Preview 3000 solved problems in chemistry by tusachduhoc

SCHAUM'S SOLVED PROBLEMS SERIES 3000

SOLVED PROBLEMS IN

CHEMISTRY

by David E. Goldberg, Ph.D. Brooklyn College

SCHAUM'S OUTLINE SERIES McGraw-Hill

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San Francisco

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CONTENTS Chapter 1

Chapter

MEASUREMENT

1.1 1 .4 1 .6

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Chapter

Elements, Compounds, Mixtures

Chapter

Periodic Trends

4.4

Chapter 12

Electron Diffraction I

2.3

Ionic and

The Mole; Formula Calculations

Light I

5.2

5.3

6.2

Photoelectric Effect

4.3

Empirical

5.4

Bohr Theory

Electronic Structures of Atoms and Ions

Consequences of Electronic Structure

Bond Lengths and Bond Energies I

7.4

Resonance

BONDING THEORY

8.1 Valence Bond Theory

7.5

8.2

7.2

Dipole Moment

Geometry of Molecules

Organic Nomenclature and Classification

9.4

9.2

Balancing Chemical Equations

10.2

7.3

Other Intermolecular

119

Structural Isomerism

Other Properties of Solutions I

THERMODYNAMICS 18.1 Heat, Internal Energy, Enthalpy I 18.2 Heat Capacity and Calorimetry I 18.3 Law of Dulong and Petit I 18.4 Enthalpy Change I 18.5 Enthalpies of Ions in Solution I 18.6 Free Energy Change and Entropy I

335

CHEMICAL KINETICS 19.l Rate Laws I 19.2 Half-Life I 19.3 Collision Theory I 19.4 Reaction Mechanisms

373

20.1 20.4

399

EQUILIBRIUM

Le Chatelier's Principle I

20.2 Equilibrium Constants

Thermodynamics of Equilibrium I

20.3 KP

ACIDS AND BASES 21.1 Acid-Base Theory I 21.2 Ionization Constants I 21.3 Ionization of Water I 21.4 Buffer Solutions I 21.5 Hydrolysis Equilibrium I 21.6 Polyprotic Acids and Bases I 21.7 Indicators

427

HETEROGENEOUS AND OTHER EQUILIBRIA 22.l Solubility Product Equilibria I 22.2 Competitive Reactions I 22.3 Coordination Equilibria I 22.4 Miscellaneous Applications of Equilibria I

480

ELECTROCHEMISTRY 23. l Electrical Units I 23.2 Electrolysis I 23.3 Galvanic Cells I 23.4 Nernst Equation I 23.5 Practical Applications I 23.6 Electrochemical Equilibrium and Thermodynamics I

510

NUCLEAR AND RADIOCHEMISTRY 24.1 Nuclear Particles and Nuclear Reactions I 24.2 Half-Life I 24.3 Binding Energy I 24.4 Nuclear Cross Section I 24.5 Radiochemistry I

542

NONMETALS 25. l General I 25.2 Halogens I 25.3 Group VI Elements I 25.4 Group V Elements I

563

METALS AND METALLURGY 26.1 Metallic Bonding I 26.2 Alloys I 26.3 Main Group Metals I 26.4 Transition and Inner

573

and Titration I

25.5 Groups IV and III I 25,6 Noble Gases I

Transition Metals I 26.5 Metallurgy I

27. I 27.3

27.2 Nomenclature of Coordination Compounds Isomerism of Coordination Compounds I 27.4 Valence Bond Theory of Coordination Compounds I 27.5 Crystal Field Theory I 27.6 Other Concepts I

COORDINATION COMPOUNDS

Properties of the Coordination Sphere I

584

609

To the Student The best way to ensure that you understand the concepts of General Chemistry is to solve many problems on each topic. You should attempt a large number of different problems, rather than merely reworking the same problems again and again, since you might have a tendency to memorize the solution in the latter case. Be sure to read each problem carefully, since a small difference in the wording of a problem can make a large difference in its solution. Since there is no set order of topics in general chemistry texts, you will have to consult the Table of Contents in this book to find the problems you wish to do. The problems in each section start with the more basic ones and progress to those that are more difficult. In some chapters, you may find problems based on material which you have not yet covered in your course. Do not attempt to do these before you cover the material on which they are based. For example, some texts cover equilibrium before thermodynamics, while others cover it afterward. Do not attempt the section on the

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thermodynamics of equilibrium until you have been introduced to both topics in your course. There are numerous methods to solve most problems. The solution methods are usually related to each other but may seem very different. In this book, some related problems are solved using one method, and 路some with another. Many of the problems, especially in the early chapters, use several solution methods. You should attempt to do the problems yourself before looking at the solutions given. If you get the correct answer using a reasonable method, you need not worry that you did not use the method selected here. If the method presented is clearer that the one you used, however, you might consider adopting it for future similar problems.

CHAPTER 1

UMeasurement

1.1 1.1

EXPONENTIAL NUMBERS

List the powers of ten, from 1 0- 4 to 1 06 , together with their explicit meanings. I

10 - 1 = _!__ = 0. 1 10

1 0째 = 1

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10 1 = 10 1 1 10 - 2 = - = - = 0.01 1 0 2 100

10 2 = 1 0 x 10 = 100 1 03 = 10 x 10 x 10 = 1 000 104 = 10 x 10 x 10 x 10 = 10 000

10 - 3

105 = 10 x 10 x 10 x 1 0 x 10 = 1 00 000

1 0 6 = 1 0 x I 0 x 1 0 x 10 x 10 x 10 = 000 000

1 103

= - =

-1000

= 0.001

1 1 10 - 4 = - = -- = 0.0001 1 04 1 0 000

In the expression 105, the base is 1 0 and the exponent is 5. 1.2

Express the following numbers in standard exponential form:

(a) 22 400 (e) 0.0454

(b)

7 200 000 (f) 0.000 06

(d)

(h)

0.454 0.000 000 5

Any number may be expressed as an integral power of 10, or as the product of two numbers one of which is an integral power of 10 (e.g., 300 = 3 x 1 0 2 ). (c) 454 = 4.54 x 10 2 (b) 7 200 000 = 7.2 x 10 6 (a) 22 400 = 2.24 x 1 04 (e) 0.0454 = 4.54 x 10 - 2 (d ) 0.454 = 4.54 x 1 0 - 1 (f) 0.000 06 = 6 x 1 0- 5 (h) 0.000 000 5 = 5 x 10-7 (g) 0.003 06 == 3.06 x 10 - 3

Moving the decimal point one place to the right is equivalent to multiplying a number by 1 0; moving the decimal point two places to the right is equivalent to multiplying by 1 00, and so on. Whenever the decimal point is moved to the right by n places, compensation can be achieved by dividing at the same time by 1 0"; the value of the number remains unchanged. Thus 0.0325 =

3.25 = 3.25 x 10 - 2 10 2

Moving the decimal point one place to the left is equivalent to dividing by 10. Whenever the decimal point is moved to the left n places, compensation can be achieved by multiplying at the same time by 10"; the value of the number remains unchanged. For example, 7296 1.3

Evaluate: (a) (d )

a3 x a5 107 x 10- 3

(b) 10 2 x 103 (e) (4 x 1 04)(2 x 10 - 6 )

72.96 x 10 2 = 7.296 x 103

10 x 10 (f) (2 x 105)(3 x 1 0 - 2 ) (c)

In multiplication, exponents of like bases are added. (a) a3 x a5 = a3 + 5 = a 8 _ (b) 10 2 x 103 = 10 2 + 3 = 105 + 1 1 (d) 101 x 10- 3 = 107 - 3 = 1 04 (c) 10 x 1 0 = 10 = 1 02 (e) (4 x 104 )(2 x 10 - 6 ) = 8 x 104 - 6 = 8 x 10 - 2 (f) (2 x 105)(3 x 1 0 - 2) = 6 x 1 05 - 2 = 6 x 1 03

1.4

C H APTER 1

Evaluate: a5 (a) -3 a

(b)

10 2 10 5

8 x 10 2 2 x 10- 6

(c)

(d)

5. 6 x 10- 2 1.6 x 104

In division, exponents of like bases are subtracted. 10 2 a5 5 - 3 = a2 = 10 2 - 5 10- 3 (a ) = a (b) 10 5 a3 5.6 x 10 - 2 5.6 8 x 10 2 � 2 4 = 3.5 2 6 = 4 x ios x 10- = x 10 + (d) = (c) 1 .6 x 104 1.6 2 x 10- 6 2

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1.5

1.6

1.8

1.9

1.10

1.1 1

10- 6

Evaluate the following expressions: ( a ) a 0 (b) 10 ° (c ) (3 x 10) 0 (d) 7 x 10 ° (e) 8.2 (d) 7 x 10 ° = 7 (e) 8.2 (c) (3 x 10) 0 = 1 (b) 10 ° = 1 I (a ) a0 = I Express as radicals: (a ) 10 2 1 3 (b) 10 3 ' 2 (c) 10 11 2 (d) 4 3 1 2 (c) 10 11 2 = JiO I (a ) 10 2 1 3 = {/102 (b) 10 3 1 2 = J103 I (a )

(l0 3 ) 2 = 10 3x 2 = 10 6

(b)

( 10 - 2 ) 3 = 1 0 - 2 x 3

x x

10 ° 10 ° = 8.2

(d) 4 3 1 2 = J43 = .j64

10- 6

Take the square root of each of the following numbers, using exponential notation as an aid: (a) 90 000 (b) 3.6 (c) 4.9 x 10- 5 Take the cube roots of the following numbers: (d) 8 x 109 (e) 1 .25 x 10 5

10 3

I To extract the square root of a power of 10, divide the exponent by 2. If the exponent is an odd number it should be increased or decreased by 1, and the coefficient adjusted accordingly. To extract the cube root of a power of 10, adjust so that the exponent is divisible by 3; then divide the exponent by 3. The coefficients are treated independently. (a) y'9oOoO = � = J9 x Jl04 = 3 x 10 2 or 300 (b) J3.6 x 10 3 = J36 x 10 2 = J36 x .[102 = 6 x 10 1 or 60 (c) J4.9 x 10- 5 = J49 x 10- 6 = J49 x Jlcf6 = 7 x 10- 3 or 0.007 (d) � = � x M = 2 x 10 3 or 2000 (e) </1.25 x 10 5 {/1 25 x 10 3 = eft2s x </103 = 5 x 10 or 50 =

Calculate the value of

(4.0

10- 1 00 ) + (2.0 x 10- 1 0 1 ) 2_0 x 10 2 00 _

You must know the rules for handling exponents, because not all electronic calculators do this type of problem. (4.0 x 1 0 - 1 00 ) + (2.0 x 1 0 - 1 0 1 ) (4.0 x 1 0 - 1 00) + (0.20 x 1 0 - 1 00) 4.2 x 1 0 - 1 00 2. 1 x 10 1 00 2.0 x 10- 2 00 2.0 x 1 0 - 2 00 2.0 x 10- 2 00 =

Units can also be expressed with exponents, both positive and negative. Show the dimensions of the result two ways when a mass in grams (g) is divided by a volume in cubic meters (m 3 ). I g/m 3 or g · m - 3 Compute and state the answers in standard exponential form: (a ) (2.0 x 10 1 3 ) + (1.5 x 10 1 4) (b) (8.0 x 10- 1 4)/(4.0 x 10 - 1 3 ) 4 17 (d) (6.6 x 10 1 5 ) - (3.0 x 10 1 6 ) (c) (5.0 x 10 )(2.0 x 1 0 - ) I ( a ) 1 .7 x 10 1 4 (b) 2.0 x 1 0 - 1 (c) 10 x 10 1 3 1.0 x 10 1 4 =

1.12

(d) -2.3

101 6

10- 2 (b) 104/10- 3 (c) 2.5 x 10 7 x 4.0 x 10 3 I ( a ) In multiplication, add exponents algebraically: 10 2 (b) In division, subtract exponents algebraically: 10 7 (c) Multiply coefficients and exponential parts separately and simplify: 10 x 10 1 0 10 1 1 . Evaluate: ( a ) 104

M EASU R E M ENT

1.13

Simplify the following expressions:

48 000 000 1 200 3 (c) (4 x 1 0 - )(5

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(b)

(a)

(e) (J4.0

0.0078 120 (6 000 000)(0.00004) 4 (d) (800) 2 (0.0002) 3 (/) ( {/ 6.4 x 1 0 - 2 )(\/27 000)({/ 2. 1 6

104) 2

10- 6 )(J 8. l

10 3 )(J0.0016)

10- 4)

