is the mass of the molecule, a is the length of the crystal face, and nx and ny are the quantum numbers for the x and y coordinates. The lowest set of degenerate states is (nx , ny ) = (2, 1), (1, 2) , because E(2, 1) = E(1, 2). The only state with lower energy is (1,1), which is unique. As with the three-dimensional box, there are no states where either nx or ny are zero, because the corresponding wavefunctions are zero everywhere. 60

2.51 To solve the degeneracy for the three-dimensional box in terms of the energy, we first needed an equation for the energy. We’ll need a different equation for the two-dimensional box. The math is almost identical, but if we start from Eq. 2.40, we only add terms for x and y if the box is two-dimensional: π 2 h̄2 Enx ,ny = 2m

n2y n2x + a2 b2

π 2 h̄2 2 n + n2y 2mA x

≈

h2 n2 π 2 h̄2 n2 = 2mA 8mA 2 ≡ n ε0 , ≡

where ε0 = h2 /(8mA), and where the box is initially given length a along x and b along y, but we have assumed a2 and b2 each to be roughly equal q to the area A for simplicity. We have also created the variable n, which in this case is equal to n2x + n2y . Now we derive the degeneracy. Beginning with the energy level density, we draw a circle of radius n, and stipulate that since all states have energy proportional to n2 , the area within a distance dn of the circumference (2πn) of this circle at a given radius n contains the points (nx , ny ) for all the states that have the same energy. Since all valid states must have positive nx and ny , we only count the points in the quadrant where nx and ny are both positive, which is one fourth of the entire circumference: πn 2πn dn = g(E) = 4 2

1 2n

π . 4

In the limit of large n, the average degeneracy approaches a constant a little less than one. The density of quantum states W (E) is approximately W (E) = =

= R n2 π

n dn n1 2

R E2 E1

g(E)

E2 − E1

(n22 − n21 )ε0 n2

π 22 − 21 = 2ε0 n22 − n21 π . = 4ε0 For the one-dimensional box, the density of states W decreased with energy; for the three-dimensional box, W increases as E 1/2 . Our result here predicts that W for the two-dimensional box is a constant. Try it out; you’ll find, for example, that the number of states with energies between 16 and 30 in units of h2 /(2mA2 ) is the same as the number between 31 and 45. 2.52 The energy is all kinetic energy, because the potential energy inside the box is zero. The speed along the z axis will determine the kinetic energy contribution from motion along z, and that’s what we want to maximize: π2 2 nx + n2y + n2z 2 24 = n2x + n2y + n2z .

12π 2 =

If the sum is 24, then nz must be less than 5. Setting nz = 4, we can satisfy the above equation if 61

nx = ny = 2. Now we solve for vz from the kinetic energy along z; mvz2 π2 2 = n = 8π 2 Eh = 3.44 · 10−16 J 2 2 z r 1/2 2Kz 2(3.44 · 10−16 J) vz = = = 2.75 · 107 m s−1 . me 9.109 · 10−31 kg

Kz =

2.53 The number of nodes in a wavefunction of the particle in a one-dimensional box is equal to n − 1. For the three-dimensional box, the wavefunction along each coordinate obeys the same rule, using the corresponding quantum number. So if there is one node along z, then nz = 2. That means there is a node every 0.5 Å along each of the other coordinates also, meaning 3 nodes along x so nx = 4 and 7 nodes along y so ny = 8. 2.54 This problem demonstrates how, within limits, one kind of potential energy curve can be used to estimate parameters of another potential energy curve. The particle in a box and the Coulomb potentials do not have the same shape, but the particle in a box may still be used to obtain a rough idea of the dimensions of the electron distribution around a nucleus. The problem requires only writing the energy level expression for each of the two potentials, and determining the conditions under which the transition energy is equal. h2 2 h2 (2 + 12 + 12 ) − (12 + 12 + 12 ) = 3 2 8ma 8ma2 2 4 2 4 1 1 Z me e 3 Z me e . − 2 = E2 − E1 = 2 2 1 2 4 2(4πǫ0 )2 h̄2 2h̄

E2,1,1 − E1,1,1 =

box atom

If E2,1,1 − E1,1,1 = E2 − E1 , then 3 h2 3 Z 2 me e 4 = 8 me a 2 8 (4πǫ0 )2 h̄2 a2 =

(4πǫ0 )2 h2 h̄2 Z 2 m2e e4

2(4πǫ0 )πh̄2 Z 2 m2e e2

a0 =

4πǫ0 h̄2 m2e e2

2π a0 . Z

For Z = 1, this is a box roughly 3 Å on a side, compared to the 2a0 = 1.06 Å diameter of the electron orbit in the Bohr model of the hydrogen atom. 2.55 Will there be more states or fewer between E = 61 and 65 in these units than between 55 and 60? We expect more, because the density of quantum states generally increases as E increases. However, there are still some fluctuations at this low energy range, and we can’t be sure before checking. In fact, we will get the same number of states. To get to E = 65π 2 h̄2 /(2mV 2/3 ), we do not have to consider any states for which nx , ny , or nz > 7, because with any ni = 8, n2i + n2j + n2k ≥ 66. The possible 62

combinations are these: E 61 62 65

g 6 12 6

states (6, 4, 3) (6, 5, 1), (7, 3, 2) (6, 5, 2)

2.56 This system now has six coordinates: three for each of the two particles. We start with the lowest energy state, and see what happens as we add one quantum of energy at a time to any or any combination of the three coordinates: π 2 h̄2 (n2 + n21y + n21z + n22x + n22y + n22z ) (2ma2 ) 1x 2 2 π h̄ g states E 2ma 2 E=

6 9 12 14 15

1 6 15 6 20

(111)(111) (211)(111), (121)(111), . . . (221)(111), (212)(111), (211)(211), . . . (311)(111), (131)(111), . . . (222)(111), (221)(211), (221)(121), . . .

2.57 These are tiny particles but a fairly big system. Which end of the correspondence principle are we at: the classical or the quantum? As a rough rule, it is enough that any one of the criteria be satisfied for the system to behave classically. To be certain, check the de Broglie wavelength against the domain of the system. In this case, even though the particle mass is very low and the energy is unspecified, it will be essentially impossible for us to slow the electrons enough to get λdB values close to 2 cm. The large volume is sufficient for the electron to behave classically. 2 2 π h̄ where, in For a particle in a three-dimensional box, ∆En,n+1 is equal to one to three times 2ma 2 this case, a = 2 cm. Therefore, ∆E ≈

2π 2 (1.055 · 10−34 J s)2 = 3.01 · 10−34 J 2(9.109 · 10−31 kg)[(2 cm)(0.01 m cm−1 )]2

= 1.52 · 10−9 cm−1 .

The electronic energy levels are essentially continuous.

Chapter 3 3.1 The answer is in the geometry. If I move a short distance δr from the nucleus, it doesn’t matter what angle I’m at from the x or z axes; the distance I travel is still δr. However, if I change angle by an amount δθ, the actual distance I travel depends on my distance from the nucleus, r. A very small excursion of δθ in angle is a change in distance of rδθ, so the distance is proportional to r. The same is true for motion along φ, except in that case the distance traveled also depends on the value θ. If we travel a complete circle, from 0 to 2π, when θ is close to zero, we describe a small circle close to the z axis. At the same value of r, but with θ = π/2, we would describe a much bigger circle. The distance traveled for a change δφ = r sin θ δφ. 3.2 No, stationary particle 1 doesn’t lose energy, at least not in the same sense as the accelerating particle 1. Briefly, the accelerating charge loses energy by first converting kinetic and/or potential energy into photons, which propagate away from the system and which somewhere else can be expressed 63

as a potential energy to do work. In contrast, the stationary particle 1 exerts a force on particle 2 by virtue of a potential energy that the system already has. The trick here is not to let the phrasing of the question mislead you. Energy is not “lost” in chemical systems; we simply transfer it from one place or form to another. In the case of the accelerating particle, the particle’s kinetic energy is converted to potential energy (in the form of the photon electromagnetic field), which is transmitted elsewhere at the speed of light. When those photons run into particle 2, then the photon energy can be converted back into kinetic energy by accelerating particle 2. Because particle 1 does not interact directly with particle 2 in this scenario, we say that particle 1 has lost energy by transferring it through photons into particle 2. So how about the stationary particle 1? In this case, the energy used to accelerate particle 2 is obtained from the potential energy U = q1 q2 /(4πǫ0 r), which is already present as potential energy in the system. If particle 1 remains fixed, then it starts with zero kinetic energy and it stays at zero kinetic energy. The complication is that the potential energy U of the system is a shared function of particles 1 and 2, because it depends on their relative positions; it doesn’t really belong to either particle alone. As particle 2 moves, the potential energy for the system of both particles changes, while an equal and opposite change in particle 2’s kinetic energy allows the total energy to stay conserved. What’s important for our purposes is the value of the potential energy for the interaction between the two particles. A deeper significance to this question may lie in the fact that the fundamental forces in physics are described theoretically by exchanges of particles, and the electromagnetic force is presumed to act by the exchange of photons between the interacting particles. In other words, even our stationary particle example here involves energy passing from particle to particle by means of photons. The two scenarios perhaps do not differ as much as they seem. 3.3 The 5s state is higher energy and its density is more widely distributed in space; one would therefore expect the 5s electron to behave more classically. However, because the potential energy function levels off at large r, the average kinetic energy of the electron actually decreases as the overall energy of the state increases (see Problem 1.14). In the limit that you put just enough energy into the electron to ionize it, the average kinetic energy approaches zero. As one can see from the radial wavefunctions in Fig. 3.7, the de Broglie wavelengths lengthen considerably as one moves outward from the nucleus. Consequently, the answer is perhaps best put this way: the 5s electron should act more classically than the 2s in the region close to the nucleus that they both occupy, but measured properties of the 5s electron at large r will show much more quantum-like behavior, even more so than the 2s. How this appears in the lab will depend on the specifics of the experiment, but one application that is particularly sensitive to these differences is molecular beam scattering, where atoms and/or molecules are bounced off each other and the angles of deflection are carefully measured. These experiments reveal patterns of rainbow and glory scattering, which have analogs in the behavior of scattered light, and are thus indications of the wave-like nature of the electrons. 3.4

a. What is the value of l? 2 b. What are all the possible values of ml ? −2, −1, 0, 1, 2 c. What are all the possible values of ms ? −1/2, 1/2 d. How many orbitals are there? 5

3.5

a. What are the values of n and l, and what are the possible values of ml ? n = 6, l = 1, ml = −1, 0, 1 b. How many radial nodes does the wavefunction have? n − l − 1 = 4 64

c. How many angular nodes does the wavefunction have? l = 1 d. How many orbitals are there in the subshell? number of values of ml = 3 e. What is the probability density at r = 0? If l = 1, then Rn,l (r = 0) = 0. 3.6 To begin, just to be clear, there cannot be any angular momentum in the one-dimensional box because one coordinate doesn’t allow any angular motion. Mathematically, the ~r and p~ vectors are always parallel, so their cross product (proportional to the angular momentum) is zero. But a classical particle in a three-dimensional box can certainly have orbital motion and angular momentum. Quantum mechanics can predict classical results, as long as we’re willing to go to the trouble to drag the math along as we climb up in energy towards the classical limit. Therefore, if we know that a classical particle can have angular momentum, we should be able to find it in the quantum mechanics. Figure 3.6 is a logical place to look for how angular momentum comes into your wavefunction. In those graphs, only one of the wavefunctions pictured has zero angular momentum: the l = 0 function. What distinguishes this function from all the others? It has no angular nodes, for one thing, but the critical point is that it is spherically symmetric. If our wavefunction depends on angle at all, then the motion of the particle is angle-dependent and the L̂2 term in the Hamiltonian cannot be zero. All of our three-dimensional box wavefunctions vary with angle. We know this because the wavefunctions must all go to zero at the walls, but the distance from the center of the box to the wall is shorter if we travel along a coordinate axis, x or y or z, than if we travel diagonally towards one of the corners. Therefore, all of our three-dimensional box wavefunctions have angular momentum, even the ground state (1,1,1). However, unlike the one-electron atom wavefunctions, these wavefunctions are not eigenstates of L̂2 , so the angular momentum is not precisely known for any of the three-dimensional box states, and this is why L played no role in our discussion of the particle in a box. In case you’re interested, the average values of L2 in the lowest energy states of the three-dimensional box, obtained from a numerical integration, are 0.43h̄2 for (1,1,1) (E = 3E0 ), 3.12h̄2 for (2,1,1) (E = 6E0 ), 8.05h̄2 for (2,2,1) (E = 9E0 ), 6.48h̄2 for (3,1,1) (E = 11E0 ), and 15.2h̄2 for (2,2,2) (E = 12E0 ). One last point: even though our eigenstates all have angular momentum, this doesn’t mean that a wavefunction with zero angular momentum is impossible in the three-dimensional box. In fact, because the eigenstates are a complete set of functions, there must be some way to combine them to get a spherically symmetry wavefunction, so that the angular momentum is zero. However, this would be achieved by adding together different wavefunctions from many different energies, and the resulting L = 0 wavefunction would not be an eigenfunction of the Hamiltonian. In other words, it would not have a well-defined energy, and therefore would rapidly settle into a stationary state with constant energy and non-zero angular momentum. 3.7 We can write wavefunctions for the particle in a one-dimensional box that include directiondependence, but those functions are different in an important respect from the direction-dependent angular momentum functions eiml φ . If we tried going by the same logic that underlies Fig. 3.13, we would put the direction-dependence into the function by adding an imaginary term as follows: ψ(x) = sin x ± i cos x = −ie±ix

0 ≤ x < a.

Here, for example, we can imagine that a sine wave running in the +x direction would look like − cos x after moving one quarter wavelength, and the wavefunction of a particle traveling in the −x direction would shift towards + cos x, giving rise to the two different signs of the cos x term. These wavefunctions with general form e±ikx are called plane waves, because the value of the wavefunction at any value of x is the same at all points along the yz plane. They are often used in theoretical studies of optics, solid 65

state electronics, and other problems involving the propagation of energy, where the direction of motion often matters. As written, however, these are not correct solutions to the one-dimensional box problem, because the probability density does not approach zero at the boundaries x = 0 and x = a. Because the probability density must be zero outside the walls, the probability density therefore becomes a discontinuous function, violating one of our requirements for physically realistic behavior. The critical difference between these two complex functions proposed for the particle in a box and the iml φ e phase factors for the atomic wavefunctions is what happens when we average the system over time. Just like the classical particle rolling without friction between two walls, the quantum particle would change direction upon hitting one wall and move towards the opposite wall. Its wavefunction would have to change from, say, sin x + i cos x to sin x − i cos x at regular intervals, and the imaginary term would average over time to zero. To show the direction-dependence of a particle in a one-dimensional box, it would be necessary to write a time-dependent wavefunction, and by definition that is not a stationary state, nor is it an eigenfunction of our time-independent Hamiltonian. No reversal of direction is necessary for circular motion, so the direction of rotation does not vanish upon averaging over time. 3.8 Yes, this is consistent with the correspondence principle. Greater mass corresponds to a more classical limit, and should be accompanied by more nearly continuous energy levels—so the gaps between levels for the heavier nucleus should be smaller than those for electron–electron interactions. 3.9 [Thinking Ahead: What coordinate or coordinates best parametrizes this motion? Angular motion in the xy plane is described by the angle φ, so that is probably the most appropriate coordinate to use for this system.] There are no kinetic terms for motion in r or θ because motion is confined to constant r and θ. Therefore, the only kinetic term must be the one that involves the derivative with respect to φ. The potential energy term is unchanged. Ĥ = −

Ze2 ∂2 h̄2 2 ∂φ2 − 4πǫ r . 2 2me r sin θ 0

Furthermore, we can set θ = π/2, because it is only at this value that motion along φ describes a circle with the nucleus in the middle. (At lower values of θ, for example, motion along φ would generate a circle centered on some positive value of z, above the nucleus.) So we can simplify the final equation to Ĥ = −

h̄2 ∂ 2 Ze2 − . 2me r2 ∂φ2 4πǫ0 r

The eigenfunctions of this Hamiltonian are the φ-dependent parts of our spherical harmonics, eiml φ and the eigenvalues are m2 h̄2 Ze2 Eml = l 2 − . 2me r 4πǫ0 r These energies are r-dependent because r is a constant that we can choose freely. h̄ ∇2 and the potential 3.10 The Hamiltonian consists of two parts: the kinetic energy operator K̂ = − 2m energy U . The potential energy term is given in the problem, so we just need to write K̂ correctly. From our experience with the Cartesian and spherical polar forms of K̂, we know to expect three terms corresponding to motion along each of the three cylindrical coordinates. Because z is the Cartesian coordinate, we can just use the z term of the Cartesian form of ∇2 :

K̂z = −

h̄ ∂ 2 . 2m ∂z 2

Similarly we can borrow the r- and φ-dependent parts of the polar form of ∇2 , but what do we do about the factor of sin θ that appears in the φ-dependent part? Because r is measured from the z axis, 66

not from the origin, the ~r vector we’re using is always perpendicular to z. That means that in this coordinate system, θ is fixed at π/2 and sin θ = 1: h̄ 1 ∂ 2 ∂ 1 ∂2 K̂r + K̂φ = − . r + 2m r2 ∂r ∂r r2 ∂φ2 The overall Hamiltonian is the sum of these terms (including U ): Ĥ = Kz + Kr + Kφ + U 2 h̄ C ∂ 1 ∂ 2 ∂ 1 ∂2 =− − . + r + 2 2 2 2 2m ∂z r ∂r ∂r r ∂φ r 3.11 We begin with our explicit expressions for the Laplacian operator in polar coordinates, from Eq. 3.3, ∇2 =

∂ 1 ∂ 1 ∂2 1 ∂ 2 ∂ r + sin θ + r2 ∂r ∂r r2 sin θ ∂θ ∂θ r2 sin2 θ ∂φ2

and the wavefunction, from Tables 3.1 and 3.2, ψn,l,ml =2,1,1 = AR2,1 (r)Y11 (θ, φ) Zr =A e−Zr/(2a0 ) sin θ eiφ a0 ∂ 2 AZr −Zr/(2a0 ) ∂ 1 ∂ 1 1 ∂ 2 ∂ 2 ∇ ψ2,1,1 = 2 r e sin θ eiφ + sin θ + r ∂r ∂r r2 sin θ ∂θ ∂θ r2 sin2 θ ∂φ2 a0 Z 1 ∂ 2 AZ −Zr/(2a0 ) AZr −Zr/(2a0 ) sin θ eiφ e e − = 2 r r ∂r a0 a0 2a0 ∂ AZr −Zr/(2a0 ) 1 iφ e (cos θ) e sin θ + 2 r sin θ ∂θ a0 1 AZr −Zr/(2a0 ) iφ + 2 2 e sin θ(−e ) r sin θ a0 AZ AZr2 −Z 1 −Zr/(2a0 ) e + e−Zr/(2a0 ) = 2 (2r) r a0 a0 2a0 3AZ 2 r2 −Zr/(2a0 ) AZ 2 r3 −Z −Zr/(2a0 ) − e − e sin θ eiφ 2a20 2a20 2a0 AZr −Zr/(2a0 ) 1 − sin2 θ + cos2 θ eiφ e + 2 r sin θ a0 AZr −Zr/(2a0 ) 1 − 2 2 e sin θ eiφ r sin θ a0 2 Z 1 cot2 θ csc2 θ 3Z Z2 AZr −Zr/(2a0 ) e sin θ eiφ 2 − − − + 2 − 2 + = a0 r 2a0 r 2a0 r a0 r r2 r2 (cos2 θ − 1) 6= constant × ψ211 2 2 r sin θ sin2 θ 1 =− 2 2 =− 2 r r sin θ 2 −2Z Z = + 2 ψ211 a0 r a0 ψ211 is not an eigenfunction of ∇2 . 67

3.12 [Thinking Ahead: How will I recognize the angular motion in my expression for K? We will replace all the x, y, z-dependence in K by dependence on r, θ, φ. The motion along θ and φ will show how those angles change, so the angular motion terms will be the ones that depend on θ̇ and φ̇.] What we’ll need to do is break the Cartesian velocity into its polar components, and then examine the angular terms. The equations linking the two coordinate systems are in Eq. A.6, and we can easily make the substitutions, get comfortable, and then start in on the derivatives: x = r sin θ cos φ

ẋ = sin θ cos φ ṙ + r cos θ cos φ θ̇ − r sin θ sin φ φ̇

y = r sin θ sin φ

ẏ = sin θ sin φ ṙ + r cos θ sin φ θ̇ + r sin θ cos φ φ̇

z = r cos θ

ż = cos θ ṙ − r sin θ θ̇

Now we square the velocity components and plug them into the kinetic energy expression: m 2 K= ẋ + ẏ 2 + ż 2 2 m 2 sin θ cos2 φ ṙ2 + r2 cos2 θ cos2 φ θ̇2 + r2 sin2 θ sin2 φ φ̇2 = 2 + 2r sin θ cos θ cos2 φ ṙ θ̇ − 2r sin2 θ sin φ cos φ ṙ φ̇ − 2r2 sin θ cos θ sin φ cos φ θ̇φ̇ + sin2 θ sin2 φ ṙ2 + r2 cos2 θ sin2 φ θ̇2 + r2 sin2 θ cos2 φ φ̇2

+ 2r sin θ cos θ sin2 φ ṙ θ̇ + 2r sin2 θ sin φ cos φ ṙ φ̇ + 2r2 sin θ cos θ sin φ cos φ θ̇ φ̇ + cos2 θ ṙ2 + r2 sin2 θ θ̇2 − 2r sin θ cos θ ṙ θ̇ and now combine terms with common factors, recalling sin2 α + cos2 α = 1: mh 2 sin θ ṙ2 sin2 φ + cos2 φ + r2 cos2 θ θ̇2 sin2 φ + cos2 φ + r2 sin2 θ φ̇2 sin2 φ + cos2 φ 2 i +2r sin θ cos θ ṙ θ̇ sin2 φ + cos2 φ + cos2 θ ṙ2 + r2 sin2 θ θ̇2 − 2r sin θ cos θ ṙ θ̇ i mh = sin2 θ + cos2 θ ṙ2 + r2 sin2 θ + cos2 θ θ̇2 + r2 sin2 θ φ̇2 2 i mh 2 = ṙ + r2 θ̇2 + r2 sin2 θ φ̇2 . 2 =

So the final kinetic energy expression is actually rather simple, even in polar coordinates. Furthermore, it breaks down quite clearly into the radial term mṙ2 /2 and the angular terms mr2 θ̇2 /2 and mr2 sin2 θφ̇2 /2. Just to be clear that these last two terms depend only on the angular velocity, it is worth pointing out how the angular velocity depends on these derivatives θ̇ and φ̇. We need the velocity component perpendicular to the vector ~r; this will be the angular velocity vang , and its magnitude is the speed that we would plug into our equation for the angular momentum, L = mvang r. We can break vang down into two components, vθ and vφ , but these are not exactly the same as θ̇ and φ̇ (for one thing, the units are not equal). The total angular velocity can be computed from vθ and vφ by vector sum (note that the vectors ~vθ and ~vφ are always perpendicular to each other): 2 vang = vθ2 + vφ2 .

The angular velocity components are related to the angle derivatives as follows. If we travel at a constant speed vθ along θ, then we will complete one orbit in time τ , where vθ τ = 2πr. This corresponds to a change in the value of θ of 2π, and so we can write the same time in terms of the rate of change of θ: θ̇τ = 2π. 68

Equating the times, we find vθ = rθ̇. The solution for vφ is similar, except that φ is measured only in the xy plane, so the distance traveled for a complete circuit through φ is 2π(x2 + y 2 ), or 2πr2 sin2 θ. Therefore, vφ = r sin θφ̇. Finally, we may combine these results to show: i mh 2 K= ṙ + r2 θ̇2 + r2 sin2 θ φ̇2 2 m 2 ṙ + vθ2 + vφ2 = 2 2 mvang mṙ2 = + 2 2 2 m2 vang r2 mṙ2 = + 2 2mr2 L2 mṙ2 , + = 2 2mr2 so the angular momentum contributes to the kinetic energy in exactly the same way for both the classical and quantum cases. 3.13 Looking ahead, the uncertainty principle will set a limit on β, which then gives the maximum value of Lz . For a circular orbit, the angular momentum L can be written in terms of the r and p values because L = mrv, as discussed in the Bohr model. Let’s start by combining the uncertainties: δz δpz =

h̄ 1 rp sin2 β = . 2 2

I am using the equality here because I want the minimum possible value of β (the least uncertainty in z and pz , so the least tilt). Now I know something about β, and for a circular orbit I also can require that L = rp. Okay, let’s carry on: rp sin2 β = h̄ h̄ h̄ sin2 β = = rp L

from above L for circular motion

Lz = L cos β ≡ lh̄ lh̄ L= cos β l2 h̄2 l2 h̄2 L2 = = 2 cos β 1 − sin2 β l2 h̄2 = 1 − (h̄/L)

from problem, Lz = lh̄ divide by cos β trig identity from sin2 β above

L2 − h̄L = l2 h̄2 i p 1h h̄ ± h̄2 + 4l2 h̄2 L= 2

multiply by denominator quadratic formula

Select the “+” sign in front of the square root because we define the magnitude L to be a positive number. i p h̄ h 1 + 1 + 4l2 2" # r 1 h̄ 1 + 2l +1 = 2 4l2

factor out h̄ factor out 2l 69

h̄ 1 ≈ 1 + 2l +1 2 8l2 h̄ h̄ = + + lh̄ 2 8l

binomial theorem

This p may not look like such an interesting answer at first, but in fact it’s very close to the result that 2 L = h̄ l(l + 1) from Section 3.1. A Taylor series expansion of (1 + x)1/2 to three terms is 1 + x2 − x8 , and to that order p h̄ h̄ h̄ l(l + 1) = lh̄ + − . 2 8l

Our answer for the tilted circular orbit predicts the correct value of L in the three-dimensional atom to within roughly h̄/(4l), which is an 18% error for l = 1 and decreases rapidly as l increases.

3.14 The eigenvalue of ŝ2 gives us the magnitude of the vector (squared, that is), and the eigenvalue of ŝz gives us its projection onto the z axis. With those two values, a little trigonometry is sufficient to tell us the angle of the vector, θ: 2

1 = 2

1 3 + 1 h̄2 = h̄2 2 4

|s | = s(s + 1)h̄ r 3 h̄ |s| = 4 h̄ sz = ± 2 h̄/2 sz 1 = ±p cos θ = = ±√ s 3 3/4h̄ 1 θ = arccos ± √ = 54.7◦, 125.3◦ 3

s for the electron is 1/2

So, if we wanted the arrows we draw to represent the actual orientations of the electron spin, we would draw them more than 50◦ from vertical. That’s not why we draw the arrows, however. We draw them partly just to obtain a simplified, visual picture of the angular momentum. All the same, if you can just remember this important result from quantum mechanics, it may prevent some confusion when we get to the vector model of the atom in Chapter 4. 3.15 Start by following the instructions and rewriting the angular wavefunction in Cartesian coordinates. This turns out to be not so bad: ψ2,1,1 = R2,1 (r) Y11 (θ, φ) = R2,1 (r) A1,1 sin θ e

by Eq. 3.33 iφ

by Table 3.1

= R2,1 (r) A1,1 sin θ (cos φ + i sin φ) ! ! p x y x2 + y 2 p +p = R2,1 (r) A1,1 p x2 + y 2 + z 2 x2 + y 2 x2 + y 2 ! x + iy = R2,1 (r) A1,1 p 2 x + y2 + z 2

by Eq. 2.8

cancel

p x2 + y 2 .

I used the Euler formula (Eq. 2.8) because I had equations for sin φ and cos φ given in the problem. 70

The rest is differentiation and algebra. We write the operator using the explicit differential form of the momentum, and get started: h̄ ∂ ∂ x R2,1 (r) Y11 (θ, φ) −y i ∂y ∂x h̄ ∂R2,1 (r) ∂R2,1 (r) = x Y11 (θ, φ) − y Y11 (θ, φ) i ∂y ∂x 1 1 ∂Y (θ, φ) ∂Y (θ, φ) h̄ x R2,1 (r) 1 . − y R2,1 (r) 1 + i ∂y ∂x

(xp̂y − y p̂x )ψ2,1,1 =

At this point, it might occur to you that the radial term shouldn’t be playing a big role in an angular momentum problem, and you’d be right. The terms inpthe first set of square brackets cancel, because x and y only appear in Rn,l (r) as coordinates in r = x2 + y 2 + z 2 and because the derivatives have this feature: ∂f (r) ∂r ∂f (r) x ∂f (r) = = ∂x ∂r ∂x ∂r r and similarly ∂f∂y(r) = ∂f∂r(r) yr . Therefore, when you use our operator on R2,1 (r), you just get h̄ ∂R2,1 (r) ∂R2,1 (r) x −y i ∂y ∂x ∂R2,1 (r) y ∂R2,1 (r) x h̄ x −y = i ∂r r ∂r r h i h̄ ∂R2,1 (r) xy yx = − i ∂r r r

(xp̂y − y p̂x )R2,1 (r) =

= 0.

So we only need to look at the effect of the operator on the spherical harmonic: ∂Y11 (θ, φ) ∂Y11 (θ, φ) h̄ x R2,1 (r) − y R2,1 (r) (xp̂y − y p̂x )ψ2,1,1 = i ∂y ∂x " # x+iy x+iy ∂A1,1 r ∂A1,1 r h̄R2,1 (r) = x −y i ∂y ∂x h̄A1,1 R2,1 (r) i 1 x(x + iy) y(x + iy) = − − x − y i r r3 r r3   =

  h̄A1,1 R2,1 (r)   ix − y − xy(x + iy) − yx(x + iy)     3 i r r{z } |

group by denominator

ix − y ir x + iy = h̄A1,1 R2,1 (r) r = h̄A1,1 R2,1 (r)

−1/i = i

= h̄R2,1 (r)Y11 (θ, φ) = h̄ψ2,1,1 .

put in eigenvalue form

So the eigenvalue is just h̄, which is what we expected because this should be the eigenvalue ml h̄ (Eq. 3.12) of the operator L̂z for ml = 1. But isn’t it more convincing to see in Cartesian coordinates, at least this once? 71

3.16 L̂z and Ĥ share the same set of eigenfunctions, so we can plug in the eigenvalues: L̂z Ĥ − Ĥ L̂z ψ = L̂z (Ĥψ) − Ĥ(L̂z ψ) = L̂z (Eψ) − Ĥ(ml ψ) = Eml ψ − ml Eψ = 0 = 0. If the two operators share the same eigenfunctions, then they commute and the eigenvalue of the commutator [L̂z , Ĥ] must be zero. 3.17 Our wavefunctions include the s orbitals, for which there is no angular term and no angular contribution to the kinetic energy (because the derivatives with respect to θ and φ give zero). Therefore, the maximum radial kinetic energy at given n is for the s orbital, for which all the kinetic energy comes from radial motion: Z2 Z 2 me e 4 = Kr,max = K = Eh . 2n2 2(4πǫ0 )2 n2 h̄2 The expression for K is taken from Eq. 1.11 from the Bohr model. The minimum radial kinetic energy corresponds to maximum l, when as much motion is in the angular coordinates as possible. The maximum value of l is n − 1. The contribution to the kinetic energy from angular motion is given by h̄2 n(n − 1) −2 h̄2 l(l + 1) −2 L2 r = r . = 2me r2 2me 2me To express r−2 as a function of n and Z, we need to solve the quantum average value theorem for the l = n − 1 radial wavefunctions: n−1 Zr Rn,n−1 (r) = An,l=n−1 e−Zr/(na0 ) by inspection of Table 3.2 a0 Z ∞ r−2 = (r−2 )Rn,n−1 (r)2 r2 dr by Eq. 2.10 Z0 ∞ = Rn,n−1 (r)2 dr cancel r2 0

Z ∞ 0

A2n,n−1

= A2n,n−1

Z a0

Zr a0

2n−2

2n−2 Z ∞

e−2Zr/(na0 ) dr

from Rn,n−1 (r) above

r2n−2 e−2Zr/(na0 ) dr

factor out constants

2n−2

(2n − 2)! [2Z/(na0 )]2n−1 2n−2 Z (2n − 2)! 22n = 2n+2 n (2n − 1)! a0 [2Z/(na0 )]2n−1 2 2 Z = . a0 n3 (2n − 1)

= A2n,n−1

Z a0

So the maximum angular kinetic energy becomes 2 2 2n(n − 1) Z h̄ Kθφ,max = 2 2me a0 n3 (2n − 1) Z 2 (n − 1) = 2 Eh . n (2n − 1)

by Table A.5 An,n−1 given in problem simplify

by eq. for

L2 2me r2

above

h̄2 /(me a20 ) = Eh

To find the minimum radial kinetic energy, we subtract this angular kinetic energy from the total, using K = −En as given in the problem: Kr,min = K − Kθφ,max

Z2 Z 2 (n − 1) E − Eh h 2n2 n2 (2n − 1) n−1 Z2 . Eh = 2 1− n 2n − 1

When n = 1, we still get the result for the 1s orbital, that Kr,min = K because there is no angular motion. In the limit that n becomes large, the minimum possible radial contribution to the kinetic energy approaches half of the total kinetic energy. 3.18 This turns out to be a straightforward application of the separation of variables. The radial and angular wavefunctions in Tables 3.1 and 3.2 are each normalized already: Z π Z 2π 0

Y ∗ (θ, φ)Y (θ, φ) sin θdθ dφ = 1 Z ∞ R∗ (r)R(r)r2 dr = 1. 0

SO when we combine these to make the overall wavefunction, ψn,l,ml (r, θφ) = Rn,l (r)Ylml (θ, φ), then we can break up the normalization integral for ψn,l,ml (r, θφ) into integrals over the angular and radial coordinates: Z Z ψ ∗ ψdτ = [R∗ (r)Y ∗ (θ, φ)][R(r)Y (θ, φ)]dτ Z π Z 2π Z ∞ Y ∗ (θ, φ)Y (θ, φ) sin θ dθ dφ R∗ (r)R(r)r2 dr = 0

= 1 × 1 = 1.

3.19 This problem is essentially a rite of passage: you’ve seen the solutions to the one-electron atom, but you should verify that those solutions actually work in a few cases. Here we’ve selected the 2p radial wavefunction, and you should be able to show that it satisfies the Schrödinger equation and gives the correct energy as its eigenvalue. First we need the wavefunction: Zr e−Zr/(2a0 ) . R21 (r) = A21 a0 And now we’re ready to charge ahead with the math: h̄2 l(l + 1) h̄2 ∂ 2 ∂ Zr Zr Ze2 h̄2 l(l + 1) −Zr/(2a0 ) Ĥr + A21 e = A21 − e−Zr/(2a0 ) r − + 2me r2 a0 2me r2 ∂r ∂r 4πǫ0 r 2me r2 a0 h̄2 Z ∂ 2 ∂ −Zr/2a0 = A21 − re r 2me r2 a0 ∂r ∂r Z 2 e2 −Zr/2a0 h̄2 l(l + 1)Z −Zr/2a0 . + e e − 4πǫ0 a0 2me ra0 73

Examining the first term alone: ∂ 2 ∂ Zr −Zr/2a0 ∂ 2 −Zr/2a0 −Zr/2a0 − e )= r (re r e ∂r ∂r ∂r 2a0 ∂ Z 3 −Zr/2a0 2 −Zr/2a0 = − r e r e ∂r 2a0 Zr2 −Zr/2a0 e = 2re−Zr/2a0 − 2a0 3Z 2 −Zr/2a0 Z 2 3 −Zr/2a0 − + r e r e 2a0 (2a0 )2 Z 2 Z2 3 −Zr/2a0 2r − 2 r + 2 r . =e a0 4a0 ∂ 2 ∂ Plugging this in for ∂r r ∂r (re−Zr/2a0 ) gives

h̄2 2Z 2 h̄2 l(l + 1) Z Z2 3 R21 (r) = A21 − 2r − Ĥ + r + 2 r 2me r2 2me r2 a0 a0 4a0 2 2 2 2h̄ Z Z e e−Zr/2a0 + − 4πǫ0 a0 2me ra0 h̄2 Z 2h̄2 Z 2 = A21 − + me a0 r 2me a20 Z 2 e2 h̄2 Z h̄2 Z 3 r − e−Zr/2a0 + − 8me a30 4πǫ0 a0 me a 0 r 2 2 h̄2 Z 3 h̄ Z − r = A21 me a20 8me a30 h̄2 Z 2 −Zr/2a0 − e me a20 Zr h̄2 Z 2 h̄2 Z 3 −Zr/2a0 e−Zr/(2a0 ) =− A21 re A21 =− 8me a30 8me a20 a0 = −

l(l + 1) = 2

a0 =

4πǫ0 h̄2 me e 2

Z 2 e2 R21 (r). 8(4πǫ0 )a0

R21 (r) is an eigenfunction of Ĥ = [Ĥr + (h̄2 l(l + 1))/(2me r2 )] with eigenvalue −

Z2 Z 2 e2 =− Eh = En . 8(4πǫ0 )a0 2n2

3.20 The 2s wavefunction has quantum numbers n = 2, l = 0, ml = 0, and we set Z = 2 for He+ . We only need to examine the radial part of the wavefunction: Zr e−Zr/(2a0 ) = 0 for radial node Rn,l=2,0 (r) = 1 − 2a0 Zr =1 2a0 2a0 2(0.529 Å) r= = = 0.529 Å. Z 2 74

3.21 We can find the critical points of a function, including maximum values, by taking the derivative and setting it equal to zero. Because we are looking for the maximum along r, we can ignore the angular part of the wavefunction. Zr R21 (r) = A21 e−Zr/(2a0 ) from Table 3.2 a0 2r Z = 2 for He e−r/a0 = A21 a0 ∂ r 2A21 −r/a0 at maximum − e−r/a0 = 0 e R21 (r) = ∂r a0 a0 r =0 1− a0 r = a0 = 0.529 Å. This is the same value of r at which the 2s function goes to zero (Problem 3.20). Despite having the same average potential energy, which is a function of r, the 2s and 2p functions have very different shapes. 3.22 All we need to do to find the angular nodes is find those values of θ that cause Y20 (θ, φ) to be zero: r 5 (3 cos2 θ − 1) = 0 6π 3 cos2 θ − 1 = 0 3 cos2 θ = 1

1 cos θ = ± √ 3

θ = 0.955, 2.186 rad = 54.7◦ , 125.3◦

(Other values of θ will satisfy the first equation, but θ should be restricted to values between 0 and π.) The angular nodes along θ of the functions Plml (θ) appear in Figure 3.6 where the curves converge to the center. If you check, you should find that the values above are consistent with the figure. 3.23 [Thinking Ahead: How many nodes does this imply there are along each coordinate? For these functions, nodes occur between the critical points (where the derivative is zero), not including the limits of integration (such as r = 0 and r = ∞). The exception is s states, but those have a non-zero derivative at r = 0, and we don’t have an s state in this case. Therefore, it appears there will be no nodes along r (only one critical point between the integration limits), one node along θ (two critical points), and one node along φ (one critical point between 0 and 2π). That alone is enough to tell us that l = 2 (because there are two angular nodes) and n = 3 (because the total number of nodes is n − 1).] The real part of the wavefunction reaches a critical point along φ at 0 and at π. Since taking the real part of the wavefunction always changes eiml φ to cos(ml φ), this means |ml | = 1, because any higher magnitude of ml would have more minima and maxima between 0 and π, and ml = 0 wouldn’t have any. The wavefunction reaches a critical point (minimum or maximum) along θ at π/4 and at 3π/4. One way to solve this is to look for a wavefunction in Fig. 2.3 that has lobes pointing along θ = 45◦ and 135◦ and that has |ml | = 1. This works only for l = 2. (If you are already familiar with the Cartesian form of the d orbitals, you may also recall that this shape gives the dxz and dyz functions, which require l = 2 and ml = ± 1.) Because this is not an s function, the radial wavefunction is zero at r = 0, and it reaches a critical point at only one spot between r = 0 and r = ∞. Therefore, it never crosses zero and there are no radial nodes. Given that the number of radial nodes n − l − 1 is 0 and given that l = 2, n must be 3. 75

To solve for Z, take the derivative of R3,2 (r) at the critical point r0 = a0 and solve for Z: # " 2 2 d Z Zr Z −Zr/(3a0 ) −Zr0 /(3a0 ) A 2r0 − = 0 =A e e dr a0 a0 3a0 r0

6r0 = 6. a0

3.24 For He+ , Z = 2. If there is only one node along r, then we are looking for a radial function Rn,l with a first-order polynomial in r. These are the functions with n − l − 1 = 1 or n = l + 2, and Zr , where k is some looking at Table 2.2 we find that the polynomial in these cases is written 1 − ka 0 integer. To have a node at r = 6a0 , we need 1 −

Z(6a0 ) = 0 ka0

k = 6Z = 12,

which is the function for n = 4, l = 2. To get the one node along θ, we also need a function of θ in the angular part that crosses zero once between θ = 0 and π. For l = 2, this is true only for ml = ± 1. The function sin θ does not cross 0 at all between θ = 0 and π, and cos2 θ crosses zero twice. So n, l, |ml | = 4,2,1. 3.25 The radial distribution function is obtained by multiplying the radial wavefunction by the rdependent part of the volume element, r2 (dropping the factor of dr). That’s because the volume element must contain the information that the sampled volume increases as r2 when we integrate over r. Similarly, the factor of sin θ appears in the volume element because a particular value of θ corresponds to a cone centered on the z axis, and the volume occupied by the wall of that cone will decrease as θ approaches zero and the cone converges on the z axis. So our polar distribution function is simply the square of the θ-dependent part of the wavefunction, multiplied by sin θ. The n quantum number being 3 is irrelevant, because we are only examining the angular wavefunction. The normalization constants Al,ml are those for Ylml (θ, φ) in Table 3.1, but dropping a factor of (2π)−1/2 (before squaring), which comes from the integral over φ: 1 sin θ 2 3 2πA21,0 P2pz (θ)2 sin θ = cos2 θ sin θ 2 5 2πA21,0 P2px (θ)2 sin θ = (3 cos2 θ − 1)2 sin θ 8π 2πA21,0 P2s (θ)2 sin θ =

(3s)

(3p0 )

(3d0 )

3.26 The Euler formula lets us write e3iφ = cos 3φ + i sin 3φ, so the real part of e3iφ is cos 3φ. The total number of angular nodes is equal to l, 3. The three nodes all correspond to values of φ. There is no separate angular node at θ = 0, because any node in the angle 76

φ also forces the wavefunction to be zero at θ = 0. The orbital graphed as a function of φ at θ = π/2 is shown next.

3.27 We can derive the ml value from the φ-dependent part of the wavefunction, l from the θ-dependent part, and n from the r-dependent part: 4 Zr R(r) = An,l n=5 e−Zr/(5a0 ) = An,l Lln (r)e−Zr/(na0 ) a0 |ml |

Y (θ, φ) = Al,ml sin3 θ cos θ e3iφ = Al,ml Pl

(θ)eiml θ

ml = 3

For l = n − 1, Lln (r) =

Zr a0

and therefore |ml |

(θ) = sinl−1 θ cos θ

l=4.

3.28 This problem is asking you to recognize some of the relationships among the one-electron wavefunctions. For the designation 5g, we have n = 5 and l = 4. 3/2 l Zr For l = n − 1, Rn,l (r) = An,l aZ0 e−Zr/(na0 ) a0 For ml = ±l, Ylml (θ, φ) For ml = ±(l − 1) Ylml (θ, φ)

So, choosing ml = +4, An,l,ml 3.29

Z a0

3/2

Zr a0

= Al,ml sinl θeiml φ = Al,ml sinl−1 θ cos θeiml φ

e−Zr/(4a0 ) sin4 θe4iφ .

a. [Thinking Ahead: What is the condition for tunneling, and how can we relate one of the parameters in that condition to the distance? Tunneling occurs where the potential energy is greater than the energy of the particle. Because the potential energy depends only on position (as opposed to kinetic energy, which depends only on velocity), we use our expression for the potential energy as a function of r.] The tunneling region begins where the potential energy becomes greater than the system energy, in other words at distances r > rt where U (rt ) = En . So we solve for rt for Z = 2 (because the atom is He+ ) and n = 2 (because the electron is in a 2p state): Z 2 me e 4 Ze2 = En = − 4πǫ0 rt 2n2 h̄2 2n2 (4πǫ0 )h̄2 2n2 h̄2 = rt = (Ze2 ) 2 Z me e 4 Zme e2 2 4(4πǫ0 )h̄ = = 4a0 = 2.12 Å. me e 2

U (rt ) = −

b. To get the fraction of the electron’s mass that lies in the tunneling region, we need to evaluate the probability density of the electron at r > 4a0 and multiply this by the total mass of the electron. The wavefunction is ψ2,1,1 (r, θ, φ) = R2,1 (r) Y11 (θ, φ) " # "r # 3/2 1 3 2 2r −2r/(2a0 ) iφ = √ e sin θ e , a0 8π 24 a0 where Z has been set equal to 2 for helium. The complex conjugate ψ ∗ differs only in a change in the φ-dependent phase factor to e−iφ . When we multiply ψ ∗ and ψ, these two phase factors cancel, so they can be left out of the integral. In fact, it is possible to remove the entire angular contribution to ψ ∗ ψ and integrate only over r, since only r-dependent information is sought. We are evaluating the probability density over a limited range of r, so we choose the integral limits accordingly. Factoring the constants out of the integral, we have:

8 24a30

Z ∞ 4a0

2r a0

e−2r/(a0 ) r2 dr.

This integral can be evaluated on paper or by a program such as Maple to obtain a value of 891e−8a30 /2. Multiplying this value by the other constants yields a mass in the tunneling region of Z ∞ 2 8 2r me e−2r/(a0 ) r2 dr = 0.050me . 24a30 a0 4a0 Maple. The command to evaluate the density is: (8/(24*a0ˆ3))*int(2*(rˆ4/a0ˆ2)*exp(-2*r/a0),r=4..infinity) Mathematica. The command to evaluate the density is: (8/(24*a0ˆ3))*Integrate[2*(rˆ4/a0ˆ2)*Exp[-2*r/a0],{r,4*a0,Infinity}] 3.30 The total number of nodes is 3 = n − 1 n = 4 . The number of angular nodes is l l = 2 . The other two quantum numbers are not specified by the information in the problem, so any of the following 1 values yields a valid solution: ml = 0, ±1, ±2, ms = ± . 2 3.31 We are looking for any values of the coordinates that set the overall wavefunction equal to zero. First we need the wavefunction: ψ42−1 (r, θ, φ) = R42 (r)ψ2−1 (θ, φ) " # # "r 2 Z 3/2 Zr Zr 15 iφ −Zr/(4a0 ) = sin θ cos θe . 1− e √ 3/2 a0 12a0 8π 64 5a 0

For He+ , Z = 2. For the real part only, replace e−iφ with cos φ. At the nodes, ℜ [ψ42−1 (r, θ, φ)] = 0: 2 2r r 1− e−r/(2a0 ) sin θ cos θ cos φ = 0 a0 6a0 There is one radial node, at r =0 1− 6a0

r = 6a0 . 78

The zeros along the angles are at

sin θ cos θ cos φ = 0

θ=

π . 2

φ=

π . 2

There are three nodes total: one radial and two angular. We don’t count the zero value at θ = 0 as a node, because the function does not cross zero at theta = 0. In fact, the φ = π2 nodal plane contains the points that correspond to θ = 0, so it would not count as a distinct node either. Maple. This is as good a time as any to look at one way of generating atomic orbital wavefunctions using Maple. We can generate the spherical harmonics using the LegendreP function and the exponential term. The radial wavefunction similarly uses the LaguerreL function to generate the polynomial. In both cases, I eliminate some complex but elegant expressions for the normalization constants by integrating the unnormalized functions: • Aa:=(l,m)->sqrt(1/(2*Pi*int(LegendreP(l,m,cos(theta))ˆ2*sin(theta),theta=0..Pi))); Y:=(l,m)-> Aa(l,m)*LegendreP(l,m,cos(theta))*exp(m*I*phi); Note that Maple will write these expressions for Ylml (θ, φ) entirely in terms of cos θ, which it can do by replacing the sin θ in our formulas by (1 − cos2 θ)1/2 . • assume(Z>0); Runnorm := (n, l, r) -> LaguerreL(n-l-1, l, Z*r/n)*rˆl*exp(-Z*r/n); Ar := (n, l) -> sqrt(1/(int(Runnorm(n, l, r)ˆ2*rˆ2, r = 0 .. infinity))); R := (n, l, r) -> Ar(n, l)*Runnorm(n, l, r); This formula for the radial wavefunction assumes that the distance r is expressed in units of a0 (i.e., I have set a0 = 1). Note that we have to specify Z > 0 for the normalization constant to be evaluated. • solve(simplify(Y(2, -1, theta, phi)) = 0, theta); This will also return the values 0 and π, which we will exclude because they lie at the limits of the allowed range for θ. • solve(simplify(cos(-phi)) = 0, phi); To get the nodes in the real part of the wavefunction, we need to isolate the real part of e−iφ , which is equal to cos(−φ). • Z := 2; solve(simplify(R(4, 2, r)) = 0.*r); We need to specify the value of Z to find the r-dependent nodes. Mathematica. • To generate the atomic orbitals in Mathematica, (with thanks to library.wolfram.com), we combine the radial wavefunction: RadialWavefunction[r ,n Integer,l Integer]/;(n>0&&l>=0&&l<n) := Zˆ(3/2) * 2ˆ(l+1) * Exp[-Z*r/n] * nˆ(-2-l) * (Z*r)ˆl * Sqrt[(n-l-1)!/((n+l)!)] * LaguerreL[n-l-1,2*l+1,2*Z*r/n]; with the spherical harmonic for the angular wavefunction: AngularWavefunction[theta ,phi ,l Integer,m Integer]/;(l>=0&&m>=-1&&m<=l) := SphericalHarmonicY[l,m,theta,phi]; Here we used the correct analytical formula for the normalization constant of the radial wavefunction. See the Maple syntax above for an example where we normalize the wavefunction on the fly. If we needed to (for this problem, we don’t), we could multiply the radial and angular wavefunctions to get the full, three-dimensional wavefunction: 79

Psi[r ,theta ,phi ,n Integer,l Integer,m Integer]:=RadialWavefunction[r,n,l] * AngularWavefunction[theta,phi,l,m];

• We can find the nodes of the angular and radial wavefunctions separately, using the Solve function and each of the three coordinates. Starting with the radial part, we get: Solve[RadialWavefunction[r,4,2]==0,r] gives r = 0 and r = 6. The results are in units of a0 . We ignore the value r = 0, because we have defined nodes as regions where the wavefunction changes sign, crossing zero. Since r can never be negative, the wavefunction reaches zero at r = 0 but it does not cross zero, so r = 0 does not correspond to a node. • The nodes along θ can be found the same way: Solve[AngularWavefunction[theta,phi,2,-1]==0,theta] This gives solutions at θ = 0, −π/2, π/2. We neglect the solution at θ = 0 because (again) this is at the end of the domain, so the wavefunction does not cross zero at this value of θ. • The φ-dependent part is more complicated, because we first need to extract the real part of the wavefunction. The Re function does that, but the additional function ComplexExpand is needed to put the real part of the wavefunction into a form that can be treated by Solve: Solve[ComplexExpand[Re[AngularWavefunction[theta,phi,2,-1]]]==0,phi] This gives the solutions φ = ±π/2 + 2πC, where C is any integer. These values of φ all correspond to the yz plane, so the result is really just a single node. 3.32 ψn,l,ml =3,1,−1 = R3,1 (r)Y1−1 (θ, φ) Zr = A3,1,−1 L13 (r)e−Zr/(3a0 ) P11 (θ)e−iφ P11 (θ) = sin θ a0 r √ ! 3/2 4 2 Z Zr Zr 3 √ = 1− e−Zr/(3a0 ) sin θ e−iφ 8π 27 3 a0 a0 6a0 3/2 2 Z Zr Zr = √ 1− e(−Zr)/(3a0 ) sin θ e−iφ 27 π a0 a0 6a0 3.33 To get one angular node we need l = 1; to get one radial node, we need n − l − 1 = 1 or n = 3. Any of the 3p orbitals will satisfy the requirements, and the simplest is the ml = 0 orbital: ψ3,1,0 (r, θ, φ) = R3,1 (r)Y10 (θ, φ) r 3/2 √ Zr 4 2 Zr 3 Z −Zr/(3a0 ) √ 1− e cos θ = a0 6a0 4π 27 3 a0 3/2 Z Zr Zr 2 √ 1− e−Zr/(3a0 ) cos θ = 27 π a0 a0 6a0 R 3.34 Normalizing any function ψ means evaluating the integral all space A2 ψ ∗ ψ dτ and adjusting A so that the integral is exactly equal to one. In this case, the function is the electronic wavefunction 80

obtained by taking the product of the radial part Rn,l and the angular part Ylml . ψn,l,ml = Rn,l (r) Ylml (θ, φ) |ml |

= An,l Al|ml | Lln (r)Pl Zr L12 (r) = a0 0 P1 (θ) = cos θ.

(θ)e−Zr/(na0 ) eiml φ

So we set An,l Al,|ml | = A and determine the value of A that normalizes ψ210 : Z ∞ 0

r2 dr

Z π

sin θ dθ

Z 2π 0

dφ|ψ210 |2 =

Z ∞ 0

r2 dr "Z

Z π

sin θdθ

Z 2π 0

2 Zr cos2 θe−2Zr/(2a0 ) ] dφ [A2 a0 # Z

π Zr e−Zr/a0 cos2 θ sin θdθ a0 0 0 3 "Z ∞ 4 # a0 1 Zr Zr 2 −Zr/a0 3 π = 2πA − d e cos θ| 0 Z3 a0 a0 3 0 2 2π 2 a30 [24] = A 3 Z3 3 3 a0 = 32π A2 Z3

= 2πA2

∞

r2 dr

eiml φ = 1, because ml = 0 2 a 3 Zr Zr 0 · d . r2 dr = a0 a0 Z For normalization: Z

all

space

| ψ210 |2 dτ = 1.

We’ve just shown: Z

all

space

| ψ210 | dτ = 32π

a30 Z3

A2 .

So 32π

a30 Z3

A2 = 1

Z3 . 32πa30

3.35 We set n, l, ml = 3, 1, 0 and Z = 2, and combine the radial wavefunction from Table 2.2 with the angular wavefunction from Table 2.1: r √ 3/2 2 2r 3 2r 4 2 −2r/(3a0 ) 0 1− e cos θ. ψ3,1,0 (r, θ, φ) = R3,1 (r)Y1 (θ, φ) = √ a a 6a 4π 27 3 0 0 0 81

This function has n − l − 1 = 1 radial node, which we find by setting the Laguerre polynomial to zero and solving for r: 2r 1− =0 r = 3a0 . 6a0 There is l = 1 angular node, at θ = π/2, or 90◦ , where cos θ = 0. 3.36 The angular wavefunction for l, ml = 3, 1 is r 21 sin θ(5 cos2 θ − 1) eiφ . Y31 (θ, φ) = 64π The real part of this is obtained by replacing the complex term eiφ = cos φ + i sin φ by its real part, cos φ, and then we find the values of θ and φ that zero the wavefunction: r 1 21 sin θ(5 cos2 θ − 1) cos φ = 0 ℜ Y3 (θ, φ) = 64π if any of the following is true:

cos φ = 0 :

at φ = π/2 and 3π/2

sin θ = 0 : at θ = 0, π r 1 at θ = 0.352π, 0.648π = 63.4◦ , 116.6◦ cos θ = ± 5 Of these, the two values of φ correspond to the same nodal plane, and θ = 0, π are not nodes because those are the limits of θ (the wavefunction does not cross zero at those values). The nodes are at φ = 90◦ , θ = 63.4◦ , 116.6◦. As expected, there are l = 3 distinct angular nodes. 3.37

a. The energy: Solution: Set n = 7 and Z = 3, and use the general equation for the energy of the one-electron atom: En = −

32 9 Z2 Eh = − Eh = − Eh = −4.00 · 10−19 J. 2 2n 2 · 72 98

b. The value of the orbital angular momentum: Solution: For a d orbital, l = 2. The eigenvalue p √ of the L̂2 operator is L2 = h̄2 l(l + 1), so the value of L is h̄ l(l + 1) = 6h̄, or 2.58 · 10−34 J s. c. The projection of the orbital angular momentum onto the z axis: Solution: The eigenvalue of L̂z is ml h̄, which for ml = −1 is −h̄ or −1.05 · 10−34 J s. ~ 3.38 p The idea here is that we know the magnitude of the angular momentum vector L from L = h̄ l(l + 1) and the projection of that vector onto the z axis from Lz = h̄ml (p. 109). That means we ~ and the z axis: can find the angle, call it α, between L h̄ml Lz 2 1 = p =√ =√ L 12 3 h̄ l(l + 1) 1 α = arccos √ = 54.7◦ , 3

cos α =

~ and the xy plane is the where we plugged in ml = 2 and l = 3 for an f orbital. The angle between L complement of α: β = 90◦ − α = 35.3◦ . 82

3.39 The l = n − 1 radial wavefunctions have no Laguerre polynomial (n − l − 1 = 0) so these have the relatively simple form n−1 Zr Rn,n−1 (r) = Anl e−Zr/(na0 ) . a0 We take the general form for the radial probability density, R(r)2 r2 , plug in the expression for Rn,n−1 (r), and then take the derivative to find the maximum:

Z a0

2(n−1)

r2(n−1)+2 e−2Zr/(na0 ) = A′ r2n e−2Zr/(na0 ) d 2Z 2n−1 −2Zr/(na0 ) ′ 2 2 2n −2Zr/(na0 ) e − =0 Rn,n−1 (r) r = A 2nr r e dr na0 2Z r2n = 2nr2n−1 na0 2 2

Rn,n−1 (r) r

= A2nl

3.40

2n2 a0 = 2Z

n2 a0 . Z

combine constants d(uv) = udv − vdu divide out A′ , e−2Zr/(na0 )

solve for r

a. Having φ-component equal to e−iφ : ml = −1. The φ-component is always eiml φ . b. Having a pure real wavefunction: ml = 0. Only ml = 0 has no imaginary component because then eiml φ = 1. c. having a complex conjugate equal to the ml = 2 wavefunction: ml = −2. The complex conjugate of an orbital ml is the same function except that the φ-component becomes e−iml φ . d. Having the greatest density in the xy plane: ml = 3, −3. The largest magnitude of ml means the greatest angular momentum parallel to z. The plane of rotation is perpendicular to the angular momentum, and therefore the greatest density in the xy plane is also for the largest magnitude of ml . e. Having the greatest number of nodes along θ: ml = 0. Having nodes along θ means motion along θ, so we now want the lowest magnitude of ml , in other words, zero.

3.41 The ψ2,1,1 and ψ2,1,−1 functions are already normalized, and they are orthogonal. We can show they are orthogonal in this case by looking only at the φ-part (the only term in which the two functions differ): Z 2π

−iφ ∗ iφ

) e dφ =

Z 2π

iφ

(e )e dφ =

Z 2π

iφ

Z 2π

(e2iφ )dφ

(cos 2φ + i sin 2φ)dφ

2π

1 [sin 2φ − i cos 2φ] =0. 2 0 83

Therefore, normalizing ψa and ψb does not require us to evaluate any explicit integrals: ψa′ ≡ Aψa Z ′ 2 2 |ψa |2 dτ |ψa | dτ = A all all space space Z ∗ 2 =A [ψ2,1,−1 + ψ2,1,1 ] [ψ2,1,−1 + ψ2,1,1 ] dτ Z ∗ ∗ ∗ ∗ ψ2,1,−1 ψ2,1,−1 + ψ2,1,−1 ψ2,1,1 + ψ2,1,1 ψ2,1,−1 + ψ2,1,1 ψ2,1,1 dτ = A2 Z Z Z Z ∗ ∗ ∗ ∗ = A2 ψ2,1,−1 ψ2,1,−1 dτ + ψ2,1,−1 ψ2,1,1 dτ + ψ2,1,1 ψ2,1,−1 dτ + ψ2,1,1 ψ2,1,1 dτ

= A2 [1 + 0 + 0 + 1]

= 2A2 = 1

if ψa′ is normalized, so

1 √ . 2

Therefore, √ ψa′ = [ψ2,1,1 + ψ2,1,−1 ] / 2, and using ψb∗ = −i(ψ2,1,1 + ψ2,1,−1 ), ψb′ =

i(ψ2,1,1 − ψ2,1,−1 ) √ . 2

Which is 2px and which 2py ? We need to draw what we mean by 2px and 2py : y

y 2px

2py φ x

These are functions like the 2pz , but centered on the x and y axes instead. Because they change shape along the angle φ, unlike the 2pz function, they must have some φ-dependent term giving them this shape. To get the same shape as the 2pz (which varies as cos θ) we need a function cos φ for the px , and sin φ for the py . We find the φ dependence of ψa and ψb by expanding the eiφ term: ψa = ψ2,1,1 (r, θ)eiφ + ψ2,1,−1 (r, θ)e−iφ = ψ2,1,1 (r, θ) eiφ + e−iφ

= ψ2,1,1 (r, θ)[(cos φ + i sin φ) + (cos φ − i sin φ)] ψa′ = 2px

= ψ2,1,1 (r, θ)[2 cos φ]

ψb = i(ψ2,1,1 (r, θ)eiφ − ψ2,1,−1 (r, θ)e−iφ ) = iψ2,1,1 (r, θ)(eiφ − e−iφ )

= iψ2,1,1 (r, θ)[(cos φ + i sin φ) − (cos φ − i sin φ)] 84

= iψ2,1,1 (r, θ)[2i sin φ] ψb′ = −2py

= −2ψ2,1,1 (r, θ) sin φ

(Note that in some references, the normalization constants for ψ2,1,1 and ψ2,1,−1 would have opposite signs, which would change the definitions of px and py .) 3.42 Let’s first make sure we’re clear on what is meant by the subscripts having the same nodes and phases as the orbital. For example, the dx2 −y2 orbital is zero when x = y (which is at φ = π/4 and 3π/4), positive where |x| > |y|, and negative where |y| > |x|, just like the function x2 − y 2 . These orbitals, like the px and py , are real, because they are constructed to describe the location of the electron rather than the angular momentum. So we need to combine the l = 2 wavefunctions in such a way that the imaginary components vanish. There’s only one way to do this: the ml = 0 function is real already, the ml = ±1 functions have imaginary terms ±i sin φ, and the ml = ±2 functions have imaginary terms ±i sin(2φ). To exactly cancel the imaginary parts, we must combine ml = +1 with ml = −1, and ml = +2 with ml = −2, and leave ml = 0 alone. We factor out the Rn,l (r) term, because it is the same for each wavefunction in the d subshell. Because the Ylml functions in Table 3.1 are already normalized and are orthogonal to each other, the normalization proceeds as shown in Problem 3.41. Here are the possibilities: a. b. c. d. e.

Y20 = A2,0 (3 cos2 θ − 1) √ √ −i(Y21 + Y2−1 )/ 2 = 2A2,1 sin θ cos θ sin φ √ √ (Y21 − Y2−1 )/ 2 = 2A2,1 sin θ cos θ cos φ √ √ −i(Y22 + Y2−2 )/ 2 = 2A2,2 sin2 θ sin(2φ) √ √ (Y22 − Y2−2 )/ 2 = 2A2,2 sin2 θ cos(2φ)

To get the z dependence, we look at θ, because θ < π/2 is positive z and θ > π/2 is negative z. The functions cos2 θ and sin2 θ are the same whether θ is less than or greater than π/2, so these are independent of the sign of z. On the other hand, cos θ has the same sign as the value of z. To get the x and y dependence, we look at φ, because φ = 0 is the x axis and φ = π/2 is the y axis. The function sin φ points along the y axis and has the same sign as the y coordinate, whereas cos φ points along the x axis and has the same sign as x; sin(2φ) has nodes at the x and y axes, and cos(2φ) has nodes at |x| = |y|. Combining these results, we see that cos θ sin φ behaves like the function yz, cos θ cos φ behaves like xz, sin(2φ) behaves like xy, and cos(2φ) behaves like x2 − y 2 . That leaves Y20 for the dz2 orbital, and indeed (3 cos2 θ − 1) has the same sign whether z is positive or negative. So (a) above is the dz2 , (b) is the dyz , (c) is the dxz , (d) is the dxy , and (e) is the dx2 −y2 . 3.43 We only need to consider that part of the orbital wavefunction that depends on ml , and that is the angular part. The ml quantum number affects both the Legendre polynomial P(θ) and the phase factor eiml φ . Let’s pick two wavefunctions a and b, where the ml values, ma and mb , are unequal: ψa = Ra (r)Pa (θ)eima φ ψb = Rb (r)Pb (θ)eimb φ Z

ψa∗ ψb dτ = 0 Z Z Z 2π 2 = Ra (r)Rb (r)Pa (θ)Pb (θ)r dr sin θdθ e−ima φ e+imb φ dφ 0

= =

Z Z

Ra (r)Rb (r)Pa (θ)Pb (θ)r2 dr sin θdθ

Z 2π

ei(mb −ma )φ dφ

Ra (r)Rb (r)Pa (θ)Pb (θ)r2 dr sin θdθ

Z 2π 0

Z Z

[cos(mb − ma )φ + i sin(mb − ma )φ] dφ

Ra (r)Rb (r)Pa (θ)Pb (θ)r2 dr sin θdθ [sin(mb − ma )φ − i cos(mb − ma )φ]2π 0 Z Z = Ra (r)Rb (r)Pa (θ)Pb (θ)r2 dr sin θdθ {[0 − i] − [0 − i]}

= 0 if ma 6= mb . 3.44

a. Find an equation for a in this wavefunction in terms of m, E, and U . ∂ ∂ 2 ψ(x) = −aAe−ax = a2 Ae−ax = a2 ψ(x) 2 ∂x ∂x 2m = − 2 (E − U )ψ(x) h̄ 1/2 2m a = − 2 (E − U ) h̄ b. How can you tell that this wavefunction is invalid everywhere except in the tunneling region? In the tunneling region, E − U is negative. Elsewhere it is positive, so the solution to a would require taking the square root of a negative number, yielding an irrational (and invalid) solution.

3.45 The complete wavefunction is found by combining the expressions for Y1−1 (θ, φ) in Table 2.1 and R3,1 (r) in Table 2.2: " r # √ 3/2 3 Z 4 2 Zr Zr −Zr/(3a0 ) ψ3,1,−1 = √ − sin θ e−iφ 1− e a0 6a0 8π 27 3 a0 The only imaginary contribution to the wavefunction comes from e−iφ , which is equal to cos φ − i sin φ. If we leave out the imaginary term −i sin φ, we get the real part of the wavefunction: " r # √ 3/2 3 4 2 Z Zr Zr −Zr/(3a0 ) − sin θ cos φ. ℜ(ψ3,1,−1 ) = √ 1− e a0 6a0 8π 27 3 a0 The nodes occur where the wavefunction crosses zero, so we look for in ψ3,1,−1 that cross zero. terms

This means we can ignore all the constants, as well as the factors

Zr a0

and e−Zr/(3a0 ) (which go to

zero at r = 0 and r = ∞, respectively, but which do not cross zero). That leaves the following three terms: Zr 1− sin θ cos φ. 6a0

The first term is zero at r = 3a0 (using Z = 2 for He+ ). The second term is zero only at θ = 0 and θ = π, and because θ has a range of only 0 to π, this means the wavefunction does not cross zero at these values. The third term goes to zero at φ = π/2 and at φ = 3π/2, but since these correspond to the same plane we count them as one node. Therefore, there are two nodes altogether. 3.46 We assemble the proper dml =−2 angular wavefunction from Table 3.1. We don’t need the radial wavefunction because the question pertains only to the angular distribution. Take the square modulus in order to obtain the probability density, and then integrate it over the volume specified: Z 2π Z π Z 2π Z π Z 15 15 π 15 dφ sin θdθ sin2 θ e+2iφ sin2 θ e−2iφ = dφ sin5 θdθ = sin5 θdθ. 32π 0 32π 0 16 π/2 π/2 π/2 86

3.47 Despite some of the problems in this text, modern quantum calculations are rarely carried out by calculus on paper. The most important part of the problem is setting up the correct integrals; 2 computers can then readily solve them numerically. The probability density is |ψ| = ψ ∗ ψ, and we evaluate that with Z = 2: ψ32ml = A32 R32 (r)Y2ml (θ, φ) Z 2 Å Z π Z 2π 2 P= r dr sin θdθ dφ =

8 · 27 812 · 15a30

Z 2 Å

= 0.244.

