Preview 1000 mcqs for a level chemistry by tusachduhoc

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Section I: PHYSICAL CHEMISTRY Topic 1

Topic 2

Topic 3

Atoms, Mol ecules and Stoichiometry ----------------------------------------- 14 Atomic Structure

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Chemical Bonding

Topic 4

T he gaseous State

Topic 6

Electrochemistry - -

Topic 5

Topic 7

Topic

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Chemical Energetics ----------------------------------------------------------------- 66 --------

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Equilibria -------------------------------------------------------------------------------- 97 -

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Reaction Kinetics --------------------------------------------------------------------- 114

Section II: INORGANIC CHEMISTRY Topic 9

Topic 10 Topic 11

The Periodic Table: Chemical Periodicity---------------------------------- 129 Group II

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14 2

Group VII --------------------------------------------------------------------------------- 154

Topic 12

An Introduction to the Chemistry of T ransition Elements------------- 165

Topic 13

Introductory Topics of Organic Chemistry

Section III: ORGANIC CHEMISTRY Topic 14

Topic 15 Topic 16

Topic 17

Topic 18

Topic 19

Hydrocarbons

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Halogen Derivatives

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Hydroxy Compounds Carbonyl Compounds

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17 4

182

195

- 2 09

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Carboxylic Acids and Derivatives Nitro gen Compounds

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221

238

255

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TABLES OF CHEMICAL DATA

Important values, constants and standards

= 8.31 J 1\1 mor1

= 6.02 x 1023 mor1

=3.00 x 108 m s-1

rest mass of proton, : H

= 1.67 x 10- 27 kg

rest mass of neutron, �n

2 = 1.67 x 10- 7 kg

= 9.11 x 10- 31 kg

=-1.60 x 10- 1 9 c

molar gas constant the Faraday constant " the Avogadro constant the Planck constant speed of light in a vacuum

rest mass of electron,

-�e

electronic charge

F h

= 9.65 x 104 c mor1

= 6.63 x 10--34 J s

Vm = 22.4 dm3 mor1 at s.t.p. Vm = 24 dm3 mor1 under room conditions (where s.t.p. is expressed as 101 kPa, approximately, and 273 K (0 C)) = 1.00 x 10-1 4 mol2 dm- s ionic product of water molar volume of gas

(at 298 K (25 C]) °

specific heat capacity of water

= 4.18 kJkf11\1 (=4.18 Jg- 1\1 )

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Standard electrode potential and redox potentials, F at 298 K (25 °C)

For ease of reference, two tabulations are given:

(a) an extended list in alphabetical order; (b) a shorter list in decreasing order of magnitude, i.e. a redox series. (a)

F in alphabetical order E"'7N

Electrode reaction Ag++ eA'f3+ + 3eBa2+ + 2eBr2 + 2eCa2++ 2eCZ2 + 2e2HOCZ + 2H++ 2eCo2++ 2eCo3++ e[Co(NH3)s]2++ 2eCr2+ + 2eCr3++ 3eCr3++ eCr20/- + 14H++ 6eCu++ eCu2++ 2eCu2++ e[Cu(NH3)4f + 2eF2 + 2eFe2• + 2eFe3• + 3eFe3• + e[Fe(CN)s]3- + eFe(OHh + e2H• + 2e12 + 2eK• + eLt+ e2 Mg • + 2eMn2• + 2eMn3• + eMn02 + 4H• + 2eMn04- + eMn04- + 4H• + 3eMn04- + 8H• + 5eN03- + 2H• + e-

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

Ag AZ Ba 2BrCa 2czCZ2 + 2H20 Co Co2+ Co + 6NH3 Cr Cr Cr2+ 2Cr3++ 7H20 Cu Cu cu· Cu + 4NH3 2FFe Fe Fe2• [Fe(CN)s)4Fe(OH)2 + OW H2 2r K Li Mg Mn Mn2• Mn2• + 2H20 Mno/Mn02 + 2H20 Mn2• + 4H20 N02 + H20

+0.80 -1.66 -2.90 +1.07 -2.87 +1.36 +1.64 -0.28 +1.82 -0.43 -0.91 -0.74 -0.41 +1.33 +0.52 +0.34 +0.15 -0.05 +2.87 -0.44 -0.04 +0.77 +0.36 -0.56 0.00 +0.54 -2.92 -3.04 -2.38 -1.18 +1.49 +1.23 +0.56 +1.67 +1.52 +0.81

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Atoms, Molecules

a n d Stoichiometry Section A The relative atomic mass of boron, which consists

of the isotopes

;0s and ;1s is 1 0.8. What is the

percentage of

;1s atoms in the isotopic mixture?

A c

0.8% 20%

B D

8.0% 80%

Helping CMtcejd,; CH3CH20H + 30 2

% 02

�

2C0 2 + 3H 2 0

CH3CHO +

CH3CH3 +

Use of the Data Booklet is relevant to this question.

f 02

2C0 2 + 2H2 0

� �

2C0 2 + 3H 2 0

Helping concejt!;; Let

x be the percentage of ;1s atoms in the isotopic

mixture.

1 0( 1 - x) + 1 1x 1 0.8 x 0.8 (or 80%) =

What is the number of molecules in oxygen under room conditions?

A c

In the absence of a catalyst, ammonia burns in an excess of oxygen to produce steam and nitrogen. What is the volume of oxygen remaining when 60 cm 3 of ammonia is burnt in 1 00 cm3 of oxygen, all volumes being measured at the same tempera ture and pressure? A

35 cm 3 45 cm 3

B D

Helping ccinc,ejdJ

%x 60

60 cm 3 of N H 3 requires :. Volume of excess 02

Helping concejtu

U n der roo m co n d i t i o n s ,

24000 cm 3 •

45 c m of 0 2 .

1 00 - 45 = 55 cm

For complete oxidation, 1 mol of an organic compound requires 3 mol of oxygen gas. What could be the formula of the compound? B D

1000

chemistr�

CH3CHpH CH3C02H

rf/c<j with H ELPs

1 .34 x 1022 3.0 x 1026

1 m o l e of gas occ u p ies

n =

Which of these samples of gas contains the same number of atoms as 1 g of hydrogen (M,: H2, 2 ) ? B

� = 2.083 x 1 0-2 mol 24000 2 23 N = 2.083x 1 0- x 6.02 x 1 0 22 = 1 .25x1 0

Hence, :.

40 cm 3 55 cm 3

1 .25 x 1 022 3.0 x 1 022

500 cm3 of

22 g of carbon dioxide (M,: C02, 44) 8 g of methane (M,: CH4, 1 6) 20 g of neon (M,: Ne, 20) 8 g of ozone (M,: 03 , 48)

Helping CMte,</lu A:

n =

22 x 3 44

1 .5 mol page 15

Topic l Atoms, Molecules & Stoichiometry

Frequently Examined

Helping co-ncejtti

[Relative atomic mass: Hg, 200]

Let x be the number of C atoms per molecule of X. Therefore, 1 mole of X will produce x moles of C02 when completely burnt in 02• Amount of X used = 0·1 1 2 22.4 = 5.00 x 1 0-3 mol

� ��1

6) 12 = 2.0 x 1 0-2 mol . . (5.00 x 1 0-3 )x = 2.0 x 1 0-2 => x = 4

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Amount of C02 produced =

1 6. Ethanedioate ions, C204 -, are oxidised by hot acidified, aqueous potassium manganate(Vll) according to the following equation .

2Mn0 4 - (aq) + 5C 2 o / - (aq) + 1 6 W (aq) 2 2Mn + (aq) + 1 0C0 2 (g) + 8H 2 0(C)

�

What vo l u me of 0 . 02 0 m o l dm-s potass i u m manganate(Vl l ) i s required t o oxidise completely 1 .0 x 1 0-s mol of the salt KHC2 0•. H2cp.? 20 ems 50 ems

Questions

B D

40 ems 1 25 ems

2.5 x 1 0-9 g 1 .o x 1 0-7 g

B D

4.0 x 1 0-8 g 1 .0 x 1 0-5 g

Helping cCincejdJ Amount of lipoyl groups in 5.0 kg of body liquid = (1 .0 x 1 0-8 ) x 5.0 = 5.0 x 1 0-8 mol Mass of Hg = (5.0 x 1 o-8 ) x 200 = 1 .0 x 1 0-5 g

1 8.