48 OOO OOO 48 x 10 = 4.0 x 106 - 2 = 4.0 x 104 or 40 000 1 200 1 2 x 10 0.0078 7.8 x 10- 3 (b) = 6.5 x 10- 5 or 0.000 065 1 20 1.2 x 10 2 2 2 2 (c) (4 x 1 0 - 3 )(5 x 1 04) = (4 x 1 0 - 3 )(5 x 1 0 8 ) 4 x 5 x 10- 3 + 8 = 1 00 x 10 5 = I x 10 7 4 4 6 5 (6 000 000)(0.000 04) (6 x 10 )(4 x 1 0 - ) 6 x 44 10 6 x 1 0 - 2 0 x (d) = = 2 4 2 2 2 3 3 (8 x 10 ) (2 x 10- ) 8 x 2 3 104 x 1 0 - 1 2 (800) (0.000 2) 10- 1 4 6 x 256 10 6 - 2 0 6 x = 3 x = 2 4 1()-=8 = 3 x 1 0 64 x 8 10 - 1 (e) (J4.0 x 1 0 - 6 )(J 8. l x 10 3 )(J0.0016) = (J4.0 x 10- 6 )(J 8 l x 10 2 )(J1 6 x 1 0 - 4 ) = (2.0 x 10- 3 )(9.0 x 10 1 )(4.0 x 1 0 - 2 ) = 72 x 10- 4 = 7.2 x 1 0 - 3 or 0.0072 2 (/) ({/ 6.4 x 1 0 - )({/2 7000)({/2. 1 6 x 1 0 - 4 ) = ({/ 64 x 1 0 - 3 )({/27 x 10 3 )({/ 216 x 1 0 - 6 ) = (4.0 x 10 - 1 )(3.0 x 10 1 )(6.0 x 10- 2 ) = 72 x 10- 2 or 0.72 (a )

1.14

1.15

1.16

Find the logarithm of each of the following numbers: (a) 4.56 (b) 1 .70 (c) 9.75

I (a)

0.6590

(b) 0.2304

(d)

0.0294

(e) 0.4997

(d)

(/)

Find the antilogarithm of each of the following: (a) 0.4502 (b) 0.8579 (c) 0.7042 I (a) 2.82 (b) 7.21 (d) 4.055 (e) 9.322 (c) 5.06

(d)

1 .07 (e) 3. 1 6 (/) 1 .00

0.0000

0.6080 (e) 0.9695

Determine the antilog of each of the following: (a) 2.6 1 70 (b) 7.42 (c) - 2.0057 (d) - 0.4776 I (a ) 4. 14 x 10 2 414 (c) 9.87 x 1 0 - 3 (b) 2.6 x 107 (d) 3.33 x 1 0 - 1 = 0.333 =

1.2 1.17

METRIC SYSTEM

Show the relationships among the metric units of length, volume, and mass and the common English units of these quantities.

1 meter = 100 cm = 1000 mm = 0.001 km = 39.37 in. 1 kilogram = 1000 g = 2.2 lb 1 inch = 2.54 cm = 0.0254 m = 25.4 mm = 2.54 x 107 nm 1 foot = 1 2 in. = 12 x 2.54 cm = 30.48 cm = 0.3048 m = 304.8 mm 1 liter = 1 dm 3 = 10- 3 m 3 = 1 .06 quarts 1 yard = 3 ft = 9 1.44 cm = 914.4 mm = 0.9144 m 1 mile = 5280 ft = 6.336 x 104 in. = 1.609 x 10 5 cm 1.609 x 10 3 m = 1 .609 1 pound = 0.4536 kg = 453.6 g = 4.536 x 10 5 mg 1 1 1 ounce = !6 lb = x 453.6 g = 28.35 g = 0.02835 kg 16 1 metric ton = 1000 kg = 10 6 g =

1.18

10 6 mm

(a) Express 3.69 m in kilometers, in centimeters, and in millimeters. (b) Express 36.24 mm in centimeters and in meters.

M EASU R E M ENT

1 .61

The General Sherman tree, located in Sequoia National Park, is believed to be the most massive of living things. If the overall density of the tree trunk is assumed to be 850 kg/m3, calculate the mass of the trunk by assuming that it may be approximated by two right conical frustra having lower and upper diameters of 1 1.2 and 5.6 m, and 5.6 and 3.3 m, respectively, and respective heights of 2.4 and 80.6 m (see Fig. 1-1). A frustrum is a portion of a cone bounded by two planes, both perpendicular to the axis of the cone. The volume of a frustrum is given by nh(d + r� + r 1 r 2 )/3, where h is the height and r 1 and r 2 are the radii of the cone at the bounding planes. I V1 tn(80.6 m) [ l .65 2 + 2.8 2 + ( 1 .65)(2.8)] = 1281 m 3 V2 = tn(2.4 m)[5.6 2 + 2.8 2 + (5.6)(2.8)] = 1 38 m3 850 kg . So V 1419 m 3 and m = (1419 m3) � = 1.21 x 10 6 kg = 1.21 x 103 metnc tons =

(

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)

l.65m

80.6m

2.4m

Fig. 1.1

1.62

1.63

Calculate the percent sodium in a breakfast cereal which is advertised to contain 1 10 mg of sodium per 100 g of cereal. 1 10 x 1 0 - 3 g Na I x 100% = 0. 1 10% Na 100 g total (a) Calculate the mass of pure HN0 per mL of the concentrated acid which assays 69.8% by weight HN0 and 3 3 has a density of 1 .42 g/mL. (b) Calculate the mass of pure HN0 3 in 60.0 mL of concentrated acid. (c) What volume of the concentrated acid contains 63.0 g of pure HN0 3 ?

I (a)

1 .00 mL of acid has a mass of 1 .42 g. Since 69.8% of the total mass of the acid is pure HN0 3 , then the number of grams of HN0 3 in 1 .00 mL of acid is 0.698 x 1 .42 g = 0.99 1 g. (b) Mass of HN0 3 in 60.0 mL of acid = 60.0 mL x 0.99 1 g/mL = 59.5 g HN0 3 . (c) 63.0 g HN0 3 is contained in 63 · 0 g 0.99 1 glmL

1.64

63.6 mL acid

The density of gold is 19.3 g/cm3. Calculate the diameter of a solid gold sphere having a mass of 422 g. I

The result sought is a diameter, D, in centimeters. The data are a mass of gold (in grams), its density, d (in grams per cubic centimeter), and the shape of the object (spherical). The radius, r, of a sphere is related to its volume, V, by r �3 V/4n, and the volume of this sphere is 422 g/(19.3 g/cm3). Thus =

1 .65

3(422 g) = 2r = 2 3 = 3.47 cm 4n(l 9.3 g/cm3)

The density of aluminum is 2.70 g/cm3. An irregularly shaped piece of aluminum weighing 40.0 g is added to a 100-mL graduated cylinder containing exactly 50.0 mL of water. To what height in the cylinder will the water level rise?

C HAPTE R 5

UModern Structure of the Atom

5.1

PHYSICAL BACKGROUND

Unit conversions: 1 J = 1 kg·m 2/s2 = ergs l J = l V x l C 1 eV = 1. 60 x 10-19 J If a particlof eelebeictron ng accel by aispotential a charge in magnitude number voltserated of energy numericalhasly equal to theequalpotential in volts. to the charge on an electron, the 1 eV = energy of 1 electron being accelerated by V eV = energy of 1 electron being accelerated by V

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107

5.1

Convert 1 atomic mass unit (u) to kg. 2C has a mass of 0.0 120000 kg. Since there are 6.03 x 102 3 atoms in 1 mol, each of By definition, 1 mol of 1 mass 12.00 u, 0.0 120000 kg kg 12.00 u = 0.6.00120000 3 x 102 3 or u = 12.00(6.03 x 102 3) = 1.66 x 10 - 2 7 kg What did Rutherford's alpha-particle-scattering experiment prove? Atoms contain massive, positively charged centers-the nuclei. experiment (inIn anarbitrary units):to measure the charge on the electron, the following values of charge were found on oil droplets - 1.6 x l0-19 -2.4 x 10-1 9 -4.0 x 10-1 9 What value of the electronic charge would be indicated by these results (in the same units)? -0.8 x 10-19, the largest common factor of the values given. An elcharge ectron onandthea body on it are 2.0 m apart. Calculate the force of attraction between them. The electronwithisa -1.1.06Cxcharge o- 1 9 C. = k qd21 q 2 = (9 0 x 109 N·m 2/C 2) ((-l. 6 x (2.100-m)19 2C)(l.O C) ) = -3. 6 x 10-1 0 N The negative sign implies an attractive force. How muchto aenergy wilof2.be 7rel6 Aeased when a sodium ioofapproach n and a chloriindaesodium ion, original ly atcrystal infinite)? distance, arex 10-1 brought 0 together distance (the shortest distance chloride 1 A = 1 Assume that the ions act as point charges, each with a magnitude of 1. 60 x 10-1 9 C (the electronic charge). m. 9.0 x 1092 J•m)((l.60 x 10-190 C)2 ) - 8. 3 x 10 _ 19 J E - k q 1dq 2 - ( c 2 .7 6 x 10 -1 m ion pairs. Ofl ascourse in a between crystal ofionsNaCl, therecharge;are This energybetween corresponds toion119andkcalseveral per molionsofofNaopposic1te-charge attractions a given as wel repulsions of like hence for a mole of ion pairs in a crystal the energy of attraction is 185 kcal/mol, approximately 1. 5 times as great. I

5.2

5.3

I 5.4

5.5

•

T ,

5.6

What experiment negated Thomson's model of the atom as an intimate mixture of negative and positive particles? Rutherford's experiments with large-angle scattering of alpha particles by a metal foil.

68 5.7

5.8

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5.9

5.1 0

5.1 1

C H A PTER 5

The magnitude of the charge on the electron is 4.8 x 1 0 - i o esu. What is the magnitude of the charge on the proton? on the nucleus of a helium atom? . I Since the charge on the proton is equal in magnitude to that on the electron, the charge is also 4.8 x 1 0 - 1 0 esu. Since a helium nucleus contains 2 protons, its charge is 9.6 x 1 0 - 1 0 esu.

Calculate the force between two bodies 2.00 cm apart, each having a 1 .0 x 1 0 - 5 C charge. ( 1 .0 x 1 0 - s C) 2 I q 1 q2 f = k --;f2 = (9.0 x 10 9 J · m/C 2 ) = 2.2 x 10 3 J/m = 2.2 x 10 3 N (0.0200 m) 2

Two 1 .0-g carbon disks 1 .00 cm apart have opposite charges of equal magnitude such that there is a 1 .00 x 10 - 5 N force between them. Calculate the ratio of excess electrons to total atoms on the negatively charged disk. I

hence = kq 1 q 2 /r 2 q 1 q 2 = fr 2 /k q 1 = q 2 = .JJr 2 /k = .J( l .00 x 10- s N)(0.0100 m) 2 /(9.0 x 10 9 J · m/C 2 ) = 3.3 x 10 - 1 ° C on each disk (3.3

( 1 .601 electron ) = 2. 1 I 0 19 C 6.02 10 2 3 atoms ) 5.0 ( 1 .0 g) ( 1 2.0 g 10 1 0 C) _

2. 1

109 excess electrons = 4.2 5. 0 x 10 22 atoms

109 electrons

10 22 atoms

1 0 - 1 4 electron/atom

Notice the relatively low number of electrons which can make a measurable force. By no means can the carbon atoms be considered ions. Calculate the energy required to move a 1 .0 x 10- i o C negatively charged body from infinite distance to a point (a) 1 .0 cm from a 1 .0 C negatively charged body, (b) 0. 10 cm from a 1 .0 C negatively charged body, (c) from position (a) to position (b). (d ) How would the answer to (a) be changed if the first body were positively charged?

(a) (b)

(c)

(d )

E = (9.0

10 1 1 J · cm/C 2 )

E = (9.0

10 1 1 J · cm/C 2 )

( 1 .0

q q E=k 1 2 d

lO- t o C)(l .O C) = 90 J 1 .0 cm 10 - i o C)( l .O C) = 900 J 0. 10 cm

�E = (900 J) - (90 J) = 8 10 J The 8 1 0 J of energy would be liberated instead of required.

(a) Would i t take more energy to move a negatively charged body from 3.0 t o 2.0 cm from a second negatively charged body, or from 2.0 to 1 .0 cm from the second body? Explain. (b) Would it take more energy to move a charged body from midway between two charged parallel plates 1 .0 cm toward the positive plate or from this second (off center) position to a position 1 .0 cm closer to the positive plate? Explain.

I (a)

The energy required to move the charged body from infinite distance to a given distance may be expressed in terms of the distance d by the equation E = constant/d

The energy in moving the charged body from 3.0 cm to 2.0 cm from a second charged body is merely the difference:

(

)

I 1 constant constant = constant - - = constant/6.0 2 .0 3.0 2.0 cm 3.0 cm Similarly for the difference between 2.0 and 1 .0 cm:

constant constant = constant/2.0 1 .0 cm 2 . 0 cm

M O D E R N STR U CTU RE OF T H E ATO M

Since t is greater than j;, the energy required to move the charged body from 2.0 cm to 1 .0 cm is greater than that to move from 3.0 cm to 2.0 cm. (b) In contrast, a charge in a uniform electric field experiences a force which is not dependent on its position in the field; hence both of the moves described require equal energies. 5.12

What, if any, is the effect of a magnetic field on (a) a static electric charge in the field? moving through the field?

(b)

an electric charge

I (a )

None. (b) An electric charge in motion generates its own magnetic field which the external magnetic field will interact with, creating a force perpendicular to its original direction of motion.

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5.13

Why does the charge-to-mass ratio of positive rays depend on the residual gas in the discharge tube? Why is the charge-to-mass ratio of all cathode rays the same? I

Cathode rays, no matter what their source, are composed of electrons - all of which have the same charge-to-mass ratio. The ions remaining after the loss of the electrons might have the same magnitude of charge, but different masses. Hence they will have different charge-to-mass ratios.