8 · 23 812 · 15a30

2r a0

e−2r/(3a0 )

Y2ml (θ, φ)∗ Y2ml (θ, φ)

4 r −4r/(3a0 ) 2 r dr 4 e a0

Y ’s normalized

The result can be obtained analytically or by a symbolic math program: • octave: function xr = x(r); xr= (8/(81ˆ2*15))*rˆ2*2ˆ7*rˆ4*exp(-4*r/3); endfunction quad(”x”,0,2.0/0.529188) • Maple: int((8/(81ˆ2*15))*rˆ2*2ˆ7*rˆ4*exp(-4*r/3),r=0..(2/0.529188)) • Mathematica: Integrate[(8/(81ˆ2*15))*rˆ2*2ˆ7*rˆ4*Exp[-4*r/3],{r,0,2/0.529188}] Roughly one quarter of the electron density is at distances less than 2.00 Å. 3.48 The condition for orthogonality is Z

all space

ψ1∗ ψ2 dτ = 0.

The rest is getting the limits, integrand, and volume element correct: r # 3/2 Z 2π Z π Z ∞ " 1 Z Zr 1 −Zr/(2a0 ) √ 1− e 2a0 4π 2 a0 0 0 0 " √ # r 3/2 2 2 2 15 Z Zr 2 2iφ −Zr/(3a0 ) √ × sin θ e r2 dr sin θ dθ dφ e a a 32π 81 15 0 0 3.49 The average value of a parameter for a quantum mechanical system is evaluated by integrating the parameter’s operator over the square modulus of the wavefunction: Z hLx i = ψ ∗ L̂x ψdτ Z ∞ Z π Z 2π Zr Zr (−Zr)/(3a0 ) +iφ 2 1− e sin θ e = ih̄ r dr sin θdθ dφ A a0 6a0 0 0 0 ∂ Zr Zr ∂ (−Zr)/(3a0 ) −iφ × sin φ A 1− e sin θ e + cot θ cos φ ∂θ ∂φ a0 6a0 where √ 3/2 r Z 3 4 2 A= √ 8π 27 3 a0 87

3.50 We apply the quantum average value theorem, Eq. 2.10, to the kinetic energy operator and the n = 3, l = 1, ml = 1 wavefunction: h̄2 ˆ 2 ∇ 2me ψ = R31 (r)Y11 (θ, φ) 2 r r = √ 1− e−r/(3a0 ) sin θ eiφ 27 π a0 6a0 Z hKi = ψ ∗ K̂ψdτ K̂ = −

2 Z 2π Z ∞ Z π 2 h̄2 2 r r √ r2 dr 1− e−r/(3a0 ) sin θ e−iφ − sin θdθ ∇ dφ 27 π a0 6a0 2me 0 0 0 r r 1− e−r/(3a0 ) sin θ e+iφ . × a0 6a0

3.51 The term in the Laplacian that gives the radial kinetic energy is the one that depends on the derivatives of r: h̄2 1 ∂ 2 ∂ r . K̂r = − 2me r2 ∂r ∂r This operator will only act on the radial part of the wavefunction, so I don’t need the angular term Y11 (θ, φ), but I do need R21 (r). Plugging in Z = 2 for He: Z ∞ hKr i = R21 (r)K̂r R21 (r)r2 dr 0 2 3 Z ∞ 2r 1 ∂ 2 ∂ 2r 2 h̄ −r/a0 −r/a0 r2 dr. e e r =− 2me 24a30 a0 r2 ∂r ∂r a0 0 3.52 For lithium, Z = 3, and we need the wavefunction ψ2,1,1 = R2,1 (r)Y11 (θ, φ). Then we use the average value theorem, remembering to take the complex conjugate of ψ2,1,1 for the first term: # r 3/2 Z ∞ Z π Z 2π " µs µI (3 cos2 θ − 1) 3 Z 1 Zr −Zr/(2a0 ) −iφ √ = e sin θ e r3 a0 8π 24 a0 0 0 0 # r 3/2 " 3 Z µs µI (3 cos2 θ − 1) 1 Zr √ × e−Zr/(2a0 ) sin θ eiφ r2 dr sin θ dθ dφ r3 a0 8π 24 a0 Z ∞ Z π Z 2π 243µs µI r2 e−3r/(a0 ) (3 cos2 θ − 1) sin3 θ dr dθ dφ. = 64πa50 0 0 0 3.53 The 1s ground state has quantum numbers n = 1, l = 0, ml = 0, and for hydrogen Z = 1. Because we want the probability density within a certain range of r values, we will find that we can neglect the angular part of the integration altogether: ψ100 (r, θ, φ) = R10 ψ00 (θ, φ) # "r # " 3/2 Z 1 −r/a0 e = 2 a0 4π 3/2 1 Z = √ e−r/a0 π a0 Z 2π Z π Z a0 |ψ|2 r2 dr sin θdθdφ P(r < a0 ) = 0

 Z 2π Z π = 0

1 4π



sin θdθ dφ

4 a30

Z a0 0

e−2r/a0 r2 dr

a0 r 2 a0 r 3 1 −2r/a0 a0 r (− cos θ)|π0 (φ)|2π − e + + = 0 4π 2 2 4 a0 a0 r 2 4 −2r/a0 2 2 × 3e − r+ − a0 2 (−2/a0 )2 (−2/a0 )3 0 3 a a3 4 a3 a3 = 3 e−2 − 0 − 0 − 0 − 1 0 − 0 − 0 a0 2 2 4 4 3 1 4a 1 = 30 −e−2 1 + = 0.323. a0 4 4 2

3.54 The charge in a region a < r < b is given by the following integral: −e

Z b a

|Rn,l (r)|2 r2 dr.

It is not necessary to include the angular part of the wavefunction because the property sought does not depend on angle (although we could include the Ylml part of the wavefunction and integrate over angles too, and we’d get the same answer). Based on this, the problem is actually asking us to solve for r0 in the following equation: −e

Z r0 0

|R1,0 |2 r2 dr = −0.5e.

Divide both sides by −e, set Z = 1 for H, and here we go: Z r0 0

#2 Z r0 " 3/2 1 −r/a0 2 |R1,0 | r dr = r2 dr e a 0 0 Z r0 4 e−2r/a0 r2 dr = 3 a0 0 r0 2 4 r a0 ra2 2a3 = 3 − − 0 − 0 e−2r/a0 a0 2 2 4 0 2 2 4 2a3 r0 a0 r0 a0 2a30 −2r0 /a0 = 3 + 0 = 0.5 e − − − a0 2 2 4 4 2 2

The integral in this equation was solved by the standard method Z

u dv = uv −

v du,

but it leaves us with a transcendental equation; we cannot write a simple algebraic equation to solve for r0 . However, we can use numerical programs (with a programmable calculator or computer) or successive approximation (on paper; see Chapter A) to get the result: r0 = 0.320 a0 = 0.17 Å. 3.55 The function we were given is just R(r). Because we are looking for the average value of an 89

r-dependent property, we only need the r-dependence of the total wavefunction. Z ∞ R(r)rR∗ (r)r2 dr hri = 0 √ 2 16 2 12r ∗ R(r) = R (r) = p 3 e−4r/a0 27 5a0 a0 √ !2 Z ∞ 2304 2 = e−8r/a0 r4 r3 dr √ 7/2 0 27 5a0 Z ∞ Z 53.97 ∞ −8r/a0 7 n! = e−cx xn dx = n+1 r dr e 7 a0 c 0 0 8! 53.97 = 0.130 a0 . = a70 (8/a0 )8 R 3.56 The integral has the form ψ ∗ Âψ dτ , where the integral is over all space and the operator Â is −h̄2 ∇2 . This operator is the momentum squared operator p̂2 , and the integral evaluates the mean square momentum. Examining the part of the integrand in square brackets, this is the original wavefunction (although the constants have been factored out). The φ-dependent term is eiφ , so ml = 1. The angular part of the integral is first-order in sin θ, and the rl term in the radial wavefunction is first-order, so l = 1. The Laguerre polynomial 1 − ar0 in the radial wavefunction is first-order

in r, so n − l − 1 = 1, and n = 3. Finally, to get the Z value, set the exponential e−2r/a0 equal to e−Zr/(na0 ) . This gives 2 = Z/n, so Z = 6.

3.57 The 3pz orbital is the 3p orbital aligned with the z axis, having ml = 0, so the state has wavefunction ψ = R31 (r)Y10 (θ, φ). The region we’re looking at defines an octant (one-eighth of the space), where θ goes from 0 to π/2 (to keep z positive) and where φ goes from 0 to π/2 (to keep x and y positive). So this integral only depends on the angular wavefunction: Z π/2 Z π/2 Z ∞ P= |R31 (r)Y10 (θ, φ)|2 r2 dr sin θdθdφ 0 0 0  !2 "Z # Z r Z ∞ π/2 π/2 3 2 2 cos θ sin θdθdφ dφ  = R31 (r) r dr 4π 0 0 0 {z } | =1 because R is normalized. So you could stop here for the integral. To solve the integral: " # h π i 3 Z π/2 2 = [1] cos θ sin θdθ 2 4π 0 π/2 cos3 θ 3 0 1 3 − = = − − = 8 3 8 3 3 0

1 8

You would guess this answer, because the pz orbital is cylindrically symmetric about the z axis, so if you look in this octant you’ll find exactly 1/8 of the density. 3.58 The atomic number is Z = 1 and the wavefunction is ψ4,3,−3 (r, θ, φ) = R4,3 (r)Y3−3 (θ, φ) # # "r " 3/2 3 Z 35 Zr 1 √ sin3 θ e−3iφ . e−Zr/(4a0 ) = a0 64π 768 35 a0 90

The integrand in this case depends on r and on the angles, so we can’t simplify by integrating over only radial or over only angular coordinates. However, the final integral is simplified by the fact that in evaluating |ψ|2 , the factors of e−3iφ and (e−3iφ )∗ = e+3iφ cancel to give a factor of 1. The rms value of x is therefore: Z ∞ Z π Z 2π 1/2 p hx2 i = |R4,3 Y3−3 (θ, φ)|2 (r sin θ cos φ)2 r2 sin θ dθ dφ 0

Z ∞ 0

2 R4,3 r4 dr

1 7682 · 35a30

Z π 0

Z ∞ 0

r a0

|Y3−3 (θ, φ)|2 sin3 θ dθ 6

−r/(2a0 ) 4

r dr

Z 2π

35 64π

Z π 0

1/2 cos2 φdφ 9

sin θ dθ

Z 2π 0

)1/2 cos φ dφ . 2

3.59 The kinetic energy term (classical or quantum) depends only on the speed v, and the potential energy term depends on only the distance r: U =−

Ze2 . 4πǫ0 r

And it’s the energies of the states that the Bohr model gets right. From the equation for vBohr , we 2 can see that Bohr’s kinetic energy me vBohr /2 will be equal to the expectation value hKi = me v 2 /2. Because E and K are correctly predicted by Bohr’s model, so is U : Ze2 Ze2 1 hU i = − =− 4πǫ0 r 4πǫ0 r 2 Ze =− 4πǫ0 rBohr 1 1 = rBohr r rBohr =

−1 1 . r

3.60 We use the average value theorem here, with Z = 1, n = 2, l = 0. 3/2 Z Zr 1 e−Zr/2a0 R20 (r) = A20 1 − A20 = √ 2a0 2 a0 r

= A220

Z ∞

R20 (r)r2 R20 (r)r2 dr

2 Z ∞ r 1− e−Zr/2a0 r4 dr 2a 0 0 Z Z ∞ Z ∞ ∞ 1 1 2 r5 e−r/a0 dr + 2 r6 e−r/a0 dr = A20 r4 e−r/a0 dr − a0 0 4a0 0 0 1 1 = A220 24a50 − 120a60 + 2 (720a70 ) a0 4a0 84a50 = 42a20 = A220 (84a50 ) = 2a30 = A220

1/2

= 6.48a0 . 91

Note the difference between this value and the Bohr model value of 4a0 for r. Our root mean square (rms) value for the real orbital wavefunction is greater than the Bohr value of r for a couple of reasons. First, the Bohr r value is an accurate predictor of h1/ri for the one-electron atom (see Problem 3.59), which tends to be dominated by low values of r (because low r means a relatively high value of 1/r). In contrast, the rms value of r—which depends on r2 rather than /1r—is dominated by the high values of r. Second, in the three-dimensional atom, the larger values of r sample a much larger volume of space than the small values of r, and so again tend to contribute more to our rms value of r than small values. This is why, for example, the radial probability distributions in Fig. 3.11 appear shifted to higher r values than the wavefunctions in Fig. 3.7. 3.61 [Thinking Ahead: The coordinates of these wavefunctions are separable. Could that simplify the work needed here, the way it often simplifies the integrals for calculating probabilities or expectation values? The operator given in the problem is separable in the same way: it has an r-dependent part multiplied by a θ-dependent part. Therefore, the entire integrand can be divided into three parts, dependent on r, θ, and φ, and these can then be integrated separately.] This problem demonstrates that the principles discussed up to the present section are sufficient for a much more sophisticated treatment of spectroscopy; the math is longer but no harder than what has already been encountered. If the transition is forbidden, it only means that there is (nominally) zero transition strength; the intensity integral should be zero. The transitions indicated both violate the ∆l = ±1 selection rule, and the first also violates a ∆ml = 0 selection rule for radiation polarized along the z axis. Because the integrals will be zero, the normalization constants do not matter. a. ψ211 = A211 R21 (r)P11 (θ)eiφ ψ210 = A210 R21 (r)P10 (θ) Z ∞

r dr

Z π

sin θdθ

Z 2π

∗ dθ ψ211 er

Z 2π 0

cos θ ψ210 = e

Z ∞ 0

e−iφ dφ =

2 R21 (r)r3 dr

Z 2π

(cos φ − i sin φ)dφ

Z π 0

P11 (θ)P10 (θ) cos θ

sin θ dθ

−iφ e dφ A211 A210

= (sin φ + i cos φ)|2π 0 =i − i = 0

The electric dipole transition strength is therefore zero, and this transition is forbidden for radiation polarized along the z axis. Because the φ term is determined only by the quantum number ml , this must be due to violation of a selection rule for ml . It is sufficient for one selection rule to be violated if the transition strength is to be zero. In this case, a second selection rule (for ∆l) is also violated, and the θ term in the integral will also vanish. b. ψ310 = A310 R31 (r)P10 (θ) Z ∞ 0

r2 dr

Z π 0

sin θdθ

Z 2π 0

∗ dφ ψ310 er cos θ ψ210 = e

× 92

Z ∞

R31 (r)R21 (r)r3 dr

Z 2π 0

Z π 0

dφ A310 A210

P10 (θ)2 cos θ sin θdθ

P10 (θ) = cos θ

Z π 0

P10 (θ)2 cos θ sin θdθ =

Z π

cos3 θ sin θdθ

1 | cos4 θ|π0 4 1 = − (1 − 1) = 0 4

=−

This transition is also forbidden, this time because the θ-dependent term vanishes. This makes sense, because there is no violation of ∆n or ∆ml selection rules, so the r and φ terms should not vanish. Only the ∆l = ±1 selection rule is violated, and l determines the θ-dependent part of the wavefunction. 3.62 The transition strength is proportional to Z 2 ∗ ψ µ̂ ψ dτ final t initial all space

The transition is forbidden if this integral is zero. Z 2π Z π Z ∞ = A100 A21−1 | dφ sin θdθ r2 drR21 (r)Y1−1 (θ, φ)∗ (er cos θ cos φ)R10 (r)Y00 (θ, φ)|2 0

1 = 4π

Z 2π

dφ

Z π

sin θdθ

Z ∞ 0

r dr

r a0

sin θ e+iφ (er cos θ cos φ)(1)(1)

3.63 This is similar to Problem 3.61, and demonstrates how transition strengths in spectroscopy are calculated from the wavefunctions. a. For ψ210 → ψ100 , the transition integral is: Z ∞ Z π Z 2π ∗ r2 dr sin θdθ dφ ψ100 er cos θ ψ210 0 0 0 Z ∞ 3 = eA100 A210 R10 (r) R21 (r)r dr 0

Z π

Z 2π P00 (θ)P10 (θ) cos θ sin θdθ dφ 0 0 " 3 # Z π a 4 Z ∞ Zr Zr Zr 0 −3Zr/(2a0 ) 2 · e d cos θ sin θdθ = 2πeA100 A210 Z a0 a0 a0 0 0 " 4 # a 4 2 5 Z ∞ 3Zr 3Zr 0 −3Zr/(2a0 ) = 2πeA100 A210 e d Z 3 2a0 2a0 0 1 × − cos3 θ|π0 3 " # a 4 2 5 2 0 = 2πeA100 A210 (24) Z 3 3 s Z3 Z3 · Z3 210 · 3πea40 A100 A210 where A100 A210 = = √ = 3 3 6 4 2 3 Z 32π a0 a0 4 2πa30 ×

√ 128 2 ea0 243 Z 93

1 0.745 e a0. Z

b. For ψ210 → ψ200 : Z ∞ Z π Z 2π ∗ r2 dr sin θdθ dφ ψ200 er cos θψ210 0 0 0 Z ∞ R20 (r)R21 (r)r3 dr = eA200 A210 0

Z π

−

Z 2π P00 (θ)P10 (θ) cos θ sin θdθ dφ 0 0 " 3 # Z π a 4 Z ∞ Zr Zr Zr Zr 0 −Zr/a0 1− · d e cos2 θ sin θdθ = 2πeA200 A210 Z a0 2a0 a0 a0 0 0 "Z " ## Z 4 5 ∞ ∞ a 4 1 2 Zr Zr Zr Zr 0 − d d e−Zr/a0 e−Zr/a0 = 2πeA200 A210 Z a a 2 a a 3 0 0 0 0 0 0 s 24 · 32 πea40 Z3 Z3 · Z3 =− = A200 A210 where A200 A210 = 3 3 4 2 3Z 8 · 32π a0 a0 16πa30 3 ea0 . Z

The (2, 1, 0) → (2, 0, 0) transition is transition. 3.64

√3 128 2/243

= 16 times stronger than the (2, 1, 0) → (1, 0, 0)

a. List the possible values of ml . For l = 2, the permitted ml values are −2, −1, 0, +1, +2 . b. List the possible values of j. If there is only one electron, then s = 21 . Combining this with l = 1 for any d orbital, we can find the possible values of the resultant j:

j = |l − s|, |l − s| + 1, . . . l + s =

3 5 , 2 2

c. Write the number of radial nodes. The number of radial nodes is given by n−l−1 = 4−2−1 = 1. 3.65 For the 3d orbital, n = 3, l = 2, and with one electron we must have s = 21 . The nuclear spin I = 1 is given. The number of ml values is 2l + 1 = 5. The number of ms values for each ml state is 2s + 1 = 2. The number of hyperfine levels for each ml , ms state is then given by 2I + 1 = 3. The total is 5 × 2 × 3 or 30 . 3.66 Let’s take a look at the nature of the spin and orbital angular momenta in each of our options. • (c): s state: l = 0, so there is no spin-orbit energy. ASO l · s = 0. • (a) and (b): the orbital is smaller if Z is larger, so (a) > (b). • (b) and (d): the orbital is smaller if n is smaller, so (b) > (d). The correct sequence is

c d b a. 94

Chapter 4 4.1

1. Which term(s) make positive contributions to the energy? A,B,E Keep in mind that the kinetic energy (A and B) is always positive.

2. Which term(s) make an exact solution to the Schrödinger equation impossible? E This is the term that makes Ĥ non-separable. 3. Which term(s) have the greatest magnitude? C,D The electron–nucleus attractions must always dominate the energy, because otherwise the electrons will not stay bound to the nucleus. 4. Which term(s) appear in the Hartree–Fock Hamiltonian for calculating orbital energy ε1 ? A,C,E These are the only terms that involve electron 1. 4.2 To progress from sodium to potassium, we add eight protons to the nucleus, and to keep the atomic charge neutral we also add eight electrons. Each added proton increases the real atomic number, and each added electron partially shields that proton’s charge from the outermost electron. Therefore, the effective atomic number is higher for the valence electron in potassium, because the eight added electrons only partially shield the charge of the added protons. Why, then, does potassium have a 2 greater atomic radius? Recall that for the one-electron atom, rn = nZ a0 (Eq. 1.13). Therefore, the higher principal quantum number (n = 4) of the valence electron in potassium raises the effective orbital radius by a factor of about (4/3)2 = 1.8 compared to the n = 3 valence electron of sodium. This is partly offset by the increase in effective atomic number, so the actual difference in atomic radius is a factor of only about 1.22. 4.3 Does the quantum mechanics allow the atom in this state to respond in exactly the “wrong” direction to an applied force? The answer to that question, in principle, is yes: but the particular example given doesn’t work out that way. Quantum mechanics does not forbid the excited state of a system from extracting energy from a force field, such as an external electric field, to rise in energy. Hypothetically, the excited state of an atom in an electric field could be polarized in a direction opposite to the force applied. This is a remarkable discrepancy between quantum and classical mechanics, but it doesn’t violate the conservation laws. The reason it would not be observed in the example given is that the question neglects the contributions from higher energy excited states. The commonly observed, lower energy excited states can always borrow character from these higher energy states so that they too become more stable in the presence of an electric field.[2] There are, however, other examples of quantum systems for which certain excited state wavefunctions lie primarily in regions of high potential energy, even though the forces at work would appear to be pushing the other way. The bottom line is that the distribution of quantum particle wavefunctions is based only partly on their response to classical forces; the wave nature of the particle also plays a role. The ground state particle in a box, for example, is found mostly near the middle of the box, even though there are no forces pushing it there. 4.4 In the many-electron atom, the energy spacing between orbitals of the same n will tend to decrease at higher n. There are several ways to explain this, and what follows is just one version. The shielding of the low-l electrons is much less effective at high n than at low n. You can think of this as arising either from the lowered repulsion between core and valence electrons when the interaction occurs over a larger volume (and therefore with larger average distances between the electrons), or because the penetration of the low-l electrons towards the nucleus is poorer when the density is distributed over a larger range of r. Either way, the effective atomic number is more strongly affected at low n than at high n. 95

4.5 The value of

e2 4πǫ0 r12

calculated using Eq. 4.17 is for the 1s2s helium atom. In the 1s2s state, the

regions of space occupied by the two electrons do not overlap as much as in the 1s2 state, and therefore the repulsion term is smaller. a. How many electron kinetic energy terms appear? 4 because there are 4 electrons.

4.6

b. How many electron–electron repulsion terms appear? 6 because there are 6 distinct pairs of interacting electrons: 12,13,14,23,24,34. 4.7 The electrons fill every occupied subshell in the noble gases, so all the ml and ms values cancel: L = 0 and S = 0. 4.8

a. Which has the greatest amount of shielding? B+ because the higher nuclear charge will bring the electrons closer together on average, and that will increase the electron–electron repulsion energy. b. Which has the lowest energy 2s electrons (having the highest ionization energy)? B+ because the higher nuclear charge will stabilize the electrons more than the lower nuclear charges in Be and Li− .

4.9 We can attack this methodically, starting from the Aufbau electron configuration of the neutral sodium atom, Na: 1s2 2s2 2p6 3s removing one electron to form the monocation, Na+ : 1s2 2s2 2p6 and finally promoting any of the 2p electrons to the lowest energy unfilled orbital (the 3s): Na+∗ : 1s2 2s2 2p5 3s. One has to be careful, however, because the lowest excited state configuration is often not so conclusively identified by inspection. If the highest energy occupied atomic orbital X is only partly filled, it is unclear whether the lowest excited state will be formed by moving an electron out of orbital X to a higher energy orbital, or by moving an electron into orbital X from the orbital below. 4.10

a. In our look at perturbation theory, the zero-order energy is calculated treating the individual electrons as though they don’t interact at all, so we can just add together the one-electron energies of two 1s electrons and one 2s electron: Z2 1 1 1 E0 = − + 2+ 2 2 12 1 3 1 = −4.5 2 9 = −9.5 Eh . b. The first-order correction is calculated using only the perturbation Hamiltonian, which in our case is the electron–electron repulsion term. This has no negative contributions to the overall energy, so it will raise the total energy when added to E0 .

4.11 Z

a. The orbitals are mutually orthogonal, meaning Z Z 1s(1)2s(1)dτ1 = 0 1s(2)2s(2)dτ2 = 0 α(1)β(1)dω1 = 0 96

α(2)β(2)dω2 = 0.

The orbitals are also all normalized, meaning Z Z 1s(i)2 dτi = 2s(i)2 dτi = 1

α(i)2 dωi =

β(i)2 dωi = 1.

Therefore, if we integrate the square modulus of this wavefunction, all of the cross terms integrate to zero, and all the others integrate to one. The result is: Z |Ψ|2 dτ1 dτ2 dω1 dω2 = A2 (1 + 1 + 1 + 1)(1 + 1 + 1 + 1) = 16A2 = 1 all space

1 4.

b. Although the electrons are indistinguishable (each has equal probability of being found in 1s or 2s, α or β orbitals) and is antisymmetric under exchange of labels 1 and 2 or 3 and 4, it is symmetric under exchange of labels 1 and 3 or 2 and 4 and non-symmetric under exchange of labels 1 and 4 or 2 and 3. To be correct, the spin-spatial wavefunction must be antisymmetric under exchange of any two electrons. 4.12

a. Write a Hamiltonian to describe three electrons in a one-dimensional box of length a along the x axis. 2 e2 ∂2 ∂2 1 1 ∂ 1 h̄2 + , + + + + Ĥ = − 2me ∂x21 ∂x22 ∂x23 4πǫ0 x12 x13 x23

where x12 = |x1 − x2 |, and so on.

b. Write the terms from your Hamiltonian that you would treat as the perturbation if you had to solve the Schrödinger equation for this system using perturbation theory. 1 1 1 e2 + + Ĥ ′ = 4πǫ0 x12 x13 x23 c. Write the equation for the zero-order energy E0 in the ground state of this system. We cannot put three electrons in the same state n, because that would violate the Pauli exclusion principle. But we can put two electrons in n = 1 with opposite spins, and the third in n = 2. For the zero-order energy, we neglect the repulsion between the electrons and just use Eq. 2.31 for the energy of a particle in a one-dimensional box: E0 = En=1 (1) + En=1 (2) + En=2 (3) =

π 2 h̄2 2 3π 2 h̄2 (1 + 12 + 22 ) = 2 2me a me a 2

4.13 C3+ has 3 electrons and atomic number Z = 6, so we have 3 kinetic energy terms, 3 electron– nucleus potential energy terms, and 3(3 − 1)/2 = 3 electron–electron repulsion terms: Ĥ

h̄2 2 ∇1 + ∇22 + ∇23 − 2me

−

6e2 1 1 1 + + 4πǫ0 r1 r2 r3

e2 1 1 1 . + + 4πǫ0 r12 r23 r13

4.14 These terms are the electron–nuclear potential energy for electron 3 (with Z = 5, boron) and the kinetic energy operators for electrons 1 and 2. We need both of these kinds of terms for each electron, and at a minimum we have 3 electrons, making this Ĥ for the B2+ ion. We then also need the electron–electron repulsion terms, one for each pair of electrons. The complete Hamiltonian is therefore: Ĥ = −

5e2 5e2 h̄2 2 h̄2 2 h̄2 2 e2 e2 e2 5e2 − − − ∇1 − ∇2 − ∇3 + + + (4πǫ0 )r1 (4πǫ0 )r2 (4πǫ0 )r3 2me 2me 2me (4πǫ0 )r12 (4πǫ0 )r23 (4πǫ0 )r13 97

4.15 Knowing the terms that contribute to the two-electron atom, one can extrapolate to atoms with more electrons. The interactions are all the same; the only change is in the number of interactions. Ze2 h̄2 ∇21 + ∇22 + ∇23 − 2ml 4πǫ0 2 e 1 1 1 + + + 4πǫ0 r12 r23 r31

Ĥ = −

1 1 1 + + r1 r2 r3

∇i = Laplacian for electron i; i = 1, 2, 3

ri = distance from nucleus to electron i

rij = distance between electrons i and j Z = atomic number e = fundamental charge me = electron mass h̄ = Planck’s constant/2π 4.16 Coordinates: –e

–e

r12 r2

r23 r13

+Ze

r3 +e (positron)

Our approach to setting up a Hamiltonian is completely general for systems involving the Coulomb force, no matter how unconventional the particles are. All we need to know is the number and mass of each particle, so that we can write the kinetic energy operator, and charge on each particle, so that we can write the Coulomb potential energy. Ĥ = K̂ + Û =−

h̄2 Ze2 Ze2 e2 e2 e2 Ze2 − + + − − . ∇21 + ∇22 + ∇23 − 2me 4πǫ0 r1 4πǫ0 r2 4πǫ0 r3 4πǫ0 r12 4πǫ0 r23 4πǫ0 r13

4.17 This is not even as straightforward a question as it may first appear, because “shielding” is not strictly quantified for wavefunctions that are not solutions to the Schrödinger equation. Let’s look at the problem two different ways and you can decide. None of these functions has any angular dependence, because the angular term is given by Y00 (θ, φ), which is a constant. Therefore, we will look only at the radial part of these wavefunctions. Function a is a flat function out to a0 , where it ends. Function b is like our 1s orbital, sharply peaked at the nucleus. Function c decreases in a straight line from the nucleus out to a0 , where it reaches zero. Function d is like our 3d orbital, peaking gradually at some distance from the nucleus and then tailing off. According to the traditional perspective on this effect, the shielding decreases the more the electron penetrates through the shielding electrons to get to the nucleus. Function d, the only function that is zero at the nucleus (at r = 0), therefore sees the greatestpshielding. it decreases from that point on as The value of the radial part of function b at r = 0 is 4/a30 , and p r increases. Function a is slightly lower at r = 0, having a value of 3/a30 , and stays constant out to 98

p r = a0 . Function c has the highest amplitude at the nucleus, 30/a30, dropping slowly as r increases, and therefore has the least shielding. MOST: d LEAST: c The only problem with this is that, for exactly the same reasons, the electron–electron repulsion is greatest for ψc and least for ψd . The core electron is close to the nucleus, so when one of these wavefunctions penetrates towards low r, it experiences more of the nuclear charge but also more of the electron–electron repulsion. 4.18 The hyperfine effect in the hydrogen atom is due to a single unpaired electron in a 1s orbital. The helium atom has no hyperfine effect, because both electrons are paired in a 1s orbital, leaving no net angular momentum. The hyperfine effect in the lithium atom is due to a single unpaired electron in a 2s orbital. Spin effects of the two 1s electrons cancel (to very high precision). The hyperfine effect is stronger when the electron is in the vicinity of the nucleus. The 1s electron is, on average, closer to the nucleus than a 2s electron in the same atom. Lithium has a greater nuclear charge than hydrogen, which would pull the electrons in closer, but this is largely shielded from the 2s electron by the 1s electron. The largest hyperfine energy will be found for 1 H. 4.19 The maximum value is obtained when the shielding is completely ineffective: Zeff (max) = Z = 11. The minimum value is obtained when all the other ten electrons are completely shielding the nucleus, balancing out the charge of one proton each: Zeff (min) = 11 − 10 = 1. A reasonable estimate will assume the 1s electron shielding is essentially 100% effective (bringing Zeff (max) down to 9) and the 3s electron shielding is completely ineffective (bringing Zeff (min) up to 2). Assuming that the remaining n = 2 electrons are only partly effective at shielding another n = 2 electron, a reasonable range is 3 < Zeff < 8 . The actual value, computed from the orbital energy, is 4.7. 4.20 increase effect Z + number of 2p electrons − number of 3p electrons − nuclear spin 0 l for 3s from 0 to 1 − Increasing Z will increase the electron–electron repulsion (and therefore the shielding) but not as much as it increases the attraction of the electron for the nucleus, so the net effect is to increase Zeff . Adding more electrons, no matter where, always increases the electron–electron repulsion and decreases Zeff . The nuclear spin doesn’t affect the shielding. The p orbitals are more heavily shielded than the corresponding s orbitals, so Zeff drops in the last case. 4.21 99

increase Zeff for the 2s electrons total electron repulsion energy for the 2s electrons ǫi for the 2s electrons total zero-order energy E0 L for the ground term state

effect from S to Cl+ + Z increases while the number of electrons stays the same. + With the increase in Z, the volume occupied by the 2s electrons will drop. − The greater nuclear charge stabilizes the 2s electrons, so the energy drops. − All the orbitals are more stable, so the overall energy is lower. 0 With the same electron configuration, Cl+ will have same vector model as S.

a. Write the ground state electron configuration of sodium (Na). 1s2 2s2 2p6 3s

4.22

b. Which subshell (1s, 2p, 3d, etc.) in this configuration has the electrons that are most strongly shielded? 3s c. Which subshell in this configuration has the smallest average distance hri from the nucleus? 1s d. Which subshell in this configuration has the electrons with the greatest orbital angular momentum? 2p 4.23 The first question is: what’s this equation even trying to do? Since the reduction of nuclear charge to an effective nuclear charge is the result of electron–electron repulsion, that repulsion should be what this expression tries to represent. What’s going on in this equation is that the attraction energy of electron j for the nucleus, −Ze2 /(4πǫ0 rj ), is being estimated by integrating e2 /(4πǫ0 rj ) over the distance of j from the nucleus, and at each point reducing the atomic number Z by the contribution from electrons 1 through j − 1. If we write the first two terms as follows: Z ri Rn2 i ,li (ri )ri2 dri Ze − e 0

we see that this represents the nuclear charge Ze minus the charge on electron i from ri = 0 to ri . Well, it would make more sense to measure the density of electron i out to the point where electron j is; in other words, integrate ri from 0 to rj . (It doesn’t make sense to integrate a variable x from 0 to x anyway.) Similarly, this should then be added up over all the electrons other than j (which is done), and then the result integrated over all values of rj from 0 to ∞. Finally, the resulting equation predicts the attraction energy of electron j for the nucleus, not the effective atomic number, in other words we’ve estimated Zeff e2 /(4πǫ0 rj ) instead of Zeff (just check the units). So the last thing we do is divide by e2 /(4πǫ0 rj ) (this is what the denominator is supposed to be doing). An improved equation is Zeff ≈

j−1 ∞X i=1

Z−

Z rj 0

Rn2 i ,li (ri )ri2 dri

Rn2 j ,lj (rj )

e2 4πǫ0 rj

rj2 drj

)

/e2

Z ∞ 0

Rn2 j ,lj (rj )rj drj .

4.24 The more classical system in each case is: a. Valence electrons : valence electrons have higher energy, more nodes, greater spatial distribution, less wave-character. 100

b. Core electrons in iron atoms: velocity ∝ Z for core, so p = me v ∝ Z, so λdB = h/p is smaller for bigger atoms. c. Nitrogen atoms at 2 Å : attractive forces mean higher kinetic energy (there is a lower potential energy than at 10 Å, and the total energy is the same); higher momentum, smaller λdB . 4.25 The electron configuration is the lowest excited configuration obtained from the ground 1s2 2s configuration. Just looking at the lowest available unfilled orbitals, we could either promote a 1s electron to the 2s orbital or the 2s electron to the 2p, but clearly the second option requires less energy and therefore gives the lowest energy excited state. So we write the electron configuration for 1s2 2p (we can choose the value of ml since it was not specified, here I’m using ml = 0): " r # " 3/2 r # 3/2 1 1 4 4 −4r1 /a0 −4r2 /a0 ψ(1, 2, 3) = 2 2 e e a0 4π a0 4π # " r 3/2 4 4r3 3 1 e−2r3 /a0 cos θ3 × √ a0 4π 24 a0 4.26 The setup for this problem is just a little complex, because you need to integrate over the Zeff (r2 ) values for electron 2, and each value of Zeff (r2 ) is itself an integral over r1 . But the integral over r1 has an analytical solution, and that will make it possible to find a general solution for the integral over r2 . We replace the factors of Z/a0 by s to simplify the notation. Z ∞ Rn2 ,l2 (r2 )2 Zeff (r2 ) r22 dr2 hZeff in2 ,l2 = 0 Z r2 R1,0 (r1 )2 r12 dr1 Zeff (r2 ) = Z − 0 Z r2 e−2sr1 r12 dr1 = Z − 4s3 0 r2 2 1 r1 r1 −2sr1 3 + + = Z − 4s −e 2s 2s2 4s3 0 2 1 r2 1 r2 3 −2sr2 = Z − 4s . − e + + 4s3 2s 2s2 4s3 = Z − 1 − e−2sr2 2s2 r22 + 2sr2 + 1 . The integral above is solved by taking advantage of a general solution (Table A.5) Z s 0

xn e−ax dx =

n! an+1

− e−as

n X

n!sn−i . ai+1 (n − i)! i=0

Now we can solve the integral for hZeff i of the 2s or 2p electrons. First, for the 2s: Z ∞ hZeff i2,0 = R2,0 (r2 )2 Zeff (r2 ) r22 dr2 0 Z ∞ 3 s sr2 2 −sr2 −2sr2 =Z − 1 + 1− e 2s2 r22 + 2sr2 + 1 r22 dr2 e 2 2 0 3 Z ∞ s2 r22 2 2 2sr2 s −3sr2 2s r2 + 2sr2 + 1 r22 dr2 + 1− e =Z − 1 + 2 0 2 4 3 Z ∞ 2 2 s 2sr2 s r2 −3sr2 =Z − 1 + 2s2 r22 + 2sr2 + 1 r22 dr2 1− + e 2 0 2 4 101

Z s3 ∞ −3sr2 2 4 2s r2 + 2sr23 + r22 − 2s3 r25 − 2s2 r24 − sr23 e 2 0 s3 r25 s2 r24 s4 r26 r22 dr2 + + + 2 2 4 Z s3 ∞ −3sr2 s4 r26 3s3 r25 s2 r24 =Z − 1 + − + + sr23 + r22 dr2 e 2 0 2 2 4 s3 s4 3s3 s2 6! 5! 4! 3! 2! =Z − 1 + − + + s + 2 2 (3s)7 2 (3s)6 4 (3s)5 (3s)4 (3s)3 1 1 720 3 120 1 24 6 2 =Z − 1 + − + + + = Z − 1 + 11 35 . 7 6 5 4 2 2 3 2 3 4 3 3 33

=Z − 1 +

The same can be done (a little more easily) for the 2p orbital: Z ∞ hZeff i2,1 = R2,1 (r2 )2 Zeff (r2 ) r22 dr2 0 Z ∞ 3 s =Z − 1 + (sr2 )2 e−sr2 e−2sr2 2s2 r22 + 2sr2 + 1 r22 dr2 24 0 3 Z ∞ s =Z − 1 + e−3sr2 2s4 r26 + 2s3 r25 + s2 r24 dr2 24 0 s3 6! 5! 4! 3 2 =Z − 1 + 2s4 + 2s + s 24 (3s)7 (3s)6 (3s)5 1440 240 24 1 + + = Z − 1 + 11 =Z − 1 + 35 . 24 37 36 35 So we get the same effective atomic number for both orbitals this way. The point is that the penetration of the 2s orbital towards the nucleus does not lead to a lower average effective atomic number for the whole 2s orbital, relative to the 2p. One should be cautious about reading too much into this, because any argument about the nature of the relative shielding seen by different orbitals of the same shell is prone to circularity. Certainly, if the 2s orbital is more stable than the 2p, then it must be possible to phrase an explanation in terms of the relative exposure of the nucleus to electrons in these orbitals, because the electron–nucleus attraction is the only stabilizing influence (negative potential energy term) in the system. The critical point to understand is that we write the potential energy as the sum of the electron–electron repulsions and the electron–nucleus attractions. Shielding is just a combination of these two terms used to make the effect more intuitive. For higher n values, the effective atomic number calculated this way does depend on l, with higher values of l corresponding to slightly lower values of Zeff . However, the 2s and 2p subshells are where we would expect to see the greatest effect, because they experience the greatest splitting of any two subshells in the same shell. By the way, this is not to say that penetration of s orbitals to regions near the nucleus is unimportant. The high electron density of s orbitals at the nucleus, compared to l 6= 0 orbitals, is responsible for special terms in the hyperfine interaction and the Lamb shift, a difference in the 2s and 2p orbital energies in atomic hydrogen (in addition to the spin-orbit and relativistic corrections) with implications for the structure of the proton [3]. Maple. One way to generate the atomic orbital functions is shown in the solution to Problem 3.31, so let’s use that radial wavefunction for this problem: • Zeff := r2 -> Z-(int(R(1, 0, r1)ˆ2*r1ˆ2, r1 = 0 .. r2)); • Zeffavg := (n, l) -> int(R(n, l, r2))ˆ2*Zeff(r2)*r2ˆ2, r2 = 0 .. infinity); 102

• Zeffavg(2,0); Zeffavg(2,1); Mathematica. One way to generate the atomic orbital functions is shown in the solution to Problem 3.31: RadialWavefunction[r ,n Integer,l Integer]/;(n>0&&l>=0&&l<n) := Zˆ(3/2) * 2ˆ(l+1) * Exp[-Z*r/n] * nˆ(-2-l) * (Z*r)ˆl * Sqrt[(n-l-1)!/((n+l)!)] * LaguerreL[n-l-1,2*l+1,2*Z*r/n]; Let’s just use that radial wavefunction for this problem: • Zeff[r2 ] := Z-(Integrate[R[r1,1,0]ˆ2*r1ˆ2, {r1,0,r2}]); • Zeffavg[n , l ] := Integrate[(RadialWavefunction[r2,n,l])ˆ2*Zeff[r2]*r2ˆ2, {r2,0,Infinity}]; • Zeffavg[2,0]; Zeffavg[2,1]; 4.27 The magnetic fields of the electrons in the filled orbitals all cancel, so we need not worry about them. • The single 2p electron has l = 1 and s = 1/2, and therefore gives rise to two distinct j states: j = 1/2 and 3/2. • These each interact with the nuclear spins to produce two new states for j = 1/2 (F = I − 1/2 and I + 1/2) or four new states for j = 3/2 (F = I − 3/2, I − 1/2, I + 1/2, and I + 3/2). If we label our states (I, j, F ) and keep in mind that the two isotopes will give different energies, our states are now these: (3/2,1/2,1) (3/2,1/2,2) (3/2,3/2,0) (3/2,3/2,1) (3/2,3/2,2) (3/2,3/2,3) (3,1/2,5/2) (3,1/2,7/2) (3,3/2,3/2) (3,3/2,5/2) (3,3/2,7/2) (3,3/2,9/2). The total number of distinct states is 12. 4.28

a. Each of the N electrons may be paired with any of the remaining N − 1 electrons, which gives a total of N (N − 1) pairings. This counts each pair twice, however (for example, we would look at all the pairs involving electron 1 first, then all the pairs involving electron 2, and so on, so that the pairs 1,2 and 2,1 would both be counted), so we divide this number in half. The total number of distinct pairs is N (N − 1)/2. P b. The total energy EHF has each repulsion term once, whereas the sum N i=1 ǫi contains each PN repulsion term twice, so the difference i=1 ǫi − EHF is the total electron–electron repulsion energy. Dividing this by the number of electron pairs gives us the number we want: atom He Li Be B C

ǫi ( Eh ) −1.834 −5.152 −10.08 −16.70 −24.94

EHF ( Eh ) −2.860 −7.432 −14.57 −24.44 −37.69

N (N − 1)/2 1 3 6 10 15

PN ( i=1 ǫi − EHF )/[ Eh N (N − 1)/2] 1.03 0.76 0.75 0.77 0.85

The average repulsion is greatest in He and C, which have the smallest effective volume of this set of atoms. 4.29 Carbon has a lower nuclear charge than nitrogen and similar shielding, so C > N. N∗ promotes an electron to higher n, reducing shielding of 2s electrons, so N > N∗ . N+ removes that electron altogether, reducing shielding even more, so N+ < N∗ .

∗

< N

N+ N∗ N C

< N < C

103

α β −1.6229, −1.3060 −1.363, −1.033 −1.1638, −0.7269 −0.8296, −0.5846

Eh Eh Eh Eh

4.30 For helium, Z = 2. Ψ = 1s(1)2s(2) − 2s(1)1s(2) r 3/2 2 1 1s(i) = 2 e−2ri /a0 4π a0 r 3/2 2 1 1 ri √ 2s(i) = 1− e−ri /a0 4π 2 a0 a0 2 Z ∞ Z π Z 2π Z ∞ Z π Z 2π 1 e2 e 1 2 2 = r1 r2 sin θ1 sin θ2 dr1 dr2 dθ1 dθ2 dφ1 dφ2 . |Ψ(1, 2)|2 4πǫ0 r12 4πǫ0 0 r 12 0 0 0 0 0 4.31 No, the wavefunction is not valid. According to the Pauli exclusion principle, for the Li atom with two electrons in the 1s orbital, the spin wavefunctions cannot be the same—so they cannot all be α. 4.32

a. This problem is just an extension of the Pauli symmetry principles to a three-electron atom. Don’t be intimidated by the extra electron, the problem is still the same as it was for helium. Change the labels of electrons 1 and 2, and see if the function is equal to ±1 times the original function. P̂213 ψ(1, 2, 3) = ψ(2, 1, 3) h √ P̂213 1/ 6[1s(1)1s(2)2s(3)α(1)β(2)α(3) + 2s(1)1s(2)1s(3)α(1)α(2)β(3) + 1s(1)2s(2)1s(3)β(1)α(2)α(3) − 1s(1)2s(2)1s(3)α(1)α(2)β(3)

−1s(1)1s(2)2s(3)β(1)α(2)α(3) − 2s(1)1s(2)1s(3)α(1)β(2)α(3)]] √ = 1/ 6[1s1s2sβαα + 1s2s1sααβ + 2s1s1sαβα

− 2s1s1sααβ − 1s1s2sαβα − 1s2s1sβαα] √ = 1/ 6 [−1s1s2sαβα − 2s1s1sααβ − 1s2s1sβαα + 1s2s1sααβ + 1s1s2sβαα + 2s1s1sαβα] = − ψ(1, 2, 3)

antisymmetric

b. One term in the Li wavefunction given is: 1s1s2sαβα which has no orbital angular momentum (all ml , l = 0), so L = 0, ML = 0. From the spin functions we see that there are two α electrons (MS = + 21 ) and one β electron (ms = − 21 ), so MS = + 21 + 21 − 12 = + 12 . Two of these electrons are in the same orbital, so their spins cancel, and S = 21 : L=0 ML = 0

S = 12 MS = 12

c. The wavefunction is determined by the quantum numbers n, l, ml , and ms of the electrons. We cannot change the shells (n) or subshells (l) without increasing the energy of the atom. The ml values must all be zero, because l is equal to zero for each electron. However, we can change the spin. The 1s electrons must be opposite in spin, but the 2s electron may be α or β. Therefore, there are only two possible distinct wavefunctions, and the other wavefunction is: 1 √ [1s1s2sαββ + 2s1s1sβαβ + 1s2s1sββα 6 − 1s2s1sαββ − 1s1s2sβαβ − 2s1s1sββα]. 104

d. This problem involves all three electrons in the electron configuration, but is still quite similar to the helium atom symmetrization problem. Switch the labels as indicated by the permutation operator, and try to put it in the form ± 1 times the original function. P̂312 ψ(1, 2, 3) = ψ(3, 1, 2) h √ P̂312 1/ 6[ 1s 1s2sαβα + 2s1s 1sααβ + 1s2s1sβαα −1s2s 1sααβ − 1s 1s2sβαα − 2s1s1sαβα]] √ = 1/ 6 [2s1s1sααβ + 1s2s1sβαα + 1s1s2sαβα

− 1s1s2sβαα − 2s1s 1sαβα − 1s2s 1sααβ] √ = 1/ 6 [ 1s1s2sαβα + 2s1s 1sααβ + 1s2s1sβαα − 1s2s1sααβ − 1s1s2sβαα − 2s1s 1sαβα] = ψ (1, 2, 3)

symmetric

4.33 The Coulomb and exchange integrals are the two contributions to the electron–electron repulsion energy, which appears in the helium atom Hamiltonian as e2 /r12 . Therefore, if we know the repulsion energy, we can use it (under our assumptions) to estimate the separation r12 between the two electrons. e2 = (0.420 ± 0.044) Eh = (1.83 ± 0.19) · 10−18 J 4πǫ0 r12 1 (1.113 · 10−10 C2 J−1 m−1 )[(1.83 ± 0.19) · 10−18 J] = r12 (1.602 · 10−19 C)2 = 87.6 · 108 m−1 (singlet)

71.0 · 108 m−1 (triplet)

hr12 i ≈ h1/r12 i−1 = 1.14 Å (singlet)

1.41 Å (triplet)

4.34 To start this problem, let’s begin with the definitions of χsym and χanti . If α and β are orthogonal and normalized, then: Z Z ∗ α (1)α(1)dτ1 = β ∗ (1)β(1)dτ1 = 1 Z Z α∗ (1)β(1)dτ1 = β ∗ (1)α(1)dτ1 = 0 1 χsym = √ [α(1)β(2) + β(1)α(2)] 2 1 χanti = √ [α(1)β(2) − β(1)α(2)] 2 Z

χ∗sym χsym dτ1 dτ2 = 12

[α(1)β(2) + β(1)α(2)]∗ [α(1)β(2) + β(1)α(2)]dτ1 dτ2

[α∗ (1)β ∗ (2) + β ∗ (1)α∗ (2)][α(1)β(2) + β(1)α(2)]dτ1 dτ2 Z Z α∗ (1)α(1)β ∗ (2)β(2)dτ1 dτ2 + α∗ (1)β(1)β ∗ (2)α(2)dτ1 dτ2 = 21 Z Z + β ∗ (1)α(1)α∗ (2)β(2)dτ1 dτ2 + β ∗ (1)β(1)α∗ (2)α(2)dτ1 dτ2

= 21

105

Z Z

Similarly,

∗

α (1)α(1)β (2)β(2)dτ1 dτ2 =

α (1)α(1)dτ1

∗

β (2)β(2)dτ2

= [1] × [1] = 1 Z Z α∗ (1)β(1)β ∗ (2)α(2)dτ1 dτ2 = α∗ (1)β(1)dτ1 β ∗ (2)α(2)dτ2 = [0] × [0] = 0 etc.

χ∗sym χsym dτ1 dτ2 = 12 [1 + 0 + 0 + 1] = 1,

so χsym is normalized.

χ∗anti χanti dτ1 dτ2 Z 1 = 2 (α(1)β(2) − β(1)α(2))∗ (α(1)β(2) − β(1)α(2))dτ1 dτ2 Z α∗ (1)β ∗ (2)α(1)β(2)dτ1 dτ2 = 21 Z − α∗ (1)β ∗ (2)β(1)α(2)dτ1 dτ2 Z − β ∗ (1)α∗ (2)α(1)β(2)dτ1 dτ2 Z + β ∗ (1)α∗ (2)β(1)α(2)dτ1 dτ2 = 12 [1 − 0 − 0 + 1] = 1,

And

∗

so χanti is normalized.

χ∗anti χsym dτ1 dτ2 Z

[α(1)β(2) − β(1)α(2)]∗ [α(1)β(2) + β(1)α(2)]dτ1 dτ2 Z α∗ (1)β ∗ (2)α(1)β(2)dτ1 dτ2 = 12 Z + α∗ (1)β(1)β ∗ (2)α(2)dτ1 dτ2 Z − β ∗ (1)α(1)α∗ (2)β(2)dτ1 dτ2 Z − β ∗ (1)β(1)α∗ (2)α(2)dτ1 dτ2 = 12

= 12 [1 + 0 − 0 − 1] = 0, so χanti and χsym are orthogonal. 4.35

a. 2

P̂21 ψa = e−(r2 −r1 )/a0

2 2 2 r12 r2 = −ψe2(r2 −r1 )/a0 − a20 a20

non-symmetric b. P̂21 ψb = (cos θ2 + cos θ1 )e−r2 /a0 e−r1 /a0 symmetric 106

4.36 Let’s apply the permutation operator that exchanges labels 2 and 3 and see if we can put it back in the form of our original wavefunction: P̂132 ψ = [1s 2p 2s ααβ + 2p 2s 1s αβα + 2s 1s 2p βαα]. None of these terms is identical to any term in ψ, so P̂132 ψ 6= ±ψ; the wavefunction is asymmetric under the permutation, and therefore cannot be a correct wavefunction. 4.37 E g 12 3

states (2, 1, 1)(1, 1, 1)(1, 1, 1), (1, 2, 1)(1, 1, 1)(1, 1, 1), . . .

15 9

(2, 2, 1)(1, 1, 1)(1, 1, 1), (2, 1, 2)(1, 1, 1)(1, 1, 1), . . .

17 3

(3, 1, 1)(1, 1, 1)(1, 1, 1), (1, 3, 1)(1, 1, 1)(1, 1, 1), . . .

18 17

(2, 2, 2)(1, 1, 1)(1, 1, 1), (2, 2, 1)(2, 1, 1)(1, 1, 1), . . .

• For the first level, there is one n value of 2, and there are only three places to put it (along x, y, or z) so the degeneracy is 3. The same logic applies to the third state. • For the second level, there are two 2’s. We can either put them both into the state for one electron (three possibilities: 211, 121, and 112) or we can split the 2’s over two different electrons (six possibilities: 211 211, 211 121, 211 112, 121 121, 121 112, 112 112). Remember that it doesn’t matter which electron has a given state, any combination of states can only be counted once; hence 121 112 111 is the same as 112 121 111. The total number of possibilities in this case is 3+6 = 9. • For the E = 18 level, the same method gives one possibility for all the 2’s in one electron state, nine if split over two electrons, and ten if split over three. Then remove the three of those possibilities that put all three electrons in the same state. 4.38 With total spins of 1, these atoms will act as bosons. Therefore, the overall function must be symmetric with respect to exchange of any two labels. The spatial and spin parts of this example function are both antisymmetric with respect to exchange of these labels. For example: P̂ (21)[abc + cab + bca − cba − acb − bac] = [bac + acb + cba − bca − cab − abc]

= −[abc + cab + bca − cba − acb − bac]

Therefore, the combined wavefunction is symmetric: P̂ (21)ψ(1, 2, 3)χ(1, 2, 3) = [−ψ(1, 2, 3)][−χ(1, 2, 3)] = ψ(1, 2, 3)χ(1, 2, 3). The function is valid. 4.39 Electrons 1 and 2 have changed places in the first and second terms, and then again in the third and fourth terms. The relative signs of the two terms in each group must be opposite for the function to change sign under P̂21 : Ψ(1, 2) = exp(−r1 /a0 ) exp(−2r2 /a0 )α(1)β(2) ⊖ exp(−2r1 /a0 ) exp(−r2 /a0 )β(1)α(2) ⊕ exp(−r1 /a0 ) exp(−2r2 /a0 )β(1)α(2) ⊖ exp(−2r1 /a0 ) exp(−r2 /a0 )α(1)β(2) It is also okay if the third and fourth terms have − and + respectively; it only matters (for this problem) that they have opposite signs. 107

4.40 (i) For the electrons to be indistinguishable, each electron must be equally represented in the four states 1sα, 1sβ, 2sα, and 3sα. There are 6 terms with electron 1 in the 1sα state, 6 terms where it’s in the 1sβ, and 6 terms with electron 1 in the 3s orbital. So we’re missing a term with electron 1 in the 2s orbital. (ii) If electron 1 is in the 2s, then electron 2 has an equal chance of being in the 1sα state, the 1sβ, or the 3s. There is only one term with 2s(1)3s(2), so we need one more. The one we have is 2s3s1s1sαααβ, so the missing term has the other arrangement for electron 3 and 4, and looks like 2s3s1s1sααβα. (iii) For Ψ to be antisymmetric under exchange of two electron labels, there must be an equal number of terms with and without minus signs in front. Of the 5 terms shown with electron 1 in the 2s, 2 have minus signs and 3 don’t. So the missing term has a leading minus sign. The missing term is −2s3s1s1sααβα. 4.41 The He 1s2s example is intended to show that we need one term for each possible arrangement of the electrons into the available spatial and spin orbitals. In the case of He 1s2s it is possible to have both electrons with the same spin, so you only need two terms to show the two electrons in either a 1s2sαα or 2s1sαα configuration. For ground state 1s2 2s2 Be, there are four one-electron, spin-spatial states available: 1sα, 1sβ, 2sα, and 2sβ. For the complete wavefunction to reflect the indistinguishability of the four electrons, it must include every combination of those four electrons in four states. There are 4! such combinations, for a total of 24 possible terms. 4.42

a. P̂12 [2s1s3s + 3s2s1s + 1s3s2s − 3s1s2s − 1s2s3s − 2s3s1s]ααα = −Ψ eigenvalue = −1. b. P̂23 [1s3s2s + 2s1s3s + 3s2s1s − 1s2s3s − 2s3s1s − 3s1s2s]ααα = −P si eigenvalue = −1.

c. P̂12 P̂23 Multiply the two eigenvalues above together: (−1) × (−1) = 1. Is the wavefunction a valid, symmetrized wavefunction? yes. 4.43 The function should be antisymmetric with respect to exchange of any two electrons. This one is antisymmetric only with respect to exchange of 1 and 2, or 3 and 4. For example, χ = αβαβ − αββα − βααβ + βαβα

P̂23 χ = [α(1)β(3) − β(1)α(3)][α(2)β(4) − β(2)α(4)] = ααββ − αββα − βααβ + ββαα 6= −χ

4.44 No. This is a symmetric spin function, and 1s2 2s2 will only have a symmetric spatial function. The spin-spatial wavefunction would therefore be symmetric, which violates the symmetrization principle. 4.45 Any two-particle wavefunction that is symmetric under exchange of the labels for both particles is acceptable, because these particles are bosons. Therefore, we can combine a symmetric spatial part 108

with a symmetric spin part, or an antisymmetric with an antisymmetric. For example: 1 Ψ(1, 2) = √ 1s(1)1s(2) [α(1)β(2) + β(1)α(2)] 2 1 Ψ(1, 2) = √ [1s(1)2s(2) + 2s(1)1s(2)] γ(1)γ(2) 2 1 Ψ(1, 2) = [1s(1)2s(2) − 2s(1)1s(2)] [α(1)γ(2) − γ(1)α(2)] 2

symmetric spatial symmetric spatial antisymmetric spatial

4.46 Normalizing a wavefunction of this type is actually very simple, because all the terms are orthogonal. For example, Z Z Z Z Z Z ∗ 2 (1s1s2s) (1s2s1s)dτ1 dτ2 dτ3 = |1s(1)| dτ1 1s(2)2s(2)dτ2 2s(3)1s(3)dτ3 all space | {z }| {z } 0

because all the one-electron atomic wavefunctions are orthogonal. Therefore, Z Z Z Z Z Z 2 2 |ψ | dτ dτ dτ = A [|1s1s2s|2 + |1s2s1s|2 + |2s1s1s|2 ]dτ1 dτ2 dτ3 A2 sym 1 2 3 all space

= A2 [1 + 1 + 1] = 3A2 = 1

1 . 3

For the three-electron atom, the electron–electron repulsion term of the Hamiltonian has three terms: e2 e2 e2 + + . 4πǫ0 r12 4πǫ0 r23 4πǫ0 r31 The exchange integral therefore has the form: Z Z Z 2 e2 e2 e2 ± [1s(1)1s(2)2s(3)1s(1)2s(2)1s(3) + + 3 4πǫ0 r12 4πǫ0 r23 4πǫ0 r31

+ 1s(1)1s(2)2s(3)2s(1)1s(2)1s(3) + 1s(1)2s(2)1s(3)2s(1)1s(2)1s(3)]

2 =± 3

Z Z Z

e2 e2 e 1s(1)2s(1)1s(2)2s(2) + 1s(2)2s(2)1s(3)2s(3) + 1s(3)2s(3)1s(1)2s(1) . 4πǫ0 r12 4πǫ0 r23 4πǫ0 r31 2

4.47 This will still be a fermion, and therefore subject to the Pauli exclusion principle. No two may have the same four quantum numbers n, l, ml , ms ; but now ms = − 23 , − 21 , + 21 , + 32 , so we can put four electrons into each s orbital, twelve into each p orbital, etc: Z = 10 :

1s4 2s4 2p2 .

4.48 Helium has two electrons, and therefore the spin S must be an integer, and 2S + 1 must be odd, eliminating (a) and (c). Choice (d) is not a valid answer at all, because J cannot be 1 if S and L are both zero. The correct answer is (b). The 3 D3 term will be the lowest energy term, for example, arising from the 1s3d excited state of atomic helium. RRRR 4.49 a. Ĥ1 (xxyzαβαα) (xxyzαβαα) dτ1 dτ2 dτ3 dτ4 6= 0. RRRR R b. Ĥ1 (xxyzαβαα) (xxyzβααα) dτ1 dτ2 dτ3 dτ4 = 0. The spin integral β(2)α(2)dτ1 vanishes. 109

4.50

RRRR c. Ĥ1 (xxyzαβαα) (zxxyαβαα) dτ1 dτ2 dτ3 dτ4 = 0 because the spatial integral vanishes. Ĥ1 is the sum of several e2 /(4πǫ0 rij ) terms, and each term can only influence the integral over two electrons i and j, which will always leave two other electrons unaffected by that Ĥ1 term. For this case, at least one of those two electrons always gives an orthogonal pair of functions. RRRR d. Ĥ1 (xxyzαβαα) (yxxzαβαα) dτ1 dτ2 dτ3 dτ4 6= 0. The x(2)β(2)z(4)α(4) spatial integral survives. RRRR e. Ĥ1 (xxyzαβαα) (xyxzααβα) dτ1 dτ2 dτ3 dτ4 = 0. The spin integral β(2)α(2) vanishes. Z Z

e2 1s(1)1s(1)1s(2)1s(2) dτ1 dτ2 r12 Z Z 2 e I2 ≡ 1s(1)1s(1)2p(2)2p(2) dτ1 dτ2 r12 Z Z 2 e I3 ≡ 1s(1)2p(1)1s(2)2p(2) dτ1 dτ2 r12 Z Z 2 e 2p(1)2p(1)2p(2)2p(2) dτ1 dτ2 I4 ≡ r12

I1 ≡

= 1.250 = 0.485 = 0.034 = 0.390

We need to add the electron–electron repulsion terms to the zero-order energy for the 1s2 and 1s2p electron configurations. For the 1s2 configuration, the repulsion energy is estimated from integral (I1 ) above; there are no cross terms to distinguish between Coulomb and exchange integrals. For the 1s2p configuration, the repulsion energy is the Coulomb integral (Ib ) plus or minus the exchange integral (Ic ). Subtracting the exchange integral gives the energy of the triplet state (as for the 1s2s configuration), and adding gives the singlet state energy. 22 22 − = −4.000 2 2·1 2 · 12 22 22 E0 (1s2p) = − − = −2.500 2 · 12 2 · 22 E1 (1s2 1 S) = E0 (1s2 ) + (I1 ) = −2.750 E0 (1s2 ) = −

E1 (1s2p 1 P ) = E0 (1s2p) + (I2 ) + (I3 ) = −1.981

E1 (1s2p 3 P ) = E0 (1s2p) + (I2 ) − (I3 ) = −2.049

∆Es = E(1s2p 1 P ) − E(1s2 1 S) = 0.769 Eh ∆Et = E(1s2p 3 P ) − E(1s2 1 S) = 0.701 Eh

4.51 We only need the r values for the two electrons, because the s orbitals have no angular dependence. The singlet state, as in the case of the 1s3s configuration, uses the unique antisymmetric spin wavefunction [αβ − βα] and the symmetric spatial wavefunction: 1 1 Ψ = √ [1s(1)3s(2) + 3s(1)1s(2)] √ [αβ − βα]. 2 2 We can rewrite the spatial part with Z = 2 for helium: ψ1s =

2 · 23/2 −2r/a0 e 3/2 a0

1 √ 4π

110

! 2Zr 2Z 2 r2 2 · 23/2 1 −2r/(3a0 ) √ 1− ψ3s = √ e + 3/2 3a0 27a20 4π 27a0 2 4r2 8r2 8 e−2r1 /a0 e−2r2 /(3a0 ) 1 − + ψ=√ 3a0 27a20 27πa30 4r1 8r12 −2r2 /a0 −2r1 /(3a0 ) 1− e +e [α(1)β(2) − β(1)α(2)] . + 3a0 27a20 4.52 Here we need a five-electron wavefunction for boron (Z = 5), but in the zero-order approximation we can write this as the product of the appropriate orbital wavefunctions: ψtotal = ψ(1) ψ(2) ψ(3) ψ(4) ψ(5). The ground state electron configuration is 1s2 2s2 2p. ψ(1s) = ψn,l,ml =1,0,0 = A1,0,0 e−5r/a0 5r −5r/(2a0 ) 1− ψ(2s) = ψ2,0,0 = A2,0,0 e 2a0 5r ψ(2p0 ) = A2,1,0 e−5r/(2a0 ) cos θ 2a0 Other values of ml may be used for 2p. Continuing, we have 5r4 5r3 −5r4 /(2a0 ) −5r1 /a0 −5r2 /a0 −5r3 /(2a0 ) 1− e 1− e e ψtotal = A e 2a0 2a0 5r cos θα(1) β(2) α(3) β(4) α(5). × e−5r5 /(2a0 ) 2a0 To symmetrize this (not required by the problem), we would prepare a Slater determinant of the possible spin-spatial orbitals. 4.53 For a three-electron atom: h̄2 Ze2 e2 1 1 1 1 1 1 + + + + + ∇21 + ∇22 + ∇23 − 2me 4πǫ0 r1 r2 r3 4πǫ0 r12 r23 r13 2 e 1 1 1 Ĥ = Ĥ0 + Ĥ ′ , where Ĥ ′ = + + 4πǫ0 r12 r23 r13 Ĥ = −

ψ0∗ Ĥ ′ ψ0 dτ = A2 ×

e2 4πǫ0

Z ∞

Z π

dθ

1 1 1 + + r12 r23 r13

(

Z 2π

dφ

Z ∞

r22 dr2

−3r1 /a0

−3r2 /a0

−3r3 /(2a0 )

2 3r3 1− 2a0

where Z ∞ 0

Z π 0

dθ

Z 2π 0

dφ =

Z ∞ 0

r12 dr1

Z ∞ 0

r32 dr3

Z π

sin θ1 dθ1 . . .

Z 2π

dφ1 . . . .

4.54 If the Zeeman energy is a small correction to the electronic energy, then the Zeeman Hamiltonian corresponds to the perturbing term Ĥ ′ in perturbation theory. Ψ = ψχ Z

ψ ∗ Ĥ0 ψdτ =E

(0)

111

χ∗ ĤZ χdτs Copyright c 2014 Pearson Education, Inc.