A sample of 1 0 dms of polluted air is passed through lime water so that all the carbon dioxide present is precipitated as calcium carbonate. The mass of calcium carbonate formed is 0.05 g. What is the percentage, by volume, of carbon dioxide in the air sample? [Relative atomic masses: C, 1 2; 0, 1 6; Ca, 40; 1 mol of gas under experimental conditions has a volume of 24 dms.] A c

0.03% 0.1 2%

B D

0.05% 0.3%

Helping C<Y1wejitJ

In 1 mole of KHC 2 0•. H2cp•• there are 2 moles of c20.2-. 3 nc o.2- = 2.0 x 1 0- mol 2

n Mno.-

VMn

= �5 x nC2042- = 8 x 1 0-4 mol

4 - = 8 x 1 0- = 4 x 1 0-2 dm3 = 40 cm3 CMn 0.020 04

nMn O•

1 7. Nervous disorders resulting from mercury poison

ing occur because mercury forms a 1 : 1 complex with lipoyl groups which are vital for glucose me tabolism. If the average concentration of lipoyl groups in the body fluid is 1 .0 x 1 o-s mol kg-1 , what mass of mercury could complex all the lipoyl groups in a human containing 5.0 kg of body fluid?

1 000

cbemist�

o.o5 40 + 1 2 + 3 x 1 6 = 5.0 x 1 0-4 mol -2 . . % by volume of C0 2 = 1 ·2 x 1 0 x 1 00 = 0.1 2 10 Amount of CaC0 3 precipitated =

rf!cq with HELPs

1 9. A mixture of 1 0 ems of methane and 1 O ems of ethane

was sparked with an excess of oxygen. After cooling to room temperature, the residual gas was passed through aqueous potassium hydroxide. What vol ume of gas was absorbed by the alkali? A

1 5 ems 25 ems

B D

20 ems 30 ems

page 18

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2 Atomic Structure

Topic

Frequently Examined

Gallium has the electronic configuration [Ar] 3d 1 0 4s2 4p ' , where [Ar] represents the configuration of ar gon.

1 9.

In which order are the electrons lost in forming the Ga•• ion? 1 st 2nd 3rd 4th A

4p 3d 3d

4s 4p

Questions

Helping ccvncejl tJ In 0, there are 44 electrons and its electronic configu ration is [Ar] 3d ' 0 4s2 4p6 4d6 5s2 • 2 Hence, 0 + has 42 electrons and its electronic configu ration is [Ar] 3d ' 0 4s2 4p6 4d6 • Note: The 2 electrons lost in forming 0 2 + are from the out

ermost 5s orbital.

Helping c,ona>ftlJ

22.

Use of the Data Booklet is relevant to this question.

In the gas phase, aluminum and a transition ele ment require the same amount of energy to form one mole of an ion with a 2+ charge.

Electrons furthest away from the nucleus are removed first. Hence, the electrons in Ga are removed in the order: 4p 4s 4s 3d.

Whal is the transition element? A

Use o f the Data Booklet is relevant to this question.

20.

The successive ionisation energies, in kJ mo1-1 , of an element X are given below. 870

1 800

3000

3600

5800

7000

1 3200

What is X? A c

:r.iAs

531

52Te

Helping co-nee/di

2 Al(g) � Al + (g) + 2e-

�H = 577 + 1 820 = 2397 kJ mo1- 1

2 Co(g) � Co + (g) + 2e

�H = 757 + 1 640 = 2397 kJ mo1 - 1

Helping co-na>ftlJ There is a large difference between the 6th and 7th ionisation energies, indicating that the 6th and 7th elec trons are from different principal quantum shells. Hence, X is a G roup VI element, i.e. 0 or Te. However, from the data booklet, the data do not fit into 0. Hence, X is Te.

23.

Whal is the electronic configuration of the atom of the element which is isoelectronic with H 2 S? 2 2 6 2 1 s 2s 2p 3s A B

2 2 2 6 2 1 s 2s 2p 3s 3p

4 2 2 6 2 1 s 2s 2p 3s 3p

21.

2 An ion 0 + contains 44 protons. 2

What is the electronic configuration 0 + ? [The symbol [Ar] represents 1 s2 2s2 2p6 3s2 3p 6 .J A

C D

1 000

5 5 [Ar] 3d1 0 4s2 4p 4d [Ar] 3d ' 0 4s2 4p6 4d2 [Ar] 3d ' 0 4s2 4p6 4d6 8 6 [Ar] 3 d ' 0 4s2 4p 4d

cbemistr�

mc<j with HELPs

6 2 2 6 2 1 s 2s 2p 3s 3p

Helping c,o,n,ceftlJ The electronic configuration of a sulfur atom is 6 2 4 2 2 1 s 2s 2p 3s 3p However, in H 2 S , there are 2 more electrons, each from a hydrogen atom in forming covalent bonds. Hence, the electronic configuration of S in H 2 S is 1 s2 2s2 2p6 3s2 3p6 (octet configuration).

page 3 1

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Topic

24.

2 Atomic Structure

Frequently Examined

Which of the following formulae represents a par ticle with the composition 1 proton , 1 neutron and 2 electrons? 2 [D represents deuterium, H] A

H-

Questions

Helping concejt,t; Given a nuclide

,;; x.

where n = mass number or nucleon number = number of protons and neutrons m = atomic number or proton number = number of protons

Helping con<>eft /<; The presence of only 1 proton means that the species is hydrogen (H or D). With 1 neutron, it is D. An excess of 1 electron over the proton indicates that it is o- .

Since both ig Ar and ig K have the same value of n, the� have the same nucleon number.

species protons neutrons electrons

1 1 1 2

1 2 2 2

1 1 0 2

27. A

The second ionisation energy of calcium is 1 1 50 kJ mo1-1 • Which of the following correctly represents this statement? 2 Ca(g) � Ca + (g) + 2e - f',.H" = +1 1 50 kJ mor 1 ca+ (g)

25.

Use of the Data Booklet is relevant to this question.

The s u nl ight-induced p hotolysis of water is being investigated as a u seful sou rce of the poll ution-free fuel hydrogen .

2Hp

2H2 + 02

�

It has been found that anatase, one of the three crystalline forms of the ionic compound Ti02, is a good catalyst for this reaction . How many electrons are associated with each ti tanium ion in the anatase lattice? A c

18 20

ca+(g)

Ca(s)

� � �

2 ca + (g) + e -

f',.H" = + 1 1 50 kJ mor 1

f',.H" = -1 1 50 kJ mor 1

Ca + (g) + e2 Ca + (aq) +

f',.H" = + 1 1 50 kJ mor 1

Helping CMu--ejiu Second I.E. is defined as the energy required (endother mic) to remove one mole of electrons from one mole of 2 gaseous M• ion to form one mole of gaseous M • ions.

I.E. + Second I.E.

First

f',.H" should b e positive.

19 22

B D

28. In 1 999,

Helping CMtcefdJ 22Ti4+

Number of electrons = 22 - 4 =

chemists claimed to be the first to identify atoms of a new element of proton number 1 1 4. This was produced by bombarding atoms of plutonium, Pu, with atoms of an isotope of a G roup I I element, X. The reaction taking place is the following. 2;1 Pu + X � m [new element) + 3 neutrons (ci n)

What is X?

26.

Which property i s the same for the two nuclides ig Ar and ig K?

1000

the number of electrons

B C

the number of neutrons the number of nucleons

the number of p rotons

cbemistr�

rf!c<j with HELPs

Helping concejil<;

The total number of protons and neutrons must balance. Hence, the proton number of X is hence, X is Ca.

1 1 4 - 94 = 20 and page 32

Topic 3 Chemical Bonding 7.

Which of the following molecules is not planar? B boron trifluoride A benzene D phosphorus trichloride C ethene

Helping ctvneejt/J

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Which equation defines the lattice energy of the ionic compound XY? B

X(s) + Y(s) --) XY(s) X(g) + Y(g) --) XY(s)

X+ (s) + y- (s) --) XY(s)

x+ (g) + y- (g) --) X Y(s)

Helping C<YJWejt£J

By definition, lattice energy is the energy released when 1 mole of an ionic compound is formed from its constitu ent gaseous ions (infinitely apart) combine together.

B C

Questions

density ethalpy change of vaporisation number of isomers vapour pressure

Helping c,o,nc,ejdi 0 �·f, c'fl Cl

PCl3 is trigonal pyramidal.

Frequently Examined

Which of the following best describes the change in the bond angle in water when the ion H30• is formed? A decreases to approximately 90° B decreases to approximately 1 09° C increases slightly D increases to approximately 1 20°

Helping <X»�

In Hp, the bond angle is 1 04.5° (2 lone pairs, 2 bond pairs). In H30+, there are 3 bond pairs and only 1 lone pair. Since bond pair exerts less repulsion, the bond angle in H30• become bigger (but still less than 1 09.5°).

1 0. As the number of carbon atoms in the homologous series of alane molecules increases, for which

As number of carbon atoms of the homologues increases, the van der Waals' forces increases so that the homo logues vaporise less readily. Hence, vapour pressure decreases.