5.14

State at least four ways in which positive (canal) rays differ from cathode rays. I

5.15

canal rays

Sign of charge e/m Mass Magnitude of charge

cathode rays

Positive Variable, depends on ions Variable, depends on ions Mostly 1 + , but sometimes 2 + , 3 + , . . .

Negative Definite value Definite value Always 1 -

Calculate the ratio of protons in the atomic nuclei to sodium atoms in a sample of NaCl. Is it necessary to know the size of the sample, the total number of atoms in the sample, or the total number of protons in the sample? To determine the charge-to-mass ratio of the electron, is it necessary to know either the charge or the mass of the electron? Under what conditions do the strengths of the magnetic and electric fields applied in a cathode ray tube permit the determination of the charge-to-mass ratio of the electron? I

Per mo! of NaCl there are

( 1 1 + 1 7) mo! of protons 1 mo! of Na atoms

28 1

------- = - mole

. ratio

The size of the sample is immaterial to determining the mole ratio; for example, if one doubles the sample size, one doubles both the number of protons and the number of sodium atoms, and the ratio remains the same. The e/m ratio for electrons was known before either e or m; obviously, it is not necessary to know the individual values to determine e/m. The e/m ratio is determined using equal magnetic and electric forces on the moving electrons. 5.16

An electron volt (eV) is the energy necessary to move an electronic charge (e) through a potential of exactly 1 V. Express this energy in (a) J (b) kcal per mole of electrons (c) kJ per mole of electrons I (a) 1 .00 eV ( 1 .60 x 10- 1 9 C)( l .00 V) = 1 .60 x 10- 1 9 J 1 kcal 6.023 x 10 23 = 23.0 kcal/mo! (b) (1 .60 x 1 0 - 1 9 J) mo! 4184 J 1 kJ 6.023 x 10 23 = 96.4 kJ/mol (c) (1 .60 x 1 0 - 1 9 J) mo! 10 3 J

5.17

(--)( ( )(

) )

In an oil drop experiment, the following charges (in arbitrary units) were found on a series of oil droplets: 2.30 x 10- 1 5 , 6.90 x 10- 1 5 , 1 .38 x 10- 1 4, 5.75 x 1 0 - 1 5 , 3.45 x 1 0 - 1 5 , 1 .96 x 1 0 - 1 4 . Calculate the mag足 nitude of the charge on the electron (in the same units). I 1 . 1 5 x 10- 1 5 , which is the largest number which divides all the listed charges evenly. (The smallest of the charges listed is 2.30 x 1 0 - 1 5 , but this charge does not divide into all the others an even number of times; hence 2.30 x 1 0 - 1 5 must represent the charge (in arbitrary units) of 2 electrons.)

70 5.1 8

C H A PTER 5

Assuming that the oil was sufficiently nonvolatile, would i t b e possible t o perform Millikan's oil drop experiment in an evacuated apparatus? Explain. I

No, the oil droplets must reach a terminal velocity to be able to estimate their weights by Stokes' law. In the ab sence of air, there would be no good way to determine their weights, and the force of the electric field would then not be precisely known.

5.19

In an oil drop experiment, the terminal velocity of an oil droplet was observed to be 1 .00 mm/s. The density of the oil is 0.850 g/cm 3 , and the viscosity of air (17) is 1 .83 x 1 0 - 5 N · s/m 2 . Calculate the mass and the radius of the oil droplet.

mg = (!ndr3)g 2 dr2g m -m m = !ndr3 d = -= v = -6n17r 6n17r 9 1'/ V !nr3 v17 9(0.00l OO m/s)( l.83 x 10 - 5 N · s/m 2 ) (%r, r = 2d = 2(0.850 x 10 3 kg/m 3 )(9.80 m/s 2 ) = 3 · 14 x 1 0 _ 6 m y 9 m = !ndr3 = ( 1 .33)(3.14)(0.850 x 10 3 kg/m 3)(3.14 x 1 0 - 6 m) 3 = 1 . 1 0 x 10 - 1 3 kg

---

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5.20

5.21

5.22

Although alpha particles have a larger charge, beta particles are deflected more than alpha particles in a given electric field. Explain this observation. I

Beta particles are much Jess massive, and therefore their charge-to-mass ratio is larger despite their lower charge.

The radius of a nucleus, in cm, can be estimated by R � 1 .4 x 1 0 - 1 3 A l / 3 where I

is the mass number of the atom. Calculate the approximate density of a polonium-2 1 0 nucleus. R � ( 1 .4 x 1 0 - 1 3 )(210) 1 1 3 = 8.3 x 1 0 - 1 3 cm

The characteristic x-ray wavelengths for the lines of the K. series in magnesium and chromium are 9.87 and 2.29 A, respectively. Using these values, determine the constants a and b in Moseley's equation and predict the wavelengths of x-rays in this series for strontium and for chlorine . I

.Jv = a(Z - b) c 3.00 x 1 0 8 m/s v = = Mg A. 9.87 A Ver = 1 . 3 1 X 10 1 8 Hz � = 5.51 x 1 08 .;;;; = 1 . 1 4 x 109 5.5 1 x 10 8 = a(1 2 - b) 1 . 14 x 10 9 = a(24 - b)

(

)(-1 0 1 0 A) = 3.04 x 10 1 7 /s = 3.04 x 1 0 1 7 Hz m

Dividing the second of these equations by the first to eliminate a yields 1 . 14 x 10 9 24 - b = 2.07 = 5.51 x 10 8 1 2 - b 24 - b = 2.07( 12 - b) = 24.8 - 2.07b b = 0.75 Substitution of this value of b into the first equation above yields 5.5 1 x 10 8 = a( 1 2 - 0.75) = l l.25a 5.5 1 x 10 8 = 4.90 x 107 a = 1 1 .25 The values may be checked using the second equation 1.14 x 109 = 4.90 x 10 7 (24 - 0.75)

= 1 . 14 x

10 9

STO I C H I O M ETRY

1 73

Mole Method

190 x 10 3 g CaCO 3 = 1 .90 x 10 3 mol CaC0 3 100 g CaC0 3 /mol CaC0 3 n(CaO) = n(CaC0 3 ) = 1 .90 x 10 3 mol Cao 3 m(CaO) = ( l .90 x 10 mol Ca0)(56. 1 g CaO/mol CaO) = 107 x 10 3 g Cao = 107 kg CaO n(CaC0 3 ) =

Factor-Label Method

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Amount of CaO 1 1 .46

(

l mol CaC0 3 ( 1 90 x 10 3 g CaC0 3 ) 100 g CaC0 3 = 107 x 10 3 g Cao = 107 kg Cao

)( l mol CaO )(56. l g CaO ) I

mol CaC0 3

1 mol Cao

Most commercial hydrochloric acid is prepared by heating NaCl with concentrated H2S0 4 . How much sulfuric acid containing 90.0% H2S0 4 by weight is needed for the production of 2000 kg of concentrated hydrochloric acid containing 42.0% HCI by weight? I

( 1 ) Amount of pure HCI in 2000 kg of 42.0% acid is (0.420)(2000 kg) = 840 kg. (2) Determine the amount of H2S0 4 required to produce 840 kg HCI. 2NaCI +

H2S0 4

----+

(I mo! = 98. 1 g)

From the equation,

Na2S0 4 +

(2 mo!

2HC1

2(36.46) = 72.92 g)

98. 1 . 1 g HCI requires g H 1 S0 4 72.92 98. 1 840 kg HCI requires (840) kg = 1 1 30 kg H2S0 4 . 72.92

72.92 g HCI requires 98. 1 g H2S0 4 , 1 kg HCI requires

98. 1 kg H2S0 4 72.92

and

(

)

(3) Finally, determine the amount of sulfuric acid solution containing 90.0% H2S0 4 that can be made from 1 1 30 kg of pure H2S0 4 . Since 0.900 kg of pure H2S0 4 makes 1 .00 kg of 90.0% solution, 1 1 30 kg of pure H2S0 4 will make 1 1 30 kg H2S0 4 ------ = 1 260 kg soIn 0.900 kg H2S0 4/kg soln

1 1 .47

A 1.00 g total sample of a mixture of CaC0 3 (s) (FW = 1 00) and glass beads liberates 0.220 g of C02 (MW = 44.0) upon treatment with excess HCl (MW = 36.5). (Glass does not react with HCI. ) (a) Calculate the weight of CaC0 3 in the original mixture. (b) Calculate the weight percent of calcium in the original mixture.

1 1 .48

----+

CaC0 3 + 2HC1 C02 + H20 + CaC12 1 mo! C02 1 mo! CaC0 3 100 g CaC0 3 = (0.00500 mo! CaC0 3 ) = 0.500 g CaC0 3 (0.220 g C02) 44.0 g C02 mol C O 2 mo l Ca CO 3 40.0 g Ca 0.200 g Ca 1 mol Ca = 0.200 g Ca (0.00500 mo! CaC0 3 ) ( 1 00%) = 20.0% Ca mo! CaC0 3 mol Ca l .OO g total A particular 100-octane aviation gasoline used 1 .00 cm 3 of tetraethyllead, (C2H 5 )4 Pb, of density 1 .66 g/cm 3 , per liter of product. This compound is made as follows: 4C2H 5 Cl + 4NaPb --> (C2H 5 )4 Pb + 4NaCI + 3Pb. How many g of ethyl chloride, C2H 5 Cl, is needed to make enough tetraethyllead for 1.00 L of gasoline? I The mass of 1 .00 cm 3 (C2H 5 ) 4 Pb is ( 1 .00 cm 3 ) ( 1 .66 g/cm 3 ) = 1 .66 g; this is the amount needed per liter. In terms of moles,

(

)(

)

( (

)

1 .66 g . = 0.005 1 4 mol Number of mol (C2H 5 )4 Pb needed = 3 2 3 gImo 1

)

The chemical equation shows that 1 mol (C2H 5 )4 Pb requires 4 mol C2H 5 CI. Hence 4(0.005 14) = 0.0206 mol C2H 5 CI is needed. m(C2H 5 Cl) = (0.0206 mol)(64.5 g/mol) = 1 .33 g C2H 5 CI

1 74

1 1 .49

C H A PTER 1 1

A carbonate, MC0 3 , of an unknown metal is treated with HCI to produce C02 • Upon treating 1 .00 g of MC0 3 with excess HCI, 0.522 g of C02 is produced. Find the weight percent of the metal in MC0 3 •

(0.522 g C02)

(��o� ��:)(1 :�� 3 ) = O.ot 186 mo! MC0 3

(84.3 g/mol MC0 3 ) - (60.0 g/mol CO� - ) = 24.3 g/mol M i +

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1 1.50

1 1 .51

1 1 .52

(

1.00 g MC0 3 = 84.3 g/mol MC0 3 0.01 186 mo! MC0 3 24.3 g/mol M 2 + 0 (l OO0%) = 28 · 8% 84.3 g/mol MC0 3

)

How many kg of pure H2S0 4 could be obtained from 2.00 kg of pure iron pyrites (FeS2) according to the following reactions? I

First it should be noted that there is no by-product loss or other permanent loss of sulfur, so that it is not even necessary to balance the equations or to use them further. Each incoming atom of sulfur produces an outgoing molecule of H2S0 4 . (One formula unit of FeS2 contains 2 atoms of S, and one molecule of H2S0 4 contains 1 atom of S.) Hence n(FeS2) =

2000 g n(H2S0 4) = 2n(FeS2) = 2 x 1 6.7 = 33.4 mo! H2S0 4 = 1 6.7 mo! FeS2 120 g/mol m(H2S0 4) = (33.4 mol)(98.0 g/mol) = 3270 g = 3.27 kg H2S0 4

Another Met hod 1 mo! FeS2 ( 1 20 g FeS2) yields 2 mol H2S0 4 (2 x 98 = 196 g H2S0 4). Thus

1 20 kg FeS2 yields 196 kg H2S0 4

and

1 kg FeS2 yields

�� = 1 .63 kg H2S04

2 kg FeS2 yields 2 x 1.63 = 3.26 kg H2S0 4

When 2.86 g of a mixture of 1-butene, C 4 H 8 , and butane, C 4 H 1 0 , was burned in excess oxygen, 8.80 g of C02 and 4. 14 g of H20 were obtained. Calculate the percentage by mass of butane in the original mixture. I

(8.80 g C02)

1 mo! C02 (44.0 ) = 0.200 mo! C02 g C02

(4. 14 g H20)

( 11 8.0mo!g H20 ) = 0.230 mo! H20 H20

The mixture contains 0.200 mo! of carbon atoms and 0.460 mo! of hydrogen atoms. Let x = number of moles of C 4H 8 and y = number of moles of C 4 H 1 0 . Then 4(x + y) = 0.200 8x + l Oy = 0.460 (doubling the first equation) 8x = 0.400 - 8y (0.400 - 8y) + lOy = 0.460 (substituting the last equation into the second) 2y = 0.060 y = 0.030 mo! C 4H 1 o

In one process for waterproofing, a fabric is exposed to (CH 3 )2SiCl2 vapor. The vapor reacts with hydroxyl groups on the surface of the fabric or with traces of water to form the waterproofing film [(CH 3 )2Si0]0, by the reaction n(CH 3 )2SiC12 + 2n OH - ---+ 2n CI - + n H20 + [(CH 3 )2Si0]0

262

C H A PTER 1 4

TABLE 14.2 interactions Fig. 14.18

type

c-c a-a c-a c-c a-a c-a c-c a.- a c-c c-a c-c

(b) (c)

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(e)

14.66

14.67

(f) (h)

(i)

(j)

(

J6 r

.JS r )§ r J0, r

per unit

{31 8planes others {

li}

9 planes 9 planes 8 cubes 3 planes 9 others 9 planes

4 4 4

1 2 rect. prisms

9 planes 6 prisms 1 cube

n8 4

2 4 4

total

54 36 36 32 27

18 24 4

energy contribution

- 54e 2 /r + 36e 2 /J2 r + 36e 2 /J2 r - 32e 2 /./3 r + 27e 2 /2r

- 72e 2 /./5 r + 48e 2 /J6 r

+ 18e 2 /.)8 r - 24e 2 /3r + 4e 2 /J0, r

�)

LIQUIDS

A liquid of density 850 kg/m 3 having a surface tension of 0.055 N/m, will rise how far in a glass capillary of 1 .40 mm inside diameter? 2(0.055 kg/s 2 ) I 2y h=-= = 0.01 9 m = 19 mm y = irhdg 3 rdg (0.70 x 10- m)(850 kg/m 3 )(9.80 m/s 2 )

(a ) Distinguish between the triple point and the freezing point of a pure substance. (b) For most pure substances, which is apt to be higher, the triple point or the freezing point?