Ψ∗ Ĥ0 Ψdτ dτs Z Z = χ∗ χdτs ψ ∗ Ĥψdτ Z = ψ ∗ Ĥψdτ Z E (1) = E (0) + Ψ∗ ĤZ Ψdτ dτs Z Z (0) ∗ =E + ψ ψdτ χ∗ ĤZ χdτs Z = E (0) + χ∗ ĤZ χdτs . E

(0)

4.55 The first-order correction is a matter of substituting the right values: E1′ =

2 1 (9.27 · 10−24 J T−1 ) (0.20 T)(5.034 · 1022 cm−1 / J) = 0.031 cm−1 . 3 2

For the second-order correction, the integral has a numerical value: 1 3

" 2 #1/2 2 1 3 − (2.00 − 1.00)(9.27 · 10−24 J T−1 )(0.20 T)(5.034 · 1022 cm−1 / J) = 0.044 cm−1. 2 2

We plug the integral provided into the numerator in Eq. 4.21 and the spin-orbit energy difference into the denominator: (0.044 cm−1 )2 E2′ = = −0.0053 cm−1. −0.65 cm−1

The effect of the second-order correction is to pull the j, mj = 12 , 12 state to lower energy, pulling it away from the j, mj = 23 , 21 state that it interacts with. The second-order correction is about one sixth the first-order correction, indicating that at this field, the energy obtained from perturbation theory will converge well. However, the second-order correction increases as BZ2 while the first-order correction increases in direct proportion to BZ . That means that at higher fields the perturbation power series expansion converges poorly, and the calculated energies will become inaccurate. 4.56 (a) The first-order wavefunction is given by Eq. 4.20: ∞ R X ψ0 Ĥ ′ ψ0,k dτ ψ1′ = ψ0 + ψ0,k . E0 − E0,k k=1

R From this we see that the question may be rephrased: why is the integral ψ0 Ĥ ′ ψ0,k dτ equal to zero when ψ0 = 1s2 and ψ1,k = 1s2p? Recall that the atomic orbital wavefunctions are each orthogonal to any other. In the case of 1s and 2p, this can be seen because the 2p function has equal shares of negative and positive phase, while 1s is spherically symmetric, and the product 1s2p therefore also has equal shares of negative and positive amplitude, which cancel when integrated over all space. Our integral is different, however, in that the integrand includes a factor of Ĥ ′ = e2 /(4πǫ0 r12 ). The critical insight in this problem is that the likelihood of a particular value of r12 is exactly the same in the region where 1s2p is positive (θ2p < π/2) as where 1s2p is negative (θ2p > π/2). Therefore, the integrand is positive in the region θ2p < π/2 and negative, with the same magnitude, in the region θ2p > π/2. These two regions have equal size, so again the integrand cancels when integrated over all space. 112

R (b) The integral ψ0 Ĥ ′ ψ0,k dτ vanishes unless the ml values for electrons 1 and 2 are both zero, R and unless the l values are even. Therefore, a 2s2 configuration has a non-zero value of ψ0 Ĥ ′ ψ0,k dτ but the 2p2 configuration does not. 4.57

a. Write the zero-order Hamiltonian Ĥ0 for this system for the region inside the box (0 ≤ x < a). Solution: Our zero-order system is the one-dimensional box without the barrier, so the h̄2 ∂ 2 Hamiltonian is Ĥ0 = − 2m ∂x2 . b. Write the perturbation Hamiltonian Ĥ ′ for this system. Solution: The barrier contributes a fixed potential energy ǫ between x = a/4 and 3a/4, and doesn’t show up in the rest of the box, so Ĥ ′ = ǫ for a/4 ≤ x < 3a/4 and Ĥ ′ = 0 for x < a/4 and x ≥ 3a/4. c. Write the integral that must be solved to find the first order correction to the energy E1′ for the ground state wavefunction. Solution: For this we just want to solve the average value theorem for Ĥ ′ over the n = 1 (ground state) particle in a box wavefunction: E1′ =

4.58

Z 3a/4 a/4

ψ1∗ Ĥ ′ ψ1 dx =

2 a

Z 3a/4 a/4

sin2

πx a

ǫ dx

a. Ua (x) is a straight line with negative slope, from energy E1 at x = 0 to 0 at x = a. b. E1 is a horizontal line from the top of Ua (x). c. The first-order correction is: Z a Z 2E1 a 2 nπx x 2E1 a E1 E1PT = ψ(x)2 Ua (x) dx = 1− dx = = . sin a 0 a a a 4 2 0 We add the first-order correction to the zero-order energy to get the first-order energy:

E (Eh )

E0PT + E1PT = E1 +

E1 3 = E1 = 0.150 Eh. 2 2

0.150

first order

0.100

zero order

d. e. The wavefunction, like the case in Example 4.2, leans in the direction of lower energy. , then what is the similar 4.59 a. If the first-order correction to the wavefunction is ∂ψ ∂λ λ=0

expression for the third order correction to the wavefunction, in terms of λ? Solution: This is asking for the the third-order term in the Taylor series expansion, which is 1 ∂3ψ ψ3 ≡ . 6 ∂λ3 λ=0 113

b. The part of the Schrödinger equation that depends on λ to first order reads Ĥ0 ψ1 + Ĥ ′ ψ0 = E0 ψ1 + E1 ψ0 . Write the part of the Schrödinger equation that is proportional to λ3 . Solution: We want all the terms that combine to give λ to the third power. Since Ĥ has only two contributions, Ĥ0 + λĤ ′ , there will be only two terms on the lefthand side that go as λ3 . On the righthand side, any combination of the energy and the wavefunction where the orders add up to 3 will appear: Ĥ0 ψ3 + Ĥ ′ ψ2 = E0 ψ3 + E1 ψ2 + E2 ψ1 + E3 ψ0 . 4.60 The electron configuration of ground state Li+ is the same as helium 1s2 . The lowest excited state is 1s2s. Therefore, the zero-order wavefunction is "

ψ(1, 2) = ψ100 (1)ψ200 (2) = 2

Z a0

3/2

−Zr1 /a0

1 4π

# "

1 × √ 2

Z a0

r # 3/2 Zr2 1 −Zr2 /(2a0 ) 1− e 2a0 4π

For the zero-order energy, we add the one-electron energies for each of the orbitals occupied (here, just the 1s and the 2s), using Z = 3 for lithium: N X Z2 32 1 1 − 2 Eh = − E0 = + 2 Eh = −5.625 Eh. 2ni 2 12 2 i=1 4.61

a. The electrons are not indistinguishable, and (as a result of this) the wavefunction is not antisymmetric under exchange of labels 1 and 2. b. The zero-order energy for this state is the same as for the singlet and triplet states obtained in the text, because the wavefunction still consists of one 1s electron and one 2s electron. So we only need to look at the first-order correction, which comes from the electron repulsion term: Ea = E0 + = E0 +

Z Z

(1s2sαβ − 1s2sβα)2 1s2 2s2

× dτ1 dτ2 dω1 dω2

e2 dτ1 dτ2 dω1 dω2 4πǫ0 r12

e2 α(1)2 β(2)2 + β(1)2 α(2)2 + 2α(1)β(2)β(1)α(2) 4πǫ0 r12

All of the orbital wavefunctions we are using are mutually orthogonal, and that goes for the spin wavefunctions α and β as well as the spatial wavefunctions. That means the third term in R square brackets above vanishes, because the integral α(1)β(1)dω1 is zero. The other spin terms integrate to one because α and β are normalized wavefunctions, so our final energy depends only on the integral over the spatial part: Ea = E0 +

Z Z

1s2 2s2

e2 dτ1 dτ2 = (Es + Et )/2. 4πǫ0 r12

This wavefunction gives the energy that would result if the exchange integral is left out of the singlet or triplet energies, which is halfway between the two values. 114

4.62 e Zr3 1− e ψ0 = 1s(1)1s(2)2s(3) = Ase e−Zr3 /a0 2a0 e e Zr3 −Zr1 /a0 −Zr2 /ae0 e−Zr3 /a0 1− ψvar = A(1 + c1 r12 )(1 + c2 r23 )(1 + c2 r31 )e e 2a0 −Zr1 /ae0 −Zr2 /ae0

where c1 and c2 are inter-electron interaction constants and Z1 is the effective atomic number for the 1s electrons, and Z2 is the effective atomic number for the 2s electron. Note that the 1s and 2s electron interactions are to a first approximation identical for both 1s electrons, but the unpaired spin of the 2s electron may result in different interactions between electrons 1 and 3 compared to 2 and 3. In other words, a constant c3 such that (1 + c3 r31 ) is included instead of (1 + c2 r31 ) is possible, but would not be the first choice. 4.63 [The first printing of the textbook asks for you to optimize two coupled constants, c1 and c2 , but only the ratio c1 /c2 can be optimized. The function and energy expression should instead be given as ψ(r) =

30 c5

1/2

(c − r) if 0 ≤ r < c

ψ(r) = 0 if r ≥ c Z c hEi = φ(r)Ĥ φ(r)r2 dr.] 0

In this problem, we have left out the fundamental constants h̄, me , and 4πǫ0 . This is equivalent to stipulating that the properties of the system will all be evaluated in atomic units. Therefore, the energy we get at the end will be also in atomic units, in other words hartrees. The partial derivative we will need is: −

1 1 ∂ 1 ∂ 1 1 ∂ 2 ∂ r (c − r) = − 2 r2 (−1) = 2 (r2 ) = 2 (2r) = . 2 2r ∂r ∂r 2r ∂r 2r ∂r 2r r

The variational energy is: Z c

φ(r)Ĥ φ(r)r2 dr Z c 30 1 1 ∂ 2 ∂ = (c − r)r2 dr r − (c − r) − c5 2r2 ∂r ∂r r 0 Z c 30 1 (c − r) 2 = − r dr (c − r) c5 r r 0 Z c 30 (c − r) r − cr + r2 dr = 5 c Z0 c 30 = cr − c2 r + cr2 − r2 + cr2 − r3 dr 5 c 0 Z c 30c32 = (c − c2 )r + (2c − 1)r2 − r3 dr 5 c1 0 c 2 r3 r4 30 2 r (c − c ) + (2c − 1) − = c5 2 3 4 0 2 3 30 c c4 2 c (c − c ) + (2c − 1) − = c5 2 3 4 3 4 4 3 4 30 c c 2c c c = − + − − c5 2 2 3 3 4

hEi =

115

3 c c4 − 6 12 5 5 = 2− c 2c =

30 c5

At the minimum variational energy, the derivative with respect to our variational parmater c equals zero: 10 5 5 ∂ 5 ∂ hEi =− 3 + 2 =0 − = 2 ∂c ∂c c 2c c 2c 10 5 = 2 c3 2c c3 10 · 2 = 2 5 c c=4 5 5 5 5 hEi = 2 − − = = −5/16, c 2c 42 2·4 where this solution gives the variational energy in hartrees. The correct result is −1/2, so this is not a very good variational function, obtaining only a little more than half of the correct energy. However, you can see from this exercise how we could improve the wavefunction by adding higher order powers of r to the expression. 4.64

a. Which parameter has the greatest magnitude? c1 The contribution from the n = 1 state will still be greatest. The variational ground state function should still have no nodes. b. Which parameter is zero? c2 The n = 2 state has opposite phase in each half of the box, so adding this in will shift the wavefunction to one side or the other. This is no benefit when the potential looks the same in both halves of the box, so the variational function will not use the n = 2 state.

U(x)

c. Sketch the variational wavefunction.

a/2 x −U0

The ground state function should have no nodes, and will look like a combination of (mostly) n = 1 plus some n = 3, which will cause it to peak in the middle where the potential energy reaches its minimum value. 4.65 Solution: We want to vary a to get the lowest possible value for hEi, so we take the derivative 116

to find the minimum of hEi as a function of a: h̄2 a ∂ hEi ∂ h̄2 a2 3k 3k = = + − 2 =0 ∂a ∂a 2m 2a m 2a 1/3 h̄2 a 3km 3k 3km a3 = = 2 a = m 2a 2h̄2 2h̄2 " 1/2 1/3 # 3km 3km exp − r ψ(r) = 2 2 2h̄ 2h̄2 2/3 2 1/3 2 3k h̄ 3km 2h̄ + hEi = 2 2m 2 3km 2h̄ 5/3 2 2 1/3 3 k h̄ = 2 m 4.66 What we need to get eventually is an equation for µ of the box when this external electric field E is applied. From that we will get the α value when we just divide by E. Our first-order n = 1 wavefunction will be (Eq. 4.20) R (0) ′ (0) ψ1 Ĥ ψ2 dx (0) (1) (0) ψ2 , ψ1 = ψ1 + (0) (0) E1 − E2 where we have simplified the equation so that only the n = 2 state contributes to the correction. To (0) calculate this, we will need the En values—which are just the energies of the n = 1 and n = 2 states in the unperturbed box—and we will need the integral over Ĥ ′ : # "r # Z d/2 "r Z πx 2πx 2 2 1 ′ (−eEx ) dx cos sin ψ0 Ĥ ϕ(i)dτ = 4πǫ0 −d/2 d d d d Z d/2 πx 2eE 2πx =− sin x dx. factor out constants cos (4πǫ0 )d −d/2 d d Let’s pause here to look at the integral, because we will need it again below. This is an opportunity to show off a couple of tricks for doing the math on paper. Let’s simplify the notation a bit by replacing πx/d with z, which changes the limits of the integral from x = ±d/2 to z = ±π/2: 2 Z π/2 Z d/2 πx 2πx d sin x dx = cos(z) sin(2z) z dz. cos d d π −π/2 −d/2 Don’t forget the factors of d/π out front; one is needed to convert x to z and the other to convert dx to dz. First, we make the arguments of the sine and cosine functions the same using a simple trig identity: sin(2z) = 2 cos(z) sin(z). Next, the combination of x and the trig functions can be obtained through integrating by parts: Z Z cos(z) sin(2z) z dz = 2 cos2 (z) sin(z) z dz Z 1 1 cos3 (z) − cos3 (z) dz integrate by parts = 2 −z 3 3 z 1 1 =2 − cos3 (z) − z cos3 (z) − sin(z) + sin3 (z) 3 3 3 1 2 z cos3 (z) − sin(z) + sin3 (z) . =− 3 3 117

R R The integration by parts is u dv = uv − v du where u = z, du = dz, dv = cos2 (z) sin(z)dz, and v = −(1/3) cos3 (z). So the definite integral can now be evaluated: 2 Z π/2 π/d 2 1 d 2d 3 3 z cos (z) − sin(z) + sin (z) cos(z) sin(2z) z dz = − π 3π 2 3 −π/2 −π/2 2 2 1 1 4 2π 2d − 1− − −1 + − =− 3π 2 3 3 3d2 3 8d2 = . 9π 2 Okay, with that we get back to the physics and complete the integration over Ĥ ′ , replacing z in the integral we just solved by πx/d: Z Z d/2 πx 2πx 2eE sin x dx. ψ0 Ĥ ′ ϕ(i)dτ = − cos (4πǫ0 )d −d/2 d d 2 2eE 8d =− (4πǫ0 )d 9π 2 16edE =− 2 9π (4πǫ0 )

Now we’re ready to write the first-order wavefunction: r 2πx −16edE/(9π 2 ) πx 2 (1) sin cos − . ψ1 = d (4πǫ0 )d d −3π 2 h̄2 /(2me d2 ) (1)

Call it ψ1 = A(ψ1 + bψ2 ) for now, where b = 32me ed3 E/[27π 3 (4πǫ0 )h̄2 ]. Don’t go away yet—we’re getting close to finishing now. With what we have we can calculate the dipole moment µ. To do this, take special note of the symmetry: ψ1 (x) is an even function, symmetric about x = 0; ψ2 (x) is odd, and x is odd. Only integrals of even functions will be non-zero, so we don’t actually have that much work to do: Z d/2 µ=e |ψ(x)|2 x dx −d/2

= A2 e

Z d/2

−d/2

2 ψ1 x + 2bψ1 ψ2 x + b2 ψ22 x dx

Z d/2 πx 2πx 2 sin x dx = 2bA e cos d d d −d/2 2 2 8d = 2bA2 e d 9π 2 32bA2 ed = . 9π 2 2

solved above

There’s one step left—we have to normalize the first-order wavefunction to find out what A2 is, but this is straightforward whenever we work with orthonormal functions like the particle-in-a-box functions: Z d/2 (1) |ψ1 |2 dx 1= −d/2

= A2

Z d/2

−d/2

2 ψ1 − 2bψ1 ψ2 + b2 ψ22 dx 118

= A2 1 − 2b(0) + b2 (1) = A2 (1 + b2 ) 1 A2 = (1 + b2 ) So our dipole moment is µ= =

32bA2 ed 32bed = 2 2 9π 9π (1 + b2 ) 322 me e2 d4 E = 9 · 27π 5 (4πǫ0 )h̄2 (1 + b2 )

4 3π

me e2 d4 E . h̄ (4πǫ0 )(1 + b2 ) 2

The factor of (1 + b2 ) in the denominator is close to 1 as long as the applied field is not large compared to the gap between the n = 1 and n = 2 states (b is roughly the ratio of the electric field energy to E2 − E1 ; if it is not small, then perturbation theory is not valid anyway). So finally, our polarizability is µ/E: 5 5 4 4 4 me e2 d4 d α= = . 2 3π 3π a0 (4πǫ0 )h̄ This result would be exactly the same for a three-dimensional box, because the applied electric field along the x axis would not affect the shape of the wavefunction along y and z. However, the three coordinates are not separable in real molecules, and the polarizability in general will depend on the orientation of the molecule in the electric field. Note also the large dependence on the size of the box. This correctly predicts that polarizability is very sensitive to the size of the atom (or molecule). The polarizability has units of volume, and if you try out a value of 1 Å for d, this equation predicts an α value of 3 · 10−24 m3 , which is much larger than polarizabilities of isolated atoms and molecules, where the potential for the electrons more strongly resists their being pushed around by external fields. 4.67 A detailed solution to the polarizability by perturbation theory is worked out for the particle in a one-dimensional box in Problem 4.66. This time, let’s show a convenient simplification. If you recognize that the operators Ĥ = −eEz and µ̂ = ez are the same to within a constant factor, then it becomes R clear that you really just need to solve one integral, of the form ψ1 ψ2 z dτ . The polarizability to first order may then be written this way: 1 hµi EZ 1 (1) (1) = ψ1s (ez)ψ1s dτ E  R Z X ψ (0) (−eEz)ψ (0) dτ Z (0) 1 (0) (0) (0) 1s i +2 ψ (ez)ψ dτ =  ψ1s (ez)ψi dτ 1s 1s E E − E i 1s i>1 {z } | =0 

α=

(0)

ψj (−eEz)ψi dτ (E1s − Ei )(E1s − Ej ) i>1 j>1 | {z }

≈ −2e2

X i>1

≈0

(0)

ψ1s (z)ψi dτ Ei − E1s

  (0) (0) ψj (ez)ψi dτ   

119

We are further simplifying by restricting the sum to the lowest excited states, with n = 2. Of these, only the 2pml =0 state will contribute, because the 2s wavefunction and the 2p wavefunctions with ml = ±1 are symmetric about the z axis, so when multiplied by z they become antisymmetric about z and the integral sums to zero. (Other ways to see this: µ is the same operator used for electric dipole transitions, so these integrals will obey the selection rules ∆l = ±1, ∆ml = 0 for an electric field along the z axis; R also the integral with 2s vanishes because the angular part of the integral will just be cos θ sin θ dθ, which is zero, and the 2p states with ml = ±1 give zero integrals because the φ part of the integral can R 2π be factored out of the problem, and will yield orthoganality integrals of the form 0 e±iφ , which are zero.) So let’s evaluate the integral from the last equation above, now writing in that the state i is the 2pml =0 state, and remembering that z can be written as r cos θ: Z ∞ Z (0) (0) R1,0 (0) R2,1 (r) (r)r2 dr ψ1s (z)ψ2p(ml =0) dτ = 0

Z π Z 2π 0

Y00 (θ, φ) Y10 (θ, φ) cos θ sin θ dθ dφ

3 Z Zr 2 √ e−3Zr/(2a0 ) r3 dr a0 24 a0 0 Z π Z 2π √ ! 3 × cos2 θ sin θ dθ dφ 4π 0 0 4 Z 4! 2 = √ (3Z/2a0 )5 24 a0 √ ! π 3 1 (− cos3 θ) × (2π) 4π 3 0 ! √ √ (2 · 4! · 25 ) a0 28 2a0 2 3 √ = = . Z 2 3 35 Z 35 24 =

Z ∞

xn e−ax dx =

n! an+1

cos θ sin θdθ = − cos2 θ

And finally we plug this back into our equation for α:

α ≈ −2e2 = −2e2

(0)

ψ1s (z)ψ2p(ml =0) dτ

E2p − E1s 2 8√ 2 2a0 /(35 Z)

[(−Z 2 /8) − (−Z 2 /2)]me e4 /h̄2 218 a20 /(310 Z 2 ) =− Z 2 (−3/8)me e2 /h̄2 221 a3 = 11 04 3 Z 11 a30 = 34 2Z 4 .

a0 = h̄2 /(me e2 )

For the hydrogen atom (Z = 1), this works out to 11.8a30 or 1.75 · 10−30 m−3 , which is much larger than the exact value of 4.5a30 [4]. The discrepancy is because our neglect of higher states in the perturbation theory is a much poorer assumption here than in the case of the one-dimensional box, and the contributions of higher energy levels will partially cancel the contribution from the 2pz orbital alone. In fact, almost a third of the polarizability is due to states of the free electron plus H+ ion. Still, this relatively simple calculation gets us in the right ballpark right away. 120

4.68 [Thinking Ahead: What sign should each result have? The radial kinetic energy—like any kinetic energy—should be positive. The electron–nuclear potential energy is always an attractive interaction, so that contribution should be negative. What should the result be if we set Zeff = Z? By setting Zeff equal to the real atomic number Z, we set the wavefunction back to the one-electron 1s orbital wavefunction, and so we should get the zero-order energy for helium, −4.00 Eh.] a. For the kinetic energy, we need to solve the average value integral, using psi1 for the wavefunction and K̂r for the operator. Let us break up the two one-electron terms in the wavefunction, setting ψ1 (1, 2) = ψ(1)ψ(2), where 3/2 Zeff e−Zeff ri /a0 . ψ(i) = 2 a0 Similarly, let’s write K̂r = K̂r (1) + K̂r (2), because the energy contributions from electrons 1 and 2 are independent (and also equal). Then we can solve the average value integral for a single electron. There is no angle dependence, so we integrate only over the radial coordinate r: Z ∞

K̂r (i) r2 dr ( ) 3 Z ∞ h̄2 Zeff −Zeff r/a0 1 ∂ 2 ∂ −Zeff r/a0 2 =− e 4 r dr r e 2me a0 r2 ∂r ∂r 0

hKr (i)i =

Let’s break down the radial integral and solve it alone first: Zeff 1 ∂ h 2 −Zeff r/a0 i r e a0 r2 ∂r Zeff Zeff 1 −Zeff r/a0 2 −Zeff r/a0 − 2re r e =− a0 r2 a0 Zeff 2 Zeff =− e−Zeff r/a0 . − a0 r a0

1 ∂ 2 ∂ −Zeff r/a0 =− r e r2 ∂r ∂r

Z ∞ 0

1 ∂ ∂ e−Zeff r/a0 2 r2 e−Zeff r/a0 r ∂r ∂r

Z ∞ Zeff Zeff −Zeff r/a0 2 e−Zeff r/a0 r2 dr r dr = − − e a0 r a0 0 Z ∞ Zeff Zeff r2 e−2Zeff r/a0 dr. 2r − =− a0 a0 0

At this point we can use the general analytical solution for the exponential integrals from Table A.5: Z ∞ n! xn e−ax dx = n+1 , a 0 yielding Z ∞ 1 1 ∂ ∂ 2 Zeff Zeff 2 e−Zeff r/a0 2 r2 e−Zeff r/a0 r2 dr = − − r ∂r ∂r a0 (2Zeff /a0 )2 a0 (2Zeff /a0 )3 0 2 Zeff a0 2a30 Zeff =− − 2 3 a0 2Zeff a0 8Zeff a0 1 1 a0 = =− − + . Zeff 2 4 4Zeff Finally, substituting this into our original equation for hKr (i)i, we get the radial kinetic energy 121

of one electron: h̄2 hKr (i)i = − 2me 2 = 12 Zeff

( 4

Zeff a0

3 ) 2 h̄2 a0 Zeff − = 4Zeff 2me a20

h̄2 2 = 12 Zeff Eh . me a20

The kinetic energy term is positive, as it should be. b. The potential energy term is simpler. Again we evaluate the integral for a single electron: Z Ze2 ∞ 1 hU (i)i = − ψ1 (i)2 r2 dr 4πǫ0 0 r 3 Z ∞ 2 Ze Zeff =− 4 e−2Zeff /a0 r dr 4πǫ0 a0 0 3 Ze2 4Zeff 1 =− 4πǫ0 a30 (2Zeff /a0 )2 3 2 Ze2 4Zeff a0 =− 3 2 4πǫ0 a0 4Zeff 2 e2 Zeff Ze = −ZZeff = −ZZeff Eh =− 4πǫ0 a0 4πǫ0 a0 As required, the potential energy contribution is negative, and larger in magnitude than the kinetic energy term (because Z cannot be less than Zeff ). c. Combining the kinetic and potential energies for electrons 1 and 2, we arrive at Eq. 4.24: 2 Zeff Eh . − ZZeff Eh = −2Zeff Z − hE0 i = 2 12 Zeff 2 If we set Z = Zeff , we find ourselves back at the zero-order energy, hE0 i = −Z 2 Eh .

4.69 Z Zeff ǫ1 ( Eh ) EHF ( Eh ) zero-order energy ( Eh )

He 2 1.69 −0.917 −2.862 −4.00

H− 1 < > > −1.00

first-order correction ( Eh )

+1.25

the atomic number for H is 1 no matter what charge take away one proton from He, and Zeff must decrease weaker e-nuc attraction, so H− orbital energy is greater and so too must the total energy be greater. The zero order energy is the energy of two H electrons, each with energy − Eh /2 The e-e repulsion is less in H− because the electrons not so close to the nucleus

4.70 By comparison to values in Table 3.1 (or if by remembering the rules for ionization energies in general chemistry), we find that the valence electron in the alkali metals has a higher HF energy (i.e., a lower ionization energy) as we move down the periodic table. For example, in Table 3.1 the Na 3s orbital is higher energy (less negative) than the Li 2s orbital. Similarly, we expect that the K 4s will be higher than the Na 3s. Moving across the third row of the periodic table from Na to Cl, we expect the 3s orbital to get more stable, because the shielding from adding more electrons does not completely balance the increased stability from adding more protons to the nucleus. The analogous orbital in the second row, the F 2s orbital, will again be more stable (because of the lower n quantum number). And finally, the 2s in F+ will be more stable than the 2s in F because the removal of one electron reduces the electron–electron repulsion. The final ordering is e < b < a < c < d. 122

4.71

a. What is the ionization energy of the 2s electron? 0.196 Eh b. What is the total ionization energy to form Li3+ ? 7.433 Eh c. Write the electron configuration for the lowest excited state of Li. 1s2 2p1 . d. Is the total Hartree–Fock energy higher (less negative) or lower (more negative) in the lowest excited state than in the ground state? e. Is the 1s orbital energy higher (less negative) or lower (more negative) in the lowest excited state than in the ground state?

4.72

a. Ground state Ca. Solution: The electron configuration 1s2 2s2 2p6 3s2 3p6 4s2 has six different subshells, all completely filled so that there are no magnetic or exchange repulsion effects to worry about, leaving 6 energy levels. b. Ground state K. Solution: The electron configuration 1s2 2s2 2p6 3s2 3p6 4s has six different subshells, and the unpaired 4s electron means that all the α spin electrons in the first 5 subshells will have slightly different energies from the β spin electrons, for a total of 2 × 5 + 1 = 11 energy levels. c. Lowest excited state Mg. Solution: The lowest excited electron configuration moves one 3s electron to the 3p subshell, giving 1s2 2s2 2p6 3s3p. This has 5 different occupied subshells, and the unpaired 3s and 3p electrons will split the α and β spin energies in the other subshells, giving at least 2 × 3 + 2 = 8 energy levels. In addition, however, the single unpaired 3p electron must go into some specific orbital, e.g., the 3pz . This means the two energies of the 2p subshell split one more time, because the 2pz and 3pz electrons will have a different interaction than the 2px or 2py with the 3pz . This gives a total of 10 energy levels.

4.73

a. The higher energy the orbital, the closer to ionization, so this is the same as asking to put the atoms in order of decreasing first ionization energy. Following the periodic trends, we would arrange them as follows: He<Ar<K, meaning that helium has the most stable valence electrons (lowest energy, hardest to ionize). b. The core electrons in a neutral atom have a single trend: the energy becomes more negative as the nuclear charge increases, regardless of what n value the valence electrons have: K<Ar<He. Both of these trends appear in the Table of HF energies in Table 4.1.

4.74 b a d c. (The actual HF energies in Eh are roughly −9.0, −58.5, −0.7, and −5.4 for a–d, respectively.) d < c because P differs from O by 6 protons (which will stabilize the 2p electrons) and 6 electrons (which will destabilize it by shielding). The effect of each new electron never completely cancels the contribution of the new proton, so a given subshell is generally more stable in a heavier element. a < d because P has a lower nuclear charge than S, and because the 2p electrons are more shielded than the 2s electrons. The 2s electrons in S have more shielding Finally, you can safely assume the 1s orbital on Al to be the lowest energy of all. Even though Al does not have as high an atomic number Z as P or S, and in the absence of shielding the change in energy from Z = 13 to 15 and 16 is much smaller than the change from doubling n from 1 to 2. Furthermore, the 1s electrons are much less shielded than the n = 2 electrons, so the effect of the added electrons matters less. 4.75 Going from N to O, we add one proton to the nucleus and one electron to the valence shell. The result on each orbital energy is that the destabilizing electron–electron repulsion increases, but the 123

nuclear-electron attraction increases more, so that overall any given orbital in the atom becomes more stable in O than it was in N. Because this trend applies to any given orbital, it also applies to the total electronic energy of the atom. property from N to O 1s orbital energy − 2s orbital energy − total repulsion energy + total kinetic energy + total energy − 4.76 The calcium ground state electron configuration is 1s2 2s2 2p6 3s2 3p6 4s2 . At this point, all the shells are filled, so each subshell corresponds to exactly one energy value. That changes once we make the lowest excited state. Because there are no partially-filled orbitals, the lowest energy excited state must be obtained by promoting one of the highest energy 4s electrons into the lowest energy available orbital, the 3d. This means the 3d and the remaining 4s electrons no longer occupy the same region of space, and their spin interactions with all the other electrons no longer cancel (even if they still have opposite spins). This splits the HF orbital energies of the α and β electrons, so the 1s, 2s, 2p, 3s, and 3p orbitals each get split into two different energies, one energy for the α electron(s) and a different energy for the β. That would be ten distinct energy values for the core orbitals, plus one each for the 4s and 3d valence electrons for a total of 12. However, there’s one more effect to consider. The 3d electron will have a particular ml value, which identifies how strongly aligned the orbital is with the z axis. This will lead to an additional splitting of the p subshell energies into the ml = 0 and the ml = ±1 groups. The ml = −1 and ml = +1 orbitals will continue to have the same energy (if the electron spin is the same) because the sign of ml determines the direction of the orbital motion but does not affect the spatial distribution of the electron. That means our 12 orbital energies increase to a total of 14 after we include the splitting of the 2p and 3p subshells. 4.77 a. In the C5+ ion (one-electron atom with Z = 6), the energies are given by −Z 2 /2n2 = −18/n2 Eh . E(C) E(C5+ ) 1s −11.3475 −18.0 −11.3021 −18.0 2s −0.8296 −4.5 −0.5840 −4.5 2p −0.4388 −4.5 −0.4388 −4.5

∆E 6.6525 6.6979 3.6704 3.9160 4.0612 4.0612

The shift is largest where the electron-electron repulsion is strongest: between the two 1s electrons that occupy a small volume and experience much mutual repulsion, and in the 2p orbitals, where the electrons are effectively shielded from the nuclear charge by 1s and 2s electrons. b. In the 3 P term, S = 1 so the two 2p electrons have spins oriented in the same direction. In the 1s and 2s orbitals, the only difference between the two electrons is the value of the quantum number ms , the direction of the electron spin. Therefore, the energy splitting which is seen in the 1s and 2s subshells but not in the 2p subshell must be an electron spin effect. This is also evident from the other atoms listed in Table 4.1. The splitting of subshell energies into two values is seen only for those atoms that have unpaired electrons (Li, B, C, Na), not for states in which all the electron spins are paired (He, Be, Ne). When electron spins are unpaired, the atom has a net electron spin; in other words, S 6= 0. But how can this change the energies 124

of electrons in filled orbitals? How can the unpaired electrons in the carbon 2p subshell split the energies of the 1s and 2s electrons? Let’s say that the unpaired electrons in the atom have α spin. These unpaired electrons interact with the α electrons in filled shells differently than they interact with the β electrons in the filled shells, because the spins are different. The splitting in the orbital energies is zero for the 2p electrons, because they share the same spin (they differ only in ml value, and there is nothing in the atom to give different ml values different energies). The 2s splitting is large because these electrons share the same volume as the 2p electrons and have strong spin interactions with them. The 2sα electron has spin magnetic moment aligned with the 2pα electrons and is shifted to lower energy, while the 2sβ electron is shifted to higher energy. The 1s electrons are subject to the same shifts, but the effect is reduced because the 1s and 2p orbitals do not overlap as much as the 2s and 2p. (While this answer is qualitatively accurate in its prediction, we must point out that the Hartree– Fock calculations do not explicitly include spin interactions, and the resulting split in the α and β electron energies is actually the result of terms such as the exchange integral in Eqs. 4.49 and 4.50. The spin interaction is accounted for by symmetrization, not by direct calculation of the magnetic energy.) c. 3 P → 5 S ∆E = −37.5957 − (−37.6906) = 0.0949 Eh = 249 kJ mol−1 . This promotion is likely if it allows the formation of a 400 kJ mol−1 bond that could not form otherwise. 4.78 Fluorine has an electron configuration 1s2 2s2 2p5 . The 1s electrons will have an energy between −11.3 Eh (C 1s) and −32.8 Eh (Ne 1s). We may be able to improve our estimate by finding the effective atomic number Zeff,i experienced by electron i in these other atoms, where r

Ei . Eh

5 3.92 1.99 1.59

6 4.76 2.38 1.87

Zeff,i ≈ ni

−2

We find the following: H Z

1 1.00

He 2 1.35

Li 3 2.23 1.25

Be 4 3.08 1.57

Ne 10 8.09 3.92 2.59

Na 11 9.00 4.73 3.49 1.21

orbital 1s 2s 2p 3s

For atomic numbers between 5 and 11, Zeff (1s)/Z ranges from 0.78 to 0.82, so we can reasonably guess that for the F 1s electrons, this ratio is about 0.81. Similarly, Zeff (2s)/Z ranges from 0.40 to 0.43, and Zeff (2p)/Z is constant at 0.32. Using our expression for Zeff,i to obtain the energy, we estimate −26.6 Eh, −1.79 Eh, and −1.04 Eh for the 1s, 2s, and 2p orbitals, respectively. The actual calculated values (averaging over energies within an orbital) are −26.4, −1.58, and −0.73 Eh. 4.79 The carbon ground state electron configuration is 1s2 2s2 2p2 . The C+ ground state electron configuration is 1s2 2s2 2p. The angular momenta in the 1s2 2s2 electrons net zero. All angular momentum is from the 2p electron. 125

l = 1, s = 21

ml = +1 0 ↑ ↑ ↓

↓

 −1 ML MS   1    ML (max) 1 2   1   0 2  MS (max) 1 ↑ −1 L 2   1 − 12  S   1   0 −2  2S + 1   ↓ −1 − 21

=1 = 21 =1 = 12 =2

J = |L − S|, . . . , L + S = 21 , 32 . All states are accounted for. The term states are 2 P 1 , 2 P 32 . 2

4.80 To get a term with L = 4, we have to be able to get an ML value of 4. We can’t get any orbital angular momentum with only s electrons, and the highest ML we can get with a p subshell is L = 2, if we place two electrons in the ml orbital. That takes us to at least period 4 of the periodic table, where we can use d electrons. We can get ML = 4 if we put two electrons in the 3dml =2 orbital, and therefore that electron configuration must have an L = 4 term state. This first becomes possible with the ground state electron configuration [Ar]4s2 3d2 , the electron configuration for titanium, Ti. In order to get both 3d electrons into the same orbital, they must have canceling spins, which means that this G term has zero spin and a multiplicity of 2S + 1 = 1. The 3d subshell is less than half full, so the lowest energy is for the minimum J value, given by |L − S| = 4 − 0 = 4. In fact, since S = 0, this is the only J value possible. 4.81 120. We need to consider only the electrons in the partly filled 4d subshell, which contains five orbitals. There are two groups of combinations: those where two electrons are paired and the third is unpaired, and those where all three electrons are unpaired. In each case, we can then ask how many ways are there of placing the electrons in the orbitals, and then how many ways are there of arranging the spins: a. 2 paired, 1 unpaired. The pair can be in any of the 5 orbitals, and for each of those the unpaired can be in any of the four remaining orbitals, giving 5 × 4 = 20 possible arrangements for the ml quantum numbers. Furthermore, the unpaired electron may have ms = 21 or − 12 for each ml arrangement, giving a total of 40. b. All 3 unpaired. There are at least two ways to count the number of possible arrangements of the electrons in the orbitals, neglecting ms for now. You can just count them, because there aren’t that many, or you can show that this is the same as the number of ways of arranging two empty orbitals. Again there are 5 × 4 = 20 ways of placing two things in five slots, but because there’s no difference between the two empty orbitals we divide 20 by 2 to get 10 distinct arrangements of the electrons in the orbitals. Then, because there are three independent electrons, there are 23 = 8 possible ways of arranging the ms values in each of these 10 ml arrangements, resulting in 80 distinct combinations. Add this to the previous 40 to get a grand total of 40 + 80 = 120. 4.82 Many possible solutions, but for each of them: (i) we only need the 6d and 5f electrons, because all the other subshells (including the 7s) are full and contribute zero to the spin and orbital angular momenta; (ii) the maximum MS is 3/2; (iii) the maximum ML is 8. Here are a few examples: 1

6d 6d1 6d1 6d1 6d1

ml = −2 −1 ↑

2 2

↑

↑ ↑ ↑

5f 5f 2 5f 2 5f 2 5f 2

ml = −3 −2 −1 ↑↓

↓ 126

0 1

↑ ↓

↑↓ ↑ ↑ ↑

ML −8 +8 +7 +7 0

MS +1/2 +1/2 +3/2 +1/2 +1/2

4.83 This is tractable using the vector model (20 microstates), but can also be managed by seeing that we have one s electron with s = 1/2 and l = 0, and 9 d electrons, for which all the angular momenta must cancel except for the s and l of one electron (or more accurately, one electron “hole”). The one uncanceled electron has s = 1/2 and l = 2, so we are coupling two spins of 1/2 (which give 0 and 1 as possible results) and have only the net l = 2 of the uncanceled d electron. Therefore, the terms are 1 D and 3 D, and applying Hund’s rules we get the ranking 3 D3 <3 D2 <3 D1 <1 D2 . 4.84 First, we ignore all the electrons in filled subshells. The overall L value must be 2, because the 5s electron has no orbital angular momentum, and the d electron has l = 2, so the vector sum of the l’s is 2. The only possible values of S are 0 (if the two electron spins cancel) and 1 (if they add up). So the two terms resulting are the 1 D and 3 D. For the 1 D, the only J value is |L − S| = L + S = 2. For the 3 D, the possible J values are 1, 2, and 3. The higher-spin 3 D state is the lowest in energy: 3 D1,2,3 <1 D2 . 4.85 This electron configuration contains 5 electrons in the partly filled subshells. Thus, the total spin S must be half-integer; the possible values will be 1/2 (if four electron spins cancel), 3/2, and 5/2 (if all five electrons have “parallel” spins). That means that the multiplicities 2S + 1 must be 2, 4, or 6. Because S is half-integer in this case and L is always an integer, J = |L − S|, |L − S| + 1, . . . , L + S must also be half-integer. That leaves four options: 4 F1/2 , 6 D1/2 , 2 F7/2 , and 2 S1/2 . Of these, the 4 F1/2 can be dropped because a J value of 1/2 cannot be obtained from S = 3/2 and L = 3 (the minimum J is |3 − 3/2| = 3/2). The remaining three are arranged in order of decreasing spin, and then decreasing L value:

D1/2 , 2 F7/2 , 2 S1/2 .

4.86 first term : 2 D second term : 4 P third term : 2 P fourth term : 2 S

L=2 S = 12 10 states L=1 S = 32 12 states L=1 S = 21 6 states L=0 S = 21 2 states

The missing state is

ML = 0; MS = + 21 .

4.87 ML 3 2 1 1 0 0 4

MS 1 1 1 1 1 1 0

term 3 F 3 F 3 F 3 P 3 F 3 P 1 G

increasing energy:

ML 3 3 2 2 2 1 1 3

MS 0 0 0 0 0 0 0

term 1 G 3 F 1 G 3 F 1 D 1 G 3 F

ML 1 1 0 0 0 0 0

MS 0 0 0 0 0 0 0

term 3 P 1 D 1 G 3 F 3 P 1 D 1 S

F,3 P,1 G,1 D,1 S

4.88 Z = 4. The ground state is

1s2 2s2 ; all orbitals are filled, so L = 0, S = 0, J = 0: 127

S0 .

The lowest excited state is

1s2 2s2p ; the vector model:

2s 2p ml = 0 −1 0 +1 ↑ ↑ ↑ ↓ ↑ ↑ ↑ ↓ ↑ ↑ ↑ ↓ ↓ ↑ ↓ ↓ ↓ ↑ ↓ ↓ ↓ ↑ ↓ ↓

The term states are

P0,1,2

and

ML −1 −1 0 0 +1 +1 −1 −1 0 0 +1 +1

MS 1 0 1 0 1 0 0 −1 0 −1 0 −1

P P 3 P 3 P 3 P 3 P 1 P 3 P 1 P 3 P 1 P 3 P 3

P1 .

4.89 3

S1 → S = 1 : two unpaired electrons L = 0 : no net orbital angular momentum

Be has four electrons, so two are paired and two are unpaired. One possibility places all electrons in s orbitals (since L = 0): 1s2 2s3s . This has electronic term states 1 S0 and 3 S1 , as we can determine by applying the complete vector model:

2s ↑ ↑ ↓ ↓

3s MS ML ↑ +1 0 ↓ 0 0 ↑ 0 0 ↓ −1 0

S S 1 S 3 S 3

S=1 S=0

L=0

Other answers are possible. 128

4.90 1s 3d ml = 0 ml = −2 −1 0 ↑ ↑ ↑ ↑ ↑

↓

↑

↓

+1 +2 ML MS −2 1 −1 1 0 1 +1 1 ↑ ↑ +2 1 −2 0 −1 0 0 0 +1 0 ↓ +2 0 ↓ −2 0 −1 0 0 0 ↑ +1 0 +2 0 ↑ −2 −1 −1 −1 0 −1 ↓ +1 −1 ↓ +2 −1

↓

↑

↓

D D 3 D 3 D 3 D 3 D 3 D 3 D 3 D 3 D 1 D 1 D 1 D 1 D 1 D 3 D 3 D 3 D 3 D 3 D 3

Lmax = 2, Smax = 1, J = 1, 2, 3, 3 D1,2,3 . This is the largest value of S. For an orbital less than half-full, 3

lower J is lower energy. The ground state is

D1 .

4.91 1s2 contributes no angular momentum because it is a filled s-shell. ML = (1) (2) (3) (4) (5) (6)

−1 0 ↑ ↑

↓

+1 ML MS −1 + 21 0 + 12 ↑ +1 + 12 ↓ +1 − 12 0 − 12 −1 − 12

These are the only possible orientations of one electron in the 2p orbital.  L=1⇒P    L = ML (max) = 1 2 P1/2,3/2 1  S = MS (max) = 2   2S + 1 = 2

This accounts for all the MS and ML values allowed so the only term symbols are 4.92

P1/2,3/2 .

a. Because this atom has 22 electrons, an even number, S must be an integer, not a halfinteger. Consequently, 2S + 1 must be an odd number, not 4. b. An F term has L = 3, and therefore must have an ML = 3 state. This cannot be achieved when the only unpaired electrons are in a 4s4p3 configuration—there is no way then to get ML = 3. Note that J = 7/2 is consistent with S = 3/2, L = 3. 129

4.93 Mg : 1s2 2s2 2p6 3s2

ml =

3s 0 ↑ ↑ ↑ ↑ ↑ ↑ ↓ ↓ ↓ ↓ ↓ ↓

Mg∗ : 1s2 2s2 2p6 3s3p

3p −1 0 +1 ML MS ↑ −1 +1 ↑ 0 +1 ↑ +1 +1 ↓ −1 0 ↓ 0 0 ↓ +1 0 ↑ −1 0 ↑ 0 0 ↑ +1 0 ↓ −1 −1 ↓ 0 −1 ↓ +1 −1

P P 3 P 3 P 3 P 3 P 1 P 1 P 1 P 3 P 3 P 3 P

Smax = 1,

L = 1,

J = 0, 1, 2 :

P0 , 3 P1 , 3 P2

These account for (2S + 1)(2L + 1) = 9 states. The remaining three states have S = 0, L = 1, and J = 1: 1

P1 .

4.94 Using only those electrons, the highest value of S we can obtain is 4/2 = 2 when all four valence electrons have parallel spins. This is possible if we promote one of the 5s electrons to the 4d subshell, generating a configuration [Kr]5s4d3 . The lowest energy term will be the one with the highest L and ML value for that configuration when the three 4d electrons are in separate orbitals. The highest ML value attainable is then ML = 3, for one electron each in ml = 2, 1, 0. That corresponds to L = 3 or an F term, and the lowest energy J value for this configuration (half-filled subshells or less) when L = 3 and S = 2 is J = 1. The term symbol is 5 F1 . 4.95 Rather than solving the whole vector model, we can determine the states by carefully looking at the configuration. The unpaired s electron can only affect the spin of the term states; it cannot affect the L value. To find the possible L values, we look only at the f 13 part of the configuration. This is a nearly full f subshell, lacking only one electron, and therefore gives rise to the same term as an f 1 configuration. The f 1 configuration yields a 2 F term only: only one electron is in an unfilled subshell and so the total L and S are the same as for that one electron (l = 3, s = 1/2). If we now add the unpaired s electron, the possible spin values become S = 0 (if the s and f electron spins cancel) or 1 (if they don’t cancel). The possible terms are therefore 1 F or 3 F , and the J values are 3 for the 1 F and 2, 3, and 4 for the 3 F (3 − 1, 3, 3 + 1). According to Hund’s rules, the higher spin is lower energy, and because the f subshell is more than half full, the higher J is lower energy: 3

4.96

F4 , 3 F3 , 3 F2 , 1 F3 .

• Maximum S is 2, if the ground state (which must have maximum S) has 2S + 1 = 5. That means a maximum of four unpaired electrons (or four “holes,” but that is not an option for the low-energy excited state of Ti).

• Maximum L is 6. Even if you’re not sure that I means L = 6, the spin multiplicity is 1, which means S = 0, and therefore J = L. • All four electrons are in one subshell, such that the maximum sum of their ml values is 6. An s subshell can’t hold four electrons. If the subshell is a p, the maximum ML value is 2. If the subshell is a d, the maximum ML value is 6. 130

Therefore, the configuration is the

[Ar]3d4

configuration.

4.97 We’re starting from the ground electron configuration of Ti, [Ar]4s2 3d2 , and looking for the term state from this configuration that comes just above the 3 F ground term. (Note that it is not generally necessary to change the electron configuration to reach the first excited term state.) We neglect the filled subshells, leaving us with only the 3d2 electrons. It isn’t necessary to solve the whole vector model to answer this problem, but with 45 combinations it’s not that difficult to do the whole thing. Here’s one other way to look at it first. The lowest energy term state is 3 F because the highest MS value we can obtain for two electrons is 1, which has to come from putting the two electrons in different orbitals, so the maximum ML we can get for MS =1 is ML =3 (for one electron in ml = 2 and one in ml = 1). This corresponds to a term state with L = 3 and S = 1, or 3 F . If there are any other term states with S = 1, they are next in order of increasing energy according to Hund’s first rule. So we can start by only looking for the combinations with MS = 1, because these will have the greatest spin: ml = −2

↑ ↑ ↑ ↑

−1 ↑ ↑ ↑ ↑

0 ↑ ↑ ↑ ↑

1 ↑

↑ ↑ ↑

2 ML ↑ +3 ↑ +2 ↑ +1 ↑ 0 +1 0 −1 −1 −2 −3

MS 1 1 1 1 1 1 1 1 1 1

F F 3 F 3 F 3 P 3 P 3 P 3 F 3 F 3 F 3

Three of these have been assigned to a 3 P term because the 3 F only accounts for seven of the combinations. There are no other combinations with MS = 1, and there are no combinations with higher MS . Therefore, there must be a term with L = 1, S = 1 to account for the three remaining combinations. This is the only other term with S = 1, and therefore is the lowest excited term state. The 3d subshell is less than half full, so the lowest value of J is the lowest energy. That would be J = |L − S| = 0. If one works through the whole vector model, one finds there are also 1 G, 1 D, and 1 S terms.

4.98 This 3d2 subshell gives rise to all the same combinations in the vector model as obtained for Ti, but now we will get twice as many possible combinations because the 4s electron may have ms = 1/2 or −1/2. The singlet states in Ti have S=0, and with the addition of this unpaired electron those all become S=1/2 states (so 2S + 1 becomes 2): 1 G, 1 D, 1 S become 2 G, 2 D, 2 S. The triplet states in Ti have S=1, and when combined with the spin of the 4s electron these may become either S = 1/2 states (if the spins partially cancel) or S = 3/2 states (if the spins add together): 3 F , 3 P become 4 F , 4 P , 2 F , 2 P. We list these in order of increasing energy putting highest S values first, and (for given S) the highest L values first. J varies from |L − S| to L + S, and because the d subshell is less than half full, in order of increasing J: 4 F3/2 , 4 F5/2 , 4 F7/2 , 4 F9/2 , 4 P1/2 , 4 P3/2 , 4 P5/2 , 2

G7/2 , 2 G9/2 , 2 F5/2 , 2 F7/2 , 2 D3/2 , 2 D5/2 , 2 P1/2 , 2 P3/2 , 2 S1/2 .

4.99 The valence electron configuration is 6s2 4f 14 5d2 . We only need to consider the 5d electrons. There are 45 states total: 5 with the two electrons paired in ml = −2, −1, 0, 1, and 2; and 40 with the 131

electrons in different ml states (10 ↑↑, 10 ↑↓, 10 ↓↑, and 10 ↓↓). The ML and MS values are as follows: ML MS ML MS ML MS −3 −1 −3 0 −3 0 −2 −1 −2 0 −2 0 −1 −1 −1 0 −1 0 −1 −1 −1 0 −1 0 0 −1 0 0 0 0 0 −1 0 0 0 0 1 −1 1 0 1 0 1 −1 1 0 1 0 2 −1 2 0 2 0 3 −1 3 0 3 0

ML MS −3 1 −2 1 −1 1 −1 1 0 1 0 1 1 1 1 1 2 1 3 1

ML MS −4 0 −2 0 0 0 2 0 4 0

First Term: S = 1 L = 3 J = 2, 3, 4 3 F2 3 F3 3 F4 This accounts for (2 · 1 + 1)(2 · 3 + 1) = 21 states Second Term: S = 1 L = 1 J = 0, 1, 2 3 P0 3 P1 3 P2 This accounts for (2 · 1 + 1)(2 · 1 + 1) = 9 states Third Term: S = 0 L = 4 J = 4 1 G4 9 states Fourth Term: S = 0 Fifth Term: S = 0

L=2 J =2 L=0 J =0

5 states 1 state

4.100 Finding term symbols using the vector model can be a completely mechanical exercise, following the recipe in the chapter. ML = −1 0 1 ML MS ↑↓ ↑↓ ↑ −1 + 21 ↑↓ ↑↓ ↓ −1 − 12 ↑↓ ↑ ↑↓ 0 + 12 ↑↓ ↓ ↑↓ 0 − 12 ↑ ↑↓ ↑↓ +1 + 12 ↓ ↑↓ ↑↓ +1 − 12 L = ML (max) = +1 2 P S = MS (max) = + 21 This accounts for (2L + 1)(2S + 1) = 3 × 2 = 6 states. There are only six states (or “combinations”)

total, so

is the only term. Degeneracy = 6.

4.101 The highest spin yields the lowest energy. The highest spin is obtained when all three valence electrons have spins in the same direction. S = 12 + 21 + 21 =

3 2

2S + 1 = 4

For S = 3/2, the highest L value will have the lowest energy. In the vector model, we look for the state with MS = 3/2 that has the largest ML : 4s

↑ ml = 0 3d ml = −2 −1

↑ +1

↑ +2

ML = 3 ⇒ L = 3. Note that we can’t get ML = 4 except when MS = ±1/2. L = 3 ⇒ F , S = 3/2, L = 3 gives J values of 3/2, 5/2, 7/2, and 9/2. The 3d subshell is less than half-filled, so the lowest J is the lowest energy. The term is

F3/2 . 132

4.102 The electron configuration of F+ is 1s2 2s2 2p4 ; it is isoelectronic with the O atom. There are four electrons in 2p orbital: ml = +1 0 −1 ML MS ↑↓ ↑↓ +2 0 ↑ ↑↓ ↓ 0 0 ↑↓ ↑↓ −2 0 ↑↓ ↓ ↑ +1 0 ↑↓ ↑↓ 0 0 ↓ ↑ ↑↓ −1 0 ↑↓ ↑ ↑ +1 1 ↑ ↑↓ ↑ 0 1 ↑ ↑ ↑↓ −1 1 ↑↓ ↑ ↓ +1 0 ↓ ↑↓ ↑ 0 0 ↑ ↓ ↑↓ −1 0 ↑↓ ↓ ↓ +1 −1 ↓ ↑↓ ↓ 0 −1 ↓ ↓ ↑↓ −1 −1

D D 1 D 3 P 3 P 3 P 3 P 3 P 3 P 1 D 1 S 1 D 3 P 3 P 3 P

ML (max) = 2 MS (max) for ML = 2 is 0 2S + 1 = 1

L=2 S=0 1

ML (max) = 1 MS (max) for ML = 1 is 1 2S + 1 = 3

L=1 S=1 3

ML (max) = 0 MS (max) = 0 2S + 1 = 1

D term

P term

L=0 S=0 1

S term

The lowest energy term has the largest value of S (Hund’s first rule), so 3 P . The lowest energy J state is the highest value of J for more than half-full shell (2p4 is more than half-full). J = |L − S| , . . . , L + S = 0, 1, 2 for 3 P The lowest energy state is 3 P2 MJ EZeeman = gs µB JZ BZ = 2(9.274 · 10

= −J, −J + 1, . . . , J = −2, −1, 0, +1, +2. JZ = M J −24

−1

)MJ (1.0 T)

= (27980 MHz)MJ

1MHz 6.626 · 10−28 J

MJ EZeeman (MHz) +2 +55960 +1 −27980 0 0 −1 −27980 −2 −55960 4.103 There is no orbital angular momentum here, so no L or ML values to worry about. We just look at all the possible ways we can add the four individual nuclear spins. There are five sets: all four spins up, three up and one down, two up and two down, one up and three down, and all four down. I’ll just draw one of each and count the number of possible configurations for each: MI

# configurations

↑↑

1 config

4 configs

↑↑

↑↓ ↓↓

6 configs

↓↓

−1

4 configs

↑↑ ↑↓ ↓↓

↓↓

−2 133

1 config Copyright c 2014 Pearson Education, Inc.

That’s a total of 16 configurations. The maximum MI is 2, so we have one nuclear spin state with I = 2. That state will have five components, with MI = −2, −1, 0, 1, 2, each of which contribute to the list of configurations above. Of the 11 configurations remaining, the maximum MI value is 1, and there are three configurations with that value, so we have three distinct I = 1 states, each with MI = −1, 0, 1. Those account for nine more configurations, leaving two configurations with MI = 0, and hence two I = 0 states.

Chapter 5 5.1 Since the electron experiences only attractive forces from the two nuclei, its potential energy is negative everywhere. 5.2 Fbinding (R, ri , θi ) is not a function of the electronic coordinate φ, the angle about the internuclear axis. This is because the electric field of the nuclei is cylindrically symmetric about this axis. As in atoms, which also see no change in potential energy along φ, only the kinetic term 1 1 ∂2 r2 sin2 θ ∂φ2 is φ-dependent in the Hamiltonian. The solution for the φ-dependent part of the electronic wavefunction in a diatomic molecule is therefore the same as in atoms: eiλφ . 5.3 Qualitatively, we’d expect the normalization constant to increase as the two 1s atomic orbitals become more widely separated. Because the probability density is proportional to the square modulus of the wavefunction, regions of constructive overlap where the functions add together give a proportionately larger contribution to the probability density. The more constructive overlap there is, the smaller the normalization constant needs to be in order to keep the overall integrated probability density equal to one. To see this mathematically, consider two simple limits. In the limit that R → 0, ψ+ becomes the same as two 1sA orbital wavefunctions added together, because centers A and B are in exactly the same place: Z Z 2 |ψ+ (R = 0)|2 dτ = C+ (0)2 [2(1sA )] dτ Z = 4C+ (0)2 (1sA )2 dτ = 1 r 1 1 = . C+ (0) = 4 2 In the opposite limit, as R → ∞, the two functions are sufficiently separated that the overlap integral R 1sA 1sB dτ vanishes: Z Z |ψ+ (R = 0)|2 dτ = C+ (∞)2 [(1sA + 1sB )]2 dτ Z Z Z 2 2 2 = C+ (∞) (1sA ) dτ + 2 (1sA )(1sB )dτ + (1sB ) dτ = C+ (∞)2 [1 + 2(0) + 1] = 2C+ (∞)2 = 1 r 1 C+ (∞) = . 2 134

In this particular case (two 1s orbitals with the same phase), we can be certain that the constructive overlap decreases steadily as R increases. For more complicated functions, the trend in C(R) will not be so straightforward. 5.4 The big difference between our look at electrons around atoms and our H2 dissociation problem is that, in the second case, the possible locations of the electrons are well-separated in space: once we measure the location of an electron on atom A, it cannot spontaneously move to atom B. For the atom, the same electron can be detected in multiple locations, even at the same time, because the electron’s de Broglie wavelength is (generally) a large fraction of the distances over which we will be able to detect the electron. Our analysis of Eq. 5.22 uses a convenient statistical interpretation of quantum mechanics, in which the probability density represents the distribution of results obtained from a large number of measurements on a system. This is particularly appropriate when (as in this H2 dissociation) there are distinct possible results and the average value of the measurement may not correspond to any actual state of the system. For example, a mixture of ground state 1s and excited state 2s hydrogen atoms will yield an average energy between −0.500 Eh and −0.0.125 Eh . A mixture that has a probability 2 2 distribution 0.96[ψ1s ] + 0.04[ψ2s ] would have an average energy hEi = −0.485 Eh , but no single atom in the entire sample will have an energy of −0.485 Eh . The probability distribution in that case tells us that 96% of the atoms are in the 1s state and 4% in the 2s state. Similarly, Eq. 5.22 describes the set of possible measurement outcomes, but can no longer describe a detected state of the single H2 molecule after dissociation. 5.5 The Born–Oppenheimer approximation effectively separates the electronic and nuclear coordinates. For the electrons, we may use Cartesian or spherical coordinates measured relative to some origin (such as the molecular center of mass), or we may measure the distances and the angles θ with respect to the nuclei A and B:   x1 , y1 , z1 , x2 , y2 , z2 rA1 , rB1 , θA1 , θB1 , φ1 , φ2 electronic coordinates or or  r1 , θ1 , φ1 , r2 , θ2 , φ2 , rA2 , rB2 , θA2 , θB2 , φ1 , φ2

Similarly, the positions of the nuclei may be specified using either Cartesian or spherical coordinates, but in either case these are usually relative to some lab-fixed coordinate system: nuclear coordinates {R, Θ, Φ

5.6

or X, Y, Z

a. R < 0.1 Å: Ĥeff is dominated by repulsive terms in Unuc−nuc and Uelec−elec b. R = Re ; Uvib (R) is negative at this distance, meaning that Uvib (Re ) < Uvib (∞), because the Unuc−elec term dominates and this is the only negative contribution to the energy. c. With very large R, the atomic terms dominate: K̂elec , Unuc−elec, Uelec−elec; not Unuc−nuc .

5.7

a. K̂elec + kinetic energy is always positive b. K̂nuc + kinetic energy is always positive c. Ûelec−elec + a repulsion term increases E d. Ûnuc−elec − an attraction term decreases E e. Ûnuc−nuc + a repulsion term increases E 135

5.8 The answer to the first question is no. The cylindrical coordinate system is terrific for the kinetic energy operator, because the three coordinates are mutually perpendicular (just as they are in Cartesian and spherical polar coordinates). This feature allows the Laplacian to be separated into three terms, one for the second derivative of each coordinate: h̄2 1 ∂ ∂ ∂2 1 ∂2 K̂ = − + 2 . s + 2me s ∂s ∂s s2 ∂φ2 ∂z However, the potential energy is the sum of two contributions that depend on rA and rB (with φ a third independent variable), and it is not separable in cylindrical coordinates because rA and rB each depend on both s and z. So we cannot obtain a coordinate system for H+ 2 in which the kinetic energy operator and the potential energy are each separable, but it is still possible for a coordinate system to be found in which the overall Hamiltonian, with both kinetic and potential terms, can be divided into three distinct terms. The prolate hyperspherical coordinate system permits this, where the three coordinates are [5] ξ=

rA + rB , R

η=

rA − rB , R

φ,

which allow the wavefunction to be written as the product of three terms: ψ(ξ, η, φ) = AX(ξ)Y (η)eimφ . 5.9 The molecule has 14 electrons and 2 nuclei, based on the kinetic energy terms (which are one for each particle), and the nucleus–electron potential energy terms (which depend on distances such as rA1 ) tell us that the atomic numbers Z of the two nuclei are 6 (C) and 7 (N). The term with the factor of 42 is the nucleus–nucleus repulsion term, proportional to ZA ZB . The atomic numbers add up to 13, so there is one excess electron. Therefore, the molecule is CN− . 5.10 1

e2 (1.602 · 10−19 C)2 = 4πǫ0 a20 (1.113 · 10−10 C2 J−1 m−1 )(5.292 · 10−11 m)2 = 8.234 · 10−8 N.

5.11 If the electron is on the z axis halfway between the two nuclei, rA = rB = R/2, θA = 0, and θB = π. For fluorine, Z = 9. ZB ZA ZB ZA e2 Fbinding = 2 cos θA − r2 cos θB − R2 4πǫ0 rA B e2 9 9·9 9 = (1) − (−1) − 2 4πǫ0 (R/2)2 (R/2)2 R 2 (36 + 36 − 81)e = 4πǫ0 R2 =

−

9e2 . 4πǫ0 R2

5.12 The problem suggests that if we set θA = 0 and θB = π (which puts the electron exactly on the axis between the two nuclei) that we can simplify Eq. 5.1 to the form given. So let’s make those 136

substitutions: θA = 0 θB = π cos θA = 1 cos θB = −1 e2 1 1 1 Fbinding = 2 cos θA − r2 cos θB − (r + r )2 4πǫ0 rA A B B e2 1 1 1 Fbinding = 2 + r2 − (r + r )2 4πǫ0 rA A B B 2 2 2 2 rB (rA + rB )2 + rA (rA + rB )2 − rA rB e2 = 2 r2 (r + r )2 4πǫ0 rA B B A   0 2 2 2 2 3 4 2 rB − rA rB ) + 2rA rB + rB + rA (rA + rB )2  e2    (rA =   2 r2 (r + r )2  4πǫ0  rA A B B >0

ZA = ZB = 1

because rA , rB > 0.

5.13 This one is testing the basic math required to calculate Fbinding . Fbinding =

e2 ZH ZHe ZH ZHe cos θ − cos θ − H He 2 2 4πǫ0 rH rHe R2

√ R = 1 Å θH = π/2 θHe = 3π/4 rHe = 2 Å   0 2 e  1 2  2 Fbinding = cos(π/2) −  2 cos(3π/4) − 2 4πǫ0 1 Å2 1 Å 1 Å =

4πǫ0 (1 Å )

ZH = 1 ZHe = 2 rH = 1 Å cos(3π/4) = −.707

(.707 − 2) < 0

so antibinding. e− r @ r @ He @ θHe $ θH @ @ r @r H R He

5.14 The maximum force approaches infinity, as can be seen from the equation, for rH = 0 or rH = R. The maximum classical binding force is obtained when the electron is between the two nuclei and very close to either of them. To find the minimum along the bond and between the two nuclei, however, we actually have to do the math. The conditions stated require cos θHe = 1 and cos θH = −1 for the orientation drawn here. e− QQ rHe QrH Q Q θHe Q θH - ZHe = 2 ZH = 1 137 Copyright c 2014 Pearson Education, Inc.

R = rHe + rH rHe = R − rH

∂ ∂ e2 1 2 2 Fbinding = 2 + r 2 − R2 ∂rH ∂rH 4πǫ0 rHe H 1 2 2 ∂ e2 + − = 2 ∂rH 4πǫ0 (R − rH )2 rH R2 e2 −3 (−2) 2(R − rH )−3 (−1) + rH = 0 at maximum or minimum = 4πǫ0 2 1 R = 3 21/3 rH = R − rH rH = = 0.44R (R − rH )3 rH 1 + 21/3 e2 1 2 2 Fbinding = + − 2 2 2 4πǫ0 (0.56R) (0.44R) R 2

= 9.5 4πǫe0 R2 . This is the minimum (the maxima occur at the nuclei). 5.15 This molecule turns out to be right on the border. If we look only along the axis between the two nuclei, we find that Fbinding is always positive. Calling the He nucleus A, Eq. 5.1 gives θA = 0, θB = π : e2 1 2 2 Fbinding = 2 + r 2 − R2 . 4πǫ0 rA B 2 Because rA is always less than R between the nuclei, we know that 2/rA is bigger than 2/R2 , and 2 2 2 therefore (2/rA ) + (1/rB ) must always be greater than 2/R . Therefore, the binding force is positive everywhere along the line between the He and H nuclei. The problem is, the binding region disappears quickly as we move away from this axis, approaching a much stronger repulsive force of −2 units when the electron is far from the nuclei. This greater nuclear-nuclear repulsion prevents the molecule from being stable. H+ 2 is the only stable, one-electron diatomic molecule.

5.16 This is a matter of substitution into the binding force equation; all the coordinates are provided in the problem. At the midpoint of the chemical bond, θA = 0

cos θA = 1

θB = π 2

Fbinding =

ZA e ZB e ZA ZB e + − (4πǫ0 )(R/2)2 (4πǫ0 )(R/2)2 4πǫ0 R2

cos θB = −1

a. ZA = ZB = 1 F =

e2 7e2 7(1.602 · 10−19 C)2 2e2 − = = = 1.61 · 10−7 N. 2 2 2 4πǫ0 (R/2) 4πǫ0 R 4πǫ0 R (1.113 · 10−10 C2 J−1 m−1 )(10−10 m)2

b. ZA = ZB = 2 F =

4e2 4e2 12e2 − = = 2.77 · 10−7 N. 2 2 (4πǫ0 )(R/2) 4πǫ0 R 4πǫ0 R2 138

5.17 This problem involves substitution into the binding force equation, but also requires determining what the coordinates are (in terms of R). Even though the equation is given in terms of R, rA , rB , θA , and θB , only three of these need to be specified (the other two coordinates can always be determined from any three).

@ √ @ 2R @ @ r @r 3π/4

θA = π/2; rB =

√ 2R;

θB = 3π/4; ZA = ZB = 1 e2 ZA ZB ZA ZB Fbinding = cos θ − cos θ − A B 2 2 4πǫ0 rA rB R2 1 1 1 1 e2 √ − − (0) − = 4πǫ0 R2 2R2 R2 2 1 e2 √ −1 = 4πǫ0 R2 2 2 5.18 These angles are the same as in Fig. 5.2a: θA = 0, θB = π. In addition, we may specify that rA = rB = R/2 and ZB = 6 for carbon. The equation that we want to satisfy is 6 6ZX ZX e2 + − 2 <0 Fbinding = 4πǫ0 (R/2)2 (R/2)2 R 24 6ZX 4ZX + 2 − 2 <0 R2 R R − 2ZX + 24 < 0 ZX = 12 With ZX = 12 and C combined to give a one-electron molecule, the charge on the carbide must be 12 + 6 − 1 = 17. This molecule is MgC17+ . 5.19 First we recognize that the plane passing through the middle of the bond causes θA and θB to be related: θB = 180 − θA ; cos θB = cos(180 − θA ) = − cos θA ZB ZA ZB ZA e2 cos θ − cos θ − A B 2 2 4πǫ0 rA rB R2 ZA ZB 1 e2 (Z + Z ) cos θ − = A B A 4πǫ0 r2 R2

Fbinding =

rA = rB ≡ r cos θ =

R/2 r

ZA ZB (ZA + ZB )R3 (ZA + ZB ) > or > r3 3 2 2r R ZA ZB 1/3 ZA + ZB r< R. ZA ZB

The region is binding as long as R

If ZA = ZB = 1, then r < 21/3 R is binding. 139

5.20 Any term where the same electron is on two different nuclei goes to zero. In the examples given, that happens in integrals (ii), because it has a term 1sA (2) 1sB (2), and in integral (iv), which splits both electrons across the two nuclei: RR i) 1sA (1) 1sA (1) 1sA (2) 1sA (2) dτ1 dτ2 RR ii) 1sA (1) 1sA (1) 1sA (2) 1sB (2) dτ1 dτ2 RR iii) 1sA (1) 1sA (1) 1sB (2) 1sB (2) dτ1 dτ2 RR iv) 1sA (1) 1sB (1) 1sA (2) 1sB (2) dτ1 dτ2 . 5.21 The lowest energy MO will have the least nodes, so all three orbitals having the same phase: ψlow = (1sA ) + (1sLi ) + (1sB ). The highest energy MO will have the most nodes possible, because that minimizes the electron density in the binding regions between the nuclei. This is achieved by reversing the phase of the 1sLi orbital in the middle, so that it destructively interferes with the hydrogen 1s orbitals on either side: ψlow = (1sA ) − (1sLi ) + (1sB ). 5.22 s # "s # "s # 64 −4rA1 /a0 64 −4rA2 /a0 8 1 −rB3 /a0 2rA3 −2rA3 /a0 ψMO = + e e e 1− e πa30 πa30 πa30 a0 πa30 s # "s 8 1 −rB4 /a0 2rA4 −2rA4 /a0 + 1− e e × πa30 a0 πa30 "s

= 1sA (1) 1sA (2) [2sA (3) + 1sB (3)] [2sA (4) + 1sB (4)] . Examining the e−Zr/(na0 ) terms, it is clear that nucleus A has an atomic number ZA = 4 (Be) and nucleus B has atomic number ZB = 1 (H). Therefore, this is a BeH molecule with only four electrons— the molecule is BeH+ . 5.23 The H atom 1s orbitals are at roughly −0.5 Eh . The C 1s orbitals are at −18 Eh before shielding is taken into account; they are still at −11 Eh after shielding is included. Therefore, the lowest energy MO is made largely from the carbon 1s orbitals: ψ = A(RIJ , RJK , θIJK )[1sI (1) + 1sJ (1) + 1sk (1)]. The normalization constant now depends on two bond lengths and a bond angle. 5.24 The lowest excited VB wavefunction is the 1σg 1σu triplet state, which must have an antisymmetric ∗ spatial term: ψVB [1σg 1σu ] = [1sA 1sB − 1sB 1sA ]. This can be rewritten using the MO wavefunctions ψMO [1σg 1σu ] = [1sA (1) + 1sB (1)][1sA (2) − 1sB (2)] = 1sA 1sA + 1sB 1sA − 1sA 1sB − 1sB 1sB

ψMO [1σu 1σg ] = [1sA (1) − 1sB (1)][1sA (2) + 1sB (2)] = 1sA 1sA − 1sB 1sA + 1sA 1sB − 1sB 1sB 1 (ψMO [1σu 1σg ] − ψMO [1σg 1σu ]) 2 1 = {2 · 1sA 1sB − 2 · 1sB 1sA } = 1sA 1sB − 1sB 1sA . 2

ψVB [1σg 1σu ] =

140

5.25 Our job is to find the large-R limit of E+ (R) =

HAA (R) + HAB (R) 1 + S(R)

E− (R) =

HAA (R) − HAB (R) , 1 − S(R)

where S(R) =

1sA 1sB dτ

HAA (R) =

1sA Ĥeff 1sA dτ

HAB (R) =

1sA Ĥeff 1sB dτ.