1 1 . Which of the following isomers is likely to have the

highest boiling point? (CH3)2 CHCH(CH3) 2 B (CH3) 2 CHCH 2 CH 2 CH3 C CH3CH 2 CH(CH3)CH 2 CH3 CH3CH 2 CH 2 CH 2 CH 2 CH3 D

Helping c,o,m:ejdi

All the 5 compounds are isomers of hexane but (D) is unbranched. It has the greatest surface area of contact for VOW interaction and hence, the highest boiling point.

1 2. Which one of the following molecules will have no

permanent dipole? C2Cl4 A C C 2 H 5 CI

Helping <X»� Although there is a permanent dipole in each C-CI bonds, the effect of each cancels one another vectorically due to the symmetrical distribution of the 4 Cl atoms. Cl 'Cl

/" Cl / C = C ....,.

1 3. Which of the following molecules contains six bond

ing electrons?

property of the alkanes does the numerical value decrease?

1000

cbemist117

'ff!c </ with HELPs

page 40

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Topic 3 Chemical Bonding

Frequently Examined

Helping conc-¢1/s

Helping c0-ncejt!J

6 bonding electrons is equivalent to 3 bond pairs of electrons.

1 4. The atoms

.,..... N , I Cl

X and Y have the electronic configura

XY X2 Y

CCI, exists as discrete molecules and its molecules are non-polar. The forces operating between its molecules are id-id interactions .

tions shown below. X, 1 s2 2s 2 2p 6 3s2 3p6 3d 1 0 4s 2 ; Y , 1 s2 2s2 2p 4 Which one of the following compounds are they likely to form? C

Q uestions

B D

Helping concejttJ

XY2 XY,

has a completely filled d-subshell (d 1 0 s2 ) and it be haves like a Group II element. Y has 6 valence electrons and therefore is in G roup VI. It is expected that X will lose 2 electrons to Y so that both have stable electronic configurations. i.e. x 2 + y 2 - .

1 5. In which one of the following compounds does the underlined element not have eight electrons in the

outer shell?

1 7. Which one of the following structural features is

common to both diamond and graphite? A a carbon-carbon bond length equal to that in ethane B covalent bonds between carbon atoms C delocalised electrons D each carbon atom bonded to four others

Helping concejttJ In diamond and graphite, covalent bonds operate between carbon atoms and the extension of these bonds through out the lattices gives rise to macromolecular structure.

1 8. Which of the following is a property of a solution of

dry hydrogen chloride in dry methylbenzene? A It has a pH less than 7. B It is a non-conductor of electricity. C It reacts with magnesium to give hydrogen. D II reacts with dry copper(ll) oxide on warming to give a blue solution.

Helping cMtcejtlJ D

PCl 5

Helping cMUX!jt!J

Li has an electronic configuration of 1 s 2 2s 1 • On losing an electron to 0, u• acquires a duplet configuration (NOT octet).

1 6. Which type of bond is responsible for intermolecular

forces in liquid tetrachloromethane, CCI,? A covalent bonding B hydrogen bonding C induced dipole - induced dipole attractions D permanent dipole - permanent dipole attractions

1000

Cbemistr;g

rf!c<j with HELPs

Dry HCI does not dissociate in non-polar solvents such as methylbenzene. Since there is no free ions or elec trons, it does not conduct electricity.

1 9. What is the approximate value of the 0-C-O bond

angle in ethanoic acid? A 45° B D c 1 09"

900 1 200

Helping co-nceftlJ R

C-0

page 4 1

Topic 3 Chemical Bonding

Frequently Examined

30. Which diagram best represents the structure of solid

magnesium oxide?

Questions

Helping co-ncefi/J C H3 - C = 0 I

There is no H bonded directly to a small and highly electronegative atom, e.g. N, 0, F. B:

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CH 3 - �- H · · · · · · : �- CH 3 H H I

CH 3 - 0 : · · · · · · H - O H

H I

CH 3 I

H - N : · · · · · · H - N- H I

Helping roncejt/J

In an ionic lattice, the cations and anions occupy alter nate positions to maximise attraction and minimise repul sion. Only structure C satisfies this condition.

31 . Which set of properties could apply to a non-ionic

compound which has a giant lattice? physical state

electrical conductivity

liquid

does not conduct

liquid

does not conduct

at room temp. of the molten compound

solid

conducts well

solid

does not conduct

33. Which feature is present in the ions carbonate,

ethanoate, nitrate and phenoxide (phenate)? A all bond angles are 1 20° B dative covalent bonds C delocalised electrons D hydrogen bonds

Helping OMux'fttJ

m. p./ ° C

-1 1 4

melts over a temp. range

808 1610

Helping OMu:ejt/J

A non-ionic compound which has a giant lattice has strong covalent bonds. Thus, it has a high melting point. Since it does not contain mobile charges (e.g. electron), it does not conduct electricity.

Note: It is not a mixture and thus does not melt over a range of temperatures.

32. Which of the following molecules will not form a

hydrogen bond with another of its own molecules?

[ 0 i/�-� O' ·o

glass window. The gecko has millions of microscopic hairs on its toes and each hair has thousands of pads at its tip. The result is that the molecules in the pads are extremely close to the glass surface on which the gecko is climbing. What is the attraction between the gecko's toe pads and the glass surface? B

cbemistr;g

'Yflci with HELPs

34. The gecko, a small lizard, can climb up a smooth

1000

co-ordinate bonds covalent bonds page 44

4 The Gaseous State

Topic

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Frequently Examined

1 .6 x 1 0-3 x 8.31 x 273 Pa 3.0 x 1 0-t>

Questions

1 2. Which curve shows the correct graph of pV against for a fixed mass of an ideal gas at constant temperature? p

1 .6 x 1 o-3 x 8.31 x (273 + 273) Pa 3.0 x 1 0-6 1 .6 x 1 0-3 x 8.3 1 x 273 Pa 3.0 x 1 0-3

1 .6 x 1 o-3 x 8.3 1 x (273 + 273) Pa 3.0 x 1 0-3

Helping (XY}tce/dJ

p V = nRT nRT P=V (1 .6 x 1 0-3 mol} x 8.31 J K- 1 mo1- 1 x (273 + 273) K (3.0 x 1 0-3 ) m 3 _

1 1 . A small spacecraft of capacity 1 O m3 is connected

Helping (XY}t<:efdi Using ideal gas equation, pV = nRT, for a fixed mass of gas (n = constant) at constant temperature, pV = constant. Therefore, no matter how p varies, pV remains constant. In fact as p increases, V decreases in such a way that pV remains constant.

to another of capacity 30 m3 • Before connection, the pressu re inside the smaller craft is 50 kPa and that inside the larger is 1 00 kPa. If all measurements are made at the same tem perature, what is the pressu re in the combined arrangement after connection?

75 kPa 1 00 kPa

Helping conceft/J

87.5 kPa 1 25 kPa

�f where p1 = 50 kPa, V, = 1 0 m3 . n2 = �i where p2 = 1 00 kPa, V2 = 30 m3 •

1 3. Which graph is correct for a given mass of an ideal gas at constant pressure?

nl =

Pr VT = n = n + n = 1 ( r 1 2 RT P1V, + P2 V2 ) RT => PrVr = P1V, + P2 V2 (but Vr = V, + V2 )

P1V, + P2 V2 => Pr = �� V, + V2 (50 x 1 0 3 ) x 1 0 + (1 00 x 1 0 3 ) x 30 = 1 0 + 30 3 = 87.5 x 1 0 Pa (or 87.5 kPa)

1 000 chemistr;y

?IJl/cC/ with H ELPs

Helping co-nayt!J For an ideal gas, pV = nRT = nR(273 + t) where t is the temperature in °C. At a constant pressure, p is a constant and so are n and R. page 60

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Topic 5 Chemical Energetics Then

Frequently Examined

/:J.H� = !:J.H?(IC1 3 ) - !:J.H?(ICI) -88 = i'lH?(ICl 3 ) - i (+1 4) !:J.H?(IC1 3 ) = -81 kJ mo1- 1

23. The standard enthalpy changes of formation of iron(ll) oxide, FeO(s), and aluminium oxide, A1p3 (s), are -266 kJ mo1-1 and -1 676 kJ mo1-1 respectively.