I (a) The triple point is the point at which solid, liquid, and vapor are in equilibrium with each other, with no other substance present; the freezing point is the point at which solid and liquid are in equilibrium under l atm total pressure. (There must be some other substance present to achieve 1 atm pressure.) (b) The freezing point is higher for most substances, which have a positive slope of the liquid-solid equilibrium line in the phase diagram.

Refer to Fig. 14. 19. If one heated CC14 at a constant pressure of 1 atm starting at point D until point E is reached, what phase change(s) would be observed? Melting

Refer to Fig. 14. 19. How many variables (degrees of freedom) must be specified in order to define point E? Two: temperature and pressure

Refer to Fig. 14. 19. The triple point of CC14 is represented by which point?

14.70

J5 r

number of units

The total lattice energy for this isolated unit cell is the sum of these energy values: ei 36 � E_ 27 72 48 � 24 + + + + + 54 + U0 = J2 J2 J3 2 J5 J6 .j8 3 Jfi. r 2 2 = ( - 54 + 25.46 + 25.46 - 18.48 + 1 3.5 - 32.20 + 1 9.60 + 6.36 - 8 + 1 . 1 5)e /r = - 21 . 1 5e /r

1 4.69

{ }

(g)

14.68

J2 r J2 r ./3 r J2 r

{ }

(d )

14.65

c-a

(a)

14.3

distance

Point C

All other factors being equal, which will cool to room temperature faster - a closed container of water at 100 °C or an open container of water at 100 °C? Explain your answer.

I The water will cool faster uncovered, since it can evaporate that way. Evaporation cools, since the most ener getic molecules are the ones that are lost, and they may use energy to push tack the atmosphere.

SOLIDS A N D L I Q U I DS

263

(critical point) - - - -e - - - - -

Liquid

Solid

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- - - - -

-0-

- - -

Fig. 14.19 Phase Diagram of CCl 4 •

14.71

An astronaut in an orbiting spaceship spilled a few drops of his drink, and the liquid floated around the cabin. In what geometric shape was each drop most likely to be found? Explain.

The drops would be spherical - the shape with the smallest surface area per unit volume, in response to surface tension.

14.72

The enthalpy of fusion of H 2 0 at 0 °C is 1 .435 kcal/mo!. Calculate fl.S for the following process at 0 °C: H20(/) ¢ H 2 0(s).

14.73

Since fl.G = 0,

M =

fl. H = - 1435 caljmol 273 K T

- 5.26 calfmol · K

When a liquid in a thermostated chamber is vaporized at constant temperature by the very slow withdrawal of a piston, the vapor is continuously in equilibrium with the liquid. The heat absorbed from the bath in the process is equal to what thermodynamic property of the liquid? Write an expression for calculating the value of 8G0 for this isothermal process at constant pressure. I

The heat is absorbed at constant pressure, with no work other than expansion. Hence the heat absorbed is equal to fl. Hva v· Since the liquid and vapor are in equilibrium, fl.G = 0.

1 4.74

As heat is removed from a liquid which tends to supercool, its temperature drops below the freezing point and then rises suddenly. What is the source of the heat which causes the temperature rise?

14.75

14.76

The enthalpy of fusion

Explain why water would completely fill a fine capillary tube which is open at both ends when one end is immersed in water. I The surface tension pulls the water into the capillary. In a fine capillary, the surface tension is great enough to overcome the attraction of gravity on the water. In a measurement of surface tension by the falling drop method, 5 drops of a liquid of density 0.797 g/mL weighed 0.220 g. Calculate the surface tension of the liquid.

The mass of the average drop is

0· 220 g = 0.0440 g, and its volume is 5 0.0440 g 0 .7 97 gImL

0.0552 mL

264

C H A PTER 1 4

Assuming a spherical shape, the radius of the drop is determined from the equation V 1'irr 3 : J3V 3 3(0.0552 cm 3 ) 0_236 cm r �. {4-; 4(3. 14) mg (0.0440 g)(9 B O cm/s 2 ) 29. l dyne/cm 0.029 1 N/m y 2nr 2(3.14)(0.236 cm) The surface tension of liquid mercury is 0.490 N/m, and its density is 1 3.6 x 10 3 kg/m 3 . How far will the mercury level be depressed when a glass capillary with 0.40 mm radius is placed in a dish of mercury?

= = = =

1 4.77

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14.78

14.79

14.80

= !rhdg 2y 2(0.490 kg/s 2 ) = 0.0183 m = 1 8.3 mm h=-= 3 rdg (0.40 x 10- m)(1 3.6 x 10 3 kg/m 3 )(9.80 m/s 2 )

The equation applies to the depression of mercury in the capillary as well as to elevations of other liquids in capillaries.

A certain liquid has a viscosity of 1 .00 x 104 poise and a density of 3.2 g/mL. How long will it take for a platinum ball with a 2.5 mm radius to fall 1.00 cm through the liquid? The density of platinum is 21.4 g/cm 3 . I

The mass of the platinum ball and of the liquid it displaces are determined from the respective densities and the volume of the ball. Volume is given by V 1'irr 3 !(3. 14)(0.25 cm) 3 0.0654 cm 3 m(liq) (0.0654 cm 3 )(3.2 g/cm 3 ) = 0.21 g m(Pt) (0.0654 cm 3 )(21 .4 g/cm 3 ) 1.40 g (m m 0)g (1 .40 g - 0.2 1 g)(980 cm/s 2 ) v 2.47 x 10 2 cm/s 6nr17 6(3.14)(0.25 cm)(l .00 x 104 poise) 1 .00 cm d 40· 5 s t 2.47 x 10 2 cm/s

= =

= -;;- =

(a) On the phase diagram for water, Fig. 14.20, what feature represents the equilibrium of solid water and water vapor? (b) What phase change(s) occur(s) when a sample at point E is heated at constant pressure until point F is reached? (c) What is the temperature at which line BC intersects the 1 atm line called?

Fig. 14.20

(a) Line represents the solid-gas equilibrium. (b) When a sample at point E is heated, melting and later vaporization occurs until the temperature reaches that represented by point F. (c) The freezing point is the temper ature at which line BC crosses the P 1 atm line.

Knowing that the density of ice is less than that of water, explain why the slope of the solid-liquid equilibrium line in the phase diagram of water (Fig. 14.20) is in accord with Le Chatelier's principle. I

As the pressure increases at constant temperature, this equilibrium shifts to the liquid phase-the more dense phase. Thus a movement straight up on the phase diagram, increasing the pressure at constant temperature, causes melting as one crosses the solid-liquid equilibrium line. The slope is correct. 14.81

Substance X has its triple point at 18 °C and 0.5 atm; its normal melting point is 20 °C, and its normal boiling point is 300 °C. Sketch the phase diagram for X.

�

C HAPT E R 1 5

Oxidation and Reductio

15.1

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1 5.1

15.2

1 5.3

15.4

OXIDATION NUMBER; OXIDIZING AND REDUCING AGENTS

What is the oxidation number of sulfur in each of the following cases? (a) s 2 - (b) H 2 S0 4 (c) S 2 0� - (d) CS 2 (e) S 8 (f) Na 2 S 4 0 6 (g) S 2 Cl 2 . I (a) The oxidation number is - 2, which is equal to the charge on the monatomic ion. (b) In H 2 S0 4 , the oxidation numbers must total zero. With 2 hydrogen at + 1, 4 oxygen at - 2, sulfur at x, (c)

x = +6

The oxidation number of sulfur is + 6

In s 2 0; - , the oxidation numbers must total - 2. With 3 oxygen at - 2 and 2 sulfur at x, 2x + 3( - 2) = - 2

The oxidation number of sulfur is + 2.

(d) From just the rules stated above, any assignment of oxidation numbers which gives a net charge of zero could

be made. However, noting that sulfur is in the same periodic group with oxygen, and by analogy with C0 2 , the sul fur is usually assigned an oxidation number of - 2. Then, 2 sulfur at - 2 = - 4, carbon at + 4 = + 4, and the net charge = 0. (e) The oxidation number of any free element is zero. (f) The oxidation numbers of alkali and alkaline earth metals in their compounds are uniformly + 1 and + 2, respectively. The oxidation 'n umber of sulfur can be established on this basis: 2 sodium at + 1, 4 sulfur at x, 6 oxygen at - 2, or 2( + I) + 4x + 6( - 2) = 0

2.5

(g) The oxidation number of chlorine is - 1, as in all halogen compounds with elements other than oxygen or

another halogen.

2x + 2( - 1) = 0

Therefore the oxidation number of sulfur must be + l.

x= +1

In H 2 As04 , what is the oxidation number of (a) hydrogen (b) oxygen (c) arsenic? I (a) + l (b) - 2 ( c) + 5 State the oxidation number of the underlined element in each of the following: (a) e 2 0� - (b) �3 0 2 (c) Mn04 (e ) YOi (f) !J_O� + (g) ClO]

(d) Mno� -

The oxidation number of the underlined element is represented by x.

(a) 2x + 7( - 2) = - 4 (c) x + 4( - 2) = - 1 ( e) x + 2( - 2) = + 1

x = +5 x = +7 x = +5

3x + 2( - 2) = 0 (d) x + 4( - 2) = - 2 (g) + 5 (f) + 6 (b)

x = +! x = +6

What is the oxidation state of hydrogen in each of the following? (a) HCl (b) H + (c) NaH (d) H 2 (e) LiAIH 4

I 0

15.5

2( + 1) + x + 4( - 2) = 0

Hydrogen has a + 1 oxidation state in all its compounds except those with group IA, IIA, or IIIA metals. It has oxidation state as an element. (a) + 1 (b) + I (c) - 1 (d ) 0 (e) - 1

Which of the following is the anhydride of HN0 3 ? (a) N 2 0 (b) NO (c) N0 2

(d )

N 2 0 4 (e) N 2 0 5

N 2 0 5 is the anhydride. The easiest way to tell is to check the oxidation number of nitrogen. HN0 3 and N 2 0 5 both have nitrogen in the + 5 oxidation state.

15.6

268

Calculate the oxidation number of the underlined element (c) Na 2 Cr 2 0 1 (d) Q3 I (a) + 4 (b) o (c) + 6 (d ) 0

each of the following: (a) £0� -

OXI DAT I O N A N D R E D U C T I O N

15.7

15.8

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15.9

269

(a) What is the oxidation state of hydrogen in LiAIH 4? (b) Write an equation for the reaction of hydrogen in that oxidation state with water. I (a) - 1

What is the oxidation state of nitrogen in each of the following? (a ) NH 3 (b) HN 3 (c) N 2 H 4 (d) N0 2 (e) N 2 0 4 (/) NO :Z (g) NH 2 0H (h) NO (i) HN0 3 (j) N 2 0 (k) HCN I ( a ) - 3 (H has an oxidation state of + 1 .) (b) -t (Oxidation states need not be integral.) (c) - 2 (d) + 4 (e) + 4 (/) + 3 [x + 2( - 2) = - 1 , the charge on the ion.] (g) - 1 (h) + 2 (i) + 5 (j) + l (k) - 3 (C is + 2 in CN - and HCN, analogous to C in CO, but this value is somewhat arbitrary.)

Which one(s) of the following involve redox reactions? (a ) Burning of gasoline (b) evaporation of water (c) human respiration (d ) preparation of metals from their ores (e) production by lightning of nitrogen oxides from nitrogen and oxygen in the atmosphere (/) production by lightning of ozone (0 3 ) from 0 2 (g) reaction of H 2 S0 4 with NaOH a, c, d, e. [b is a physical change; f is the change of an element in one form (oxidation state = 0). to another form (oxidation state = O); g is the reaction of an acid with a base.]

15.10

In the process of drying dishes with a dishtowel, what may be considered the wetting agent? the drying agent? What happens to the wetting agent? to the drying agent? In an oxidation-reduction reaction, what happens to the oxidizing agent? to the reducing agent?

The wetting agent is the dish; the drying agent is the towel. (The drying agent gets wet during the process, since the water must go into it.) In a redox reaction, the oxidizing agent is reduced; the reducing agent is oxidized. (The electrons are transferred in a manner similar to the water in the drying analogy.)