At large R, the overlap integral S vanishes for the same reason the ψMO cross terms vanish at large R. We can also show that HAB also vanishes at large R. In the limit of large R, Ĥeff becomes the Hamiltonian for an isolated one-electron atom, because −ZA e2 /rA = 0 for infinite rA : Z

Ĥeff 1sB = EB 1sB Z Z 1sA Ĥeff 1sB dτ = 1sA (EB 1sB ) dτ = EB 1sA 1sB dτ = EB S = 0,

where EB is the energy of the one-electron atom B and S is the overlap integral. We can only set HAB equal to EB S when R is large, because this makes the distance to nucleus A so large that the potential terms (ZA ZB e2 )/R and −ZA e2 /rA vanish. Therefore, E+ (R → ∞) = E− (R → ∞) = HAA (R), where HAA (R) is just the energy of isolated atom A: Z Z Z 1sA Ĥeff 1sA dτ = 1sA (EA 1sA ) dτ = EA 1sA 1sA dτ = EA . For the isolated 1s H atom, EA = −0.5 Eh and this is also the value of E+ and E− at large R. 5.26 At some large distance between the nuclei, the cross terms in the diatomic wavefunction become negligible; the wavefunction is that of the separated atoms or ions. This problem requires evaluating a more explicit form of the wavefunction, so that the cross terms can be identified and subtracted off. 1 [ψMO (ground) + ψMO (excited)]2 4 1 = {[1sA (1) + 1sB (1)] [1sA (2) + 1sB (2)] 4 2 + [1sA (1) − 1sB (1)] [1sA (2) − 1sB (2)]} 1 = {1sA (1)1sA (2) + 1sB (1)1sB (2) + 1sA (1)1sA (2) + 1sB (1)1sB (2)}2 4 o 1n 2 2 4 [1sA (1)1sA (2)] + 4 [1sB (1)1sB (2)] + 8 [1sA (1)1sA (2)1sB (1)1sB (2)] = 4 2 2 = [1sA (1)1sA (2)] + [1sB (1)1sB (2)] − + − ⇒ HA + H+ B + HA + HB .

|ψs |2 =

The two final states both correspond to hydrogen ions, so the predicted outcome is 100% H+ + H− . 5.27 The basic principle is this: multiply the two functions together and integrate to show that you get zero. Both are real functions, so you don’t need to worry about taking the complex conjugate. What happens is you get canceling integrals over (1sA )2 and (1sB )2 , which each integrate to one because the 141

atomic orbital functions are normalized, and you get two canceling overlap integrals over (1sA )(1sB ). Note that the overlap integrals are not zero, but because of the signs they do cancel. Z

ψ+ ψ− dτ = A+ (R)A− (R)



 = A+ (R)A− (R)  −

5.28

(1sA + 1sB )(1sA − 1sB )dτ Z

1 Z (1sA )2 dτ + (1sA )(1sB )dτ

 Z 1  (1sA )(1sB )dτ − (1sB )2 dτ  = 0.

a. No. These are not orthogonal if R > 0, because the positive and negative regions of the 2pzB orbital will be multiplied by different amplitudes of the 1sA function, so the two regions will not cancel when integrated. The exception is that if R = 0, then the two regions of the 2pzB orbital will multiply equal shares of the 1sA , and cancel. Any pair of distinct, one-electron wavefunctions centered on the same origin is an orthogonal pair.

b. Yes, and at all R, because the positive and negative regions of the 2pxB are equally represented at every distance from nucleus A. Therefore, the two regions are multiplied by equal amplitudes of the 1sA function, so the two regions will cancel when integrated. a)

5.29 Neglecting normalization, 2pz = ψ210 =

Zr −Zr/(2a0 ) e cos θ . a0

For plots along the z-axis, θ = 0, cos θ = 1. H+ 2 , ZA = ZB = 1. 2pZA =

rA −rA /(2a0 ) e a0

2pZB =

rB −rB /(2a0 ) e a0

If we choose z = 0 to be midway between the nuclei, then at z = 0: rA = rB = R/2;

at z = −R/2: rA = 0, rB = R. 142

1.2 1

0.5

0.8 0.6

-4

-2

0.4 0.2

-0.5 -4

-1

-2

-0.2

= =

|z + (R/2)| |z + a0 |

= |z − (R/2)| = |z − a0 |

5.30 Both the electron 1 and electron 2 spatial parts of the function change sign under exchange of the labels 1 and 2, so the combined spatial function does not change sign and is symmetric. The spin function is also symmetric. Therefore, the overall wavefunction is symmetric with respect to exchange of the electron labels, which violates the symmetrization principle. 5.31 Because this is for a single electron, we need to write the wavefunction in terms of a single set of independent coordinates, such as rA , θA , and φ. ψ+ = A+ (R)(1sA + 1sB ) = A+ (R)(e−rA /a0 + e−rA /a0 ) y = rA sin θA zA = rA cos θA 2 1/2 rB = (y 2 + zB ) = (y 2 + (R − zA )2 )1/2 1/2 2 . = rA sin2 θA + (R − rA cos θA )2

e rA

rB θB

θA A

To normalize, find A+ such that 1 = A2+

Z 2π Z π Z ∞ h i2 2 2 2 2 1/2 −rA /a0 rA drA sin θA dθA dφ. + exp − rA sin θA + (R − rA cos θA ) e 0

5.32 The ground state puts both electrons in the ψ+ lowest energy MO, so the lowest excited state MO configuration is the ψ+ ψ− . Including the spin part of the wavefunction, this MO configuration gives 143

four states, analogous to the four spin-spatial wavefunctions in 1s2s excited state helium atom: Ψsinglet = [ψ+ ψ− + ψ− ψ+ ] [αβ − βα] * Ψtriplet = [ψ+ ψ− − ψ− ψ+ ] [αα]

Ψtriplet = [ψ+ ψ− − ψ− ψ+ ] [αβ + βα] Ψtriplet = [ψ+ ψ− − ψ− ψ+ ] [ββ]

The singlet state is highest energy, according to Hund’s first rule. 5.33 The way to solve this would again be to multiply the wavefunction out to get nine distinct terms, square the whole thing, then integrate in the limit that all R’s are infinite. But we already know that all the cross terms from squaring the wavefunction will vanish in the limit of large R, so we only need to look at the original nine terms and what their squares predict. And if you spot the symmetry, you really only need to see the first three terms: ψ(1, 2)2 = A2 1sA (1)2 1sA (2)2 + 1sA (1)2 1sB (2)2 + 1sA (1)2 1sC (2)2 + 1sB (1)2 1sA (2)2 + 1sB (1)2 1sB (2)2 + 1sB (1)2 1sC (2)2 +1sC (1)2 1sA (2)2 + 1sC (1)2 1sB (2)2 + 1sC (1)2 1sC (2)2 .

In each row, the two electrons are on the same nucleus (making H− + 2H+ ) in one term, and on different nuclei (making 2H+H+ ) in the other two terms. Therefore, the prediction is that we get (H− + 2H+ ) one-third of the time and (2H + H+ ) the other two thirds of the time. To calculate the final percentage, for example, H atom appears as two thirds of the product in (2H + H+ ), and these products appear two thirds of the time, so the final percentage of H atom is (2/3)(2/3) = 4/9 = 44%. The final distribution of products is: 4 2 2 = = 44% H 3 3 9 2 4 1 1 2 H+ + = = 44% 3 3 3 3 9 1 1 1 = = 11%. H− 3 3 9 5.34 The He nucleus has greater nuclear charge, leading to a larger and more rapidly decreasing electron distribution in its vicinity:

5.35 First we need some kind of shorthand. We shall write 1sA as a, 1sB as b, 2sA as c, and 2sB as d. Then we write the orbitals in order of the electron labels. Our valence bond wavefunction, for example, could be written: ψVB = [ab electron no. (1)(2)

ba] [ab (1)(2) (3)(4) 144

−

ba] [cd + (3)(4) (5)(6)

dc] (5)(6)

Multiplying this out, we obtain eight terms: ψVB = ababcd + ababdc − abbacd − abbadc + baabcd + baabdc − babacd − babadc The molecular orbital wavefunctions will all have the form: ψMO = (a ± b) (a ± b) (a ± b) (a ± b) (c ± d) (c ± d) . electron no. (1) (2) (3) (4) (5) (6) Multiplying any of these all the way out gives 64 terms. It isn’t necessary to do that. We know, for example, that electrons 5 and 6 have the form (cd + dc) in ψVB . To get this, we need to combine two MO functions as follows (by analogy with Eq. 5.26): 1 ′ ′ [ψ (1, 2, 3, 4) (c + d)(c + d) − ψA (1, 2, 3, 4) (c − d)(c − d)] 2 A 1 ′ = ψA (1, 2, 3, 4)[(cc + cd + dc + dd) − (cc − cd − dc + dd)] 2 ′ = ψA (1, 2, 3, 4)[cd + dc] , ′ where ψA (1, 2, 3, 4) represents the part of ψMO that describes electrons 1, 2, 3, and 4. Likewise, to get the correct form for electrons 1 and 2 we need

1 ′ ′ ′ [(a + b)(a + b)ψC (3, 4, 5, 6) − (a − b)(a − b)ψC (3, 4, 5, 6)] = [ab + ba]ψC (3, 4, 5, 6), 2 ′ where ψC represents the part of ψMO that describes electrons 3, 4, 5, and 6. For electrons 3 and 4, we need

1 [(ab + ba)(a − b)(a + b)(cd + dc) − (ab + ba)(a + b)(a − b)(cd + dc)] 2 = (ab + ba)(ab − ba)cd + dc) = ψVB . ′ ′ Combining these results, using ψC (3, 4, 5, 6) = (a ± b)(a ± b)(cd + dc) and ψA (1, 2, 3, 4) = (a ± b)(a ± b)(a − b)(a + b) or (a ± b)(a ± b)(a + b)(a − b), we get

1 {[[(a + b)(a + b)(a − b)(a + b)(c + d)(c + d) − (a + b)(a + b)(a − b)(a + b)(c − d)(c − d)] 8 − [(a − b)(a − b)(a − b)(a + b)(c + d)(c + d) − (a − b)(a − b)(a − b)(a + b)(c − d)(c − d)]]

ψVB =

− [[(a + b)(a + b)(a + b)(a − b)(c + d)(c + d) − (a + b)(a + b)(a + b)(a − b)(c − d)(c − d)]

− [(a − b)(a − b)(a + b)(a − b)(c + d)(c + d) − (a − b)(a − b)(a + b)(a − b)(c − d)(c − d)]]}

Eight MOs are necessary to reproduce this ψVB . 5.36 Even if you didn’t know already that π orbitals have no density along the nuclear axis (and therefore can’t be constructed from s or pz orbitals), you can see that the s and pz orbitals on the carbons are used entirely in the construction of the sp hybrid orbitals that make up the C C and C H σ bonds. Therefore, the π bonds are constructed from the remaining, unhybridized px and py orbitals. For these to be bonding orbitals, with more electron density in the region between the nuclei than outside that region, the orbitals on the two nuclei need to be combined with the same phase. As we saw for the ψ± wavefunctions, there are again R-dependent normalization constants A: ψ1 = A1 (2pxA + 2pxB )

ψ2 = A2 (2pyA + 2pyB ) . 145

5.37 We integrate the product of the two wavefunctions. If they are orthogonal, the integral will be zero. Z Z Z Z ψa ψb dτ = C1 C2 (2sA )2 dτ + (2sB )2 dτ − 2 (2sC )2 dτ Z √ Z √ Z +2 (2sA )(2sB )dτ + 2 (2sA )(2sC ) − 2 (2sA )(2sC ) √ Z √ Z + 2 (2sB )(2sC ) − 2 (2sB )(2sC ) Z = 2C1 C2 (2sA )(2sB )dτ 6= 0 So these are not orthogonal. 5.38

a. Assign each energy to the correct molecular orbital in the MO configuration. −15.682 −15.678 −1.470 −0.777 −0.632 −0.612 −0.612

1σg 1σu 2σg 2σu 3σg 1πu 1πu

b. Calculate the total electron–electron repulsion energy in Eh . Solution: These are the orbital energies, and each orbital is occupied by two electrons. The sum of the orbital energies is −35.463 Eh, and we double this because there are two electrons in each orbital, obtaining −70.926 Eh. This sum contains each electron–electron interaction twice, so we subtract the total HF energy, −108.983 Eh, to get the repulsion energy: −70.926 Eh − (−108.983 Eh) = 38.057 Eh. c. To estimate the strength of the electron–nuclear interaction, calculate the effective atomic number of an electron in the highest energy orbital. Solution: Since the 1πu electrons correlate to the 2p atomic orbitals, we should set the principal quantum number equal to 2: Zeff = n

p √ −2E/ Eh = 2 2 · 0.612 = 2.21.

5.39 The functions we have available are the 2s, 2px , 2py , and 2pz . Because we are looking only at the bonds to atom CA that lie in the xq-plane, the 2py orbitals will not contribute (since the 2py has opposite phase on opposite sides of the xz-plane, it would make the C C bonds look different above and below the plane of the carbon atoms). So we want hybrid orbitals of the form ψ± = c1 (2s) + c2 (2pz ) ± c3 (2px ). The ± sign is there because the only difference between the CA CB and CA CC bond hybrids is the direction along the x axis. Finally, we want these hybrids to reach maximum amplitude at θ = 150◦ and 210◦ . We can check that by looking at the derivative dψ± /dθ and setting it to zero. 146

3r 1 6r ψ± = c1 1 − e−3r/a0 + c2 e−3r/a0 cos θ a0 2 a0 6r 1 e−3r/a0 sin θ cos φ , ± √ C3 a0 4 2

θ x

30o

p where we have used Z = 6 for carbon, and where we’ve factored 1/ 8πa30 out of the normalization constants because c1 , c2 , and c3 can be adjusted to replace them. The φ-dependence in the 2px function is obtained from another problem, but is not necessary here. In the xz plane, φ = 0 and cos φ = 1. 6r 6r 1 dψ± 1 e−3r/a0 sin θmax ± √ C3 e−3r/a0 cos θmax = 0 = − c2 dθ θmax 2 a0 a0 4 2 √ c2 1 1 2 2 = ± cot θmax = ± cos(150◦ ) − c2 sin θmax = ± √ C3 cos θmax 2 c3 4 2 √ = ± 3 = ±1.732 All we know from the calculation are the relative sizes of c2 and c3 . However, since the CA −H hybrid orbitals cannot include contributions from the 2px orbital, we know that all the (2px ) character is in these bonds, so 1 c3 = √ 2

√ 3 c2 = 4

c1 = (1 − c22 − c23 )1/2 =

√ 5 . 4

5.40 The requirements for any two functions a1 (4s) + b1 (4px ) + c1 (4py ) + d1 (3dxy ) a2 (4s) + b2 (4px ) + c2 (4py ) + d2 (3dxy ) are the following: a21 + b21 + c21 + d21 = a22 + b22 + c22 + d22 = 1 (normalized) a1 a2 + b1 b2 + c1 c2 + d1 d2 = 0 (orthogonal). The second equation ignores cross terms, such as a1 b2 , because they correspond to integrals over two orthogonal orbitals, such as (4s) and (4px ). All of the one-electron orbitals that we derived in Chapter 3 are orthogonal to each other. 1 1 1 1 Example: (4s) + (4px ) + (4py ) + (3dxy ) 2 2 2 2 1 1 1 1 (4s) + (4p) − (4py ) − (3dxy ) 2 2 2 2 5.41 From the orientations and phases of the orbitals drawn below, we see that all the constructive interference of the functions is in the (−x, −y) quadrant, with a net negative phase. 147

-d xy

px + py

-s

5.42 If we start from (spb ) = c1 (2s) + c2 (2pz ). The squares of the coefficients need to equal 1 (for it to be normalized, so c21 + c22 = 1) and the products of p pthe spa and spa coefficients need to cancel (for the two hybrids to be orthogonal, so 2/5c1 − 3/5c2 = 0). That gives: r r 2 3 (2s) + (2pz ). (spb ) = 5 5 5.43 To find the maximum along angle φ, find where the derivative of the function becomes zero: d ψ(φ) = 0 dφ i √ 1 d h√ d 2(2s) − (2px ) + 3(2py ) (sp2b ) = √ dφ 6 dφ 1 (2s) = R2s √ 4π

(2px ) = R2p (2py ) = R2p

θ = π/2

sin θ = 1

3 sin θ cos φ 4π 3 sin θ sin φ 4π

" r # √ d 3 −d 1 √ d d 2 2 (2s) + R2p (sp ) = √ cos φ + 3 sin φ dφ b dφ 4π dφ dφ 6 r √ 1 sin φ + 3 cos φ = R2p 8π √ = 0 if sin φ = − 3 cos φ √ sin φ = tan φ = − 3 = 2π/3, 5π/3 cos φ i √ √ 1 1 h R2s 2 − 2 3R2p (sp2b )(φ = 5π/3) = √ √ 6 4π i √ √ 1 1 h 2 R2s 2 + 2 3R2p . (spb )(φ = 2π/3) = √ √ 6 4π 148

So 2π/3 is the maximum. Similarly, for (sp2c ): " r # √ d d 1 d 1 √ d 2 2 (2s) − R2p (sp ) = √ cos φ + 3 sin φ dφ c dφ 8π dφ dφ 6 r √ √ R2p 3 (sin φ − 3 cos φ) = 0, if tan φ = 3φ = √ 24π 2

= π/3, 4π/3.

The maximum occurs at φ = 4π/3. 5.44 The problem is simplified if we think first about the symmetry of these orbitals. For example, the sp3a and sp3d orbitals have the same equations except for the sign of the px and py contributions. In other words, these orbitals point in opposite directions along the xy plane, but in the same direction along the z axis. The sp3a orbital lies at the same angle from the z axis as the sp3d orbital, but on the opposite side of the z axis. Therefore, the angle between these two orbitals is twice the angle of sp3a from the z axis, which is the θ value at which the sp3a orbital reaches its maximum magnitude. With the px and py contributions equal for both sp3a and sp3d , the angles φ for the maximum magnitude must be 45◦ and 225◦, respectively. This is not essential to the results we will obtain shortly, but the two values of φ must be 180◦ apart (because the sp3a and sp3d orbitals are on opposite sides of the z axis). d d(sp3a ) = [s + px + py + pz ] dθ dθ d = [Cs + Cp sin θ cos φ + Cp sin θ sin φ + Cp cos θ] dθ = Cp cos θ cos φ + Cp cos θ sin φ − Cp sin θ 1 1 − sin θ = Cp cos θ √ + √ 2 2 h√ i = Cp 2 cos θ − sin θ = 0 at maximum magnitude √ 2 cos θ = sin θ √ tan θ = 2 θ = 54.736◦ 2θ = 109.47◦. 5.45 The two orbital equations must be of the form: ψa = c1 (3s) + c2 (3px ) + c3 (3py ) + c4 (3pz ) + c5 (3dz2 ) ψb = d1 (3s) + d2 (3px ) + d3 (3py ) + d4 (3pz ) + d5 (3dz2 ) where

5 X i=1

c2i =

5 X i=1

d2i = 1 and where

5 X

ci di = 1. Some ci = 0 is acceptable. Some simple examples

i=1

(all three of the following are mutually orthogonal): 1 a. √ [(3s) + (3px ) + (3py ) + (3pz ) + (3dz2 )] 5 1 b. √ [(3s) + (3px ) − (3pz ) − (3dz2 )] 4 1 c. √ [−(3s) − (3px ) + 4(3py ) − (3pz ) − (3dz2 )]. 20 149

5.46 The orbitals are orthogonal and normalized. (sp3a ) must be composed of the 2s, 2px , 2py , and 2pz wavefunctions. Therefore, we may write: (sp3a ) = a(2s) + b(2px ) + c(2py ) + d(2pz ), with a, b, c, and d the constants we need to determine. This hybrid must be orthogonal with the other three, which means: q q q R a. (sp3a )(sp3b ) = 16 a − 32 b + 16 d = 0 b.

(sp3a )(sp3c ) =

(sp3a )(sp3d ) =

1 6a +

1 6b +

1 6b −

q 1 c + 16 d = 0 2

From (b) and (c), c = 0. That leaves r r r r r r 1 1 1 1 2 1 a+ b+ d= a− b+ d, 6 6 6 6 3 6 which means b = 0. That leaves

1 a+ 6

1 d=0 6

a = −d.

Finally, since (sp3a ) is also normalized, we have Z (sp3a )(sp3a ) = a2 + b2 + c2 + d2 = 1 a2 + d2 = 1

a = −d

2a2 = 1

a=±

(sp3a ) = ±

1 2

1 (2s) − 2

d=∓ r

1 2

# 1 (2p2 ) . 2

5.47 CH4 q

(sp3 )a = 12 (2s) + (0)(2px ) + (0)(2py ) + 34 (2pz ) q q q 1 (sp3 )c = 21 (2s) + 12 (2px ) − 16 (2py ) − 12 (2pz )

CHCl3

q q 1 (sp3 )b = 12 (2s) + (0)(2px ) + 23 (2py ) − 12 (2pz ) q q q 1 (sp3 )d = 12 (2s) − 12 (2px ) − 16 (2py ) − 12 (2pz )

(sp )a = 0 (2s) + 0 (2px ) + 0 (2py ) + + (2pz )

(sp3 )b = 0 (2s) + 0 (2px ) + + (2py ) + + (2pz )

(sp3 )c = 0 (2s) + + (2px ) + − (2py ) + + (2pz )

(sp3 )d = 0 (2s) + − (2px ) + − (2py ) + + (2pz )

The one essential difference is that the C Cl bonds have less (2pz ) character than the C H bonds in methane. To put it another way, the θ values of b, c, and d change by the same amount, while the 150

φ values stay fixed. All the hybrids that had zero contribution from the px or py orbital keep that contribution at zero so as not to change φ (the angle in the xy plane). Because the pz coefficient is positive for a and negative for the other orbitals, we increase the magnitude of pz in a by adding to the coefficient and decrease the magnitude in the other hybrids also by adding to those coefficients. Reducing the pz contribution to b, c, and d means that the magnitudes of the px and py have to increase to keep the MOs normalized. The example in this problem is the simplest possibility—you can also alter the amount of 2s character in MO a, as long as you compensate by an opposite effect in the other orbitals. 5.48

a. What is the hybridization at each oxygen atom? Each oxygen has one double bond and two lone pairs, which is three groups, so the hybridization is sp2 . b. What is the hybridization at the carbon atom? The carbon has two double bonds and no lone pairs, so two groups and the hybridization is sp. c. Write a formula in terms of the atomic orbitals of these atoms for the wavefunction that forms the C OB bonding orbital. Assume the hybrid orbitals at each atom are all equivalent. The equivalent sp and sp2 hybrids that lie on the z axis must be r r 1 1 (sp± )C = (2s)C ± (2pz )C 2 2 r r 1 2 (2s)O ± (2pz )O . (sp2± )O = 3 3 To point the (sp± )C orbital towards OB , we want (sp+ )C , and to point (sp2± )O towards the C atom, we want (sp2− )O . To increase bonding density between the C and the O atom, we now add these two together. The phase of each hybrid is positive in the bonding region, between the two atoms, so they constructively interfere and increase the density of the electron: r r r r 1 1 1 2 (2s)C + (2pz )C + (2s)O − (2pz )O . 2 2 3 3 d. Sketch the amplitude of that wavefunction along z. ψ

p 5.49 There are infinite possibilities, but all must have a coefficient of magnitude 2/5 (giving 2/5, or p 40%, to the probability density) for the s contribution to the unique hybrid orbital, and 1/5 for each of the remaining equivalent orbitals. The simplest way—I think—to determine the p contributions is to align the unique orbital along the z axis (so omit px and pz character), and put one of the remaining three equivalent orbitals in the xz plane (so the py character is zero). The last three hybrids equally divide the pz character left over from the unique hybrid. The last two orbitals will equally divide the py character and the remaining px character. r r 2 3 3 (s) + (0)(px ) + (0)(py ) + (pz ) spa = 5 5 r r r 1 2 2 sp3b = (s) + (px ) + (0)(py ) − (pz ) 5 3 15 r r r r 1 1 1 2 (s) − (px ) + (py ) − (pz ) sp3c = 5 6 2 15 151

sp3d = 5.50

1 (s) − 5

1 (px ) − 6

1 (py ) − 2

2 (pz ) 15

a. The fractional contribution to an MO from each of the atomic orbitals is measured by the square (or, more accurately, square modulus) of the coefficient, because the probability density (rather than the wavefunction) is what we can measure directly. For each of these orbitals, we square the px and py orbital coefficients, add them, and multiply by 100 to get the percentage: percentage p (100%)0.32142 = 10.3%

ψa = (0.9470)(2s) + (0.3214)(2px) 1 ψb = (0.2272)(2s) − (0.6696)(2px) + √ (2py ) 2 1 ψc = (0.2272)(2s) − (0.6696)(2px) − √ (2py ) 2

(100%)(0.66962 + 21 ) = 94.8% (100%)(0.66962 + 21 ) = 94.8%

Note that the percentages add up to about 200%, consistent with having used up both p orbitals completely. What’s remarkable is that the ψa function ends up being 90% s orbital—which you would expect to be a very poor bonding orbital because the electron distribution is almost evenly distributed in all directions around the nucleus, instead of focused into the bonding region between the atoms. That allows the two C C bond MOs, ψb and ψc , to approach the 90◦ bond angle you would associate with two pure p orbitals. If all three sp2 MOs were equivalent, as for example in BF3 , we would expect each orbital to have 33% s character and 67% p character. b. The z axis is not involved in any of the hybrid orbitals, and therefore is the axis perpendicular to the plane of the molecule. The py orbital contributes only to the two equivalent hybrid orbitals psib and ψc (which differ only in the relative phase of the py term), and therefore that is the axis perpendicular to the carbonyl bond (because the carbonyl bonding orbital must be unique). These would not be the conventional orientation of the axes, because normally we would choose the z axis to be the symmetry axis, in this case along the carbonyl bond.

y α

O z

5.51 Let’s write the eigenvalue equation and take a look at it: 1 Ĥψa = Ĥ √ [(2s) + (2pz )] . 2 Because the Hamiltonian is a linear operator, we can separate its action on the two components of our wavefunction, and then we can see our way to the answer: 1 Ĥψa = Ĥ √ [(2s) + (2pz )] 2 i 1 h = √ Ĥ(2s) + Ĥ(2pz ) 2 1 = √ [E2s (2s) + E2p (2pz )] 2 152

E2 = √ [(2s) + (2pz )] 2 = E2 ψa , where, for the one-electron atom, the 2s and 2p orbitals have the same energy, so E2s = E2p = E2 , and the eigenvalue E2 is equal to −0.25 Eh. 5.52 Taking for granted that we only need to examine the angular part of the wavefunction, the angular node in an sp hybrid orbital appears when the s and p spherical harmonics cancel. From Table 3.1, we have the angular wavefunctions, r 1 Y00 = 4π r 3 0 Y0 = cos θ 4π and now we look for the node condition, ca1 (s) + ca2 (pz ) = ca1

1 + ca2 4π

3 cos θ 4π

and this equals zero when √ ca1 + ca2 3 cos θ = 0. Let’s assume ca1 is positive. Then the angular node will appear when cos θ is negative, which is at π/2 < θ ≤ π. To prevent the node from appearing, we need ca1 large enough that it completely cancels the contribution from the pz orbital when it is most negative, at θ = π and cos θ = −1: √ ca1 + ca2 3(−1) = 0 √ ca1 = 3ca2 . To keep the hybrid orbital normalized we also need c2a1 + c2a2 = 1 √ ( 3ca2 )2 + c2a2 = 4c2a2 = 1 1 ca2 = 2

√ 3 ca1 = . 2

With these coefficients, the angular part of spa will be positive everywhere, and there will be no angular nodes. √ The companion hybrid orbital, spb , must then have coefficients cb1 = 1/2 and cb2 = − 3/2, in order to be normalized and orthogonal to spa . The angular part of spb can then be written r √ r 1 3 3 1 1 − cos θ = √ (1 − 3 cos θ) . 2 4π 2 4π 2 4π This angular function is negative between θ = 0 and θ = 0.39π or 70◦ , but once cos θ becomes less than 1/3, then this function changes sign. So there is no node in spa with these coefficients, but spb still has an angular node. 153

5.53 The carbons are each bonded to three atoms. The two resonance structures predict that each C C bond is the average of a formal double bond (1.34 Å) and a formal single bond (1.46 Å). The average of these predicts the aromatic C3-C3 bond length, 1.40 Å, to within the 0.01 Å resolution in our table, so the agreement is to within at least 0.7%. 5.54

a. (i) H C: single, C3 H = 1.08 Å (ii) C O: double, C3 O1 = 1.22 Å

H i)

H C e− O

H ii)

C e− O H

e−

O e−

H iii)

e−O

C H

H C e− O

b. The bonding strength increases (a) the more the electron is on the bond axis between the two nuclei, and (b) within that region, the closer the electron is to either nucleus. 5.55 The chemical shift is inversely proportional to the shielding constant: δ ≈ σ0 − σ = 192.3 − 65.2 = 127.1 ppm. 5.56

a. Fill out the remaining entries in the table. Solution: The shielding constant σ = σ0 − δ = 170 ppm is unchanged when we change the reference, so we can calculate the new chemical shift as δ ′ ≈ σ0′ − σ. The local magnetic field is given by Blocal = B0 (1 − σ), so B0 − Blocal = σB0 . Keep in mind that the σ’s and δ’s are in ppm, so need to be multiplied by 10−6 when comparing to other numbers. The chemical shift is not affected by change in magnetic field. nucleus field (T) reference δ (ppm) B0 − Blocal (T)

N 9.4 NH3 99 0.0016

N 9.4 CH3 NO2 −282 0.0016

N 14.1 NH3 99 0.0024

b. Which of the following describes the δ value of nitrogen atom Na in neutral hydrozoic acid, HNN15 Na , at 9.4 T using NH3 as a reference (circle one): < 99 ppm 99 ppm > 99 ppm 3− Solution: Adding a proton to N will take electron density away from the atoms in the anion, which reduces the shielding and increases the chemical shift.

Chapter 6 6.1 Because the plane of the molecule will be a symmetry mirror plane, only point groups with reflection symmetry are possible contenders for the planar molecules, which eliminates C1 , Ci , all of the Cn , Dn , and S2n groups, and several of the cubic point groups. In addition, if there is a proper rotation symmetry element Ĉn with n > 2, then its axis must be perpendicular to the molecular plane, because rotation 154

of the molecular plane about any other axis by less than π will not bring the plane back to where it was. This eliminates the rest of the cubic groups (which have several different Ĉn axes with n > 2), all the Cnv and Dnd groups with n > 2, because the principal axes in those cases are not perpendicular to the σ̂v and σ̂d mirror planes, by definition. The C2v group can apply to planar molecules, such as H2 O, because a Ĉ2 rotation of the molecular plane can take place about an axis that lies in the plane, while still returning the plane to its original orientation. The D2d point group does not apply to planar molecules, however, because the Ĉ2′ rotation axes are neither perpendicular to nor parallel to the mirror planes, and therefore cannot return the molecular plane to its original orientation. Our final list therefore consists of Cs , C2v , Cnh , Dnh . From this last set, we should drop D∞h , because it applies only to linear molecules. 6.2 This is the workhorse problem of Chapter 6: if you can’t identify a molecule’s point group, you can’t take advantage of the, well, advantages that molecular symmetry can offer. a. The symmetry elements are Ê, Ĉ43 , Ĉ42 = Ĉ2 , Ĉ4 : containing the axial Cl F bond; 2σ̂v : in the plane of the paper and perpendicular to the paper; 2σ̂v′ or 2σ̂d : containing the Ĉ4 axis and two equatorial F-atoms. The group is C4v . b. The group is D2h .

@ @

Ĉ2′ @ @ H

H Ĉ2′

The symmetry elements are Ê; Ĉ2 (perpendicular to page, through the center of the molecule); 2Ĉ2′ (shown); 2σ̂v (perpendicular to the page, containing one of the Ĉ2′ axes each); σ̂h (plane of the page). c. The symmetry elements are Ê, Ĉ2 , σ̂v , σ̂v′ . The group is C2v .

- y

σ̂yz , σ̂xz

x d. NH+ 4 Td

Ĉ2

e. HO2 Cs f. N− 3 D∞h g. 1,1-dichloroethene C2v 155

h. SO2 C2v

z A 6 -y A A A A A x A A A A Ĉ2

i. Ê, Ĉ2 , σ̂xz , σˆyz ,

C2v

σ̂xz

ˆ The point group is D2h . j. 3Ĉ2 axes (one perpendicular to the page); 3σ̂v ; I.

@ HH @

D3h H H @ @ Cl k. Cl H l. chair: Ĉ3 , no σ̂h , 3Ĉ2 → D3d m. boat: Ĉ2 , no σ̂h , no perpendicular Ĉ2 ’s Ĉ2v n. cyclohexadiene, C6 H8 . There is only a Ĉ2 axis, bisecting the two unique single bonds. The point group is C2 . o. There is a Ĉ3 axis through the carbon-atom chain, and three mirror planes that each contain the Ĉ3 axis and one C-H bond. The point group is C3v . 6.3 From the flowchart (Fig. 6.7) or by examining the character tables, one finds that the S2n groups are very similar to the Cn groups, differing only by the addition of the improper rotation symmetry elements. Whenever the improper rotation axis coincides with a proper rotation axis, the index of the improper rotation axis must be twice the index n of the proper rotation axis. We can show this by considering the two alternatives: • If the indices were the same (so you had Ŝn and Ĉn symmetry elements sharing the same axis), then the operator σ̂h would be a member of the group in addition to the rotation operators. That’s because Ŝn = σ̂h Ĉn , and we can add a reverse rotation Ĉn−1 to each side: Ŝn Ĉn−1 = σ̂h Ĉn Ĉn−1 = σ̂h . 156

If the two operators Ŝn and Ĉn−1 are elements of the group, then σ̂h must be also. (If Ĉn is in the group, then Ĉn−1 must also be.) If we have symmetry elements Ê, Ĉn , Ŝn , and σ̂h , and nothing else, then the group is Cnh . • The other possibility is that there are symmetry elements Ŝm and Ĉn with the same axis, but m 6= n and m 6= 2n. That won’t work, because two consecutive improper rotations Ŝm Ŝm must always yield a proper rotation: Ŝm Ŝm = (σ̂h Ĉm )(σ̂h Ĉm ) 2 = σ̂h2 Ĉm ,

where σ̂h and Ĉm commute because they operate on different coordinates; σ̂h affects only the z value and Ĉm affects only x and y, 2 = Ĉm Ĉm ≡ Ĉ2m . = Ĉm

Two Ĉm rotations is the same as one rotation by 2π/(2m). If Ŝm is a member of the group, then Ĉn must also be a member of the group, where 2m = n. The Sn group with n an odd integer would be equivalent to the group Cnh , and therefore is not listed separately. 6.4 The answer is in the statement from Section 6.1 (p. 329), “Degeneracy appears in the irreducible representations whenever the group contains an operator that can be carried out in opposite directions with different results.” What singles out D2d from the other groups with n = 2 is that it has a four-fold axis, the axis for the improper rotation Ŝ4 . The clockwise improper rotation by π/2, Ŝ4 , is a distinct operation from the counterclockwise improper rotation by π/2 (which we would label Ŝ43 ). This is suggested by the appearance of Ŝ4 in the character table as two operators (i.e., the heading to the column is “2Ŝ4 ”). The highest order proper rotation axis in D2d is n = 2 for the three Ĉ2 operators, so the name of the group is given with n = 2, but it is the only n = 2 group that also has a higher order improper rotation axis. 6.5 These problems take explicit functions that represent very simple electronic wavefunctions. Functions (a) and (d) are the 1s and 2p1 atomic wavefunctions. The 1s function is spherically symmetric, and is unchanged under each of these symmetry operations. Functions (b) and (c) represent σ bonding orbitals in homonuclear and heteronuclear molecules, respectively. (i) Inversion. ˆ Iψ(x, y, z) = ψ(−x, −y, −z) p p ˆ = (−x)2 + (−y)2 + (−z)2 = x2 + y 2 + z 2 = r Ir ! ! −z z ˆ ˆ = arccos p = θ + π or π − θ Iθ = I arccos p x2 + y 2 + z 2 x2 + y 2 + z 2 y −y −y ˆ ˆ = arctan = arctan =φ Iφ = I arctan x −x −x 2

ˆ −r /a = e−r /a a. Ie

symmetric 157

b. i h 2 2 2 2 x p Iˆ e−(z−R/2) /a + e−(z+R/2) /a 2 x + y2 i h 2 2 2 2 −x p = e−(−z−R/2) /a + e−(−z+R/2) /a x2 + y 2 i h 2 2 2 2 −x p = e−(z+R/2) /a + e−(z−R/2) /a x2 + y 2

antisymmetric

i h 2 2 c. Iˆ e−(z−R/2)/a + e−(z+R/2)/b (x2 + y 2 )

i h 2 2 = e+(z+R/2)/a + e+(z−R/2)/b (x2 + y 2 )

2 2 Iˆ re−r /a cos θ eiφ 2

= re−r /a cos θ − eiφ

non-symmetric

= re−r /a (− cos θ)(−eiφ )

where cos(π − θ) = − cos θ symmetric

ei(φ+π) = cos(φ + π) + i sin(φ + π) = − cos φ − i sin φ = −eiφ (ii) Reflection in the xy plane and with respect to reflection in the xz plane. σ̂h ψ(x, y, z) = ψ(x, y, −z) reflection in xy plane σ̂h r = r σ̂v ψ(x, y, z) = ψ(x, −y, z) 2

σ̂v e

−r 2 /a2

σ̂h φ = φ

reflection in xz plane σ̂v r = r

a. σ̂h e−r /a = e−r /a

σ̂h θ = π − θ

σ̂v θ = θ

σ̂h φ = −φ

symmetric

−r 2 /a2

symmetric i h 2 2 2 2 b. σ̂h e−(z−R/2) /a + e−(z+R/2) /a √ 2x 2 ix +y h −(−z−R/2)2 /a2 −(−z+R/2)2 /a2 √ x = e + e x2 +y 2 h i 2 2 −(z+R/2) /a −(z−R/2)2 /a2 √ x = e + e x2 +y 2 i h −(z−R/2)2 /a2 −(z+R/2)2 /a2 √ x + e σ̂v e 2 2 i x +y h −(z−R/2)2 /a2 −(z+R/2)2 /a2 √ x = e + e 2 2 x +y

h i 2 2 2 2 c. σ̂h e−(z−R/2) /a + e−(z+R/2) /b (x2 + y 2 ) i h 2 2 2 2 = e−(−z−R/2) /a + e−(−z+R/2) /b (x2 + y 2 ) i h 2 2 2 2 = e−(z+R/2) /a + e−(z−R/2) /b (x2 + y 2 ) i h 2 2 2 2 σ̂v e−(z−R/2) /a + e−(z+R/2) /b (x2 + y 2 ) h i 2 2 2 2 = e−(z−R/2) /a + e−(z+R/2) /b (x2 + y 2 ) 158

symmetric

non-symmetric

symmetric Copyright c 2014 Pearson Education, Inc.

d. i h 2 2 2 2 σ̂h re−r /a cos θ eiφ = −re−r /a cos θ eiφ cos(π − θ) = − cos θ

anti-symmetric

i h 2 2 2 2 σ̂v re−r /a cos θ eiφ = re−r /a cos θ e−iφ

non-symmetric

e−iφ = cos(−φ) + i sin(−θ) = cos φ − i sin φ 6= ± eiφ (iii) Rotation by π around the z axis. Ĉ2 ψ(x, y, z) = ψ(−x, −y, z) rotation by π around z axis Ĉ2 r = r 2

Ĉ2 θ = θ

Ĉ2 φ = φ + π

a. Ĉ2 e−r /a = e−r /a symmetric i h 2 2 2 2 b. Ĉ2 e−(z−R/2) /a + e−(z+R/2) /a √ 2x

x +y 2

i h 2 2 2 2 −x = e−(z−R/2) /a + e−(z+R/2) /a p x2 + y 2

c. Ĉ2 e−(z−R/2)/a + e−(z+R/2)/b (x2 + y 2 ) i h = e−(z−R/2)/a + e−(z+R/2)/b (x2 + y 2 ) d. Ĉ2

i h 2 2 re−r /a cos θ eiφ 2

= re−r /a cos θ ei(φ+π) = −re−r /a cos θ eiφ

antisymmetric

symmetric

antisymmetric

6.6 These are essentially algebra problems involving the symmetry operators. These can always be solved, although not always easily, by explicitly writing the effect of one operator on the coordinates of the function. Another approach is to consider the effect of these operations on an object with no symmetry—for example, a cube with all of its faces labeled. The first method is used here. a. Ŝ4 = σ̂h Ĉ4 Ĉ4 ψ(x, y, z) = ψ(x′ , y ′ , z) ′

σ̂h ψ(x, y, z) = ψ(x, y, −z)

′

σ̂h Ĉ4 ψ(x, y, z) = ψ(x , y , −z) = Ĉ4 σ̂h ψ(x, y, z) σ̂h Ĉ4 ψ = ψ(x′ , y ′ , −z) = Ĉ4 σ̂h Ŝ4 Ŝ4 = Ŝ4 Ŝ4 = σ̂h Ĉ4 σ̂h Ĉ4 = σ̂h σ̂h Ĉ4 Ĉ4 = σ̂h2 Ĉ42 σ̂h2 = Ê

Ĉ42 = Ĉ2 ,

so Ŝ4 Ŝ4 = Ê Ĉ2 = Ĉ2 . 159

b. ˆ Iψ(x, y, z) = ψ(−x, −y, −z)

σ̂xy ψ(x, y, z) = ψ(x, y, −z)

σ̂yz ψ(x, y, z) = ψ(−x, y, z)

σ̂xz ψ(x, y, z) = ψ(x, −y, z)

σ̂xy σ̂yz σ̂xz ψ(x, y, z) = σ̂xy (σ̂yz (σ̂xz ψ(x, y, z))) = σ̂xy (σ̂yz ψ(x, −y, z)) = σ̂xy ψ(−x, −y, z) = ψ(−x, −y, −z) ˆ = Iψ(x, y, z)

6.7 One way of doing this is to use the Cartesian coordinates: = σ̂xy f (−y, x, z) = f (−y, x, −z)

Ŝ4 (z) f (x, y, z) = σ̂xy Ĉ4 (z) f (x, y, z) ˆ (x, y, z) = f (−x, −y, −z). If

So Â is the operator that changes f (−y, x, −z) to f (−x, −y, −z), which is just Ĉ4 (z). This can also be shown graphically. a

b A

d ^ S4

c ^ C4

B c

d B

^ I

6.8 This one has four symmetry elements to contend with. Because Ê is the same as doing nothing, that leaves only the nine entries that don’t involve Ê to figure out. All of the remaining operators are binary, in that a second application undoes the work (e.g., IˆIˆ = Ê), so the diagonal elements of the table are all Ê. The remaining six elements can be solved algebraically. For example, ˆ y, z) = Ĉ2 ψ(−x, −y, −z) = ψ(x, y, −z) = σ̂h ψ(x, y, z). Ĉ2 Iψ(x, C2h Ê Ĉ2 Iˆ

Ê Ê Ĉ2 Iˆ

σ̂h

Ĉ2 Ĉ2 Ê σ̂h Iˆ

Iˆ Iˆ σ̂h Ê Ĉ2

σ̂h σ̂h Iˆ Ĉ2 Ê

6.9 All of these can be determined by drawing a square with labeled corners, and rotating or reflecting it, but a few terms can be determined more quickly. The direction of rotation for Ĉ2 doesn’t matter, and the direction of rotation for Ĉ4 also determines the direction for Ŝ4 , so the multiplication table is unaffected by these choices. Each symmetry element is unaffected when combined with Ê, so the first row and first column give unchanged operators. The next most straightforward set is the squared operators along the diagonal of the multiplication table. Two 180◦ rotations in the same direction return you to the starting point, and so do two inversions. Two 90◦ rotations is the same as one 180◦ rotation, so Ĉ42 = Ĉ2 . 160

D4h Ê Ĉ4 (z) Ĉ2 (w) Iˆ 6.10

Ê Ê Ĉ4 (z) Ĉ2 (w) Iˆ

Ĉ4 (z) Ĉ4 (z) Ĉ2 (z) Ĉ2 (y) Ŝ43 (z)

Iˆ Iˆ

Ĉ2 (w) Ĉ2 (w) Ĉ2 (x) Ê σ̂vz

Ŝ43 (z) σ̂vz Ê

a. Ŝ4 Ĉ4 Iˆ = σ̂h Ĉ4 Ĉ4 Iˆ = σ̂h Ĉ2 Iˆ because Ŝ4 = σ̂h Ĉ4 and Ĉ42 = Ĉ2 . Choose the Ĉ2 axis to be the z axis, then σ̂h = σ̂xy ˆ y, z) = σ̂xy Ĉ2 ψ(−x, −y, −z) = σ̂xy ψ(x, y, −z) = ψ(x, y, z) σ̂xy Ĉ2 Iψ(x, Ŝ4 Ĉ4 Iˆ = Ê.

b. Label the faces of a triangle A and B, and the corners a, b, and c:

b C3

b I

b B a

I C = S5 3

or, mathematically, 2π Ĉ3 ψ(r, z, φ) = ψ r, z, φ + 3 ˆ Iψ(r, z, φ) = ψ(r, −z, φ + π) 5π 5π ˆ = σ̂h ψ r, z, φ + = σ̂h Ĉ65 ψ(r, z, φ) I Ĉ3 ψ(r, z, φ) = ψ r, −z, φ + 3 3 = Ŝ65 ψ(r, z, φ)

161

c. σ̂xy Ê IˆĈ2 (z) ψ(x, y, z) = σ̂xy Ê Iˆ ψ(−x, −y, z) = σ̂xy Ê ψ(x, y, −z)

= σ̂xy ψ(x, y, −z)

= ψ(x, y, z) = Ê ψ(x, y, z) σ̂xy Ê IˆĈ2 (z) = Ê d. IˆĈ4 σ̂h ψ(x, y, z) = IˆĈ4 ψ(x, y, −z) ˆ = Iψ(y, −x, −z) = ψ(−y, x, +z) = Ĉ43 ψ(x, y, z) IˆĈ4 σ̂h = Ĉ43 6.11 The D3h symmetry elements are Ê, Ĉ3 , Ĉ32 , 3Ĉ2 , σ̂h , Ŝ3 , Ŝ32 , 3σ̂v . There are many possible solutions: b L −L σ̂ v a c > ′ Ĉ Ĉ2 2 b a σ̂v′ σ̂v @ L L +L +L a c c b Ĉ2′′ σ̂v′′

Z Z ~ Z a

6.12

c L +L

F B F

F σv’

σv

a L +L

Ĉ2 σ̂h

Ĉ2′′ Ŝ3 σ̂v′′ Ĉ3 σ̂v′ Ĉ32 σ̂h Ĉ2 Ŝ3 Ĉ2 Ŝ32 Ĉ2′′

Ĉ2′ Ŝ32 Ĉ3 σ̂v′ Ĉ32 σ̂v′′

Ĉ3 σ̂v ψ(r, θ, φ) = Ĉ3 ψ(r, θ, −φ) 2π = ψ r, θ, −φ + 3 2π Âσ̂v Ĉ3 ψ(r, θ, φ) = Âσ̂v ψ r, θ, φ + 3 2π = Âψ r, θ, −φ − 3 2π = ψ r, θ, −φ + 3 4π = Ĉ32 ψ(r, θ, φ) Âψ(r, θ, φ) = ψ r, θ, φ + 3 162

Â = Ĉ32

6.13 Ĉ2 (s) a

a z

C2(x)

u d C2(s)

top

r s

C (z)

top

top x

f e

Because the two operations we perform are proper rotations, which you can do by hand with a real model of the system, the resulting operation must also be a proper rotation.

6.14 a

3 top

C^2(3)

bottom

b 2

C^2(2)

E^ a

C^6(z) top

top

6.15 a

y x z

A b

c a

y x z

a b

2 C^3 ( z )

A b

S^3 ( z ) x

B b

I^ x

a c

C^3 ( z )

A c

6.16 163

(a) d

(b) d

f C2(y)

c b

f S4(x)

a b

6.17 Ê commutes with any operator because it does not affect the function it operates on. Iˆ and Ĉ2 commute because they only affect the Cartesian coordinates by factors of −1: ˆ ˆ IˆĈ2 (z)ψ(x, y, z) = Iψ(−x, −y, z) = ψ(x, y, −z) = Ĉ2 (z)Iψ(x, y, z).

The only one remaining that really needs checking is the first one, and a picture shows that these do not commute, with a remaining Ĉ3 rotation separating ÂB̂ and B̂ Â: Â B̂ Ĉ3 σ̂xz Ĉ3 (z) Ê

Ŝ4 (z)

Iˆ

Ĉ2 (z)

6.18 The Ĉ3 (y) operator changes the magnitudes of the x and z coordinates, but leaves y unchanged. Therefore, only an operator that has exactly the same effect on the x coordinate as on the z coordinate will commute with Ĉ3 . From the D2h group, those operators are Ĉ2 (y), σ̂xz , and Iˆ . For example, if Ĉ3 (y) ψ(x1 , y1 , z1 ) = ψ(x2 , y1 , z2 ), then Iˆ Ĉ3 (y) ψ(x1 , y1 , z1 ) = ψ(−x2 , −y1 , −z2 )

164

= Ĉ3 (y) ψ(−x1 , −y1 , −z1 ) = Ĉ3 (y) Iˆ ψ(x1 , y1 , z1 ). Copyright c 2014 Pearson Education, Inc.

On the other hand, Ĉ2 (z) Ĉ3 (y) ψ(x1 , y1 , z1 ) = ψ(−x2 , −y1 , z2 )

= Ĉ3 (y)2 ψ(−x1 , −y1 , z1 )

= Ĉ3 (y)2 Ĉ2 (z) ψ(x1 , y1 , z1 ) 6= Ĉ3 (y) Ĉ2 (z) ψ(x1 , y1 , z1 ). 6.19 Yes, the point group operators are associative. It may be plain to you from the fact that we require our operators to be executed sequentially from right to left (so it doesn’t matter if we repackage two of them under a single name, the same manipulations would be carried out in the same order). For a more rigorous proof, let ÂB̂ = D̂, another operator in the point group. The question then is: is it always true that (ÂB̂)Ĉψ = D̂Ĉψ = Â(B̂ Ĉ)ψ? For every operator Â, there is also an inverse operator Â−1 in the point group that undoes that operation, such that Â−1 Âψ = ψ. ˆ each operator is its own inverse, in that Â2 ψ = ψ.) Then it must be (For the operators Ĉ2 , σ̂, and I, the case that Â−1 D̂ = Â−1 ÂB̂ = B̂. If we apply Â−1 to the left of both sides of the equation for associativity, we have (Â−1 D̂)Ĉψ = (Â−1 Â)(B̂ Ĉ)ψ (Â−1 D̂)Ĉψ = (B̂ Ĉ)ψ B̂ Ĉψ = B̂ Ĉψ. 6.20 We are looking for a single operator Ŝ3−1 that undoes Ŝ3 = σ̂h Ĉ3 . Since we break the improper rotation up into two steps, lets use two steps to unwind it. In other words, let Ŝ3−1 = ÂB̂. We can find a form for this operator be solving the equation Ŝ3−1 Ŝ3 ψ = ψ ÂB̂σ̂n Ĉ3 ψ = ψ We can undo the reflection with another reflection: Â(σ̂h σ̂h )Ĉ3 ψ = ψ ÂĈ3 ψ = ψ. To undo the 120◦ rotation, we need a rotation by −120◦, which is the same as a 240◦ rotation, or a Ĉ32 operation, so Â = Ĉ32 . Combining Â and B̂, we have Ŝ3−1 = ÂB̂ = Ĉ32 σ̂h = Ŝ32 . Note that rotation always commutes with the corresponding horizontal reflection, because the horizontal reflection affects only the coordinate value perpendicular to the plane, while the rotation affects only the coordinate values parallel to the plane. 165

6.21 What you need to remember here is that the angular momentum is itself a function of two vectors: the position ~r = (x, y, z) and the momentum p~ = (px , py , pz ). Therefore, in order to find out what ~ we first need to establish the effect of Ŝ2 on these other vectors. Although Ŝ2 is written happens to L, ˆ as an improper rotation, it is equivalent to the inversion operator I: Ŝ2~r = (−x, −y, −z) = −~r Ŝ2 ~ p = (−px , −py , −pz ) = −~ p

~ = ~r × p~ = (ypz − zpy , zpx − xpz , xpy − ypx ) L

~ = (−~r) × (−~ p) = (ypz − zpy , zpx − xpz , xpy − ypx ) Ŝ2 L

~ . =L

The eigenvalue is just 1, whereas for an ordinary position or momentum vector, the eigenvalue is −1. The symmetry of the angular momentum under an improper rotation is opposite to the symmetry of ordinary vectors, and for this reason the angular momentum is classed as a pseudovector. The same is true for the magnetic field dipole, which is a vector proportional to the angular momentum of the charged particle. This is a significant distinction when symmetry arguments are used to understand physical systems, because it is easy to forget that pseudovectors do not behave like ordinary vectors.

6.22 The point group for the rigid conformation drawn is Cs ; Ê and σ̂xy are the only symmetry elements. J '$ H J y J 6 P H ZP J x ZH J &% J z However, in most applications the methyl group rotates quickly enough that the symmetry properties are essentially those of a C2v symmetry molecule.

6.23

a. A rectangular table. The four table legs are equivalent, so we look for operations that can exchange those, such as Ĉ2 rotation about an axis perpendicular to the tabletop, and the two vertical mirror planes containing that axis and parallel to the edges of the table. However, we cannot have any operations in the group that flip the table over onto the floor, so no inversion, no horizontal reflection, no Ĉ2 around an axis parallel to the floor, and no improper rotations. The point group is C2v .

b. A simple chair. The only symmetry element (besides Ê) is the reflection through the plane that bisects the back and seat of the chair, so the point group is Cs .

6.24 The structure has an eight-fold axis, but different structures above and and below the center of that axis, so no horizontal reflections, improper rotations, or inversion, and no dihedral Ĉ2 axes. In fact, the Ĉ8 rotations are the only symmetry elements. The point group is C8 .

6.25 There are five distinct possibilities. Other substitutions fall into one of these five point groups: 166

C2v

D2h

D3h

D6h

6.26 It’s not linear, there’s a Ĉ4 proper rotation axis through the axial B-D bonds, a σh mirror plane through the equatorial B-H bonds, and four vertical mirror planes (two that contain the equatorial bonds, two that cut between the equatorial bonds). The point group is D4h . 6.27

a. Is the transition from a b2 orbital to an e orbital allowed by electric dipole selection rules? Yes. By Raman? Yes, because b2 ⊗ e = e, which corresponds to (x, y) and (xz, yz). b. In some point groups, your choice of the x and y axes can affect which representation label you assign to a particular function. Is this one of them? No, because none of the symmetry elements depend on the location of the x and y axes. Also, x and y have the same representation (e) and therefore cannot be distinguished by the symmetry operations of this group. c. What is the minimum possible number of atoms in a molecule that belongs to this point group? 5 The molecule cannot be linear, so there must be an atom that lies off the principal rotation axis. Because that atom must be moved by the Ĉ4 rotation to another equivalent atom, there must be at least four atoms off the axis. There must also be at least one more to prevent the molecule from being planar (four equivalent atoms in the same plane will have D2h or D4h symmetry).

6.28 [Thinking Ahead: What property makes a set of operators a group? The point group contains every operator that can be generated from any combination of these two operations, so one way to approach this problem is to carry out combinations of these operations and see what we get, until we stop getting new operators.] The four successive Ĉ4 rotations should be straightforward: 90◦ , 180◦ , 270◦ , and 360◦ (the last one reminds you to include Ê as an operator in the group). Carrying out inversion twice just gives you back Ê, so you need at most to check what happens when you combine Iˆ with each of the successive Ĉ4 rotations. The combinations IˆĈ4 and IˆĈ43 are the corresponding improper rotations Ŝ43 and Ŝ4 . The combination IˆĈ2 is just reflection through the xy plane (change sign on x and y, then on x, y, and z). The point group is C4h , which contains Ĉ42 (z) = Ĉ2 (z), Ĉ43 (z), Ê, σ̂(xy), Ŝ4 , Ŝ43 . 6.29

a. What is the point group for the box? D2h b. What is the symmetry representation for the ground state wavefunction? ag The ground state wavefunction is always the totally symmetric wavefunction. 167

c. What is the symmetry representation for the lowest excited state (nx , ny , nz ) = (1, 1, 2) wavefunction? b1u With nz = 2, there is a node halfway across the box along the z axis, and the whole function changes sign there. That means the wavefunction is antisymmetric under the σ̂xy reflection, but still symmetric under σ̂xz and σ̂yz . d. Write the selection rule for ∆nz for electric dipole–allowed transitions in this system; in other words, what are the allowed values for ∆nz , assuming ∆nx = ∆ny = 0? ∆nz must be odd. As nz increases, the symmetry of the wavefunction alternates between two possible symmetries (the specific representations depending on the values of nx and ny ). The inversion symmetry also alternates with each step in nz . So an even value of ∆nz does not change the inversion symmetry, which means those transitions are electric dipole forbidden. And if ∆nz = 1 is allowed, than all other odd ∆nz transitions are allowed (because those final states will all have the same symmetry). 6.30 Superimposing a molecule on its mirror image requires that after reflection we be able to carry out a proper rotation of the molecule to obtain the original form (proper rotation is the only one of our symmetry operations that can be carried out on a model without taking it apart and putting it together again). Reflection followed by proper rotation is improper rotation, so any group without any Ŝn operations can apply only to chiral molecules. But note that the family of Ŝn operations includes ˆ Examples of chiral point groups are C1 , C2 , C3 , . . . , D2 , D3 , . . . . Ŝ1 ≡ σ̂ and Ŝ2 ≡ I). ˆ This is the 6.31 Ironically, the I point group differs from the Ih group in that it lacks the operator I. same way that, for example, the C2 and C2h groups differ. Look what happens in that case. Imagine that we start with a C2h molecule such as trans-difluoroethene. Inversion Iˆ is one of the symmetry elements. Now let us deform the molecule in such a way that the Ĉ2 operator remains a symmetry element but Iˆ ceases to be a symmetry element; for example, we could push the two F atoms a little below the plane of the other atoms. In doing so, we’ve spoiled the thing that distinguished the au representation from ag , and the bu from the bg , while preserving the rotational symmetry that distinguishes between a and b representations. We not only eliminate the Iˆ operator in this case, we also eliminate the σ̂h operator from the list, because the planar symmetry was broken at the same time as the inversion symmetry. The net result is that we combine every pair of representations xg and xu into one representation x, and we lose any symmetry elements that distinguished between xg and xu . (In the conventional arrangement of the character table, it will always be the second half of the symmetry elements that are eliminated when the inversion symmetry is spoiled.) The same happens when we spoil the inversion symmetry of the Ih group to form the I group: I A(ag )

Ê 1

T1 (t1 )

T2 (t2 )

G(g)

H(h)

12Ĉ5 1 √ 1 (1 + 5) 2 √ 1 (1 − 5) 2 −1 0

Functions 12Ĉ52 20Ĉ3 15Ĉ2 1 1 1 x2 + y 2 + z 2 √ 1 0 −1 (Rx , Ry , Rz ), (x, y, z) (1 − 5) 2 √ 1 0 −1 (1 + 5) 2 −1 1 0 (2z 2 − x2 − y 2 , 0 −1 1 x2 − y 2 , xy, yz, zx)

6.32 This is a long question with a relatively short answer, but you have to be comfortable first with the parity problem of the original wavefunction. [Thinking Ahead: The spherical harmonics, YNMN , are not necessarily symmetric with respect to inversion. For example, s functions are spherically symmetric, so they are symmetric under inversion, but p functions are antisymmetric under inversion. You may want to take a look at Fig. 3.6 to see this.] The spherical harmonics oscillate in parity with N : even 168

N are symmetric under inversion, while odd N are antisymmetric. The eiΛφ phase factor is also not symmetric under inversion. The angle φ changes direction under inversion, and so eiΛφ becomes e−iΛφ . If we have that straight, then let’s just operate with Iˆ on the proposed parity-conserving wavefunction: n o Iˆ ψΛ,N,MN ,p = Iˆ ψ0 YNMN eiΛφ + p(−1)N e−iΛφ = ψ0 (−1)N YNMN e−iΛφ + p(−1)N eiΛφ = ψ0 YNMN (−1)N e−iΛφ + p eiΛφ = p ψ0 YNMN p(−1)N e−iΛφ + eiΛφ = p ψΛ,N,MN ,p .

I was able to factor out p and leave a p behind on the first term because p2 is always 1. So we’ve shown ˆ you get back the same function times the parity, either −1 that if you operate on the function with I, or +1. 6.33 The point group is D2h . The first part of the problem revisits the vector model, but this time applied to nuclear spins. Let’s label the four protons on the molecule as drawn here.

Ha C Hc

Hb C

The vector model for total MI values of 1 and 2 looks like this: label αααα βααα αβαα ααβα αααβ

Ia ↑ ↓ ↑ ↑ ↑

Ib ↑ ↑ ↓ ↑ ↑

Ic ↑ ↑ ↑ ↓ ↑

Id ↑ ↑ ↑ ↑ ↓

MI 2 1 1 1 1

The MI = 2 wavefunction αααα (call it ψ0 ) is all by itself, and we can already assign its representation. Because all four nuclei are equivalent, having the same nuclear spin, ψ0 is symmetric under any operation that moves the nuclei around. The representation is therefore Ag . For the MI = 1 functions, we have to make linear combinations, so that we end up with functions consistent with the point group symmetry. We may start with one function that borrows equally from all four functions, because this is sure to be symmetric: ψ1 =

1 (βααα + αβαα + ααβα + αααβ) . 2

Because we start with four functions, we should form four linear combinations, all orthogonal to each other. So now we just need to find three other functions orthogonal to this one, which we can do by changing the signs; we’ve seen examples of this in Chapter 5: 1 (βααα + αβαα − ααβα − αααβ) 2 1 ψ3 = (βααα − αβαα + ααβα − αααβ) 2 1 ψ4 = (βααα − αβαα − ααβα + αααβ) . 2

ψ2 =

169

Now we need to test these functions with the symmetry operators to find the corresponding represenˆ Looking back at our drawing, we tations. It is sufficient to look at the operators Ĉ2 (z), Ĉ2 (y), and I. see that the rotations and inversion permute the locations of the nuclei as follows: Ĉ2 (z)(abcd) = (dcba) Ĉ2 (y)(abcd) = (badc) Ĉ2 (x)(abcd) = (cdab) ˆ I(abcd) = (dcba) Notice that the operation of Ĉ2 (z) is the same as Iˆ for this purpose. So now we can check what happens to our four MI = 1 wavefunctions: 1 (ααβα + αααβ + αβαα + βααα) = ψ1 2 1 Ĉ2 (y)ψ1 = (αβαα + βααα + αααβ + ααβα) = ψ1 2 1 Ĉ2 (z)ψ2 = (ααβα + αααβ − αβαα − βααα) = −ψ2 2 1 Ĉ2 (y)ψ2 = (αβαα + βααα − αααβ − ααβα) = ψ2 2 1 Ĉ2 (z)ψ3 = (ααβα − αααβ + αβαα − βααα) = −ψ3 2 1 Ĉ2 (y)ψ3 = (αβαα − βααα + αααβ − ααβα) = −ψ3 2 1 Ĉ2 (z)ψ4 = (ααβα − αααβ − αβαα + βααα) = ψ4 2 1 Ĉ2 (y)ψ4 = (αβαα − βααα − αααβ + ααβα) = −ψ4 2 Ĉ2 (z)ψ1 =

where you can simplify all this by seeing that only the minus signs change between the four sets of equations. In any case, we see that ψ1 is completely symmetric, as we might have expected from all the “+” signs, and therefore also belongs to the Ag representation. Using the Ĉ2 (z) symmetry to also identify the inversion symmetry, we see that the other gerade function is ψ4 , but because it is antisymmetric under Ĉ2 (y) it must belong to the B1g representation. Similarly, ψ2 has B2u symmetry and ψ3 has B3u symmetry. So the nuclear spin states we’ve found are: label MI ψ0 2 ψ1 1 ψ2 1 ψ3 1 ψ4 1

Γns Ag Ag B2u B3u B1g

The forbidden transitions that change parity will be between the set psi0 , ψ1 , ψ4 and the set ψ2 , ψ3 . R 6.34 If φ1 and φ2 are orthogonal, then φ1 φ2 dτ = 0. All representations except the totally symmetric representation Γs integrate to zero because they must be antisymmetric under some symmetry operation, and therefore they must have equal parts positive and negative phase. So φ1 and φ2 are therefore 170

orthogonal if Γ1 ⊗ Γ2 6= Γs . a1 ⊗ a2 a1 ⊗ b 1 a1 ⊗ b 2 a2 ⊗ b 1 a2 ⊗ b 2 b1 ⊗ b2

 = a2     = b1     = b2 6= a1 ; therefore if Γ1 6= Γ2 , the integral is always zero. = b2     = b1     = a2

6.35 The function must be symmetric with respect to any of the Ĉ2 rotations but antisymmetric with respect to inversion and the mirror reflections. This simplifies the problem, because in Cartesian coordinates, σ̂ changes the sign on one of the three coordinates, Ĉ2 changes the sign on two of the three coordinates, and Iˆ changes the sign on all three coordinates. Therefore, we want a function that changes sign if any one or all three coordinates change sign, but remains the same if any two coordinates change sign. Because two sign changes can cancel each other out, a straightforward example of this is the function xyz. 6.36 We obtain the characters for the direct product by going to the character table: D3h a′1 a′′2 e′′ a′1 ⊗ a′′2 ⊗ e′′

Ê 1 1 2 2

2Ĉ3 3Ĉ2 σ̂h 1 1 1 1 −1 −1 −1 0 −2 −1 0 2

2Ŝ3 3σ̂v 1 1 −1 1 1 0 −1 0

These are the characters for the representation E ′ . 6.37 In this solution, I’ve used the less rigorous solve-by-examination approach to reducing a representation. A2g ⊗ T2g ⊗ Eu in Oh A2g T2g Eu A2g ⊗ T2g ⊗ Eu

Ê 1 3 2 6

Ĉ3 Ĉ2 Ĉ4 1 −1 −1 0 1 −1 −1 0 0 0 0 0

Iˆ σ̂h 1 1 3 −1 −2 −2 −6 2 is reducible.

