Download Answers to Selected Exercises

Answers to Selected Exercises
Chapter 1
1.1 (a) Pure element: i, v (b) mixture of elements: vi (c) pure
compound: iv (d) mixture of an element and a compound: ii, iii
1.5 7.5 cm, 2 significant figures (sig figs) 1.6 (a) The path of
the charged particle bends because it is repelled by the
negatively charged plate and attracted to the positively charged
plate. (b) negative (c) increase (d) decrease 1.8 The particle
21.10 Formula: IF5 ;name: iodine pentafluoride;
is an ion. 32
16S
the compound is molecular. 1.12 (a) Heterogeneous mixture
(b) homogeneous mixture (heterogeneous if there are
undissolved particles) (c) pure substance (d) homogeneous
mixture. 1.14 (a) S (b) K (c) Cl (d) Cu (e) Si (f) N (g) Ca
(h) He 1.16 (a) Lithium (b) aluminium (c) lead (d) sulfur
(e) bromine (f) tin (g) chromium (h) zinc 1.18 C is a
compound; it contains carbon and oxygen. A is a compound; it
contains at least carbon and oxygen. B is not defined by the
data given; it is probably a compound because few elements
exist as white solids. 1.19 Physical properties: silvery white;
lustrous; melting point = 649°C; boiling point = 1105°C;
density at 20°C = 1.738 g cm3; pounded into sheets; drawn
into wires; good conductor. Chemical properties: burns in air;
reacts with Cl2 . 1.20 (a) Chemical (b) physical (c) physical
(d) chemical (e) chemical 1.21 (a) Add water to dissolve the
sugar; filter this mixture, collecting the sand on filter paper and
the sugar water in a flask. Evaporate water from the flask to
reproduce solid sugar. (b) Heat the mixture until sulfur melts,
then decant the liquid sulfur. 1.23 (a) 2.55 * 10-2 g
(b) 0.40 nm (c) 575 mm 1.25 (a) 1.59 g cm3. Carbon
tetrachloride, 1.59 g cm3, is more dense than water,
1.00 g cm3; carbon tetrachloride will sink rather than float on
water. (b) 1.609 kg (c) 50.35 cm3 1.27 (a) Calculated density
= 0.86 g cm3. The substance is probably toluene, density =
0.866 g cm3. (b) 40.4 cm3 ethylene glycol (c) 1.11 * 103 g
nickel 1.28 Exact: (c), (d) 1.30 (a) 3 (b) 2 (c) 5 (d) 3 (e) 5
1.32 (a) 1.025 102 (b) 6.570 * 105 (c) 8.543 * 10-3
(d) 2.579 * 10-4 (e) -3.572 * 10-2 1.34 (a) 21.11 (b) 237.4
(c) 652 (d) 7.66 * 10-2 1.36 Postulate 4 of the atomic theory
states that the relative number and kinds of atoms in a
compound are constant, regardless of the source. Therefore,
1.0 g of pure water should always contain the same relative
amounts of hydrogen and oxygen, no matter where or how
the sample is obtained. 1.38 (a) 0.5711 g O>1 g N;
1.142 g O>1 g N; 2.284 g O>1 g N; 2.855 g O>1 g N (b) The
numbers in part (a) obey the law of multiple proportions.
Multiple proportions arise because atoms are the indivisible
entities combining, as stated in Dalton’s atomic theory.
1.40 (1) Electric and magnetic fields deflected the rays in the
same way they would deflect negatively charged particles.
(2) A metal plate exposed to cathode rays acquired a negative
charge. 1.42 (a) If the positive plate were lower than the
negative plate, the oil drops ‘coated’ with negatively charged
electrons would be attracted to the positively charged plate and
would descend much more quickly. (b) The more times a
measurement is repeated, the better the chance of detecting and
compensating for experimental errors. Millikan wanted to
demonstrate the validity of his result via its reproducibility.
1.43 (a) 0.19 nm; 1.9 * 102 or 190 pm (b) 2.6 * 106 Kr atoms
(c) 2.9 * 10-23 cm3 1.45 (a) Atomic number is the number of
protons in the nucleus of an atom. Mass number is the total
number of nuclear particles, protons plus neutrons, in an atom.
(b) mass number 1.47 (a) 40Ar: 18 p, 22 n, 18 e (b) 65Zn: 30 p,
35 n, 30 e (c) 70Ga: 31 p, 39 n, 31 e (d) 80Br: 35 p, 45 n, 35 e
(e) 184W: 74 p, 110 n, 74 e (f) 243Am: 95 p, 148 n, 95 e
1.49
Symbol
52
55
Cr
Protons
Neutrons
Electrons
Mass no.
112
Mn
24
28
24
52
25
30
25
55
222
Cd
207
Rn
48
64
48
112
Pb
86
136
86
222
82
125
82
207
TO BE
UPDATED
12
84
75
24
1.50 (a) 196
78Pt (b) 36Kr (c) 33As (d) 12Mg 1.56 (a) 6C
(b) Atomic weights are average atomic masses, the sum of the
mass of each naturally occurring isotope of an element times
its fractional abundance. Each B atom will have the mass of
one of the naturally occurring isotopes, while the ‘atomic
weight’ is an average value. 1.55 (a) Cr (metal) (b) He
(nonmetal) (c) P (nonmetal) (d) Zn (metal) (e) Mg (metal)
(f) Br (nonmetal) (g) As (metalloid) 1.57 An empirical
formula shows the simplest mole ratio of elements in a
compound. A molecular formula shows the exact number and
kinds of atoms in a molecule. A structural formula shows
which atoms are attached to which. 1.59 (a) AlBr3 (b) C4H 5
(c) C2H 4O (d) P2O 5 (e) C3H 2Cl (f) BNH 2 1.61 (a) 6 (b) 6
(c) 12
H
1.63 (a) C2H6O,
H
C
H
O
H
H
H
H
(b) C2H6O,
C
H
H
C
C
H
H
O
H
H
(c) CH4O,
H
C
O
H
(d) PF3 ,
F
P
F
F
H
1.65
Symbol
Protons
Neutrons
Electrons
Net charge
59
Co3ⴙ
27
32
24
3+
70
Se2ⴚ
34
46
36
2-
192
Os2ⴙ
76
116
74
2+
200
Hg2ⴙ
80
120
78
2+
A-1
A-2
Answers to Selected Exercises
1.67 (a) Mg 2+ (b) Al3+ (c) K + (d) S 2- (e) F - 1.69 (a) GaF3 ,
gallium(III) fluoride (b) LiH, lithium hydride (c) AlI 3 ,
aluminium iodide (d) K 2S, potassium sulfide 1.71 (a) ClO 2 (b) Cl - (c) ClO 3 - (d) ClO 4 - (e) ClO - 1.73 (a) Magnesium
oxide (b) aluminium chloride (c) lithium phosphate
(d) barium perchlorate (e) copper(II) nitrate (cupric nitrate)
(f) iron(II) hydroxide (ferrous hydroxide) (g) calcium acetate
(h) chromium(III) carbonate (chromic carbonate) (i) potassium
chromate (j) ammonium sulfate 1.75 (a) Al(OH)3 (b) K 2SO 4
(c) Cu 2O (d) Zn(NO3)2 (e) HgBr2 (f) Fe2(CO 3)3 (g) NaBrO
1.77 (a) Bromic acid (b) hydrobromic acid (c) phosphoric acid
(d) HClO (e) HIO 3 (f) H 2SO 3 1.79 (a) Sulfur hexafluoride
(b) iodine pentafluoride (c) xenon trioxide (d) N2O 4 (e) HCN
(f) P4S 6 1.81 (a) ZnCO 3, ZnO, CO 2 (b) HF, SiO 2 , SiF4 , H 2O
(c) SO 2 , H 2O, H 2SO 3 (d) PH 3 (e) HClO 4 , Cd, Cd(ClO 4)2
(f) VBr3 1.82 Composition is the contents of a substance;
structure is the arrangement of these contents.
1.86 (a) 4.67 g cm3 (b) 2.56 dm3 (c) 1.95 * 106 g. The sphere
weighs 1950 kg; the thief is unlikely to be able to carry it.
1.88 (a) 60.6% Au by mass (b) 15 carat gold
1.91 (a) 2 protons, 1 neutron, 2 electrons (b) Tritium, 3H,
is more massive. (c) A precision of 1 * 10-27 g would
be required to differentiate between 3H and 3He.
1.92 (a) 5.90 * 1022 Au atoms in the cube (b) 3 pm 1.94 168O,
17
18
8O, 8O (b) All isotopes are atoms of the same element,
oxygen, with the same atomic number, 8 protons in the nucleus
and 8 electrons. We expect their electron arrangements to be
the same and their chemical properties to be very similar. Each
has a different number of neutrons, a different mass number
and a different atomic mass. 1.98 (a) An alkali metal: K
(b) an alkaline earth metal: Ca (c) a noble gas: Ar (d) a
halogen: Br (e) a metalloid: Ge (f) a nonmetal in group 1: H
(g) a metal that forms a 3+ ion: Al (h) a nonmetal that forms
a 2- ion: O (i) an element that resembles Al: Ga
1.100 (a) Nickel(II) oxide, 2 + (b) manganese(IV) oxide, 4+
(c) chromium(III) oxide, 3 + (d) molybdenum(VI) oxide, 6+
1.103 (a) Sodium chloride (b) sodium bicarbonate (or sodium
hydrogen carbonate) (c) sodium hypochlorite (d) sodium
hydroxide (e) ammonium carbonate (f) calcium sulfate
Chapter 2
2.1 Equation (a) best fits the diagram. 2.3 The empirical
formula of the hydrocarbon is CH 4 .
2.6
N2 + 3 H 2 ¡ 2 NH 3 . Eight N atoms (4 N2 molecules)
require 24 H atoms (12 H 2 molecules) for complete reaction.
Only 9 H 2 molecules are available, so H 2 is the limiting
reactant. Nine H 2 molecules (18 H atoms) determine that
6 NH 3 molecules are produced. One N2 molecule is in excess.
2.7 (a) Conservation of mass (b) Subscripts in chemical
formulae should not be changed when balancing equations,
because changing the subscript changes the identity of the
compound (law of constant composition). (c) (g), (l), (s), (aq)
2.9 (a) 2 CO(g) + O2(g) ¡ 2 CO2(g)
(b) N2O5(g) + H2O(l) ¡ 2 HNO3(aq)
(c) CH4(g) + 4 Cl2(g) ¡ CCl4(l) + 4 HCl(g)
(d) Al4C3(s) + 12 H2O(l) ¡ 4 Al(OH)3(s) + 3 CH4(g)
(e) 2 C5H10O2(l) + 13 O2(g) ¡ 10 CO2(g) + 10 H2O(l)
(f) 2 Fe(OH)3(s) + 3 H2SO4(aq) ¡ Fe2(SO4)3(aq) + 6 H2O(l)
(g) Mg3N2(s) + 4 H2SO4(aq) ¡ 3 MgSO4(aq) + (NH4)2SO4(aq)
2.10 (a) CaC2(s) + 2 H2O(l) ¡ Ca(OH)2(aq) + C2H2(g)
¢ "
(b) 2 KClO3(s)
2 KCl(s) + 3 O2(g)
(c) Zn(s) + H2SO4(aq) ¡ ZnSO4(aq) + H2(g)
(d) PCl3(l) + 3 H2O(l) ¡ H3PO3(aq) + 3 HCl(aq)
(e) 3 H2S(g) + 2 Fe(OH)3(s) ¡ Fe2S3(s) + 6 H2O(g)
2.11 (a) Determine the formula by balancing the positive
and negative charges in the ionic product. Ionic compounds
are usually solids at room temperature.
2 Na(s) + Br2(l) ¡ 2 NaBr(s) (b) The second reactant
is O2(g). The products are CO2(g) and H2O(l).
2 C6H6(l) + 15 O2(g) ¡ 12 CO2(g) + 6 H2O(l)
2.13 (a) Mg(s) + Cl2(g) ¡ MgCl2(s)
¢ "
(b) BaCO3(s)
BaO(s) + CO2(g)
(c) C8H8(l) + 10 O2(g) ¡ 8 CO2(g) + 4 H2O(l)
(d) C2H6O(l) + 3 O2(g) ¡ 2 CO2(g) + 3 H2O(l)
2.15 (a) 2 Al(s) + 3 Cl2(g) ¡ 2 AlCl3(s) combination
(b) C2H4(g) + 3 O2(g) ¡ 2 CO2(g) + 2 H2O(l) combustion
(c) 6 Li(s) + N2(g) ¡ 2 Li3N(s) combination
(d) PbCO3(s) ¡ PbO(s) + CO2(g) decomposition
(e) C7H8O2(l) + 8 O2(g) ¡ 7 CO2(g) + 4 H2O(l) combustion
2.17 (a) 108.0 u (b) 159.6 u (c) 149.0 u (d) 162.1 u (e) 150.3 u
(f) 399.9 u (g) 535.6 u 2.20 (a) 79.2% (b) 63.2% (c) 64.6%
2.21 (a) 6.022 * 1023 (b) The formula mass of a substance in u
has the same numerical value as the molar mass expressed in
grams. 2.23 23 g Na contains 1 mol of atoms; 0.5 mol H 2O
contains 1.5 mol atoms; 6.0 * 1023 N2 molecules contains 2 mol
of atoms. 2.25 (a) 32.5 g CaH 2 (b) 0.0361 mol Mg(NO 3)2
(c) 1.84 * 1022 CH 3OH molecules (d) 1.41 * 1024 C atoms
2.27 (a) molar mass = 162.3 g (b) 3.08 * 10-5 mol allicin
(c) 1.86 * 1019 allicin molecules (d) 3.71 * 1019 S atoms
2.28 (a) 2.500 * 1021 H atoms (b) 2.083 * 1020 C6H 12O6
molecules (c) 3.460 * 10-4 mol C6H 12O 6 (d) 0.06227 g
C6H 12O6 2.30 3.2 * 10-8 mol C2H3Cl dm3;
1.9 * 1016 molecules dm3 2.32 (a) C2H 6O (b) Fe2O 3
(c) CH 2O 2.34 (a) CSCl2 (b) C3OF6 (c) Na 3AlF6
2.36 (a) C6H 12 (b) NH 2Cl 2.38 (a) C7H 8 (b) The empirical
and molecular formulae are C10H 20O. 2.40 x = 10;
Na 2CO3 # 10 H 2O 2.42 If the equation is not balanced, the
mole ratios derived from the coefficients will be incorrect and
lead to erroneous calculated amounts of products.
2.44 (a) 2.4 mol HF (b) 5.25 g NaF (c) 0.610 g Na 2SiO3
2.46 (a) Al(OH)3(s) + 3 HCl(aq) ¡ AlCl3(aq) + 3 H2O(l)
(b) 0.701 g HCl (c) 0.855 g AlCl3 ;0.347 g H 2O (d) Mass of
reactants = 0.500 g + 0.702 g = 1.202 g; mass of products =
0.846 g + 0.346 g = 1.202 g. Mass is conserved, within the
precision of the data. 2.48 (a) 2.25 mol N2 (b) 15.5 g NaN3
(c) 542 g NaN3 2.50 (a) The limiting reactant determines the
maximum number of product moles resulting from a chemical
reaction; any other reactant is an excess reactant. (b) The
Answers to Selected Exercises
limiting reactant regulates the amount of products because
it is completely used up during the reaction; no more product
can be made when one of the reactants is unavailable.