1 5. 1 1

Determine the oxidation state of chlorine in (a ) Clo - (b) CIO:Z ( c) CI03 ions.

oxidation state

(a)

(b) (c) (d)

15.12

CI04 . (e) Name each of these

(e) name

Hypochlorite ion Chlorite ion Chlorate ion Perchlorate ion

+l +3 +s +7

Determine the oxidation number of (a ) N in HN 3 (b) 0 in K0 2 . (c) Draw electron dot diagrams for both of these species. I

15.13

CJo CIO:Z c103 CI04

(d )

( a ) -t

(b)

K+ :o:o:-

Given that in these compounds the oxidation state of hydrogen is + 1, of oxygen - 2, and of fluorine - 1 , determine the oxidation states of the other elements. (a) PH 3 (b) H;S (c) CrF 3 (d) H 2 S0 4 (e) H 2 S0 3 (/) Al 2 0 3 •

( a ) 3H represent an oxidation-state sum of + 3 ( + 1 for each of the three hydrogens). Then the oxidation state of P must be - 3, since the algebraic sum of the oxidation states of all atoms in a compound must equal zero. (b) Oxidation-state sum of 2H is + 2; then oxidation state of S is - 2. (c) Oxidation-state sum of 3F is - 3; then oxidation state of Cr is + 3. (d ) Oxidation-state sum of 2H is + 2 and of 40 is - 8. or a total of - 6; then oxidation state of S is + 6. The sulfate ion, SO� - , has a charge of - 2, since the total oxidation state of two H atoms is + 2. (e) Oxidation-state sum of 2H is + 2 and of 30 is - 6, or a total of - 4; then oxidation state of S is + 4. The sulfite ion, SO� - , has an ionic charge of - 2, since the total oxidation state of 2H is + 2. (/) Oxidation-state sum of 30 is - 6; then o � idation-state sum of 2AI is + 6, and oxidation state of Al is !( + 6) = + 3.

15.14

Determine the oxidation state of the underlined element in (d ) lCI (e) Ba 2 Xe0 6 (/ ) QF 2 (g) Ca(CI0 2 ) i .

1 5. 1 5

(a) + 5

(b)

( c) + 5

(d )

What is the oxidation number of Cl in Ba(CI0 3 ) i ? The c103 ion has Cl in the + 5 oxidation state.

( e) + 8

(/) + 2

(g) + 3

270

15.16

C H A PTER 1 5

What are the maximum and minimum oxidation states of nitrogen in its compounds?

15.17

+ 5 (equal to the group number) and - 3 (equal to the group number minus 8)

State the oxidation number of the underlined element: (a ) Mn04 (b) �_Q 2

15.18

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15.20

15.21

15.22

15.23

15.24

(b)

(a ) + 4

(b)

(c)

(d) - 2

(c)

(d) + 7

Seo� - (d) K 2 �

State the maximum positive oxidation state for each of the following elements in any of their compounds (all of their compounds): (a ) Ti (b) Zn (c) C (d) Cl.

15.19

(a ) + 7

(c)

Considering oxidation states, predict the formulas of the compounds which are likely to be produced by the combination of (a ) Cs and I (b) S and F.

I (a ) -

Csl (b) SF 6 (The highest oxidation number of sulfur is + 6; the negative oxidation number of fluorine is 1 Hence the compound is SF 6. Others could be SF4 and SF 2 . ) .

Name the four most important oxidation states exhibited by sulfur.

+ 6, + 4, o, - 2

What is the oxidation state of the underlined element in each of the following: (a ) Cr 2 0� - (b) Na 2� 2

I (a )

(b)

- l (in s� - ion)

Predict the highest and lowest possible oxidation states of each of the following elements: ( a ) Ta (b) Te (c) Tc (d)Ti (e) Tl.

The maximum oxidation state is the group number. The minimum oxidation state for the metals is zero; for the nonmetals it is equal to the group number minus 8. ( a ) + 5, 0 (b) + 6, -2 (c) + 7, 0 (d) + 4, 0 (e) + 3, 0.

Which of the following are examples of disproportionation reactions? What criteria determine whether a reaction is a disproportionation? (a )

Ag(NH 3 )1° + 2H + ___... Ag + + 2NH,t CaC0 3 - Cao + C0 2 Cu (e) 2 0 + 2H + ___... Cu + Cu 2 + + H 2 0 (g) 2HCuC1 2 dilute with H,o Cu + Cu 2 + + 4Cl + 2H + ( c)

(d ) 2Hg0 ___... 2Hg + 0 2 (/) CuS + 0 2 - cu + S0 2

Equations b, e, and g represent disproportionations, in that in each reaction the oxidizing agent and the reducing agent are the same.

In the following reaction, identify the species oxidized, the species reduced, the oxidizing agent, and the reducing agent. The Fe 2 + is oxidized to Fe 3 + by the nitrate ion, which is the oxidizing agent. Nitrogen is reduced from nitro gen(V) in the nitrate to nitrogen(IV) in N0 2 , by the reducing agent Fe 2 + .

15.25

15.26

State the oxidation number of the underlined element(s) in each of the following: (a ) H 2 �0 4 (b) Y(Br0 2 ) 2 ( c) KC10 3 (d ) Br03 I (a ) 6 ( c) 5 (d ) 5 (b) vn, Br111 (deduced from Br02)

Name the compounds in each of the following sets: (a) FeC1 2 , FeC1 3 (b) VC1 2 , VC1 3 , VC1 4 (c) UC1 3 (d) Cu 2 0, CuO (e) HN0 2 , HN0 3 I ( a ) Iron(II) chloride and iron(IJI) chloride. (b) Vanadium(II) chloride, vanadium(III) chloride, and vanadium(IV) chloride. The use of roman numerals is specific and easy; one does not have to remember that VC1 2 is hypovana dous chloride, VC1 3 is vanadous chloride, and VCl 4 is vanadic chloride. ( c) Uranium(III) chloride. (d ) Copper(!)

278

C H A PTER 1 5

( + 5 -> - 3) = - 8 Zn + HN0 3 ---> Zn(N0 3 ) 2 + H 2 0 + NH4N0 3 (0 -> + 2) = + 2

Balance the atoms which change oxidation state: Balance the rest:

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1 5.53

4Zn + HN0 3 ---> 4Zn(N0 3 ) 2 + H 2 0 + NH4N0 3 4Zn + 10HN0 3 ---> 4Zn(N0 3 ) i + 3H 2 0 + NH4N0 3

Mn04 ---> Mn 2 + Mn04 ---> Mn i + + 4H 2 0 8H + + Mn04 ---> Mn 2 + + 4H 2 0 5e - + 8H + + Mn04 ---> M n i + + 4H 2 0 5C 2 0i - ---> 10C0 2 + lOe lOe - + 1 6H + + 2Mn04 ---> 2Mn 2 + + 8H 2 0 5C 2 01 - + 1 6H + + 2Mn04 ---> 10C0 2 + 2Mn 2 + + 8H 2 0 5Na 2 C 2 04 + 2KMn04 + 8H 2 S04 ---> IOC0 2 + 2MnS04 + 8H 2 0 + K 2 S04 + 5Na 2 S0 4 C 2 01 - ---> C0 2 C 2 01- ---> 2C0 2 + 2e -

2(0 -> - l) = - 2 CdS + 1 2 ---> Cd 2 + + 21 - + S ( - 2 -> 0) = + 2 i or CdS + 1 2 ---> Cd + + S + 21 -

(2 -> 7) = + 5 (4

-> 2) = - 2

2Mn0 + 5Pb0 2 ---> 2 Mn0 4 + 5Pb 2 + 2Mn0 + 5Pb0 2 ---> 2Mn04 + 5Pb 2 + + 4H 2 0 2Mn0 + 5Pb0 2 + 8H + ---> 2Mn04 + 5Pb 2 + + 4H 2 0 2Mn0 + 5Pb0 2 + 10HN0 3 ---> 2HMn04 + 5Pb(N0 3 ) 2 + 4H 2 0

Cr 3 +(i - ) J + K + oH - + Cl 2 ---> (K + ) i cro1 - + K + J04 + K + c1 - + H 2 0 (Note that both the chromium and the iodide are oxidized in this reaction.) I

or I

2(3 -> 6) + 6( - l -> + 7) = 54

2Crl 3 + KOH + Cl 2 ---> Cr 2 0� - + 6I04 + H 2 0 + 2c1 2(0 -> - l) = - 2 2 C rl 3 + 27Cl 2 ---> Cr 2 0� - + 6104 + 54Cl 2Crl 3 + 620H - + 27Cl 2 ---> Cr 2 0� - + 6I04 + 3 1 H 2 0 + 54c1 2Crl 3 + 62KOH + 27Cl 2 ---> K 2 Cr 2 0 7 + 6K104 + 3 1 H 2 0 + 54KCI

H 3 AS04 ---> H 3 As0 4 + 2e 3HAso� - + 3H 2 0 ---> 3H 3 AS 0 4 + 6e-

HAso � HAso� - + H 2 0

--->

Br03 Br03 + 6H +

OTH E R CONCENTRAT I O N U N I TS

299

16.27

(21 .0 mL)(0.800 N) = 1 6.8 meq H + 2H + + CaO -----> Ca i + + H 2 0 470 mg 28 0 mg (100%) = 42.0% CaO = 470 mg ( 1 6.8 meq Ca0) · 1 120 mg meq By the Kjeldahl method, the nitrogen contained in a foodstuff is converted into ammon;a. If the ammonia from 5.0 g of a foodstuff is just sufficient to neutralize 20 mL of 0. 100 N acid, calculate the percent nitrogen in the foodstuff.

(

(20 mL)(0. 100 N) = (2.0 meq)

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(34 mg NH 3 ) 16.28

16.29

(

)

( 1 714mgmgNHN 3 ) = 28 mg N

7 mg ( 1meq ) = 34 mg NH 3 N ) (100%) = 0.56% N ( 500028mgmgsamp le --

What is the purity of concentrated H2S04 (density 1 .800 g/mL) if 5.00 mL is neutralized by 84.6 mL of2.000 N NaOH?

(49.0meqmg ) = 8.29 103 mg = 8.29 g H2 S04 8.29 g 1 .800 g . (5.00 mL) (--) = 9.00 g solution ( . ) ( 1 00%) = 92. 1% . g so 1 ut1on 9 00 mL

(84.6 mL)(2.000 N) = ( 1 69.2 meq H 2 S04)

A 10.0 mL portion of (NH4) i S0 4 solution was treated with excess NaOH. The NH 3 gas evolved was absorbed in 50.00 mL of 0. 1000 N HCI. To neutralize the remaining HCl, 2 1 .50 mL of 0.0980 N NaOH was required. What is the molar concentration of the (NH 4 ) z S0 4 ? How many g (NH4) z S04 are in 1 .00 L solution?

(50.00 mL)(0. 1000 N HCl) = 5.000 meq HCl (21 .50 mL)(0.0980 N NaOH) = 2.107 meq NaOH for excess HCl 2.893 meq HCl used, therefore 2.893 meq NH 3 Net: (NH 4) i S0 4 + 20H - - 2NH 3 + so� - + 2H20 1 meq (NH 4) i S0 4 1 mmol (NH 4) z S0 4 = 1 .45 mmo I ( NH 4) 2 SO 4 (2.893 meq NH 3 ) 1 meq NH 3 2 meq (NH 4) z S0 4 14 mol ( 1.00 L) · = 1 9. 1 g/L = 0. 145 M

)(

( \�om:zi

16.30

)

) (° � )(1��n

Exactly 50.0 mL of a solution of Na 2 C0 3 was titrated with ? 5.8 mL of 3.00 N HCI. co� - + 2H + ---> C0 2 + H20

If the density of the Na2C0 3 solution is 1.25 g/mL, what percent Na 2 C0 3 by weight does it contain?

16.31

What is the equivalent weight of an acid 1 .243 g of which required 3 1 .72 cm 3 of 0. 1923 N standard base for neutralization?

16.32

(3.00mLmeq ) = 197 meq HCI, which yields 197 meq Na2C0 3 53.0 mg Na 2 C0 3 ) = 10 500 mg = 10.5 g Na2 C0 3 (197 meq Na 2 C0 3 ) ( meq Na 2 CO 3 g Na 2 C0 3 1 .25 g (50.0 mL) (--) = 62.5 g solution ( 10.5 . ) (100%) = 1 6.8% Na2C0 3 mL 62.5 g solution (65.8 mL)

(3 1 .72 cm 3 )

(0. 1923cm 3meq )

6. 100 meq base,

1 .243 g = 203 · 8 g/eq 0.006 100 eq

hence 6. 100 meq acid

The molecular weight of an organi-.:: acid was determined by the following study of its barium salt. 4.290 g of the salt was converted to the free acid by reaction with 21 .64 mL of 0.954 N H2S04 . The barium salt was known to contain 2 mo! water of hydration per mo! Ba 2 +, and the acid was known to be monoprotic (monobasic). What is the molecular weight of the anhydrous acid?

�

C HA PT E R 1 7

Properties of Solution

Note: Before working the problems of this chapter, you should familiarize yourself with Table 1 7. 1, p. 3 1 6, Table 1 7.2,

p. 330, Tables 1 7.3 and 1 7.4, p. 332, and Table 1 7.5, p. 333.