Possible ways to get 6 for Ê: T ⊕T T ⊕E⊕A T ⊕ 3A

E⊕E⊕E E⊕E⊕A⊕A E ⊕ 4A 6A

To get 0 for Ĉ3 , the only remaining options are T ⊕T

T ⊕E⊕A

E⊕E⊕A⊕A

ˆ these must all be “u” representations: To get −6 for I, Tu ⊕ Tu

Tu ⊕ Eu ⊕ Au

Eu ⊕ Eu ⊕ Au ⊕ Au

To get +2 for σ̂h , these must be Tu ⊕ Tu ; to get 0 for Ĉ2 , these must be T1u ⊕ T2u . 171

6.38

a. D3h E′ E ′′ ′ E ⊗ E ′′

Ê 2Ĉ3 3Ĉ2 2 -1 0 2 -1 0 4 1 0

σ̂h 2 −2 −4

E ′ ⊗ E ′′ = E ′′ + A′′1 + A′′2 Ê ⇒ 4 means

E + E, E + A + A, or A + A + A + A

2Ĉ3 ⇒ + 1 means

E + A + A, otherwise this would be 0 or ± 2

3Ĉ2 ⇒ 0 means

E ′′ + A′′1 + A′′2

E ′′ + A′′ + A′′

σ̂h ⇒ − 4 means

b. a1 ⊗ b2 ⊗ e ⊗ e

Ê = 4 2Ŝ4 = 0 Ĉ2 = 4 2Ĉ2′ = 0 2σ̂d = 0

To get 4 for Ê, we need E ⊕ E, E ⊕ A ⊕ A or 4A (where each A could actually be either A or B). To get 4 for Ĉ2 , we need 4A. To get 0’s for Ŝ4 , Ĉ2′ and σ̂d , we can only have: A1 ⊕ A2 ⊕ B1 ⊕ B2 . 6.39 The characters are 4, 1, 0, 4, 1, 0. Because E ′′ is antisymmetric with respect to σ̂h , its square must be symmetric, so only x′ representations are possible. To get a 4 for the first character, the reduction must be of the form 2E ′ , E + 2A′ , or 4A′ . Of these, only E ′ + 2A′ gives 1 for the second character. To get 0 for the third character, the two A’s are A1 and A2 , so the result is:

A′1 ⊕ A′2 ⊕ E ′ 6.40 D6h a) A1g ⊗ E1g

b) A1u ⊗ B1u

c) B2u ⊗ E2g d) E2u ⊗ E1g ⊗ B2g

Ê 2Ĉ6 2Ĉ3 Ĉ2 2 1 −1 −1

3Ĉ2′ 0

3Ĉ2′′ 0

Î 2

= E1g

= B1g

−1 −2 1 4

0 0

−1

−2 −4

= E1u

2 4

−1

1 1

1 −1

0 0

In row (d), 4 means the sum of degeneracies is equal to 4; −4 means that all components are u.

a. To get the characters 1, 1 for Ĉ6 and Ĉ3 , must be a sum of one E-state, two non-degenerate states.

b. To get 3Ĉ2′ , 3Ĉ2′′ characters = 0, 0, must be either B1u + B2u or A1u + A2u (since this must be u-states). c. To get Ĉ6 character = +1, given the previous conclusions, the non-degenerate representations are A1u + A2u , the E-state is E2u . Therefore, (d) E2u ⊗ E1g ⊗ B2g = A1u + A2u + E2u . 172

6.41 Ŝ4 (x2 − y 2 ) = σ̂h Ĉ4 (x2 − y 2 )

= σ̂h (y 2 − x2 ) = (y 2 − x2 ) = −(x2 − y 2 ).

This is sufficient, because B is the only representation with character −1 for Ŝ4 .

(x1 , y1 )

(x2 , y2 ) C4

x2 = + y

y = x 1 2 x Ĉ4 ψ(x, y, z) = ψ(y, −x, z)

σ̂h ψ(x, y, z) = ψ(x, y, −z)

6.42 The following is a model of water, with the coordinate system arranged so the z axis coincides with the Ĉ2 axis and so the molecule lies in the xz plane.

z y x Rx

The rotation around each of the three Cartesian axes is shown, and the in-plane rotation is the Ry shown in the middle. We can see that if we rotate the drawing, including the rotation arrows, about the z axis, then the arrows change direction, so this rotation is antisymmetric under Ĉ2 . Upon reflection through the xz plane, nothing happens (because the arrows lie in the xz plane), but the arrows change sign when reflected through the yz plane. Therefore, the Ry rotation has b1 symmetry. If you drew the molecule differently, you may end up with a different symmetry, but the idea is the same. If your rotation is Rz , then the rotation arrows stay the same after Ĉ2 but change sign under either reflection, so Rz has a2 symmetry. On the other hand, Rx is antisymmetric under Ĉ2 and σ̂xz , so has b2 symmetry. 6.43 The C C σ-bond occupies the region along the axis of symmetry between the two carbons, and is symmetric under all the operations of the point group, so its representation is Γtot sym = a1g . 6.44 Let’s do this one by drawing pictures to see the symmetry of the resulting combinations.

a. σg bonding

1sA + 1sB + +

173

b. σ ∗ antibonding

c. πg or πg∗ antibonding

1sA + 2pzB 6x + − + z y 2pxA − 2pxB 6x + − z − + y

6.45 For this function, λ = 0 because (i) there is no imaginary term and (ii) the only node is at z = 0; there are no angular nodes that contain the z axis. Next we check the inversion symmetry: √ ˆ = ψ(−x, −y, −z) = −5z(x2 + y 2 )e− x2 +y2 +z2 = −ψ. Iψ Therefore, ψ is a σu anti-bonding orbital. That it is anti-bonding can be seen by the antisymmetry upon reflection through the center of the bond: σ̂xy ψ = ψ(x, y, −z) = −ψ, which requires the electron density to vanish at the node in the middle of the bonding region. 6.46 This is a matter of finding a spatial wavefunction for H2 and a two-electron spin wavefunction, such that the overall permutation symmetry is antisymmetric. ψMO [(σg 1s)(σg 2s)] = {[1sA (1) + 1sB (1)][2sA (2) + 2sB (2)] + [1sA (2) + 1sB (2)] ×[2sA (1) + 2sB (1)]} [α(1)β(2) − β(1)α(2)]

This is an antisymmetric spin function, since we are using a symmetric spatial function. Other answers are possible. 1sA (1) = Ae−rA1 /a0 rA1 e−rA1 /2a0 2sA (1) = A 1 − 2a0 h i rA2 rB2 1− ψ = e−rA1 /a0 + e−rB1 /a0 e−rA2 /2a0 + 1 − e−rB2 /2a0 2a0 2a0 i h rB1 rA1 e−rA1 /2a0 + 1 − e−rB1 /2a0 1− + e−rA2 /a0 + e−rB2 /a0 2a0 2a0 × [α(1)β(2) − α(2)β(1)]

6.47

a. The molecule is not linear, has a Ĉ3 principal rotation axis perpendicular to the plane of the molecule (but no other Ĉn with n > 2), has a horizontal mirror reflection (through the plane of the nuclei), and has three vertical mirror reflection planes. Therefore, the molecule belongs to the point group D3h . 174

b. H+ 3 has two electrons, so we only need one MO, and the lowest energy MO is constructed from the lowest energy atomic orbitals overlapping so as to maximize the constructive interference. In this case, that means adding together the 1s orbitals on each of the three atoms, using the same phase for each (meaning no minus signs, to avoid destructive interference of the wavefunctions): ψ+ ≡ C+ (1sA + 1sB + 1sC ). The overall wavefunction puts both electrons in this spatial orbital but with opposite spins (α and β) to obey the Pauli exclusion principle: Ψ(1, 2) = ψ+ (1)ψ+ (2)α(1)β(2). c. The spatial part of this wavefunction already reflects the indistinguishability of the electrons, but we need a symmetrized spin part. The spatial part is symmetric under P̂21 , so we need the antisymmetric two-electron spin function χanti : √ Ψ(1, 2) = ψ+ (1)ψ+ (2)[α(1)β(2) − β(1)α(2)]/ 2. d. The orbital is symmetric under all of the symmetry rotations and reflections. The symmetry representation will be that of the totally symmetric representation in D3h , namely a′1 . 6.48

F F

F F F

F y

a1 g

F b1 g

C’2

x z

Cl C Cl Cl C Cl Cl C Cl Cl C Cl Cl C Cl H

The XeF4 molecule belongs to the point group D4h , and CH2 Cl2 belongs to C2v , with the z principal axis perpendicular to the page as drawn above. To distinguish between the b1g and b2g orbitals of XeF4 , it is necessary to label the Ĉ2′ axis as either along the Xe F bond axes or between them. To distinguish between the b1 and b2 orbitals of CH2 Cl2 , it is necessary to label the x and y axes. The MOs are drawn by examining all the equivalent bonds simultaneously (all four Xe F bonds together, the two C H bonds together, and the two C Cl bonds together), and then varying the relative phases and examining the symmetry of the result. The only complication occurs for XeF4 , because the two single-node MOs form a degenerate eu pair, in the same way that the single-node C4 H4 MOs in Fig. 5.14 constitute a doubly degenerate eg pair (those being π-bond MOs, they have a different inversion symmetry). 6.49 In order of increasing energy, ag : no nodes, bonding everywhere; bu : no node in the molecular plane, but node perpendicular to the molecular plane; au : node in the molecular plane, so weaker bonding; bg : node in the molecular plane and perpendicular to the molecular plane, so most antibonding. 175

6.50

a. All three coordinate axes are Ĉ2 symmetry axes for this molecule, and the xy, yz, and xz planes are all mirror planes. Therefore, the molecule belongs to the point group D2h . b. The answer depends on your choice of coordinate axes; since any of the axes above can be considered the principal axis, there is no single best choice for which axis to call z. This π orbital will be antisymmetric with respect to reflection through the plane of the page, but neither of the other two reflections. That means it must be one of the bu representations. If you chose the plane of the page to be the xy plane, the orbital is b1u ; if you chose that plane to be xz, the orbital is b2u ; and if you chose that plane to be yz, the orbital is b3u .

6.51 The point group is C2h , with representations ag , au , bg , bu . The C Cl bonds are in the molecular plane, so they must be symmetric with respect to σˆh , which is only true of the ag and bu representations. To get electron pairs at each bond requires the sums ag + bu and ag − bu , so both MOs are used: ag and bu . H

Q Q

Cl QH

6.52 The molecule is in the C2h point group. Test these MOs with respect to various C2h symmetry elements: Iˆ Ĉ2 σˆh a) 1 b) −1 c) 1 d) −1 e) 1 f) −1

1 1 ag 1 −1 bu −1 −1 au −1 1 bg −1 −1 au −1 1 bg

6.53 We assign the z axis to the internuclear axis by convention. These are each one of a pair of degenerate MOs, because they can be rotated about the z axis to give distinct MOs with identical energies, or (to look at it another way) because they can change sign under Ĉ∞ (z). For example, (a) is antisymmetric with respect to Ĉ4 (z) (a 90◦ rotation), and looking at the character table we see that the simplest representation where the character cos 2φ for φ = 90◦ gives −1 is the δ representation. Hence (a) has δ symmetry. Similarly, (b) is antisymmetric with respect to Ĉ2 (z), hence it has π symmetry. Iˆ Ĉ∞ 2 cos 2φ −2 2 cos φ −2

δu πu

6.54 All of these are proportional to sin θ, which is at a maximum when θ = π/2, i.e., in the xy plane. Therefore, all of these must be symmetric under the σ̂h reflection. The differences among these arise from the φ part of the function. ψ1 = A sin θ

a1g

ψ2 = A sin θ cos φ

ψ3 = A sin θ sin φ

eu 176

ψ4 = A sin θ cos 2φ

b1g or b2g

|ψ1 |2

a1g

|ψ2 |2 + |ψ3 |2

a1g

|ψ4 |

ψ1 has no φ-dependence and is positive everywhere, so belongs to the totally symmetric representation, a1g . ψ2 and ψ3 are the orbitals with one node each (cos φ has a node at φ = π/2 and sin φ at φ = π), so they constitute a doubly degenerate pair. Since they are also symmetric under σ̂h , the representation must be eu . ψ4 has two nodes, at φ = π/4 and 3π/4, so it is symmetric under Ĉ2 and Iˆ (so g) but antisymmetric under Ĉ4 (so b). The probability densities must all have the same symmetry as the nuclei, i.e., symmetric under every operation, so a1g . Note that the degenerate pair has to be treated simultaneously to get the correct symmetry. 6.55 : NA CB CC ND : (the lone pairs must go on the N atoms to keep formal charges to 0). There are 26 electrons in this molecule; we can divide them as follows: no. electrons local bonding model 4 N 1s2 core 4

C 1s2 core

C − N σ bonds

2 4

C − C σ bond N : lone pairs

C − N π bonds

MO model approximate MO σg (1sA + 1sD ) σu (1sA − 1sD ) σg (1sB + 1sC ) σu (1sB − 1sC ) σg (σAB + σCD ) σu (σAB − σCD ) σg (spA + spC ) σg (spA + spB ) σu (spA − spB ) πu (πAB + πCD ) πg (πAB − πCD )

The last column is only to give some idea what the MOs might look like. For example, the C C σ bond MO will be σg because it is composed of two sp hybrids on atoms B and C overlapping in the middle of the molecule in phase: -

O C

+ C

O-

The result is: 10 σg electrons 8 σu electrons 4 πu electrons 4 πg electrons 6.56

a. Ĉ2 and Ĉ4 are rotations through φ of π and π2 respectively, so: Ĉ2 ψ(r, θ, φ) =

ψ(r, θ, φ + π) π ψ r, θ, φ + 2

Ĉ4 ψ(r, θ, φ) = 177

These act only on the φ-dependent term of ψ(r, θ, φ), so we only need to look at that term: Ĉ2 eiφ = ei(φ+π) = cos(φ + π) + i sin(φ + π) = − cos φ − i sin φ = −eiφ

= ei(φ+ 2 ) = cos(φ + π

antisymmetric Ĉ4 eiφ

= − sin(φ) + i cos φ = ieiφ

π π ) + i sin φ + 2 2

non-symmetric

6= ±eiφ

6 y

φ+π ' q (x, y) H j H φ -x q (−x, −y)

6 x

θ q (−x, −y, −z)

q (x, y, z) θ -z H Y H % π−θ

ˆ Iψ(x, y, z) = ψ(−x, −y, −z) ˆ Iψ(r, θ, φ) = ψ(r, π − θ, φ + π)

= (e−rA /a0 + e−rB /a0 ) cos(π − θ)ei(φ+π) = (e−rA /a0 + e−rB /a0 )(− cos θ)(−eiφ ) = ψ(r, θ, φ)

symmetric

c. Antisymmetric with respect to Ĉ2 , non-symmetric with respect to all higher Ĉn rotations, because 2π 2π iφ + i sin φ + Ĉn e = cos φ + n n 6= ±eiφ except for n = 1, 2.

Therefore, λ = 22 = 1; this is π MO. Symmetric with respect to inversion; therefore, this is a πg MO, which is antibonding. To prove πg is antibonding for this case, we use the reflection operator σ̂xy : σ̂xy ψ(r, θ, φ) = ψ(r, π − θ, φ) = −ψ(r, θ, φ).

cos(π − θ) = − cos θ Antisymmetric

There is a nodal plane between the nuclei; the MO is antibonding. Another way to do the whole problem is to draw the real part of the angular wavefunction: |ψ(θ, φ)| = cos θeiφ = cos θ cos φ 178

− +

cos θ, cos φ − - z +

+ cos φ −

θ cos − + - z

6.57 au The point group is C2h . The MO is antisymmetric under inversion (so has u symmetry) but symmetric under the Ĉ2 rotation (so has a symmetry). 6.58 The point group is C2h , with the principal axis perpendicular to the page. There are six C C bond electrons, in three pairs: one pair in a σ orbital (symmetric under all operations because it lies at the center of the molecule) and two pairs in a π symmetry orbitals (one in-plane and one out-of-plane, having opposite symmetries with respect to σ̂h ). There are also two pairs of electrons associated with the two single bonds, which can be described by σ symmetry orbitals that either have the same phase or ˆ Altogether, we have the following have opposite phase, and therefore have opposite symmetry under I. five orbitals: Iˆ +1

C≡C σ-bond

σ̂h representation +1 ag

C≡C in-plane π-bond −1 C≡C out-of-plane π-bond −1 C−C same phase +1

+1 −1 +1

bu au ag

−1

C−C opposite phase 6.59 NH3 has 7 + 3 = 10 electrons. .. N

AE J EA J E A H H

number 2 2 6

local bond model symmetry N core N lone pair NH σ-bonds

a1 a1 a1 or e

MO (1a1 )2 (2a1 )2 (3a1 )2 (1e)4

6.60 The representation for the direct product has characters 4,4 cos φ cos 2φ,0 for the operations Ê, Ĉ2π/φ , and σˆv , respectively: Ê Ĉ2π/φ σ̂v Π 2 2 cos φ 0 ∆ 2 2 cos 2φ 0 Π ⊗ ∆ 4 4 cos φ cos 2φ 0

There is no representation in the character table for C∞v that has these characters, so this must be a reducible representation. We check the character under each symmetry element and find the unique combination of irreducible representations whose characters will add up to give the characters we have. There are three ways to get a 4 under the Ê operation: by adding four Σ functions (1+1+1+1=4), by adding two Σ functions and one non-Σ function (1+1+2=4), and by adding two non-Σ functions (2+2=4). Of these choices, which could lead to a character of 4 cos φ cos 2φ for the rotation? We use a trigonometric identity to put this character in the form of a sum: 4 cos φ cos 2φ = 2 cos φ + 2 cos 3φ. 179

This sum makes it clear that we can get the right character for rotation by adding together the Π and Φ representations. The direct sum of these two irreducible representations is the same as the direct product of the Π and ∆ representations. Π Φ Π⊗∆

Ê Ĉ2π/φ 2 2 cos φ 2 2 cos 3φ 4 2 cos φ + 2 cos 3φ

σ̂v 0 0 0

Π ⊗ ∆ = Π ⊕ Φ. 6.61 The non-degenerate a1g orbital is full with two electrons, but the doubly degenerate eu is only partly full, so we take the direct product eu ⊗ eu : D4h eu eu ⊗ eu

Ê 2 4

2Ĉ4 0 0

Ĉ2 2Ĉ2′ 2Ĉ2′′ Iˆ 2Ŝ4 σ̂h −2 0 0 −2 0 2 4 0 0 4 0 4

2σ̂v 2σ̂d 0 0 0 0

The order of the group Q is 16. Now we need to decompose the reducible representation. The characters ˆ and σ̂h . Therefore, we can ignore all the other terms in the sums that are all zero except for Ê, Ĉ2 , I, we use to find the coefficients for each irreducible representation. Another simplification is that we know only g representations can result from the product of two u functions, so we can ignore half the irreducible representations: eu ⊗ eu = a1 A1g ⊕ a2 A2g ⊕ a3 B1g ⊕ a4 B1g ⊕ a5 Eg 1 [(4)(1) + (4)(1) + (4)(1) + (4)(1)] = 1 a1 = 16 1 a2 = [(4)(1) + (4)(1) + (4)(1) + (4)(1)] = 1 16 1 a3 = [(4)(1) + (4)(1) + (4)(1) + (4)(1)] = 1 16 1 a4 = [(4)(1) + (4)(1) + (4)(1) + (4)(1)] = 1 16 1 [(4)(2) + (4)(−2) + (4)(2) + (4)(−2)] = 0 a5 = 16 eu ⊗ eu =

A1g ⊕ A2g ⊕ B1g ⊕ B2g

6.62 The point group is C2v , and the π bonds are antisymmetric with respect to σ̂xz . With the two π bonds having the same phase, the MO is symmetric with respect to σ̂yz , and is b2 . With the two π bonds having opposite phase, the MO is antisymmetric with respect to σ̂yz , and is a2 . q q q q 6.63 O C O q q q q a. Lowest energy MOs must be 1s2 core electrons on C and both O’s 3 1s2 cores = 6 electrons = 1σg2 1σu2 2σg2 .

b. Next lowest must be C O σ-bond electrons (stronger bond than π-bond) 2 C O σ-bonds = 4 electrons = 2σu2 3σg2 . 180

c. Next lowest are probably the lone pairs (no attractive force from bond, but high s character means stronger nuclear attraction than π electrons) 4 O-atom lone pairs = 8 electrons = 1πu4 3σu2 4σg2 . (We can also show this by forming sp2 hybrids for lone pairs.) d. Highest is probably C O π-bond electrons 2 C O π-bonds = 4 electrons = 1πg4 . 6.64 (1a1 )2 (2a1 )2 (1b1 )2 (3a1 )2 (4a1 )2 (2b1 )2 (5a1 )2 (3b1 )2 (6a2 )2 (1b2 )2 (2b2 )2 H H B B C A A A C C H

cyclopropene

z 6 -x J H

local bond model MO model type of number of type of electrons electrons electrons C-atom cores 6 a1 or b1 6 a1 or b1 C C σ-bonds 8 in-plane: a1 or b1 ; out:a1 or b2 C H σ-bonds 2 a2 or b2 C C π-bonds 22

MO 1a1 , 2a1 , 1b1 3a1 , 4a1 , 2b1 5a1 , 3b1 , 6a1 , 1b2 2b2

The C H bonds that lie in the xz plane must be either a1 or b1 because they cannot have a node at the xz plane. The C H bonds in the CH2 group cannot have a node at the yz plane because they are σ bonds, but they may have a node at the xz plane, and therefore may be either a1 or b2 . 6.65 a′′2 ⊗ e′ = E ′′ ,

which corresponds to the Raman functions xz and yz. Therefore the transition is electric dipole–forbidden and Raman–allowed. 6.66 The ground state has symmetry 1 A′ , because all the electrons are paired and all the MOs are filled, and A′ is the totally symmetric representation. The lowest excited state has symmetry e′ ⊗ a′′2 = E ′′ . We check that the transition is allowed by symmetry: A′ ⊗ E ′′ = E ′′ is again allowed by Raman selection rules, but if that makes you happy, then you should stop reading. The lowest excited state produced from this configuration is not the 1 E ′′ , but the 3 E ′′ . Both are 3 possible from an e′ a′′2 MO configuration, and the higher spin triplet state has the lower energy, for the same reason that the triplet state is lower energy than the singlet in 1s1 2s1 helium atom (Section 4.3). Therefore, the transition is forbidden by the spin selection rule. Chapter 7 discusses the spin selection rule and this pairing of states in molecules. 181

6.67 According to Fig. 6.19, the HOMO for cyclobutadiene is a b1g orbital and the LUMO is a b2g orbital. The product b1g ⊗ b2g is equal to B3g , which is the representation for yz. The spin selection rule is satisfied because I have specified that we only consider the singlet states. Therefore, the transition is Raman–allowed and electric dipole–forbidden. 6.68 Benzene belongs to point group D6h . A1g is the totally symmetric representation for D6h . The transition operator representations are given by the last column in the character table: Γµ =A2u

E1u

x, y 2

x + y , z2

A1g E1g

xz, yz 2

(x − y 2 ), xy

E2g

For the electric dipole–allowed transitions, Γµ corresponds to x, y, or z; the other representations are for Raman transitions. The transition is allowed if Γfinal ⊗ Γµ ⊗ Γinitial = A1g

Γfinal ⊗ Γµ ⊗ A1g = A1g

→ Γfinal ⊗ Γµ = A1g

only if Γfinal = Γµ . Therefore, (a) electric dipole selection rules: Γfinal = A2u , E1u ; (b) Raman selection rules: Γfinal = A1g , E1g , E2g . 6.69 First off, the spins of the two states in the transition are the same, so the ∆S = 0 spin selection rule is not violated. Next we have to check the symmetry. T1 T1 T1 ⊗ T1

Ê 8Ĉ3 3Ĉ2 6Ŝ4 6σ̂d 3 0 −1 1 −1 3 0 −1 −1 1 9 0 1 −1 −1

This is an illustration that we can often decompose a reducible representation by a qualitative analysis, although we always have the methodical technique described in the text to fall back on if needed. To get the 9 under Ê, we could combine 3 T representations (3+3+3=9) or 2 T ’s, an E, and an A (3+3+2+1=9), and so on. However, since we have a 0 under Ĉ3 and A1 and A2 both have a character of 1 under Ĉ3 while E has a character of −1, there cannot be too many A representations involved in the direct sum. The only sums that are consistent with a 9 under Ê and a 0 under Ĉ3 are T + T + T , T + T + E + A, T + E + E + A + A, and E + E + E + A + A + A. If we look at the very next character, the 1 under Ĉ2 , we see that only the form T + T + E + A will give the right value. We still don’t know what combination of T1 and T2 to use, and whether to use A1 or A2 . Finally, to get the −1’s under Ŝ4 and σ̂d , we find the only combination that works is T1 ⊕ T2 ⊕ E ⊕ A2 . Since this contains T2 (which is the representation for x, y, and z), the transition is electric dipole–allowed, and it is also Raman–allowed because E and T2 are representations for second order Cartesian functions such as xz and x2 − y 2 . An alternative solution, using the real math: The order of the group is 1+8+3+6+6 = 24. T1 ⊗ T2 = a1 A1 ⊕ a2 A2 ⊕ a3 E ⊕ a4 T1 ⊕ a5 T2 . 182

1 [(9)(3) + 8(0)(0) + 3(1)(−1) + 6(−1)(1) + 6(−1)(−1)] 24 =1

a4 =

Therefore, T1 is contained in the reducible representation and the transition is allowed. 6.70 The point group is D5h . There are five atomic p orbitals, which must combine to form five π MOs: the lowest energy will have no nodes (ignoring the node through the molecular plane), the second and third MOs form a degenerate pair with one node each, and the fourth and fifth form another degenerate pair with two nodes each. There are also five electrons, which fill two of the lower energy orbitals but only half fill the third. The HOMO therefore has one node, while the LUMO has two nodes. The symmetry Γµ of the HOMO→LUMO transition moment must have the right symmetry to introduce a node—a change in sign—perpendicular to the existing node, without changing the other symmetry properties of the π orbitals. That Γµ must have the same symmetry as the function x or y (because those functions change sign as they cross the z axis), and therefore corresponds to an electric dipole–allowed transition. 6.71 Electric dipole (ED): Γµ = Au , Bu ; Raman (R): Γµ = Ag , Bg . Because the direct product of two g functions or two u functions is g, and the direct product of a u and a g function is u, then Γu ↔ Γu ; Γg ↔ Γg will be Raman–allowed and Γg ↔ Γu will be electric dipole–allowed. C2h Ag Bg Au Bu

Ag R R ED ED

Bg R R ED ED

Au ED ED R R

Bu ED ED R R

6.72 Electric dipole (ED): Γµ = A1 (z), E(xy). Raman: Γµ = A1 (x2 + y 2 , z 2 ), E(x2 − y 2 , xz, xy, yz). Γfinal ⊗ Γµ ⊗ Γinitial = Γs = A1 ; if the transition is allowed, then Γinitial ⊗ Γfinal = Γµ . Γinitial ?

A2 → A1 :

A2 → A2 :

Γfinal

A2 → E :

⊗

forbidden

allowed by ED and Raman

⊗

allowed by ED and Raman

⊗

Γµ

6.73 A1u = Γinitial . Γfinal ⊗ Γµ ⊗ Γinitial = Γs = Ag . Γµ (Raman) = Ag , B1g , B2g , B3g . The final states are A1u , B1u , B2u , B3u . Γfinal = Γµ ⊗ Γinitial :

A1g ⊗ A1u = A1u

B1g ⊗ A1u = B1u

B2g ⊗ A1u = B2u

B3g ⊗ A1u = B3u 6.74 The point group is D3h . ′

E ⊗ E

′

Ê = 4 = E′ +

Ĉ3 1 A′1 +

Ĉ2 0 A′2

σ̂h 4

Ŝ3 1

σ̂v 0

The transition is allowed by both electric dipole and Raman selection rules. 183

6.75 [Thinking Ahead: How does λ correspond to the atomic quantum numbers? λ is equivalent to ml in the atomic case; it gives the projection of ~l onto the z axis. So for electric dipole selection transitions, the selection rule is the same as for electric dipole transitions in atoms.] a. for electric-dipole transitions? ∆λ = 0, ±1 b. for Raman transitions? ∆λ = 0, ±1, ±2 Because Raman is like a 2-photon process, λ can change twice, so in that case a change of ±2 is also possible. (You can also tell this by checking the selection rules for σ ↔ σ, σ ↔ π, σ ↔ δ in C∞v .) 6.76 Γinitial ⊗ Γfinal 6= Γt . For D2h , only Au does not correspond to an electric dipole or Raman transition moment operator. The transition between any two states that give this direct product is forbidden. For example, Ag ← → / Au , B1g ← → / B1u , B2g ← → B2u , B3g ← → / B3u .

Chapter 7 7.1 The high Z elements differ from the low Z elements in two ways that are relevant to the correlation diagram: • the core electrons are more tightly bound to the nucleus, because of the greater nuclear charge, and • for two elements that occupy different rows of the periodic table, the valence electrons are at a greater distance from the more highly charged nucleus, because they occupy higher values of n. Both of these features tend to leave the atomic energy levels of high Z atoms less affected by the formation of a chemical bond than a low Z atom. The core electrons are oblivious to the chemical bond, and their energy levels shift little from the isolated atom. The valence electrons that participate in the bonding are at relatively large distances from the nucleus for a high Z free atom, and therefore their energy levels are not so strongly perturbed when they do form a bond. For both reasons, the high Z atom electronic energy levels are less affected by the formation of a bond, and therefore the high Z atoms lie closer to the separated atom limit of the correlation diagram. 7.2 Homonuclear diatomics have inversion symmetry (g or u). States from a single MO configuration are either all g or all u. Transitions between these states violate the g ↔ u selection rule. 7.3 The N2+ 2 ion does have a σ bond, and the explanation may be phrased in terms of the correlation diagram as follows. The N2+ 2 ion may be made by removing the two highest energy electrons from N2 . These are bonding electrons, however, and therefore the bond weakens. The bond length increases, and our line on the correlation diagram shifts to larger R, where the 3σg energy is lower than the 1πu energy. In fact, this is a rather complicated system, because the energy difference between the 3σg and 1πu orbitals is similar to the energy contribution from the spin-dependent exchange integral. Computations predict that the ground state is a 3 Πu state with a bond length of 1.24 Å, but that a 1 Σ+ g state lies only 0.25 eV higher in energy with a bond length of 1.14 Å. 7.4

a. With one electron, the total electron spin is the spin of one electron, which is always 1/2. Therefore, for any electronic state, 2S + 1 = 2. b. A closed-shell molecule is one in which all the MOs are filled.To fill an orbital, the electrons must all be paired, and therefore the spins all sum to zero. 2S + 1 = 1. 184

c. An atom or molecule with an even number of electrons can only have unpaired electrons in multiples of two: zero unpaired electrons (singlet state), two unpaired electrons (triplet state), 4 unpaired electrons (quintet state) and so on. Similarly, an atom or molecule with an odd number of electrons can only have an odd number of unpaired electrons: one unpaired electron (doublet spin), three unpaired electrons (quartet spin), and so on. The only way to go from an odd spin multiplicity (2S + 1 = 3) to even multiplicity (2S + 1 = 2) is to add or subtract an odd number of electrons.

7.5 An excimer is only stable in its excited states. The ground state is dissociative. The excimer formed in the AΠ state will fluoresce or phosphoresce (depending on the spin values for the X and A states) to the ground state, whereupon it immediately dissociates.

7.6

a. In the point group D2h , b2g ⊗ au = B2u . The representation au has the effect of inverting the inversion symmetry.

b. The degeneracy of the λ = ±4 γ orbital of a linear molecule is 2, because any linear molecule orbital is doubly degenerate except for the non-degenerate σ orbitals.

c. The lowest energy molecular orbital in H2 has symmetry representation σg . This is the representation of the ψ+ MO in our discussion of H+ 2 in Chapter 5. d. The ground electronic state of benzene has symmetry representation A1g . All the MOs are filled in benzene’s ground state, and the direct product obtained for filled molecular orbitals is always the totally symmetric representation of the point group.

e. In the correlation diagram for homonuclear diatomics, the 1πu molecular orbital is formed by combining two atomic 2p orbitals with ml = ±1, because ml remains a well-defined quantum number when going from atoms to linear molecules, and π orbitals are those with ml (now called λ) equal to ±1. f. The complete list of MO representations obtained from two 3d orbitals in the separated atom limit is σg , πg , δg , σu , πu , δu . The σ MOs are created from ml = 0 3d orbitals, the π’s from ml = ±1, and the δ’s from ml = ±2. The inversion symmetry depends on the relative phase of the two atomic orbitals being combined, and either g or u can always be obtained.

7.7 [Thinking Ahead: How would this process have an effect on the quantum states of the product? This question concerns the coupling of different kinds of energy in a collision involving a molecule and a free electron, so we want to examine the scales of the energies interacting.] a. The HF+ is almost always formed in an excited vibrational state, because the bond length of HF+ , and for the very rapid interaction of the free electron with the molecule, the nuclei will not have time to respond until after the collision is over: 185

HF E

In this respect, the transition is similar to photoionization. The result could be predicted on the basis of vibrational and electronic energy scales being similar, and therefore easily coupling. b. This is a harder one to predict because the energy scales are much different, and the interaction will be weaker—but will it be so weak that the HF+ will be formed in the J = 0 ground state? A significant fraction will be in the ground state, but probably most will be in excited rotational states because even a very weak coupling between the huge energy of the free electron and the closely spaced rotational states leave enough energy behind to populate J = 1 or higher. 7.8

a. O2 has sixteen electrons. σ orbitals take two, all others take four. Using the Aufbau principle, near united atom limit (lefthand side), in the correlation diagram, the MO configuration would be 1σg2 2σg2 1σu2 1πu4 3σg2 2σu2 2πu2 . b. On the opposite side, in separated atom limit, the ordering predicts an MO configuration 1σg2 1σu2 2σg2 2σu2 3σg2 1πu4 1πg2 . The actual MO configuration is closer to the separated atom limit: 1σg2 1σu2 2σg2 2σu2 1πu4 3σg2 1πg2 .

7.9 Be2 has eight electrons, so we count up the correlation diagram MOs until we use up all eight. a. Ground state 1σg2 1σu2 2σg2 2σu2 ; Bond order = 12 (4 − 4) = 0 Excited state 1σg2 1σu2 2σg2 2σu 3σg ; Bond order = 12 (5 − 3) = 1 b. Ground state 1σg2 2σg2 1σu2 1πu2 ; Bond order = 12 (6 − 2) = 2 Excited state 1σg2 1σu2 2σg2 1πu 3σg ; Bond order = 12 (6 − 2) = 2. 7.10

a. The united atom limit Solution: There are ten electrons; we follow the list of orbitals on the lefthand side of the diagram until you use them up: 1σg2 2σg2 1σu2 1πu4 . b. The separated atom limit Solution: And now the righthand side: 1σg2 1σu2 2σg2 2σu2 3σg2 .

7.11 The lowest energy excited MO configuration is 1σg2 1σu2 2σg2 2σu2 1πu4 3σg 1πg , which gives a term state symmetry of σg ⊗ πg = Πg . 186

The possible spins when there are two unpaired electrons are 0 and 1, so the multiplicities (2S + 1) are 1 and 3, respectively. The only term symbols that result from this configuration are 3 Πg and 1 Πg . 7.12 The ground state MO configuration of N2 is 1σg2 1σu2 2σg2 2σu2 1πu4 3σg2 . The excited state configuration is 1σg2 1σu2 2σg2 2σu2 1πu3 3σg2 4σg . The 4σg is the second lowest unoccupied MO; the 1πg is the lowest. The electronic states arising from this MO configuration we find from evaluating the possible values of the spin (all but two electrons are paired, so S = 0 or 1) and from the direct product for πu3 and σg : πu ⊗ σg = Πu πu3 will have the same symmetry as πu ; therefore the electronic state is 3 Πu , 1 Πu . 7.13 a. N+ 2 , 13 electrons

s= 1σg2 1σu2 2σg2 2σu2 1πu4 3σg

Bond order =

2s + 1 = |Λ| =

5 1 9−4= 2 2

b. N− 2 , 15 electrons

1 2 2 0

Σ+ g

1 2 2s + 1 = 2 |Λ| = +1 s=

1σg2 1σu2 2σg2 2σu2 1πu4 3σg2 1πg1

Bond order =

5 1 10 − 5 = 2 2

Πg

7.14 Following the dashed line for O2 in the correlation diagram from the bottom up gives the MO configuration: 1σg2 1σu2 2σg2 2σu2 3σg2 1πu4 1πg4 4σg2 3σu2 × 2πu4 5σg2 4σu2 2πg4 1δg2 or 1δg4 to get as high as the first δ MO. For the molecule to be charge-neutral, it must have 38 electrons. The neutral homonuclear diatomic with 38 electrons is formed from element 19, K. The molecule is K2 . 7.15 A δu MO must be antisymmetric with respect to a 90◦ rotation about the z axis and antisymmetric with respect to inversion. This means we need two nodes that contain the z axis (these are nodes for particular values of φ, the angle of rotation about the z axis) and we need a node through the horizontal mirror plane (which is a node at θ = 90◦ ). Looking back at Chapter 3, we find that to get two nodes in φ, we need an ml = ±2 atomic orbital. That’s because the real part of e±2φ is cos 2φ, which has nodes at φ = 45◦ and 135◦ . (Note also that cos 2φ is the same form as the character for rotation about the z axis for a δ orbital.) The symmetry for z axis rotation does not change from the united to the separated atom limit, so in either limit we are working with ml = ±2 orbitals. The lowest energy atomic orbitals that allow ml = ±2 are the 3d orbitals, and in the separated atom limit we can arrange the right inversion symmetry just by fixing the relative phases. In the united atom limit, we have a single atomic orbital with three angular nodes—two in φ and one in θ. That means l = 3, so the orbital is a 4f orbital with ml = ±2: 187

7.16 If Ne2 is so weakly bound, we can safely expect its bond order to be zero, which (using the correlation diagram) means the 3σu antibonding orbital is lower in energy than the 4σg bonding orbital, so the Ne2 configuration is 1σg2 1σu2 2σg2 2σu2 3σg2 1πu4 1πg4 3σu2 . To get NeF, we remove one electron, and label the orbitals by the corresponding C∞v representations (which just means we lose the inversion symmetry, so no g or u labels anymore): 1σ 2 2σ ∗2 3σ 2 4σ ∗2 5σ 2 1π 4 2π ∗4 6σ ∗ , where the *’s have been used to indicate antibonding orbitals. This predicts that there are ten bonding electrons and nine antibonding electrons, so the bond order is (10 − 9)/2 = 1/2. 7.17 For n = 5, there are atomic orbitals for l = 0, 1, 2, 3, 4. These correlate with molecular orbitals having |λ| ≤ l. For example, the l = 1 atomic orbital gives rise to σ and π molecular orbitals. The inversion symmetry of the MO is the same as for the corresponding atomic orbital: g if l is even, u if l is odd. γg ϕg δg

πg σg ϕu δu

πu σu δg πg

σg πu

σu σg

7.18 Li2 has six electrons and Be2 has eight. Filling the MOs each with two electrons from the bottom up, we get these ground state MO configurations: Li2 1σg2 1σu2 2σg2 Be2 1σg2 1σu2 2σg2 2σu2 The Li2 MO configuration has two full bonding orbitals and one full antibonding orbital. For Be2 we put two more electrons into the 2σu antibonding orbital, and therefore can reasonably expect it to be a less stable chemical bond. In fact, Be2 is too weak a bond to be normally considered a chemical bond. 188

2 2 1 7.19 Li+ 2 has five electrons, and the MO configuration is 1σg 1σu 2σg . There is one unpaired electron,

for a total spin of S = 1/2, so 2S + 1 = 2. The overall symmetry is the symmetry of the unpaired electron MO: 2 Σ+ g. 7.20 F2 has 18 electrons. Walking up the correlation diagram predicts a configuration 1σg2 1σu2 2σg2 2σu2 3σg2 1πu4 1πg4 . All the electrons are paired: S = 0; all λ’s cancel: Λ = 0. There are an even number of g and an even number of u electrons; so the overall inversion symmetry is g, and all λ 6= 0 MOs are filled. Therefore, the reflection symmetry is +. In fact, because all the MOs are filled, the result must be the totally symmetric representation anyway. The state is 1 Σ+ g. 2 2 2 2 7.21 B2 has ten electrons, B2+ 2 has eight electrons. Therefore, the MO configuration is 1σg 1σu 2σg 2σu . 1 The bond order is (4 − 4) = 0. 2

7.22 C2 has 12 electrons, of which eight are valence electrons. The possible valence MO configurations between O2 and H2 in the correlation diagram for eight electrons are: #bonding #antibonding a) 2σg2 2σu2 3σu2 1πu2 4 4 2 2 4 b) 2σg 2σu 1πu 6 2 1 1 (4 − 4) = 0 for (a) and (6 − 2) = 2 for (b). The correct lowest 2 2 1 3 13 energy valence MO configurations are πu4 Σ+ g (ground state), πu σu Πu (a low-lying excited state), and So the predicted bond orders are 3

πu2 σu2 Σ− g (at about 0.7 eV higher) [6], which are consistent with configuration (b). 7.23 FO has 9 + 8 = 17 electrons, so the MO configuration is 1σ 2 1σ ∗2 2σ 2 2σ ∗2 3σ 2 1π 4 1π ∗3 , by analogy to F2 : 1σg2 1σu2 2σg2 2σu2 3σg2 1πu4 1πg2 . The bond order is (10 bonding − 7 antibonding) = 2 All MOs are filled except three electrons in π ∗ :

λ=

π ∗3 −1 +1 ↑↓ ↑↓ ↑ ↓

↑ ↓ ↑↓ ↑↓

3 . 2

Π.

−1 −1 +1 +1

1 2 − 21 1 2 1 −2

Λ=± ⇒ Π Ms = ± 21 ⇒ S = 12 2S + 1 = 2

7.24 CF has 6 + 9 = 15 electrons. O2 : 1σg2 1σu2 2σg2 2σu2 3σg2 1πu4 1πg2 CF: 1σ 2 1σ ∗2 2σ 2 2σ ∗2 3σ 2 1π 4 1π ∗1 189

Using the vector model given below, we find that Λ = 1, S = 12 , and because the point group is C∞v ,

there is no inversion symmetry. The state is 2 Π.

λ=

1π ∗ +1 ↑ ↓

−1

↑ ↓

+1 +1 −1 −1

1 2 − 21 1 2 − 21

7.25 N2 has 14 electrons. According to the ordering for N2 MOs in the correlation diagram, the MO configuration is 1σg2 1σu2 2σg2 2σu2 1πu4 3σg2 . The next, lowest unoccupied MO is 1πg , and the configuration for that state would be 1σg2 1σu2 2σg2 2σu2 1πu4 3σg 1πg . This leaves two open MOs: 3σg (|λ| = 0, S = 1/2) and 1πg (|λ| = 1, S = 1/2). The lowest state for this configuration has highest spin, S = 1/2 + 1/2 = 1, 2S + 1 = 3, and Λ = 1. Therefore, the lowest state is labeled 3 Πg . 7.26 Ne2 has 20 electrons, and a ground state MO configuration 1σg2 1σu2 2σg2 2σu2 3σg2 1πu4 1πg4 3σu2 . 1 The bond order is (10 − 10) = 0; so the molecule is predicted to be dissociative (i.e., unstable). All 2 of the MOs are filled, so the ground state is 1 Σ+ g. 1 The lowest excited state will have the configuration [N2 ]1πg4 3σu 4σg ), with bond order of (11 − 9) = 2 1, which suggests the excited state will be bound. By analogy to 1σg 1σu H2 (or by the vector model), 3 + the corresponding electronic states are 1 Σ+ u and Σu . For every bonding orbital, there is a corresponding higher energy antibonding orbital. In the noble gas dimers, the number of electrons is such that these exactly balance, giving a bond order of zero. Therefore, the lowest excited state in these molecules is always formed by moving an electron from an antibonding orbital to a bonding orbital, so the bond order is always 1. 7.27 We need to realize that, because the π representation is doubly degenerate (containing both π+ and π− ), the MO configuration does not tell us exactly which orbitals are occupied, and this is why more than one overall wavefunction is possible. In order to obtain a final state with Σ symmetry, we need the values of λ for the two electrons to cancel, so let’s just consider the possibilities if we put one electron in the π+ and one electron in the π− , imposing the necessary symmetrization constraints: Ψa = [π+ (1)π− (2) + π− (1)π+ (2)] (αβ − βα) = P AB (1)P AB (2) eiφ1 e−iφ2 + e−iφ1 eiφ2 (αβ − βα)

Ψb = [π+ (1)π− (2) − π− (1)π+ (2)] (αβ + βα) = P AB (1)P AB (2) eiφ1 e−iφ2 − e−iφ1 eiφ2 (αβ + βα) .

In these equations, Ψa uses a symmetric two-electron spatial function, and therefore requires the antisymmetric two-electron spin wavefunction, while Ψb uses the antisymmetric spatial function and symmetric spin function. Now we operate on these two wavefunctions with one of the σ̂v operators, and for simplicity let’s select σ̂xz , which reverses the sign of y but leaves x and z unchanged. Recalling that 190

we may use Eqs. A.6 to convert between the polar coordinate φ and the Cartesian coordinates, we find that σ̂xz eiφ = σ̂xz (cos φ + i sin φ) x iy = σ̂xz + r sin θ r sin θ iy x − = r sin θ r sin θ

by Eq. 2.8 by Eqs. A.6 change sign of y

= (cos φ − i sin φ) = e−iφ .

by Eqs. A.6

The reflection operator does not alter the values of r or θ, which do not depend on the sign of y, and does not operate on the spin wavefunction at all. Operating on our two spin-spatial wavefunctions therefore yields σ̂xz Ψa = P AB (1)P AB (2) e−iφ1 eiφ2 + eiφ1 e−iφ2 (αβ − βα) = Ψa σ̂xz Ψb = P AB (1)P AB (2) e−iφ1 eiφ2 − eiφ1 e−iφ2 (αβ + βα) = −Ψb .

Both overall wavefunctions have Σ symmetry, because if we rotate either function by any amount δφ about the z axis, the effect is canceled; for example, e−i(φ1 +δφ) ei(φ2 +δφ) = e−iφ1 e−δφ eiφ2 eδφ e−iφ1 eiφ2 .

However, Ψa is symmetric under σ̂v and Ψb is antisymmetric under σ̂v . We have shown that by combining the two π orbitals, we are able to obtain an overall wavefunction that has Σ− symmetry, as well as a second state with Σ+ symmetry. Furthermore, we’ve shown that in this case the Σ− state is associated with the triplet spin function, whereas the Σ+ state accompanies the singlet spin wavefunction. 7.28 Our selection rules are ∆S = 0; ∆Λ = 0, ±1; and g ↔ u. The allowed transitions are 3 X 3 Σ− g ↔ D Πu

a1 ∆g ↔ e 1 Πu ↔ b1 Σ+ g.

7.29 The states shown below 10 eV are: 3 + 3 3 ′ 3 − 1 7 + 1 5 + X 1 Σ+ Σu , a ′ 1 Σ− g , A Σu , ∆u , B Πg , B u , a Πg , Σu , ∆u , Σg .

The selection rule ∆S = 0 means ∆(2S + 1) = 0, so we should consider only transitions between states of the same spin multiplicity (singlet ↔ singlet and triplet ↔ triplet): X 1 Σ+ g X 1 Σ+ g a ′ 1 Σ− u X 1 Σ+ g a ′ 1 Σ− u a 1 Πg

↔ a ′ 1 Σ− u ↔ a 1 Πg ↔ a 1 Πg 1 ↔ ∆u 1 ↔ ∆u 1 ↔ ∆u

A 3 Σ+ u A 3 Σ+ u 3 ∆u 3 ∆u B 3 Πg

↔ a 1 Πg 1 ↔ ∆u ′3 ↔ B Σu

A 3 Σ+ u 3 ∆u

3 ↔ ∆u ↔ B 3 Πg ↔ B 3 Πg ↔ B ′ 3 Σ− u ↔ B ′ 3 Σ− u

Of this set, the Σ ↔ ∆ transitions do not obey the selection rule ∆Λ = 0, ± 1. 1 The selection rule g ↔ u eliminates X 1 Σ+ g ↔ a Πg . 1 + ′ 1 − The selection rule ± ↔ ± eliminates X Σg ↔ a Σu . The allowed transitions remaining are: a ′ 1 Σ− u a 1 Πg B 3 Πg

191

↔ B 3 Πg ↔ B 3 Πg

7.30 FO2 has 9 + 2(8) = 25 electrons. All orbitals below 2b1 are filled; all above 2b1 are empty. The electronic state has B1 symmetry. With one unpaired electron, the spin must be 1/2, so that 2S +1 = 2. 2

B1 .

7.31 b2

πu

E σu

a1 a1

σg

You should find it possible to draw the correct lines just by looking at the trends in the shapes of the molecular orbitals as the bond angle is straightened out, but there is a more methodical technique. The symmetry operations that stay the same from one side of the diagram to the other include the C2v mirror plane reflection σ̂xz (which is equivalent to the σ̂v reflection in D∞h ) and the Ĉ2 proper rotation. The σg and a1 orbitals are symmetric under both operations, so the lowest energy of these are correlated. The σu , on the other hand, is antisymmetric with respect to the Ĉ2 and symmetric with respect to the σ̂v , which corresponds to b2 in the C2v limit. The other two orbitals in the C2v limit must therefore correspond to the πu degenerate pair in the D∞h limit. 7.32 BH2 has seven electrons: 1a21 2a21 1b22 3a11 . There is one unpaired electron: s=

2px

1b1 ( (( ( 3a1 2pz 2px

@ 2p y @

1 , 2s + 1 = 2 → 2 A1 2 @ @ 2p B

1s H + 1s H

1b2

1s H − 1s H

2a1

2s B

1a1 ( ( (( (

1s B

B @ @ @ H

z 6

separated atoms

united atom

7.33 H2 CO has 2 + 6 + 8 = 16 electrons and 12 valence electrons. We put them into the correlation diagram in the lowest excited state. The excited electron is much more stable when the molecule is non-planar; the other electron energies do not shift as much. The valence MO configuration is 2

1a′ 2a′ 1a′′ 3a′ 2a′′ 4a′ 5a′ . 192

The excited state is non-planar. b1 a’ 0

a’ a" a’ a"

b1 b a

2 1

a’

H C

planar

C O H non-planar H

0.4

E (har) 0.8

1.2

1.6

7.34 The MOs are sketched in the correlation diagram that follows, to show how the symmetry is obtained. There is one ambiguity: the D4h b representation may be b2g if the definitions of the mirror planes are reversed. The core 1s2 orbital on the carbon atom is the orbital that runs along the bottom of the correlation diagram, essentially unchanged in energy as the bonding orbitals move around. The e orbitals in D2d can be recognized by that representation because they are antisymmetric under the Ĉ2 (z) rotation, a characteristic that only the e representation has. The important thing from the standpoint of the correlation diagram is that the valence orbital energies rise continuously from the Td structure to the D4h structure, because the electron–electron repulsion is increasing in each case. The e orbitals remain degenerate through the D2d geometries to the D4h limit because they have one node each but in different planes. The D4h b1g orbital is substantially higher in energy because it has two nodes. H H H

H C

b1 g

H H

H e’u

H C

C H

H HH a1

H a1

H C

a1 g

H H

H Td

D2 d a1

D4 h a1 g

7.35 As the linear molecule bends, it becomes C2v , but symmetry with respect to Ĉ2 (z) rotation and 193

σ̂xz reflection will be unchanged: D∞h σg σu πu (z) πu (y) πg (z) πg (y)

Ĉ2 (z) σ̂xz +1 +1 −1 +1 +1 +1 −1 −1 −1 +1 +1 −1

C2v a1 b1 a1 b2 b1 a2

D3h a′1 e′ e′ a′′2 e′ e′′

7.36 eg

e’’

a2 u

a’’ 2

e’ a’1

a1 g

a’1

a1 g

D3 h

D3 d

The preserved symmetry elements throughout the motion are the Ĉ3 rotation and the three perpendicular Ĉ2 rotations. The horizontal and vertical mirror planes are lost, however, so the point group is D3 . Since the Ĉ3 symmetry holds throughout, what is an e representation in D3h stays e in D3d , and likewise the a representations remain a’s. Because the Ĉ2 symmetry holds, a1 correlates with a1 and a2 with a2 . Finally, because inversion reverses positions along the rotation axis, and so does horizontal reflection, xg correlates with x′ and xu correlates with x′′ . So a1g ↔ a′1 , a2u ↔ a′′2 , eg ↔ e′ , and eu ↔ e′′ . 7.37 The lowest energy MO will always allow maximum possibility for bonding, therefore having no nodes leaving the totally symmetric representation Γs . In D3h , this is the representation a′1 . 7.38 The ground state MO configuration is [ ]σ 2 or [ ]π 4 , and so on. The lowest excited MO configuration should therefore have a form such as [ ]σ 1 σ 1 or [ ]σ 1 π 1 or [ ]π 3 σ 1 . One electron is missing from the highest occupied MO (HOMO), and one electron in the lowest (normally) unoccupied MO (LUMO). Spin vectors for both electrons, ~s1 and ~s2 , can either sum together (S = 1) or cancel (S = 0). Therefore, there are both singlet and triplet states (2S + 1 = 1, 3) in the excited MO configuration. Fluorescence and phosphorescence can occur following excitation. 7.39 Given that Si2 , like C2 , will be capable of forming πu and σg bonding orbitals from the p valence electrons, let’s consider how those orbitals could result in the observed electronic states. To get the 3 Σ− g ground state, two electrons must be in the πu orbital, because two unpaired λ 6= 0 electrons are needed to get the σ̂v antisymmetry indicated by the “−.” Si2 has eight valence electrons, four of which will go into (3s)σg and (3s)σu orbitals that correlate with the 3s orbitals. If only two electrons are left to go into the πu orbital, that suggests that the (3p)σg orbital is lower in energy. The 3 Πu state therefore arises from a σg πu3 MO configuration. The problem with this σg2 πu2 Si2 ground state configuration is that, if we want to form HSiSiH, the available orbital for the H atoms to bond to is the πu , which does not have any density along the 194

molecular axis and therefore tends to form a weaker bond. C2 , on the other hand, has a ground state valence MO configuration 2σg2 2σu2 1πu4 (see Problem 7.22). This leaves the σg orbital formed from the 2p atomic orbitals available for bonding to the H atoms to form a linear molecule. By the way, a trans-bent form of HSiSiH is predicted to be stable [7], but could not be easily observed in the gas phase. However, two hydrogen-bridged isomers, corresponding roughly to SiH2 Si and SiHSiH, have had their geometries firmly established by elegant experiments [8, 9]. 3

7.40 The multiplicity S = 1/2 and the symmetry of e′ is the same as e′ , so the term is 2 E ′ . 7.41 The direct product is ag ⊗ au ⊗ bg . All MOs below 2au are filled. The point group is C2h . au ⊗ b g ⇒

au ⊗ bg

Ê 1 1

σ̂h Ĉ2 Iˆ +1 −1 −1 −1 +1 −1

−1 −1 +1

= Bu

Actually, two states are possible, because the electron spins in the au and bg MOs may be aligned or anti-aligned, yielding 1 Bu and 3 Bu states. 7.42 The point group is D3h ; the electronic state is the same as for (1e′ )1 . The representation = E ′ ; 1 there is one unpaired electron, so the spin: S = , 2S + 1 = 2 and 2 E ′ . 2 7.43 S = 32 means there are three unpaired electrons. NH2 has 7 + 2 = 9 electrons. The ground state is (1a1 )2 (2a1 )2 (1b2 )2 (3a1 )2 (1b1 )1 (4a1 )0 The quartet state is (1a1 )2 (2a1 )2 (1b2 )2 (3a1 )1 (1b1 )1 (4a1 )1 . The complete term symbol is 4 a1 ⊗ b 1 ⊗ a1 = b 1 . B1 . 7.44

a. Write the missing MO symmetry representations. Solution: The point groups are C2h (trans), C2 (transition state), and C2v (cis). The lower MO of the cis is b2 if the molecule lies in the xz plane, and b1 if in the yz plane. F

F F

0.20

E (har) 0.0

−0.20 au

b 2 or b 1

−0.40 F F trans

F F

F cis

195

b. What is the likely point group for the lowest excited state of the molecule? Solution: With two electrons in these two orbitals (the two π-bond electrons of the ethene), the lowest excited state promotes one of those electrons from the lower MO to the upper, antibonding MO. The much lower energy of the antibonding MO in the C2 geometry suggests that this would be the excited state geometry. 7.45 The MO configuration of CO in the ground state is 1σ 2 1σ ∗2 2σ 2 2σ ∗2 3σ 2 1π 4 and in the lowest excited configuration is 1σ 2 1σ ∗2 2σ 2 2σ ∗2 3σ 2 1π 3 1π ∗ . The ground state bond order is 3, but the excited state configuration bond order is 2, so the bond is weaker, and the excited state bond length is therefore longer. 7.46

a. What is the term symbol of the ground electronic state (X)? Solution: The only unfilled orbital has σ symmetry and one electron, so the term state symmetry is Σ+ and the spin S is 1/2, making the spin multiplicity 2S + 1 equal to 2. The term symbol is 2 Σ+ . b. What is the term symbol of the lowest excited electronic state (A)? Solution: The only unfilled subshell has π symmetry and three electrons, which will have the same term states as a π subshell with only one electron. Therefore, S = 1/2 again and 2S + 1 = 2, and the symmetry is Π. The term symbol is 2 Π. c. Is the X → A transition allowed by electric dipole selection rules? Solution: First we verify that the spin selection rule ∆S = 0 is satisfied (it is). Then we take the direct product Σ+ ⊗ Π = Π, which corresponds to the functions x, y. Therefore, the transition is allowed by electric dipole selection rules: yes. d. Is the X → A transition allowed by Raman selection rules? Solution: And since Π corresponds also to the functions xz and yz, the transition is also allowed by Raman selection rules: yes.

7.47

H C H

b. The point group is D2h . There are two b3u orbitals, one for the 1sA − 1sB core orbital and another for one of the C H σ-bond MOs. The b3u representation is symmetric with respect to Ĉ2 (x), so the x axis must be the axis containing the two carbon atoms. The b2g must correspond to the C H σ-bond MOs that are symmetric for rotation in the molecular plane, Ĉ2 (y), but antisymmetric under the other rotations, so the molecular plane is the xz plane. The two lowest energy MOs must be for the 1s core electrons, The π-bonding MO is unique, in that it is the only orbital that will be antisymmetric with respect to reflection through the plane of the molecule, ˆ xz , so that representation is the 1b2u . sigma C 2b23u A 1a2g 2 2 1b3u C 1b1u A 2a2g D or A 1b22g A 2 2 3ag A or D 1b2u B 196

7.48

H C C H

.C C

. H

From the linear to the bent structure, the symmetry elements of C2h are conserved. One of the σ̂v planes in D∞h becomes the σ̂h plane in C2h . One of the Ĉ2 axes in D∞h (perpendicular to the internuclear axis) becomes the principal Ĉ2 axis in C2h , and Iˆ is unchanged. Therefore, we can just keep track of the symmetry under Ĉ2 and Iˆ to find the correlating representation in C2h : group C H C H C C C C

Iˆ Γ(C2h ) Γ(D∞h ) Ĉ2 σg 1 1 ag σu −1 −1 bu σg 1 1 ag πu in-plane −1 −1 bu πu out-of-plane 1 −1 au

7.49 There are 7+2(8)=23 electrons in the molecule. They separate into groups that either lie in the xz plane, the plane of the molecule (core, σ bond, and lone pair electrons), or above and below the plane (π bond electrons, the unpaired electron). Those that lie in the xz plane must be symmetric under σ̂xz and are either a1 (no nodes) or b1 (a node along the yz plane). The others are either b2 (no yz node) or a2 (with the yz node). Given the choice, the lowest energy one will be the one without the yz node. They break down as follows: group # electrons Γ’s group # electrons Γ’s O 1s2 core 4 a1 ,b1 O lone pairs 8 a1 ,b1 ,a1 ,b1 N O π bond 2 b2 N 1s2 core 2 a1 4 a1 ,b1 O unpaired e− 1 a2 N O σ bonds N lone pair 2 a1 7.50

a. The point group is D2 . b. The lowest energy MO must have the symmetry of the totally symmetric representation, a. c. The molecule has an even number of electrons and no degenerate orbitals. Therefore, the ground MO configuration has all orbitals filled; the spin is 0 and the symmetry is again that of the totally symmetric representation: 1 A.

7.51

a. Biphenyl chromium is Cr(C6 H6 )2 and has an even number of electrons (106). The highest occupied MO has a2u symmetry, which makes it non-degenerate. The ground state of Cr(C6 H6 )2 will have two electrons in this orbital, and they must be paired (spins must cancel). This electronic state is therefore S = 0 and A1g (totally symmetric representation, since all MOs are filled), which we write 1 A1g . The point group is D6h . The lowest excited state will have an MO configuration . . . a1g a2u , so the spatial symmetry is given by a1g ⊗ a2u = A2u . The two unpaired electrons can have S = 0 or S = 1, and the lowest energy will be for S = 1, according to Hund’s first rule. Therefore, the lowest excited state is 3 A2u . b. Γi = a1g Γf = a2u Γi ⊗ Γf = a1g ⊗ a2u = A2u A2u is the representation for the function z. Therefore, the transition from the a1g to the a2u MO is allowed by electric dipole selection rules, but not by Raman. The transition between the electronic states 1 A1g and 3 A2u is forbidden by both electric dipole and Raman selection rules because the ∆S = 0 rule is violated. 197

7.52

a. 0 unpaired spins would give only S=0 (2S+1) = 1 2 unpaired spins would give S=0 and 1 (2S+1) = 1, 3 4 unpaired spins would give S=2 also. (If there were an odd number of unpaired spins, (2S+1) would need to be even.) Therefore, there must be two unpaired electrons. b. Only Λ = 0 and Λ = 2, so both electrons are in π MOs. c. Overall, ψ has u symmetry, so unpaired electrons must be in different MOs with opposite symmetry. MO configuration is πu1 πg1 .

7.53 We know that (2S + 1)max = 4 and therefore Smax = 3/2, so we must have 3 electrons in unfilled MOs. Λmax = 2 from 3 electrons, must mean that the representations of the unfilled orbitals are σ, σ, δ or σ, π, π. For the Σ electronic states we have Λmin = 0, which is not possible for σ + σ + δ, so the MOs are σ, π, π. The lowest excited state must be either σ 1 π 2 or π 2 σ 1 ; all terms are “g,” so must be σg but can be either πg2 or πu2 . The highest MOs are either σg1 πg2 , σg1 πu2 , πg2 σg1 , or πu2 σg1 . 7.54 The point group is C3v and the molecule is closed shell; therefore, S = 0 and Γ = Γs = A1 . The term symbol is 1 A1 . 7.55 HCO: 1 + 6 + 8 = 15 electrons; HCO+ : 15 − 1 = 14 electrons. Each MO holds up to two electrons. HCO’s highest occupied MO is 7a′ ; state is 2 A′ . HCO+ ’s highest occupied MO is 1a′′ ; state is 1 A′ . A′ is the totally symmetric representation, which is known because it is the representation of the lowest energy molecular orbital. 7.56

1. Find the molecule’s point group. Solution: The molecular plane must be a symmetry plane, but there are no other symmetry elements (except for Ê of course), so the point group is Cs .

2. Determine the term symbol for the ground state of the molecule. Solution: All the molecular orbitals are filled except for the in-plane orbital of the unpaired electron. This must be an a′ orbital (symmetric under reflection through the mirror plane) because it lies in the molecular plane, so the term state representation is A′ . The overall spin must be 1/2 because there is only one unpaired electron, so 2S + 1 = 2, and the term symbol is therefore 2 A′ . 7.57

a. Write the term symbol for the ground electronic state. Solution: All the orbitals are full, so the spin S is 0 (a singlet state) and Γ is the totally symmetric representation A1 : 1 A1 . b. Write the term symbol(s) for the excited electronic state(s). Solution: The spin can be either 0 or 1, because the two highest energy electrons are in separate orbitals and can either add up or cancel. To get the spatial symmetry, we take the direct product of the two partially filled MOs, keeping in mind that anything multiplied by a1 is unchanged: e ⊗ a1 = E. Therefore, the two possible term symbols are 1 E and 3 E. c. Write the term symbol for any of the excited states accessible by an allowed transition and indicate which selection rules are obeyed (“ED,” “R,” “EDR”). Solution: First, we must satisfy the spin selection rule ∆S = 0, so from the singlet 1 A1 ground state we cannot get to the 3 E excited state by an allowed transition. To see if we can get to 1 E, we find the direct product a1 ⊗e = e, which is the representation for the functions x, y, xz, and yz. Therefore, this transition is allowed by both electric dipole and Raman selection rules: 1 E, EDR. 198

7.58 There are 7 + 2 − 1 = 8 electrons: 1a21 2a21 1b22 3a21 .

1 NH+ A1 , all paired spins; so S = 0, 2S + 1 = 1. 2 ground state: First excited state: 1a21 2a21 1b22 3a11 1b11 Γtot = a1 ⊗b1 = b1 , because a1 is the totally symmetric representation. There are two unpaired electrons 1 1 1 1 in different orbitals, so S = − = 0 or S = + = 1. The excited states are 1 B1 and 3 B1 . 2 2 2 2

7.59 1a21 2a21 3a21 4a21 1b21 5a21 2b21 1b22 . H2 CO has 2 + 6 + 8 = 16 electrons; the configuration shown has fourteen electrons. The two highest energy electrons are either lone pair electrons or the π-bond electrons. The π-bond electrons will be the only ones antisymmetric with respect to σ̂xz and Ĉ2 – they will be b2 electrons. This representation is not in the MO configuration, so it must be the one left out. All the orbitals are filled, so the electronic state is 1 A1 . x H 6 qq @ -z Oq q C y H 7.60 The point group is C2v , with nine electrons. Two core electrons in carbon must be a1 ; two C D σ-bond electrons must be a1 , b1 for a given structure. The unpaired electron must be in py -orbital (since carbon has sp2 hybridization) = b2 . a21 a21 a21 b21 b2 . 7.61

1. the C C σ-bonding electrons. Solution: This is the toughest because there are 9 C-C σ bonds: the three that take part in the double bonds and the six in the single bonds. The first three split into an a1 orbital (no nodes) and a doubly degenerate pair of e orbitals (one node each) as for the BH3 example in Fig. 7.5. Similarly, the six others split into a1 (no nodes), e (one node), e (two nodes), and a2 (three nodes). The MO configuration is a21 e4 a21 e4 e4 a22 .