2.52 NaOH is the limiting reactant; 0.925 mol Na 2CO3 can be
produced; 0.075 mol CO 2 remains. 2.54 (a) NaHCO 3 is the
limiting reactant. (b) 0.524 g CO 2 (c) 0.238 g citric acid remains
2.56 0.00 g AgNO3 (limiting reactant), 1.94 g Na 2CO3 , 4.06 g
Ag2CO3 , 2.50 g NaNO 3 2.58 (a) The theoretical yield is 60.3 g
C6H 5Br. (b) 94.0% yield 2.59 (a) C4H8O2(l) + 5 O2(g)
¡ 4 CO2(g) + 4 H2O(l) (b) Ni(OH)2(s) ¡ NiO(s) + H2O(l)
(c) Zn(s) + Cl2(g) ¡ ZnCl2(s) 2.60 (a) 1.25 * 1022 C atoms;
0.208 mol C. (b) 2.77 * 10-3 mol C9H8O4; 1.67 * 1021 C9H8O4
molecules 2.63 C10H 12N2O 2.64 C6H 5Cl 2.66 (a) In a
complete combustion reaction, all H in the fuel is transformed
to H 2O in the products. The reactant with most (mol H per mol
fuel) will produce the most H 2O. A 1.5 mol sample of C3H 8
or CH 3CH 2COCH 3 will produce the same amount of
H 2O, 1.5 mol C2H 5OH will produce less H 2O.
2.69 6.46 * 1024 O atoms 2.72 (a) S(s) + O2(g) ¡ SO2(g);
SO2(g) + CaO(s) ¡ CaSO3(s) (b) 190 tonnes CaSO3 per day
Chapter 3
3.1 Diagram (c) represents Li 2SO4 3.3 (a) AX is a nonelectrolyte. (b) AY is a weak electrolyte. (c) AZ is a strong
electrolyte. 3.5 Solid A is NaOH, solid B is AgBr and solid C
is glucose. 3.7 (b) NO 3 - and (c) NH 4 + will always be
spectator ions. 3.9 Reaction (a) is represented by the
diagram. 3.10 No. Electrolyte solutions conduct electricity
because the dissolved ions carry charge through the solution
from one electrode to the other. 3.12 Although H 2O
molecules are electrically neutral, there is an unequal
distribution of electrons throughout the molecule. The partially
positive ends of H 2O molecules are attracted to the anions in
the solid, while the partially negative ends are attracted to the
cations. Thus, both cations and anions in an ionic solid are
surrounded and separated (dissolved) by H 2O.
3.14 (a) ZnCl2(aq) ¡ Zn2+(aq) + 2 Cl-(aq)
(b) HNO3(aq) ¡ H+(aq) + NO3 -(aq)
(c) (NH4)2SO4(aq) ¡ 2 NH4 +(aq) + SO4 2-(aq)
(d) Ca(OH)2(aq) ¡ Ca2+(aq) + 2 OH-(aq) 3.16 HCHO2
molecules, H + ions and CHO 2 - ions; HCHO2(aq) Δ H+(aq)
+ CHO2 -(aq) 3.18 (a) Soluble (b) insoluble (c) soluble
(d) insoluble (e) insoluble 3.20 (a) Na2CO3(aq) +
2 AgNO3(aq) ¡ Ag2CO3(s) + 2 NaNO3(aq) (b) No precipitate
(c) FeSO4(aq) + Pb(NO3)2(aq) ¡ PbSO4(s) + Fe(NO3)2(aq)
3.22 (a) sodium, sulphate (b) sodium, nitrate (c) ammonium,
chloride
3.24
Compound
Ba(NO3)2 Result
NaCl Result
AgNO3(aq)
CaCl2(aq)
Al2(SO4)3(aq)
No ppt
No ppt
BaSO4 ppt
AgCl ppt
No ppt
No ppt
This sequence of tests would definitely identify the bottle
contents. 3.25 LiOH is a strong base, HI is a strong acid, and
CH 3OH is a molecular compound and non-electrolyte. The
strong acid HI will have the greatest concentration of solvated
protons. 3.27 (a) A monoprotic acid has one ionisable (acidic)
H, whereas a diprotic acid has two. (b) A strong acid is
A-3
completely ionised in aqueous solution, whereas only a
fraction of weak acid molecules are ionised. (c) An acid is an
H + donor, and a base is an H + acceptor. 3.29 (a) Acid,
mixture of ions and molecules (weak electrolyte) (b) none of
the above, entirely molecules (non-electrolyte) (c) salt, entirely
ions (strong electrolyte) (d) base, entirely ions (strong
electrolyte) 3.32 (a) CdS(s) + H2SO4(aq) ¡ CdSO4(aq) +
H2S(g); CdS(s) + 2 H+(aq) ¡ H2S(g) + Cd2+(aq)
(b) MgCO 3(s) + 2 HClO4(aq) ¡ Mg(ClO4)2(aq) + H2O(l) +
CO2(g); MgCO3(s) + 2 H+(aq) ¡ H2O(l) + CO2(g) + Mg2+(aq)
3.34 (a) In terms of electron transfer, oxidation is the loss of
electrons by a substance and reduction is the gain of electrons
(LEO says GER). (b) When a substance is oxidised, its
oxidation number increases. When a substance is reduced, its
oxidation number decreases. 3.35 Metals in region A are
most easily oxidised. Nonmetals in region D are least easily
oxidised. 3.37 (a) +4 (b) +4 (c) +7 (d) +1 (e) 0 (f) -1
3.39 (a) Ni ¡ Ni 2+, Ni is oxidised; Cl2 ¡ 2 Cl -, Cl is
reduced (b) Fe 2+ ¡ Fe, Fe is reduced; Al ¡ Al3+, Al is
oxidised (c) Cl2 ¡ 2 Cl -, Cl is reduced; 2 I - ¡ I 2 , I is
oxidised (d) S 2- ¡ SO 4 2-, S is oxidised; H 2O 2 ¡ H 2O;
O is reduced 3.41 (a) Fe(s) + Cu(NO3)2(aq) ¡ Fe(NO3)2(aq)
+ Cu(s) (b) NR (c) Sn(s) + 2 HBr(aq) ¡ SnBr2(aq) +
H2(g) (d) NR (e) 2 Al(s) + 3 CoSO4(aq) ¡ Al2(SO4)3(aq) +
3 Co(s) 3.43 (a) i. Zn(s) + Cd2+(aq) ¡ Cd(s) + Zn2+(aq); ii.
Cd(s) + Ni2+(aq) ¡ Ni(s) + Cd2+(aq) (b) Cd is between Zn
and Ni on the activity series. (c) The position of Cd in the
activity series could be determined more precisely by
observing what happens to metals between Zn and Ni in
the series (e.g., Fe, Co) when placed in CdCl2 solution.
3.45 (a) 0.0863 M NH 4Cl (b) 0.0770 mol HNO3 (c) 83.3 cm3
of 1.50 M KOH 3.47 16 g Na+(aq) 3.49 (a) 4.46 g KBr
(b) 0.145 M Ca(NO 3)2 (c) 20.3 cm3 of 1.50 M Na 3PO 4
3.50 (a) 0.15 M K 2CrO4 has the highest K + concentration.
(b) 30.0 cm3 of 0.15 M K 2CrO 4 has more K + ions.
3.53 (a) 1.69 cm3 14.8 M NH 3 (b) 0.592 M NH 3 3.55 1.398 M
C3H4O2 3.57 0.117 g NaCl 3.59 (a) 38.0 cm3 of 0.115 M
HClO 4 (b) 769 cm3 of 0.128 M HCl (c) 0.408 M AgNO3
(d) 0.275 g KOH 3.61 27 g NaHCO3 3.62 1.22 10–2 M
Ca(OH)2 solution; the solubility of Ca(OH)2 is 0.0904 g
in 100 cm3 solution. 3.64 (a) NiSO4(aq) +
2 KOH(aq) ¡ Ni(OH)2(s) + K2SO4(aq) (b) Ni(OH)2
(c) KOH is the limiting reactant. (d) 0.927 g Ni(OH)2
(e) 0.0667 M Ni2+(aq), 0.0667 M K+(aq), 0.100 M SO4 2-(aq)
3.66 91.40% Mg(OH)2 3.68 The precipitate is CdS(s). Na+(aq)
and NO3 -(aq) are spectator ions and remain in solution, along
with any excess reactant ions. The net ionic equation is
Cd2+(aq) + S2-(aq) ¡ CdS(s). 3.70 (a) All three reactions
are redox reactions. (b) In the first reaction, N is oxidised and
O is reduced. In the second reaction, N is oxidised and O is
reduced. In the third reaction, N is both oxidised (NO2 to
HNO3) and reduced (NO2 to NO). 3.73 1.42 M KBr
3.75 (a) 2.2 10–9 M Na+ (b) 1.3 * 1012 Na+ ions cm3
3.78 0.521 M H 2O2 3.79 4.793% Ba by mass
3.81 (a) Na2SO4(aq) + Pb(NO3)2(s) ¡ PbSO4(s) + NaNO3(aq)
(b) Pb(NO 3)2 is the limiting reactant. (c) 0.200 M Na +, 0.0725 M
NO3 -, 0.064 M SO4 2- 3.84 1.766% Cl - by mass
3.85 2.8 * 10-5 g Na 3AsO4 in 1.00 dm3 H 2O
A-4
Answers to Selected Exercises
Chapter 4
4.1 As the book falls, potential energy decreases and kinetic
energy increases. At the instant before impact, all potential
energy has been converted to kinetic energy, so the book’s total
kinetic energy is 85 J, assuming no transfer of energy as heat.
4.4 (a) The temperatures of the system and surroundings will
equalise, so the temperature of the hotter system will decrease
and the temperature of the colder surroundings will increase.
The sign of q is (-); the process is exothermic. (b) If neither
volume nor pressure of the system changes, w = 0 and
¢E = q = ¢H. The change in internal energy is equal to the
change in enthalpy.
4.5 (a)
(b) ¢H = 0 for mixing ideal gases. ¢S is positive, because the
disorder of the system increases. (c) The process is spontaneous
and therefore irreversible. (d) Since ¢H = 0, the process does
not affect the entropy of the surroundings. 4.7 (a) At 300 K,
¢H = T¢S, ¢G = 0 and the system is at equilibrium.
(b) The reaction is spontaneous at temperatures above 300 K.
4.9 (a) 84 J (b) As the ball hits the sand, its speed (and hence
its kinetic energy) drops to zero. Most of the kinetic energy is
transferred to the sand, which deforms when the ball lands.
Some energy is released as heat through friction between the
ball and the sand. 4.11 The energy source of a 100-watt light
bulb is electrical current from household wiring. Energy is
radiated in the form of heat and visible light. The energy
source for an adult person is food. After a series of complex
chemical reactions that release the potential energy stored in
chemical bonds, some energy is transferred as electrical
impulses that trigger muscle action and become kinetic energy.
Some is released as heat. In both cases, energy must travel
through a network and undergo several changes in form before
it is in the correct location and form to be useful. 4.12 (a) The
system is the well-defined part of the universe whose energy
changes are being studied. (b) A closed system can exchange
heat but not mass with its surroundings. 4.13 (a) Work is a
force applied over a distance. (b) The amount of work done is
the magnitude of the force times the distance over which it is
applied. w = F * d. 4.16 (a) In any chemical or physical
change, energy can be neither created nor destroyed; energy
is conserved. (b) The internal energy (E) of a system is the
sum of all the kinetic and potential energies of the system
components. (c) Internal energy increases when work is done
on the system and when heat is transferred to the system.
4.18 (a) ¢E = 56 kJ, endothermic (b) ¢E = 0.84 kJ,
endothermic (c) ¢E = -71.0 kJ, exothermic 4.20 (a) A state
function is a property that depends only on the physical state
(pressure, temperature, etc.) of the system, not on the route
used to get to the current state. (b) Internal energy is a state
function; heat is not a state function. (c) Work is not a state
function. The amount of work required to move from one state
to another depends on the specific series of processes or path
used to accomplish the change. 4.22 (a) ¢H is usually easier
to measure than ¢E because at constant pressure ¢H = q.
The heat flow associated with a process at constant pressure
can easily be measured as a change in temperature, while
measuring ¢E requires a means of measuring both q and w.
(b) The process is exothermic. 4.24 The reactant, 2 Cl(g), has
the higher enthalpy. 4.27 (a) -13.1 kJ (b) -1.14 kJ (c) 9.83 J
4.29 At constant pressure, ¢U = ¢H - P ¢V. The values of
either P and ¢V or T and ¢n must be known to calculate
¢U from ¢H. 4.31 ¢U = -97 kJ; ¢H = -79 kJ
4.32 (a) ¢H = 726.5 kJ (b) ¢H = -1453 kJ (c) The exothermic
forward reaction is more likely to be thermodynamically
favoured. (d) Vaporisation is endothermic. If the product were
H2O(g), the reaction would be more endothermic and the
magnitude of ¢H would decrease. 4.33 (a) J/°C or J K1
(b) J/g °C or J g1 K1 (c) To calculate heat capacity from
specific heat, the mass of the particular piece of copper pipe
must be known. 4.34 (a) 4.18 J g1 K1 (b) 75.2 J mol1 °C
(c) 773 J/°C (d) 903 kJ 4.35 ΔH 41.7 kJ mol1 NaOH
4.37 ΔrxnE –25.5 kJ g1 C6H 4O2 or - 2.75 * 103 kJ mol1
C6H 4O2 4.38 (a) Heat capacity of the complete calorimeter =
14.4 kJ>°C (b) 5.40°C (c) The heat capacity of the calorimeter
would decrease. 4.39 Hess’s law is a consequence of the fact
that enthalpy is a state function. Since ¢H is independent
of path, we can describe a process by any series of steps that
add up to the overall process. ¢H for the process is the sum
of ¢H values for the steps. 4.40 ¢H = -1300.0 kJ
4.41 ¢H = -2.49 * 103 kJ 4.44 (a) Standard conditions for
enthalpy changes are P = 1 bar and some common
temperature, usually 298 K. (b) Enthalpy of formation is the
enthalpy change that occurs when a compound is formed
from its component elements. (c) Standard enthalpy of
formation ΔfH° is the enthalpy change that accompanies
formation of one mole of a substance from elements in their
standard states. 4.46 (a) 12 N2(g) + 32 H2(g) ¡ NH3(g),
ΔfH° –46.19 kJ (b) S(s) + O2(g) ¡ SO2(g), ΔfH° –296.9 kJ
(c) Rb(s) + 12 Cl2(g) + 32 O2(g) ¡ RbClO3(s),
Hf° –392.4 kJ (d) N2(g) + 2 H2(g) + 23 O2(g) ¡ NH4NO3(s),
ΔfH° –365.6 kJ 4.52 ΔrxnH° –847.6 kJ
4.49 (a) ΔrxnH° –196.6 kJ (b) ΔrxnH° 37.1 kJ
(c) ΔrxnH° –433.7 kJ (d) ΔrxnH° –68.3 kJ
4.50 ΔfH° –248 kJ 4.52 ΔfH° –924.8 kJ 4.54 (a) Fuel value
is the amount of heat produced when 1 g of a substance (fuel)
is combusted. (b) 5 g of fat 4.55 436 kJ/serving
4.56 250 kJ 4.57 (a) ¢Hcomb = -1850 kJ mol1 C3H 4 ,
- 1926 kJ mol1 C3H 6 , - 2044 kJ mol1 C3H 8
(b) ¢Hcomb = -4.616 * 104 kJ kg1 C3H 4 , - 4.578 * 104 kJ kg1
C3H 6 , - 4.635 * 104 kJ kg1 C3H 8 (c) These three substances
yield nearly identical quantities of heat per unit mass, but
propane is marginally higher than the other two.
4.58 Spontaneous: b, c, d; non-spontaneous: a, e
4.60 (a) NH4NO3(s) dissolves in water, as in a chemical cold
pack. Naphthalene (mothballs) sublimes at room temperature.