17.1 17.1

RAOULT'S LAW AND VAPOR PRESSURE LOWERING

At 40 °C the vapor pressure, in torr, of methyl alcohol-ethyl alcohol solutions is represented by P = 1 19x + 1 3 5

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where x i s the mole fraction o f methyl alcohol. What are the vapor pressures o f the pure components at this temperature?

1 7.2

17.3

17.4

If x = 0, If x = 1 ,

P = 1 35 torr P = 1 19 + 1 3 5 = 254 torr

Ethylene bromide, C 2 H4Br 2 , and 1 ,2-dibromopropane, C 3 H 6 Br 2 , form a series of ideal solutions over the whole range of composition. At 85 °C the vapor pressures of these two pure liquids are 1 73 and 1 27 torr, respectively. (a) If 10.0 g of ethylene bromide is dissolved in 80.0 g of 1 ,2-dibromopropane, calculate the partial pressure of each component and the total pressure of the solution at 85 °C. (b) Calculate the mole fraction of ethylene bromide in the vapor in equilibrium with the above solution. (c) What would be the mole fraction of ethylene bromide in a solution at 85 °C equilibrated with a 50: 50 mole mixture in the vapor? 1 mo i C 2 H4Br 2 I (a) (10.0 g C 2 H 4 Br 2 ) = 0.0532 mol C 2 H4Br 2 1 88 g C 2 H4Br 2 1 mol C 3 H 6 Br 2 (80.0 g C 3 H 6 Br 2 ) = 0.396 mol C 3 H 6 Br 2 202 g C 3 H 6 Br 2 0.0532 = 0. 1 1 8 x(C 2 H4Br 2 ) = 0.0532 + 0.396 P(C 2 H4Br 2 ) = P0x(C 2 H4Br 2 ) = ( 1 73 torr)(0. 1 18) = 20.4 torr P(C 3 H 6 Br 2 ) = P0x(C 3 H 6 Br 2 ) = ( 1 27 torr)(0.882) = 1 12 torr P(total) = 1 32 torr 20.4 torr = 0. 1 55 (b) x(C 2 H4Br 2 ) = 1 32 torr (c) P(C 2 H4Br 2 ) = P(C 3 H 6 Br 2 ) = 1 73x(C 2 H4Br 2 ) = 1 27(1 - x(C 2 H4Br 2 ) ) 300x = 127 x = 0.423

( (

) )

The vapor pressure of pure liquid solvent A is 0.80 atm. When a nonvolatile substance B is added to the solvent, its vapor pressure drops to 0:60 atm. What is the mole fraction of component B in the solution?

xP0

x(A) =

!_ 0.60 atm = 0.75 = P0 0.80 atm

x(B) = 1 - x(A) = 1 - 0.75

0.25

The vapor pressure of pure water at 26 °C is 25.21 torr. What is the vapor pressure of a solution which contains 20.0 g glucose, C 6 H 1 2 0 6 , in 70 g water?

20.0 g glucose 1 80 g/mol

0. 1 1 1 mol glucose

70.0 g H 2 0 1 8.0 g/mol

3.89 mol H 2 0

3.89 x(H 2 0) = l l l = 0.972 O. + 3.89 P = P0x(H 2 0) = (25.21 torr)(0.972) = 24.5 torr

1 7.5

312

The vapor pressure of pure water at 25 °C is 23.76 torr. The vapor pressure of a solution containing 5.40 g of a nonvolatile substance in 90.0 g water is 23.32 torr. Compute the molecular weight of the solute.

P R O P E RTI ES OF SOLUTI O N S

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17.6

313

p = pox 5.00 23.32 torr = 0.98 1 = x= 5.00 + z 23.76 torr (5.00 + z)(0.98 1) = 5.00 z = 0.0968 mol 5.40 g = 55'8 g/mol 0.0968 mol

---

At 20 °C the vapor pressure of methyl alcohol (CH 3 0H) is 94 torr and the vapor pressure of ethyl alcohol (C 2 H 5 0H) is 44 torr. Being closely related, these compounds form a two-component system which adheres quite closely to Raoult's law throughout the entire range of concentrations. If 20 g of CH 3 0H is mixed with 100 g of C 2 H 5 0H, determine the partial pressure exerted by each and the total pressure of the solution. Calculate the composition of the vapor above the solution by applying Dalton's law.

In an ideal solution of two liquids, there is no distinction between solute and solvent, and Raoult's law holds for each component of such solutions. Hence, when two liquids are mixed to give an ideal solution, the partial pressure of each liquid is equal to its vapor pressure multiplied by its mole fraction in the solution. The molecular weights of CH 3 0H and C 2 H 5 0H are 32 and 46, so 1¥ mo! CH 3 0H = (94 torr)(0.22) = 2 1 torr Partial pressure of CH 3 0H = (94 torr) lQ 1 C mo H 32 3 OH + 1QQ 4 6 mo 1 C 2 H 5 0 H 14°6° mo! C 2 H 5 0H Partial pressure of C 2 H 5 0H = (44 torr) 2 0 1 00 mo 1 C 2 H 5 0 H = (44 torr)(0.78) = 34 torr 31 mo 1 C H 3 0H + 46 The total pressure of the gaseous mixture is the sum of the partial pressures of all the components (Dalton's law): Total pressure of solution = (2 1 + 34) torr = 55 torr. Dalton's law also indicates that the mole fraction of any component of a gaseous mixture is equal to its pressure fraction, i.e., its partial pressure divided by the total pressure.

[ [

J J

---

partial pressure of CH 3 0H 2 1 torr . . of CH 0H m Mole fract10n vapor = = = 0.38 3 total pressure 55 torr partial pressure of C 2 H 5 0H 34 torr Mole fraction of C 2 H 5 0H in vapor = = --- = 0.62 total pressure 55 torr

Since the mole fraction for (ideal) gases is the same as the volume fraction, we may also say that the vapor consists of 38% CH 3 0H by volume. Note that the vapor is relatively richer in the more volatile component, methyl alcohol (mole fraction, 0.38), than is the liquid (mole fraction of CH 3 0H, 0.22). 1 7.7

At 30 °C, pure benzene (molecular weight 78. 1 g/mol) has a vapor pressure of 121.8 torr. Dissolving 1 5.0 g of a nonvolatile solute in 250 g of benzene produced a solution having a vapor pressure of 1 20.2 torr. Determine the approximate molecular weight of the solute. I

Let W be the molecular weight of the solute.

250 g = 3.20 mol benzene 78. 1 g/mol . 1 5.0 g = 1 5.0 mol solute Number of moles of solute m

. 250 g = Number of moles of benzene m

Substituting in the relation vp solution = (vp pure solvent)(mole fraction solvent),

[

3 · 20 m_o_l __ 120.2 torr = ( 1 2 1 .8 torr) ----( 1 5.0/W) mol + 3.20 mol

_ _ _

(

3·20W 1 20.2 = ( 1 2 1 .8) 1 5.0 + 3.20 W .

)

Solving, W = 350. Note that the accuracy of the calculation is limited by the term 1 2 1 .8 - 120.2 that appears in the expansion. The answer is significant only to 1 part in 1 6. 17.8

The vapor pressure of water at 28 °C is 28.35 torr. Compute the vapor pressure at 28 °C of a solution containing 68 g of cane sugar, C 1 2 H 22 0 1 1 , in 1000 g of water.

318

C H A PTER 1 7

First method First compute the molality from the freezing point equation.

M1 = K 1 m 0.372 °C = ( 1.86 °C · kg/mol)m 0.372 °C m= = 0.200 mo I/kg 1 .86 0C . kg/mo!

From the definition of molality compute the number of moles of solute in the sample.

n(solute) = (0.200 mo] solute/kg solvent)(0. 125 kg solvent) = 0.025 mol solute

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Then

17.27

17.28

1 7.29

Second method For the given concentration, I

. = 4.50 g solute Molecular weight = 1 80 g/mol 0. 025 mo 1 so 1 ute

. 4.50 g H 2 0 contams g solute 125

( )

4.50 . 1000 g H 2 0 contams 1000 = 36.0 g solute 1 25

Now, since 1 .86 °C lowering is produced by 1 mo! solute in 1000 g H 2 0, 0.372 °C lowering is produced by 0.372 = 0.200 mol solute 1 .8 6

in 1000 g H 2 0. Thus 0.200 mol solute is equivalent to 36.0 g solute, and

. = 36.0 g Molecular weight = 1 80 g/mol 0.200 mol

C 6 H 6 freezes at 5.5 °C. At what temperature will a solution of 10.0 g of C 4 H 1 0 in 200 g of C 6 H 6 freeze? The molal freezing point depression constant of C 6 H 6 is 5.12 °C/m.

(10.0 g C 4 H 1 0)

( 1 mol C4H 1 0 ) = 0. 1 72 mol C4H 1 0

0. 1 72 mol C 4 H 1 0 = 0.860 m 58.0 g C 4 H 1 0 0.200 kg C 6 H 6 M = (5. 1 2 °C/m)(0.860 m) = 4.40 °C t = (5.5 - 4.4) °C = 1 . 1 °C

An aqueous solution boils at 100.50 °C. What is the freezing point of the solution?

m = Mb/K b = M1/K1 = (0.50 °C)/(0.5 1 °C/m) = MJl( l .86 °C/m) M1 = 1 .82° from which t = - 1.82 °C

Explain by use of a phase diagram why the normal boiling point of water is raised by addition of a nonvolatile solute while the freezing point is lowered. Label all pertinent points explicitly.

I See Fig. 1 7.2. The vapor pressure is lowered so that the temperature at which it reaches 1 atm is higher (point A). The intersection of the curve with the solid-vapor curve (which is not affected by solute) is at a lower temperature (point B). /

Fig. 17.2 1 7.30

The freezing point of pure camphor is 1 78.4 °C, and its molal freezing point constant, K 1 , is 40.0 °C · kg/mol. Find the freezing point of a solution containing 1 .50 g of a compound of molecular weight 1 25 in 35.0 g of camphor.

420

20.71

CHAPTER 20

When 1 .00 mo! of pure ethyl alcohol is mixed with 1 .00 mo! of acetic acid at room temperature, the equilibrium mixture contains f mo! each of ester and water. (a) What is the equilibrium constant? (b) What is !l.G0 for the reaction? (c) How many mo! of ester are formed at equilibrium when 3.00 mo! of alcohol is mixed with 1 .00 mo! of acid? All substances are liquids at the reaction temperature. I (a)

alcohol acid C2H50H(l) + CH3COOH(l )

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(1) n at start: (2) Change by reaction: (3) n at equilibrium:

-t 1 -t=t

�

ester water CH3COOC2H5(l) + H20(l )

0 +i t

-t 1 -t=t

0 +t 2 3

This tabular representation is a convenient way of doing the bookkeeping for equilibrium problems. Under each substance in the balanced equation an entry is made on three lines: ( 1 ) the amount of starting material; (2) the change (plus or minus) in the number of mo! due to the attainment of equilibrium; and (3) the equilibrium amount, which is the algebraic sum of entries ( 1 ) and (2). The entries in line (2) must be in the same ratio to each other as the coefficients in the balanced chemical equation. The equilibrium constant can be found from the entries in line (3). Let V = L of mixture and choose, as usual, 1 moljL as the standard-state concentration. Then 2/3 2/3 V V [ester] [water] = =4 K= [alcohol] [acid]

( )( ) (1�)(1�)

Note that the concentration of water is not so large compared with the other reaction components that it remains constant under all reaction conditions; the concentration of water must therefore appear in the expression for K, along with the concentrations of the reactants and of the other product. (b) !l.G0 = - 2.303RT log K = - (8.3 14 J/K)(298. 1 K)[(2.303)(log 4)] = - 3440 J = - 3.44 kJ Because K is independent of the choice of standard-state concentration (the sum of the exponents in the numerator equals the sum of the exponents in the denominator), so also is 1!,.G0• (c) Let x = moles of alcohol reacting. at start: Change by reaction: n at equilibrium: n

K=4=

3.00 -x 3.00 - x

1 .00 -x 1 .00 - x

0 +x x

%) = 3 - 4xx2 + x2 x l .OO - x 3-( 00 -( �)( )( V ) V

0 +x x

-----

Then x 2 = 4(3 - 4x + x 2 ) or 3x 2 - 1 6x + 12 = 0. Solving by the quadratic formula, + 10.6 - b ± Jb 2 - 4ac + 16 ± .)( 1 6) 2 - 4(3)(12) 16 X= = 4.4 or 0.90 2(3) 2a 6 Of the two roots to the quadratic equation, only one has physical meaning. The applicable root can generally be selected very easily. In this problem, we started with 3.00 mo! of alcohol and 1 .00 mo! of acid. We can see from the reaction equation that we cannot form more than 1 mo! of ester, even if we use up all the acid. Therefore the correct root of the quadratic is 0.90. Consequently, 0.9 mo! of ester is formed at equilibrium. Note that the number of mo! of ester formed is greater than the number of moles of ester in equilibrium in (a) (0.90 compared with 2/3). This result was to be expected because of the increased concentration of alcohol, one of the reactants. A further addition of alcohol would lead to an even greater yield of ester, but in no case could the amount of ester formed exceed 1 mo!, since 1 mo! would represent 100% conversion of the acid. In practice, the actual selection of the excess concentration to be used may depend on economic factors. If alcohol is cheap compared to acid and ester, a large excess of alcohol might be used to assure a high percent conversion of the acid. If, on the other hand, alcohol costs more per mo! than acid, it would be more sensible to use an excess of acid and aim for a high percentage conversion of the alcohol.