What is the enthalpy change under standard con ditions for the following reaction? 3FeO(s) + 2Al(s) � 3Fe(s) + Al 2 0 3 (S) A

21 . Given the following enthalpy changes l 2 (g) + 3Cl 2 (g) l 2 (S) � l 2 (Q)

�

21Cl 3 (S) /:J.H 9 = -2 1 4 kJ mo1- 1 /:J.H 9 = +38 kJ moi- 1

What is the standard enthalpy change of formation of iodine trichloride, ICl 3 ? A

+1 76 kJ mo1-1 -88 kJ mo1-1

�12(s)

; (38) kJ

� l2 (g)

B D

Questions

+878 kJ mo1-1 -1 942 kJ mo1-1

-878 kJ mo1-1 -2474 kJ mo1-1

B D

Helping a;,ncefuJ !:J.H9 =

L "'H?(products) - I !:J.H?(reactants)

= !:J.H?(Al 2 03 ) - 3/:J.H?(FeO) = -1 676 - 3(-266) = -878 kJ mo1- 1

+1 38 kJ mo1-1 -1 38 kJ mo1-1

24. When steam condenses, 44 kJ mo1·1 of heat en thalpy is evolved.

What is the entropy change when 54 g of steam condenses at 1 00 °C?

:. !:J.H? = i (38) + i (-21 4) = -88 kJ mo1- 1

-354 J K-1 mo1-1 1 1 8 J K-1 mo1-1

B D

-1 1 8 J K-1 mo1-1 354 J K-1 mo1-1

Helping c<nu:ejtu

22. The standard enthalpy change of combustion of but-

When steam condenses at equilibrium, !:J.G9 = 0.

1 -ene, C 4H 8 (g), is x kJ mo1-1 • The standard enthalpy change of the reaction 2C2H4 (g) � C4H8 (g) is y kJ mol-1 •

What is the standard enthalpy change of combus tion of ethene, C2 H4(g)? A

�2 + y kJ mor 1

x+ Y kJ mor 1 2

x + l'._2 kJ mo1- 1

x - y kJ mor 1 2

/:J.G 9 = /:J.H9 - T/:J.S 9 O = -44 kJ mo1-1 - (273 + 1 OO)!:J.S 9 /:J.S 9 = -0.1 1 8 kJ K - 1 mo1- 1 nH2 0

54 = - = 3 mol 18

:. Entropy change = -3 x 0.1 1 8 = -0.354 kJ K- 1 mor 1 = -354 J K- 1 mol - 1

Helping C01teejtti

25. The enthalpy change of reaction between calcium and water can be measured in the laboratory. Ca(s) + 2H20(£)

�

Ca(OHh(s) + H2 (g)

What information, other than that obtained in this experiment, is needed to calculate a value for the enthalpy change of formation of Ca(OHMs)? 1 000 Cbemistr�

'ff!ctj

with HELPs

page 71

Topic 5 Chemical Energetics enthalpy change of atomisation of calcium enthalpy change of combustion of hydrogen first and second ionisation energies of calcium lattice energy of calcium hydroxide

C D

Ca(s) 2•

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Frequently Examined

2H20(C) � Ca(OH)i(s)

AH,[H,(g)]

[Ca(s)

white grey

H2(g)

02(g)j

H2(g)

. . li H !iH1[Ca(OHh (s)] - 2 x 1i Hc[H2 (g)] =

Ii G 9

!iG 9 = -2.09 - 1 2 (+7.3)

!iG 9 = -2090 - 1 2 (+7.3)

liG 9 = -2090 - 285 (-7.3)

C D

Less energy is required to remove one electron from the calcium atom than to remove two electrons. More energy is released in forming chloride ions from chlorine molecules in the formation of CaCl 2 (s) than in the formation of CaCl(s). The lattice energy of CaCl(s) is less exother mic than that of CaCl 2 (s). When CaCl(s) is formed from its elements, more energy is released than when CaCl2 (s) is formed from its elements.

Helping cwu:ejtti

In the Born-Haber cycle, one of the steps involves the lattice energy. For CaCI, Ca exists in Ca+ and it has a smaller charge and bigger size than Ca2+ in CaCl 2 • JL.E.J oc

I 1 + r_ q+q-

= -2.09 - 285 (-7.3)

Helping co-ncejta d.G 9 (grey) - !iG9 (white) 3 = [-2.09 x 1 0 - (273 + 1 2)(44.1)] - [0 - (273 + 1 2)(51 .4)] = -2090 - 285(-7.3) =

26. Which statement helps to explain why calcium and A

51 .4 44.1

0 -2.09

d.G 9

chlorine form CaCl 2 rather than CaCI?

d.H,9 I kJ mor 1 S0 I J K- 1 mor 1

What is the expression for d.G 9 , in J mo1-1 , for the formation of grey tin from white tin at 1 2 °C?

AH,[Ga(OH�(')j

H2(g)

Questions

Note: !iG 9 = !i H 9 - T!iS 9

28. Phosphine reacts with hydrogen iodide to form phosphonium iodide in the reaction shown.

PH 3 (g) + Hl(g) � PH/r (s) liH9 = -1 01 .8 kJ mor 1 Given that d.H,9 for PH 3 (g) = +5.4 kJ mo1-1 , and li H,9 for Hl(g) = +26.5 kJ mo1-1 , what is the stan dard enthalpy change of formation of phosphonium iodide? A

-1 33.7 kJ mo1-1 +69.9 kJ mo1-1

Helping e<»ic,ejttJ

-69.9 kJ mo1-1 +1 33.7 kJ mo1-1

From the relationship, it can be easily seen that the lattice energy for CaCI would be less exothermic. This makes the formation of CaCl(s) less favourable.

27. At temperatures below 1 3 •c, shiny, ductile metallic

tin, known as 'white tin', changes slowly into a grey powder which is brittle. Data for each form of tin are given in the table.

1 000 cbemistr:g

rf/cCJ with HELPs

[

liH,9 PH4 + 1- (s J

]

d.H,9( PH3 (g)] + li H,9[H l(g)] + !iH9 = +5.4 + 26.5 - 1 01 .8 = -69.9 kJ mo1- 1 =

page 72

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Topic 6 Electrochemistry

23.

Frequently Examined

Use of the Data Booklet is relevant to this question.

When aqueous hydrogen peroxide, Hp2 , is mixed with acidified potassium dichromate(VI), there is a colour change from orange to green. When aque ous hydrogen peroxide is added to acidified potas sium iodide solution, there is a colour change from colourless to brown. The oxidation number of oxygen in Hp2 is -1 . What are the oxidation numbers of oxygen after the reactions with potassium dichromate(VI) and potassium iodide?

after reaction with potassium dichromate(VI) -2

after reaction with potassium iodide -2

-2

Therefore,

mass of Cl 2 = _!_2 x 1 000 x 71 = 35500 g = 35.5 kg mass of H2 = _!_2 x 1 000 x 2 = 1 000 g = 1 kg

mass of NaOH = 1 000 x 40 = 40000 g = 40 kg

25.

Use of the Data Booklet is relevant to this question.

If iron is heated separately with chlorine, bromine and iodine, what are the likely products? A B c

chlorine FeCl 2 FeCl 3 FeCl 3 FeCl 3

H2 02 is oxidised by K2 Crp1• Hence, the oxidation number of 0 increases (from -1 to 0). The oxidation number of Cr decreases from +6 to +3. Hp2 oxidises Kl to 1 2 • The oxidation number of 0 decreases (from -1 to -2). The oxidation number of I increases from -1 to 0.

Helping concef£J

i Cl 2

Use of the Data Booklet is relevant to this question.

In the commercial electrolysis of brine, the products are chlorine, hydrogen and sodium hydroxide. What is the maximum yield of each of these prod ucts w h e n 5 8 . 5 kg of sod i u m c h l o ride a re electrolysed as brine?

A B c

yield of yield of yield of chlorine I kg hydrogen I kg sodium hydroxide I kg 35.5 1 40 35.5 2 40 71 1 40 71 1 80

�

FeCl 3

E�011 = 1 .36 - (-0.04) = +1 .40 V > O Fe +

i Br2

�

FeBr3

E�eu = 1 .07 - (-0.04) = +1 . 1 1 V > O Fe + 1 2

24.

iodine Fel 2 Fel 2 Fel2 Fel2

bromine FeBr2 FeBr2 FeBr3 FeBr3

Fe +

Helping c-Mtcejiti

Questions

�

Fe1 2

E�eu = 0.54 - (-0.44) = +0.98 V > 0 Fe3• can oxidise 1- to give 1 2 : Fe 3+ + 1- � Fe 2 +

.!2 1 2

E�011 = 0.77 - 0.54 = +0.23 V > 0

26. The diagram shows the apparatus needed to mea sure E9 for the reaction below.

....--- v �--.

Helping cMU:e�u

The overall reaction: 2Na0H(aq) + H2 (g) + Cl 2 (g) 58.5 x 1 0 3 = 1 000 mol Amount of NaCl used = 23 + 35.5 2NaCl(aq) + 2H2 0(C)

1 000 chemist�

�

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hydrochloric acid

page 8 6

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Topic 6 Electrochemistry

Frequently Examined

What are the identities of x

A Fe

Fe c Pt D Pt 8

and Y?