(b) Melting of a solid is spontaneous above its melting point but
non-spontaneous below its melting point. 4.61 (a) Endothermic
(b) at or above 100°C (c) below 100°C (d) at 100°C
4.62 (a) For a reversible process, the forward and reverse changes
occur by the same path. In a reversible process, both the system
and the surroundings are restored to their original condition by
exactly reversing the change. A reversible change produces the
maximum amount of work. (b) There is no net change in
the surroundings. (c) The vaporisation of water to steam is
reversible if it occurs at the boiling temperature of water for a
specified external (atmospheric) pressure, and only if the heat
Answers to Selected Exercises
needed is added infinitely slowly. 4.64 No. ¢E is a state
function. ¢E = q + w; q and w are not state functions. Their
values do depend on path, but their sum, ¢E, does not.
4.66 (a) An ice cube can melt reversibly at the conditions of
temperature and pressure where the solid and liquid are in
equilibrium. (b) We know that melting is a process that
increases the energy of the system even though there is no
change in temperature. ¢E is not zero for the process.
4.67 (a) At constant temperature, ¢S = qrev>T, where qrev is the
heat that would be transferred if the process were reversible.
(b) No. ¢S is a state function, so it is independent of path.
4.70 (a) For a spontaneous process, the entropy of the universe
increases; for a reversible process, the entropy of the universe
does not change. (b) For a reversible process, if the entropy of the
system increases, the entropy of the surroundings must decrease
by the same amount. (c) For a spontaneous process, the entropy
of the universe must increase, so the entropy of the surroundings
must increase, or decrease by less than 42 J K1.
4.71 ¢S = 0.762 J K1 4.72 (a) The higher the temperature, the
more microstates are available to a system. (b) The greater the
volume, the more microstates are available to a system. (c) Going
from solid to liquid to gas, the number of available microstates
increases. 4.74 (a) ¢S is positive. (b) ¢S is positive for Exercise
17.8(a) and (c). 4.75 S increases in (a) and (c); S decreases in (b)
4.77 (a) The entropy of a pure crystalline substance at absolute
zero is zero. (b) In translational motion the entire molecule moves
in a single direction; in rotational motion the molecule rotates or
spins around a fixed axis. In vibrational motion the bonds within
a molecule stretch and bend, but the average position of the atoms
does not change.
(c) H
H
Cl
H Cl
Translational
Cl
Rotational
H
Cl
H Cl
Vibrational
4.80 (a) ¢S 6 0 (b) ¢S 7 0 (c) ¢S 6 0 (d) ¢S 7 0
4.81 (a) C2H6(g) (b) CO2(g) 4.82 For elements with similar
structures, the heavier the atoms, the lower the vibrational
frequencies at a given temperature. This means that more
vibrations can be accessed at a particular temperature,
resulting in greater absolute entropy for the heavier elements.
4.83 (a) ¢S° = -120.5 J K1. ¢S° is negative because there are
fewer moles of gas in the products. (b) ¢S° = +176.6 J K1.
¢S° is positive because there are more moles of gas in the
products. (c) ¢S° = +152.39 J K1. ¢S° is positive because the
product contains more total particles and more moles of gas.
(d) ¢S° = +92.3 J K1. ¢S° is positive because there are more
moles of gas in the products. 4.84 (a) ¢G = ¢H - T¢S
(b) If ¢G is positive, the process is non-spontaneous, but the
reverse process is spontaneous. (c) There is no relationship
between ¢G and rate of reaction. 4.86 (a) Exothermic
(b) ¢S° is negative; the reaction leads to a decrease in
disorder. (c) ¢G° = -9.9 kJ (d) If all reactants and products
are present in their standard states, the reaction is
spontaneous in the forward direction at this temperature.
A-5
4.88 (a) ¢G° = -140.0 kJ, spontaneous (b) ¢G° = +104.70 kJ,
non-spontaneous (c) ¢G° = +146 kJ, non-spontaneous
(d) ¢G° = -156.7 kJ, spontaneous
4.89 (a) C6H12(l) + 9 O2(g) ¡ 6 CO2(g) + 6 H2O(l)
(b) The number of moles of gas is reduced in the reaction,
so ΔS° is almost certainly negative. Because ¢S° is negative,
¢G° is less negative than ¢H°. 4.90 (a) The forward
reaction is spontaneous at low temperatures, but becomes
non-spontaneous at higher temperatures. (b) The reaction is
non-spontaneous in the forward direction at all temperatures.
(c) The forward reaction is non-spontaneous at low
temperatures, but becomes spontaneous at higher
temperatures. 4.92 ¢S 7 +76.7 J K1 4.94 (a) T = 330 K
(b) non-spontaneous 4.95 (a) ¢H° = 155.7 kJ,
¢S° = 171.4 J K1. Since ¢S° is positive, ¢G° becomes more
negative with increasing temperature. (b) ¢G° = 18.5 kJ. The
reaction is not spontaneous under standard conditions at 800 K
(c) ΔG° –15.7 kJ. The reaction is spontaneous under standard
conditions at 1000 K. 4.96 (a) C2H2(g) + 52 O2(g) ¡
2 CO2(g) + H2O(l) (b) -1299.5 kJ of heat produced/mol C2H2
burned (c) wmax = -1235.1 kJ mol1 C2H2 4.100 (a) ¢G
becomes more negative. (b) ¢G becomes more positive.
(c) ΔG becomes more positive. 4.101 ΔH° 269.3 kJ,
ΔS° 0.1719 kJ K1 (a) PCO2 6.1 1039 bar
(b) PCO2 1.6 104 bar 4.103 The spontaneous air bag
reaction is probably exothermic, with - ¢H and thus -q. When
the bag inflates, work is done by the system, so the sign of w is
also negative. 4.104 ¢H = 38.95 kJ; ¢E = 36.48 kJ
4.106 4.88 g CH 4 4.109 (a) ¢H° = -631.1 kJ (b) 3 mol of
acetylene gas has greater enthalpy. (c) Fuel values are 50 kJ g1
C2H2(g), 42 kJ g1 C6H6(l). 4.112 (a) ¢H 7 0, ¢S 7 0
(b) ¢H 6 0, ¢S 6 0 (c) ¢H 7 0, ¢S 7 0 (d) ¢H 7 0, ¢S 7 0
(e) ¢H 6 0, ¢S 7 0 4.115 (a) For all three compounds listed,
there are fewer moles of gaseous products than reactants in
the formation reaction, so we expect ΔfS° to be negative. If
ΔfG° ΔfH° – TΔfS° and ΔfS° is negative, –TΔfS° is positive and
ΔfG° is more positive than ΔfH°. (b) In the formation of CO(g)
from C(s) and O2(g) in their standard states, each O2(g) molecule
that reacts with C(s) gives rise to two CO(g) molecules—this
increases the number of gas phase molecules making ΔfS°
positive and consequently ΔfG° more negative.
4.117 (a) K = 4 * 1015 (b) An increase in temperature will
decrease the mole fraction of CH 3COOH at equilibrium.
Elevated temperatures must be used to increase the
speed of the reaction. (c) K = 1 at 836 K or 563°C.
4.119 (a) ¢G = 8.77 kJ (b) wmin = 8.77 kJ. In practice,
a larger than minimum amount of work is required.
4.121 (a) 1.479 * 10-18 J>molecule (b) 1 * 10-15 J>photon.
The X-ray has approximately 1000 times more energy than
is produced by the combustion of 1 molecule of CH4(g).
4.126 (a) Acetone, ΔvapS° 88.4 J mol1 K; dimethyl ether,
ΔvapS° 86.6 J mol1 K; ethanol, ΔvapS° 110 J mol1 K; octane,
ΔvapS° 86.3 J mol1 K; pyridine, ΔvapS° 90.4 J mol1 K.
Ethanol does not obey Trouton’s rule. (b) Hydrogen bonding
(in ethanol and other liquids) leads to more ordering in the
liquid state and a greater than usual increase in entropy upon
vaporisation. Liquids that experience hydrogen bonding are
probably exceptions to Trouton’s rule. (c) Owing to strong
hydrogen bonding interactions, water probably does not obey
Trouton’s rule. ΔvapS° 109.0 J mol1 K. (d) ΔvapH for
A-6
Answers to Selected Exercises
C6H5Cl L 36 kJ mol1 4.128 (a) For any given total pressure,
the condition of equal moles of the two gases can be achieved
at some temperature. For individual gas pressures of 1 atm and
a total pressure of 2 atm, the mixture is at equilibrium at
328.5 K or 55.5°C. (b) 333.0 K or 60°C (c) 374.2 K or 101.2°C
(d) The reaction is endothermic, so an increase in the value of K
as calculated in parts (a)–(c) should be accompanied by an
increase in T.
Chapter 5
5.1 (a) No. Visible radiation has wavelengths much shorter
than 1 cm. (b) Energy and wavelength are inversely
proportional. Photons of the longer 1 cm radiation have less
energy than visible photons. (c) Radiation with l = 1 cm is in
the microwave region. The appliance is probably a microwave
oven. 5.3 (a) n = 1, n = 4 (b) n = 1, n = 2 (c) In order of
increasing wavelength and decreasing energy: (iii) 6 (iv) 6
(ii) 6 (i) 5.6 (a) In the left-most box, the two electrons cannot
have the same spin. (b) Flip one of the arrows in the left-most
box, so that one points up and the other down. (c) Group 16
5.7 (a) Metres (b) 1>second (c) metres/second 5.9 (a) True
(b) False. The frequency of radiation decreases as the
wavelength increases. (c) False. Ultraviolet light has shorter
wavelengths than visible light. (d) False. Electromagnetic
radiation and sound waves travel at different speeds.
5.11 (a) 3.14 * 1011 s -1 (b) 5.45 * 10-7 m = 545 nm
(c) The radiation in (b) is visible, the radiation in (a) is not.
(d) 1.50 * 104 m 5.13 (a) Quantisation means that energy can
only be absorbed or emitted in specific amounts or multiples of
these amounts. This minimum amount of energy is equal to a
constant times the frequency of the radiation absorbed or
emitted; E = hn. (b) In everyday activities, macroscopic objects
such as our bodies gain and lose total amounts of energy much
larger than a single quantum, hn. The gain or loss of the
relatively minuscule quantum of energy is unnoticed.
5.15 (a) 4.54 * 10-19 J (b) 4.47 * 10-21 J (c) 6.92 * 10-8 m
69.2 nm; ultraviolet 5.17 (a) 6.11 * 10-19 J>photon
(b) 368 kJ mol1 (c) 1.64 * 1015 photons 5.19 (a) The
' 1 * 10-6 m radiation is in the infrared portion of the spectrum.
(b) 8.1 * 1016 photons>s 5.21 (a) Emin = 7.22 * 10-19 J
(b) l = 275 nm (c) The excess energy of the 120 nm photon
(E120 = 1.66 * 10-18 J) is converted into the kinetic energy of
the emitted electron, Ek 9.34 10–19 J. 5.23 When applied
to atoms, the notion of quantised energies means that only
certain values of ¢E are allowed. These give rise to the lines
in the emission or absorption spectra of excited atoms.
5.25 (a) Emitted (b) absorbed (c) emitted
5.26 E2 = -5.45 * 10-19 J; E6 = -0.606 * 10-19 J;
¢E = 4.84 * 10-19 J; l = 410 nm; visible, violet.
5.28 (a) Ultraviolet region (b) ni = 6, nf = 1
5.30 (a) l = 5.6 * 10-37 m (b) l = 2.65 * 10-34 m
(c) l = 2.3 * 10-13 m 5.32 4.14 * 103 m s1
5.34 (a) The uncertainty principle states that there is a limit to
how precisely we can simultaneously know the position and
momentum (a quantity related to energy) of an electron. The
Bohr model states that electrons move about the nucleus in flat
circular orbits of known radius and energy. This violates the
uncertainty principle because the electron would have no
uncertainty in position or momentum in the z-direction.
(b) De Broglie stated that electrons demonstrate the properties
of both particles and waves, that each particle has a wave
associated with it. A wave function is the mathematical
description of the wave associated with a particle.
(c) Although we cannot predict the exact location of an electron
in an allowed energy state, we can determine the probability of
finding an electron at a particular position. This statistical
knowledge of electron location is the probability density and is
equivalent to ° 2, the square of the wave function °.
5.36 (a) n = 4, l = 3, 2, 1, 0 (b) l = 2, ml = -2, -1, 0, 1, 2
5.38 (a) 3p: n = 3, l = 1 (b) 2s: n = 2, l = 0 (c) 4f: n = 4,
l = 3 (d) 5d: n = 5, l = 2 5.40 (a) impossible, 1p (b) possible
(c) possible (d) impossible, 2d 5.42 (a) The hydrogen atom 1s
and 2s orbitals have the same overall spherical shape, but the
2s orbital has a larger radial extension and one more node than
the 1s orbital. (b) A single 2p orbital is directional in that its
electron density is concentrated along one of the three Cartesian
axes of the atom. The dx2 - y2 orbital has electron density along
both the x- and y-axes, while the px orbital has density only
along the x-axis. (c) The average distance of an electron from
the nucleus in a 3s orbital is greater than for an electron in
a 2s orbital. (d) 1s 6 2p 6 3d 6 4f 6 6s 5.44 (a) In the
hydrogen atom, orbitals with the same principal quantum
number, n, have the same energy. (b) In a many-electron atom,
for a given n value, orbital energy increases with increasing l
value: s 6 p 6 d 6 f 5.46 (a) + 12 , - 12 (b) a magnet with a
strong non-homogeneous magnetic field (c) they must have
different ms values; the Pauli exclusion principle 5.48 (a) 6
(b) 10 (c) 2 (d) 14 5.50 (a) Each box represents an orbital.
(b) Electron spin is represented by the direction of the halfarrows. (c) No. In Be, all the boxes are full, so Hund’s rule is
not needed. 5.52 (a) Cs, [Xe]6s 1 (b) Ni, [Ar]4s 2 3d 8 (c) Se,
[Ar]4s 2 3d 10 4p4 (d) Cd, [Kr]5s 2 4d 10 (e) Ac, [Rn]7s 2 6d 1 (f) Pb,
[Xe]6s 2 4f 14 5d10 6p 2 5.54 (a) Mg (b) Al (c) Cr (d) Te
5.56 (a) The fifth electron would fill the 2p subshell before
the 3s. (b) Either the core is [He], or the outer electron
configuration should be 3s 2 3p 3. (c) The 3p subshell would fill
before the 3d. 5.62 (a) The Paschen series lies in the infrared.
(b) ni = 4, l = 1.87 * 10-6 m; ni = 5, l = 1.28 * 10-6 m;
ni = 6, l = 1.09 * 10-6 m 5.64 v = 1.02 * 107 m s1
5.67 (a) l (b) n and l (c) ms (d) ml 5.69 If ms had three
allowed values instead of two, each orbital would hold three
electrons instead of two. Assuming that there is no change in
the n, l and ml values, the number of elements in each of the
first four rows would be: 1st row, 3 elements; 2nd row,
12 elements; 3rd row, 12 elements; 4th row, 27 elements
5.72 1.7 * 1028 photons 5.75 (a) Bohr’s theory was based on
the Rutherford nuclear model of the atom: a dense positive
charge at the centre and a diffuse negative charge surrounding
it. Bohr’s theory then specified the nature of the diffuse
negative charge. The prevailing theory before the nuclear model
was Thomson’s plum pudding model: discrete electrons
scattered about a diffuse positive charge cloud. Bohr’s theory
could not have been based on the Thomson model of the atom.
(b) De Broglie’s hypothesis is that electrons exhibit both particle
and wave properties. Thomson demonstrated that electrons
have mass—a particle property, while cathode rays behave as
waves. De Broglie’s hypothesis rationalises these two seemingly
contradictory properties of electrons.
Chapter 6
6.1 The billiard ball has a definite ‘hard’ boundary, while a
quantum mechanical atom has no definitive edge. Billiard balls
Answers to Selected Exercises
can be used to model nonbonding interactions, such as Ne
atoms colliding, because there is no penetration of electron
clouds. They are an inappropriate model for bonding
interactions, because bonding electron clouds do penetrate
and the bonding atomic radius of an atom is smaller than its
nonbonding radius. 6.3 The energy change for the reaction
is the ionisation energy of A plus the electron affinity of A.