EQU I L I B R I U M

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20.72

42 1

In an experiment at 490 °C, the following equilibrium composition was obtained for the reaction H 2 (g) + 1 2 (g) � 2Hl(g) + heat [HI] = 1 .02 x 1 0 - 2 mol/L [H 2 ] = 8.62 x 1 0 - 4 mol/L [1 2 ] = 2.63 x 10 - 3 mol/L (a) Evaluate the equilibrium constant. (b) Calculate l'lG0• ( c) Predict the effect of increasing the volume on the equilibrium. (d ) Predict the effect of increasing the temperature on the equilibrium. ( l .02 x 10 - 2 ) 2 I [HIJ 2 = 45.9 = (a ) K = [H 2 ][I 2 ] (8.62 x 1 0 - 4)(2.63 x 1 0 - 3 ) (b) l'lG 0 = - 2.30R T log K = - (2.30)(1 .99 cal/K)(763 K)(log 45.9) = - 5.80 x 10 3 cal = - 5.80 kcal

(c) There should be no effect. Increasing the volume decreases the partial pressure of each component. But since there are equal numbers of mol of gas on each side of the equation, there will be no shift of the equilibrium. (d ) The equilibrium will shift left.

20.73

Calculate l'lE0 for the reaction 2A(g) + B(g) --+ A 2 B(g) for which l'lS0 = 5.0 1/K, K = 1 .0 T = 300 K.

1 0 - 1 0 , and

The change in the number of moles of gas, l'ln, is - 2.

l'lG0 = - 2.30R T log K = - 2.30(8.3 1 J/K)(300 K)( - 10) = 57.3 kJ l'lH° = l'lG0 + T l'lS0 = 57.3 x 10 3 J + (298 K)(5.0 J/K) = 58.8 kJ

Assuming ideal gas behavior, l'l(P V) = (l'ln)R T, and

l'lE0 = l'lH° - l'l(P V) = l'lH° - (l'ln)R T = (58.8 kJ) - ( - 2)(8.3 14

20.74

10- 3 kJ/K)(300 K) = 63.8 kJ

Calculate the value of the equilibrium constant for the following reaction at 300 K and constant pressure: A(g) + B(g) � C(g) + D(g) + E(g)

using the ideal gas law and the following data: I

20.75

l'lE0 = - 90.0 kcal

l'lS0 = 100 cal/K

l'ln = 1 l'lH° = l'lE0 + l'l(P V) = l'lE0 + (l'ln)R T = l'lE0 + R T = - 90 000 + 2.0(300) = - 89 400 cal l'lG0 = l'lH° - T l'lS0 = - 89 400 - (300)(100) = - 1 19 400 cal - l'lG0 = 2.30R T log K + 1 19 400 log K = K = 3 x 10 8 6 = 86.5 (2.30)(2.00)(300)

Assuming that the heat capacities of H 2 (g), N 2 (g), and NH 3 (g) do not vary with temperature and further assuming that l'lS is independent of temperature for the reaction estimate the minimum temperature at which this reaction will occur spontaneously with all reactants at unit activity.

At the minimum temperature at which the reaction occurs spontaneously, l'lG = l'lH - T l'lS = 0. (Any higher temperature would make l'lG negative; any lower would make it positive.) Then l'lH = T l'lS and T = l'lH/l'lS. From Tables 1 8.2 and 1 8.5,

20.76

For the reaction

l'lH = 21'lH/(NH 3 ) = 2( - 1 1 .04) = - 22.08 kcal = - 92.38 kJ l'lS = 2S0(NH 3 ) - 3S0(H 2 ) - S0(N 2 ) = 2(192.5) - 3(1 30.57) - 191 = - 197.7 J/K - 92.38 kJ T= = 467 K - 0. 1 977 kJ/K C 2 H 6 (g) � C 2 H 4(g) + H 2 (g)

l'lG0 is 5.35 kcal at 900 K. Calculate the mole percent of hydrogen present at equilibrium if pure C 2 H 6 is passed over a suitable dehydrogenation catalyst at 900 K and 1 .00 atm pressure.

ELECTROC H E M I STRY

531

6Mn04 + 48H + ( 1.00 M) + 10Cr 3 + + 35H 2 0 --> 6Mn z + + 24H 2 0 + 5Cr 2 0� - + 70H + (0.70 M) 0.0592 (0.20) 6 (0. 10) 5(0. 70) 7 0 E = 0 1 8 - -- 1 og = 00 1 9 V 30 (0. 10)6 (0.20) I O 0

(

)(

)

The results would have been identical if more significant figures had been used.

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23.108

Calculate the potential of an indicator electrode, versus the standard hydrogen electrode, which originally contained 0. 100 M Mn04 and 0.800 M H + and which has been treated with 90% of the Fe 2 + necessary to reduce all the Mn04 to Mn 2 + . E0 = 1.51 V I Mn04 + SH + + 5e - --> Mn 2 + + 4H 2 0 J 2 5Fe + --> 5Fe + + 5eMn04 + 8H + + 5Fe 2 + --> 5Fe 3 + + Mn 2 + + 4H 2 0 [H + J = 0.0800 [Mn 2 + ] = 0.0900 [Mn04 ] = 0.010 [Fe 3 + J = 0.450 Since the Mnv11/Mn11 and Fe111/Fe11 systems are both in the same vessel, they must both experience the same half-cell potential. [Mn 2 + ] 0.0592 0.0592 (0.0900) = l . 5l - -- log E = l . 5 l - -- log = 1 .39 V [Mn04 ][H + J 8 (0.0100)(0.0800) 8 5 5

(

23.109

)

(

)(

)

To perform an analysis of a mixture of metal ions by electrodeposition, the second metal to be deposited must not begin plating out until the concentration ratio of the second to the first is about 10 6 • What must be the minimum difference in standard potential of two metals which form di positive ions in order for such an analysis to be feasible?

23.110

)(

The E value is the same for both metals, since both are present in the same solution.

(

)

0.0592 0 = E 1 - E 2 = E01 - E20 - -- (log 10 6) 2

(0.0592)

E� - E� = -- (6.00) = 0.1 8 V

At what potential should a solution containing 1 M CuS0 4 , 1 M NiS0 4 , and 2 M H 2 S0 4 be electrolyzed so as to deposit essentially none of the nickel and all of the copper, leaving 1.0 x 1 0 - 9 M Cu 2 +? I E0 = 0.34 V Cu 2 + + 2e - --> Cu E = 0.34 -

(0·0�92 )(1og [C:2 + ] ) = 0.34 - (0.0296)(9.00)

= 0.34 - 0.266 = 0.07 V

Electrolysis at 0.07 V would leave 1 .00 x 1 0 - 9 M Cu 2 + without reducing the Ni i + concentration. [At 0.07 V, the concentration of nickel(II) ion could theoretically be as high as 6 x 10 1 0 M. The calculation follows.] E0 = - 0.25 V Ni 2 + + 2e - --> Ni

(

0.0592 0.07 = - 0.25 - -2

23.1 1 1

) (log

1 [Ni 2 + J

)

0.32 = (0.0296)(log [Ni i + ] )

10 1 ° M

From the data fo r the two calomel, Hg 2 Cl 2 , half-reactions i n Table 23. 1 , calculate the concentration of saturated KC! at 25 °C.

Treat the E0 value for the saturated calomel half-cell as an E value for the standard calomel half-cell. Hg2Cl2 + 2e - --+ 2Hg + 2Cl - (sat KCl) E0 = 0.270 V E = 0.244 V 0 92 0.244 = 0.270 - (0.0592)(log [Cl -]) E = E0 (log [CI - ] 2 ) [Cl - ] = 2.8 M

(°" � )

23.1 1 2

[Ni 2 + J = 6

For the cell Pt I H 2 (0.75 atm) I HCI (0.25 M) I I Sn 2 + (1.50 M), Sn4 + (0.60 M ) I P t (a) write the half-cell reactions (b) write the cell reaction (c) calculate the cell potential (d ) calculate the ratio of concentrations of tin(II) to tin(IV) which would cause the potential to be zero.

I (a)

E0 = 0.00 V 2 4 E0 = 0. 1 3 V Sn + + 2e - --> Sn + E0 = 0. 1 3 V (b) H 2 + Sn 4 + ------> 2H + + Sn 2 + 0.0592 (1.50)(0.25) 2 = 0. l 5 V (c) E = 0 . l 3 - -- 1 og (0.60)(0.75) 2

(

)(

)

C HAPTE R 24

\:\

Nuclear and Radiochemistry� Note:

Before working the problems of this chapter, you should familiarize yourself with Table 24. 1 , p. 546, Table 24.2, p. 555, and Table 24.3, p. 559.

24.1

NUCLEAR PARTICLES AND NUCLEAR REACTIONS The designations for small particles are:

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24.1

24.2

24.3

24.4

24.5

d IX

lH iH

proton deuteron alpha particle beta particle (electron)

gamma ray (photon)

neutron positron

1He - ? /3

Select from the following list of nuclides (a) the isotopes (b) the isobars (c) the isotones: i� Ar, i� K, 1?Sc, i�Sc, �gzr

I (a) Isotopes have the same atomic numbers: 1?Sc and 1iSc.

and 1?Sc.

(b) Isobars have the same mass numbers: i�Ar

Show that a mass of 1 .00 u is equivalent to 932 MeV of energy. I

1 1 g= kg 6.02 x 1 0 2 6 6.02 x 1 0 2 3 1 kg (3 x 10 8 m/s) 2 = 1 .49 E = mc 2 = 6.02 x 1 0 2 6 1 u=

Since

(

)

1 MeV = 10 6 eV = ( 1 0 6)( 1 .6

But

E = ( 1 .49

and so

[A more precise value is 93 1 . 5 MeV/u.J

1 0 - 1 0 J)

1 0 - 1 9 C)( l .O V) = 1.6

C ��;�1 ) .6

10- io J x

10- 1 3 J

= 932 MeV

Do the designations lH and &n for the proton and neutron imply that these two particles are of equal mass ( 1 u)?

I No. Mass numbers reflect actual masses only to the nearest u. The precise masses are 1 .007 276 5 u for the proton

and 1 .008 665 0 u for the neutron.

What are the numbers of protons, electrons, and neutrons in 2 39Pu?

I Protons: 94, from the periodic table. Electrons: 94, since the atom is uncharged. Neutrons:

239 - 94 = 145.

Write the complete nuclear symbols for natural fluorine and natural arsenic. Each has only one naturally occurring isotope.

I Since each element has only one naturally occurring isotope, the mass number is the integer closest to the atomic

weight.

24.6

1 �F, nAs

How many protons, neutrons, and electrons are there in each of the following atoms:

(a) 3He

(b) 1 2 C

I (a ) From the periodic table, we see that the atomic number of He is 2; therefore the nucleus must contain 2 protons.

Since the mass number of this isotope is 3, the sum of the numbers of protons and neutrons is 3; therefore there is 1 neutron. The number of electrons in the atom is the same as the atomic number, 2. (b) The atomic number of carbon is 6; hence the nucleus must contain 6 protons. The number of neutrons is 12 - 6 = 6. The number of electrons is the same as the atomic number, 6. ( c) The atomic number of lead is 82; hence there are 82 protons in the nucleus. The number of neutrons is 206 - 82 = 1 24. There are 82 electrons.

542

N UCLEAR A N D R A D I O C H E M I STRY

24.7

Determine the number of (a) protons, (2) 72Ge. (3) 9Be. (4) 2 3 5U.

I (1)

24.8

(a) 32 (b) 38 (c) 32

(2)

(b) neutrons,

(a) 32 (b) 40 (c) 32

543

(3) (a) 4 (b) 5 (c) 4

(4) ( a) 92 (b) 1 43 (c) 92

Classify each of the following nuclides as "probably stable," "beta emitter," or "positron emitter": igCa 2 i�Pb � B 1 ��Ho igAI 1 �gsn j�Kr

1 ��Hg

I The nuclides near the "belt of stability" are probably stable; those above that belt are beta emitters; those below,

positron emitters. Thus the stable nuclides are 2 0 8 Pb and 1 2 0 Sn; the beta emitters are 49Ca, 3 0 Al, and 94Kr; the positron emitters are 1 95 Hg, 8 B, and 1 5 0 Ho.

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24.9

24.10

Of the three isobars 1 i�Cd, 1 iiln, and 1 �6Sn, which is likely to be radioactive? Explain your choice.

1 iiln, which has an odd number of protons and an odd number of neutrons.

Which of the following nuclides is least likely to be stable? (e) ns

( a ) igCa

(b) igAl

(d) � � Mn

I 3 0Al, which has an odd number of protons and an odd number of neutrons.