2. the C C π-bonding electrons. This is exactly analogous to the first three σ bonds above. (The fact that these are π bonds does not influence the symmetry representation in this case. The two sides of the π bond are not equivalent in this molecule; it would have to have more symmetry for that.) The MO configuration is a21 e4 . 3. the unpaired electrons. The orbital representations again are analogous to the first three σ bonds, but this time each orbital is only half filled. The MO configuration is a1 e2 . 7.62 This problem is asking for the energy differences between the ground state and various higher energy states. This corresponds to the distance along the energy axis in the potential energy diagram for H2 . a. from 0 eV to bottom of (σg 1s)(σg 2s) ≈ 11.0 eV b. from 0 eV to bottom of H+ 2 ground state ≈ 15.9 eV c. to H + H+ ≈ 18.7 eV 7.63 H2 has MO configuration 1σg2 and a bond order of 1. H− 2 , according to the correlation diagram, will have an MO configuration 1σg2 2σg , which has a bond order of 3/2. Therefore, we should expect H− 2 to have a smaller equilibrium bond length and deeper potential well than H2 . Furthermore, since the electron affinity for H atom is known (0.75 eV), we know the spacing between the H2 and H− 2 is the 199

electronic state for the 1σg2 2σg configuration. The total electron spin must be S = 2 + product rule gives Σ+ g for the symmetry, so the overall state term symbol is Σg :

1 , and the direct 2

E (eV) 8 b 3Σ + u 6

0.75 eV

H2 X 1Σ+g 2

0 0.4

0.8

1.2

1.6

2.0

2.4

H2 X 1Σg+

7.64

a. The O2+ molecule has 16 − 2 = 14 electrons: four in the 1s core orbitals, four more in 2 the 2σg and 2σu orbitals arising from the 2s atomic orbitals, leaving six to fill the three bonding orbitals (3σg and 1πu ) that come from the 2p orbitals. According to the general chemistry rules, the 3σg lies below the 1πu at the O2 position, although that rule becomes increasingly unreliable as we get further from the neutral molecules assumed by our diatomics correlation diagram. The MO configuration is therefore 1σg2 1σu2 2σg2 2σu2 3σg2 1πu4 . Because all the orbitals are full, the electronic state has singlet spin and has the symmetry of the totally symmetric representation in D∞h , 1 + Σg . This configuration and term symbol are the same as for the very stable, triply bonded N2 , + and this in large part explains the comparative stability of the O2+ 2 dication. Because O has the same number of electrons as for neutral atomic nitrogen, it has the same electron configuration, 1s2 2s2 2p3 , and it has the same ground state term symbol: 4 S3/2 .

b. The Coulomb potential energy is qA qB /(4πǫ0 rAB ), and for the example given that comes to Frep =

e2 (1.602 · 10−19 C)2 = −10 4πǫ0 R (1.113 · 10 C2 J−1 m−1 )(1.1 · 10−10 m)

= 2.097 · 10−18 J = 13.1 eV.

Going by Fig. 7.8, the attractive potential of N2 at R = 1.1 Å is about −10.0 eV. The sum of the attractive potential and the additional repulsion of the ionic charges is therefore approximately 3.1 eV. The actual value is 3.6 eV [10]. 7.65

a. What is the spin S of this state? Solution: 2S + 1 = 7, so S = 3. b. What is the minimum number of unpaired electrons in this electronic state? Solution: To get a total spin of 3 from electrons with spins of 1/2 each, there must be at least 6 unpaired electrons. (There could be more, in principle, because two unpaired electrons in different orbitals may have canceling spins.) c. What are the term symbols for the N atoms in the large-R limit? Solution: This state correlates with the ground states of the atoms, because it converges to the same energy at large R as the ground and lowest excited states. The ground state term of nitrogen is 4 S. d. What is the MO configuration for this state? Solution: To obtain a spin of 3, we need all six valence electrons to occupy separate orbitals: (1σg )2 (1σu )2 (2σg )2 (2σu )2 (1πu )2 (3σg )1 (1πg )2 (3σu )1 . 200

Each π MO is doubly degenerate, so we can put two electrons in 1πu , with one going in the πx and one in the πy orbital, for example. 7.66

a. Electric dipole selection rules only. Solution: The inversion symmetry will change with each electric dipole transition, so this can only be done in three steps. Any combination of 1 Ag → 1 Bnu → 1 Bmg → 1 Au will work as long as the two intermediate states are B representations with different numerical subscripts n and m. For example: 1 Ag → 1 B1u → 1 B2g → 1 Au . b. Any combination of Raman and electric dipole selection rules. Solution: This can be done in two steps with any combination of the form 1 Ag → 1 Bnu → 1 Au or 1 Ag → 1 Bng → 1 Au .

7.67 Photodissociation is likely to occur, because the 3 Σ+ u state is unbound. 7.68 We have two forms of the atom: He and He+ . We know the potential curve of He+ , because it is a one-electron atom: 4e2 . U (He+ ) = − (4πǫ0 )r The two-electron atom He requires some approximation, because the second electron exerts a repulsive force on the first; but we know that this will always raise the potential energy. We may approximate this using the effective atomic number Zeff = 1.69 obtained in Section 4.2. (This is very approximate, because that value changes as the distance r of our electron changes.) We know also that the ionization potential of He is the difference in energy between 1s2 He (−2.90 Eh) and 1s He+ (−2.00 Eh ). Therefore, the He curve is offset from the He+ curve so that the energy difference between these two states is 0.90 Eh:

r (a0) 2

E (har)

He + 2e-

He + e-

The arrow indicates the ionization of 1s2 He to 1s He+ . Finally, the energy difference between the two curves at large r is the ionization potential of He+ , 2 Eh . 7.69

a. Write the term symbol for this electronic state. Solution: All the orbitals are filled and all the electron spins cancel, so the representation is the totally symmetric representation, and S = 0. The point group is D∞h , so the term is 1 Σ+ g. b. The lowest excited state, A, of F2 is a 3 Πu term. Put an “X” next to any of the following, which would allow us to observe this state spectroscopically only by (iv) phosphorescence to the ground state from A. The spin must change, so no allowed transition connects the ground and A states. 201

7.70

a. requires at least three states: We label the ground state X, the excited state with the strong (allowed) transition at 6.0 eV A, and the state with phosphorescent emission (forbidden) the a state. A reasonable assumption is that the a state has a different spin from the X and A states, as this will make the phosphorescence transition strongly forbidden. b. requires a fourth, dissociative state that joins the X state at large R (because both will lead to ground state atoms).

c. requires a fifth, dissociative state that yields excited state O atom. We arrive at this state by a transition from the a state, and therefore can expect this new state to have the same spin as the a state. Combining (a), (b), and (c), we arrive at a potential energy diagram with a minimum of five states. The actual potential energy diagram follows. 4∆

A 2Σ +

N + O(1D ) N + O(3P)

a 4Π

2 X 2Π

0.8

1.2

1.6

2.0

2.4

R (A) 7.71

a. Estimate to within 1 eV the ionization energy of atomic oxygen. 14 eV b. Estimate to within 1 eV the minimum photoionization energy of NO from the X, v = 0 ground state. 9 eV c. Estimate to within 1 eV the minimum photodissociation energy from the X, v = 0 ground state. 8 eV d. If NO is formed in the b4 Σ− excited state, what are all of the likeliest processes (e.g., photoionization to X 1 Σ+ ) by which the molecule can lose its excess electronic energy? Give the term symbols for the final states of any processes you list. Solution: There are three possibilities, based on the drawn curves: (i) fluorescence to a followed by phosphorescence to X, (ii) phosphorescence directly to X, and (iii) predissociation by a level-crossing to D.

7.72

a. Fluorescence from A. 202

b. Phosphorescence from B after a level crossing. c. Predissociation, probably after a second level crossing from B to C. Other solutions are also possible.

E (eV)

C 6 4 2

X Π

0.8

1.2

1.6

2.0

2.4

2.8

R (A) 3 − 7.73 Photodissociation from 1 Σ+ u . Transition to the Σu is forbidden, so not likely. The transition to

the 1 Πu is allowed, but because the transition must occur vertically from v = 0, 8.0 eV is not enough energy to reach that state. The transition to the dissociating 1 Σ+ u state is both allowed and within the energy range of the photon. 7.74

a. fluorescence to 5 Σ+ g; b. phosphorescence to A3 Σ+ u;

c. phosphorescence to X 1 Σ+ g. There is only one state shown that could be the final state of an allowed emission, because only one state lower in energy has the same spin: the 5 Σ+ g state. Because this is an electric dipole–allowed transition, this is a fluorescence, and any path beginning with this step is more likely than a path that begins with a forbidden transition. From 5 Σ+ g , there are no allowed transitions to lower energy, because all the lower energy states are singlets or triplets, so we must violate the spin selection rule. The symmetry selection rule for electric dipole transitions will allow a Σ+ g state to connect only to a state or a Π state. There are no Π states shown, so the next transition is a phosphorescence Σ+ u u u 3 + (forbidden transition) to the A Σu , followed at last by phosphorescence to the X 1 Σ+ g. 7.75 The two curves have such strongly overlapping vibrational states that intersystem crossing from the A state back to the X is probably the most likely process. Fluorescence is also possible. 7.76 (c) is an allowed 2 Π → 2 Σ fluorescence transition. (a) and (b) both correspond to re-association of the AB+ ion with an electron, which requires the molecule to first find an electron, normally a slow process when compared to fluorescence. 7.77 The maximum kinetic energy of an ejected electron Ke will be the difference between the incident photon energy and the energy needed to ionize one molecule of the gas. Changing the speed to SI units of m s−1 , we have Ke ≥

1 (6.2 · 105 m s−1 )2 (9.109 · 10−31 kg) /(1.602 · 10−19 J/ eV) {z }| {z } 2| m

= 1.093 eV

hν = IE + (Ke )max IE = hν − (Ke )max = 3.9 eV. 7.78

a. ∆S = 0: A is a spin doublet. ∆Λ = 0, ±1: A is Σ or Π.

+ ↔ +: if A is Σ, it must be Σ+ : 2 Σ+ or 2 Π. There is no g or u term and therefore no inversion symmetry. The correct symmetry for the A state of BeF is 2 Π. 203

b. Because Re shifts to such a large value, it is difficult to excite the A state without adding enough vibrational energy that it photodissociates. Photodissociation or fluorescence are the most likely processes following excitation. 7.79 The last option in the table may look reasonable, but the A state of M has too large a bond length compared to the X state of M+ . Remember that the transitions must be represented by vertical lines, and these two states don’t allow that vertical line to connect the bound regions of the two curves.

E hν

a M hν XM

fluorescence from A M to X M fluorescence from a M to X M Raman transition from X M to X M via A M phosphorescence from A M to X M phosphorescence from a M to X M photodissociation of A M photodissociation of a M predissociation of A M via a M predissociation of X M+ via a M photoionization via A M to X M+

1 allowed going up, so allowed coming down forbidden because spin changes symmetry selection rule violated phosphorescence is for forbidden transitions the a state isn’t stable A is not a dissociating state can’t get to a directly from X 2 the fluorescence is more likely M and M+ are not the same molecule the two states are not vertically aligned

7.80 Photoionization. The only accessible curve at 10 eV above v = 0 is the X M+ curve. Because M+ is a different molecule from M, having a different number of electrons, no other interactions between X M+ and A M or B M are possible. 7.81 The function and its derivatives are U (R) = 1.5(R − 1.5)2 + 0.2(R − 1.8)4

dU (R)/dR = 3.0(R − 1.5) + 0.8(R − 1.8)3

d2 U (R)/dR2 = 3.0 + 2.4(R − 1.8)2 .

The equation we want to solve is dU (R)/dR = 0, so the Newton-Raphson method would be set up as the successive solution to the equations Rn+1 = Rn −

dU (Rn )/dR 3.0(Rn − 1.5) + 0.8(Rn − 1.8)3 = Rn − . 2 2 d U (Rn )/dR 3.0 + 2.4(R − 1.8)2

If we start from a guess Re value of 2.000, suggested by the problem, then we get the following series of projected Rn+1 values: R = 2.000, 1.464, 1.579, 1.560, 1.564, 1.563, 1.563. The series rapidly converges to the correct minimum. 7.82

a. The X state is the ground electronic state by definition (also it has Te = 0), and so it must come from the ground state MO configuration. The ground state configuration we can obtain from the correlation diagram for diatomics, filling 16 electrons in along the line drawn for O2 , or 204

from the periodic table. The electronic states are determined only by the unfilled orbitals in the configuration, in this case πg2 . What the table tells us first is that configuration B has a significantly weaker bond strength (lower ωe ) and longer bond length (lower Be ) than the ground state, while the bonds formed with configuration C are stronger and shorter than in the ground state. This implies that B has a lower bond order than A, while C has a higher bond order. Furthermore, B has the same electronic states as A except the inversion symmetry has changed from g to u. This means the configuration has the form πg πu (still two π electrons, but with 3 + different Iˆ symmetry). The configuration that yields two electronic states 1 Σ+ u and Σu we have seen before as one of the H2 excited configurations; it has the form σg σu . From the available orbitals, • A=

[Be2 ] 3σg2 1πu4 1πg2

bond order = 2,

• B=

[Be2 ] 3σg2 1πu3 1πg3

bond order = 1,

• C=

[Be2 ] 3σg2 1πu4 3σu1 4σg1

bond order = 3.

+ b. The ground state is a 3 Σ− u state. The electric dipole functions are represented by Σu and Πu in the D∞h point group.

• To avoid violating the ∆S spin selection rule, we only look at triplet states. • Σ ⊗ ∆ only gives ∆, so the ∆ states are ruled out. • g ⊗ g only gives g, so the g states are ruled out.

• Σ− ⊗ Σ+ only gives Σ− , so the Σ+ states are ruled out. No transitions are allowed. c. No transitions to the A ground configuration are allowed. The likeliest process is first a level crossing into the lowest energy state of the B configuration—the C 3 ∆u state. All of these states are bound states, because we have vibrational constants for them; therefore, no dissociation or predissociation is likely. From this state, there are no allowed transitions to the A configuration, so the molecule phosphoresces.

Chapter 8 8.1

a. D b.

Br, because the reduced mass is greater.

C N , because the bond is weaker.

c. I 1 Σ− CO, because the bond is weaker. 8.2 This is the same manipulation as we do for the MOs and even the atomic p orbitals. Because degenerate states have the same energy, by definition, any linear combination of degenerate states results in wavefunctions that still satisfy the Schrödinger equation. We’ve seen that states with motion around some symmetry axis always give rise to degenerate pairs, because the motion can be either clockwise or counterclockwise without affecting the energy. These will be eigenstates of the Hamiltonian and of the L̂z operator. However, we can simplify the physical interpretation of these two states by taking the linear combinations such that one state constrains its motion along the x axis and the other has 205

its motion along the y axis. Hence, we can form px and py orbitals from the pml =1 and pml =−1 atomic orbitals. Looking at it the other way, any linear combination of the two Cartesian bending modes (bending in the xz plane and bending in the yz plane) results in a circular or elliptical motion of the oxygen atoms in CO2 . This motion must have angular momentum. O

8.3

a. All three molecules are made from atoms with the same valence, so will have similar bonding. In that case, the more massive atoms will give the higher µ and lower k: MgO CaO CaS b. In this case, the reduced masses are similar but the k values increase from single- to double- to triple-bond: H3 CCH3 H2 CCH2 HCCH

8.4 The function has the shape shown below. The v = 1 and v = 2 excited states have 1 and 2 nodes, respectively. Because the walls are so steep, the distribution of energy levels in the well should be rather like the particle in a one-dimensional box, but because the potential energy beyond the wall is still finite, the wavefunctions tunnel into the wall (unlike the particle in a box):

8.5 Many answers are possible: a. Induce the v = 0 → 3 overtone transition in the ground state with one laser, then measure absorption of the second laser with transition to A′ Π state. This may be the most obvious way to make the measurement, but in practice it’s extremely difficult because the overtone transitions are usually very weak. That leads to a very low number of molecules in the correct initial state for the second transition. b. Induce the X → A transition directly and measure fluorescence to X v = 3. This is usually going to yield much stronger signals than the previous option, because both transitions are allowed electronic transitions. The drawback is that fluorescence occurs to a large number of vibrational levels in the X electronic state, so the final signal is still not very strong. 206

8.6

a. How many of these are translational coordinates? Solution: 3 b. How many of these are rotational coordinates? Solution: non-linear, so 3 c. How many of these are vibrational coordinates? Solution: non-linear, so 3Natom − 6 = 3 d. How many of these are electronic coordinates? Solution: three coordinates for each of ten electrons, so 30

8.7 The second derivative of the effective potential energy Eelec (R) describes the curvature of the potential—whether it’s “concave up” or “concave down” and just how steeply curved it is. The third derivative, therefore, tells us what is the change in curvature of the potential with R: 3 ∂ E ∂ ∂ 2 Eelec = . 3 ∂R ∂R ∂R2 SHO

Morse

Eelec is positive. At large R, as the At the equilibrium geometry, the potential is concave up and ∂ ∂R 2 curve asymptotically approaches zero, the potential has become concave down. Overall, the curvature for the potential decreases as R increases, so

∂3E < 0. ∂R3 But (R − Re )3 changes sign from negative (R < Re ) to positive (R > Re ). Therefore, the term in the Taylor series is 3 ∂ Eelec (R) > 0 : R < Re 3 (R − R ) e 3 < 0 : R > Re ∂R Re 8.8 increase v

change in ∆R As v increases, the vibrational energy increases and this increases ∆R. µ – Larger µ means a smaller ωe , which reduces the vibrational energy and ∆R. Re 0 By itself this does not affect ∆R; it is R − Re that matters, not Re . ωe xe + Greater anharmonicity flattens the potential curve, which increases ∆R. k – (See below) Increasing k does two things: it increases ωe (which increases ∆R) and it makes the potential curve narrower (decreasing ∆R). To see which is more important, we can solve for x = (R − Re )max , the upper classical turning point, in the harmonic oscillator: s 1/2 1 1 1 2h̄ k 2 k (v + ) x = √ (v + ) E(x) = x = ωe (v + ) = h̄ 2 2 µ 2 2 µk +

So as k increases, the upper turning point gets closer to Re , and ∆R decreases. 207

8.9 The rigorous way to figure this out is to use one of the recursion relations that appear in Table 8.1. These are equations that tell you how to get the next function in a series when you know the previous functions. In this case, we could use the equation y Hv (y) = vHv−1 (y) +

1 Hv+1 (y). 2

Taking v = 5 and v − 1 = 4, we can solve for the polynomial of v + 1 = 6: H6 (y) = 2 {yH5 (y) − 5H4 (y)} = 2 y 32y 5 − 160y 3 + 120y − 5 16y 4 − 48y 2 + 12 = 2 32y 6 − 160y 4 + 120y 2 − 80y 4 + 240y 2 − 60 = 2 32y 6 − 240y 4 + 360y 2 − 60 = 64y 6 − 480y 4 + 720y 2 − 120

On the other hand, you could also just look for the patterns. From the Hermite polynomials for v = 0 to 5, we note the following: a. Only even powers of y appear when v is even. b. The polynomial is always of order v in y. c. The terms alternate in sign. d. The coefficient of the first term is 2n . e. The coefficient of the last term repeats for v = 1 and 2, and again for v = 3 and 4, so can be expected to do so for v = 5 and 6. f. Now it gets harder. The coefficient of the second term is the first coefficient from the previous v times the v − 1 term in the triangle series 1,3,6,10,15,21,. . .. For example, the second term in v = 5 has coefficient of 16 (the first coefficient for v = 4) times 10 (the fourth term in the triangle series) = 160. So for v = 6 we can expect that coefficient to be 32 times 15 = 480. g. And even harder. In the even v, the next to last term has a coefficient larger than the last term’s by v, so for v = 6, once we know that the last term has a coefficient of 120, we can guess that the next to last term has a coefficient 6 × 120 = 720. So the solution is H6 (y) = 64y 6 − 480y 4 + 720y 2 − 120. 8.10 We need to write the v = 2 wavefunction and then integrate the square modulus to verify that the normalization constant we have yields one for the integral. 2

η2 (x) = A2 (4y 2 − 2)e−y /2 1/4 kµ y= x h̄2 2 1/4 h̄ dy dx = kµ Z ∞ Z ∞ Z ∞ Z ∞ 2 2 2 η2 (x)2 dx = A22 16y 4 e−y dx + A22 −16y 2 e−y dx + A22 4y 2 e−y dx −∞

−∞

= 4A22

2 1/4

h̄ kµ

Z ∞

−∞

208

−∞

y 2 e−y dy +

2 e−y dy .

y 4 e−y dy − 4

Z ∞

From Table A.5: Z ∞

Z ∞

3√ π 4 0 −∞ √ Z ∞ Z ∞ π 2 −y 2 2 −y 2 y e dy = 2 y e dy = 2 −∞ 0 Z ∞ Z ∞ √ 2 2 e−y dy = π e−y dy = 2 2

y 4 e−y dy = 2

−∞

Z ∞

−∞

y 4 e−y dy =

√ 1/4 √ 3√ π h̄2 4 + π π −4 kµ 4 x 2 1/4 √ h̄ 2 = A2 (8 π) = 1 if normalized kµ 1/2 1/8 kµ 1 √ . A2 = 8 π h̄2

η2 (x) dx = 4A22

8.11 From perturbation theory, Eq. 4.14 E1PT =

ψ0∗ Ĥ ′ ψ0 dτ D E = Ĥ ′

For our example, Ĥ ′ =

1 6

∂ 3 Uvib ∂R3

(R − Re )3 ,

so we would evaluate the integral Z D E 1 ∂3U vib ′ ψ0∗ (R − Re )3 ψ0 dR Ĥ = 6 ∂R3 Re The integral is zero because the wavefunctions are all either symmetric or antisymmetric about R = Re ; in other words, ψ0∗ (R − Re ) = ψ0∗ (R − Re )ψ0 (Re − R) . However, (R − Re )3 = −(Re − R)3 , so the integral can be written Z ∞ Z ∞ Z ∞ Z Re ∗ 3 ∗ 3 ∗ 3 ψ0∗ ψ0 (R − Re )3 dR ψ0 ψ0 (R − Re ) dR + ψ0 ψ0 (R − Re ) dR = − ψ0 ψ0 (R − Re ) dR + −∞

= 0.

Therefore, Ĥ ′ = 0. 8.12 The expression a(θ − θe )2 + b(θ − θe )4 describes a potential well whose walls are steeper than those of the harmonic oscillator. This will tend to make the energy level spacing increase with v, so the constant ωe xe should be negative: 2 1 1 1 − ωe xe v + . > ωe v + Evib ≈ ωe v + 2 2 2 2 1 In the limit of high v, this predicts Evib → ωe xe v + ≈ −ωe xe v 2 . This is the energy level 2 spacing predicted for the particle in a box, which has infinitely steep walls. Therefore, the ωe xe term over-compensates for the steepness of U (θ), and ωe ye will likely be negative also. 209

8.13 Classically, the vibrating atoms in a diatomic travel between a minimum and a maximum bond length (the classical turning points) limited by the values of R where the energy of the molecule E = U (R), the potential energy. Since E = K + U , the kinetic energy, and the rate of change of R, is smallest at the turning points. Since this is where R changes the slowest, the molecule will normally be found near the turning points. When R is between the turning points, the kinetic energy is high, so the molecule spends little time in this region. The wavefunction’s square modulus can also be said to be proportional to the time spent by the molecule at that point, so for a classical (high energy) wavefunction, the amplitude will be large near the turning points, and small in between.

8.14 This problem is looking for an estimate, not a detailed calculation. What ways do we have of finding momentum? Calculating hpi = hmvi precisely would be difficult here, because we would have to find the average speed of the atoms in the molecule. One way to estimate the momentum would be to estimate the average kinetic energy K as roughly Evib (v = 10)/2. The vibrational energy is split between kinetic and potential energy, and we’re estimating roughly half the energy is in each form. The average momentum can then be approximated: D√ E p hpi = 2mK ≈ mEvib (v = 10) s 1 = mωe v + 2 p = (1.66 · 10−27 kg)(2650 cm−1 )(1.986 · 10−23 J/cm−1 )(10.5) = 3.0 · 10−23 kg m s−1 .

A different estimate can be obtained by using the de Broglie equation to get the momentum, with the de Broglie wavelength given by the spacing between nodes in the vibrational wavefunction. For v = 10, there are ten nodes, roughly evenly spaced. λdB ≈

(1.80 Å − 1.20 Å) 0.60 Å = (n/2) 5

0.12 Å

λdB =

h , p

hpi =

6.626 · 10−34 J s h ≈ = 5.52 · 10−23 kg m s−1 λdB 0.12 · 10−10 m

The agreement between the two methods is fairly good for such crude estimates.

8.15 This is a simple problem on its face, but never underestimate the importance and potential dangers of coordinate transformations. The shapes of the two functions η(y) and η(R) are exactly the same. 210

The essential difference between η(y) and η(R) is the volume element that we use for the integral: dy versus dR. Because y is just a scaled version of the coordinate R, the difference in the volume element will just be a factor in the normalization constant: Z ∞ 1= |ηv (y)|2 dy −∞ Z ∞ 2/2 2 = A2v Hv (y)e−y dy dy = dR

−∞ 1/4

kµ h̄2

Z ∞

|ηv (R)|2 dR −∞ Z ∞ 2/2 2 ′2 = Av Hv (y)e−y dR

−∞

1/4 Z ∞ 2/2 2 h̄2 dy Hv (y)e−y kµ −∞ ! " #1/2 1/4 2 kµ A′v = A2v h̄2

2 = A′v

A′v = Av

kµ h̄2

1/8

8.16

1 2m

E = hK + U i = hKi + hU i 2 1 p + 12 kx2 = p2 + 12 k x2 = 2m 2m 1 ωe 1 = p2 + 21 k v + 12 = p2 + 12 ωe v + 12 = ωe v + 21 2m k 2m 2 1 1 p = 2 ωe v + 2

p2 =

2m ωe v + 12 = mωe v + 21 . 2

8.17 The equilibrium bond length is the bond length at the point where the potential energy reaches its minimum value, and the well-depth is the difference in potential energy between the equilibrium bond length and infinite bond length. It is natural, therefore, to find the equilibrium bond length first, and this is identified by taking the derivative of UM (R); the derivative will only equal zero at R = Re . h i2 d d De 1 − e−β(R−Re ) − De UM (R) = dR dR i d h = De 1 − 2e−β(R−Re ) + e−2β(R−Re ) − 1 dR h i = De 0 − 2(−β)e−β(R−Re ) + (−2β)e−β(R−Re ) − 0

= 0 at UM (R)

211

The equilibrium bond length corresponds to a critical point:

0 = 2βe−β(R−Re ) − 2βe−2β(R−Re )

e−β(R−Re ) = e2−β(R−Re ) R − Re = 2(R − Re ) so the minimum is at R = Re , Dissociation energy = well-depth

= UM (R → ∞) − UM (Re ) n o n o 2 2 = De 1 − e−∞ − 1 − De 1 − e0 − 1 n o n o 2 2 = De [1 − 0] − 1 − De [1 − 1] − 1

= De (0)2 − De (−1) = De .

8.18 We begin by solving for β from the Morse oscillator equation: 2 ∆R ≡ R − Re UM (R) = De 1 − e−β∆R − De −2β∆R −β∆R − 2e = De e r k β= . 2De This equation does not depend explicitly on thepvibrational energy; it depends only on ∆R. But at low vibrational energy, ∆R is small, so β∆R = k/(2De )∆R ≪ 1. That means we can simplify the exponential terms using the Taylor series expansion: 1 1 1 ex = 1 + x + x2 + x3 + . . . ≈ 1 + x + x2 , if x ≪ 1 2 3! 2 1 1 UM (R) ≈ De 1 − 2β∆R + (2β∆R)2 − 2 1 − β∆R + (β∆R)2 2 2 2 2 = De −1 + β ∆R k∆R2 −1 = De 2De =

1 k(R − Re )2 − De . 2

In other words, at low vibrational energy, the Morse potential energy curve has the same shape as the simple harmonic oscillator potential. 8.19 The maximum and minimum allowed values of R are those for which the energy De /2 (measured from the bottom of the well) is equal to U (R). For the simple harmonic oscillator (SHO), we can solve for δR easily in terms of k and De , because Re must lie exactly halfway between the minimum and maximum allowed R values: 2 1 1 1 1 USHO (R) = De = k(R − Re )2 = k δRSHO 2 2 2 2 r De . δRSHO = 2 k 212

For the Morse oscillator it is a little more work. The equation we use has De added to it so that the energy is referred to the bottom of the well, rather than to the dissociation limit. UMorse (R) = De

1 − e−β∆R

1 1 − e−β∆R = ± √ 2

2 o

1 De 2

1 e−β∆R = 1 ± √ 2 1 1 ∆R = − ln 1 ± √ β 2 These two values of ∆R (one for “+” and one for “−”) for the Morse oscillator correspond to Re − Rmin and Rmax − Re , respectively, so δR is obtained by subtracting one from the other: 1 1 1 δR = Rmax − Rmin = − ln 1 − √ ln 1 + √ β 2 2 r r 2De De = [.535 + 1.228] = 2.493 k k Morse 2.493 = = 1.246, SHO 2 so the Morse oscillator, because of its dissociative potential at large R, allows more freedom of motion at this energy than does the harmonic oscillator. 8.20 The initial vibrational energy is roughly ω(v + 1/2) = 3ω/2, and to this we add a photon energy hν. The molecule dissociates to A and B∗ , which lie at a combined energy DX + EB above the bottom of the ground state well. To conserve energy, the energy after adding the photon 3ω/2 + hν must be equal to the final energy DX + EB + Kex , so Kex = 3ω/2 + hν − DX − EB . 8.21 The vibrational constant increases with force constant k and decreases with reduced mass µ. The very low reduced mass of the diatomic hydrides tends to give them particularly high values of ωe , and 14 1 N H is no exception. The heavier isotope 14 N2 D will have about the same k value but a reduced mass lower by roughly 14·1

µND = 14+1 14·2 = 1.63. µNH 14+2 √ That means the 14 N2 D vibrational constant will be lower than ωe (14 N1 H) by a factor of 1/ 1.63, giving about 2600 cm−1 , still greater than 14 N2 . Of the N2 molecules, we’d expect 14 N2 to have the highest ωe , because it has the smallest reduced mass and a triple bond (so high k value). The cation N+ 2 removes an electron from a bonding orbital, so weakens the bond, reducing k and ωe . Then the added mass gives 15 N14 N+ the lowest ωe value of all. Therefore, the ordering is: a < b < e < d < c. 8.22 Properly, we should integrate the probability density from R = 1.99 Å to R = 2.01 Å. However, Gaussian functions don’t have nice analytical solutions when integrated over finite limits, so an approximation is handy here. Since the range of the integral is so small, let’s just take the value of the 213

probability density at R = 2.00 Å and multiply it by the range of the integral δR = 0.02 Å: 1/4 kµ y = (∆R) h̄2 2

ψ = AHv (y)e−y /2 # " 1/4 1/2 (∆R)2 4kµ kµ = exp − 2 πh̄2 h̄2

∆R = R − Re = 1.26 ± 0.01 Å

δR = 0.02 Å Z 1.27 Å |ψ(R)|2 dR ≈ |ψ(R)|2 δR P(∆R) = 1.25 Å " # 1/2 1/2 4kµ (1.26 Å)2 kµ = exp −2 (0.02 Å) 2 πh̄2 h̄2 k = 570 kg s−2 1 µ = (1.008 amu)(1.661 · 10−27 kg amu−1 ) = 8.367 · 10−28 kg 2 1/2 4(570 kg s−2 )(8.367 · 10−28 kg) P(∆R) ≈ π(1.055 · 10−34 J s)2 " # 1/2 (570 kg s−2 )(8.367 · 10−28 kg) −10 −20 2 × (0.02 · 10 m) exp − (10 m ) (1.055 · 10−34 J s)2 = 1.073 · 108 e−47.53 = 2.447 · 10−13 . 8.23 Given the vibrational frequency and molecular formula of a diatomic molecule, the harmonic force constant k can be calculated by solving Eq. 8.19: s k ( N m−1 ) −1 ωe ( cm ) = 130.28 µ ( amu) 2 ωe ( cm−1 ) [µ ( amu)] −1 . k(Nm ) = (130.28)2 ZrO YF

ωe (cm−1 ) µ(amu) k(N m−1 ) 937.2 13.58 703 630.9 15.65 367

8.24 ∆E ≈ 4ωe 1353.788 ωe ≈ = 338.4 cm−1 4 k = 2.54 · 10−27 kg = 1.53 amu µ= (2πcωe )2 For each atom: m = 2µ ≈ 3 amu. Since He does not form molecules with such strong bonds, the molecule is tritium molecule, 3 H2 . (The force constant is so low because the measurement is in one of the nearly dissociative excited electronic states.) 214

8.25 If the molecule is stable, there must be at least one vibrational level trapped in the well. This means that the zero-point energy must be less than the well depth, and that allows us (within the limits of the harmonic oscillator approximation) to place an upper limit on the force constant: s ωe 1 k = < 8 cm−1 2 2πc µ mHe µ= = 2.0015 amu 2 k < (32πc)2 µ = 0.030 N cm−1 . 8.26 To get the force constants from the vibrational constant, we just need to invert our Eq. 8.19 to solve for k: s s k k ( N m−1 ) 1 −1 ωe (cm ) = = 130.28 . 2πc µ µ ( amu) Ignoring whatever is attached to the carbons, the reduced mass is µ=

(12)(12) = 6 amu. 12 + 12

So we’re ready to go: k ( N m−1 ) =

C C C C C C

ωe ( cm−1 ) 130.28

ωe (cm−1 ) 1000 1600 2100

[µ ( amu)]

k ( N m−1 ) 353 905 1560

8.27 The vibrational constants for CO are ωe = 2169.82 cm−1, ωe xe = 13.29 cm−1, and ωe ye = 0.01 cm−1 . The last one, ωe ye , is so small that it can be neglected for this problem. For the transition v ′′ → v ′ , ∆E = Ev′ − Ev′′ = 4208.37 ≈ ωe (v ′ − v ′′ ) 4208.37 = 1.9 ≈ v ′ − v ′′ = 2 2169.82 Ev+2 − Ev ≈ ωe (v + 2.5) − ωe xe (v + 2.5)2 − ωe (v + 0.5) − ωe xe (v + 0.5)2 = 2ωe − (4v + 6)ωe xe

4208.37 − 2ωe = −131.27 ≈ − (4v + 6)ωe xe 131.27 = 9.88 ≈ 4v + 6 ωe xe v=1 The transition is v = 1 → 3. 8.28 The vibrational constant ωe depends on the reduced mass µ and the force constant k, s s 1 k k ( N m−1 ) = 130.28 . ωe ( cm−1 ) = 2πc µ µ (amu) Fortunately, the reduced mass is easy to calculate for a diatomic: µ=

(12.00)(14.00) m(12 C)m(14 N) = = 6.462 amu. m(12 C) + m(14 N) (12.00) + (14.00) 215

Plug this and the ωe value into the equation and solve for k: k ( N m−1 ) =

[µ ( amu)][ωe ( cm−1 )]2 [6.462][2033]2 = = 1573 N m−1 . (130.28)2 (130.28)2

8.29 The harmonic zero-point energy is ωe /2, so the zero-point energy classical turning points are the distances Rt where U (Rt ) = ωe /2. We need to solve for the separation between the Rt values. ωe 1 k(Rt − Re )2 = 2r 2 ωe (Rt − Re ) = ± k U (Rt ) =

where the positive value is for the maximum Rt and the negative value is for the minimum Rt . We’re interested in the difference: r ωe ∆Rt = 2 k s [ωe ( cm−1 )](100 cm/ m)(hc) =2 k( N m−1 ) s ωe ( cm−1 ) = 8.91 · 10−12 k( N m−1 ) r 4401.21 = 2.47 · 10−11 m = 0.25 Å ∆Rt (H2 ) − 8.91 · 10−12 570 r 2358.07 −12 ∆Rt (N2 ) − 8.91 · 10 = 9.04 · 10−12 m = 0.09 Å. 2290 The H2 , because its low reduced mass leads to a large zero-point energy and because it has a substantially lower force constant, has more than twice the ground state vibrational amplitude of N2 . 8.30 The point in this problem is that, if the shape of the potential curve (which determines k) does not change with isotopic substitution, then ωe ∝ µ−1/2 . ωe (35 Cl2 ) = 564.9 cm−1 s k 1 −1 ωe ( cm ) = 2πc µ µ(35 Cl2 ) = 17.5 amu µ(37 Cl2 ) = 18.5 amu 1/2 17.5 37 ωe (35 Cl2 ) ωe ( Cl2 ) = 18.5 = 549.4 cm−1 8.31 ωe = 2309.60 Evib (v = 2) = ground state Evib (v = 0) = Evib (v = 2) − Evib (v = 0) = = =

ωe xe = 39.36 1 − ωe 2 2 1 ωe − 2 2ωe − 2(2309.60) −

ωe ye = −0.09 2 1 ωe xe 2 + 2 2 1 ωe xe + 2 6ωe xe + 6(39.36) +

3 1 ωe ye 2 2 3 1 ωe ye 2 15.5ωe ye 15.5(−0.09)

4381.65 cm−1 216

8.32 From Table 8.2, looking only at molecules similar to ICl: Molecule 35 Cl35 Cl 79 Br79 Br 127 79 I Br 127 127 I I 127 35 I Cl

µ (amu) 17.48 39.46 48.66 63.45 27.44

k ( N m−1 ) 320 250 210 170

ωe (cm−1 ) 560.50 325.29 268.71 214.52

ωe xe (cm−1 ) 2.90 1.07 0.83 0.61

The force constant k for ICl should be between that of Cl2 and that of IBr, probably more similar to IBr’s. Estimate k is about 250 N m−1 . That predicts s r 1 h̄ k 2.5 · 105 −1 = 393 cm−1. ≈ ωe ( cm ) = ch µ 2πc 27.44 · 1.66 · 10−24 ωe xe will also lie between the values for Cl2 and IBr, say around 1.5 ± .5, again because ICl will be more like IBr than like Cl2 . ωe = 384.18 cm−1 The actual values are ωe xe = 1.51 cm−1 . 8.33 a. H2 O: 3 atoms, non-linear; 3N − 6 = 3 vibrational modes b. C2 H2 : 4 atoms, linear; 3N − 5 = 7 vibrational modes c. PF5 : 6 atoms, non-linear; 3N − 6 = 12 vibrational modes d. C6 H12 O6 : 24 atoms, non-linear; 3N − 6 = 66 vibrational modes 8.34

a. The molecule has 14 atoms and is non-linear. Therefore, it will have 3Natom −6 vibrational modes = 36. b. Hydrogen stretches will remain hydrogen stretches in either the normal mode or local mode picture. In the local mode picture, the number of C H bond stretching modes must equal the number of C H bonds, 10. c. Similarly, the total number of stretching modes is equal to the total number of bonds, (10 C H bonds) + (3 C C bonds) = 13 stretching modes. d. Any non-stretching motions must be bends—changing the bond angles rather than bond lengths. 36 − 13 = 23 bending modes. e. The lowest value of ωe should correspond to the largest reduced mass and lowest force constant. Therefore, we need the mode that allows the most atoms to move in the direction of least resistance. This must be a bend of the C C C C backbone:

C C

217

8.35 For the reduced mass, remember that we have a C H stretch where the H atom is connected only to the carbon while the carbon atom is connected to some huge compound. Therefore, the H atom is essentially vibrating against the rest of the molecule: mH mbig ≈ mH = 1 amu. mH + mbig

µ=

For the force constant, we could estimate k ≈ 6 · 102 N m−1 by comparison to HCl or by using C H stretching frequency of 3000 cm−1 : s k ( N m−1 ) −1 ωe ≈ 3000 cm ≈ 130.28 µ ( amu) k ≈ 530 N m−1 . 8.36 Degeneracies are in parentheses: v2 = 1 v2 = 2 v2 = 3 v2 = 4

950 1900 2850 3800

v1 = 1

3337

v3 = 1

3444(2)

v4 = 1 1627 (2) v4 = 1, v2 = 1 2577 (2) v2 = 2 3527 (2) v4 = 2 3254 (3) 8.37 There are 4 atoms, so there are 3 · 4 − 6 = 6 vibrational modes total. a. The number of stretching modes equals the number of chemical bonds. Because the two C H bonds are equivalent, they must be treated simultaneously, either as symmetric or antisymmetric stretches.

.. ..

b. That leaves three that are bends. Only two of these are motions in the plane of the molecule, because if we fix the value of each of the HCO angles, the HCH bond angle is fixed. The third bend is an out-of-plane bend. Applying the usual symmetry constraints to the two equivalent hydrogens (if one moves, the other must move in the equivalent direction or the opposite direction), we get these modes:

.. ..

+ .. ..

- C

H a1

H b1

H +

H b2

Note that the mode created by having one H atom move up while the other moves down is actually just a rotation about the z axis. 218

8.38 We know that for the a′2 mode, the displacement arrows must be symmetric under reflection through the molecular plane, which means the motions are all in-plane. Similarly, the a′′2 mode is antisymmetric under σ̂h , and can only involve out-of-plane motions. For the a′2 mode, we need the motion to be perpendicular to the Ĉ2 axes, so that the mode will be antisymmetric with respect to Ĉ2 . The motion drawn below for the Cl’s and H’s rocking in opposite directions accomplishes this. For this molecule, any out-of-plane motion that is symmetric under the Ĉ3 principal axis rotation will have a′′2 symmetry. H Cl

H Cl

+ Cl

−

− Cl

− H

H −

− −

Cl +

a’2

a"1

Other similar solutions exist. 8.39 Using the harmonic approximation: E(v1 , v2 , v3 ) = Eν1 (v1 ) + Eν2 (v2 ) + Eν3 (v3 ) = v1 ν1 + v2 ν2 + v3 ν3 relative to ground state. E(1, 0, 0) = (ν1 ) = 1200 cm−1 E(1, 1, 0) = (ν1 ) + (ν2 ) = 1200 + 63 = 1263 cm−1 E(0, 1, 2) = (ν2 ) + 2(ν3 ) = 63 + 2(2159) = 4381 cm−1 8.40

a. Evib = (1 × 1330.0) + (1 × 667.3) + (2 × 2349.3) = 6695.9 cm−1 b. σg ⊗ πu ⊗ σu ⊗ σu = Πu c. Πu is doubly degenerate; degeneracy = 2.

1 (3657 + 1595 + 3756) = 4508 cm−1. The bend is not degenerate and 2 is not counted twice, as it is for linear molecules. The levels below 6000 cm−1 are:

8.41 The zero-point energy is

(v1 , v2 , v3 ) E(cm−1 ) (0, 0, 0) 0 (0, 1, 0) 1595 (0, 2, 0) 3190 (1, 0, 0) 3657 (0, 0, 1) 3756 (0, 3, 0) 4785 (1, 1, 0) 5252 (0, 1, 1) 5351 8.42 In its lowest vibrational state, the molecule has some zero-point kinetic energy and is still vibrating 219

along each vibrational coordinate. E(zero−point) = E(v1 = v2 = v3 = 0) 1 1 1 1 = ωe1 + ωe2 + +ωe3 2 2 2 2 | {z } degeneracy

−1

= 2506.95 cm

8.43 Four atoms, linear 3N − 5 = 7 vibrational modes. Using local mode picture, each bond must have a vibration, so there is one C C stretch and two C H stretches. The remaining four modes must be two doubly degenerate bends. C H stretches lie in range 3000 − 3600 cm−1 1800 − 2200 cm−1 C C stretches lie in range bends lie in fingerprint region, roughly 400 − 1200 cm−1 Zero-point energy =

1 (ωe1 + ωe2 + . . . + ωe7 ) = 4700 − 7100 cm−1. The actual value is 6100 cm−1. 2

8.44 The 3444 cm−1 vibration is a hydrogen stretch, similar to the H F or H Cl stretch. Changing from reduced mass from approximately 1.0 amu to 2.0 amu, reducing ωe = p H to D will increase the √ h̄ k/µ by roughly a factor of 2. 3444 ωe (ND3 ) ≈ √ = 2435 cm−1 2

The actual value is 2564 cm−1 . 8.45 To diagonalize the 3 × 3 Hessian matrix by hand, one solves for the roots λ such that the determinant of H − λI is equal to zero. Here’s the equation to solve: |H − λ I| =

A−λ B D B C −λ B D BA − λ

= (A − λ)2 (C − λ) + 2BD2 − 2 (A − λ)B 2 − (C − λ)D2

= −λ3 + (2A + C)λ2 − (2AC + A2 − 2B 2 − D2 )λ + (A2 C + 2B 2 D − 2AB 2 − CD2 ) = 0.

One of the things we should realize at the outset is that we will get three solutions (because we have three coordinates: za , zb , and zc ), but one of these solutions will be the translational motion of the molecule along z. Because our energy equation does not depend on the absolute location of the molecule, the eigenvalue that goes with this motion must be zero. So there must be one root to our determinant equation for which λ = 0. That is only going to be a valid solution if the sum of the zero-order terms above, (A2 C + 2B 2 D − 2AB 2 − CD2 ), is also zero. These terms can be factored to (D − A)(2B 2 − AC − CD), and then it is sufficient to prove that (2B 2 − AC − CD) = 0. (By varying the coupling between bonds rab and rbc , a potential can always be constructed such that D and A are not equal, so it is not possible to require that D − A = 0.) Proving this mathematically is instructive, if you need reminding how derivatives work, but it’s also tedious, and that math has been pushed to the end of this solution. 220

Taking for granted that the zero-order terms vanish, we multiply through by −1 to get Eq. 8.31: λ3 − (2A + C)λ2 + (2AC + A2 − 2B 2 − D2 )λ = 0. One root of the equation is λ = 0, as promised. We divide the equation by λ, and at that point we need only the quadratic formula to find the remaining roots: 0 = λ2 − (2A + C)λ + (2AC + A2 − 2B 2 − D2 ) 1/2 o 1n λ= (2A + C) ± (2A + C)2 − 4(2AC + A2 − 2B 2 − D2 ) 2 1/2 o 1n = (2A + C) ± 4A2 + 4AC + C 2 − 8AC − 4A2 + 8B 2 + 4D2 ) 2 1/2 o 1n = (2A + C) ± C 2 − 4AC + 8B 2 + 4D2 ) . 2

This is correct, but we can simplify a little further, knowing that (2B 2 − AC − CD) = 0 from the logic above: 1/2 o 1n (2A + C) ± C 2 + 4D2 − 4(2B 2 − AC − CD + CD) λ= 2 1/2 o 1n (2A + C) ± C 2 − 4CD + 4D2 = 2 1/2 o 1n = (2A + C) ± (C − 2D)2 2 1 = {(2A + C) ± (C − 2D)} = A + C + D, A − D. 2

Finally, it remains to show that (2B 2 − AC − CD) = 0. (There must be an elegant symmetry argument to show that these terms vanish, but I haven’t found it.) The terms A, B, C, and D involve successive derivatives with respect to za and zb and zc , and to simplify the notation let’s use this shorthand: U (za + δz, zb , zc ) = (100) U (za , zb + δz, zc ) = (010) U (za , zb , zc + δz) = (001) U (za + 2δz, zb, zc ) = (200) and so on. According to the definition of the first derivative, as long as δz becomes infinitesimal, I may then write 1 ∂2U ma ∂za2 1 1 (200) − (100) (100) − (000) = − ma δz δz δz 1 (200) + (000) − 2(100) = ma δz 2 2 ∂ U 1 B=√ ma mb ∂za ∂zb 1 1 (110) − (100) (010) − (000) =√ − ma mb δz δz δz A=

221

1 (110) + (000) − (100) − (010) =√ ma mb δz 2 1 ∂2U mb ∂zb2 1 1 (020) − (010) (010) − (000) = − mb δz δz δz 1 (020) + (000) − 2(010) = mb δz 2 2 ∂ U 1 D=√ ma mc ∂za ∂zc 1 (101) + (000) − (100) − (001) =√ . ma mb δz 2 C=

1 1 (101) − (100) (001) − (000) =√ − ma mb δz δz δz

Because δz is vanishingly small, we should get the same values for these four parameters even if we add or subtract δz to all the za values in a sum, so (for example) (200) + (000) − 2(100) = (100) + (−100) − 2(000). Furthermore, shifting za by δz in one direction yields the same potential energy (and derivatives) as shifting zb and zc in the opposite direction, so (−100) = (011),

(111) = (000).

Lastly, the D∞h symmetry requires that (010) = (0 − 10),

(110) = (00 − 1) = (100).

These three operations allow us to put all four parameters in terms of (000), (100), (010), (011): 1 [(100) + (011) − 2(000)] ma δz 2 1 [(110) + (000) − (100) − (010)] B= √ ma mb δz 2 1 = √ [(000) − (010)] ma mb δz 2 1 1 C= [(010) + (0 − 10) − 2(000)] = [2(010) − 2(000)] mb δz 2 mb δz 2 1 [(111) + (010) − (110) − (011)] D= √ ma mb δz 2 1 = √ [(000) + (010) − (100) − (011)] . ma mb δz 2 A=

When we fold these results now into our zero-order terms from Eq. 8.31, we find they vanish: n 1 (2B 2 − AC − CD) = 2 [(000) − (010)]2 − 4 ma mb δz 2 [(010) − (000)] [(100) + (011) − 2(000) + (000) + (010) − (100) − (011)]} n o 1 2 = 2 [(000) − (010)] − 2 [(010) − (000)] [−(000) + (010)] = 0. ma mb δz 4

8.46 We can jump straight to our eigenvalues for the Hessian matrix (Eq. 8.32): λ = A − D, A + C + D, 222

and this time retaining the term in D get a new form for Eqs. 8.35 and 8.36: k −D mO (2mO + mC )k λ2 = A + C + D = + D. mO mC r k ω1 = −D mO s (2mO + mC )k + D. ω2 = mO mC λ1 = A − D =

We have the experimental values ω1 = 1333 cm−1 and ω2 = 2349 cm−1, and we have the masses, so there are two equations and two unknowns (k and D). This can be solved by standard algebra: k −D mO (2mO + mC )k ω22 = +D mO mC A mO ω12 = k − mO D mO mC mO mC B ω22 = k + D (2mO + mC ) (2mO + mC ) mO mC mO mC 2 2 D ω = −mO + A − B mO ω 1 − (2mO + mC ) 2 (2mO + mC ) (16)(12) −1 −1 = (16 amu)(1333 cm ) − amu (1333 cm ) (2 · 16 + 12) ω12 =

× (1.661 · 10−27 kg amu−1 )(2.998 · 108 m s−1 ) (16)(12) = −16 + amu D (2 · 16 + 12)

D = −0.560 N m−1 k = 33.4 N cm−1 .

What the result shows us is, well, first of all that we can expect that D is rather small compared to k. Despite all our talk about symmetry and the importance of treating equivalent bonds simultaneously, the two bonds really do function independently to a large extent. But there is some relationship between them, and in most cases – as with CO2 —the second derivative D that defines the coupling between the two bonds is negative. What this signifies is that if we lengthen one bond (for example, ∂zc > 0), then the energy derivative for shortening the other bond (∂za > 0) gets smaller (recall that as we’ve defined our coordinate system, increasing za brings atom a closer to atom b, so rab decreases). The derivative of the potential energy is the force, so by lengthening rbc it requires less force for rab to shrink. The experimental manifestation of this will be that as we break one of the C O bonds in CO2 , the other bond, on average, gets shorter. Chemically, this makes sense of course, because in the limit that we actually break one CO2 bond, the remaining CO molecule has a shorter bond length. This ratio between stretching vibrational constants can therefore tell us something about the electronic structure of the molecule. If D < 0 in a triatomic, then breaking one bond is liable to shift electrons into bonding orbitals of the remaining bond. We would expect the opposite behavior, with D > 0, if breaking one bond increased the electron density in antibonding orbitals of the remaining diatomic. 8.47 (a) For the symmetric stretch, the two oxygens have ~r position vectors in the same direction but p~ momentum vectors in opposite directions, so the cross products ~r × ~p are equal in magnitude 223

but opposite in direction, and cancel. The central atom has zero momentum vector, so its angular momentum ~r × p~ is zero as well. in the plane of your drawing, with a rotational angular momentum For the antisymmetric stretch, on the other hand, the oxygen atom momentum vectors point in the same direction, and there is a net angular momentum. For the carbon atom, the directions of both the ~r and p~ vectors are reversed, and so the angular momentum vector is parallel to those of the oxygens. The net angular momentum vector is perpendicular to the plane of the drawing, corresponding to an overall rotation of the molecule in the plane. Turning this result around means that the bending and the antisymmetric stretching modes of CO2 are coupled by molecular rotations. If the molecule is in an excited bending state, then making the molecule rotate at the same time leads to a tiny bit of antisymmetric stretch. This affects the vibrational energies and the spectroscopic transition strengths. (b) In general, for the Coriolis interaction to couple two vibrational modes νi and νj , we need the symmetry of the direct product Γi ⊗ Γj to contain the irreducible representation of the molecular rotation (functions Rx and Ry in the character table for D∞h ). The irreducible representation we need is therefore πg , so the general selection rule is Γi ⊗ Γj = πg ⊕ . . . . We can make this simpler by breaking it into one selection rule for the inversion symmetry and one for the vibrational angular momentum, l. To get the g inversion symmetry in the product, i and j must have the same inversion symmetry: g ↔ g, u ↔ u. And to get the π symmetry in the product, which corresponds to one unit of angular momentum, we need i and j to be different by one unit of angular momentum: ∆l = ±1. Mathematically, this selection rule occurs because the angle dependence of the representations for D∞h is found in the characters 2 cos λφ, and the trigonometric identity cos x cos y = [cos(x+y)+cos(x−y)]/2 ensures that you will only get π symmetry if the λ values in i and j differ by one. 8.48 We just need to put the momentum operators into derivative form, and take a first look. Notice, by the way, that we can keep this all in terms of the unitless coordinate y; there is no need to convert to x because the conversion factor c that appears in the operator x will be canceled by the same factor in the denominator of the momentum operator ∂/∂x. The equations in Table 8.1, φv−1 (y) 1 φv+1 (y) φv−1 (y) 1 φv+1 (y) dφv (y) y φv (y) = Av v + − = Av v Av−1 2 Av+1 dy Av−1 2 Av+1 allow us to write the results of these operations in terms of other harmonic oscillator functions: Z Z ∂ ∂ φv (y) dτ − φv (y) −ih̄ yφv (y) dτ hypy − py yi = φv (y) y −ih̄ ∂y ∂y Z φv−1 (y) φv+1 (y) dτ − = φv (y) y (−ih̄) Av v Av−1 2Av+1 Z ∂ φv−1 (y) φv+1 (y) − φv (y) −ih̄ Av v dτ + ∂y Av−1 2Av+1 Z 1 v y φv−1 (y) − yφv+1 (y) dτ = −ih̄Av φv (y) Av−1 2Av+1 224

∂ ∂ 1 − φv (y) φv−1 (y) + φv+1 (y) dτ Av−1 ∂y 2Av+1 ∂y Z 1 (v − 1) φv−2 (y) + φv (y) = −ih̄Av φv (y) v Av−2 2Av 1 v+1 1 − φv (y) + φv+2 (y) 2 Av 2Av+2 1 v−1 φv−2 (y) − φv (y) −v Av−2 2Av 1 v+1 1 − φv (y) − φv+2 (y) dτ 2 Av 2Av+2     v − (v + 1) + v − (v + 1) Z φv (y) φv (y) dτ = −ih̄Av  2Av   {z } | Z

   

Z Z v(v − 1) − v(v − 1) −1 + 1 φv (y) φv−2 (y) dτ + φv (y) φv+2 (y) dτ  Av−2 4Av+2  {z } | {z } | =0 =0 −2 = −ih̄. = −ih̄Av 2Av

This indeed is the expected answer for this operator, which we would know from seeing the general result from Problem 2.18. 8.49 The vibrational state symmetry representations are given by the direct product of the vibrational mode symmetries, a1 ⊗ a1 = A1 a 1 ⊗ b 1 = B1 b1 ⊗ b1 = A1 , so the only possibilities are A1 and B1 . 8.50 The point group is C2v , with the principal rotation axis passing between the two Br atoms and lying in the plane of the molecule. The vibrational motion is symmetric under Ĉ2 rotation about this axis and antisymmetric under either mirror reflection, so its representation is a2 . 8.51

a. The point group is D2h . Ĉ2 (z) Ĉ2 (y) (b) +1 +1 (c)

−1

Ĉ2 (z) Iˆ +1 +1 −1

−1

ag b1u

b. ag ⊗ b1u = B1u . 8.52

a. Ĉ2 ψv = ψv

σ̂xz ψv = ψv

b. Ĉ2 ψv = ψv

σ̂v ψv = −ψv

c. Ĉ2 ψv = ψv

σ̂v ψv = ψv

225

d. Ĉ2 ψv = −ψv

σ̂v ψv = −ψv

e. Ĉ2 ψv = −ψv

σ̂v ψv = ψv

f. Ĉ2 ψv = ψv

σ̂v ψv = ψv z y x

(a)

(b)

C C

a1 (d)

-C +

(f)

C C

+ H

a2 (e)

(c)

8.53 The point group is C2v . Both motions are antisymmetric with respect to reflection through the yz plane. Mode (a) is also symmetric under σ̂xz , whereas (b) is antisymmetric. The representations are (a) b1 and (b) a2 .

8.54 The number of vibrational modes is 3Natom − 6 = 6.

H O

O ag

H O

O bu

O H

H O

O ag

H O

O ag

O bu

O H

O au

8.55 226

C’2

C’’ 2

D6 h

D3 h

ˆ In the hexachlorobenzene, this mode is antisymmetric under the Ĉ6 rotation but symmetric under I, so this is either b1g or b2g , depending on how we define the perpendicular Ĉ2 axes or σ̂v planes. Defining the axes as shown, the representation would be symmetric under Ĉ2′′ but not Ĉ2′ , making it b2g . This cannot be IR active, because it is a g mode. For the trichlorobenzene, the motion is symmetric under the Ĉ3 principal axis rotation and σ̂v reflections, but antisymmetric under σ̂h , making it a′′2 . Checking the character table, we see that this corresponds to the z function, and is IR active. 8.56 D3

8.57 The molecule is in the point group C2h . The two modes have the following symmetry characters: ν1 ν2

Ê 1 1

Ĉ2 1 1

Iˆ −1 1

σ̂h −1 1

So the representations are au for ν1 and ag for ν2 . The v1 = 1 state has symmetry Au , v2 = 1 has symmetry Ag , and Au ⊗ Ag = Au , which is the representation for the function z, one of the electric dipole transition operators. Therefore, the transition Au → Ag is electric dipole–allowed. 8.58 The initial state is Γelec = A1 Γvib = B2 Γinitial = A1 ⊗ B2 = B2 . The final state is Γelec = B1 Γvib = b2 ⊗ b2 = A1 Γfinal = B1 ⊗ A1 = B1 . Γinitial ⊗ Γfinal = Γµ . B2 ⊗ B1 = (1 1 − 1 − 1) = A2 . The transition is electric dipole–forbidden , but is Raman–allowed. + + 8.59 The symmetry representations of the states are (a) Σ+ g , (b) Σu , (c) Σg ⊕ ∆g , and (d) σg ⊗ σu . The initial state (0,1,0) has Πu symmetry, which has allowed transitions only to Σg , Πg , and ∆g states. Therefore, only (a) and (c) are valid upper states for an allowed transition.

8.60 The transition energy depends on ωe and ωe xe . Therefore, to get ωe xe from the transition energy, we need to know the value of ωe . We can get ωe from µ and k, so we’re all set: (1.008)(12.00) = 0.9299 amu 1.008 + 12.00 s r k ( N m−1 ) 259.0 N m−1 −1 = 130.28 = 2174.3 cm−1 ωe ( cm ) = 130.28 µ ( amu) 0.9299 amu ∆E = ωe (v ′ + 21 ) − (v ′′ + 12 ) − ωe xe (v ′ + 21 )2 − (v ′′ + 21 )2 µ=

= ωe [(3/2) − (1/2)] − ωe xe [(9/4) − (1/4)] = ωe − 2ωe xe = 2046.3 cm−1

ωe xe =

(2174.3 cm−1) − (2046.3 cm−1) ωe − ∆E = = 64.0 cm−1 . 2 2 227

8.61

a. The walls near the equilibrium position are steeper than the walls of the harmonic oscillator near Re , so we expect a larger zero-point energy then the harmonic oscillator. However, the walls rapidly become flatter than the walls of the harmonic oscillator, which will tend to collapse the levels, so n = 2 and n = 3 are closer together than n = 1 and n = 2. b. The n = 1 state will have no nodes, the n = 2 will have one, and therefore the n = 3 should have two nodes. All of these wavefunctions should tunnel into the potential wall because the wall is not infinitely steep.

n= 3 n=2 n= 1 8.62 n=6 n=5 n=4 n=3

n=2

n=1

8.63 The wavefunction ψ(x) ∝ sin x is characteristic of the constant potential energy U (x), as either a free particle or a particle in a box. Because this is q normalized, this is a particle in a box (the integral R∞ 2 |ψ| dx diverges for the free particle). ψ(x) = a2 sin nπx , where n is a quantum number, a is a −∞ the length of the box. 1 r − 2 (2.0 Å) 2 = a 1/2 2 a = 2 Å = 4 Å.

The potential energy surface is a one-dimensional box 4 Å long.

nπx( Å) nπx = a 4 n = 4 · 8 = 32.

8πx( Å) =

8.64 The states can be divided into the n = 1 and n = 2 states, which are confined to a relatively small domain and which show some tunneling, and the n = 3 and n = 4 states, which have exactly the same limits and apparently do not tunnel (because the wavefunctions fall straight to zero instead of leveling off before reaching zero). This means that the potential must have walls that widen abruptly 228

at an energy between E2 and E3 . At that point, the walls become infinitely steep, therefore preventing tunneling in states n = 3 and n = 4.

n=4 n=3 n=2

n=1

Chapter 9 9.1 This is just a comparison of moments of inertia µRe2 . As the second atom goes from H to F to Br to I, the reduced mass and the bond length increase, so Be decreases. The trickiest is the isotope pair 85 Rb81 Br and 87 Rb79 Br, but assuming the same Re value, the reduced mass is found to be (85)(81)/(85+81) = 41.476 for 85 Rb81 Br and (87)(79)/(87+79) = 41.404 for 87 Rb79 Br. Having the larger reduced mass, 85 Rb81 Br must have the lower Be value. (Values are from Ref. [11].) 9.2

a. The isotopes determine the reduced mass, and therefore the vibrational constant ωe and the rotational constant B. b. The electronic state generally determines the bond strength, Re , and other parameters that affect both ωe and Be . c. Although the density may affect the intensities of the transitions, it will not affect the frequencies unless it is changed so much as to change the chemical environment of the molecules (for example, when a gas sample density is increased to the point the sample liquefies).

9.3 They are all forbidden, for the following reasons: a. 1 D2 →1 S0 . All atomic transitions have a selection rule ∆l = ±1. The electron configuration must change for an electric dipole–allowed transition. The transition indicated here amounts to a flip of one of the electronic orbital angular momenta, changing L. This is a change only in the magnetic field of the atom, not the electric field, and therefore is not induced by the electric field of the radiation. b. 1 Σg →3 Πu . The simplest, and perhaps most important, selection rule is ∆S = 0, which is violated by this S = 0 → 1 transition. c. v = 0 → 1 of the symmetric C2 H2 stretch. This is a σg stretch, and has zero dipole derivative. By group theory (σg → σg ) and by classical arguments (no change in electric field), this transition is forbidden. d. J = 0 → 2 in CO. This violates the ∆J = ±1 selection rule for pure rotational transitions. e. J = 1 → 2 in N2 . This is forbidden because N2 has no dipole moment, and therefore its rotations are unaffected by the radiation’s electric field. 229

9.4 Remember that the angular momenta in a molecule tend to couple together, just like the L and S angular momenta in the atoms in Section 4.4. In our selection rules for the perpendicular transition, J represents the total angular momentum, not just the rotational angular momentum. That’s because, as a practical matter, the magnetic field generated by molecular rotations is often so weak that it couples immediately to any magnetic fields from vibrational or electronic motion that may be present, making it difficult to experimentally distinguish the effects of rotation from those of other angular momenta. Therefore, when the vibrational angular momentum changes by one unit, J can also change, but by no more than one unit. For an example, let’s label the rotational and vibrational angular momentum ~ and G, ~ respectively. If we start from a state where G ~ = 0, then in that initial state J~ = R. ~ A vectors R ~ 6= 0. If G ~ is roughly perpendicular transition will then give us some vibrational angular momentum G ~ ~ ~ parallel to the original rotational angular momentum R, then J will increase. If G is roughly anti~ then J~ will decrease. And if G ~ is roughly perpendicular to the original R, ~ parallel to the original R, then J~ can stay the same. 9.5 µ=

Be ( cm−1 ) =

1 mS 2 = (31.97 amu) = 15.98 amu 2mS 2 16.858 16.858 = 2 2µ( amu) Re ( Å) (15.98 amu)(1.892 Å)2

= 0.295 cm−1 . 9.6 We take advantage of the fact that isotopic substitution changes the reduced mass of a compound but has little effect on the potential energy curve (and therefore the Re value). The masses of the isotopes are 34.97 amu for 35 Cl and 36.97 amu for 37 Cl, and the value of Be for 35 Cl2 comes from Table 9.1: Be =

h̄2 2µRe2

Be [35 Cl2 ] = 0.2441 cm−1 35 µ( Cl2 ) 37 35 Be [ Cl2 ] = Be [ Cl2 ] µ(37 Cl2 ) 34.97/2 = (0.2441) = 0.2309 cm−1 . 36.97/2 9.7 From Table 9.1, Be (X) = 60.8536 cm−1 Be (B) = 60.864 cm−1 1 Be ( J) Be (cm−1 ) = (100)hc h̄ = 400πcµRe2 1/2 h̄ Re = 400πcµBe ( cm−1 ) 1/2 1.055 · 10−34 J s Be ( cm−1 )−1/2 = 400π(2.998 · 108 m s−1 )(8.3683 · 10−28 kg) = (5.7824 · 10−10 )Be ( cm−1 )−1/2 230

Re (X) = 0.74125 · 10−10 m = 0.74125 Å Re (B) = 0.74119 · 10−10 m = 0.74119 Å. 9.8 The mass of 58 Ni is actually 57.93 amu, and the mass of 1 H is 1.008 amu, so the reduced mass of the molecule is (57.93)(1.008)/(57.93 + 1.008) = 0.9908 amu. The most convenient equation to draw on for this problem is Eq. 9.5, which solves the unit conversions to get Be ( cm−1 ) =

16.858 . µ( amu) Re ( Å)2

Solving for Re , we get

Re =

16.858 (0.9908)(7.700)

1/2

= 1.486 Å.

9.9 The rotational constants B0 and Be are not identical: Be is the rotational constant of a fictional rigid molecule with bond length equal to Re , and B0 is the observed rotational constant of the v = 0 molecule (and therefore includes effects from zero-point motion). However, the two values differ only by αe /2, and are typically within 1% of each other. Therefore, the bond length of a molecule can be estimated from B0 alone. 16.858 B0 ( cm−1 ) ≈ Be ( cm−1 ) = µ( amu) Re ( Å)2 s 16.858 . Re = µ( amu) B0 ( cm−1 ) The reduced masses are mC mN /(mC + mN ) = 6.444 amu for CN and m2I /(2mI ) = 63.45 amu.

CN I2

B0 ( cm−1 ) µ( amu) 1.9013 6.444 0.037389 63.45

R( Å) 1.173 2.666

9.10 Here we need to calculate the moments of inertia using Eqs. 9.12 and 9.13: rC = 0.7565 Å rN = 1.9035 Å mC = 12.000 amu mN = 14.003 amu

h̄ Be ( cm−1 ) = 4πcIa

1.513/2

1.147+1.513/2

com

2 2 Ia = 2mC rC + 2mN rN 2

= 115.209 amu Å 16.858 = 0.146 cm−1 . Be = 115.209 231

9.11 We can interpolate between the diatomics F2 and Cl2 : k(F2 ) = 4.4 · 102 N m−1

k(Cl2 ) = 3.2 · 102 N m−1 R(F2 ) = 1.418 Å

R(Cl2 ) = 1.988 Å µ(FCl) = 12.3 amu. Assume the force constant and bond length of FCl are the averages of the values above: k(FCl) ≈ 3.8 · 102 N m−1 and R(FCl) = 1.703 Å: r

3.8 · 102 N m−1 = 724 cm−1 12.3 amu 16.858 = 0.47 cm−1 Be ( cm−1 ) ≈ (12.3 amu)(1.703 Å)2 −1

ωe ( cm

) ≈ 130.28

The actual value is 0.51 cm−1 . 9.12 The a moment of inertia is by definition the smallest, and therefore the a axis is the axis to which the atomic masses are closest. The c axis is the axis from which the masses are most distant. This allows us to label the a and c principal axes in the molecules, and the b axis is the one left over. For example, in SF4 , the axis parallel to the axial FSF chain runs very close to those three atoms, and only the two equatorial F atoms are at any great distance. (Remember that distance is more important than mass in determining the moment of inertia.) Therefore, this axis is the a axis. For the other two axes, all four F atoms are significantly distant, but the equatorial FSF bond angle being greater than π/2 means those atoms are closer to the out-of-page axis than to the C2v rotation axis.