6.5 Mendeleev placed elements with similar chemical and
physical properties within a family or column of the table. If no
element with the correct chemical and physical properties of an
appropriate atomic weight was known, he left a blank space.
He predicted properties for the ‘blanks’ based on properties
of other elements in the family and on either side.
6.7 (a) Effective nuclear charge, Zeff , is a representation of the
average electrical field experienced by a single electron. It is
the average environment created by the nucleus and the other
electrons in the molecule, expressed as a net positive charge at
the nucleus. (b) Going from left to right across a period,
effective nuclear charge increases. 6.9 (a) Z = 19, S = 18,
Zeff = Z - S = 1. (b) The valence electron in K is a 4s electron.
Because s electrons have finite probability of being close to the
nucleus, shielding of them is never perfect and calculated
values of Zeff reflect this. 6.11 Atomic radii are determined
by distances between atoms in various situations. Bonding
radii are calculated from the internuclear separation of two
atoms joined by a chemical bond. Nonbonding radii are
calculated from the internuclear separation between
two gaseous atoms that collide and move apart but do
not bond. 6.13 1.37 pm 6.15 (a) Decrease (b) increase
(c) F 6 S 6 P 6 As 6.16 (a) In moving up or down a group
the principle quantum number (n) changes, while in
moving left or right within a period n remains the same and
only Zeff changes, resulting in a relatively smaller variation.
(b) S 6 Si 6 Se 6 Ge 6.17 (a) Be 6 Mg 6 Ca
(b) Br 6 Ge 6 Ga (c) Si 6 Al 6 Tl 6.19 (a) Electrostatic
repulsions are reduced by removing an electron from a neutral
atom, effective nuclear charge increases and the cation is
smaller. (b) The additional electrostatic repulsion produced by
adding an electron to a neutral atom decreases the effective
nuclear charge experienced by the valence electrons, and
increases the size of the anion. (c) Going down a column,
valence electrons are further from the nucleus, and they
experience greater shielding by core electrons. The greater
radial extent of the valence electrons outweighs the increase
in Z. 6.21 The red sphere is a metal; its size decreases on
reaction, characteristic of the change in radius when a metal
atom forms a cation. The blue sphere is a nonmetal; its size
increases on reaction, characteristic of the change in radius
when a nonmetal atom forms an anion. 6.23 (a) An
isoelectronic series is a group of atoms or ions that have
the same number of electrons. (b) (i) N 3-: Ne (ii) Ba2+: Xe
(iii) Se 2-: Kr (iv) Bi 3+: Hg 6.25 (a) Se 6 Se 2- 6 Te 2(b) Co3+ 6 Fe 3+ 6 Fe 2+ (c) Ti 4+ 6 Sc 3+ 6 Ca
(d) Be 2+ 6 Na + 6 Ne 6.27 B(g) ¡ B+(g) + e-;
B+(g) ¡ B2+(g) + e-; B2+(g) ¡ B3+(g) + e- 6.29 (a) The
electrons in any atom are bound to the atom by the electrostatic
attraction to the nucleus. Therefore, energy must be added in
order to remove an electron from an atom. The ionisation
energy, ¢E for this process, is thus positive. (b) F has a greater
first ionisation energy than O because F has a greater Zeff and
A-7
the outer electrons in both elements are approximately the
same distance from the nucleus. (c) The second ionisation
energy of an element is greater than the first because more
energy is required to overcome the larger Zeff of the 1 + cation
than that of the neutral atom. 6.31 (a) The smaller the atom,
the larger its first ionisation energy. (b) Of the non-radioactive
elements, He has the largest and Cs the smallest first
ionisation energy. 6.33 (a) Ar (b) Be (c) Co (d) S (e) Te
6.35 (a) Si 2+, [Ne]3s 2 (b) Bi 3+, [Xe]6s 2 4f 14 5d10 (c) Te 2-,
[Kr]5s 2 4d10 5p6 or [Xe] (d) V 3+, [Ar]3d 2 (e) Hg 2+, [Xe]4f 14 5d10
(f) Ni 2+, [Ar]3d 8 6.37 Ionisation energy:
Se(g) ¡ Se+(g) + e-; [Ar]4s 2 3d 10 4p 4 ¡ [Ar]4s 2 3d 10 4p 3;
electron affinity: Se(g) + e- ¡ Se-(g);
[Ar]4s 2 3d 10 4p4 ¡ [Ar]4s 2 3d 10 4p 5 6.39 Li + e - ¡ Li -;
[He]2s 1 ¡ [He]2s 2; Be + e - ¡ Be -;
[He]2s 2 ¡ [He]2s 2 2p 1. Adding an electron to Li completes
the 2s subshell. The added electron experiences essentially the
same effective nuclear charge as the other valence electron;
there is an overall stabilisation and ¢E is negative. An extra
electron in Be would occupy the higher-energy 2p subshell.
This electron is shielded from the full nuclear charge by the
2s electrons and does not experience a stabilisation in energy;
¢E is positive. 6.41 Ionisation energy of F -: F-(g) ¡
F(g) + e -; electron affinity of F: F(g) + e- ¡ F-(g). The two
processes are the reverse of each other. The energies are equal
in magnitude but opposite in sign. I 11F -2 = - E(F). 6.43 The
smaller the first ionisation energy of an element, the greater the
metallic character of that element. 6.45 Ionic: MgO, Li 2O,
Y2O3 ;molecular: SO 2 , P2O5 , N2O, XeO3 . Ionic compounds are
formed by combining a metal and a nonmetal; molecular
compounds are formed by two or more nonmetals.
6.46 (a) An acidic oxide dissolved in water produces an acidic
solution; a basic oxide dissolved in water produces a basic
solution. (b) Oxides of nonmetals, such as SO2 , are usually
acidic; oxides of metals, such as CaO, are basic.
6.48 (a) BaO(s) + H2O(l) ¡ Ba(OH)2(aq)
(b) FeO(s) + 2 HClO4(aq) ¡ Fe(ClO4)2(aq) + H2O(l)
(c) SO3(g) + H2O(l) ¡ H2SO4(aq)
(d) CO2(g) + 2 NaOH(aq) ¡ Na2CO3(aq) + H2O(l)
6.50 (a) Na, [Ne]3s 1; Mg, [Ne]3s 2 (b) When forming ions, both
adopt the stable configuration of Ne; Na loses one electron and
Mg two electrons to achieve this configuration. (c) The
effective nuclear charge of Mg is greater, so its ionisation
energy is greater. (d) Mg is less reactive because it has a higher
ionisation energy. (e) The atomic radius of Mg is smaller
because its effective nuclear charge is greater. 6.52 (a) Ca is
more reactive because it has a lower ionisation energy than
Mg. (b) K is more reactive because it has a lower ionisation
energy than Ca. 6.53 (a) 2 K(s) + Cl2(g) ¡ 2 KCl(s)
(b) SrO(s) + H2O(l) ¡ Sr(OH)2(aq) (c) 4 Li(s) +
O2(g) ¡ 2 Li2O(s) (d) 2 Na(s) + S(l) ¡ Na2S(s)
6.55 H, 1s 1; Li, [He]2s 1; F, [He]2s 2 2p5. Like Li, H has only one
valence electron, and its most common oxidation number is
+1. Like F, H needs only one electron to adopt the stable
electron configuration of the nearest noble gas; both H and F
can exist in the - 1 oxidation state. 6.56 (a) F, [He]2s 2 2p 5;
Cl, [Ne]3s 2 3p 5 (b) F and Cl are in the same group, and both
adopt a 1- ionic charge. (c) The 2p valence electrons in F
are closer to the nucleus and more tightly held than the 3p
electrons of Cl, so the ionisation energy of F is greater.
(d) The higher exothermic electron affinity of F makes it more
A-8
Answers to Selected Exercises
reactive than Cl toward H 2O. (e) While F has approximately
the same effective nuclear charge as Cl, its small atomic radius
gives rise to large repulsions when an extra electron is added,
so the overall electron affinity of F is less exothermic than that
of Cl. (f) The 2p valence electrons in F are closer to the nucleus
so the atomic radius is smaller than that of Cl. 6.58 (a) 4s
(b) Because s electrons have a greater probability of being
close to the nucleus, they experience less shielding than p
electrons with the same n-value. Zeff = Z - S and Z is the
same for all electrons in As. If S is smaller for 4s than 4p, Zeff
will be greater for 4s electrons and they will have lower
energy. 6.62 Moving across the representative elements,
electrons added to ns or np valence orbitals do not effectively
screen each other. The increase in Z is not accompanied by a
similar increase in S. Zeff increases and atomic size decreases.
Moving across the transition elements, electrons are added
to 1n - 12d orbitals, which can approach more closely to the
nucleus and thus do significantly screen the ns valence
electrons. The increase in Z is accompanied by a
corresponding increase in S. Zeff increases more slowly and
atomic size decreases more slowly. 6.64 I1 through I4
represent loss of the 2p and 2s electrons from the outer shell
of the atom. Z is constant while removal of each electron
reduces repulsion between the remaining electrons, so Zeff
increases and I increases. I5 and I6 represent loss of the 1s core
electrons. These electrons are closer to the nucleus and
experience the full nuclear charge, so the values of I5 and I6
are significantly greater than I1 - I4 . I6 is larger than I5
because all electron-electron repulsion has been eliminated.
6.67
O, [He]2s22p4
2s
2p
O2, [He]2s22p6 [Ne]
2s
O 3-, [Ne]3s 1
2p
The third electron would be
added to the 3s orbital, which is
farther from the nucleus and
more strongly shielded by the
[Ne] core. The overall attraction
of this 3s electron for the oxygen
nucleus is not large enough for
O 3- to be a stable particle.
6.69 (a) The group 12 metals have complete 1n - 12d
subshells. An additional electron would occupy an np subshell
and be substantially shielded by both ns and 1n - 12d
electrons. This is not a lower energy state than the neutral atom
and a free electron. (b) Group 11 elements have the generic
electron configuration ns11n - 12d10. An additional electron
would complete the ns subshell and experience repulsion with
the other ns electron. Going down the group, size of the ns
subshell increases and repulsion effects decrease, so effective
nuclear charge increases and electron affinities become more
negative. 6.74 (a) Li, [He]2s 1; Zeff L 1 + . (b) I1 L
5.45 * 10-19 J>atom L 328 kJ>mol (c) The estimated value of
328 kJ mol1 is less than the Table 6.4 value of 520 kJ mol1.
Our estimate for Zeff was a lower limit; the [He] core electrons
do not perfectly shield the 2s electron from the nuclear charge.
(d) Based on the experimental ionisation energy, Zeff = 1.26.
This value is greater than the estimate from part (a), which is
consistent with the explanation in part (c).
Chapter 7
7.2 (a) No. A1 and A2 have like charges and repel each other.
(b) A1Z1 . Lattice energy increases as ionic charge increases and
decreases as inter-ionic distance increases. These ions all have the
same magnitude of charge; A1 and Z1 have the smallest radii and
their combination leads to the largest lattice energy. (c) A2Z2 has
the smallest lattice energy, because it has the largest inter-ionic
distance. 7.5 (a) Moving from left to right along the molecule,
the first C needs 2 H atoms, the second needs 1, the third needs
none and the fourth needs 1. (b) In order of increasing bond
length: 3 6 1 6 2 (c) In order of increasing bond enthalpy:
2 6 1 6 3 7.7 (a) Valence electrons are those that take part in
chemical bonding. This usually means the electrons beyond the
core noble-gas configuration of the atom. (b) A nitrogen atom has
5 valence electrons. (c) The atom (Si) has 4 valence electrons.
7.9 P, 1s 2 2s 2 2p 6 3s 2 3p 3. A 3s electron is a valence electron; a 2s
(or 1s) electron is a nonvalence electron. A 3s or 3p electron is a
valence electron; a 2s, 2p or 1s electron is a nonvalence electron.
7.10 (a) Al
(b) Br
(c) Ar
(d) Sr
7.12 Mg O
Mg2 O
2
7.14 (a)AlF3 (b) K 2S (c) Y2O3 (d) Mg3N2 7.16 (a) Sr 2+,
[Ar]4s 2 3d 10 4p 6 = [Kr], noble-gas configuration (b) Ti 2+,
[Ar]3d 2 (c) Se 2-, [Ar]4s 2 3d 10 4p 6 = [Kr], noble-gas
configuration (d) Ni 2+, [Ar]3d 8 (e) Br -, [Ar]4s 2 3d 10 4p 6 = [Kr],
noble-gas configuration (f) Mn3+, [Ar]3d 4 7.18 (a) Lattice energy
is the energy required to totally separate one mole of solid ionic
compound into its gaseous ions. (b) The magnitude of the lattice
energy depends on the magnitudes of the charges of the two ions,
their radii and the arrangement of ions in the lattice. 7.21 The
large attractive energy between oppositely charged Ca2+ and
O 2- more than compensates for the energy required to form Ca2+
and O 2- from the neutral atoms. 7.22 The lattice energy of
RbCl(s) is +668 kJ mol1. This value is smaller than the lattice
energy for NaCl because Rb + has a larger ionic radius than Na +
and therefore cannot approach Cl - as closely as Na + can.
7.24 (a) A covalent bond is the bond formed when two atoms
share one or more pairs of electrons. (b) Any simple compound
whose component atoms are nonmetals, such as H 2 , SO2 and
CCl4 , are molecular and have covalent bonds between atoms.
(c) Covalent, because it is a gas at room temperature and below.
7.26 (a) O O (b) A double bond is required because there are
not enough electrons to satisfy the octet rule with single bonds
and unshared pairs. (c) The greater the number of shared
electron pairs between two atoms, the shorter the distance
between the atoms. An O “ O double bond is shorter than an
O ¬ O single bond. 7.28 (a) Electronegativity is the ability of an
atom in a molecule to attract electrons to itself. (b) The range of
electronegativities on the Pauling scale is 0.7–4.0. (c) Fluorine is
the most electronegative element. (d) Cesium is the least
electronegative element that is not radioactive. 7.30 (a) Br (b) C
(c) P (d) Be 7.32 The bonds in (a), (c) and (d) are polar. The
more electronegative element in each polar bond is: (a) F
(c) O (d) I 7.34 The calculated charge on H and F is 0.41e.
7.36 (a) MnO2 , ionic (b) P2S 3 , covalent (c) CoO, ionic
(d) copper(I) sulfide, ionic (e) chlorine trifluoride, covalent
(f) vanadium(V) fluoride, ionic
A-9
Answers to Selected Exercises
7.38 (a)
(b) C
H
H
Si
(c) F
O
S
F
(c)
N
N
N
N
N
N
H
N
H
(d)
(e)
O
S
O
O
O
Cl
O
(f) H
N
O
H
H
H
(d)
(e)
Cl
Cl
C
H
F
N, 3; O, 2
0
P
Cl
1
Cl
0
Cl
0
P, 5; Cl, 1; O, 2
1
(d)
O
2
1
Cl
O
3
O
Cl
O
1
O
O
H
0
0
1
O
Cl, 5; H, 1; O, 2
1
Cl, 7; O, 2
7.44 (a) O
N
O
O
N
O
(b) NOF is isoelectronic with NO 2 -; both have 18 valence
electrons. (c) Since each N ¬ O bond has partial double-bond
character, the N ¬ O bond length in NO 2 - should be shorter
than in species with formal N ¬ O single bonds. 7.46 (a) The
octet rule states that atoms will gain, lose or share electrons
until they are surrounded by eight valence electrons. (b) The
octet rule applies to the individual ions in an ionic compound.
For example, in MgCl2 , Mg loses 2 e - to become Mg 2+ with the
electron configuration of Ne. Each Cl atom gains one electron
to form Cl - with the electron configuration of Ar. 7.48 The
most common exceptions to the octet rule are molecules with
more than eight electrons around one or more atoms.