Questions

What will be observed when a few drops of acidified aqueous hydrogen peroxide are added to an ex cess of aqueous potassium iodide?

Fe 3+ (aq) Fe 2+ (aq) + Fe 3+ (aq) Fe 3+ (aq) 2 Fe + (aq) + Fe 3+ (aq)

A B

C D

Helping CMu:ejU;

3 In the Fe •/Fe2• half cell, an inert electrode (e.g. Pt or C) is required.

The solution turns brown and effervescence occurs. The solution turns brown without effervescence. The solution does not change colour and effer vescence occurs. The solution turns purple and effervescence occurs.

Helping c(lnceftu The overall reaction: 21- (aq) + H2 0 2 (aq) + 2 W

�

2H 2 0(t) + l 2 (aq)

E�011 = + 1 .77 - 0.54 = 1 .23 V

27. Sodium thiosulfate is used in the textile industry to remove an excess of chlorine from bleaching pro cesses by reducing it to chloride ions. S 20 32 - + 4Cl2 + 5H 2 0 � 2HS0 4 - + aw + ac1-

The solution turns brown as a result of the formation of 12 • However, there is no effervescence since no 02 gas is evolved.

ln this reaction, how many moles of electrons are supplied per mole of thiosulfate? A c

Helping wncejl/;;

29. Sulfur dioxide gas is converted into sulfate ions when

2 a

it is bubbled into aqueous manganate(Vll) ions. 2 Mn0 4 - (aq) + a W (aq) + 5e- � Mn + (aq) + 4H2 0( () S02 (g) + 2H2 0( ( ) � so/- (aq) + 4W (aq) + 2e

Balancing the half equation sp32-/HS04-: s2 0 32 - + 5H 2 0 � 2H s o 4 - + aw + ae-

Which graph shows how the pH changes as sulfur dioxide is bubbled at a constant rate into a well stirred solution of manganate(Vll) ions until its colour just fades?

Step 1 : Balance S S 2 0 32 - � gHS0 4 -

B pH

Step 2: Balance O by adding Hp S2 0 3 2 - + �H 2 0 � 2HSO 4 -

Step 3: Balance H by adding H• S 2 0/- + 5H 2 0

�

2HS04 - + aw

Step 4: Balance charge by adding electrons s 2 0/-

5H 2 0

�

2Hso 4 - + aw + ae-

time

c pH

Half equation

28. 1 2 (aq)

2e- � 2 r (aq)

2W (aq) + 0 2 (g) + 2e- � H2 0 2 (aq)

1000

cbemistr�

7!f/c<J with HELPs

£9 / V + 0.54

+0.6a

time

Helping ccwicef14

Overall reaction:

2Mn0 4 - + 580 2 + 2H2 0

�

2 2Mn + + 5SO/- + 4H + pa ge 8 7

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Topic 7 Equilibria

Frequently Examined

Helping cOJuxfi/J

AgBr(s) � Ag+ (aq)

Ksp = [Ag+ ][Bn = 5 x 1 0-1�

X (g) � 2X(g); 2

moi2 dm-6

32. When 0.20 mol of hydrogen gas and 0.1 5 mol of iodine gas are heated at 723 K until equilibrium is established, the equilibrium mixture is found to contain 0.26 mol of hydrogen iodide.

The equation for the reaction is as follows. +

l 2 (g) � 2Hl(g)

What is the correct expression for the equilibrium constant K0? A

2 x 0.26 0.20 x 0. 1 5

(0.26)2 0.07 x 0.02

(2 x 0.26)2 0.20 x 0. 1 5 (0.26)2 0. 1 3 x 0. 1 3

2 HI

0.2 0.2 - 0. 1 3 = 0.07

�6 2 2 Kc = [Hl] = ( 0 ) [H ][1 ) ( 0i7 )( 0i2 ) 2 2 --

__,___

0 0.26

0. 1 5 0. 1 5 - 0. 1 3 = 0.02

lc== :� o

273

T/ K

Helping CMweft,U Since !lH > 0, as temperature increases, there will be greater dissociation. The pressure becomes greater than expected. At extreme high temperature, when dissocia tion is near to complete, the pressure would be double than expected (since n is double).

34. In an acid-base titration, a 0. 1 0 mol dm-3 solution

Helping concejt!J initial amt I mol eqm amt / mol

!lH positive

Which graph best represents the p- T relation ship of the gas?

Br- (aq)

1 :. [Ag + ) = (5 x 1 0- 1 3 ) 2 = 7. 1 x 1 0-7 mol dm 3

H2 (g)

Questions

of a base is added to 25 cm3 of a 0. 1 0 mol dm-3 solution of an acid.

The pH value of the solution is plotted against the volume, V, of base added as shown in the diagram. 12

(0.26)2 0.07 x 0.02

8 6 4

-I - - - ----- -- -

33. The graph shows the pressure-temperature (p-T)

relationship of a 1 mol sample of helium in an en closed volume. --

0 p

---

273

.. T I K

Similar pressure-temperature measurements were made for a 1 mol sample of a gas which dissoci ates. 1000

cbemistr�

rf/cCf with HELPs

v/cm3

This diagram could represent a titration between A B C

CH 3C0 2 H(aq) and NH 3 (aq). CH 3C0 2 H(aq) and KOH(aq). HCl(aq) and KOH(aq). HCl(aq) and NH 3 (aq). page 1 05

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Topic 7 Equilibria

Frequently Examined

��

Helping crinc,ejdJ The graph starts with a pH of 1 , indicating that the acid is a strong acid. Since there is only 1 equivalence point, the acid should be monobasic. The equivalence point is at about pH = 5, showing that the base is a weak base. hence, the answer is (D) because HCI is a strong monobasic acid while NH 3 , is a weak base.

35. Stomach juices have a pH of 1 .0. Aspirin is 4 a monobasic acid represented by HA (K. = 1 0- mol dm-3 ) which dissociates into ions H• and k. What are the relative concentrations of H•, A- and HA when aspirin from a tablet enters the stomach? A

[ W ] > [HA] > [K ]

[HA) > [ W ] = [K ]

[W] > [K ] > [HA)

[ W ] = [K ) > [HA]

Helping cMicejUJ

A B c D

smallest pH pH pKa pKw

pKa pK w pKw PKa

Questions

largest p Kw pKa pH pH

Helping e<»iceflLJ

pH = -lg(1 o-7 ) = 7 pKw = - lg(1 0- 1 4 ) = 1 4 Ka =

[W ][OW ] [H 20J

[H 2 O]Ka = [H+ ][OW ] = Kw

Hence, K a = � [H 2 0] - lgK a = - lg K w + lg[H 2 0J PKa = pK w + lg 1 000 18 = 1 4 + 1 .74 = 1 5.74

At pH = 1 .0, [H•] = 0.1 mol dm-3 • When aspirin dissolves, HA � W + K

concentration

[HA] = a :. [W] = 0.1 + x and therefore, [ W ] > [K ). (K ) = X

37. Two diatomic gases, X2 and Y2 , react as follows. X2 (g) + Y2 (g) � 2XY(g) A mixture containing 0.5 moles each of X2 and Y2 is heated in a closed container and the reaction allowed to reach equilibrium. The graph shows how the number of moles of each gas varies with time.

Since K. is small and there are H+ present, dissociation of HA is low, i.e. [HA] > [A-J. Hence, [W ], [HA] > [K ]. Amount of aspirin consumed cannot be very high.

0.6 0.5

moles 0.4 0.3

:. [W ] > [HA] > [K ).

0.2 0. 1

time

Use of the Data Booklet is relevant to this question.

36.

Water dissociates as shown. H2 0 � H+ + OW

At 25 °C, the equilibrium value of [H•J is 1 0-1 mol mol dm-3 . dm-3 ; (H 2 0) =

1��0

Which is the value of the equilibrium constant Kc for this reaction? A c

1 .5 9

B D

3 18

What is the order of increasing numerical value of pH, pK. and pKw for this equilibrium at this tempera ture? [pKw = -logKwl 1 000

cbemistr,y

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page 1 06

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Topic 8 Reaction Kinetics

Frequently Examined

What is the function of the hydrochloric acid in the reaction mixture? A B C

to increase the reaction rate by catalytic action to ensure that the reaction reaches equilibrium to maintain a constant pH during the reaction to suppress ionisation of the ethanoic acid formed

or a backward reaction ( :. (D) is wrong). However, it does not affect the equilibrium constant (only affected by temperature) nor the yield of product.

Helping concejtt4 HCI increases the rate of hydrolysis and is not con sumed at the end of the reaction. Therefore, it is acting as a catalyst.

Which one of the following curves would be ob tained if the rate of reaction was plotted against time for an auto-catalytic reaction (i.e. reaction in which one of the products catalyses the reaction)?