24.11

Using the approximate equation for the radius of the nucleus, r = 1 .4 A 1 1 3 x 1 0 - 1 3 cm, the nucleus of 1 ��Ag and compare it with the density of metallic silver ( 1 0.5 g/cm3) . I r = 1 .4( 1 07) 1 1 3 x 1 0 - 1 3 cm = 6.65 x 1 0 - 1 3 cm

calculate the density of

The volume, mass, and density of a single nucleus are calculated from the radius by the following equations:

The nuclear density is 1 .4 24.12

V = tnr3 = (tn)(6.65 x 1 0 - 1 3 cm)3 = 1 .23 x 1 0 - 3 6 cm3 107 g/mol = 1 .78 x 1 0 - 2 2 g m= 6.02 x 1 0 2 3 atoms/mo! m 1 .78 x 1 0 - 2 2 g = 1 .4 x 10 1 4 g/cm3 d=-= V 1 .23 x 1 0 - 3 6 cm 3 1 0 1 3 times that of silver metal.

Antimatter consists of particles which have properties and characteristics similar to particles in known atoms but opposite charges. For example, an electron and a positron are two particles identical in every respect except for the sign of their charges. Speculate on the existence of a world composed of negative antiprotons and positrons (along with an array of other antiparticles) and neutrons. Would the chemistry in such a world have formulas similar to those of compounds on earth? What would happen if material from earth encountered material from the "antiworld"?

I One can imagine that everything would seem "natural," with "hydrogen" atoms consisting of one antiproton and

one positron, "carbon" atoms having six antiprotons and six neutrons in the nucleus and six positrons outside the nucleus, etc. The "carbon" and "hydrogen" could form "CH " by the "usual" rules of covalent bond formation. 4 Nuclear reactions in which electrons were produced would lead to annihilation reactions: /r + p + -+ energy. Similarly, reaction of material from the two worlds, e.g., ordinary hydrogen and "hydrogen" from the antiworld, would also lead to annihilation reactions: Proton + antiproton -+ energy

24.13

When the pure nuclide ��Cr is bombarded with alpha particles, two reactions occur and neutrons and deuterons are observed as product particles . Write equations showing the formation of the possible product nuclides.

I ��Cr(cx, n)��Fe

��Cr(cx, d)� �Mn (The sum of the subscripts must be the same on both sides of the equation, as must the sum of the superscripts.)

24.14

Complete the following nuclear equations: 1 �N + iHe ---+ 1 �O + · · · (b) �Be + iHe ---+ 1 �C + (c) �Be(p, ex) · · · (d ) i�P ---+ f�S + · · · (e) iH ---+ �He + · · · (/) igCa(cx, . . . }i¥Sc

(a )

I (a) The sum of the subscripts on the left is

7 + 2 = 9 . The subscript of the first product on the right is 8. Hence the second product on the right must have a subscript (nuclear charge) of 1 . The sum of the superscripts on the left is 14 + 4 = 1 8. The superscript of the first product on the right is 1 7. Hence the second product on the right

COOR D I NAT I O N C O M POU N DS

27.39

589

(a) Calculate the freezing point of the solution containing 24.8 g solute per kg water for each of the following solutes: (i) [Co(NH 3 ) J (N0 2 ) J ] (ii) [Co(NH 3 )4(N0 2 ) 2 ] [Co(NH 3 ) 2 (N0 2 )4] (iii) [Co(NH 3 ) 5 (N0 2 )][Co(NH 3 ) z (N02)4] 2 (b) Write empirical formulas for the compounds described in (a) and in Problem 17.19. (c) For which set of com pounds is the term polymerization isomerism appropriate? Explain. (d ) Explain why the compounds of one of these sets cannot be distinguished by freezing point depression methods, while those of the other set can be dis tinguished in this way. (e) Suggest an experimental technique which can be used to distinguish between the poly merization isomers.

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(a)

(i) (24.8 g) M

1 mo! ) ( 248 = 0. 100 mo! g

= (0. 100 m)(l .86 °C/m) = 0. 1 86 °C ; t = - 0. 186 °C 1 mol compound 2 mol ions = O. l OO mo! ions (ii) (24.8 g) mol compound 496 g M = 0. 186 °C; t = - 0. 1 86 °C 1 mo! compound 3 mol ions (iii) (24.8 g) = 0. 100 mol ions mo 1 compound 744 g �t = 0. 1 86 °C; t = - 0. 1 86 °C (b) CoN 6 H 9 0 6 and CH. (c) Polymerization isomerism applies to the coordination compounds, since the freezing points cannot distinguish between the compounds of differing formula weights. (d ) The value of i is (almost) proportional to the number of ions per formula unit for the complex compounds, whereas i = 1 for both of the (covalent) organic compounds. (e) Conductance.

)(

( (

27.40

)

)(

)

Describe the types of isomerism which can be associated with carbonatoaquotetraamminecobalt(III) chloride monohydrate.

Geometric, hydrate, and coordination isomerism are all possible, including the possibility of the carbonate group being mono- or bidentate. The following compounds results from these types of isomerism: [Co(NH 3 )4(C0 3 )]C1 · 2H 2 0 and cis and trans isomers of each of the following: [Co(NH 3 )iC0 3 )Cl] ' 2H 2 0, [Co(NH 3 )4(H 2 0) 2 ]C0 3 · Cl, [Co(NH 3 )iH 2 0)CIJC0 3 H 2 0, [Co(NH 3 )4(C0 3 )(H 2 0)]CI · H 2 0 ·

27.41

Two compounds have empirical formulas corresponding to Cr(NH 3 ) J (N0 2 ) J . In aqueous solution, one of these is a nonelectrolyte, while the other conducts electricity. What is the lowest possible formula weight of the conducting reagent? What is the highest possible formula weight for the nonconducting reagent? I

The lowest conducting polymerization isomer is one of the dimers (compare answer to Problem 27.38), with a formula weight 482 g/mol. The highest nonconducting isomer is the monomer [Cr(NH 3 ) J (N0 2 ) J ], with a formula weight 241 g/mol. 27.42

The compound Co(NH 3 h(H 2 0)(N0 2 )(0Cl)Br can exist in a variety of isomeric forms. Write structural formulas for at least 10 isomeric forms of this composition, and indicate the types of isomerism shown.

I Geometric, optical, linkage, coordination, and hydrate isomerism must be considered. In addition to the five isomers shown in Fig. 27.7, attachment of the nitro group through the oxygen atom instead of the nitrogen atom would yield five additional isomers. Exchange of the N0 2 group with Br yields five more.

[ [

NH3 H3N -t--O H2 I / ( - ) H , N --,-N02 0 (+)

J l

OH2 H3N +-- NH3 I H3N-+- o/ NOz Cl

27.43

[

�

H3N NH I / H ,N--f-OH2 N02

Write the formulas for all possible polymerization isomers of [Pt(NH 3 ) 2 (N0 2 ) 2 ].

[Pt(NH 3 )4] [Pt(N0 2 )4] [Pt(NH 3 ) J (N0 2 )] 2 [Pt(N0 2 )4]

Fig. 27.7

[Pt(NH 3 ) J (N0 2 )][Pt(NH 3 )(N0 2 h] [Pt(NH 3 )4] [Pt(NH 3 )(N0 2 ) 3 ] 2

590

27.44

C H A PTER 27

Sketch the mirror image of the structure indicated in Fig. 27.8 and state if it is optically active.

Fig. 27.9

Fig. 27.8

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27.45

27.46

See :Fig. 27.9. The two isomers are optically active.

Some ligands are multifunctional; that is, they have two or more atoms which can bind to the central metal atom or ion. Each binding site occupies a different corner on the coordination surface. Ethylenediamine (abbreviated en) is such a ligand; the two binding atoms are nitrogens, and the two binding sites must be cis to each other. How many geometrical isomers of [Cr(en) z C1 2 ] + should exist, and which isomer(s) might display optical activity?

Two geometrical isomers exist, cis and trans (Fig. 27. 10). Each en can be represented by an arc terminating at the two binding sites. By drawing other arrangements of the arcs while preserving the positions of the chlorines, one can see that (b) is a distinct mirror image of (a), whereas there is no mirror image of (c) except (c) itself. In other words, only the cis isomer can be optically active; the trans isomer is its own mirror image. Cl

en en

(a)

(b)

c is

Fig. 27.10

Predict whether [Ir(en) J ] 3 + should exhibit optical isomerism. If so, prove by diagrams that the two optical isomers are not simple rotational aspects of the same compound.

There are two optical isomers, which may be represented as in Fig. 27. 1 1 . If you make models with cardboard, sticks, and pipe cleaners, you can rotate either one as much路 as you like and never get the other (without breaking bonds).

27.47

Fig. 27.1 1

The [Co(en)Br 2 1 2 ] - ion has two optical isomers (nonsuperimposable mirror images). Draw the structures of these two isomers. Do not draw any other isomers. I See Fig. 27. 1 2.

COO R D I NAT I O N C O M PO U N DS

591

, Ll=D

(;�"' N� 27.48

Fig. 27.12

To demonstrate that optical activity is not related to the presence of carbon atoms in a molecule, Werner prepared [Co{(H0) 2 Co(NH 3 )4 } J J 6 + . Draw the structure of this complex ion, and show how Werner explained its optical activity.

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See Fig. 27. 13. This complex, like Co(en)� + , has a mirror image which is not superimposable; hence it is optically active.

27.49

The complex ion [M(CN)(N0 2 )(H 2 0)(NH 3 )] + was found to be optically active. What does this finding signify about the configuration of the coordination sphere?

The coordination sphere cannot be planar. By implication it may be tetrahedral, since most simple 4-coordinate complexes are either planar or tetrahedral, but it does not prove that the complex is tetrahedral.

27.50

Draw all geometric and optical isomers of Co(en)Cl 3 Br - .

See Fig. 27. 14. There are no optical isomers, but there are two geometric isomers with two Cl atoms trans to each other or Cl trans to Br.

27.51

Cl uh � Cl

Cl rhCl �l Br

Fig. 27.14

State the type(s) of isomerism possible for all isomers corresponding to each of the following : [Cr(NH 3 ) 4Br 2 ]N0 2 (c) [Cr(en) 2 Cl 2 ] + (d) [Cr(en} J ] 3 + (e) [Cr(NH3) 4C!Br]Br

(b)

(a)

I (a)

[Cr(NH 3 )4 Br 2 ] +

See Fig. 27. 1 5. There is geometric (cis-trans) isomerism. (b) There is geometric isomerism with two bromine atoms in the coordination sphere, with one bromine atom and a nitro ligand, and with one bromine atom and a nitrito ligand. Thus there are also coordination isomers and linkage isomers. The nitrito ligand is bonded through oxygen; the nitro ligand is bonded through nitrogen. A total of six isomers correspond to the formula. (c) There i s geometric isomerism. and in the cis i somer t here also exists the possibility of optical .isomerism. A total of th ree isomers can exist. (d) Two optical isomers are possible. (e) Geometric and coordination isomers exist-cis-trans chloro-bromo and cis-trans dibromo isomerism make a total of four possible isomers.

Fig. 27.15

COO R D I NATI O N C O M POU N DS

60 1

CFSE for the d6 weak field ion is - 0.40�. The �Hhyd is higher because the energy of the complex is lowered by that much, - 1 1.4 kcal/mo!. CFSE

27.100

- 1 1.4 kcal/mo I

0.40�

�

28.5 kcaljmol

Using the observed lattice energies tabulated below, estimate the magnitude of � in octahedrally coordinated crystals of VO, MnO, and FeO. Oxide

kJ/mol

TiO

Cao

- 3882

- 3465

MnO

- 39 1 7

FeO

- 38 1 3

NiO

coo

- 3923

- 3992

- 4076

ZnO

- 4035

Plot (Fig. 27.21) the values of U versus atomic number. The Ca 2 + , Mn 2 + (weak field), and Zn 2 + ions, d 0 , d5 , and d1 0 , are expected to have CFSE 0. Draw a smooth curve through these points, and estimate the CFSE of v i + and Fe 2 + from the difference between the experimental points and those on the smooth curve. CFSE - 1.2� - 226 kJ/mol � 1 88 kJ/mol For v 2 + d 3 , 2 CFSE - 0.4� - 67 kJ/mol � 1 67 kJ/mol For Fe + d6 ,

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There is no way to estimate � for Mn 11, since CFSE same range.

0 no matter what the value of �. but it probably is in the

4050 3950 0

3 8 50

_§_

-, -"

3750

:::)'

3650 3 5 50 3450 22

27.101

Fig. 27.21

Explain why platinum(II) and palladium(II) form square planar complexes almost exclusively but only a few nickel(II) complexes are square planar.

Platinum(II) and palladium(II) complexes have greater values of �- The greater � values imply greater stabiliza tion of the square planar configuration, because the d8 ions have CFSE = - 2.456� vs. - l .20� for octahedral coordination spheres. The greater the magnitude of �, the greater is the difference in CFSE. When the difference in CFSE is sufficient to overcome the loss of bonding energy in two bonds (6 to 4), the square planar complex is

stabilized.

27.102

The M-0 bond distances in a series of 6-coordinate crystal oxides are tabulated below. Plot these data against the atomic numbers of the metals, and explain the observed trends. Cao

Oxide

2.4

d, A I

TiO

2.05

2. 1

MnO 2.2

FeO 2. 1 8

NiO 2.08

ZnO 2.1

See Fig. 27.22. Except for iron(II), the points fall on the now familiar double-humped curve. The curve is inverted, because as the interactions between atoms get stronger the distances get shorter. 2.4

2.3 d

2.2 2. 1 2.0

m�

........_ l!l

Fig. 27.22