7.50
2
O
F
10 electrons around Sb
7.40 (a) Formal charge is the charge each atom in a molecule
would have assuming all atoms have the same
electronegativity. (b) Formal charges are not actual charges.
They are a bookkeeping system that assumes perfect covalency,
one extreme for the possible electron distribution in a
molecule. (c) Oxidation numbers are a bookkeeping system
that assumes perfect ionic bonding where the more
electronegative element holds all electrons in a bond. The true
electron distribution is some composite of the two extremes.
7.42 Formal charges are shown on the Lewis structures;
oxidation numbers are listed below each structure.
(a) N O (b)
1
O
0
1
(a)
F
F
H
1
N
Sb
H
O
(c)
F
N
S
O
O
Other resonance structures
that violate the octet rule
can be drawn, but the
single best structure, the
one shown here, obeys
the octet rule
(b) H
Al
H
H
6 electrons around Al
7.52 (a) Cl
(b) Cl
1
Be
0
0
Be
Cl
2
Cl this structure violates the octet rule.
0
Cl
1
Cl
Be
2
0
2
Cl
2
Be
2
Cl
0
(c) Since formal charges are minimised on the structure that
violates the octet rule, this form is probably most important.
7.54 (a) ¢H = -304 kJ (b) ¢H = -82 kJ (c) ¢H = -467 kJ
7.56 (a) ¢H = -321 kJ (b) ¢H = -103 kJ (c) ¢H = -203 kJ
7.57 (a) Exothermic (b) The ¢H calculated from bond enthalpies
(-97 kJ) is slightly more exothermic (more negative) than that
obtained using ΔfH° values (-92.38 kJ). 7.58 The average
Ti ¬ Cl bond enthalpy is 430 kJ mol1. As the bond enthalpies
increase steadily as the oxidation state of Ti increases, the
average Ti Cl bond enthalpy found, has a high degree of
uncertainty. 7.60 (a) Six (nonradioactive) elements. Yes, they
are in the same group, assuming H is placed with the alkali
metals. The Lewis symbol represents the number of valence
electrons of an element, and all elements in the same group
have the same number of valence electrons. By definition of a
group, all elements with the same Lewis symbol must be in the
same group. 7.62 (a) 779 kJ mol1 (b) 627 kJ mol1
(c) 2195 kJ mol1 7.63 (a) A polar molecule has a measurable
dipole moment while a nonpolar molecule has a zero net dipole
moment. (b) Yes. If X and Y have different electronegativities,
the electron density around the more electronegative atom will
be greater, producing a charge separation or dipole in the
molecule. (c) The dipole moment, m, is the product of the
magnitude of the separated charges, Q, and the distance
between them, r. m = Qr 7.68 (a) +1 (b) -1 (c) +1
(assuming the odd electron is on N) (d) 0 (e) +3
7.71 ¢H is +42 kJ for the first reaction and -200 kJ for the
second. The latter is much more favourable because the
formation of 2 mol of O ¬ H bonds is more exothermic than
the formation of 1 mol of H ¬ H bonds.
7.73 (a)
O
O
N
H
H
O
C
C
N
N
N
O
H
N
C
H
H
O
N
H
O
(b) Each terminal O “ N ¬ O group has two possible
placements for the N “ O; this leads to 8 resonance structures
with the carbon–nitrogen framework shown in (a). Structures
with N “ N can be drawn, but these do not minimise formal
A-10
Answers to Selected Exercises
charge. There are 7 resonance structures with one or
more N “ N, for a total of 15 resonance structures.
(c) C3H6N6O6(s) ¡ 3 CO(g) + 3 N2(g) + 3 H2O(g)
(d) According to Table 7.4, N ¬ N single bonds have the
smallest bond enthalpy and are weakest. (e) The enthalpy
of decomposition for the resonance structure in (a) is
-1668 kJ mol1 C3H 6N6O6 . For the resonance structure with
3 N “ N and 6 N ¬ O bonds, ¢H = -1632 kJ mol1. The actual
value is somewhere between these two values. The enthalpy
change for 5.0 g of RDX is then in the range 37–38 kJ.
7.76 (a) 178 pm (b) 178 pm (c) The resonance structures for SO2
indicate that the sulfur–oxygen bond is intermediate between a
double and single bond, so the bond length of 1.43 pm should be
significantly shorter than the single bond distance of 1.75 pm.
(d)
S S
S S
O
O
S
S
S
S
S
S
S
S
S
S
S
S
In order to rationalise the sulfur–oxygen bond distance of
1.48 pm, a resonance structure with S “ O must contribute to
the true structure. The sulfur atom attached to oxygen in this
resonance structure has more than an octet of electrons and
violates the octet rule. 7.78 (a) A, In 2S; B, InS; C, In 2S 3
(b) A, In(I); B, In(II); C, In(III) (c) In(I), [Kr]5s 2 4d 10; In(II),
[Kr]5s 1 4d10; In(III), [Kr]4d10. None of these are noble-gas
configurations. (d) In(III) in compound C has the smallest
ionic radius. Removing successive electrons from an atom
reduces electron repulsion, increases the effective nuclear
charge experienced by the valence electrons and decreases the
ionic radius. (e) Lattice energy increases as ionic charge
increases and as inter-ionic distance decreases. Only the charge
and size of In changes in the three compounds. In(I) in
compound A has the lowest charge and largest ionic radius, so
compound A has the smallest lattice energy and lowest melting
point. Compound C has the largest lattice energy and highest
melting point. 7.81 (a) Azide ion is N 3-. (b) Resonance
structures with formal charges are shown.
N
N
1
1
N
1
N
0
N
1
N
2
N
N
N
2
1
0
(c) The left structure minimises formal charges and is probably
the main contributor. (d) The N ¬ N distances will be equal, and
have the approximate length of a N ¬ N double bond, 1.24 pm.
Chapter 8
8.3 (a) Square pyramidal (b) Yes, there is one nonbonding
electron domain on A. If there were only the 5 bonding domains,
the shape would be trigonal bipyramidal. (c) Yes. If the B atoms
are halogens, A is also a halogen. 8.6 (a) i, Formed by two s
atomic orbitals; ii, s-type MO; iii, antibonding MO (b) i, formed
by two p atomic orbitals overlapping end-to-end; ii, s-type MO;
iii, bonding MO (c) i, formed by two p atomic orbitals
overlapping side-to-side; ii, p-type MO; iii, antibonding MO
8.8 (a) Yes. The bond angle specifies the shape and the bond
length tells the size. (b) Yes. The only possible bond angles in this
arrangement are 120° angles. 8.9 (a) An electron domain is a
region in a molecule where electrons are most likely to be found.
(b) Like the balloons in Figure 8.5, each electron domain occupies
a finite volume of space, so they also adopt an arrangement
where repulsions are minimised. 8.11 (a) Trigonal planar
(b) tetrahedral (c) trigonal bipyramidal (d) octahedral
8.13 The electron-domain geometry indicated by VSEPR
describes the arrangement of all bonding and nonbonding
electron domains. The molecular geometry describes just the
atomic positions. In H 2O there are 4 electron domains around
oxygen, so the electron-domain geometry is tetrahedral. Because
there are 2 bonding and 2 nonbonding domains, the molecular
geometry is bent. 8.15 (a) Tetrahedral, trigonal pyramidal
(b) trigonal planar, trigonal planar (c) trigonal bipyramidal,
T-shaped (d) tetrahedral, tetrahedral (e) trigonal bipyramidal,
linear (f) tetrahedral, bent 8.17 (a) i, trigonal planar; ii,
tetrahedral; iii, trigonal bipyramidal (b) i, 0; ii, 1; iii, 2 (c) N and
P (d) Cl (or Br or I). This T-shaped molecular geometry arises
from a trigonal bipyramidal electron-domain geometry with 2
nonbonding domains. Assuming each F atom has 3 nonbonding
domains and forms only single bonds with A, A must have 7
valence electrons and be in or below the third row of the periodic
table to produce these electron-domain and molecular
geometries. 8.19 (a) 1–109°, 2–109° (b) 3–109°, 4–109°
(c) 5–180° (d) 6–120°, 7–109°, 8–109° 8.21 Each species has
4 electron domains, but the number of nonbonding domains
decreases from 2 to 0, going from NH 2 - to NH 4 +. Since
nonbonding domains exert greater repulsive forces on adjacent
domains, the bond angles expand as the number of nonbonding
domains decreases. 8.23 (a) Although both ions have 4 bonding
electron domains, the 6 total domains around Br require
octahedral domain geometry and square planar molecular
geometry, while the 4 total domains about B lead to tetrahedral
domain and molecular geometry. (b) CF4 will have bond angles
closest to the value predicted by VSEPR because there are no
nonbonding electron domains around C. In SF4 the single
nonbonding domain will enter into more repulsive interactions,
‘push back’ the bonding domains and lead to bond angles that
are non-ideal. 8.25 (a) In Exercise 8.17, molecules (ii) and (iii)
will have non-zero dipole moments. Molecule (i) has no
nonbonding electron pairs on A, and the 3 A ¬ F bond dipoles
are oriented so that they cancel. Molecules (ii) and (iii) have
nonbonding electron pairs on A and their bond dipoles do not
cancel. (b) In Exercise 8.18, molecules (i) and (ii) have a zero
dipole moment. 8.27 (b) NH 3 , (c) SF4 and (e) CH 3Br are polar.
8.29 (a) Orbital overlap occurs when valence atomic orbitals on
two adjacent atoms share the same region of space. (b) In
valence-bond theory, orbital overlap allows two bonding
electrons to mutually occupy the space between the bonded
nuclei. (c) Valence-bond theory is a combination of the atomic
orbital concept and the Lewis model of electron-pair bonding.
8.31 (a) H ¬ Mg ¬ H, linear electron domain and molecular
geometry (b) The Mg atom has 2 electrons in its 2s orbital; these
electrons are paired. Without promotion, Mg has no unpaired
electrons available to form bonds with H. (c) The linear electron
domain geometry in MgH 2 requires sp hybridisation.
(d)
8.32 (a) sp, 180° (b) sp2, 120°
H
Mg
(c) sp3 d 2, 90° and 180° (d) sp3 d, 90° and 120° (e) sp3 d, 90°, 120°
and 180° 8.34 (a) B, [He]2s 2 2p 1. One 2s electron is
Answers to Selected Exercises
promoted to an empty 2p orbital. The 2s and two 2p orbitals
that each contain one electron are hybridised to form three
equivalent hybrid orbitals in a trigonal planar arrangement.
(b) sp2
(c)
A-11
Be2 + has a bond order of 0.5 and is slightly lower in energy
than isolated Be atoms. It will probably exist under special
experimental conditions. 8.45 (a, b) Substances with no
unpaired electrons are weakly repelled by a magnetic field. This
property is called diamagnetism. (c) O 2 2-, Be2 2+ 8.47 (a) B2 +,
s2s 2s*2s 2p2p 1, increase (b) Li 2 +, s2s 1, increase (c) N2 +,
B
s2s 2s*2s 2p2p 4p2p 1, increase (d) Ne2 2+, s2s 2s*2s 2p2p 4p*2s 4, decrease
(d) A single 2p orbital is unhybridised. It lies perpendicular to
the trigonal plane of the sp2 hybrid orbitals. 8.35 (a) sp2 (b) sp3
(c) sp (d) sp3 d (e) sp3 d 2 8.37 (a) Both atomic and molecular
orbitals have a characteristic energy and shape; each can hold a
maximum of two electrons. Atomic orbitals are localised around
a single atom, and their energies are the result of interactions
between an electron and other subatomic particles (electrons and
nuclei) in a single atom. Molecular orbitals can be delocalised
over several atoms, and their energies result from interactions
between an electron and all the subatomic particles in the
molecule. (b) There is a net lowering in energy that accompanies
bond formation, because the electrons in H 2 are strongly
attracted to two H nuclei, rather than one H nucleus as in the
H atom. (c) 2
8.39 (a)
s*1s
s*1s
1s
1s
s1s
s1s
H2
(b) There is one electron in H 2 +. (c) s1s 1 (d) BO = 12 (e) Fall
apart. If the single electron in H 2 + is excited to the s*1s orbital,
its energy is higher than the energy of an H 1s atomic orbital and
H 2 + will decompose into a hydrogen atom and a hydrogen ion.
8.41
s*2p
(a)
2pz
2pz
s2p
p*2p
(b)
8.49 CN, s2s 2s*2s 2p2p 4s2p 1, bond order = 2.5, paramagnetic;
CN +, s2s 2s*2s 2p2p 4, bond order = 2.0, diamagnetic; CN -,
s2s 2s*2s 2p2p 4s2p 2, bond order = 3.0, diamagnetic 8.51 (a) 3s,
3px , 3py , 3pz (b) p3p (c) 4 (d) If the MO diagram for P2 is similar
to that of N2 , P2 will have no unpaired electrons and be
diamagnetic. 8.54 SiF4 is tetrahedral, SF4 is seesaw, XeF4 is
square planar. The shapes are different because the number of
nonbonding electron domains is different in each molecule, even
though all have four bonding electron domains. Bond angles and
thus molecular shape are determined by the total number of
electron domains. 8.56 (a) Two sigma, two pi (b) two sigma,
two pi (c) three sigma, one pi (d) four sigma, one pi
8.59 The compound on the right has a non-zero dipole moment.
8.64 s2s 2s*2s 2p2p 4s2p 1p*2p 1 The bond order of N2 in the ground
state is 3; in the first excited state it has a bond order of 2.
Owing to the reduction in bond order, N2 in the first excited
state would be more reactive, have a smaller bond energy and
a longer N ¬ N separation. 8.67 (a) C O (b) The bond
order of 3.0, predicted using Figure 8.36, agrees with the triple
bond in the Lewis structure. (c) According to Figure 8.36, the
highest energy electrons in CO would occupy a p2p molecular
orbital. If the highest energy electrons occupy a s-type orbital,
the order of p2p and s2p orbitals must be reversed. Figure 8.32
shows how the interaction of the 2s orbitals on one atom and
the 2p orbitals on the other atom can cause this reversal.
(d) The p2p bonding MOs of CO will have a greater
contribution from the more electronegative O atom, because
the 2p orbitals of O are lower in energy than the 2p orbitals of C
(compare Figure 8.36). 8.68 (a) HNO2 (b) O N O H
(c) The geometry around N is trigonal planar.
(d) sp2 hybridisation around N (e) three sigma, one pi
8.72 From bond enthalpies, ¢H = 5364 kJ; according to Hess’s
law, ¢H° = 5535 kJ. The difference in the two results, 171 kJ, is
due to the resonance stabilisation in benzene. The amount of
energy actually required to decompose 1 mol of C6H6(g) is greater
than the sum of the localised bond enthalpies. 8.74 (a) 3dz2
(b)
s*
3d
2px
2px
p2p
(c) s2p 6 p2p 6 p*2p 6 s*2p 8.43 (a) When comparing the same
two bonded atoms, bond order and bond energy are directly
related, while bond order and bond length are inversely related.
When comparing different bonded nuclei, there are no simple
relationships. (b) Be2 is not expected to exist; it has a bond order
of zero and is not energetically favoured over isolated Be atoms.
3dz2
3dz2
s3d
The ‘donuts’ of the 3dz2 orbitals have been omitted from the
diagram for clarity.
(c) A node is generated in s*3d because antibonding MOs are
formed when atomic orbital lobes with opposite phases
interact. Electron density is excluded from the internuclear
region and a node is formed in the MO.
A-12
Answers to Selected Exercises
(d)
s*3d
3dz2
3dz2
s3d
s*4s
4s
4s
s4s
(e) The bond order of Sc 2 is 1.0.
Chapter 9
9.1
(a) V2 5
3
(b) V2 V1
300 K, V1
500 K, V2
1
2
V1
1 bar, V1
2 bar, V2
9.4 Over time, the gases will mix perfectly. While the atoms will
continue to move from one bulb to the other, on average each
bulb will contain 4 blue and 3 red atoms. The ‘blue’ gas has the
greater partial pressure after mixing, because it has the greater
number of particles at the same T and V as the ‘red’ gas.