If the rate of decay of a radioactive isotope de creases from 200 counts per minute to 25 counts per minute after 24 hours, what is its half-life? A C

2 hours 6 hours

B D

<ate

4 hours 8 hours

. . 3 x 1112 = 24 1112 = 8 hours

Alternatively, let x be the half-life. 24 24 x = -1 x = 25 � ( 1 ) 200 x ( 1 ) 2 2 8 24 ..!. ..!. � x = 8 hours in = In � x 2 8 -

Helping C(J,:1u:ejUJ

A catalyst incr-eases the rate of a reaction by increasing the rate constant, regardless of whether it is a forward

chemist�

time

In the beginning, as the reactants react, the product formed catalyses the reaction. Therefore, the reaction rate increases. A point is reached whereby the reaction rate decreases due to the excessive depletion of the reactants (as oppose to the catalytic effect of the prod uct).

Which statement about the effect of a catalyst on a reversible reaction is correct? A It increases the equilibrium constant for the forward reaction. B It increases the yield of product in an equilib rium. C It increases the rate constant for both the for ward reaction and the reverse reaction. D It increases the rate constant for the forward reaction but not that of the reverse reaction.

1 000

<ate

Helping con,cejt/J

lL_ �

Helping concejtt4 200 � 1 00 � 50 � 25

Questions

rf/cCJ with H ELPs

The acid-catalysed iodination of propanone may be investigated by reacting dilute aqueous iodine with solutions contain i n g known concentrations of propanone and acid. The rate can be followed using a colorimeter. Why is a large excess of propanone used? A B C

to buffer the acid concentration to give a convenient rate of reaction to keep the rate of reaction constant to keep the propanone concentration effectively constant

Helping CMtcejtl;

CH3COCH3 + 1 2 � CH3COCH21 + W + R

= k[CH 3COCH3 t(12 JY [W ]z page 1 1 6

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Topic 9 The Periodic Table: Chemical Periodicity

Frequently Examined

Help i ng <XVnC.e/l-14

21 . Scrap metal often consists of a mixture of aluminium,

iron, chromium and copper. After removing the iron magnetically, the aluminium is removed from the other metals by a physical rather than chemical method.

Si02 is macromolecular, PbO is ionic; Co is a transition metal. A,C: no transition metal B: no macromolecular oxide

What property of aluminium enables it to be sepa rated in this way? It It It It

B C

1 9. The diagram shows part of the Periodic Table.

Na Mg K

& I Ti l v l D l � l � l � I M l � I �

Br Kr

Ga Ge As Se Br

K Sc

Help i ng � Among the elements in the given Periodic Table, K has the highest number of shells of electrons (3) and the lowest number of protons.

20. Which statement concerning only the elements in the third period, sodium to argon, is correct?

B C

The element that has exactly four atoms in its molecule is sulfur. The element with the highest electrical conduc tivity is aluminium. The element with the highest melting point is aluminium. The element with the largest anion is chlorine.

Help i ng � A: B:

Sulfur has 8 atoms per molecule. Al has 3 valence electrons per atom (Na has 1 , Mg has 2). So it has the highest electrical conductivity.

Si has the highest melting point.

s2- > Cl-

p 3-

1 000

chemist�

has a low density. has a low melting point. is a poor conductor of electricity. is non-magnetic.

�

Which element in the diagram has the largest atomic radius? A

Questions

'Yf!c'I with HELPS

Help i ng crnwejttJ Both Cr and Cu are transition elements and they have higher melting points than Al which is in the main group. Al can be more easily melted and siphoned off where Cr and Cu remain as solids.

22. The species Ar, K• and Ca2• are isoelectronic (have the same number of electrons). In what order do their radii increase? smallest -> largest A Ar Ca2• K• B Ar K• Ca 2• 2 K• C Ca • Ar D Ca 2• K• Ar

Help i ng conaytlo

Nuclear charge: Ca2• > K• > Ar The valence electrons are the most strongly bound in Ca2• due to the highest nuclear charge and smallest ionic size.

23. An excess of cold water was added to 0.3 mol of a chloride of the third period of the Periodic Table . at room temperature. 0.6 mol of HCI was forme_d. Which chloride was treated?

page 1 3 3

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Topic l 0 Group II

Frequently Examined

Questions

Helping wnco/ttJ

ILE.I oc

I 1 '+ q

++qr-_

Thus, A H501n becomes less exothermic down the group and the solubility decreases mainly due to the change in AHhyd ·

Down the group, r. increases. Hence, both IL.E.I and l 8 Hhyd l decrease. The solubility of the sulfate decreases down the group. 8 Hso1n =IL E . l - l 8 Hhyd l

Due to the large size of the anion ( r_ ) , IL.E.I decreases marginally (as ( r. +r_) increases marginally). However,

27.

1 �� 1 is small and hence the decrease in 1 �: 1 causes a

Gallstones can form ih the gall bladder and are very painful. The inorganic part of gallstones is calcium ethanedioate which is insoluble in water. The cor responding magnesium ethanedioate is soluble in water. What factor accounts for the difference in solubility between calcium ethanedioate and magnesium ethanedioate?

significant decrease in l 8 Hhyd l · Consequently, l 8 H501nl decreases significantly and the solubility decreases.

B C

26.

The solubilities of the Group II metal sulfates de crease as the proton number of the metal increases.

What factor affects this trend? A

the atomic radius of the metal atom

B c

the enthalpy change of formation of the sulfate the enthalpy change of hydration of the metal ion the first ionisation energy of the metal

Calcium ethanedioate has a higher solubility product than magnesium ethanedioate. Calcium ethanedioate has a numerically higher lattice energy than magnesium ethanedioate. Calcium ions have a lower enthalpy change of hydration than magnesium ions. Calcium is more electropositive than magne sium.

Helping �

2 M +(g) + X2 (g)

Helping ooncejltJ

8 Hso1n = 8 Hhyd - LE.

Ca2• is larger than Mg 2•. Hence, it has a lower charge density and attracts less H2 0 molecules. Less energy is released compared to Mg 2 • during hydration i.e.

8 Hsoln

8 Hhyd(Ca2+ ) is less exothermic.

l 8HL.E. l - 1 8 Hhyd l

SO/- is large. A change of M 2• in size does not affect 8 HL.E. significantly.

1 8 HL.E. I

I 1 r+q++qf-_

On the other hand, when M 2+ increases in size down the group, l 8 Hhydl decreases significantly. 1 000

chemistr;g

'ff!c <J with HELPs

Consequently, 8 H501n (CaX) is less exothermic and CaX is less soluble compared to MgX.

28.

Barium sulfate occurs naturally as barite, which is a solid ore. Magnesium sulfate, however, occurs mainly in solution. Why is this? page 1 48

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Topic 1 1 Group VII

Frequently Examined

Which graph correctly describes a trend found in the halogen group?

20.

Questions

The white precipitate is Agel, which dissolves readily in aqueous NH 3 to give a colourless .s olution containing diaaminesilver(I) complex (Tollens' reagent!).

A 22.

What helps t o explain this trend?

L� f� j �

As a more powerful ligand, NH 3 can displace er- ions and Br ions, but not 1- ions. B e1- ions and B r ions form complexes with NH 3 (aq), but 1- ions do not. C The value of the solubility product of the silver halides decreases from Agel to Agl. D The covalent bonding between Ag and the halogen atom increases in strength from Agel to Agl. A

.8 0

c� Br2 12

H-CL H-Br H·I

Helping wncejta

With increasing number of electrons in the molecule and hence increasing the molecular size, induced dipole-in duced dipole (id·id) interaction (van der Waals' forces) increases.

A compound

21 .

pale yellow precipitate

Kl(aq)

reacts in the following ways. white

W(aq)

precipitate

For the three halides, K..ab remains the same since x is not involved in the reaction. The minimum [Ag•] re quired to form the complex is the same. However, for a smaller K.P (e.g. Agl), this minimum [Ag•] is still high enough such that [Ag+ ][ r ] > Kw Hence, Agl does not dissolve in aqueous NH 3 • The solubility is thus governed by the K.P of the respective AgX. Ksp(Agel) > Ksp(AgBr) > Ksp (Agl) Hence, the solubility decreases from Agel to Agl .

solution

What could compound C

Helping concejl£J

colourless

precipitate does not dissolve

The solubility of the silver halides in aqueous ammonia decreases from Agel to Agl .

be?

AgN0 3 Pb(N0 3) 2

23.

Aqueous chlorine is added t o aqueous sodium bromide and the mixture is shaken with an equal volume of trichloroethane. Which observation would be made?