Cl H F

H C

F 102 F

H Cl

c a

a or b a

C Cl H

b or a

9.13 For the homonuclear diatomic X2 , the reduced mass µ is equal to m2X /(2mX ) = mX /2. h̄ 16.858 = 4πcµRe2 µ ( amu) Re ( Å)2 16.858 1 16.858 µ ( amu) = = 64.96 amu = mX = (0.0396)(2.56)2 2 Be ( cm−1 ) Re ( Å)2

Be ( cm−1 ) =

mX = 129.9 amu. This is the atomic mass for 130 Te, so the molecule is Te2 . 232

9.14 Choose one inertial axis to coincide with the Ĉ3 principal axis. y

Ha x

120

B Hc

1.210 A Our atoms have these coordinates:

B Ha Hb Hc

x 0 0 −1.048 +1.048

y 0 1.210 −0.605 −0.605

z 0 0 0 0

From Eq. 9.12: Izz = 3(1.008 amu)(1.210 Å)2 = 4.427 amu Å2 Ixx = (1.008 amu)[(1.210 Å)2 + 2(−0.605 Å)2 ] = 2.214 amu Å2 Iyy = (1.008 amu)[(−1.048 Å)2 + (1.048 Å)2 ] = 2.214 amu Å2 Check that the off-diagonal elements are zero (Eq. 9.13): Ixy = −(1.008 amu)[0 + (−1.048)(−.605) + (1.048)(−.605)] = 0 Iyz = −(1.008 amu)[0 + 0 + 0] = 0

Izx = −(1.008 amu)[0 + 0 + 0] = 0 So these are principal inertial axes. Since Izz is the largest, z is the c inertial axis, and because Ixx = Iyy , the molecule is an oblate symmetric top: 16.858 16.858 = = 3.806 cm−1 Izz 4.427 16.858 = 7.614 cm−1 . A=B= Ixx C=

9.15 The origin of our coordinate system is the carbon atom, the center of mass. A quick way to tell that the carbon is indeed the center of mass is to notice that it lies on at least two different proper rotation axes. Proper rotation axes must always pass through the center of mass, so the intersection of any two proper rotation axes is the center of mass. We choose the z axis to lie along one of the CCl bonds, which conforms to our usual convention of lining the z axis up with a principal symmetry axis. For simplicity, we’ll choose the x axis to lie in one of the ClCCl planes, and label the Cl atoms as shown. 233

Cl a

Cl d

C Cl d Cl c

Cl b

Cl c

Cl b

120 o y

It’s not too much work to find the coordinates of each of these atoms in this coordinate system, given the symmetry. The C atom is at the origin. The atom Cla is on the z axis one bond length from the origin. The Clb atom is in the xz plane, so yb = 0. Some trigonometry is required for the remaining coordinates, but keeping the symmetry in mind speeds this along. It’s useful to notice, for example, that all three Cl atoms b, c, and d have the same z coordinate, and that c and d have the same x coordinate and equal but opposite y coordinates. zb = R cos θ

xb = R sin θ

yb = 0 ◦

zc = zb

xc = xb cos(120 )

yc = xb sin(120◦ )

zd = zb

xd = xb cos(120◦ )

yd = −xb sin(120◦ )

The atomic coordinates in Å are given in the following table:

C Cla Clb Clc Cld

x 0.0 0.0 1.665 −0.833 −0.833

y 0.0 0.0 0.0 −1.442 1.442

z 0.0 0.0 −0.589 −0.589 −0.589

With the masses 12.00 amu for 12 C and 34.969 amu for 35 Cl, we can now calculate the moments of inertia along our coordinate axes: Izz = (12.00)(02 + 02 ) + (34.969)[(02 + 02 ) + (1.6652 + 02 ) + 2(0.8332 + 1.4422)] 2

= 290.9 amu Å

Ixx = (12.00)(02 + 02 ) + (34.969)[(02 + 1.7662 ) + (02 + 0.5892) + 2(1.4422 + 0.5892)] 2

= 290.9 amu Å

Iyy = (12.00)(02 + 02 ) + (34.969)[(02 + 1.7662 ) + (1.6652 + 0.5892) + 2(0.8332 + 0.5892 )] 2

= 290.9 amu Å

Izx = −(12.00)(0)(0) − (34.969)[(0)(1.766) + (1.665)(−0.589) + 2(−0.833)(−0.589)] =0

Ixy = −(12.00)(0)(0) − (34.969)[(0)(0) + (1.665)(0) + (−0.833)(−1.442) + (−0.833)(1.442)] =0

Iyz = −(12.00)(0)(0) − (34.969)[(0)(1.766) + (0)(−0.589) + (−1.442)(−0.589) + (1.442)(−0.589)] = 0.

234

This is a spherical top molecule; all the diagonal moments of inertia are equal, and it doesn’t matter which we call the a, b, or c axis. A=B=C= =

h̄2 2Izz 2

(1.055 · 10−34 J s)2

−1

2(290.9 amu Å )(1.661 · 10−27 kg amu−1 )(10−10 m Å

)

= 1.150 · 10−24 J = 0.0579 cm−1 .

9.16 Ethene is a prolate top, because it is possible to run the x axis through the two heavy carbon atoms, leaving only the light H atoms off the axis. The moments of inertia about the other two axes must be much smaller, because they will have contributions from the carbon atoms. So the a axis is the x axis. Between the y and z axes, we can see that the carbons are the same distance from both axes but the hydrogens are a little further from the z axis, so Iz will be a little larger than Iy , and therefore the c axis is the z axis. The bond lengths we need are the C3 C3 length of rC = 1.34 Å, and the C3 H length of rH = 1.08 Å. Arranging the coordinate system as given, the moments of inertia are Ix = 4mH [rH sin(60◦ )]

= 3.53 amu Å h r i2 rC 2 C + 4mH + rH cos(60◦ ) Iy = 2mC 2 2 2

= 16.7 amu Å 2 rC 2 rC 2 Iz = 2mC + 4mH + rH cos(60◦ ) + (rH sin(60◦ )) 2 2 2

= 17.6 amu Å .

From these values, we can get the corresponding rotational constants using Eq. 9.5: 16.858 = 4.78 cm−1 Ix 16.858 B= = 1.01 cm−1 Iy 16.858 C= = 0.96 cm−1 . Iz A=

B−A = A−B = −1. 9.17 For a prolate top, B = C, so k = 2B−A−B A−B 2B−B−C B−C For an oblate top, A = B, so k = B−C = B−C = +1. Therefore, the closer the value of k is to ±1, the more the molecule is like a symmetric top.

CH2 Cl2 COCl2 CH3 F C4 H4 O H2 O

k −0.982 −0.614 −1.000 +0.916 −0.491

near-prolate very asymmetric or near-prolate prolate near-oblate very asymmetric

9.18 The most obvious difference between the predicted and observed rotational constants is the much lower observed value of the A constant. The theoretical value is so high (900.0 GHz) because that 235

geometry is nearly linear, with only the light H atom far off the heavy-atom axis. Therefore, the observed value is consistent with the CCCO chain being bent rather than straight. The straight CCCO chain in the predicted structure indicates sp hybrids on the two righthand carbon atoms. So, if we look for structures that introduce an sp2 hybrid to one of these carbons, we’ll get a bent CCCO chain. The sp2 hybrid cannot go onto the middle carbon while keeping the formal charges to zero, but it can move from the first carbon to the third, giving the structure drawn below. This explains not only the much smaller observed A rotational constant, but also the slight increase in B and C constants as the heavy atoms move a little closer to the center of mass.

O 9.19 Because the a and c inertial axis labels change when we go from a prolate to an oblate top, let’s just use the Cartesian axis labels, with z the Ĉ3 symmetry axis. As θ changes, the molecule will remain a symmetric top, so Ix will always be equal to Iy . The value of θ we’re looking for is the one that makes Ix = Iz . So we need these two moments of inertia in terms of θ. The equation for Iz is relatively straightforward, because all three H atoms are always the same distance from the z axis, and the N atom lies on the axis so doesn’t contribute: Iz = 3mH (rNH sin θ)2 . Ix , on the other hand, is painful to calculate by hand, because we need to measure the distance of the N atom from the x axis. The x axis goes through the center of mass, not through the N atom, so we need the distance zN of the N atom from the center of mass: X mi z i = 0 i

0 = mN zN + 3mH (rNH cos θ − zN ) 3mH rNH cos θ zN = . mN − 3mH

Now, with a bit of trigonometry, we can write an equation for Ix in terms of the θ-dependent zN : io n h 2 2 2 2 . Ix = mN zN + mH (rNH cos θ − zN ) + 2 (rNH cos θ − zN ) + (rNH sin θ sin(π/3))

Setting the two moments of inertia equal to each other and solving for θ numerically yields θ = 0.924 radians or 53◦ . Maple. A set of Maple commands to solve this problem follows: • mH:=1.0079; mN:=14.01; rNH:=1.012; • rN:=theta->3*mH*rNH*cos(theta)/(mN-3*mH); • iz:=theta->3*mH*(rNH*sin(theta))ˆ2; • ix:=theta->mN*rN(theta)ˆ2 + mH*( (rNH*cos(theta) - rN(theta))ˆ2 + 2*( (rNH*cos(theta) - rN(theta))ˆ2 + (rNH*sin(theta)*sin(Pi/3))ˆ2 ) ); • solve(ix(theta)=iz(theta),theta); Mathematica. A set of Mathematica commands to solve this problem follows: 236

• mH=1.0079; mN=14.01; rNH=1.012; • rN[theta ] := 13*mH*rNH*Cos[theta]/(mN-3*mH); • iz[theta ] := 13*mH*(rNH*Sin[theta])ˆ2; • ix[theta ] := 1mN*rN[theta]ˆ2 + mH*( (rNH*Cos[theta] - rN[theta])ˆ2 + 2*( (rNH*Cos[theta] - rN[theta])ˆ2 + (rNH*Sin[theta]*Sin[Pi/3])ˆ2 ) ); • Solve[ix[theta]==iz[theta],theta]; 9.20 We’ll use Eq. 9.22 for the energy, and the ∆J = ±1, ∆Kc = 0 selection rules: Erot = Kc2 (C − B) + BJ(J + 1)

′ ′′ ∆Erot = Erot − Erot

= Erot (J + 1, Kc) − Erot (J, Kc )

= [Kc2 (C − B) + B(J + 1)(J + 2)] − [Kc2 (C − B) + B(J + 1)]

= B[J 2 + 3J + 2 − J 2 − J] = B(2J + 2)

= 2B(J + 1). This equation is the same for symmetric tops as for linear molecules, which in some sense are a special case of the symmetric top. 9.21 For the cases we are considering, predicting the spectrum of a molecule is a matter of knowing the molecular constants (e.g., B0 and D0 ) and the relevant equation. For a pure rotational transition, only the rotational constants are necessary. D0 is given in the problem, and B0 is available from Table 9.1. hν = E ′ − E ′′ = BJ ′ (J ′ + 1) − D[J ′ (J ′ + 1)]2 − BJ ′′ (J ′′ + 1) − D[J ′′ (J ′′ + 1)]2

J = 0 → 1,J ′′ = 0, J ′ = 1

hν = 2B − 4D − 0 = 2B − 4D

J = 10 → 11,J ′′ = 10, J ′ = 11 hν = 132B − (132)2 D − 110B − (110)2 D = 22B − 5324D

CN: B(MHz) = B( cm−1 )c/106 = 56999MHz D = 0.0078MHz I2 : B(MHz) = 1120.9 MHz D = −1.36 · 10−4 MHz CN I2

J = 0 → 1 J = 10 → 11 113998MHz 1253936MHz 2241.8MHz 24661MHz

9.22 This is a prolate top, because the point group is C3v and the heaviest atoms *C and F) lie on the symmetry axis, so B = C and Erot = (A − B)Ka2 + BJ(J + 1): ′ Erot ′′ Erot

= =

∆Erot

(154000 − 25536)22 (154000 − 25536)02 513856

+ +

(25536)(3)(3 + 1) (25536)(2)(2 + 1)

153216 237

= 667072 MHz

9.23 The lowest transitions are J = 0 → 1 and J = 1 → 2. ∆Erot = E(J + 1) − E(J) = B(J + 1)(J + 2) − D[(J + 1)(J + 2)]2 − BJ(J + 1) − D[J(J + 1)]2 = 2J ′ B − 4J ′ 3 D

∆E(1 → 2) = 4B − 32D = 4950.312 MHz ∆E(0 → 1) = 2B − 4D = 2481.384 MHz

∆E(1 → 2) − 2∆E(0 → 1) = −24D = −12.456 MHz −12.456 D= = 0.51900 MHz. −24

9.24 The R(0) and R(1) in a rovibrational spectrum are spaced by approximately 2Be : ∆E[R(1)] − ∆E[R(0)] = 503.065 − 502.589 cm−1 = 0.476 cm−1 ≈ 2Be

0.476 = 0.238 cm−1 = 7135 MHz 2 The v = 0 pure rotational transitions are measured at 14467.9 and 28935.7 MHz. Since the rotational energy is roughly Be J(J + 1), the rotational transition energy in any vibrational state is roughly Be ≈

∆Erot ≈ Be [(J + 1)(J + 2) − J(J + 1)]

= Be [J 2 + 3J + 2 − J 2 − J] = Be (2J + 2),

where the transition is from J to J + 1. ∆Erot 14467.9 MHz ≈ Be 7135 MHz = 2.03 ≈ 2 = 2J + 2 if J = 0. This is the J = 0 → 1 transition. 28935.7 MHz ∆Erot ≈ Be 7135 MHz = 4.06 ≈ 4 = 2J + 2 if J = 1 This is the J = 1 → 2 transition. Therefore, we can find B0 : B0 =

14467.9 ∆Erot = = 7233.95 MHz (2J + 2) 2

if the distortion is neglected. We can check this using the J = 1 → 2 transition, which would have a different distortion correction: ∆Erot 28935.7 B0 = = = 7233.93 MHz (2J + 2) 4 so our neglect of distortion is good to one part in 105 , which is not surprising because distortion is not 1 a large effect at low J. Knowing B0 gives us one equation: B0 = Be − αe = 7233.9 MHz. Knowing 2 the spacing between R(0) and R(1) gives us another; from Eq. 9.27: ∆E[R(1)] = ωe − 2ωe xe + 4Be − 8αe ∆E[R(0)] = ωe − 2ωe xe + 2Be − 3αe ∆E[R(1)] − ∆E[R(0)] = 0 − 0 + 2Be − 5αe 238

= = =

503.065 cm−1 502.589 cm−1 0.476 cm−1

v ′ = 1, J ′ = 2 v ′ = 1, J ′ = 1 = 14270MHz

1 These are our two equations: Be − αe = 7233.9 MHz and 2Be − 5αe = 14270 MHz. Multiplying the 2 first equation by 2 and subtracting the second leaves: 4αe = 197.8 MHz αe = 49.45 MHz = 1.65 · 10−3 cm−1 . 9.25 This is an example of a more difficult problem in spectroscopy: assigning the upper and lower state quantum numbers to the transitions, and determining the molecular constants. It helps in such a problem to recognize the important terms in the equation for the transition energies; for example, D0 is not going to contribute more than a fraction of a percent to the rotational transition energy at low J. The molecular constants can not be determined until the quantum numbers are known—until the spectrum is assigned. The quantum numbers are determined by exploiting the fact that they are restricted to certain values. In this case, we know that the rotational quantum numbers must be whole numbers, and that in a pure rotational spectrum the selection rule is ∆J = +1. ′′ ′ J → J + 1 : Erot − Erot = [B(J + 1)(J + 2) − D[(J + 1)(J + 2)]2 ]

− [BJ(J + 1) − D[J(J + 1)]2 ]

= B[(J + 1)(J + 2 − J)] − D[(J + 1)2 ((J + 2)2 − J 2 )]

= 2B(J + 1) − D[(J + 1)2 (J 2 + 4J + 4 − J 2 )]

= 2B(J + 1) − 4D(J + 1)3

≈ 2B(J + 1)

4 2B(J1 + 1) J1 → J1 + 1 : 12835.20 = 1.331 ≈ = J2 → J2 + 1 : 9640.68 3 2B(J2 + 1) J1 = J2 = 12835.20 = 9640.68 =

3 → 2 →

4 3

8B − 4(43 )D = 8B − 256D(×6) 6B − 4(3)3 D = 6B − 108D(×8)

−114.24 = 0.17 MHz −672 −672D = (12835.20)6 − (9640.68)8 = −114.24MHz D=

B = (12835.20 + 256D)/8 = 1609.84 MHz 9.26 First, let’s review those definitions: fundamental: v = 0 → 1 P (1): J = 1 → 0 R(0): J = 0 → 1 . Now we’ll get the transition energies by adding the rotational and vibrational contributions to the upper 239

and lower states of each transition: hν = E ′ − E ′′

′ ′ ′′ ′′ = (Evib + Erot ) − (Evib + Erot )

Evib = ωe (v + 1/2) − ωe xe (v + 1/2)2

Erot = Be J(J + 1) − αe J(J + 1)(v + 1/2)

P (1) :E ′ ⇒ ωe (3/2) − ωe xe (3/2)2 + Be (0) − αe (0)(3/2)

E ′′ ⇒ ωe (1/2) − ωe xe (1/2)2 + Be (2) − αe (2)(1/2)

hν = ωe − ωe xe (2) − Be (2) + αe

R(0) :E ′ ⇒ ωe (3/2) − ωe xe (3/2)2 + Be (2) − αe (2)(3/2)

E ′′ ⇒ ωe (1/2) − ωe xe (1/2)2 + Be (0) − αe (0)(1/2) hν = ωe − ωe xe (2) + Be (2) − αe (3)

ωc (cm−1 ) H Cl 2989.74 IBr 268.4 35

ωe xe (cm−1 ) Be (cm−1 ) αe (cm−1 ) P (1) R(0) 52.05 10.591 0.302 2864.02 2905.92 0.78 0.0559 0.0002 265.06 266.95

9.27 We need the upper and lower state quantum numbers, and then the energy equations: ∆E = E ′ − E ′′ v=0→1

J =1→2

E ′ = ωe (1 + 21 ) − ωe xe (1 + 12 )2 + Be (2 · 3) − αe (2 · 3)(1 + 21 )

E ′′ = ωe ( 12 ) − ωe xe ( 21 )2 + Be (1 · 2) − αe (1 · 2)( 12 )

∆E = ωe − (−2ωe xe ) + 4Be − 8αe

= 2309.60 − 2(39.36) + 4(3.2535) − 8(0.0608)

= 2243.41 cm−1. 9.28 The wavenumber unit, cm−1 , functions as an energy unit because it is simply the reciprocal of the photon wavelength, which is proportional to the photon energy: ∆E[R(J ′′ )] = ωe − 2v ′ ωe xe + 2J ′ Be − J ′ (J ′ + 2v ′ )αe R(4) J ′′ = 4

J ′ = J ′′ + 1 = 5

v=0 → 1

v′ = 1

∆E[R(4)] = 2989.74 − 2(1)(52.05) + 2(5)(10.5909) − (5)(5 + 2)(0.019) = 2990.88 cm−1

∆E( J) = ∆E( cm−1 ) h( J s) c( cm s−1 ) 1 λ( cm) = ∆E(cm−1 ) = 3.34350 · 10−4 cm = 3.34350 µm. 9.29 We can see that ∆J = −2 because the rotational term is roughly −4J ′′ Be . We know that the rotational lines in a ∆J = −1 spectrum decrease in energy by roughly 2Be as J ′′ increases by one, so the ∆J = −2 spectrum should have lines dropping by 4Be as J ′′ increases. We can also tell that 240

v ′ = v ′′ + 2, because the vibrational transition energy is roughly 2ωe . To find v ′′ , we need to check the ωe xe term: ∆Evib = Evib (v ′ ) − Evib (v ′′ ) 1 1 1 1 ωe (v ′′ + 2 + ) − ωe xe (v ′′ + 2 + )2 − ωe (v ′′ + ) − ωe xe (v ′′ + )2 2 2 2 2 = 2ωe − ωe xe (4v ′′ + 6) = 2ωe − 10ωe xe .

So 4v ′′ + 6 = 10, meaning v ′′ = 1, v ′ = 3. 9.30 This is the typical problem in spectroscopy we’ve encountered in earlier chapters: we can measure the energies, but then we have to find a unique assignment of the quantum numbers. The big help is that these quantum numbers must be (in this case) integers. The rest is algebra: ∆Erot = Bv [J ′ (J ′ + 1) − J ′′ (J ′′ + 1)] = Bv [7(7 + 1) − 6(6 + 1)] = 14 Bv 105.33 = 7.5236, Bv1 = 14 1 Bv = Be − αe (v + ) 2 8.3511 − 7.5236 1 − = 3 v1 = 0.226 2

102.17 = 7.2979 14 1 Be − Bv − v= αe 2 8.3511 − 7.2979 1 − = 4 v2 = 0.226 2

Bv2 =

9.31 We want the energy splitting between P (J + 1) and P (J) ≡ ∆P , and R(J) and R(J − 1) ≡ ∆R: R(J) ⇒ ∆Evib + Bv+1 (J + 1)(J + 2) − Dv+1 [(J + 1)(J + 2)]2 − (Bv J(J + 1) − Dv [J(J + 1)]2 )

R(J − 1) ⇒ ∆Evib + Bv+1 J(J + 1) − Dv+1 [J(J + 1)]2 − (Bv (J − 1)J − Dv [(J − 1)J]2 )

∆R = 0 + Bv+1 2(J + 1) − Dv + 1(J + 1)2 ((J + 2)2 − J 2 ) − Bv 2J + Dv J 2 ((J + 1)2 − (J − 1)2 )

= 2(J + 1)Bv+1 − 4(J + 1)3 Dv+1 − 2JBv + 4J 3 Dv

The expression for R(J) comes from hν = E ′ − E ′′ , with ′ E ′ = Evib + Bv′ (J + 1)(J + 2) . . . ′′ E ′′ = Evib + Bv′′ J(J + 1) . . . ′ ′′ Evib − Evib = ∆Evib ,

and where v ′ = v + 1, and v ′′ = v. The same thing for ∆P gives: P (J) ⇒ ∆Evib + Bv+1 (J − 1)J − Dv+1 [(J − 1)J]2 − (Bv J(J + 1) − Dv [J(J + 1)]2 )

P (J + 1) ⇒ ∆Evib + Bv+1 J(J + 1) − Dv+1 [J(J + 1)]2 − (Bv (J + 1)(J + 2) − Dv [(J + 1)(J + 2)]2 )

∆P = 0 + Bv+1 J(−2) − Dv + 1J 2 ((J − 1)2 − (J + 1)2 + 2Bv (J + 1) + Dv (J + 1)2 (J 2 − (J + 2)2 )

= −2JBv+1 + 4J 3 Dv+1 + 2(J + 1)Bv − 4(J + 1)3 Dv 241

9.32 The band head occurs when the spacing between two adjacent P -branch or R-branch lines becomes zero. This can be caused by the difference between B ′ and B ′′ . Assume ∆v = 1 (this is not essential, but it lets us set ∆Evib ≈ ωe for simplicity). ∆E[R(J • )] ≈ ωe + B ′ (J • + 1)(J • + 2) − B ′′ J • (J • + 1)

∆E[P (J • )] ≈ ωe + B ′ (J • − 1)J • − B ′′ J • (J • + 1) The spacing between adjacent lines in each branch is ∆R ≡ R(J • + 1) − R(J • )

= B ′ [(J • + 2)(J • + 3) − (J • + 1)(J • + 2)] − B ′′ [(J • + 1)(J • + 2) − J • (J • + 1)]

∆P ≡ P (J • + 1) − P (J • )

= B ′ [J • (J • + 1) − (J • − 1)J • ] − B ′′ [(J • + 2)(J • + 1) − J • (J • + 1)]

∆R = 2B ′ (J • + 2) − 2B ′′ (J • + 1)

∆P = 2B ′ J • − 2B ′′ (J • + 1)

At the band head, this spacing becomes zero: a. 2B ′ (J • + 2) = 2B ′′ (J • + 1) J • (2B ′ − 2B ′′ ) = 2B ′′ − 4B ′ For the band head in R-branch,

b. 2B ′ J • = 2B ′′ (J • + 1)

J• =

B ′′ − 2B ′ . B ′ − B ′′

J • (2B ′ − 2B ′′ ) = 2B ′′

For the band head in P -branch,

J• =

B ′′ . B ′ − B ′′

J • > 0, so the band head is in the R-branch when B ′ < B ′′ and in the P -branch when B ′ > B ′′ . 9.33 ′ ′′ ′ ′′ hν = ∆E = E ′ − E ′′ = (Evib − Evib ) + (Erot − Erot )

9.34

Q-branch : J ′ = J ′′ v ′ = 2 v ′′ = 1 " 2 !# 5 3 5 3 hν = ωe − ωe − ωe xe − ωe xe 2 2 2 2 5 3 + Be (J(J + 1) − J(J + 1)) − αe J(J + 1) − J(J + 1) 2 2 2 2 2 2 − D2 J (J + 1) − D1 J (J + 1) 1 ωe − 4ωe xe − J(J + 1)αe − J 2 (J + 1)2 (D − 2 − D1 ) ν= h ′ ′′ ∆E = ∆Evib + Erot − Erot

∆E(J → J) = ∆Evib + [B1 J(J + 1) − D1 [J(J + 1)]2 ] − [B0 J(J + 1) − D0 [J(J + 1)]2 ]

= ∆Evib + (B1 − B0 )J(J + 1) − (D1 − D0 )[J(J + 1)]2 = E[Q(J)] 242

Q(J + 1)

⇒ ∆Evib

Q(J)

⇒ ∆Evib

Q(J + 1) − Q(J)

(B1 − B0 )(J + 1)(J + 2) − (D1 − D0 )[(J + 1)(J + 2)]2 (B1 − B0 )J(J + 1) − (D1 − D0 )[J(J + 1)]2 (B1 − B0 )2(J + 1) − (D1 − D0 )(J + 1)2 ((J + 2)2 − J 2 ) 2(J + 1)(B1 − B0 ) − 4(J + 1)3 (D1 − D0 )

9.35 2 1 1 1 E = ωe v + − ωe xe v + + Be J(J + 1) − αe J(J + 1) v + 2 2 2 5 2 35 30 2 J + 2 J+ 2

z }| { J 2 +5J+6 2 z }| { 1 1 1 ′ − ωe xe 2 + Be (J + 2)(J + 3) −αe (J + 2)(J + 3) 2 E = ωe 2 2 2 2 2 1 1 1 + ωe xe + Be J(J + 1) − αe J(J + 1) E ′′ = ωe 2 2 2

∆E = 2ωe − 6ωe xe + Be (4J + 6) − αe (2J 2 + 12J + 15) 9.36

a. The P -branch is always lower in energy than the R-branch, and the J values appear consecutively beginning with the neighboring P (1) and R(0) transitions. The transitions in the problem are (from left to right): P (6), P (5), P (4), P (3), P (2), P (1), R(0), R(1), R(2), R(3), R(4). b. The vibrational constant ωe is given roughly by the frequency halfway between the P (1) and R(0) transitions; the splitting between adjacent transitions is approximately 2Be . ′ ′′ ′ ′′ E ′ − E ′′ = [Evib − Evib ] + [Erot − Erot ] = ωe + [BJ ′ (J ′ + 1) − BJ ′′ (J ′′ + 1)]

for R(0):

→ J + 1, J = 0 → 1

E ′ − E ′′ ≈ ωe + 2B;

for R(2):

E ′ − E ′′ = ωe + 6B.

J = 2 → 3,

R(2) − R(0) = 4B = 0.243 cm−1

B = 0.243/4 = 0.06075 cm−1

ωe = R(0) − 2B = 185.3865 cm−1

c. h 1 = 100hc 800π 2 cµR2 h I = µR2 = = 4.608 · 10−45 kg m2 800π 2 cB(cm−1 )

B(cm−1 ) = B( J)

= 275 amu Å2 . 243

9.37 2 1 1 1 ′ ′ ′ ′ ′ ′ ′ ∆E = ωe v + − ωe xe v + + Be J (J + 1) + αe J (J + 1) v + 2 2 2 " 2 1 1 − ωe xe v ′′ + + Be J ′′ (J ′′ + 1) − ωe v ′′ + 2 2 1 ′′ ′′ ′′ + αe J (J + 1) v + 2 2

= ωe [v ′ − v ′′ ] − ωe xe [v ′ − v ′′ + v ′ − v ′′ ] + Be [2 − 0] − αe (2v ′ + 1)

R(0)J ′′ = 0, J ′ = 1 v ′′ → v ′ =

0 1

→ →

1 2

= ωe − ωe xe (2) + 2Be − 3αe = 1324.830 cm−1 = ωe − ωe xe (4) + 2Be − 5αe = 1310.853 cm−1 2ωe xe + 0 + 2αe = 13.977 cm−1 1 ωe xe = (13.977 − 2(5.5364 · 10−3 )) = 6.983 cm−1 2

∆E0→1 −∆E1→2

ωe = 1324.830 cm−1 + 2ωe xe − 2Be + 3αe = 1337.240 cm−1 . 9.38 The problem asks for the k and Re values, which we can obtain from the vibrational and rotational constants, respectively, once we know the reduced mass: mSi mO 16 · 28 = = 10.18 amu mSi + mO 16 + 28 R(J) : v, J → v ′ , J + 1 µSiO =

′ ′ ′′ ′′ ∆E[r(J)] = Evib + Erot − Evib − Erot

= [Evib (v = 1) − Evib (v = 0)] + [Erot (J + 1) − Erot (J)]

= [hν0 ] + [B(J + 1)(J + 2) − BJ(J + 1)] = hν0 + 2B(J + 1)

R(1)1232.47cm1 = hν0 + 4B −R(0)1231.04cm1 = hν0 + 2B

1.43cm1 = 2B 1.43 B= cm−1 = 0.715 cm−1 ≈ Be 2 h̄2 Be ( J) = 2µRe2 Be ( J) Be ( cm−1 ) = 100hc h̄ = 400πµcRe2 1/2 h̄ Re = 400πµcBe (cm−1 ) 1/2 1.055 · 10−34 J s = 400π(10.18)(1.661 · 10−27 kg amu−1 ) × (2.998 · 108 m s−1 · 0.715 244

= 1.52 Å hν0 = 1231.04 cm−1 − 1.43 cm−1 = 1229.61 cm−1 2 hν0 k= µ h̄ 2 (1229.61 cm−1)(1.986 · 10−23 J/ cm−1 ) (10.18)(1.661 · 10−27 kg amu−1 ) = 1.055 · 10−34 J s = 914 N m−1 .

9.39 Raman transitions behave like two separate transitions, each obeying the ∆S = 0, ∆J = ±1 selection rules. Therefore, the overall selection rules for Raman: ∆S = 0 ∆J = 0, ±2. 9.40 2 Evib = ωe v + 21 − ωe xe v + 21

= (2169.82 cm−1)(0.5) − (13.29 cm−1 )(0.5)2 = 1081.59 cm−1

Erot = [Be − αe (v + 12 )]J(J + 1) − Dv [J(J + 1)]2

= [(1.9313 cm−1) − (0.0175 cm−1)(0.5)](10)(11) − (6 · 10−6 cm−1 )[(10)(11)]2 = 211.41 cm−1

Evib + Erot = (1081.59 + 211.41) cm−1 = 1293.00 cm−1 . 9.41 The rovibrational transition energy for an R branch line is obtained by taking the difference between the upper and lower state vibrational and rotational energies (see Eq. 9.26), and the result is given by Eq. 9.27: ∆E = ωe − 2v ′ ωe xe + 2(J ′′ + 1)Be − (J ′′ + 1)(J ′′ + 2v ′ + 1)αe . For the R(0) transition, J ′′ = 0, J ′ = 1, and for the R(1) transition J ′′ = 1, J ′ = 2. The vibrational quantum numbers are the same for both transitions, so if we calculate ∆E[R(1)] − ∆E[R(0)], the dependence on ωe and ωe xe cancels out. We also neglect the dependence on αe , leaving ∆E[R(1)] − ∆E[R(0)] = 2(1 + 1)Be − 2(0 + 1)Be = 2Be = 1999.90 cm−1 − 1985.34 cm−1 = 14.56 cm−1. From this and the reduced mass, µ = mSi mH /(mSi + mH ) = 0.9729 amu, we can estimate the rotational constant B and calculate the bond length: 16.858 14.56 cm−1 = 7.28 cm−1 = 2 µ, ( amu) Re ( Å)2 s s 16.858 16.858 = = 1.54 Å. Re ( Å) = µ, ( amu)Be ( cm−1 ) (0.9729)(7.28) Be =

9.42 The vibration-rotation coupling term −αe J(J + 1)(v + 1/2) effectively reduces the value of the rotational constant as v increases, so the spacing between the lines gets smaller. We use the equation for the transition energies in the R-branch to get the spacing ∆ between R(0) and R(1) as a function of v ′ : ∆E[R(0)] = ωe − 2v ′ ωe xe + 2(1)Be − (1)(2v ′ + 1)αe

∆E[R(1)] = ωe − 2v ′ ωe xe + 2(2)Be − (2)(2v ′ + 2)αe ∆(NO+ ) = ∆E[R(1)] − ∆E[R(0)]

= [4 − 2]Be − [(4v ′ + 4) − (2v ′ + 1)]αe = 2Be − (2v ′ + 3)αe = ∆(CO) 245

∆(CO) = 2(1.9313 cm−1) − (5)(0.0175 cm−1) = 3.7751 cm−1 1 2Be − ∆(CO) 1 2(1.9982) − 3.7751 v′ = −3 = − 3 = 4.3. 2 αe 2 0.0190 Rounding to the nearest integer gives v ′ = 4. For the v = 3 → 4 transition in 14 N16 O+ , ∆ = 3.79 cm−1 , compared to 3.78 cm−1 for 12 C16 O. 9.43 Erot = BJ(J + 1) − D[J(J + 1)]2 J = 2 : B(2)(3) − D[(2)(3)]2 = 6B − 36D

= 6(24584.35) − 36(0.040)

= 147504.66 MHz 2

J = 20 : B(20)(21) − D[(20)(21)] = 420B − 176400D

= 420(24584.35) − 176400(0.040)

= 10318371.0 MHz 9.44 2 1 1 E = Evib + Erot = ωe v + − ωe xe v + 2 2 1 + Be J(J + 1) − αe J(J + 1) v + 2 2 1 1 1 + Be (0)(1) − αe (0)(1) 2 + − ωe xe 2 + Ev,J = E2,0 = ωe 2 + 2 2 2 5 25 = ωe − ωe xe 2 4 = 9783.37 cm−1 1 1 1 E2,1 = ωe (2 + ) − ωe xe (2 + )2 +Be (1)(2) − αe (1)(2) 2 + 2 {z 2 } 2 | =

9783.37 cm−1

2Be −

5αe

= 9821.40 cm−1

9.45 2 1 − ωe xe v + + Be J(J + 1) E = Evib + Erot = ωe 2 1 − αe J(J + 1) v + 2 25 5 5 = ωe − ωe xe + 12Be − 12 · αe 2 4 2

1 v+ 2

= 3011.44 cm−1 h Be ( cm−1 ) = 2 8π cµRe2 1/2 h B ( cm) Re ( cm) = e 8π 2 cµ( g) (14)(32) µ= = 9.74 amu = 1.618 · 10−23 g 14 + 32 246

Re = 1.496 · 10−8 cm = 1.496 Å 9.46 Because rotations and translations are so similar (no potential energy), we can guess that the energy gaps will be similar when the box is about the same size as the molecule itself, since this will make the distance scale for translation the same as for rotation. Our energy levels are given by: En =

n2 π 2 h̄2 2ma2

EJ = BJ(J + 1).

So our transition energies are ∆En=1→2 =

(22 − 12 )π 2 h̄2 3π 2 h̄2 = 2ma2 2ma2

∆EJ=1→2 = B(6 − 2) = 4B.

The mass of one 14 N2 molecule is 28.00 amu = 4.650 · 10−26 kg. Set the two transition energies equal and solve for a: 3π 2 h̄2 2ma2 2 2 1/2 3π h̄ = a= 8mB

4B =

3π 2 (1.055 · 10−34 J s)2 8(4.650 · 10−26 kg)(1.9987 cm−1 )(1.986 · 10−23 J/cm−1 )

1/2

= 1.49 Å.

As expected, the box is roughly the size of the molecule. (Incidentally, this means that an accurate calculation would have to incorporate the volume of the molecule into the calculation for En .) 9.47

a. ψn=1 =

πx 2 cos a a r

ψrot (J = 1, MJ = 0) = Y10 (Θ, Φ) = r r 2 3 = a 4π a=

3 cos Θ 4π

8π 3

πx =Θ a Θ=

3 x 8

b. The similarity between the two systems is that the potential energy is zero for motion inside the one-dimensional box and for free rotation. However, unlike the one-dimensional box wavefunctions, the rotational wavefunction must have the same value and same derivatives at the beginning and end of a rotation through an angle of 2π; it must be ready to exactly repeat itself. As a result, the rotational problem requires an even number of nodes, because if the wavefunction changes sign once, it must change sign back to where it was originally before it completes one cycle. The 247

rotational states therefore correspond to the one-dimensional box states with odd n, each of which has an even number of nodes, n − 1. If we set the odd value of n equal to 2k + 1, where k is a whole number, then the energies of these one-dimensional box states are E2k+1 = n2 E1 = (2k + 1)2 E1 = (4k 2 + 4k + 1)E1 , where E1 = (π 2 h̄2 )/(2ma2 ) is the energy of the n = 1 state. These energies have the same pattern as the rotational energy levels, Erot = BJ(J + 1) but with an offset: E2k+1 = (4k 2 + 4k + 1)E1 = (4E1 )(k 2 + k) + E1 = (4E1 )k(k + 1) + E1 . So the relationship between the two sets of energies may be written E2k+1 − E1 = Erot , where B = 4E1 =

2π 2 h̄2 . ma2

9.48 The v = 1 level may appear near (3/2)ωe , like a normal harmonic oscillator, but thereafter there is no barrier to the internal rotation, and the system behaves more like a rotating molecule; v = 2 is further from v = 1 than v = 1 is from v = 0. v=2 E

v=1

v=0

180

360

9.49 This is one example of a breakdown in the Born–Oppenheimer approximation, and it is found in the vibrational spectra of bending modes of any molecule with a degenerate electronic state, and strongest in Π electronic states. Born–Oppenheimer breaks down here because the vibrational dynamics vary depending on the relative orientation of the nuclear and electronic wavefunctions, so the nuclear and electronic coordinates cannot be separated in modeling this problem mathematically. No doubt you’re looking for the answer to the actual question, though. The K = 0 states correspond to Λ = −l, so you could start by just plugging that into our equation for ψ. What you will be looking for is the relative angle between the π MO plane and the bending plane, φ − ξ: i h ψΛ,l,K,N,MN ,p = ψ0 YNMN ei(Λφ+lξ) + p(−1)N e−i(Λφ+lξ) i h = ψ0 YNMN ei(Λφ−Λξ) + p(−1)N e−i(Λφ−Λξ) = ψ0 YNMN [cos[Λ(φ − ξ)] + i sin[Λ(φ − ξ)]] + p(−1)N [cos[Λ(φ − ξ)] − i sin[Λ(φ − ξ)]] ψ0 YNMN 2 cos(φ − ξ) : p = (−1)N = MN ψ0 YN 2iΛ sin(φ − ξ) : p = −(−1)N There are two possible kinds of K = 0 wavefunctions: those proportional to cos(φ − ξ) and those proportional to sin(φ − ξ). The cos(φ − ξ) wavefunctions reach maximum amplitude when φ − ξ is 248

zero or π, i.e., when the molecule bends in the plane of the π MO. Similarly, the sin function is at maximum amplitude when |φ − ξ| is π/2, so the bending plane is perpendicular to the π MO plane. The experimental consequence of this is that the K = 0 states are divided into two groups: a lower energy group for the bending in one plane, and a higher energy group for bending in the perpendicular plane. (Details of the particular electronic state determine whether the in-plane or out-of-plane bending is lower energy.) The two groups have opposite parity: (−1)N for the cosine functions and −(−1)N for the sine functions. 9.50 Because the point group is C3v we know that NH3 is a symmetric top, and because B 6= C we know that it is an oblate top. For oblate tops (Eq. 9.22): Erot = Kc2 (C − B) + BJ(J + 1). B = 298000 MHz = 298 GHz; C = 189000 MHz = 189 GHz. Kc 0

J 0 1 2 3 1 2 3 2 3

Erot (GHz) 0 596 1788 3576 487 1679 3467 1352 3140

Chapter 10 10.1

a. Because hydrogen bonds between HCl and CO2 are stronger than bonds between CO2 and CO2 . T because there is H-bonding b. The strongest dispersion forces is for CO2 +CO2 . T because the highest α c. If HCl bonds to HCl, there is no induced dipole moment contribution to the bonding. F but it’s not the dominant force

10.2

a. The dipole–dipole force will be greatest for (HF)2 , because HF has the greater µ; this dimer is also a better candidate for hydrogen bonding (because F is more electronegative than Cl). b. The dipole–induced dipole force will also be greatest for (HF)2 , which we find by comparing 3

µ2 α = 8.2 D2 Å for (HF)2 to µ2 α = 3.2 D2 Å for (HCl)2 . c. The dispersion force will be greatest for (HCl)2 , because HCl has the higher value of α. 10.3

a. a large ωe and large Be ; b. a large ωe and small Be ; c. a small ωe and large Be ; 249

d. a small ωe and small Be . Solution: d, both small. 10.4

a. AB b. AB

AB positive

BA positive but near zero

10.5 No, this is not a pure quadrupole field. There is no monopole component, because the total charge is zero. However, a plane can be drawn such that the two positive charges are on one side and the two negative charges on the other, so there must be a net dipole moment across that plane. Once we subtract out the dipole moment component, so that both sides of the plane have zero charge, there will be a roughly quadrupolar distribution of charge remaining. 10.6 This is a distribution composed of four charges where the spacing d+ is shrunk to zero. Another way to look at it is as the sum of a quadrupole and two, mutually canceling dipoles. There is no net charge (so no monopole); there is a center of inversion (so no dipole). The dominant term is therefore quadrupole.

10.7 [Thinking Ahead: What sort of attractive interaction do we have at the start? The forces binding Ar to NH3 are primarily dipole–induced dipole, but the interaction will have some of the features of a hydrogen bond. In particular, the H atoms are relatively unshielded. Therefore, we are starting from a relatively strong intermolecular bond for argon.] a. + The ion–dipole (or monopole–dipole) interaction is nonetheless much stronger than neutral molecule interactions, and therefore ionization will increase the bond strength. b. 0 The bond strength is not affected by temperature, because temperature is a measure of the total energy available—not the potential energy. c. + A more massive atom (all else being equal) vibrates with a lower zero-point energy without affecting the shape of the potential energy curve. Hence the energy gap between v=0 and the dissociation limit increases. d. − Although dispersion forces will increase with replacement of the H atom by F, the replacement of the small H atom by a larger F atom is the principal effect, leading to a larger molecular separation and lowering the bond strength. (The charge distribution within the molecule also changes, and general chemistry arguments would predict that NF3 should have a greater dipole moment than NH3 because the N F bond is more polar than the N H bond. However, it turns out that NF3 is actually less polar than NH3 , because the electron density of the N-atom lone pair counters the concentration of charge at the fluorines.) e. − Not only is the H atom lost with the change to N2 , but any dipole–induced dipole attraction is eliminated as well, leaving only the dispersion force, which for a molecule as small as N2 we expect to be fairly weak. The bond strength drops. 10.8 In order of increasing bond strength: 250

geometry

binding forces

dispersion

not much choice in geometry

dispersion

T-shape allows maximum overlap with CO2 π-system.

dipole–induced dipole

The linear geometry gives the best dipole alignment. The H is towards the Ar, because it provides the advantage of a weak hydrogen bond.

monopole–induced dipole

Again, not much choice as to geometry. The positive charge of the H+ polarizes the argon atom.

notes

Ar O= C = O

H-C

10.9 H2 O is the only molecule with a dipole moment, and so will have the strongest bond to walls, and be the least likely to be pumped away. 10.10 In addition to weak dispersion forces, there are localized dipole–dipole couplings, involving the carbonyl groups, and other functional groups. O @

C E Ea a

Most of all, the larger size allows many more possible van der Waals interactions available per molecule, such that distortion of the liquid structure requires breaking more van der Waals bonds than in a smaller molecule. 10.11

2.4 A 3.5 A

The square well has an attractive term while the hard sphere does not, and it will deflect the first molecule towards the second. 10.12 [Thinking Ahead: What field is going to induce the induced dipole? This is simpler than the dipole–induced dipole interaction, because this time the electric field that induces the induced dipole is the field due to a single point charge, q/R2 .] We can use the equation for the monopole–dipole energy, 251

and replace the dipole moment µA with the induced dipole moment µ∗A arising from the electric field of the ion qB : µ∗A = αA E = µ∗A qB R2 αA q 2 = − 4B R

αA ∐B R∈

u1−2∗ (R) = −

10.13 u2−2 = −

2µA µB 1 cos θ cos θ − sin θ sin θ cos(0) A B A B (4πǫ0 )R3 2

µA = µB = 1 D = 3.3357 · 10−30 C m

(3.3357 · 10−30 C m)2 µA µB = −3 (4πǫ0 )R( Å)3 (1.113 · 10−10 C2 J−1 m−1 )(10−30 m3 Å )R ( Å)3 1 = (10−19 J) R( Å)3 −2 · 10−19 1 u2−2 ( J) = sin θ sin θ cos θ cos θ − A B A B 2 R( Å)3 u( J) u( cm−1 ) = 1.986 · 10−23 J/ cm 1 −1.007 · 104 cos θ cos θ − sin θ sin θ = A B A B 2 R( Å)3 R ( Å) θA θB 2.0 0 0 3.0 0 0 ◦ 3.0 90 0 3.0 180◦ 0 3.0 0 90◦ 3.0 90◦ 90◦ ◦ 3.0 180 90◦ ◦ 3.0 180 180◦ 4.0 0 0

u( cm−1 ) −1260 −373 0 +373 0 +186 0 −373 −157

orientation → → → → ↑ → ↑ → ↑ ↑ ↑ ↑ ↑ ↑ ← ← → →

10.14 [Thinking Ahead: Can you break this down into a combination of systems we’ve looked at in the chapter? The monopole is as simple as a charge distribution gets, but we can make a quadrupole field by adding together two dipoles.] So one solution to this problem takes advantage of our potential energy function u1−2 (R) for the interaction between a monopole and a dipole: u1−2 (R) = −

µA qB . R2

We can express the quadrupole charge distribution as the sum of four dipole moments: 252

The magnitude of each dipole is the same; call it µA . We can sum the dipoles on the right √ to get √ one resultant 2µA , and the dipoles on the left to get a second resultant, also of magnitude 2µA , but in the opposite direction:

+ 2 µA

−

d d + 4 4

2 µA

−

+ q

√ √ 2µA qB 2µA qB u1−4 (R) = − + 2 d 2 R− 4 R + d4 √ d (R−2 )dR = 2µA qB dR d d dR = 2 = 4 2 √ 2µA qB d u1−4 (R) ≈ − R3 2 √ 2µA qB d ∝ R−3 =− R3

10.15 Adding up the charges on A+ , we find that it averages to a monopole of charge +e. B has no net charge, so no net monopole. To get the dipole moments, we see that the separation between positive and negative charges for A+ and for B are aligned along the z axis between the two molecules. Projecting the charges onto this axis, the charges on A+ are +1.6e and −0.6e, and for B the charges are 0.4e and −0.4e. For the dipole moment, we need a symmetric charge distribution (because the dipole is based on separated charges that are equal in magnitude but opposite in value). This is all set for B, which gives a dipole moment magnitude of (0.4e)(4.0a0 ) = 1.6 ea0. For A+ , the net charge is +e, and so we need to subtract +e from the charges before calculating the dipole moment. Subtracting the charge from the net +1.6e on the left gives charges of +0.6e and −0.6e, separated by 2.0a0 , predicting a dipole of (0.6e)(2.0a0) = 1.2 ea0. 253

2.0a0

4.0a0

+0.8

−0.2

+0.2 −0.6

4.0 a0

4.0a0 +0.2

+0.8

−0.2

20.0a0 1.2 ea0

1.6 ea0

+1.0 e

qA µB (1.0 e)(1.6 ea0 ) =− = −0.0040e2/[(4πǫ0 )a0 ] (4πǫ0 )R2 (4πǫ0 )(20.0 a0 )2 2(1.2 ea0 )(1.6 ea0 ) 2µA µB =− = −0.00048e2/[(4πǫ0 )a0 ]. u2−2 = − (4πǫ0 )R3 (4πǫ0 )(20.0 a0 )3 u1−2 = −

10.16 The two ions will polarize the atom in the same direction, generating an induced dipole at the atom of magnitude µ∗A = 2αA E(R) = 2αA qB /R2 . We can then plug this induced dipole into our equation for the monopole–dipole interaction, u1−2 (R) ∝ −µA qB /R2 , to get u(R) = −C

2 µ∗A qB αA qB = −C . R2 R4

10.17 The distance √ qA to −qB and −qA to qB is R. Set R′ equal to the distance qA to qB and −qA to −qB . Then R′ = R2 + d2 . The rest of the analysis follows the same strategy as for obtaining the dipole–dipole interaction potential: (−qA )(−qB ) qB (−qA ) qA qB (−qA )qB 1 + √ + +√ 4πǫ0 R R R2 + d2 R2 + d2 " −1/2 # d2 2qA qB 2 2 2qA qB 1− 1+ 2 =− − +√ = 4πǫ0 R 4πǫ0 R R R2 + d2 2 d 2qA qB 1− 1− ≈− 4πǫ0 R 2R2 2 (qA d)(qB d) 2qA qB d µA µB =− =− =− 2 3 4πǫ0 R 2R 4πǫ0 R 4πǫ0 R3

u2−2 (R) =

10.18 There are two possible solutions that can be found by symmetry. The goal is to ensure that the dipole and the quadrupole are each equally influenced by the positive and negative charges on the other. x

10.19 The idea is to find an expression for udisp before we plug in the polarizabilities, and then use the expressions we know for the particle in a box to evaluate anything we need. One equation to start 254

from would be Eq. 9.30, which depends on the wavefunctions and energies of the system: udisp (R) ≈

e4 (4πǫ0 )2 R6

e4 = (4πǫ0 )2 R6

2 2 R ψB1 zB ψB2 dzB ψA1 zA ψA2 dzA (EA1 + EB1 ) − (EA2 + EB2 ) 2 Ra 2 2 R a A sin πzaA sin 2πz zA dzA sin a a a 0 0 R

πzB a

sin

2πzB a

zB dzB

(2 − 8)π 2 h̄2 /(2ma2 )

The integrals turn out to each be equal to 8a2 /(9π 2 ), so the final value is udisp (R) ≈ −

[32a/(9π 2 )]4 e4 2(32/9)4a6 me4 2(32/9)4a6 = − = − Eh . 6π 10 R6 6π 2 (4πǫ0 )2 h̄2 R6 /(2ma2 ) 6π 10 (4πǫ0 )2 h̄2 R6

add up to zero

0.8 e

0.4 e project charges onto symmetry axis

4.0 −0.8e 2.0

y(a 0)

symmetry axis

y(a 0)

10.20 We can add up the charges to get the monopole component, +2.0e. To find the dipole moment, one can vector-sum the point charges, or recognize from the symmetry that the dipole moment vector must lie along the axis x + y = 2. Note that the monopole component must be subtracted, so that a symmetric distribution of charges can be used to calculate the dipole moment magnitude as the product √ of the magnitude of the charge separation, 0.2e, and the distance between the separated charges, 4 2a0 , √ to give a dipole moment of 0.8 2ea0 = 1.13ea0.

0.8 e

4.0

subtract monopole component

0.0 e

−0.2e

4.0 1.13 ea 0

2.0

0.0

0.0 0.4 e

1.2 e

0.0

2.0 4.0 x(a 0)

0.0 1.2 e 0.0

0.2 e

2.0 4.0 x(a 0)

0.0

4.0 x(a 0)

10.21 [Thinking Ahead: If we assume the two molecules are both diatomics, how many coordinates are there altogether? There are three Cartesian coordinates for each atom, so 12 total. How many are for just the internal vibrations? Each diatomic has just one vibrational coordinate, which we can ignore.] And as with any other polyatomic system, we can also ignore the three overall translational coordinates, and the three rotational coordinates for the rotations of the entire complex. That’s 12 total, minus three translations, three rotations, and two internal vibrations, leaving four. x θA φ A

θ B R φ

There are different ways to define these coordinates. One straightforward coordinate is the separation between the centers of mass of the two molecules. Then looking at the angles, we see that two angles specify the absolute orientation of each linear molecule in space. but we don’t need both angles φ for rotation about the intermolecular axis—we only need the relative angle φ = (φA − φB ). Therefore, there are only three angles and one distance as, for example, R, θA , θB , φ. 10.22 [Thinking Ahead: How many coordinates are there for each of the different components (nonlinear, linear, atomic), if we leave out the internal molecular vibrations? There are three translational coordinates for each, three rotational coordinates for the non-linear molecules, two rotational coordinates for the linear molecules, and none for the atoms. This helps both by setting an upper limit to the number of coordinates we want, and by reminding us that the entire system (like the non-linear molecule) will have three translations and three rotations, which are not of interest to the potential 255

energy surface we’re looking for.] The easiest way to do this is probably to first count up all the intermolecular degrees of freedom and then subtract the ones we don’t want. If the molecules are rigid, then we only need to count the translations and rotations for each component of the system. Each component has three translational degrees of freedom. However, the non-linear molecules have three rotational degrees of freedom, the linear molecules have two, and the atoms have no rotations. Therefore, the total number of degrees of freedom is N (3 trans + 3rot) + L(3 trans + 2rot) + A(3 trans) = 6N + 5L + 3A. We don’t want all of these, because these include the three translational coordinates for the centerof-mass of the entire system, and the rotational coordinates. How many rotational degrees of freedom does the whole system have? The system can never become linear if there are any non-linear molecules present, but if N = 0 then the system can become linear and we are only guaranteed two rotational coordinates. If N > 0 then there are three rotational coordinates. We subtract these overall translations and rotations from the total number, and find that the total number of coordinates needed for the intermolecular potential energy surface is 6N + 5L + 3A − 6 + δN,0 , where δN,0 is the Kronecker delta function, equal to 1 if N = 0 and equal to 0 otherwise. So, for example, the interaction between two argon atoms can be described by 6(0) + 5(0) + 3(2) − 6 + 1 = 1 parameter, the distance between the two atoms, while the interaction between two water molecules involves a potential energy surface of 6(2) + 5(0) + 3(0) − 6 = 6 distinct coordinates. 10.23 ū(H2 O)2 (3 Å) = −300 cm−1 dipole–dipole: u ∝ R−3 u(R) = (−300) ū(H2 O−but) (3 Å) = −200 cm−1 dipole–induced dipole: u ∝ R

−6

u(R) = (−200)

3R R

Trans-1,3-butadiene has no permanent dipole moment. R H2 O—H2 O 3 Å −300 cm−1 6 Å −37.5 cm−1 12 Å −4.7 cm−1

H2 O—butadiene −200 cm−1 −3.1 cm−1 −0.05 cm−1

Since these are the forces that bind molecules together in the liquid phase, they partly determine the solubility. Molecules that disperse well in water must be able to compete with the water-water attraction, which is quite strong. Butadiene does not do this very well, having less than 10% the welldepth of the H2 O H2 O interaction at only 6 Å. Butadiene, in spite of a strong short-range attraction, does not dissolve in water. We assume 1,3-butadiene here. For 1,2-butadiene, there will be a dipoledipole term. 10.24 The bonds that form in UF6 and He2 are dispersion force bonds, and depend on the polarizability of the atoms. The electron-rich UF6 molecule is much more polarizable than helium, so UF6 will have a deeper well than He2 . UF6 is also larger, and therefore has a larger value of Re than He2 . 256

U(R)

R He 2 UF6

10.25 We use the rotationally averaged form of u2−2 : utot ≈ u2−2 + u2−2∗ + udisp u2−2 = − =−

2µ2A µ2B (4πǫ0 )2 3kB T R6

4 2 (1.109 D)(3.3356 · 10−30 C m/D)

(1.113 · 10−10 C2 J−1 m−1 )2 3(1.381 · 10−23 J K−1 )(300 K)(4.2 · 10−10 m)6

= −4.44 · 10−22 J = −22 cm−1.

4µ2A α (4πǫ0 )R6 2 4 (1.109 D)(3.3356 · 10−30 C m/D) (2.7 · 10−30 m3 ) =− (1.113 · 10−10 C2 J−1 m−1 )(4.2 · 10−10 m)6

u2−2∗ = −

= −2.42 · 10−22 J = −12 cm−1.

udisp ≈ utot − u2−2 − u2−2∗ = −173 cm−1 + 22 cm−1 + 12 cm−1 = −139 cm−1. 10.26 Let’s approximate the ∆E for both molecules as 7.9 eV. The individual contributions are 2µ2A µ2B (4πǫ0 )2 3kB T R6 2(0.45 D)2 (0.11 D)2 (3.3356 · 10−30 C m/D)4 =− −10 (1.113 · 10 C2 J−1 m−1 )2 (1.381 · 10−23 J K−1 )(298 K)(4.0 · 10−10 m)6

hu2−2 iN,θ,φ = −

= −2.91 · 10−24 J = −0.0018 kJ mol−1

4µ2A α (4πǫ0 )R6 4(0.45 D)2(3.3356 · 10−30 C m/D)2 (1.95 · 10−30 m3 ) =− (1.113 · 10−10 C2 J−1 m−1 )(4.0 · 10−10 m)6

u2−2∗ (R) = −

= −3.85 · 10−23 J = −0.023 kJ mol−1

4µ2A α (4πǫ0 )R6 4(0.11 D)2(3.3356 · 10−30 C m/D)2 (5.44 · 10−30 m3 ) =− (1.113 · 10−10 C2 J−1 m−1 )(4.0 · 10−10 m)6

u2−2∗ (R) = −

= −6.43 · 10−24 J = −0.0039 kJ mol−1 257

α2 ∆E 8R6 (5.44 · 10−30 m3 )(1.95 · 10−30 m3 )(7.9 eV)(1.602 · 10−19 J/ eV) =− 8(4.0 · 10−10 m)6

udisp ≈ −

= −4.10 · 10−22 J = −0.25 kJ mol−1.

Adding these together gives a total of −0.28 kJ mol−1, dominated by the dispersion term. 10.27 a. ArHCl will be linear, so as to align the HCl dipole moment with the induced dipole of the Ar. '$

Cl—H &% Ar

or '$

H—Cl &% Ar

b. ArC6 H6 will be C6v with the Ar above the C6 H6 plane, to maximize the dispersion forces by an overlap of Ar with the diffuse π-electrons of benzene.

P S PP S S S

Ar PP P

10.28 [Thinking Ahead: How is this like other cases in quantum mechanics that you may know from earlier chapters? The well itself suggests the particle in a one-dimensional box. The flat region outside the well corresponds to the dissociation limit in an ordinary diatomic molecule.] The bound states of this potential are approximately those of the particle in a one-dimensional box, but with tunneling ′ possible at R > Rsq . For a particle in a one-dimensional box,

nπR ψv (R) = A sin ′ −R ) (Rsq sq Ev =

n2 h2 . ′ − R )2 8m(Rsq sq 258

10.29 [Thinking Ahead: How can we find the minimum of any function from its algebraic form? The derivative of the function will be zero at its minimum. (Also at its maximum, but this function does not have a defined maximum.)] For the Lennard–Jones potential (Eq. 10.44), " 6 # 12 Re Re . − 2 uLJ (R) = ǫ R R To find Re , we need to find the minimum of uLJ (R). Call that distance Rmin . duLJ (R) −13 −7 = ǫ −12Re12 Rmin + 12Re6 Rmin dR R=Rmin which equals 0 at the minimum of the potential energy. −7 −13 12Re6 Rmin = 12Re12 Rmin −7 Re12 Rmin = −13 Re6 Rmin 6 Rmin = Re6

Rmin = Re . The well-depth or dissociation energy De is De = uLJ (R → ∞) − uLJ (Re ) " " 6 # 6 # 12 12 Re Re Re Re −ǫ −2 −2 =ǫ ∞ ∞ Re Re = ǫ[0 − 0] − ǫ[1 − 2] = ǫ.

10.30 [Thinking Ahead: Sketch this interaction on a graph showing the potential energy. What is the mathematical condition that holds when the two atoms reach their minimum separation? The minimum (classical) separation distance is the point at which u(R) = E.] " 12 6 # RLJ RLJ − E = 4ǫ R R " 1 6 # 3.61 3.61 = 4(190 K) 2− 3.00 3.00 = 4740 K. 259

10.31

" 3 #) γR d e e−αR − = βǫ dR R Re Re 3 R = βǫ −αe−αRe + 3γ 3 e4 = 0 Re 3γ 3 = αe−αRe Re 1/3 αRe −αRe γ= e 3 −αRe − γ 3 = −ǫ u(Re ) = βǫ e −1 β = − e−αRe − γ 3 −1 αRe −αRe −αRe −e e = 3

du dR

(

eαRe . = αRe 3 −1

10.32 [Thinking Ahead: How do we define the equilibrium bond length and well-depth mathematically? The equilibrium bond length is where the potential energy function reaches its minimum energy. That’s a critical point on the potential energy curve, which we can find by setting the derivative equal to zero. The well-depth is the potential energy in the dissociation limit (at large R) minus the minimum potential energy (at the equilibrium R value). So first we will locate the equilibrium R value, and then find the energy at that value.] 6 du 6 ζ[1−(R/Re6 )] −ζ Re6 ǫζ e +6 7 = dR R=Re6 ζ −6 ζ Re6 R R=Re6 6 0 −ζ 1 ǫζ e +6 = ζ −6 ζ Re6 Re6 6 6 ǫζ =0 − = ζ − 6 Re6 Re6 so the critical point is at Re6 . To be sure that this is a minimum, we need to check the second derivative: " # 2 6 d2 u ζ 6 ζ[1−(R/Re6 )] ǫζ Re6 e = − 42 8 dR2 R=Re6 ζ −6 ζ Re6 R R=Re6 6ζ 42 ǫζ . = 2 − R2 ζ − 6 Re6 e6 This is positive for ζ in the range described in the text (around 15), so Re corresponds to a minimum. " 6 # 6 −ζ[1−(R/Re6 )] Re6 ǫζ e − u(Re6 ) = ζ −6 ζ Re6 ǫζ 6 0 = e −1 ζ −6 ζ 1 ǫζ = −ǫ (6 − ζ) = ζ −6 ζ so the well-depth is ǫ. 260

10.33 [Thinking Ahead: What force or forces are at work, and what atomic parameters are relevant? These molecules are not chemically bound, as attested to by the extremely low well-depths ǫ. This is to be expected because Ar and Xe are closed-shell atoms. The weak forces holding the atoms together must be dispersion or London forces, because these do not depend on any permanent multipole field (although there will be weak magnetic couplings in LiAr). The dispersion force increases with the polarizability of the atom, and the polarizability increases with the size of the atom’s electron cloud.] De ( K) = ǫ( J)/(1.381 · 10−23 J K−1 )Re ( Å)

LiAr Ar2 Xe2

ǫ ( J) 0.92 · 10−21 2.0 · 10−21 3.9 · 10−21

= RLJ ( Å)(21/6 ).

De ( K) RLJ ( Å) 67 3.7 145 3.8 282 3.9

Re ( Å) 4.2 4.3 4.4

The electron cloud gets bigger as one moves down the periodic table, so the bonding force increases in the order Li < Ar < Xe, but the atoms are also increasing in size. So De increases as the atoms become more polarizable, and Re increases as the atoms get bigger. 10.34 [Thinking Ahead: How can we solve for more than one value of R? We need two turning points: the idea is that a classical particle would spend all its time between these two R values. The only condition we have is that E = u(R) at those points, but that doesn’t mean that only one value of R will solve the equation. Since the Lennard–Jones expression is polynomial with two different powers of R, we should expect that more than one value of R will solve the problem. We can just push forward and see that this works out.] This is a plug-and-chug problem. Set the energy E equal to the Lennard–Jones potential energy u for the parameters given, and solve for R. ǫ = 2.00 · 10−21 J = 100 cm−1 " 12 6 # RLJ RLJ = −70 cm−1 − uLJ (R) = 4ǫ R R

RLJ R

−

RLJ R

=−

70 = −0.175 400 0=

RLJ R

√ 1 1 ± 1 − 0.700 2

RLJ R

−

RLJ R

+ 0.175

This is a quadratic equation.

RLJ R

= 0.77386, 0.22614 = 0.9582, 0.7805

RLJ = 3.8 Å R = 4.0 Å, 4.9 Å R1 = 4.0 Å R2 = 4.9 Å 261

10.35 Let Rc be the two turning points, and set x = (Re /Rc )6 : u(Rc ) = ǫ x2 − 2x = −0.91ǫ

0 = x2 − 2x + 0.91 i p 1h x= 2 ± 4 − 4(0.91) = 0.7, 1.3 2 Rc = x−1/6 Re = 0.957 Re , 1.06 Re .

10.36 There is no tunneling to R values less than R1 (because the potential energy is infinite at R < R1 ), but there is tunneling to high R values. The spacings between the energy levels decrease at higher energy, as in the case of the one-electron atom (which shares the same exponential potential energy function). U(R) R

−ε R

10.37 [Thinking Ahead: What relation allows us to link the energy of the system to the speeds? Any kinetic energy can be written as mv 2 /2. Our job is to find out how much kinetic energy there is at the end of the collision.] Let’s label the initial and final kinetic energies Ki and Kf , respectively. D0 = 85 cm−1 Ki = 200 cm−1 1 2 Kf = 3 mAr v 2

E = 200 cm−1 − 85 cm−1 = 115 cm−1 = Kf

mAr = 40.0 amu = 6.64 · 10−26 kg 1/2 1/2 2 2 −1 22 −1 −1 −1 = Kf (115 cm )(5.034 · 10 cm J ) v= 3mAr 3(6.64 · 10−26 kg) = 151 m s−1 .

10.38 [Thinking Ahead: What conservation law is important here? Without knowing more about the collision itself, we don’t know what the angular momentum of the system is. But we do know that the total energy must be conserved.] From Table 9.1: ωe (N2 ) = 2358 cm−1 , and Be (N2 ) = 1.999 cm−1 . We find the total rovibrational energy by adding the rotational and vibrational contributions: E(v, J) ≡ v(2358 cm−1) + J(J + 1)(1.999 cm−1), which is the energy relative to E(0, 0). The maximum rovibrational energy will be obtained if all the kinetic energy of the collision is converted to rotation and vibration of a single N2 molecule. The available energy is 1 mN2 v 2 = (28.00 amu)(1.661 · 10−27 kg amu−1 )(5.15 · 102 m s−1 )2 2 2 = 1.23 · 10−20 J = 621 cm−1 . 262

There is too little energy to induce vibrational excitation, so v = 0 before and after the collision. The maximum rotational state attainable is approximately J(J + 1) =

621 cm−1 = 311. 1.999 cm−1

J ≤ 17. 10.39 a. Before the collision, the water molecules will begin rotating to align dipoles, because the dipole–dipole force is repulsive if the dipoles are pointing against each other. b. After the collision, both water molecules will be rotating. There is not enough energy to induce vibration. 10.40 [Thinking Ahead: If the exp-6 potential does behave correctly, what do we know about its second derivative at Re ? In order for the potential to reach a minimum value at Re , the second derivative must be positive at that point.] The exp-6 potential has the equation (Eq. 10.50) " 6 # 6 ζ[1−(R/Re )] Re ǫζ e − , ue6 (R) = ζ −6 ζ R which has derivatives " 7 # 6 −ζ 6 Re ζ[1−(R/Re )] e + ζ Re Re R " 8 # 2 d2 ue6 6 ζ ǫζ Re 42 . = eζ[1−(R/Re )] − 2 dR2 ζ −6 ζ Re Re R ǫζ due6 = dR ζ −6

If we evaluate the second derivative at R = Re we see what the problem could be if ζ is in the wrong range: " 8 # 2 6 ζ ǫζ 42 Re d2 ue6 ζ[1−(Re /Re )] = e − 2 dR2 R=Re ζ −6 ζ Re Re Re " # 2 ǫζ 6 ζ 42 = − 2 ζ −6 ζ Re Re 42 6ζ ǫζ − = ζ −6 Re2 Re2 At the minimum of the curve, the second derivative must be positive. Therefore, either ζ − 6 < 0 and the function in brackets is negative, or ζ − 6 > 0 and the function in brackets is positive. The function in brackets is positive when ζ > 7, and therefore ζ must be greater than 7 or less than 6.