9.7 (a) Curve A is O2(g) and curve B is He(g). (b) For the same
gas at different temperatures, curve A represents the lower
temperature and curve B the higher temperature. 9.9 (a) A gas
is much less dense than a liquid. (b) A gas is much more
compressible than a liquid. (c) All mixtures of gases are
homogeneous. Similar liquid molecules form homogeneous
mixtures, while very dissimilar molecules form heterogeneous
mixtures. 9.11 10.3 m 9.12 (a) The tube can have any crosssectional area. (b) At equilibrium the force of gravity per unit
area acting on the mercury column at the level of the outside
mercury is equal to the atmospheric pressure. (c) The column of
mercury is held up by the pressure of the atmosphere applied to
the exterior pool of mercury. 9.13 (a) If V decreases by a factor
of 4, P increases by a factor of 4. (b) If T decreases by a factor of 2,
P decreases by a factor of 2. (c) If n decreases by a factor of 2, P
decreases by a factor of 2. 9.15 (a) If equal volumes of gases at
the same temperature and pressure contain equal numbers of
molecules and molecules react in the ratios of small whole
numbers, it follows that the volumes of reacting gases are in the
ratios of small whole numbers. (b) Since the two gases are at the
same temperature and pressure, the ratio of the numbers of
atoms is the same as the ratio of volumes. There are 1.5 times as
many Xe atoms as Ne atoms. 9.17 (a) PV = nRT; P in kPa, V in
litres, n in moles, T in kelvin. (b) An ideal gas exhibits pressure,
volume and temperature relationships described by the equation
PV nRT 9.19 Flask A contains the gas with M 30 g mol1,
and flask B contains the gas with M 60 g mol1
9.21 1.6 104 kg H2 9.22 5.72 * 1022 molecules
9.24 (a) 92 bar (b) 2.3 102 dm3 9.26 (a) 30.0 g Cl2
(b) 9.60 dm3 (c) 504 K (d) 1.92 bar 9.28 For gas samples
at the same conditions, molar mass determines density. Of
the three gases listed, (c) Cl2 has the largest molar mass.
9.30 (c) Because the helium atoms are of lower mass than the
average air molecule, the helium gas is less dense than air.
The balloon thus weighs less than the air displaced by its
volume. 9.32 (a) r 1.77 g dm3 (b) M 80.1 g mol1
9.33 3.47 * 10-6 g Mg 9.35 21.4 dm3 CO2 9.37 (a) When the
stopcock is opened, the volume occupied by N2(g) increases
from 2.0 dm3 to 5.0 dm3. PN2 0.4 bar (b) When the gases
ix, the volume of O2(g) increases from 3.0 dm3 to 5.0 dm3.
PO2 1.2 bar (c) Pt 1.6 bar 9.39 (a) PHe 1.90 bar,
PNe 1.11 bar, PAr 0.365 bar (b) Pt 3.38 bar
9.40 PCO2 0.308 bar, Pt 1.248 bar 9.42 PN2 0.99 bar,
PO2 0.39 bar, PCO2 0.20 bar 9.44 2.5 mole % O2
9.46 Pt 3.08 bar 9.47 (a) Increase in temperature at constant
volume or decrease in volume or increase in pressure
(b) decrease in temperature (c) increase in volume, decrease in
pressure (d) increase in temperature 9.49 (a) Vessel A has more
molecules. (b) The density of CO is 1.25 g dm3, and the density
of SO 2 is 1.33 g dm3.Vessel B has more mass. (c) The average
kinetic energy of the molecules in vessel B is higher.
(d) uA>uB = 1.46. The molecules in vessel A have the greater rms
speed. 9.51 (a) In order of increasing speed and decreasing
molar mass: HBr < NF3 < SO2 < CO < Ne (b) uNF3 324 m s1
9.52 The order of increasing rate of effusion is
2 37
H Cl 6 1H 37Cl 6 2H 35Cl 6 1H 35Cl. 9.54 As4S 6
9.56 (a) Non-ideal-gas behaviour is observed at very high
pressures and at low temperatures. (b) The real volumes of gas
molecules and attractive intermolecular forces between
molecules cause gases to behave non-ideally. (c) According to the
ideal-gas law, the ratio PV>RT should be constant for a given gas
sample at all combinations of pressure, volume and temperature.
If this ratio changes with increasing pressure, the gas sample is not
behaving ideally. 9.58 Ar (a = 1.34, b = 0.0322) will behave
more like an ideal gas than CO 2 (a = 3.59, b = 0.427) at high
pressures. 9.60 (a) P 0.929 bar (b) P 0.908 bar
9.63 742 balloons can be filled completely, with a bit of He
left over. 9.65 (a) 3.78 * 10-3 mol O 2 (b) 0.0345 g C8H 18
9.67 Oxygen is 70.1 mole % of the mixture. 9.69 (a) As a
gas is compressed at constant temperature, the number of
intermolecular collisions increases. Intermolecular attraction
causes some of these collisions to be inelastic, which amplifies
the deviation from ideal behaviour. (b) As the temperature of a
gas is increased at constant volume, a larger fraction of the
molecules has sufficient kinetic energy to overcome
intermolecular attractions and the effect of intermolecular
attraction becomes less significant. 9.73 (a) The partial pressure
of IF5 is 52.2 kPa. (b) The mole fraction of IF5 is 0.544.
Chapter 10
10.1 The diagram best describes a liquid. The particles are
close together, mostly touching, but there is no regular
arrangement or order. This rules out a gaseous sample, where
the particles are far apart, and a crystalline solid, which has a
regular repeating structure in all three directions. 10.3 (a) about
Answers to Selected Exercises
are weak, so the large surface tension of water draws the liquid
into the shape with the smallest surface area, a sphere.
10.25 (a) Freezing, exothermic (b) evaporation, endothermic
(c) deposition, exothermic 10.27 Melting does not require
separation of molecules, so the energy requirement is smaller
than for vaporisation, where molecules must be separated.
10.29 2.2 * 103 g H 2O 10.31 (a) The critical pressure is the
pressure required to cause liquefaction at the critical
temperature. (b) As the force of attraction between molecules
increases, the critical temperature of the compound increases.
(c) All the gases in Table 10.5 can be liquefied at the temperature
of liquid nitrogen, given sufficient pressure. 10.33 (a) No
effect (b) no effect (c) Vapour pressure decreases with
increasing intermolecular attractive forces because fewer
molecules have sufficient kinetic energy to overcome attractive
forces and escape to the vapour phase. (d) Vapour pressure
increases with increasing temperature because average kinetic
energies of molecules increase. 10.35 (a) The temperature of
the water in the two pans is the same. (b) Vapour pressure does
not depend on either volume or surface area of the liquid. At
the same temperature, the vapour pressures of water in the two
containers are the same. 10.37 (a) Approximately 17°C
(b) approximately 68 kPa (c) approximately 26°C
(d) approximately 105°C 10.38 (a) The critical point is the
temperature and pressure beyond which the gas and liquid
phases are indistinguishable. (b) The line that separates the
gas and liquid phases ends at the critical point because, at
conditions beyond the critical temperature and pressure, there
is no distinction between gas and liquid. In experimental
terms, a gas cannot be liquefied at temperatures higher
than the critical temperature, regardless of pressure.
10.40 (a) H2O(g) will condense to H2O(s) at approximately
0.6 kPa; at a higher pressure, somewhat above 101 kPa, H2O(s)
will melt to form H2O(l). (b) At 100°C and 0.50 bar, water is in
the vapour phase. As it cools, water vapour condenses to the
liquid at approximately 82°C, the temperature where the
vapour pressure of liquid water is 0.50 bar. Further cooling
results in freezing at approximately 0°C. The freezing point of
water increases with decreasing pressure, so at 0.50 bar the
freezing temperature is very slightly above 0°C.
10.43 (a)
critical point
119°C,
5040 kPa
50
Pressure bar (not to scale)
0.58 bar (b) 22°C (c) 47°C 10.6 (a) 3 Nb atoms, 3 O atoms
(b) NbO (c) This is primarily an ionic solid, because Nb is a
metal and O is a nonmetal. 10.7 (a) Solid 6 liquid 6 gas
(b) gas 6 liquid 6 solid 10.9 (a) Gases are more
compressible than liquids because there is much empty space
between gas particles. (b) The solid and liquid forms of a
substance are called condensed phases because in both states
there is very little space between particles; the volume of the
substance is condensed. (c) Liquids have greater ability to flow
than solids because their average kinetic energy is in the same
order of magnitude as the average attractive energy among
particles. 10.11 (a) London dispersion forces (b) dipoledipole forces (c) hydrogen bonding 10.12 (a) Nonpolar
covalent molecule; London dispersion forces only (b) polar
covalent molecule with O ¬ H bonds; hydrogen bonding,
dipole-dipole forces and London dispersion forces (c) polar
covalent molecule; dipole-dipole and London dispersion forces
(but not hydrogen bonding) 10.14 (a) Polarisability is the
ease with which the charge distribution in a molecule can be
distorted to produce a transient dipole. (b) Te is most
polarisable because its valence electrons are farthest from the
nucleus and least tightly held. (c) in order of increasing
polarisability: CH 4 6 SiH 4 6 SiCl4 6 GeCl4 6 GeBr4 (d) The
magnitudes of London dispersion forces and thus the boiling
points of molecules increase as polarisability increases. The
order of increasing boiling points is the order of increasing
polarisability given in (c). 10.16 (a) H 2S (b) CO 2 (c) SiH 4
10.17 (a) A bond between hydrogen and a small, highlyelectronegative atom, F, O or N. (b) CH 3NH 2 and CH 3OH.
10.18 (a) HF has the higher boiling point because it can
participate in hydrogen bonding while HCl cannot. (b) CHBr3
has the higher boiling point because it has the higher molar
mass, indicating greater polarisability and stronger dispersion
forces. (c) ICl has the higher boiling point because the
molecules have similar molar masses (hence similar dispersion
forces), but ICl is polar, giving it dipole-dipole forces that are
absent for the nonpolar Br2 . 10.19 (a) Hydrogen bonding
occurs in both water and ice. In liquid water, molecules are
close together and moving. Hydrogen bonds are constantly
broken and new ones formed. In ice, molecules adopt an
ordered structure with four hydrogen bonds per molecule.
These interactions produce a network structure with more
open space between molecules than in the liquid state and a
corresponding lower density for the solid than the liquid.
(b) Raising the temperature of a substance means increasing
the average kinetic energy of its molecules. A large amount of
heat must be added to water to break its strong hydrogen
bonds and increase its average kinetic energy and temperature.
10.21 (a) As temperature increases, the number of molecules
with sufficient kinetic energy to overcome intermolecular
attractive forces increases, and viscosity and surface tension
decrease. (b) The same attractive forces that cause surface
molecules to be difficult to separate (high surface tension)
cause molecules elsewhere in the sample to resist movement
relative to each other (high viscosity). 10.23 (a) CHBr3 has a
higher molar mass, is more polarisable and has stronger
dispersion forces, so the surface tension is greater. (b) As
temperature increases, the viscosity of the oil decreases
because the average kinetic energy of the molecules increases.
(c) Adhesive forces between polar water and nonpolar car wax
A-13
solid
1
normal
melting
point
218°C
0
230
liquid
gas
normal boiling point
183°C
triple point
219°C, 0.152 kPa
190
150
110
Temperature, °C
(b) O2(s) is denser than O2(l) because the solid liquid line on
the phase diagram is normal. (c) at 1 bar, solid O2 will melt as
it is heated. 10.44 In a crystalline solid the component
particles are arranged in an ordered repeating pattern. In an
amorphous solid there is no orderly structure. Glass is an
example of an amorphous solid. 10.46 CaTiO 3
A-14
Answers to Selected Exercises
10.48 Atomic mass 55.8 g mol1 10.50 (a) 12 (b) 6 (c) 8
10.52 a = 613 pm 10.54 (a) The U 4+ ions in UO2 are
represented by the smaller spheres in Figure 10.40(c).
The ratio of large spheres to small spheres matches the
O 2- to U 4+ ratio in the chemical formula, so the small
ones must represent U 4+. (b) density = 10.97 g cm3
10.56 (a) Hydrogen bonding, dipole-dipole forces, London
dispersion forces (b) covalent chemical bonds (c) ionic
bonds (d) metallic bonds 10.58 In molecular solids
relatively weak intermolecular forces bind the molecules in
the lattice, so relatively little energy is required to disrupt
these forces. In covalent-network solids, covalent bonds join
atoms into an extended network. Melting or deforming a
covalent-network solid means breaking covalent bonds,
which requires a large amount of energy. 10.60 Because
of its relatively high melting point and ability to conduct
electricity in solution, the solid must be ionic. 10.63 (a) In
CH 2Br2 the dispersion force will be more influential than
the dipole-dipole force. Relative to CH 2Cl2 , Br is more
polarisable than Cl and the dipole moment of CH 2Br2 is
smaller. (b) Dipole-dipole forces will play a larger role
for CH 2F2 than for CH 2Cl2 because CH 2F2 has a larger
dipole moment and F is less polarisable than Cl.
10.65 (a) Evaporation is an endothermic process. The heat
required to vaporise sweat is absorbed from your body,
helping to keep it cool. (b) The vacuum pump reduces the
pressure of the atmosphere above the water until
atmospheric pressure equals the vapour pressure of water
and the water boils. Boiling is an endothermic process, and
the temperature drops if the system is not able to absorb heat
from the surroundings fast enough. As the temperature of
the water decreases, the water freezes. 10.69 The large
difference in melting points is due to the very different forces
imposing atomic order in the solid state. Much more kinetic
energy is required to disrupt the delocalised metallic
bonding in gold than to overcome the relatively weak
London dispersion forces in Xe. 10.71 The most effective
diffraction occurs when distances between layers of atoms in
the crystal are similar to the wavelength of the light being
diffracted. Molybdenum X-rays of 0.71 pm are in the same
order of magnitude as interlayer distances in crystals and are
diffracted. Visible light, 400–700 nm or 4000–7000 pm, is too
long to be diffracted effectively. 10.73 16 Al atoms, 8 Mg
atoms, 32 O atoms
10.75
Br
CH2
CH3
CH2
CH3 CH3
(i) M 44
CH2
CH
CH3 CH3
CH2
CH3
(iii) M 123
(ii) M 72
O
C
CH3
CH2
CH3
(iv) M 58
CH3
Br
CH2
(v) M 123
CH2
CH3
OH
CH2
(vi) M 60
(a) Molar mass: Compounds (i) and (ii) have similar rod-like
structures. The longer chain in (ii) leads to stronger London
dispersion forces and higher heat of vaporisation. (b) Molecular
shape: Compounds (iii) and (v) have the same chemical formula
and molar mass but different molecular shapes. The more
rod-like shape of (v) leads to more contact between molecules,
stronger dispersion forces and higher heat of vaporisation.
(c) Molecular polarity: Compound (iv) has a smaller molar mass
than (ii) but a larger heat of vaporisation, which is probably due
to the presence of dipole-dipole forces. (d) Hydrogen bonding
interactions: Molecules (v) and (vi) have similar structures.
Even though (v) has larger molar mass and dispersion forces,
hydrogen bonding causes (vi) to have the higher heat of
vaporisation. 10.79 P(benzene vapour) 13.2 kPa.
Chapter 11
11.1 The ion-solvent interaction should be greater for Li +. The
smaller ionic radius of Li + means that the ion-dipole interactions
with polar water molecules are stronger. 11.9 If the magnitude
of ¢H3 is small relative to the magnitude of ¢H1 , ΔsolnH will
be large and endothermic (energetically unfavourable) and
not much solute will dissolve. 11.12 (a) ¢H1 (b) ¢H3
11.13 (a) Since the solute and solvent experience very similar
London dispersion forces, the energy required to separate them
individually and the energy released when they are mixed are
approximately equal. ¢H1 + ¢H2 L - ¢H3 . Thus, ΔsolnH is
nearly zero. (b) Since no strong intermolecular forces prevent the
molecules from mixing, they do so spontaneously because of
the increase in entropy. 11.15 (a) Supersaturated (b) Add a
seed crystal. A seed crystal provides an interface at which
crystallisation can occur, so the dissolved particles do not have
to form this energetically unfavourable structure themselves.