Helping wncejlt> Ag+ (aq) + r(aq) � Agl(s) The yellow precipitate is Agl , which is insoluble in aque ous NH3 • Ag+ (aq) + er- (aq) � Agel(s)

purple

red - brown -

Agel(s) + 2NH3 (aq) � [Ag(NH 3 h t (aq) + er 1 000

chemist�

rf/c<J with

HELPs

page 1 59

l 2 An Introduction to the Chemistry of Transition Elements

Topic

H2 S04 serves as an acid while Ti02 functions as a base. Salt (Ti02•SO/-) and H 20 are formed.

Platinum(IV) chloride combines with ammonia to form compounds in which the co-ordination number of platinum is 6. A formula unit of one of the compounds contains a cation and only two chloride ions. What is the formula of this compound?

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A C

Pt(NH3)8C14 Pt(NH3)4Cl4

B D

Pt(NH3)p4 Pt(NH3)3C14

Nickel is oxidised by ions with E e greater than Ee of 3 Ni2 •/Ni (--0.25 V). E e of Co2 •, C r'+, fe + and Mn 2• are --0.28 V, --0.41 V, +0.77 V and -1 . 1 8 V respectively. 3 Hence, Ni reacts with fe + . 3 2 2 2Fe + + N i � 2 Fe + + Ni + E�11 = +0.77 - (-0.25) = +1 .02 v > o

Hence, the reaction is energetically feasible.

1 7. Use of the Data Booklet is relevant to this question.

Helping cMtceftti

Sir Humphrey Davy showed that the corrosion of copper hulls of sea-going ships could be prevented by placing strips of 'sacrificial' metals on the hulls.

The compound with co-ordination number 6 may be looked upon as [Pt(NH3)�,G1.J 2 •(c1-) 2 • Since Pt is in oxidation state IV, its chloride has a formula PtC14, i.e. 4CI atoms. Hence x = 2.

Which of these metals is least likely to dissolve when attached to the copper hull of a sea-going ship? A C

Use of the Data Booklet is relevant to this question.

1 5.

In the Aromas Red Sands aquifer, the drinking water source for part of California, there are high levels of soluble, toxic chromium(VI) compounds. Which compound in the aquifer's sands is most likely 1 0 be resp o n s i b l e for t h e format i o n of t h e chromium(VI) compounds from the sparingly soluble chromium(l l l)-bearing rocks? A C

Alp3 Fep3

Helping co-�

B D

CuO ZnO

The identified substance oxidises Cr( l l l) to Cr(VI). Among the 4 oxides, Fe2 03 is the most oxidising (Fe(ll) --) Fe(lll)).

1 8.

Which aqueous solution should not be stirred with a nickel spatula because a reaction could occur? 2 A Co •(aq) B Cr'+(aq) 3 fe +(aq) C D Mn 2 +(aq) 1 000

chemist�

'ff!c tt with H ELPS

magnesium zinc

When drops of NH3(aq) are added to Cu(NOJ2 (aq), a pale blue precipitate is formed. This precipitate dissolves when an excess of NH3(aq) is added, forming a deep blue solution. Which process does not occur in this sequence?

Spatulas are often made from nickel.

B D

To function as a sacrificial metal for Cu, the metal must be more reactive than Cu. From the Data Booklet, it can be seen that Sn 2•/Sn has the least E e value, i.e. it is the least reactive among the 4 metals.

B Use of the Data Booklet is relevant to this question.

iron tin

Helping <XH"l,cejUJ

1 6.

Questions

Helping c&nceftti

Helping CMU:eft/4

1 4.

Frequently Examined

dative bond formation formation of a complex ion precipitation of copper(ll) hydroxide reduction of Cu 2 • ions

Helping C<»UX!jttl NH4 · caq) + OW (aq)

Cu2 + (aq) + 20W (aq) � Cu(OHMs) Cu2 + (aq) + 4NH 3 (aq) � [Cu(NH3 )4 ]2 • (aq)

page 1 68

Topic

1 3 Introductory Topics of Organic Chemistry

This gives rise to partial double bond character in C2-C3. Hence, its bond length is shorter than a single bond but longer than a normal C=C double bond.

Which of the following amino acids contains two chiral carbon atoms?

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1 5.

Frequently Examined

Questions

Helping co4tceft/J Possible structural isomers include

0 0 • u H ' = c ..... C - OCH 3 and H ' c = c .. O - C- CH3 etc. c ' H 3C / H',... ' cH3 H

The compound is not capable of showing cis-trans isom erism and optical isomerism.

1 7.

Warfarin is used as a rat poison.

How many chiral centres are present in the War· farin molecule?

Helping ronc¢t/J

A c

H2N - c - C02H I

B D

0 2

Helping conceftu

1 3

* H - c - CH3 I

CH2CH3 I

where *

chiral centre

O ne of the chemicals used to make the hard outer covering of golf balls has the following structural formula.

1 6.

C= C

/ C - O Cf-\

' cH3

Which of the following statements about this mol ecule is correct ?

It is a cis iso mer.

It ls a trans isomer.

It has on1y one chiral centre.

It has only structural isomers.

1 000

chemist� 'ff!c<J- with HELPs

1 8.

Many drugs show optical isomerism. The diagrams show the structure of three drugs.

P H2CH3 7H2 � ?i - CH2 - y - CH3 o N - C - CH2 - N \

amohetamine

lidocaine

CH2CH3

I Qcf �=0 tt

� 'c-J t\ A \

phenobarbital

page 1 78

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l 4 Hydrocarbons

Topic

Frequently Examined

number of moles of oxygen gas

31 .

Questions

rr"YCH3 The aromatic compound � was made to - CH3 react with an excess of hot aqueous alkaline po tassium manganate(Vl l) and the product was treated with an excess of aqueous acid. What is the most likely final product? CHO

1 n umber of carbon atoms in alkane

Help i ng (XV}tCeftls C n H2n+2 +

3n2+1 0 2

aCHO

Hence, the plot of graph should be a straight line graph with a positive slope of 3/2, and a y-intercept at 1 /2.

Help i ng conc-<fiM Alkylbenzenes are oxidised to the carboxylic acid at the side chain.

Cyclohexa-1 ,4-diene is treated with a solution of bromine in tetrachloromethane.

30.

CH3

+ 6[0] ->

co�

'co�

+ 2�0

cyclohexa- 1 ,4-diene Which product is formed? Br Br

&D&

32.

When methylbenzene is treated with bromine in the presence of a catalyst, a mixture of two monobromo isomers is formed. What are the structure of these two isomers?

C H2Br

Q Br

C H3

C=C double bonds undergoes electrophilic addition with I

bromine to give - C- C-. I I Br Br

chemistr:g

C H3

Help i ng cMc-<fiM

1 000

6 0Br r ar QBr

rf/c't

with HELPs

6r I

page 1 8 9

Topic 1 5 Halogen Derivatives

Frequently Examined

What type of reaction is step 1 ?

Helping ronc¢tl4

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Cl bonded directly to the C of a benzene ring has partial C-CI double bond character due to the overlap of the p-orbital of Cl and 1l' -orbitals of the benzene ring. It is therefore difficult to cleave. Under the given conditions, no Cl will be substituted.

21 .

Which of the following chloro-compounds is least easily hydrolysed by hydroxide ion to give the product indicated? A

C2H5CI -----+ C2H50H

CH3CHCICH3

CH 3COCI

C 6H5CI

-----+

Questions

elimination free radical substitution isomerisation nucleophilic substitution

Helping c�Li In step 1 , one of the Cl atoms is substituted by F, while one of the F atoms is substituted by Cl.

23.

Which one of the following reactions is the inorganic reagent acting as a nucleophile ?

CH 3CH(OH)CH 3

CH 3C02H

C6H50H

Helping CM�

The C-CI bond in C8H 5 CI has partial double bond char acteristics due to the delocalisation of a lone pair of electrons from chlorine into the benzene ring. This strengthens the C-CI bond and makes it difficult to be broken.

C�CI

rA{ + Kl � � o C�I + KCI 0

Helping �ta.

A nucleophile is one that is nucleus seeking. It usually has a lone electron pair or is negatively charged and is able to form a dative bond through donating the lone electron pair to the electron deficient site.

Furthermore, the electron-rich benzene system makes the attack by OH- difficult since like charges repel.

22.

24.

Under the Montreal Protocol, the use of chlorofluo rocarbons is to be phased out. Fluorocarbons are often used to replace them. One chlorofluorocarbon which was widely used as a solvent is CCl2 FCCIF2 and large stocks of it remain. One process to use up these stocks is to convert it into the fluorocarbon CH 2 FCF3 by the following route. step step 3

1 000

chemist�

step 2

Bromomethane, C H 3 Br, is used as a fumigant to destroy insect pests in grain that is to be stored. It can be made by reacting methanol with hydrogen bromide. What type of reaction is this? A B C

'ff!c 'l

with HELPS

condensation electrophilic substitution free radical substitution nucleophilic substitution page 200

Topic 1 6 Hydroxy Compounds " / / C = C,

+ Br2

----+

Frequently Examined

/o"'• C H2 - C H 2 - 0 · + H2 C -- C H2

- C- C-

I 1

I I

Br Br

Phenols undergo electrophilic substitution with Br2•

-o::

Br +

3Br2 ----+

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Which set of alcohols correctly shows a primary, a secondary and a tertiary alcohol?