Chapter 11 11.1 K4 will have the higher ionization potential, because K3 ionizes to K+ 3 , with a closed-shell twoelectron configuration. 11.2 This is not an easy question at all, but based on our discussion of periodic trends in atomic properties, one knows that silicon should be substantially larger than carbon. This increases bond distances and generally weakens π-type bonds faster than σ bonds, because π bonds are formed by atomic orbitals 263

that are parallel to each other, whereas σ bonds are usually formed from atomic orbitals that point right at each other. Therefore, the structures with multiple bonds (graphite and polyacetylene) are not favorable for silicon at all, and the fullerene structure (which relies on considerable π bond character for its stability) is probably not as greatly stabilized as in the case of carbon. Small silicon structures have been little studied at present, but calculations suggest that diamond-like structures are the most stable [12]. 11.3 The attractive forces will try to hold the two clusters together to form a (H2 O)200 cluster, but the excess translational energy must be removed. A few individual water molecules may evaporate from the surface of the cluster. The typical translational energy at a temperature of 300 K is roughly equivalent to one tenth of the hydrogen-bond dissociation energy of water. 11.4 [Thinking Ahead: Should the change be positive or negative, or are there competing effects? In order to form the bonds between the two tetrahedral clusters to form the cube, we have to break some of the original bonds, so we should not expect the stability to change by a vast amount.] Each tetrahedral cluster has six interactions (the number of edges of a tetrahedron). The cubic cluster will have 12 (the number of edges of a cube). So, to a first approximation, there is no net change in the energy when the two Ar4 clusters combine. Each atom interacts with three others before and after the process. In this simple picture, ∆E = 0. However, the average distances will generally be larger in the tetrahedron because the acute corner angles cause more crowding. Therefore, the bonding in the cubic cluster will generally be somewhat stronger. 11.5 The most stable should have the most pair interactions (atoms in direct constant). The square has four interactions, the rhombohedral has five, linear has three, tetrahedral has six. Therefore, the order is tet (1), rhom (2), square (3), linear (4). 11.6 [Thinking Ahead: What approximate shape is likely for any arrangement that maximizes the stability? A three-dimensional arrangement will give more opportunity for favorable interactions than a two-dimensional arrangement, but beyond that it’s difficult to guess. We can take advantage in this chapter of seeing that for the 13-atom atomic clusters, an arrangement of five-membered rings in the icosahedral geometry promotes stability.] The most stable arrangements, in order of increasing energy (decreasing stability), are (a) a D5h pentagonal bipyramid; (b) a Cs square bipyramid with the seventh atom in the gap above one of the triangular faces; (c) a Cs triangular bipyramid with the sixth and seventh atoms each above a different triangular face; and (d) a Cs triangular bipyramid with the sixth and seventh atoms each above adjacent bonds, effectively continuing the structure of the bipyramid. If we count the total number of direct contact interactions in each of these structures, we obtain 15 for (a), 15 for (b), 14 for (c), and 14 for (d). 11.7 [Thinking Ahead: As the cluster grows from 13, where will the next spheres be added? We will add spheres to the outside of the first shell to form the second shell, filling the hollows left by the gaps between the spheres in the first shell.] As a first guess, we could just assume that each sphere in the first shell was in contact with three in the second shell, packed together. This would predict a magic cluster size of 1 (center) + 12 (first shell) + 36 (second shell) = 48. Can we do better using the method we used to estimate that the coordination number of the central atom to be 12—by finding the surface area of a sphere of an appropriate radius? In this case, if we assume that the second shell has a radius of 2d (where d is the diameter of one sphere), we arrive at an area of 4π(2d)2 = 16πd2 = 50.3d2 . But this is an overestimate, because the spheres are more tightly packed than √ that, occupying hollows left between the spheres in the first shell. This gives a diameter of about 3d, for an area of 12πd2 = 37.7d2 . If we take this to suggest about 38 spheres in the second shell, then we predict a total of 1 (center) + 12 (first shell) + 38 (second shell) = 50. The correct answer is 55. 264

11.8 First off, we count the conformations if the H atoms in each water molecule simply switch places. There are six water molecules, and we can switch H atom positions on any number of them, allowing 26 different possibilities. Then we have to look at the possibilities if we allow the water molecules to change class, as defined in the problem by the number and type of hydrogen bonds. Let’s call those three classes A, B, and C. The two water molecules in any given class are always opposite each other in the cluster: 1 and 2, 3 and 4, and 5 and 6. Call these pairs I, II, and III. We can alter the class of any of these pairs by changing the orientations of the hydrogen atoms, but we must always have two water molecules in each class. There are then six possibilities for how the three classes are distributed among our three pairs of water molecules: ABC, ACB, BAC, BCA, CAB, CBA (listing the class for each pair in order, so ABC is class A for pair I, class B for pair II, and class C for pair III). Within each of these arrangements, we have the 26 possible switches of the H atoms in each molecule. There is a total of 6(26 ) = 46656 different conformations accessible to this cluster by tunneling motions. 11.9 The value of ǫ is 120 K = 1.0 kJ mol−1 . We would expect the first bond formed between one argon atom and another to be the strongest. A third argon atom would not bond as strongly as the first two, and subsequent bonds would be weaker. This suggests that a lower limit to the number of bonds that need to be broken is 6.43/1.0, or seven bonds. Argon liquid actually has a coordination number of about eight, depending critically on how far you extend the “shell” for the coordination number. Looking ahead, if we were to integrate over the entire first peak in Fig. 12.3, we would get a number closer to 12. 11.10 [Thinking Ahead: Why might the binding energy increase in comparison to the dimer? Why might it decrease? In our model, the binding energy of each particle in the cluster increases if we can construct a many-particle wavefunction in which all the single-particle wavefunctions have the same phase. The binding energy would decrease only if there were no way to balance the phases in the three boxes. For this linear arrangement, we can always balance the phases so that the three waves are in synch, and the average binding energy increases, if only slightly.] Thankfully, we need to repeat only the first few steps of the dispersion energy derivation to figure out what new term needs to be incorporated; after that we can rely on the old math to get the exact additional contributions. Let’s label by C, B, and A our three boxes centered at −R, 0, and +R. The full perturbation Hamiltonian, by extension from Eq. 10.26 is Ĥ ′ = Ĥnuc(A)−nuc(B) + Ĥelec(A)−elec(B) + Ĥelec(A)−nuc(B) + Ĥnuc(A)−elec(B) = Ĥnuc(C)−nuc(A) + Ĥelec(C)−elec(A) + Ĥelec(C)−nuc(A) + Ĥnuc(C)−elec(A) = Ĥnuc(C)−nuc(B) + Ĥelec(C)−elec(B) + Ĥelec(C)−nuc(B) + Ĥnuc(C)−elec(B) 2 e2 e2 e2 e 1 − − + = 4πǫ0 R R + zA − zB R + zA R − zB 2 2 2 1 e e2 e e + − − + 4πǫ0 R R + zC − zA R + zC R − zA 2 2 2 1 e e2 e e + − − + 4πǫ0 2R 2R + zC − zB 2R + zC 2R − zB ( −1 e2 zA − zB = 1+ 1+ (4πǫ0 )R R zB −1 zA −1 − 1− = 1+ R R −1 zC − zA =1+ 1+ R 265

zA −1 zC −1 − 1− = 1+ R " R −1 1 zC − zB = 1+ 1+ 2 2R zC −1 zB −1 − 1+ − 1− 2R 2R as before, we Taylor expand the (1 + x)−1 quantities, and all the first-order terms cancel, leaving ≈−

zC zB e2 zA zB . z z + z z + A B C A (4πǫ0 )R3 8

From this point, the math is the same (to within a constant) as detailed in the derivation of Eq. 10.38. R (0) (0) There are three terms now, instead of only one, but the integrals of the form ψ1 zψ2 dτ are no different. The total dispersion energy for the trimer is 1 α2 ∆E dimer trimer = −2.125Edisp . 1+1+ Edisp ≈ − (4πǫ0 )2 2R6 8 The binding energy for the third box is the energy that needs to be added to pull the third box far enough away that only the dispersion energy of the dimer remains. In other words, it is the difference between the dispersion energy of the dimer and the dispersion energy of the trimer: dimer trimer dimer Ebind (3) = Edisp − Edisp = 1.125Edisp .

To second order in perturbation theory, each addition to a cluster increases the attractive forces and the binding energy, but the effect drops off rapidly as a function of distance (in our example here, the interaction between B and C was over twice the distance of the A-C interaction, but counted one eighth as much). To see the effect of the non-pairwise interactions—for example how the presence of C influences the A-B interaction—it would be necessary to go to third order in perturbation theory, and we will see the results in Eq. 12.3. 11.11 a. Each carbon atom contributes one electron in a p orbital to the surface π system of the molecule, so there are a total of 60 electrons, and 60 atomic p orbitals in that system, and these will generate 60 molecular orbitals. b. The 60 electrons will occupy these orbitals in pairs, so with the HOMO we will fill up exactly half of the available orbitals. As we will be adding nodes to the orbitals while this is going on, we will expect that by the time we get to the HOMO, every six-membered ring will be at the same point as the HOMO in benzene: one nodal plane that intersects the ring in two places (see Fig. 6.19). If that’s true for every six-membered ring, then those rings will have the same rotational symmetry as the HOMO in benzene. The HOMO in benzene is an e1g degenerate pair with a character of −1 under the Ĉ3 rotation (C60 does not have a Ĉ6 symmetry element, so we look at the 120◦ rotation instead of the 60◦ rotation). The MO in C60 will have the same character, and that is sufficient to identify it as one of the fivefold degenerate h orbitals. c. The transition is forbidden because it violates the u ↔ g selection rule. This is why an indirect method (electron loss spectroscopy) was originally used to detect the gap between the two states. 11.12 We expect the system to be most stable when there is maximum conjugation of π systems of the two phenyl rings and the central ethyne, which occurs when the two rings are coplanar (φ = 0). 266

The potential energy maximum occurs when the two rings twist in either direction by π/2 so that the p orbitals in one ring are perpendicular to those in the other. (Note, however, that both rings can remain conjugated to the ethyne in the middle at any angle φ, because the ethyne has two orthogonal π orbitals.) E (LUMO) E (HOMO)

−π/2

π/2

Assuming (correctly) that the HOMO and LUMO orbitals involve the conjugated electrons, we can anticipate the effect of φ on these two orbital energies by getting a rough idea of the orbital geometries. In the most stable, planar conformation, there are 14 p electrons conjugated together. The first MO for these electrons will have no nodes perpendicular to the molecular plane; the second and third MOs will have one such nodal plane, and so on. The LUMO is the orbital occupied by the seventh pair of these electrons, and it will have three nodes perpendicular to the molecular plane. In order to conform to the inversion symmetry of the molecule, one of these planes must pass through the middle of the ethyne triple bond, breaking the conjugation between the rings. The HOMO, on the other hand, will have an even number of nodes and by symmetry cannot have a node through the middle of the ethyne. Therefore, as the molecule becomes non-planar, we expect that the LUMO is stabilized by the decreasing interaction between p orbitals of opposite phase, while the HOMO is destabilized as the conjugation across the ethyne group is broken by the torsion. 11.13 [Thinking Ahead: What is the shape of the potential energy curve, and how will this affect the energies and wavefunctions? The curve is graphed below, and we see that it has a relatively flat bottom like the particle in a box, but the walls flatten out quickly. Therefore, we can expect a significant zero-point energy, but the spacing between the levels will quickly begin to collapse.]

The energy levels converge more slowly than in the one-electron atom potential, because the potential energy curve does not approach the infinitely steep slope seen in the 1/r potential as r goes to zero. The s orbital radial wavefunctions retain a similar shape, but with less dramatic change in the peak heights because the potential is not varying as quickly. The wavefunctions themselves will have cusps at r = 0, but the radial distribution functions R(r)2 r2 will smoothly approach zero at r = 0. Approximate wavefunctions are sketched below. 267

3s 2s 1s

11.14

r r

π/3

b α x

r’

c r’ c

β b a x

R θ

r c

In the sketch above, r is the carbon–carbon bond length and r′ is the projection of that bond length onto a plane perpendicular to the axis of the nanotube. If we start off with a perfect graphitic sheet of sp2 carbons, all the bond angles initially are π/3 (it is convenient to use radians for some of this geometry). As we roll the sheet to form a nanotube, we keep the bonds the same length, r, but the bond angles slowly decrease as the arrangement of atoms becomes non-planar. Looking at a cross-section of the tube through atoms a and c, the carbon atoms are a distance R from the center, including the projection of atom b onto this cross-section. We can then solve for the parameters r′ , θ, α, x, and finally the new CCC bond angle, β: r′ = r sin(π/3) r 2π 2π = sin(π/3) = θ= n 2πR/r′ R π−θ α= 2 x = r′ sin α = r sin(β/2) ′ r β = 2 arcsin sin α r r π− R sin(π/3) = 2 arcsin sin(π/3) sin 2 This gives values of β of 118.6◦ for R = 5.0 Å and 116.3◦ for R = 3.1 Å. 268

The percentage of sp3 character gives the fraction of p⊥ present, and we can estimate this from the angle by balancing the contributions of sp2 and sp3 hybrids according to their probability densities: β = c2sp3 (109.5◦ ) + c2sp2 (120◦ ) 1 = c2sp3 + c2sp2 120◦ − β . c2sp3 = 120.0◦ − 109.5◦

According to this equation, the R = 5.0 Å tube has 13% sp3 character, and the R = 3.1 Å tube has 35% sp3 character. The reactivity of the nanotube surface is expected to increase with the amount of pz character, so very small tubes can be quite reactive.

11.15

a. Ĥ = K̂ + U =

−

h̄2 2 ∇ − U0 e−(r−r0 )/rz 2me

b. Because this potential energy function is isotropic and three dimensional, the angular solutions are the spherical harmonics. The lowest energy excited state is comparable to the n = 2 states of the one-electron atom, and therefore may have an angular function: r 1 0 l = 0 ml = 0 :Y0 = ; 4π r r 3 3 ±1 0 cos θ; ml = ±1 : Y1 = sin θe±iφ . or l = 1 ml = 0 :Y1 = 4π 8π 11.16 [Thinking Ahead: The ionization energies depend on the energy or energies of what quantum states? The ionization energy is determined by the gap between the highest occupied valence orbital and the lowest energy unoccupied orbital. In the metal clusters, the magic numbers we discussed predict the relatively stable neutral clusters, and for the alkaline earths, we need to be careful we count the electrons correctly.] Mg will contribute two electrons per atom to the jellium orbitals. The magic numbers will still be 2, 8, 18, and 28, causing Mg, Mg4 , Mg9 , and Mg14 to be particularly stable. In general, it will be easier to ionize an electron from a larger molecule, but Mg4 will be harder to ionize than Mg3 , and so on. From Table 11.1, 7.6 eV is the IE1 for Mg; 3.7 eV is the work function for Mg metal. 7.6

IP 3.7 n 1

11.17 Based on the figure, C60 is spheroidal with a radius of about 2.8 Å, so a volume of roughly 4π 3 (2.8

Å)3 = 92 Å . Volumes of the noble gas atoms are roughly Ne 58 Å3 , Ar 102 Å3 , Xe 165 Å. Therefore, only Ne will easily be contained by C60 . 11.18 The surface area at the van der Waals radius of one La atom would be roughly 2

2 ALa = 4πrLa = 150 Å .

269

If we assume that each carbon–carbon bond length is rC = 1.40 Å, the aromatic C3-C3 bond length in Table 5.1, then the three sp2 bonds to any carbon atom will describe a triangle 2rC sin(π/3) on a side and with area 1 2 2 AC = rC (1 + sin(π/6)) (2 sin(π/3)) = 2.55 Å . 2 So we could coat the La atom with about 150/2.55 = 59 carbon atoms. The van der Waals radius of lanthanum is not definitively measured, but the C60 molecule can completely encase the La atom. 11.19 The original sp2 hybrids are r

r 1 2 (2s) + (2px ) 3 3 r r r 1 1 1 (2s) − (2px ) + (2py ) 3 6 2 r r r 1 1 1 (2s) − (2px ) − (2py ) 3 6 2 The (2pz ) hybrid becomes

√ ( 1 − 0.084)(2pz ) + 0.084(2s), p which means the sp2 hybrids pick up −( 0.084/3) (2pz ) each. The xy plane is at θ = 90◦ , and the angle of the bond is given by the amount of px , py , and pz character (since the 2s is isotropic). We can pick the first sp2 hybrid to find the new angle: p 0.084/3 ◦ p = −11.6◦. 90 − θ = arctan − 2/3

11.20 a. Graphite. We want the average mass per unit volume. Since a single sheet of graphite has this structure:

we can divide the sheet up into triangles of equal area, with one atom per triangle:

If the bond length between neighboring atoms is √ R, the area of √ each triangle is 3 4 3 R2 (the length of one side is 3R). If the distance to the next sheet above or below is R′ , then the volume per √ carbon atom is 3 4 3 R2 R′ .

120 R R 3R

270

For an aromatic ring R ≡ 1.45 Å, so ρ = 2.25 g cm−3 =

12 amu · 1.661 · 10−24 g amu−1 √ 3 3 −8 cm)2 R′ 4 (1.45 · 10

R′ = 1.554 · 10−15

1/2

= 3.94 Å

This is a reasonable van der Waals bond length. The actual value is 3.35 Å[13]. Polyethylene. A simple treatment approximates the polyethylene chains as tubes packed in a triangular arrangement as shown below. The separation between the adjacent tubes is the distance d that we’re looking for. Based on an estimated C C bond length of 1.54 Å and the trigonometry shown below, we can infer that a tube of length L = n(1.26 Å) contains n CH2 subunits, each with a mass of about 14 amu. H d

( 3/2) d 109.5

1.5

4 Ao

54.7 o H

H o

(1.54 A)sin(54.7 o) = 1.26 A

√ The faces of three adjacent tubes points of an equilateral triangle with altitude ( 3/2)d √form the and base d, to give an area of ( 3/4)d2 . Extending this triangle by a distance L along the axis of the tubes forms a triangular prism that occupies a volume of √ 2! 3d n(1.26 Å). V = AL = 4 However, the triangle slices into only one sixth of each of the three tubes, so the mass contained within the volume of this triangular prism is M=

3 n(14 amu) = (7.0n) amu. 6

Now we can solve for the distance d by setting the known density ρm of 0.95 g cm−3 equal to M/V : ρm =

M (7.0n) amu 28 amu = √ 2 n(1.26 Å) = √ = 0.95 g cm−3 3d V 3d2 (1.26 Å) 4

(28 amu)(1.661 · 10−27 kg amu−1 )

d= √ −1 3(0.95 g cm−3 )(10−3 kg/ g)(102 cm/ m)3 (1.26 Å)(10−10 m Å )

#1/2

= 4.7 · 10−10 m = 4.7 Å. the actual value is about 5.5 Å[14]. (With thanks to George Kaminski.) b. Approximate C60 as a sphere with a radius of about 2.8 Å, and therefore volume of 92 Å3 . These spheres will have some space between them when found together. Estimating these dispersion force van der Waals bonds to be similar to the 3.24 Å length of graphite, each C60 occupies a volume of roughly 500–700 Å3 , so ρ≡

60 · 12 amu · 1.661 · 10−24 g amu−1 3

−1 3

600 Å (10−8 cm Å 271

)

= 2.0 g cm−3 ;

this is similar to graphite. 11.21 Assuming all the conformations are sterically feasible (which they won’t be) and non-equivalent, we would estimate the total number of possible conformations at 350 = 7 · 1023 . To see that this is the right form (as opposed to 503 ); try just two amino acids with three conformations each: A, B, and C. The possible combinations for acids 1 and 2 are AA, AB, AC, BA, BB, BC, CA, CB, and CC; nine possibilities = 32 . At a rate of 1013 s−1 , this still requires an absolute minimum of 7 · 108 s to sample 1% of the conformations, which is more than two years. In fact, the sampling time should be much longer because the protein is more likely to sample these conformations randomly rather than trying each one in sequence before sampling any conformation twice. 11.22 The projection of the zero-point motion onto the molecular axis is ±(0.06) sin(109◦ /2). The maximum change in the molecular length from end to end is therefore 2(200)(0.06) sin(109◦ /2) = 20 Å. However, this maximum would almost never be achieved, as we cannot assume that the stretching of all the bonds would occur with the same phase. 11.23 We evaluate the potential energy contribution from these two terms, and find the minimum by setting the derivative of this contribution to zero. One way to handle the geometry is drawn in the figure below. q1

R’ R" γ

R α

θe γ

Using these parameters γ, α, R′ , and R′′ , we can write θ = α + 2γ π − θe γ= 2 θe 2 θe α α ′ ′′ R = 2R sin = 4R sin sin . 2 2 2

R′′ = 2R cos γ = 2R sin

The main point to these substitutions is to let us write the separation R′ between the charges q1 and q2 and the bending angle θ in terms of the same parameter α. In these expressions, R and θe are constants. Now we find the equation for U (α) and minimize that value: 1 1 kθ (θ − θe )2 = kθ (α + 2γ − θe )2 2 2 1 = kθ (α + π − 2θe )2 2 q1 q2 q1 q2 Uelectrostatic = = (4πǫ0 )R′ 2(4πǫ0 )R′′ sin(α/2) c q1 q2 ≡ = 4(4πǫ0 )R sin(θe /2) sin(α/2) sin(α/2) Ubend =

272

dU d = (Ubend + Uelectrostatic) dα dα c cos α2 =0 = kθ (α + π − 2θe ) − 2 sin2 α2 c cos(α/2) + 2θe − π. α= 2kθ sin2 (α/2) Since our force constant kθ is defined in units of radians, let’s use radians for all the numerical angle values for now; in particular, 109◦ = 1.902 rad. The constant c is evaluated as follows: q1 q2 4(4πǫ0 )R sin(θe /2) −(0.03)2 (1.602 · 10−19 C)2 = − − 4.25 · 10−22 J, = −10 4(1.113 · 10 C2 J−1 m−1 )(1.5 · 10−10 m) sin(1.902/2)

c≡

and therefore c −4.25 · 10−22 J = −2.56 · 10−3 . = −1 2kθ 2(50.0 kJ mol rad−2 )(6.022 · 1020 kJ mol−1 /( J))−1 If there were no electrostatic interaction, then c = 0 and α would be α0 = 2θe − π = 0.663 rad. Using this as an initial guess, the equation above for α rapidly converges to 0.634 rad, which means the equilibrium value of θ is θ = α + 2γ = α + π − θe = 1.878 rad = 107.6◦. If the charges are made much larger, this solution for α diverges because the potential approaches −∞ if the charges overlap. The actual molecular mechanics calculation will prevent this by including an excluded volume term for the atoms. 11.24 This question gets complicated right away, because the speed of the atoms moving independently is usually greater than the speed of molecule’s center of mass (which is the relevant number here), especially as the molecule gets large. But this question asks only for an estimate, so let’s use the speed given: λdB =

6.626 · 10−34 J s h ≈ mv (8000 amu)(1.661 · 10−27 kg amu−1 )(5.40 m s−1 )

= 9 · 10−12 m = 0.09 Å.

(A range of answers is acceptable.) Because the speed we used is too fast, this is a lower limit to λdB , but even an order-of-magnitude change would leave the de Broglie wavelength at only about an angstrom. The relevant distance scales for protein folding cover a wide range: proteins exhibit large-scale folding structures (you may have heard of alpha-helices and beta-pleated sheets), which extend over 100 Å, but can also have very sharp curves in the carbon chain, coiling up over a length of less than 10 Å. The larger scale structure can be treated very well with classical spring-type models of the chemical bonds (these methods are called “molecular mechanics”), and only a high degree of detail in folding at smaller scales requires some attention to quantum mechanics. 11.25 With electrons flowing from the surface to the tip, the potential energy at the tip must be lower than at the surface. The potential barrier seen by the electrons moving off of the surface (to higher z) is 273

the work function, and is higher for gold than for aluminum. Therefore, the tunneling barrier is greater for gold, and the corresponding STM current is weaker. If the polarity is reversed so the current flows from tip to surface, the work function still determines the height of the barrier (provided the voltage difference between the surface and STM probe is the same).

4.2 eV

U tip

5.4 eV

tip

Chapter 12 12.1 CO2 , H2 , CO, and N2 are weakly bound (dispersion and very weak CO dipole–dipole bonding). H2 O, HF, (CH3 )O, and heptane are strongly bound (large dipole–dipole interactions or, for heptane, dispersion forces acting over very large molecules). The same bonding forces that make the second group relatively stable as solids also make them relatively stable as liquids. Extending this with a little thermodynamics (covered in the Thermodynamics, Statistical Mechanics, and Kinetics volume) provides a more compelling justification for the increase in the stable temperature range of the liquid as the intermolecular attractions increase. To get the liquid to evaporate at atmospheric pressure, one has to provide not only enough energy to break the intermolecular bonds, but also the additional energy to expand the volume of the sample by roughly three orders of magnitude. This energy of expansion is given approximately by Eexpand ≈ P Vgas , and the ideal gas law (P V = nRT ) lets us rewrite that as Eexpand ≈ ngas RT . Stronger intermolecular attractions require higher temperatures for melting to take place. As the minimum temperature of the liquid (the melting point) increases, the energy needed for this expansion, Eexpand , also increases. Because a liquid with a high melting point needs more energy to expand into the gas, it takes a greater increase in temperature beyond the melting point before the liquid can completely vaporize. 12.2 Only N2 will be transparent, because this is the only molecule with zero dipole derivative in all its modes. 12.3 The dominant term will usually be dispersion, unless the molecules are small and have a significant low order multipole interaction available. A strong dipole–dipole interaction may be greater than the dispersion force (as in H2 O, but not in NH3 ), and a monopole–dipole or monopole–monopole interaction is generally much stronger. a. water: dipole–dipole or hydrogen-bond; b. n-pentane: dispersion; c. mercury: Coulomb or monopole–monopole (Hg ions and free electrons); d. benzene: π-electron dispersion. In order of increasing surface tension: n-pentane, benzene, water, mercury. 274

12.4 CCl4 is a spherical top with no dipole moment and no opportunity for hydrogen bonding, so has much weaker intermolecular interactions than in water. Therefore, the structure in the correlation function will vanish after one or two molecular diameters. Also, CCl4 is larger than H2 O, so the structure will be more spread out. G(R) 1

R (σ)

12.5 Surface tension generally decreases as temperature rises, because at higher temperature there is more energy in molecular motion. This means the molecules can more easily escape the potential wells that bind them to each other to create surface tension. 12.6 Infrared radiation can penetrate vacuum or air (because vibrational transitions are not allowed for N2 or O2 ), but it is efficiently absorbed by water. An infrared detector would be less useful for (b) and (c). 12.7

a. The most stable condition is the linear geometry, with two θ values equal to zero and the other θ equal to π. Because the value of θC depends on the values of θA and θB , the cosine function may be rewritten: 3 cos θA cos θB cos θC + 1 = −3 cos θA cos θB cos(θA + θB ) + 1

= −3 cos θA cos θB [cos θA cos θB − sin θA sin θB ] + 1

= −3 cos2 θA cos2 θB + 3 cos θA cos θB sin θA sin θB + 1.

This last expression is useful, because we can see that the first term is always negative. The value of the second term can be positive or negative, but we simplify this by restricting θA and θB to values between 0 and π/2. (It is possible for one angle to be greater than π/2, but only one interior angle of a triangle can be larger than π/2 and we can always let this be θC .) Over the interval between 0 and π/2, both cos θ and sin θ are always positive, so in that interval the second term is always positive. Therefore, the most stable condition is the one that maximizes the first, negative term and minimizes the second, positive term. This occurs at θA = thetaB = 0, where the first term is −3 and the second term is zero. b. Similarly, the least stable condition is when any of the angles is equal to π/2, as this eliminates the stabilizing term completely. The stabilizing term is always greater in amplitude than the second term, so the fact that this also eliminates the second term is irrelevant. c. The general condition for the correction to be zero is 3 cos θA cos θB cos θC = −1. For an isosceles arrangement, this condition is satisfied for 3 cos2 θA cos(π − 2θA ) = −1 −3 cos2 θA cos(2θA ) = −1

6 cos4 θA − 3 cos2 (θA ) − 1 = 0 √ 1 3 ± 9 + 24 cos2 θA = 12 275

only the + sign yields a positive square

θA = arccos

√ 3 + 33 = 0.55 = 31.4◦. 12

So the three-body interaction stops being favorable for bonding once the central monomer gets more than about 30◦ off the axis between the other two monomers. The term also vanishes if R goes to infinity. 12.8 For an isosceles arrangement, RCA = RBC are both equal to RAB /(2 cos θA ). The expression for u′ABC (Eq. 12.3) can then be written 3αǫ 3 cos θA cos θB cos θC + 1 4(4πǫ0 ) (RAB RBC RCA )3 3αǫ 3 cos2 θA cos θC + 1 = 3 /(2 cos θ )2 ]3 4(4πǫ0 ) [RAB A 2 3αǫ 3 cos θA cos θC + 1 (2 cos θA )6 = . 9 4(4πǫ0 ) RAB

u′ABC =

At this point, we should look for a way to reduce our two variables (θA and θC ) to just one. Since the interior angles of the triangle must add up to π, we know that θC = π − 2θA . We can then combine some trigonometric identities to rewrite cos θC in terms of cos θA : π − θA = 1 − 2 cos2 θA . cos θC = cos (π − 2θA ) = 1 − 2 sin2 2 (We could do this in terms of θC instead and still get the same answer in the end, but this turns out to be a little neater.) Now we can write u′ABC as a function of θA only: 3αǫ 3 cos2 θA cos θC + 1 (2 cos θA )6 9 4(4πǫ0 )RAB 3αǫ 3 cos2 θA 1 − 2 cos2 θA + 1 (2 cos θA )6 = 9 4(4πǫ0 )RAB 3αǫ(26 ) 3 cos8 θA − 6 cos10 θA + cos6 θA . = 9 4(4πǫ0 )RAB

u′ABC =

If we take the derivative of this expression with respect to θA , we can find the stationary points by setting the result equal to zero. Let’s absorb all the constants into a single term A, since we only care about the θ dependence for this problem. ∂u′ABC = A 24 cos7 θA − 60 cos9 θA + 6 cos5 θA (− sin θA ) ∂θA = −6A sin θA cos5 θA 4 cos2 θA − 10 cos4 θA + 1

The possible values for θA range from 0 to π/2 (because θA + θB + θC = π and θA = θB ). Setting the derivative equal to zero gives three possible solutions for θA within values from 0 to π/2: 1. If sin θA = 0, then θA = θB = 0 which corresponds to θC = π and an arrangement with particle C between particles A and B, all on a line. This is a minimum for the potential energy, as shown in the graph below. 2. If cos θA = 0 then θA = θB = π/2, which corresponds to θC = 0 and particle C infinitely far away from A and B. 276

3. Finally, if −10 cos4 θA + 4 cos2 θA + 1 = 0, then we have to solve a quadratic equation to find the angles, but it’s worth it because this is also the only interesting case. Setting x = cos2 θA and √ 2 solving the quadratic 1 = 0 yields x = (2 ± 14)/10, but only the positive √ equation −10x + 4x + √ 2 solution x = (2 + 14)/10 is valid since cos θA must be positive. Then we take the arccos of x to find θA , and the only value between 0 and π/2 is 0.711 rad or 40.7◦. That gives a θC value of 1.72 rad=98.5◦. Plotting u′ABC as a function of θA shows that this is a maximum.

12.9 [Thinking Ahead: How is the vibrational energy related to the number of vibrations per second, the vibrational frequency? The semi-classical picture of spectroscopy in Chapter 3 imagines the electric field of the radiation introducing oscillations in the charge distribution of the atom (or molecule) at the photon frequency, ν. Therefore, for an order of magnitude estimate, we can calculate the photon energy that corresponds to this temperature, 80 K.] 80 K = 80(1.381 · 10−23 ) J = = 1.667 · 106 MHz

80(1.381 · 10−23 ) MHz 6.626 · 10−28

= 1.667 · 1012 s−1

This is the frequency corresponding to vibrational energy at 80 K, and is classically the vibration frequency (as discussed for diatomics). The timescale for two argon atoms to be in the same vicinity is approximatelyone vibrational period. 12.10 The water bending frequency may decrease because van der Waals attraction to other molecules flattens the potential surface, or it may increase because the molecule is more likely to collide with another during a bend than during a stretch. Both are reasonable answers. In fact, the vibrational frequency is lower in the liquid than in the gas, but the effect is less pronounced than for the stretches. 12.11 Table 8.3 lists vibrational modes at 950, 1627, 3337, and 3444 cm−1 . For NH3 , all of these motions will be infrared active. We expect these to become greatly broadened, but only slightly shifted in the liquid. The high-frequency N H stretches will decrease in frequency because hydrogen bonding to other N atoms will stabilize the lengthening bond, whereas the low-frequency bending motions may rise in energy as the atoms bump into neighboring molecules. The rotational constants given in Table 9.2 are 6.3 and 9.4 cm−1 , compared to 10–28 cm−1 for H2 O. These rotational motions become constrained in the liquid, which boosts the transition energies, but the lower rotational constants and the weaker hydrogen bonding (because N is less electronegative than O) indicate that the librational transitions in liquid NH3 will absorb at lower energy than those in liquid water. 277

bends

stretches

librations

400

800

1200

1600

cm−1

2000

2400

2800

3200

12.12 Water, having particularly strong hydrogen bonds, we expect always to have greater surface tension than typical organic compounds. Beyond that, a subtle balance comes into play. Over any small volume, hydrogen bonding (if it is available) will tend to be a stronger attractive force than dispersion. However, when hydrogen bonding is not available over much of the volume, then dispersion—which works anywhere—becomes more important. Among the remaining molecules, 1-hexanol turns out to have the next highest surface tension because it has both good dispersion and some hydrogen bonding available. Methanol is next, because it is still relatively small, so a fairly high density of hydrogen bonds can be formed. Hexane, which is large but has no polar interactions to speak of, has the weakest surface tension. 12.13 The function correctly goes to 0 at R = 0, and peaks at R = R0 . However, it converges to 0 as R → ∞ when it should converge to 1. 12.14 Let • represent the reference atom, and a be the lattice spacing. a

site A B C D E

distance √a 2a 2a √ √5a 2 2a

C D E

Those distances are the five smallest values of R at which peaks will appear in G(R). 12.15 The function is zero between R = 0 and RLJ /2, so we integrate between R = RLJ /2 and RLJ , where the function is equal to N : C = 4πρ

Z RLJ

RLJ /2

G(R)R2 dR

RLJ

= 4πρN =

R3 3

RLJ /2

3 4πρN 7RLJ

3 = 2.5

278

12.16 Z RLJ

G(R)R2 dR 2πR G(R) = e−R/R0 cos + 1 for R > RLJ /2 RLJ C ≡ 4πρ

RLJ = 2 Å

R0 = 8 Å 1 −3 −3 −1 ρ (cm ) = (1 g cm ) (6.022 · 1023 mol−1 ) mol g 18

= 3.346 · 1022 cm−3 "Z # Z RLJ RLJ πR 2 (−R/R0 ) R2 dR C = 4πρ R dR + 4πρ e cos RLJ RLJ /2 RLJ /2

ax2 2a(a2 − 3b2 ) 2(a2 − b2 )x cos bx + − (a2 + b2 ) (a2 + b2 )2 (a2 + b2 )3 RLJ 2b(3a2 − b2 ) bx2 4abx + sin bx + − (a2 + b2 ) (a2 + b2 )2 (a2 + b2 )3 RLJ /2

eax

where R → x,

−

2π → b, RLJ

1 → a, R0

(a2 + b2 ) = 9.885

(sin bx vanishes at RLJ and RLJ /2, so only the cosine terms appear below.) 7.398 −39.42 −0.5 cos(2π) − + − = e−2/8 (9.885) (9.885)2 (9.885)3 7.398 −19.71 −0.125 cos(π) − + − e−1/8 (9.885) (9.885)2 (9.885)3 3

= 0.454 Å

Z RLJ

RLJ /2

R2 dR =

1 3 LJ 1 3 = (23 − 13 ) = 2.33 Å R 3 3 RLJ /2 3

C = 4π(3.346 · 1022 cm−3 )(0.454 Å

+ 2.33 Å )(10−24 cm3 / Å )

= 2.8 12.17 The pair correlation function is not a complete description of the whereabouts of all the molecules in the sample; effectively, it gives only the average number of molecules that will be found at a given distance from any randomly selected molecule in the sample. If we want to find the chances that some particular number of molecules (in this case, zero) are at a given distance, we need to know information that has been averaged out when obtaining G(R). We need to know the distribution of the number of molecules at that value of R, how many molecules have 1 neighbor at that distance, how many have 2 neighbors, or 0 neighbors. There are many different distributions that can give the same average, so knowing the value of G(R) at a particular value of R is not enough to solve this problem. Looking elsewhere, we would find that answer to be incomplete. In TK Chapter 4, we show that the intermolecular potential energy function can be used with the pair correlation function to obtain equations for the number of molecules at any distance R from another, essentially what we’re looking 279

for in this problem. To a good approximation, the potential energy function can be extracted from the pair correlation function, so in principle we could find the equation that’s being asked for. But first we would need to know more about how the pair correlation function is obtained (TK Section 4.3). 12.18 a. Propane in hexane: We would expect the alkane molecules to approximately align their carbon chains. This provides the best packing of the molecules, and the proximity is important for allowing the dispersion forces to bind the molecules together.

b. Argon in water: The water molecules bind more strongly to each other than to the argon, so the solvent molecules will form small cages around each Ar atom.

c. F2 in water: The water molecules can hydrogen bond to each other and to the F2 , so the solvent cage is somewhat disrupted by direct bonding to the solute.

12.19 There’s no getting away from stoichiometry, is there? To get the mole fraction from the mass percent, we take advantage of knowing that adding the mass percentages of water and FeCl3 must add up to 100%. It’s a little simpler if we use the mass fractions wH2 O and wFeCl3 , which are the mass percentages divided by 100%. Using M for the mass, M for the molar mass, and n for the number of 280

moles of each substance, we may write MFeCl3 MFeCl3 + MH2 O nFeCl3 MFeCl3 = nFeCl3 MFeCl3 + nH2 O MH2 O M H2 O wH2 O = MFeCl3 + MH2 O

wFeCl3 =

definition of w M = nM

= 1 − wFeCl3 nFeCl3 xFeCl3 = nFeCl3 + nH2 O −1 −1 nFeCl3 + nH2 O n H2 O = = 1+ nFeCl3 nFeCl3

wFeCl3 + wH2 O = 1 definition of x

This shows that it will be convenient to get the ratio nH2 O /nFeCl3 , which we can get from the w’s: M H2 O wH2 O n H2 O , = nFeCl3 wFeCl3 MFeCl3 and therefore −1 MFeCl3 wH2 O xFeCl3 = 1 + wFeCl3 M H2 O −1 MFeCl3 1 − wFeCl3 = 1+ wFeCl3 M H2 O −1 162.2 1 − 0.846 = 0.379. = 1+ 0.846 18.02 That means that the mole fraction of water is only 1 − xFeCl3 = 0.621. Because we get four ions from the complete dissociation of one FeCl3 molecule, that would suggest that under these conditions the ions outnumber the solvent molecules by a factor of 4(0.379)/0.621 = 2.4. Is that correct? The math may be correct, but the assumption of complete dissociation is not. In such highly concentrated solutions, salts like FeCl3 may remain in solution without fully dissociating. The undissociated salt molecules may be stabilized in solution by clustering with some of the dissociated ions. For example, the central iron atom can coordinate to four chlorine atoms simultaneously, producing an FeCl− 4 ion, so that only one ion needs to be stabilized instead of the five that would form if all the FeCl3 were completely dissociated. 12.20 Based on our assumption, the solubility (in terms of mole fraction) will be proportional to the sum of the surface areas of the two ions formed when these salts dissolve in water. So the ratio of the solubility of NaCl to KCl will be given by XNaCl AKCl = XKCl ANaCl 4π (1.38 Å)2 + (1.81 Å)2 = 4π (1.02 Å)2 + (1.81 Å)2 = 1.20

XNaCl ≈ (1.20)XKCl = 0.095. The actual value is 0.0998, so this result is not too far off. However, the principal extends to larger systems poorly, partly because it neglects the formation of cluster ions at high concentrations. 281

12.21 Because the A molecules are attracted to each other much more strongly than to B or C molecules, the A liquid is expected to separate spontaneously from the other two liquids. The strength of the B C interaction compared to the B B and C C interactions suggests that these two liquids remain thoroughly mixed.

Chapter 13 13.1 A higher coordination number implies the formation of more bonds. These bonds (and hence the measured ionic radius) will be longer because the bonded ions, which all have the same charge, will repel each other.

CN = 3

CN = 4, more repulsion

13.2 Radiation that is not transmitted through the crystal must be reflected or absorbed. Therefore, the polarizability (for reflection) or electric dipole moment (for absorption) must be considerably greater along one axis than along the others. 13.3 [Thinking Ahead: How would you write these operations mathematically? We have seen that inversion can be expressed by reversing the signs on the three Cartesian coordinates, but the Cartesians are not a convenient coordinate set for rotations by 120◦ . Instead, we can use a cylindrical coordinate system (s, z, φ) for this, with φ being the rotation angle around z, as usual, and s being the distance from the z axis. In this case, the rotation amounts to adding 120◦ or 2π/3 to the value of φ. The inversion will change the sign of z as before, and it will change the value of φ to π + φ. Since s is a distance, it is always positive and does not change sign.] Using cylindrical coordinates, we have: 3̄f (s, z, φ) = Iˆ Ĉ3 f (s, z, φ) ˆ (s, z, φ + 2π/3) = If

= f (s, −z, π + φ + 2π/3) = f (s, −z, φ + 5π/3)

= f (s, −z, φ − π/3)

= σ̂h f (s, z, φ − π/3)

= −Ŝ6 f (s, z, φ).

= σ̂h (−Ĉ6 )f (s, z, φ)

The extra − sign at the end does not matter, because if rotation in one direction is a symmetry element, then rotation by the same amount in the opposite direction must also be a symmetry element. 13.4 We use the inversion operator Iˆ to change the corners around (or Ĉ2 (z)σ̂xy ), this would put a at (−1, 2, 2). To move a to (1, 0, 1) would require X̂ 2 (−Ŷ )2 (−Ẑ): IˆX̂ 2 (−Ŷ )2 (−Ẑ). 13.5 Primitive monoclinic:

r r L L L +L z Lr

y 6

r r L L L L L L L L Lr Lr

x -

282

ˆ σ̂h (x, y). The symmetry elements are those of the point group Symmetry elements: Ê, Ĉ2 (z), I, C2h . 13.6 Here is the structure of two adjoining C-centered cubic cells:

In the figure on the right, dotted lines have been added to represent the edges of a primitive tetragonal unit cell. The entire lattice can be reconstructed from this tetragonal cell or from the C-centered cubic cell, but the tetragonal cell is smaller. If the cubic cell has a lattice constant of a, it has a volume of a3 . √ The height of the tetragonal unit cell is still a, but the length and width are each only a/ 2, resulting in a volume of a3 /2. The cubic cell is not the smallest repeating unit of the lattice, so the lattice’s unit cell is primitive tetragonal instead. 13.7 [Thinking Ahead: The corners are not all equivalent, but what happens when this unit cell is placed next to its neighbors in the crystal? Regardless of the positioning of the molecules relative to the lattice points, when placed next to one another in the crystal, the unit cells regenerate the structure of each molecule. If one-third of the molecule lies inside the unit cell at one corner, then the remaining two-thirds must be made up at another corner of the same unit cell, because each unit cell is identical.] As we make the monoclinic angle γ approach 90◦ , the corners become more and more rectangular, and we find that the eight corner molecules contribute on average one-eighth each to any given unit cell; this number must remain constant for all angles γ, or else we would have unit cells with non-integer numbers of molecules. The two molecules in the faces each contribute one-half. The total is 1 1 8 +2 = 2. 8 2 13.8 There are 4Ĉ3 axes and 3Ĉ2 axes. There are no Ĉ4 axes (so it is not octahedral) nor Ŝ4 (so it is not Td ). a. crystal system: cubic b. Bravais lattice: face-centered cubic c. point group: T 13.9 We consider one cell (not, as we were warned, a unit cell) of the crystal to be a cube that contains a carbon atom at its center and the four atoms to which it is bonded at four of the eight vertices. Call this cell A. The effective number of atoms contained within this cell is 3 1 corner = . NA = 1(1 center) + 4 8 2 Call the length of one edge of the cell√d and the C C bond length r. The distance between the center atom and one of the corner atoms is 3d/2, and this must equal the bond length r. Here’s the tricky part. This cannot be a unit cell (and the fractional value for the atoms per cell is a tip-off) because the bonding stoichiometry will not work out. If this were a unit cell, each of the four corner atoms would be shared simultaneously by eight identical unit cells, each of which has a center atom. That means that each of our corner atoms would need eight bonds to each of those center atoms. 283

We know that all the carbon atoms in diamond—whether at the center or at the corner of our cell A—are bonded to exactly four other atoms. Therefore, there are two kinds of cells: A (with an atom at the center and containing four C C bonds), and B (with no atom at the center). There is an equal number of each, so that any given corner atom is shared by four A cells (which account for the four chemical bonds) and four B cells (which have no center atom and therefore no chemical bonds). The B cells must also be cubes of equal volume to the A cells to fit in between all the A cells. Now each atom has four bonds. But each B cell effectively holds only 1 1 corner = NB = 4 8 2 atoms, so the total density is given by: ρ=

(NA + NB )mC 2mC = . VA + VB 2d3

We know the density, so we can solve for d, or r: √ √ 1/3 3 3 mC r= d= 2 2 ρ √ 1/3 3 1.993 · 10−24 g = = 1.544 · 10−8 cm. 2 3.513 g cm−3 The result is 1.544 Å, which is the expected result for C C single bonds. The actual unit cell is face-centered cubic with a basis of two atoms, composed of four of our A cells and four B cells joined together for a total of eight atoms per unit cell, as shown.

13.10 With all the atoms crowded as closely as possible, the volume of the unit cell is a3 = (2r)3 = 8r3

a = 2r where r is the radius of one atom. Eight atoms each contribute 18 of their volume to the unit cell, for a net of 1 atom/unit cell. 4 πr3 π occupied volume = 3 3 = = 0.524 total volume 8r 6 So the percentage of occupied space is 52%. 284

13.11 Let’s start from the unit cell, since we need that for (c) anyway. One possible unit cell is drawn here:

a. The stoichiometry can be determined by the relative contributions of atoms A, B, and C to the unit cell: 1 (corners) = 2 A :1 (middle) + 8 8 1 1 B :8 (edges) + 4 (faces) = 4 4 2 1 1 (faces) + 4 (edges) = 2 C :2 2 4 The molecular formula is AB2 C, or NiCo2 Al. b. Given the dimensions of the original figure, our unit cell has edges of lengths a = b = 8.02 Å/3 = 2.67 Å, and c = 2.65 Å. The unit cell contains net two molecules of NiCo2 Al, and has volume abc. The density is therefore: 2(58.69 + 2 · 58.93 + 26.98 amu)(1.661 · 10−24 g amu−1 ) m = 35.9 g cm−3 . = 3 −1 3 −8 V (18.9 Å )(10 Å ) c. One unit cell is already drawn before part (a). Another choice of unit cell could put the large C atoms at the lattice points. d. The top and bottom faces are square, while the sides are rectangular, so a = b 6= c. The angles α, β, and γ are all 90◦ , so the crystal system is tetragonal. This unit cell uses the A atoms as its lattice points, and has one A atom at each corner plus one in the middle. Therefore, the Bravais lattice is body-centered tetragonal. This unit cell has the highest possible symmetry within the tetragonal crystal system, having the same symmetry elements as a plain tetragonal solid: Ĉ4 , σ̂h , and 4 Ĉ2 axes perpendicular to the principal axis (as well as others). These are sufficient to identify the crystallographic point group as D4h or 4/mmm. 13.12 The number of atoms per unit cell is determined as follows: Ba: 2(1) (center) = 2 Y: 1(1) (center) = 1

Cu: 8 81 (corner) + 8 14 (edge) = 3 O: 8 12 (face) + 12 41 (edge) = 7

The formula is YBa2 Cu3 O7 . Each of the top and bottom faces is rectangular, not square, because two sides contain Cu-O-Cu chains while the other two sides contain only Cu atoms. These two lengths are clearly shorter than the length of the edge that runs parallel to the Ba-Y-Ba chain. With a 6= b 6= c and α = β = γ = 90◦ , the crystal system is orthorhombic. The lattice points are Cu atoms, but not all the Cu atoms 285

are equivalent: eight of those shown are closer to the Y atom than the eight at the corners. The second group of Cu atoms have been chosen as the lattice points, and they appear only at the corners. Therefore, the Bravais lattice is primitive orthorhombic. 13.13 This can be calculated by examining one unit cell and multiplying by the number of identical cells. (P = primitive, B = body-centered, F = face-centered, C = C-centered.) P cubic: F cubic: B cubic: P tetragonal: B tetragonal: rhombohedral: hexagonal:

6 12 + 6 = 18 8 + 6 = 14 4 4 6 6

P orthorhombic: C orthorhombic: F orthorhombic: B orthorhombic: P monoclinic: C monoclinic: triclinic:

2 2 2 2 2 2 2

13.14 body-centered tetragonal, a = 2.0 Å, c = 3.0 Å H HH H s

B B s B B c B B a a (111)

H HH a H a (110) 1 44 + 1 4 14 √ a2 2ac

a a (001)

The (111) surface cuts through the unit cell to make the following triangle: √ a2 + c2

J J J

√ a2 + c2

3.61 Å

J J

46.2◦ J J 2.50 Å J 43.8◦ J

√ 2.83 Å 2a which has area 3.53 Å, and contains a net one-half atom surface density: (001) (110) (111)

−2

0.250 Å −2 0.236 Å −2 0.142 Å

The (001) face has the highest surface density.

1 13.15 The intersections with axes are at: a axis: h = ∞; b axis: k = ; c axis: l = ∞. The Miller 2 indices are h1 , k1 , 1l = (0,2,0). We divide to get the lowest integer values (0,1,0). 13.16

a. (1,0,0) = (0,1,0) = (0,0,1): identical

b. (1,1,0) = (0,1,1) = (1,0,1): identical 286

13.17

a. Only one-fourth of each surface atom contributes to the face of a given unit cell. 1 (4(1/4) atoms/unit cell 100 face) −2 = atoms Å = 0.111 atoms Å−2 . 9 (3.0 Å)2 /unit cell 100 face (a)

u u 6@@@@@@ @@@@@@ 3 Å @@@@@@ @@100 @@@@ @@@@@@ ?@ u @@@@@u -

(b) u HH 3 Å P HHu PP u q P 110

u HH 3 u HHu √ 3 2 Å

3 Å

u u

3 Å

1 atoms (4(1/4) atoms/unit cell 110 face) √ = 0.079 atoms Å−2 = √ (3.0 Å)(3.0 2 Å)/unit cell 110 face 9 2 Å−2

1 13.18 The number of molecules/unit cell = 6( ) + 8( 81 ) = 4 2 V = (5.623 · 10−8 cm)3 2.167 g cm−3 =

r r r

(4 molecules)(× amu/molecule)(1.661 · 10−24 g amu−1 ) (5.623 · 10−8 )3

x = 58 amu − 35 amu = 23 amu, which is the atomic mass of Na, so the answer is NaCl.

13.19 The excess electrons in the n terminal region raise the energy of the highest occupied orbitals (HOMOs), while the electron deficiency in the p terminal region reduces the HOMO energy.

unbiased forward biased reverse biased For the diode to function properly, there must still be a potential energy barrier to flow of the electrons across the np junction, because otherwise current would flow spontaneously from the n region 287

to the p region. This barrier is overcome by applying a small voltage that drops the potential energy of the p region relative to the n region (forward biasing). If the same voltage is applied in the opposite direction (reverse biasing), current does not flow from p to n, because the p terminal electrons see a substantially higher barrier than the higher energy electrons of the n terminal. 13.20 The primitive cubic unit cell for CsCl, allowing the closest contact between the anions and cations, has one Cs+ ion at its center and Cl− ions at the vertices (or vice versa):

a If the length of one side is a, the major diagonal (from one corner, through the center of the opposite corner) is √ 3a = 2(rCs+ + rCl− ) = 2(1.67 Å + 1.81 Å)(1.015) = 7.06 Å Note that the coordination number is 8. a = 4.08 Å ρ=

(132.9 amu + 35.45 amu)(1.661 · 10−24 g amu−1 ) m = V (4.08 · 10−8 cm)3

4.10 g cm−3

The actual value is 3.99 g cm−3 . 13.21 Each unit cell has an average of one molecule contained within. 6.02 · 1023 unit cells with a mass equal to

Therefore, one mole has

M = 2(26.982) + 3(15.995) = 101.95 g mol−1 ; and one mole has a volume of Vm =

M = 25.71 cm3 mol−1 . ρm

Therefore, one unit cell has a volume of V = 13.22

Vm 25.71 cm3 mol−1 = = 4.27 · 10−23 cm3 . NA 6.022 · 1023 mol−1

(4 molecules/cell)(10−5 impurities/molecule) = 2 · 1017 impurities cm−3 . (5.68 · 10−8 cm)3 288

13.23 In one unit cell of the bcc lattice are one atom in the center and eight octants (1/8) of other atoms at the eight corners:

3 8 2 √ d = √ d3 3 3 3 3 3 4 π 1 4 d d Volume filled = π +8 = d3 π 3 2 8 3 2 3 √ 3π π/3 √ = = 0.680 fraction filled = 8 8/3 3

Volume of unit cell =

Because this is true for the unit cell, it is true for the whole lattice. 13.24 Drawing the unit cells with the Cl− atoms at the vertices, the face-centered cubic unit cell looks like (a) below and the primitive cubic looks like (b): w

(a)

w e fcc

e e

where and

e = Na+ w = Cl−

4 molecules/unit cell

w (b) w

w w

1 molecule/unit cell

Define three distances: R1 is the length of a side, R2 is the length of the diagonal of a face,and R3 is the length of the unit cell diagonal. A A

@ @ R2 @ @ @ @ R1 R2 =

√ 2R1

R3 = 289

A A R3 A A HH A R2 HH A

The most compressed form of any NaCl has the minimum value of one of these distances without reducing the others below their minimums. The minimum values for fcc NaCl are: R1 = 2RCl− + 2RNa+ = 5.52 Å R2 = 4RCl− = 7.24 Å R3 = 2RCl− + 2RNa+ = 5.52 Å. √ If we push the atoms √ together along the side of the unit cell, so R1 = 5.52 Å, then R2 = 2(5.52) = 7.81 Å and R3 = 3(5.52) = 9.56 Å. None of these is less than the minimum value, so this structure is acceptable. If instead we tried to compress along the diagonal, so that R2 = 7.4 Å, then R1 = √ 7.24 Å/ 2 = 5.12 Å, which is less than the minimum value of R1 , so that structure is not acceptable. The density of most compact fcc NaCl: (4 molecules/unit cell)(23 + 35) amu(1.66 · 10−24 g/amu) = 2.29 g cm−3 (5.52 Å)3 /unit cell (10−8 cm/ Å)3 Similarly, for primitive cubic NaCl, the minimum distances are R1 = 2RCl− = 3.62 Å R2 = 2RCl− = 3.62 Å R3 = 2RCl− + 2RNa+ = 5.52 Å. √ Because R2 = 2R1 , R2 6= 3.62 Å, as R1 must be at least 3.62 Å. If R1 = 3.62 Å, then R2 = 5.12 Å and R3 = 6.27 Å ; therefore, this structure is allowed. The density of this compact primitive cubic NaCl lattice is: (1 molecules/unit cell)[(23 + 35) amu](1.66 · 10−24 g amu−1 ) (3.62 Å)3 /unit cell(10−8 cm

Å

−1 3

)

= 2.03 g cm−3

The fcc lattice is more dense. The actual value for the NaCl density is 2.17 g cm−3 , because the literature values for the atomic radii of Na+ and Cl− do not predict the Na Cl bond length exactly. 13.25 When we cram the atomic ions together, the crystal will be stable as long as the attractive interactions between opposite charges are stronger than the repulsive interactions between equal charges. The critical question is this: for what relative values of r do ions of like charge, rather than opposite charge, come into contact? For cubic unit cells, and treating the ions as hard spheres, the geometry is not too complicated. All else being equal, the primitive cubic lattice will tend to be the more stable of our two choices for diatomic salt lattices, because the coordination number of cations with anions (and vice versa) is eight, rather than six for the face-centered cubic lattice. Therefore, we first consider what will destabilize the primitive cubic lattice. For the lattice structures shown, ions in the same unit cell may come into √ contact along any of three directions: along an edge (length a); along the diagonal of a face (length 2a); or along the principal diagonal that extends from one corner of the cube, through the center, to the opposite corner (length √ 3a). For the primitive cubic lattice, oppositely charged ions come in contact along the principal diagonal, but equally charged ions will come in contact along the other two directions. For the crystal to be stable, we want √ 3d = 2(rlarge + rsmall ) √ 2d > 2rlarge d > 2rlarge . 290

√ Of the last two conditions, 2d is greater than 2r if d is greater than 2r, so we only require d > 2r. In order to satisfy these constraints, the ratio of our r values must fall in a certain range: √ d = 2(rlarge + rsmall )/ 3 > 2rlarge √ rlarge + rsmall > 3rlarge √ rsmall 1+ > 3 rlarge √ rsmall > 3 − 1 = 0.732. rlarge When the radii of the cations and anions differ by more than a factor of 0.73, the crystal becomes unstable because of the proximity of similar charges. The face-centered cubic lattice is more forgiving of size, but it also has its constraints. Anions and cations √ may come into contact along either the edges (length d) or along the principal diagonals (length 3d). If the ions came into contact along the principal diagonal, then √ 3d = 2(rlarge + rsmall ) √ d = 2(rlarge + rsmall )/ 3; the length d of one side of the unit cell would be too small to accommodate an ion at each corner and one in the middle. Therefore, the oppositely charged ions can only come into contact along the edge. Along the face diagonals, only like charges will come into contact. In this case we require: √ 2d > 4rlarge d = 2(rlarge + rsmall ). This results in the following minimum value for rsmall /rlarge : √ d = 2(rlarge + rsmall ) > 4rlarge / 2 √ rlarge + rsmall > 2rlarge √ rsmall 1+ > 2 rlarge √ rsmall > 2 − 1 = 0.414. rlarge 13.26 a. In between the square and hexagonal lattices, we will have the geometry of the oblique lattice (Fig. 13.3). b. One way to calculate the packing efficiency in this geometry is to calculate the area of the parallelogram that connects the centers of four adjacent circles. We find that area Ap is equal to (2R)(2R sin η) = 4 sin ηR2 . At any value of η, this parallelogram encloses the area of exactly one circle, Ac = πR2 . The ratio gives us the packing efficiency as a function of η: Ap πR2 π = = . Ac 4 sin ηR2 4 sin η This is a continuously decreasing function from η = 60◦ to η = 90◦ , and then increases symmetrically as η increases past 90◦ , making the square lattice at 90◦ a minimum efficiency point. 3

13.27 The mass of one benzene molecule is about 78 amu, and the volume of the unit cell is abc = 514 Å . The number of molecules per unit cell is therefore ! 3 molecules 0.606 amu 514 Å 1 molecule =4 . 3 78 amu unit cell unit cell 1 Å 291

We can put one benzene at each corner of the unit cell, but this contributes only one molecule to the unit cell overall. We have to account for three more. Because the unit cell has six faces, and any molecule located at one of the faces lies half in our unit cell, we can obtain three more molecules by placing them at the faces. This is not a face-centered lattice (there is no face-centered orthorhombic Bravais lattice) because the benzenes in the middle of each face are not oriented the same way as the benzenes at the corners. For the center of each face to be a lattice point, the molecules at that site would have to be indistinguishable from those at the corners. Instead, this is a primitive lattice with a basis of four molecules. The original structure was determined in Ref. [15].

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Sources and Acknowledgments I use this space to discuss the major references that the main text draws on, so as to give credit to the many shapers of this book without intruding on the narrative itself. Presentations of detailed applications and examples relied chiefly on review papers and research papers in the scientific literature. The broader coverage of topics and organization, on the other hand, is strongly influenced by the instructors and textbooks of the courses I took as a student, and by the books I chose for my own classes as a first-time instructor. I thank the authors cited below for their insights and their example. Please note that experimental data listed in tables—such as the spectroscopic parameters in the chapters on rotations and vibrations—are often drawn from a variety of sources, sometimes the original scientific literature in print, and values may differ from more recent work and in some cases may not be scaled for subsequent changes in the fundamental constants. Where the pedagogical purpose of cited values depends upon their precision or accuracy, I have attempted to use the best values available. However, in general this text should not be considered a reference for the best values of experimentally measured molecular parameters. • Chapter 1. – The introductory material in this chapter appears in practically every physical and general chemistry textbook. For an exemplary case, however, Moore’s was the first physical chemistry textbook [16] (to the best of my knowledge) to approach the entire course from the standpoint of a single, unified model to describe how chemistry works. An outstanding book and well ahead of its time. • Chapter 2. – Liboff’s Quantum Mechanics [17] is written for an introductory physics course, and presents the basic precepts of quantum mechanics with commendable clarity, and assumes about the same level of mathematical ability as the present text. – The huge and authoritative Physical Chemistry text by Berry, Rice, and Ross [18], originally published in 1980, is perhaps the most complete single resource for physical chemistry. Berry, Rice, and Ross influenced the inclusion of the Bohr model in this chapter, although I have tried to more strongly justify its relevance to contemporary quantum mechanics. – Studies of real atoms as they approach the Bohr limit are given by [19, 20, 21]. • Chapter 3. – The solution of the one-electron atom Schrödinger equation follows the logic of Lowe’s text [22]; I have filled in and tried to clarify many of the steps. – Table 3.1 draws on Ref [23] for the associated Legendre polynomials, and Ref. [24] was used for choice of phase. 293

– Papers that discuss the angular momentum of the photon include [25, 26]. Reference [27] presents the history of the Stern–Gerlach experiment. Reference [3] discusses the Thomas precession correction to the spin-orbit constant. • Chapter 4. – For the major topics of this chapter, I drew much from Liboff’s section on perturbation theory [17], Berry, Rice, and Ross [18] on the variational principle, and Weissbluth [28] on the vector model of the atom. – The basis of the footnote showing the Hamiltonian is Hermitian in Section 4.2 is Ref. [29]. – My original reference for background on the role of radiative cooling in interstellar molecular clouds [30]. – Charlotte E. Moore at the then National Bureau of Standards assembled the indispensable reference for atomic energies [31], such as those given in Table 4.2. – My reference for the metallic properties of highly compressed hydrogen is [32]. – [33]. The orbital energies in Fig. 4.5 were determined by MCSCF/cc-pvDZ calculations on neutral B through Ne, using Gaussian 98. – [34]. The HF orbital energies in Table 4.1 were determined by UHF/cc-pvDZ calculations using Gamess. • Chapter 5. – The concise exposition of atomic and molecular physics by Karplus and Porter [35] is the chief influence for my treatment of molecular orbital theory, although I also consulted Gilbert N. Lewis’ original monograph [36], in which he finds the greatest omission of his theory to be the unexplained driving force for electrons to pair up. – Theodore Berlin’s treatment of the binding force [37] appears as the introduction to chemical bonding in Berry, Rice, and Ross [18], where I first saw it. – Jean’s text [38] provides a detailed overview of MO theory, its relation to other models, and many examples. – Ball’s text [39] is a quirky little book, full of arcane stories and philosophical tangents, and a marvelous qualitative perspective on molecular structure. – Brown and Carrington [40] wrote this thorough survey of energy levels in diatomic molecules, and one of my references for the properties of H+ 2 . Their major interest lies in the analysis of rotationally resolved spectroscopic data, with an emphasis on free radicals. – [41]. An update of Field’s previous Perturbations in the Spectra of Diatomic Molecules, this text is more strongly focused on electronic spectroscopy and the excited state dynamics of diatomics than the Brown and Carrington text. – References on the interpretation of d-orbital hybrids include [42, 43, 44]. The principal reference for Table 5.1 was Ref. [45], and the experimental bond lengths in Example 5.3 were taken from Ref. [46]. • Chapter 6. – Cotton’s textbook [47] is the definitive introduction to group theory for chemical applications. – My sources for the pseudorotation in PF5 are Refs. [48, 49], and Ref. [50] was my reference for the TiO2 –ammonia binding in Example 6.10. 294

• Chapter 7. – The books by Steinfeld [51] and Harris and Bertolucci [52], both oriented towards the interested undergraduate student or starting graduate student, were used as references on the electronic spectroscopy of molecules, The potential energy diagram for N2 is based on a similar figure in Steinfeld’s text. – The use of correlation diagrams as the basis for understanding molecular orbital symmetry and energy ordering is drawn from Berry, Rice, and Ross [18]. – Mizushima painstakingly assembled one of the most useful sets of tables for spectroscopists in his appendices to his book [11], used as the reference for our Tables 7.1, 8.2, and 9.1. – An early paper in the description of orbital angular momentum in molecules is the paper by Lamb [53]. • Chapter 8. – The textbook by Wilson, Decius, and Cross [54] is a standard reference for vibrational spectroscopists, having established many of the conventions and approaches, but I find it rather difficult to read. The Banwell and Graybeal texts [55, 56] are oriented more for students. – My reference on vibrational overtones and the color of water is [57]. • Chapter 9. – There are several fine references on rotational energy levels and spectroscopy, ranging from Kroto’s concise treatise [58], through the historically significant Townes and Schawlow text [59], to the exhaustive texts by Brown and Carrington [40] by Gordy [60]. – More information on calculating the rotational distortion constant is available in Zare’s book [61]. – My source for the characterization of H2 by rotational Raman spectroscopy is Ref. [62], and Ref. [63] describes the determination of molecular structure from the rotational constants of isotopically substituted molecules. • Chapter 10. – Several of the derivations in this section are modeled on presentations in the recommended texts by Rigby [64] and by Stone [65] (a new edition of Stone’s text appeared in 2013). Many topics related to intermolecular forces are discussed at a level suitable for advanced undergraduates. Rigby is also the source for most of the values in Table 10.2. – References [66, 67] are Pauling’s original paper defining the hydrogen bond, and the subsequent refinement by Pimentel and McClellan. – The multipole approach to describing electric fields is based on Tinkham’s Group Theory and Quantum Mechanics [68]. – Reference [69] is the principal reference for Table 10.1, with various other entries taken from more recent literature. • Chapter 11. – Eisenberg’s Physical Biochemistry [70] gives a survey of physical chemistry principles applied to biochemical systems, including some discussion of the polymer nature of biomolecules. 295

– For more on the Gregory–Newton problem, see Ref. [71]. – References by Saykally [72, 73] include a major review paper on the known structures of water clusters, and a subsequent survey of the field. – Background on the nature of water’s anomalous contraction with temperature was provided by Ref. [74], and Ref. [75] was the principal reference for the structure of small nanotubes. • Chapter 12. – Chandler [76] covers the pair correlation function in a similar fashion in his chapter on classical fluids. – The primary references for Tables 13.3 and 13.5 are Refs. [45] and [77], respectively. Reference [78] was my source o n the settling of window-glass. • Chapter 13. – At the time this book was first being developed, the textbooks by Silbey and Alberty [79] and Noggle [80] were among few that devoted entire chapters to crystal structure, and the original versions of my Tables 13.1 and 13.2 and Figure 13.5 were drawn from these references. – Buerger’s textbooks [81, 82] are excellent introductions to crystallography. Reference [81] in particular includes a methodical presentation of the fundamental mathematics of space groups. The second book [82] focuses more on the technical aspects of crystallography. • Tools of the Trade. The histories and workings of the various experimental methods described in these sections were based on numerous references, but the following stand out: [83] (LEED), [84] (AAS), [85, 86] (NMR), [87] (photoelectron spectroscopy), [88] (Raman), [89] (FTIR), [90] (molecular beams), [91] (ATM, AFM), [92] (neutron diffraction).

Acknowledgments I am grateful to Prof. George Kaminski for his extensive review of this solutions manual. I also thank the biosketch subjects for their permission to provide a glimpse of their inspiring work. This manuscript was typeset using using LATEX2e, by Leslie Lamport. LATEX2e is based on TEX by Donald Knuth. Layout and special characters were further facilitated by the LATEX2e utilities dchem.sty and wrapfig.sty by Donald Arseneau, psfig.sty by Trevor J. Darrell, answers.sty by Mike Piff, and fancyhdr.sty by Piet van Oostrum. Figures in this solutions manual were prepared using the following programs: LATEX, Xfig (Xfig.org), Gnuplot (GNU Software Foundation), GaussView (Gaussian Inc.), and Maple (Waterloo Maple Inc.). I also thank the developers of GrindEQ (www.grindeq.com), which was invaluable in the development phase of this text.

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