11.17 (a) Unsaturated (b) saturated (c) saturated (d) unsaturated
11.18 The liquids water and glycerol form homogeneous
mixtures (solutions) regardless of the relative amounts of the two
components. The ¬ OH groups of glycerol facilitate strong
hydrogen bonding similar to that in water; like dissolves like.
11.22 (a) A sealed container is required to maintain a partial
pressure of CO2(g) greater than 1 bar above the beverage.
(b) Since the solubility of gases increases with decreasing
temperature, more CO2(g) will remain dissolved in the beverage
if it is kept cool. 11.24 SHe 5.6 10–4 M, SN2 9.0 10–4 M
11.26 (a) 2.15% Na 2SO4 by mass (b) 2.86 ppm Ag
11.27 (a) XCH3OH 0.0427 (b) 7.35% CH3OH by mass (c) 2.48 m
CH 3OH 11.29 (a) 1.46 10–2 M Mg(NO 3)2 (b) 1.12 M
LiClO 4 # 3 H 2O (c) 0.350 M HNO3 11.30 (a) 4.70 m C6H 6
(b) 0.235 m NaCl 11.32 (a) 43.01% H 2SO 4 by mass
(b) XH2SO4 = 0.122 (c) 7.69 m H 2SO 4 (d) 5.827 M H 2SO 4
11.34 (a) 0.150 mol SrBr2 (b) 1.53 * 10-2 mol KCl
(c) 4.44 * 10-2 mol C6H 12O 6 11.35 (a) Weigh out 1.3 g KBr,
dissolve in water, dilute with stirring to 0.75 dm3. (b) Weigh
out 2.62 g KBr, dissolve it in 122.38 g H 2O to make exactly 125
g of 0.180 m solution. (c) Weigh 244 g KBr and dissolve in
enough H 2O to make 1.85 L solution. (d) Weigh 10.1 g KBr,
dissolve it in a small amount of water and dilute to 0.568 L.
11.37 (a) 3.82 m Zn (b) 26.8 M Zn 11.38 4.6 kPa CO2;
1.8 10–3 M CO2 11.40 Freezing-point depression, ΔTf Kfm;
boiling-point elevation, ΔTb Kbm; osmotic pressure, MRT;
vapour pressure lowering, PA XAP°A 11.42 (a) Sucrose has
a greater molar mass than glucose, so the sucrose solution will
contain fewer solute particles and have a higher vapour
pressure. 11.45 (a) XEth = 0.2812 (b) Psoln 31.7 kPa
(c) XEth in vapour = 0.472 11.46 (a) Because NaCl is a strong
electrolyte, one mole of NaCl produces twice as many
A-15
Answers to Selected Exercises
dissolved particles as one mole of the molecular solute
C6H 12O6 . Boiling-point elevation is directly related to total
moles of dissolved particles, so 0.10 m NaCl has the higher
boiling point. (b) 0.10 m NaCl: ΔTb 0.103°C, Tb 100.1°C;
0.10 m C6H12O6: ΔTb 0.051°C, Tb 100.1°C (c) Attractive
forces between water and ions in solution result in non-ideal
behaviour. 11.51 Π 1.69 kPa 11.54 Molar mass of
lysozyme 1.39 104 g 11.56 (a) In the gaseous state,
particles are far apart and intermolecular attractive forces are
small. When two gases combine, all terms in Equation 11.1 are
essentially zero and the mixture is always homogeneous.
(b) To determine whether Faraday’s dispersion is a true
solution or a colloid, shine a beam of light on it. If light is
scattered, the dispersion is a colloid. 11.58 (a) Hydrophobic
(b) hydrophilic (c) hydrophobic (d) hydrophobic. 11.60 When
electrolytes are added to a suspension of proteins, dissolved ions
form ion pairs with the protein surface charges, effectively
neutralising them. The repulsion between the protein molecules
is diminished and they coagulate, so the colloid separates into
a protein layer and a water layer. 11.63 (a) 1.2 10–4 M O2
(b) PO2 7 kPa 11.65 136 ppm K + 11.67 (a) 0.16 m Na
(b) 2.0 M Na (c) Clearly, molality and molarity are not the same
for this amalgam. Only in the instance that 1 kg solvent and
the mass of 1 litre of solution are nearly equal do the two
concentration units have similar values. 11.68 For Hg(NO 3)2 ,
the theoretical molality assuming it dissociates completely
into ions is 9.24 102 mol kg1, and the experimental molality
is only a little less, 8.71 102 mol kg1. For HgCl2, the
theoretical molality assuming it dissociates completely into ions
is 11.05 102 mol kg1, but the experimental molality is only
3.68 102 mol kg1. Hg(NO3)2 is the better electrolyte;
HgCl2 is behaving more like a molecular substance.
11.70 (a) CF4 , 1.7 * 10-4 m; CClF3 , 9 * 10-4 m; CCl2F2 ,
2.3 * 10-3 m; CHClF2 , 3.5 * 10-2 m (b) Molality and molarity
are numerically similar when kilograms solvent and litres
solution are nearly equal. This is true when solutions are dilute
and when the density of the solvent is nearly 1 g cm3, as in
this problem. (c) Water is a polar solvent; the solubility of
solutes increases as their polarity increases. Nonpolar CF4 has
the lowest solubility and the most polar fluorocarbon, CHClF2 ,
has the greatest solubility in H 2O. (d) The Henry’s law
constant for CHClF2 is 3.5 10–4 mol dm–3 kPa–1. This value is
greater than the Henry’s law constant for N2(g) because N2(g)
is nonpolar and of lower molecular mass than CHClF2 .
11.74 The process is spontaneous with no significant change in
enthalpy, so we suspect that there is an increase in entropy.
In general, dilute solutions of the same solute have greater
entropy than concentrated ones, because the solute particles
are more free to move about the solution. In the limiting case
that the more dilute solution is pure solvent, there is a definite
increase in entropy as the concentrated solution is diluted. In
Figure 11.18, there may be some entropy decrease as the dilute
solution loses solvent, but this is more than offset by the
entropy increase that accompanies dilution. There is a net
increase in entropy.
Chapter 12
12.1 (a) Product (b) The average rate of reaction is greater
between points 1 and 2 because they are earlier in the
reaction when the concentrations of reactants are greater.
12.3 (a) x = 0 (b) x = 2 (c) x = 3 12.7 There is one
intermediate, B, and two transition states, A ¡ B and
B ¡ C. The B ¡ C step is faster, and the overall reaction
is exothermic. 12.10 (a) Reaction rate is the change in the
amount of products or reactants in a given amount of
time. (b) Rates depend on the concentration of reactants, the
temperature and presence or absence of catalyst. (c) The
stoichiometry of the reaction (mole ratios of reactants and
products) must be known to relate rate of disappearance of
reactants to rate of appearance of products.
12.12
Time
Mol A (a) Mol B [A]
≤ [A]
(b) Rate
0.065
0.051
0.042
0.036
0.031
- 0.14
- 0.09
- 0.06
- 0.05
2.3
1.5
1.0
0.8
(mol dmⴚ3) (mol dmⴚ3)
(min)
0
10
20
30
40
0.000
0.014
0.023
0.029
0.034
0.65
0.51
0.42
0.36
0.31
(M sⴚ1)
*
*
*
*
10-4
10-4
10-4
10-4
(c) Δ[B]avg/Δt 1.3 10–4 M sⴚ1 12.15 (a) –Δ[O2]/Δt 0.43 mol s–1; ¢[H2O]>¢t = 0.85 mol s1 (b) Ptotal decreases by
12 Pa min–1 12.17 (a) If [A] doubles, the rate will increase
by a factor of four. The rate constant will not change, as it is a
constant at a particular temperature. (b) The reaction is second
order in A, first order in B and third order overall. (c) The
units of k will depend on the unit used for concentration in this
case. If concentration is measured in M, the units of k will be
M–1s–1 12.18 (a) Rate k[N2O5] (b) Rate 1.16 10–4 M
s1 (c) When the concentration of N2O5 doubles, the rate of
the reaction doubles. 12.19 (a, b) k 1.7 102 M–1s–1 (c) If
[OH–] is tripled, the rate triples. 12.21 (a) Rate k[OCl–][I–]
(b) k 60 M–1s–1 (c) Rate 6.0 10–5 M s–1 12.22 (a) Rate k[BF3][NH3] (b) The reaction is second order overall. (c) kavg 3.41 M–1s–1 (d) 0.170 M s–1 12.23 (a) Rate k[NO]2[Br2]
(b) kavg 1.2 104 M–1s–1 (c) 21 ¢[NOBr]>¢t = - ¢[Br2]>¢t
(d) –Δ[Br2]/Δt 8.4 M s–1 12.25 (a) [A]0 is the molar
concentration of reactant A at time 0. [A]t is the molar
concentration of reactant A at time t. t1>2 is the time required
to reduce [A]0 by a factor of 2. k is the rate constant for a
particular reaction. (b) A graph of ln[A] versus time
yields a straight line for a first-order reaction.
12.27 (a) k = 3.0 * 10-6 s -1 (b) t1>2 = 3.2 * 104 s
12.29 (a) P 20 Pa (b) t = 51 s 12.31 (a) The plot of 1>[A]
versus time is linear, so the reaction is second order in [A].
(b) k 0.040 M–1min–1 (c) t1>2 = 38 min 12.33 (a) The energy
of the collision and the orientation of the molecules when they
collide determine whether a reaction will occur. (b) At a higher
temperature, there are more total collisions and each collision
is more energetic. 12.35 f = 4.94 * 10-2. At 400 K
approximately 1 out of 20 molecules has this kinetic energy.
12.37 (a)
Ea 7 kJ
E 66 kJ
A-16
Answers to Selected Exercises
(b) Ea(reverse) 73 kJ 12.38 Reaction (b) is fastest and
reaction (c) is slowest. 12.40 (a) k = 1.1 s -1 (b) k = 4.9 s -1
12.42 A plot of ln k versus 1>T has a slope of -5.64 * 103;
Ea –R(slope) 46.9 kJ mol–1. 12.43 (a) An elementary
reaction is a chemical process that occurs in a single step.
(b) A unimolecular elementary reaction involves only one
reactant molecule; a bimolecular elementary reaction involves
two reactant molecules. (c) A reaction mechanism is a series of
elementary reactions that describe how an overall reaction
occurs and explain the experimentally determined rate law.
12.45 (a) Unimolecular, rate = k[Cl2] (b) bimolecular,
rate = k[OCl -][H 2O] (c) bimolecular, rate = k[NO][Cl2]
12.46 (a) Two intermediates, B and C. (b) three transition
states (c) C ¡ D is fastest. (d) endothermic (e) B ¡ C
is the rate-limiting step. 12.47 (a) H2(g) + 2 ICl(g) ¡
I2(g) + 2 HCl(g) (b) HI is the intermediate. (c) first step:
rate = k1[H 2][ICl], second step: rate = k2[HI][ICl] (d) If the
first step is slow, the observed rate law is rate = k[H 2][ICl].
12.49 (a) Rate = k[NO][Cl2] (b) The second step must be slow
relative to the first step. 12.51 (a) A catalyst increases the rate
of reaction by decreasing the activation energy, Ea, or
increasing the frequency factor, A. (b) A homogeneous catalyst
is in the same phase as the reactants, while a hetereogeneous
catalyst is in a different phase. 12.53 (a) Multiply the
coefficients in the first reaction by 2 and sum. (b) NO2(g)
is a catalyst because it is consumed and then reproduced in
the reaction sequence. (c) This is a homogeneous catalyst.
12.55 Use of chemically stable supports makes it possible to
obtain very large surface areas per unit mass of the precious
metal catalyst because the metal can be deposited in a very
thin, even monomolecular, layer on the surface of the support.
12.57 (a) The catalysed reaction is approximately 1.0 107
times faster at 25°C. (b) The catalysed reaction is 1.8 105
times faster at 125°C. 12.59 Δ[Cl–]/Δt 7.0 10–7 M s–1
12.61 (a) Rate k[HgCl2][C2O42–]2 (b) kavg 8.6 10–3 M–2s–1
(c) rate 4.3 10–6 M s–1 12.62 (a) k = 4.28 * 10-4 s -1
(b) [urea] 0.90 M (c) t1>2 = 1.62 * 103 s
12.65 (a) Cl2(g) + CHCl3(g) ¡ HCl(g) + CCl4(g)
(b) Cl(g), CCl3(g) (c) reaction 1, unimolecular; reaction 2,
bimolecular; reaction 3, bimolecular (d) Reaction 2, the slow step,
is rate determining. (e) Rate = k[CHCl3][Cl2]1>2 12.66 Partial
pressure of O2 40.7 kPa 12.68 (a) Use an open-end manometer,
a clock, a ruler and a constant-temperature bath. Load the flask
with HCl(aq), and read the height of the Hg in both arms of the
manometer. Quickly add Zn(s) to the flask, and record time = 0
when the Zn(s) contacts the acid. Record the height of the Hg
in one arm of the manometer at convenient time intervals such
as 5 s. Calculate the pressure of H2(g) at each time. Since
P = 1n>V2RT, ¢P>¢t at constant temperature is an acceptable
measure of reaction rate. (b) Keep the amount of Zn(s) constant,
and vary the concentration of HCl(aq) to determine the reaction
order for HCl(aq). Keep the concentration of HCl(aq) constant,
and vary the amount of Zn(s) to determine the order for Zn(s).
Combine this information to write the rate law.
(c) - ¢[H +]> ¢t = 2 ¢[H 2]> ¢t; the rate of disappearance of H +
is twice the rate of appearance of H2(g). [H 2] = P>RT (d) By
changing the temperature of the constant-temperature bath,
measure the rate data at several temperatures and calculate the
rate constant k at these temperatures. Plot ln k versus 1>T; the
slope of the line is –Ea/R. (e) Measure rate data at constant
temperature, HCl concentration and mass of Zn(s), varying only
the form of the Zn(s). Compare the rate of reaction for metal strips
and granules.
Chapter 13
13.1 (a) kf > kr (b) The equilibrium constant is greater than 1.
13.3 (a) A2 + B ¡ A2B (b) Kc = [A2B]/[A2][B] (c) Δn = –1
(d) Kp = Kc(RT)Δn 13.6 (a) Kc = 2 (b) increase
13.8 (a) Kp = Kc = 1.2 * 10-2 (b) Since kf < kr, in order for
the two rates to be equal, [A] must be greater than [B], and the
partial pressure of A is greater than the partial pressure of B.
13.10 (a) The law of mass action expresses the relationship between
the concentrations of reactants and products at equilibrium for
any reaction. Kc = [NOBr2]>[NO][Br2] (b) The equilibriumconstant expression is an algebraic equation where the variables are
the equilibrium concentrations of the reactants and products for a
specific chemical reaction. The equilibrium constant is a number;
it is the ratio calculated from the equilibrium expression for a
particular chemical reaction. (c) Introduce a known quantity of
NOBr2(g) into a vessel of known volume at constant (known)
temperature. After equilibrium has been established, measure the
total pressure in the flask. Using an equilibrium table, calculate
equilibrium pressures, concentrations and the value of Kc.
13.12 (a) Kc [N2O][NO2]/[NO]3; homogeneous
(b) Kc [CS2][H2]/[CH4][H2S]2; homogeneous
(c) Kc [CO]4/[Ni(CO)4]; heterogeneous
(d) Kc [H+][F–]/[HF]; homogeneous
(e) Kc [Ag+]2/[Zn2+] heterogeneous 13.14 Kp 1.0 10–3
13.16 (a) Kc 77 13.20 (a) Kp PO2 (b) Equilibrium-constant
expressions only contain terms for reactants and products
whose concentration can change during the reaction, and the
concentration of pure solids and liquids remains constant.