TH20H r2

H r

secondary

CH20H

CH3

H 0H T CH- y -H 2

J J

CH3

CH3 -0H Hs

T cH31-oH CH3,-0H H H H3

H I

(CH3(CH2)10 0)1 0CH2CHpH CH3(CH2)1p (OCH2C H 2)1 00H CH3(CH2)100(CH2CHp),0 H CH3(CH2)1 00(CH2CHp),pH

Help i ng e01u:ejttJ

The base polymeric chain is + C H2 - CH2

- O -t,;-

The chain is likely to abstract a H from the alcohol or bond with the 0 of the alcohol. In (B) and (D), there are extra 0 atom in the formula. Note: 0-0 bond is weak and hence not likely to be formed.

-H

H20H

30.

CH20H

C H 2 C H2 - O - CH2 - CH2 - 0 ·

C H3

TH20H H20H T CH3- -CH20H C H3-, -CHzOH

Help i ng e01tcejttJ

A: B:

rH�H HOH

HOH

B CH3 -H

tertiary

epoxyethane

After termination of the reaction with an alcohol, what is a possible formula of such a non-ionic detergent?

Br + 3 H Br

primary

----+

28.

Questions

Which of the following isomers of C5H , ,OH gives, on dehydration, the greatest number of different alkenes? A

C H3 - CH2 - C H - CH20H I

c�

CH3 CH- -0H H3

C H3 - CH2 - CH2 - C H - CH3 I

1 °, (1 °, 2°), (1 °, 2°, 1 °) 1 °, 3°, 1 ° 1 °, (1 °, 1 °), (1 °, 1 °, 1 °)

C H2 - C H - C H2 - CH3 I OH

C H3 -

CH3 - CH - C H2 - C H20H I

CH3

Help i ng co-ncejttJ 29.

Non-ionic detergents can be made by reaction of epoxyethane, in an excess, with a c,, alcohol.

A possible mechanism involves hemolytic fission of a C-0 bond in epoxyethane giving rise to a 'double ended' free radical that initiates a chain reaction. The first propagation step is as follows.

CH3CH2 C I

chemistr�

rf/c<j with HELPs

CH2

CH3

CH3CH2CH2CH = CH2 CH3CH2 H

1 000

""' /

C=C

/ ""'

CH3 H

CH3CH2 H

""'

c c""' =

C Ha

page 2 1 6

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Topic l 7 Carbonyl Compounds

'¢ CH,'¢

Frequently Examined

�N

What is formed when this compound reacts with hydrogen cyanide? CN

CH,

CH 3

HC , NC

The aldehyde and alkene are oxidised, but not the ketone.

The aldehyde does not show optical isomerism.

37.

One industrial preparation of ethanoic acid is the direct carbonylation of methanol using a · rhodium catalyst.

II 0

rhodium

Helping concejttJ

catalyst

With HCN, nucleophilic addition occurs at the C=O bond to give cyanohydrin:

CH3� u II

2HCN ----)

36.

C H 3 �N . -� HO

CH3CH2CH2CH=CHCHO and CH 3CH2CH(CH 3)COCH2CH 3 Which characteristic applies to both compounds?

B c D

Which compound could be expected to produce C02H I HC- CH2C02H by this method. I CH2C02H

OH I HC- C02H I CH20H

OH I HC- CH2C02H I C02H

The structures for two alarm pheromones for ants are given.

The aldehyde decolourises Br2(aq) since it is also an alkene, but the ketone does not. I I , /c c"/ + H20 + Br2 _____.. C- C l I Br OH =

Questions

Both can be obtained by the oxidation of alcohols. Both decolourise aqueous bromine. Both decolourise dilute alkaline potassium manganate(Vll) Both show optical isomerism.

CH20H I HC- CH2C02H I CH2C02H

OH I HC- CH2C02H I CH20H

Helping concejttJ From the given equation, the overall result is the inser tion of a carbonyl group into the C-0 bond, 0

I II I i.e. - C - OH + CO � - C - C- OH I I Therefore,

Helping CMtcejtt:J

1 000

cbemistr�

rt/cCJ with HELPs

page 2 3 0

l 8 Carboxy/ic Acids & Derivatives

Topic

What will be the product A

CH3COCH2 CONH

Frequently Examined

Q of this first stage?

-O

0 C

Option B is incorrect since -COOH does not react with HCl(g). Option D is incorrect since alcoholic -OH does not react with aqueous NaO H . Hence, only 3 mol of aqueous NaOH is needed per mol of citric acid.

o- NH �C 2 � H3 CH3COCH2co -Q- NHNH2 0- N � C

41 .

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C H3CCH2C02C H2C H3

Tho display of a digital watch needs a liquid crystal which is (i)

Helping concejt/J

stable to acidic hydrolysis;

(ii) stable to alkaline hydrolysis; (iii) chiral.

The carbonyl functional group reacts with phenylhydra zine to form a condensation product, phcnylhydrazone.

Which of the following compounds, all of which form liquid crystals, meets these requirements?

The ester functional g roup docs not react.

0- C �

,0 0

CH3CH2CH20 -

Citric acid, which causes the sharp taste of lemon juice, has the following formula.

40.

Questions

C H2 C0 2 H I HO - C - C0 2 H I C H 2C0 2H .

3 mol of PCl5(s)

4 mol of HCl(g)

4 mol of Na(s)

4 mol of NaOH(aq)

-O-O- C H2CH(CH3 )2

CH3C�CH(CH3 )CH2

Helping co-ncejt!J

What reacts completely with 1 mol of citric acid? A

C C HaO

-Q-0-cN

-Q-0- c H3

0 II Both (A) and (8) can be hydrolysed ( - C - 0 - and

-CN) ; (B) and (C) do not contain a chiral centre. (D) is a hydrocarbon and is therefore stable in both acids and alkalis.

Helping cMU:ejt !J

C H3CH2 H - C H2

C itric acid contains 3 -COOH and 1 alcoholic -OH groups. These groups react with 1 Na each:

CH3

-o-o-CH3

RCOOH + Na � R Coo-Na+ + _!_2 H 2 ROH + Na � RO-Na+ + _!_ H 2 2

Hence, a total of 4 mol of Na is needed per mol of citric acid. Option A is incorrect since -COOH and alcoholic -OH groups react with 1 PCl5 each : RCOOH ROH

PCl5

POCl3

�

R COCI RCI

POCl3

HCI

POCl3 + HCI

Hence, 4 mol of PCl5 are noodcd per molof citric acid. 1000

cbemistr:g

rf/cl/r with HELPs

42.

When a substance X is shaken with aqueous silver nitrate at room temperature, there is no immediate precipitate. However, when X is boiled under reflux for some time with aqueous sodium hydroxide, cooled, acidified with dilute nitric acid and aqueous silver nitrate added, a white precipitate readily forms. What could

X be? page 248

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i��

1 000

ChemistT'y MCQ

______

Topic

1 6. 21.

6. 1 1.

26.

2. 7. 1 2. 1 7. 22. 27.

31.

32.

36.

37.

41.

42.

46.

Topic I.

1 1. 1 6.

B D A B

26.

41.

A B

46.

36.

B B

B D

A B

2. 7.

D B

1 8.

23 .

28.

33. 38. 43.

3. 8.

D B B B

33.

43 .

37. 42.

1 2.

1 6.

1 7.

21.

22.

26.

27.

31.

32.

36.

37.

41.

42. 47.

1 8. 23 .

A B

1 3.

1 9.

20.

24 .

25.

29.

30.

34.

35.

39.

44.

45.

9. 1 4. 1 9. 24.

29.

B A D B

1 0. 1 5.

40.

1 0. 1 5. 20. 25.

c D

B B

A A

B B

30.

40.

A A

44 .

35. 45.

1 4.

1 0.

13.

38.

1 8. 23 .

A B

28.

1 4.

28.

1 3.

27.

22.

1 1.

46.

32.

1 7.

D A

1 2.

Topic 3 Chemical B onding I.

2 Atomic Structure

21. 31.

________

1 Atoms� Molecules and Stoichiometry

D A A

AN SWERS

with � elps

34. 39.

1 9. 24. 29.

43 .

D D

44.

48.

49.

33. 38.

34. 39.

20.

25.

A B

30.

35.

40.

1 5.

45. 50.

I I I I I I I I I

page

270

i I