13.21 Kc 0.0184 13.24 (a) [H2] 0.012 M,
[N2] 0.019 M, [H2O] 0.138 M (b) Kc 653.7 7 102
13.26 (a) PCO2 4.15 bar, PH2 2.08 bar, PH2O 3.33 bar
(b) PCO2 3.97 bar, PH2 1.89 bar, PCO 0.18 bar
(c) Kp 0.086 13.27 (a) A reaction quotient is the result of a
general set of concentrations whereas the equilibrium constant
requires equilibrium concentrations. (b) to the right (c) The
concentrations used to calculate Q must be equilibrium
concentrations. 13.29 (a) Q = 1.1 * 10-8, the reaction will
proceed to the left. (b) Q = 5.5 * 10-12, the reaction will
proceed to the right. (c) Q 2.19 10–10, the mixture is at
equilibrium. 13.31 (a) [Br2] 0.00767 M, [Br2] 0.00282 M,
0.0225 g Br(g) (b) [H2] 0.014 M, [I2] 0.00859 M,
[HI] 0.081 M, 21 g HI(g) 13.32 [NO] 0.002 M,
[N2] [O2] 0.099 M 13.34 The equilibrium pressure of
Br2(g) is 0.416 bar. 13.36 [IBr] 0.447 M, [I2] [Br2] 0.0267 M
13.37 (a) Shift equilibrium to the right (b) decrease the value
of K (c) shift equilibrium to the left (d) no effect (e) no effect
(f) shift equilibrium to the right 13.39 (a) No effect (b) no
effect (c) increase equilibrium constant (d) no effect
13.41 (a) Kc Kp 1.5 10–39 (b) Reactants are much more
plentiful than products at equilibrium. 13.43 Kc 0.71
13.48 (a) Q = 1.7; Q > Kp; the reaction will shift to the left.
(b) Q = 1.7; Q > K p ; the reaction will shift to the left.
(c) Q = 5.8 * 10-3; Q < Kp; the reaction will shift to the right.
13.49 The maximum value of Q is 4.1, which is less than Kp;
reduction will occur. 13.52 PH2 = PI2 = 1.50 bar; PHI = 10.35 bar
13.54 The patent claim is false. A catalyst does not alter the
Answers to Selected Exercises
position of equilibrium in a system, only the rate of approach to
the equilibrium condition. 13.55 (a) At equilibrium, the forward
and reverse reactions occur at equal rates. (b) Reactants are
favoured at equilibrium. (c) A catalyst lowers the activation
energy for both the forward and reverse reactions. (d) The ratio of
the rate constants remains unchanged. (e) The reaction is
endothermic, so the value of K will increase with increasing
temperature. 13.57 At 850°C, Kp 14.1; at 950°C, Kp 73.8;
at 1050°C, Kp 2.7 102; at 1200°C, Kp 1.7 103. Because K
increases with increasing temperature, the reaction is
endothermic.
Chapter 14
14.1 (a) HX, the H + donor, is the Brønsted–Lowry acid. NH 3 , the
H + acceptor, is the Brønsted–Lowry base. (b) HX, the electron
pair acceptor, is the Lewis acid. NH 3 , the electron pair donor, is
the Lewis base. 14.4 (a) The order of base strength is the reverse
of the order of acid strength; the diagram of the acid with the
most free H + has the weakest conjugate base Y -. The order of
increasing basicity is: Y - 6 Z - 6 X -. (b) The strongest base, X -,
has the largest Kb value. 14.7 (a) Both molecules are oxyacids;
the ionisable H atom is attached to O. The right molecule is a
carboxylic acid; the ionisable H is part of a carboxyl group.
(b) Increasing the electronegativity of X increases the strength of
both acids. As X becomes more electronegative and attracts more
electron density, the O ¬ H bond becomes weaker, more polar,
and more likely to be ionised. An electronegative X group also
stabilises the anionic conjugate base, causing the ionisation
equilibrium to favour products and the value of Ka to increase.
14.10 Solutions of HCl and H 2SO4 conduct electricity, taste sour,
turn litmus paper red (are acidic), neutralise solutions of bases
and react with active metals to form H2(g). HCl and H 2SO 4
solutions have these properties in common because both
compounds are strong acids. That is, they both dissociate
completely in H 2O to form H+(aq) and an anion. (HSO 4 - is
not completely dissociated, but the first dissociation step for
H 2SO4 is complete.) The presence of ions enables the solutions to
conduct electricity; the presence of H+(aq) in excess of 1 10–7 M
accounts for all other properties listed. 14.12 (a) The Arrhenius
definition of an acid is a substance that increases the concentration
of H ions when dissolved in water; the Brønsted–Lowry
definition is not restricted to water, and does not requre that H
ions be generated. (b) HCl is the Brønsted–Lowry acid; NH 3 is
the Brønsted–Lowry base. 14.14 (a) IO 3 - (b) NH 3 (c) HPO 4 2(d) C6H5COO 14.16
Acid
+
(a) NH4 +(aq)
(b) H2O(l)
(c) HCOOH(aq)
Base
Δ
CN-(aq)
(CH3)3N(aq)
PO4 3-(aq)
Conjugate
Acid
+
HCN(aq)
(CH3)3NH+(aq)
HPO4 2-(aq)
Conjugate
Base
NH3(aq)
OH-(aq)
HCOO -(aq)
14.17 (a) Acid: HC2O4 -(aq) + H2O(l) Δ C2O4 2-(aq) +
H3O+(aq); Base: HC2O4 -(aq) + H2O(l) Δ H2C2O4(aq) +
OH-(aq) (b) H 2C2O 4 is the conjugate acid of HC2O 4 -. C2O 4 2- is
the conjugate base of HC2O 4 -. 14.19 (a) HBr. It is one of the
seven strong acids. (b) F -. HCl is a stronger acid than HF, so F - is
the stronger conjugate base. 14.22 (a) Autoionisation is the
ionisation of a neutral molecule into an anion and a cation.
A-17
The equilibrium expression for the autoionisation of water is
H2O(l) Δ H+(aq) + OH-(aq). (b) Pure water is a poor
conductor of electricity because it contains very few ions.
(c) If a solution is acidic, it contains more H + than OH -.
14.24 (a) [H+] 2.2 10–11 M, basic (b) [H+] 1.1 10–6 M, acidic
(c) [H+] 1.0 10–8 M, basic 14.26 [H+] [OH–] 3.5 10–8 M
14.27 (a) [H +] changes by a factor of 100. (b) [H +] changes by a
factor of 3.2 14.29 (a) [H+] decreases, pH increases (b) The
concentration of H+ will be between 105 and 106 M at a pH of
5.2, and the concentration of OH between 109 and 108 M.
By calculation, [H] 105.2 6 106 M.
[OH] 1 1014 / 6 106 2 109 M.
14.30
[Hⴙ]
[OHⴚ]
pH
pOH
Acidic or
Basic
7.5 10–3 M
2.8 10–5 M
5.6 10–9 M
5.0 10–9 M
1.3 10–12 M
3.6 10–10 M
1.8 10–6 M
2.0 10–6 M
2.12
4.56
8.25
8.30
11.88
9.44
5.75
5.70
acidic
acidic
basic
basic
14.32 [H+] 4.0 10–8 M, [OH–] 6.0 10–7 M, pOH = 6.22
14.33 (a) A strong acid is completely dissociated into ions
in aqueous solution. (b) [H+] 0.500 M (c) HCl, HBr, HI
14.35 (a) [H+] 8.5 10–3 M, pH 2.07 (b) [H+] 0.419 M,
pH 1.377 (c) [H+] 0.0250 M, pH 1.602 (d) [H+] 0.167 M,
pH 0.778 14.36 (a) [OH–] 3.0 10–3 M, pH 11.48
(b) [OH–] 0.3758 M, pH 13.5750 (c) [OH–] 8.75 10–5 M,
pH 9.942 (d) [OH–] 0.17 M, pH 13.23 14.37 3.2 10–3 M
NaOH 14.39 pH 13.400 14.41 (a) HBrO2(aq) Δ
H(aq) BrO2(aq), Ka [H][BrO2]/[HBrO2];
[HBrO2(aq) H2O(l) Δ H3O(aq) BrO2(aq),
Ka [H3O][BrO2]/[HbrO2] (b) CH3CH2COOH(aq) Δ
H(aq) CH3CH2COO(aq), Ka [H][CH3CH2COO]/
[CH3CH2COOH]; CH3CH2COOH(aq)H2O(l) Δ
H3O(aq) CH3CH2COO(aq), Ka £H3O+][CH3CH2COO–]/
[CH3CH2COOH] 14.42 Ka 1.4 10–4
14.44 [H+] [ClCH2COO–] 0.0110 M, [ClCH2COOH] 0.089 M, Ka 1.4 10–3. 14.46 [H+] [C6H5COO–] 1.8 10–3 M, [C6H5COOH] 0.048 M 14.47 (a) [H+] 1.1 10–3 M, pH 2.95 (b) [H+] 1.7 10–4 M, pH 3.76
(c) [OH–] 1.4 10–5 M, pH 9.15 14.48 [H+] 2.0 10–2 M,
pH 1.71 14.50 (a) [H+] 2.8 10–3 M, 0.69% ionisation
(b) [H+] 1.4 10–3 M, 1.4% ionisation (c) [H+] 8.7 10–4 M,
2.2% ionisation 14.51 HX(aq) Δ H+(aq) X–(aq);
Ka [H+][X–]/[HX]. Assume that the percent of acid that
ionises is small. Let [H+] [X–] y. Ka y2/[HX];
y Ka1/2[HX]1/2. Percent ionisation y/[HX] 100.
Substituting for y, percent ionisation 100 Ka1/2[HX]1/2/[HX]
or 100 Ka1/2/[HX]1/2. That is, percent ionisation varies
inversely as the square root of the concentration of HX.
14.52 [H+] 5.72 10–3 M, pH 2.24, [C6H5O73–] 1.2 10–9 M. The approximation that the first ionisation is less
than 5% of the total acid concentration is not valid; the
quadratic equation must be solved. The [H +] produced from
the second and third ionisations is small with respect to that
present from the first step; the second and third ionisations can
be neglected when calculating the [H +] and pH. 14.54 All
Brønsted–Lowry bases contain at least one unshared (lone)
pair of electrons to attract H +. 14.57 (a) (CH3)2NH(aq) +
A-18
Answers to Selected Exercises
H2O(l) Δ (CH3)2NH2 +(aq) + OH-(aq); Kb = [(CH 3)2NH 2 +]
[OH -]>[(CH 3)2NH] (b) CO3 2-(aq) + H2O(l) Δ HCO3 -(aq)
+ OH-(aq); Kb = [HCO3 -][OH–]/[CO32–]
(c) HCOO -(aq) + H2O(l) Δ HCOOH(aq) + OH-(aq);
Kb [HCOOH][OH–]/[HCOO–] 14.58 From the quadratic
formula, [OH–] 6.6 10–3 M, pH 11.82.
14.59 (a) [C10H15ON] 0.033 M, [C10H15ONH+] [OH–] 2.1 10–3 M (b) Kb 1.4 10–4 14.60 (a) For a
conjugate acid/conjugate base pair such as C6H 5OH>C6H 5O -,
Kb for the conjugate base can always be calculated from Ka for
the conjugate acid, so a separate list of Kb values is not
necessary. (b) Kb 7.7 10–5 (c) Phenolate is a stronger base
than NH3. 14.63 (a) [OH–] 1.4 10–3 M, pH 11.15
(b) [OH–] 3.8 10–3 M, pH 11.58 (c) [NO2–] 0.50 M,
[OH–] 3.3 10–6 M, pH 8.52 14.64 (a) As the
electronegativity of the central atom (X) increases, the strength
of the oxyacid increases. (b) As the number of nonprotonated
oxygen atoms in the molecule increases, the strength of the
oxyacid increases. 14.66 (a) HNO 3 is a stronger acid because
it has one more nonprotonated oxygen atom and thus a higher
oxidation number on N. (b) For binary hydrides, acid strength
increases going down a family, so H 2S is a stronger acid than
H 2O. (c) H 2SO4 is a stronger acid because H + is much more
tightly held by the anion HSO 4 -. (d) For oxyacids, the greater
the electronegativity of the central atom, the stronger the acid,
so H 2SO 4 is the stronger acid. (e) CCl3COOH is stronger
because the electronegative Cl atoms withdraw electron
density from other parts of the molecule, which weakens the
O ¬ H bond and stabilises the anionic conjugate base. Both
effects favour increased ionisation and acid strength.
14.67 (a) BrO - (b) BrO - (c) HPO4 2- 14.68 (a) True (b) False.
In a series of acids that have the same central atom, acid
strength increases with the number of nonprotonated oxygen
atoms bonded to the central atom. (c) False. H 2Te is a stronger
acid than H 2S because the H ¬ Te bond is longer, weaker and
more easily ionised than the H ¬ S bond. 14.70 Yes. The
Arrhenius definition of a base, an OH-(aq) donor, is most
restrictive; the Brønsted definition, an H + acceptor, is more
general; and the Lewis definition, an electron-pair donor, is
most general. Any substance that fits the narrow Arrhenius
definition will fit the broader Brønsted and Lewis definitions.
14.72 (a) Acid, Fe(ClO 4)3 or Fe 3+; base, H 2O (b) Acid, H 2O;
base, CN - (c) Acid, BF3; base, (CH 3)3N (d) Acid, HIO; base,
NH 2 - 14.73 (a) Cu2+, higher cation charge (b) Fe 3+, higher
cation charge (c) Al3+, smaller cation radius, same charge
14.75 (a) Kw is the equilibrium constant for the reaction of two
water molecules to form hydronium ion and hydroxide ion.
(b) Ka is the equilibrium constant for the reaction of any acid,
neutral or ionic, with water to form hydronium ion and the
conjugate base of the acid. (c) pOH is the negative log of
hydroxide ion concentration; pOH decreases as hydroxide ion
concentration increases. 14.77 (a) A higher O 2 concentration
displaces protons from HbH +, producing a more acidic solution
with lower pH in the lungs than in the tissues. (b) basic
(c) A high [H +] shifts the equilibrium toward the proton-bound
form HbH +. This results in a lower concentration of HbO 2 in
the blood and impedes the ability of haemoglobin to transport
oxygen. 14.80 Ka 1.4 10–5 14.82 [H+] 0.010 M,
[H2PO4+] 0.010 M, [HPO42–] 6.2 10–8 M, [PO43–] 2.6 10–18 M 14.84 6.0 * 1013 H + ions 14.86 (a) To the
precision of the reported data, the pH of rainwater 40 years
ago was 5.4, no different from the pH today. With extra
significant figures, [H+] 3.63 10–6 M, pH = 5.440.
(b) A 20.0 dm3 bucket of today’s rainwater contains 0.02 dm3
(with extra significant figures, 0.0200 dm3) of dissolved CO 2 .
Chapter 15
15.1 The middle box has the highest pH. For equal amounts
of acid HX, the greater the amount of conjugate base X -, the
smaller the amount of H + and the higher the pH.
15.2 (a) Drawing 3 (b) Drawing 1 (c) Drawing 2
15.7 (a) The extent of ionisation of a weak electrolyte is
decreased when a strong electrolyte containing an ion in
common with the weak electrolyte is added to it. (b) NaNO 2
15.9 (a) pH increases (b) pH decreases (c) pH increases
(d) no change (e) pH decreases 15.10 (a) [H+] 1.8 10–6 M,
pH 4.73 (b) [OH–] 4.8 10–5 M, pH 9.68
(c) [H+] 1.4 10–5 M, pH 4.87 15.12 In a mixture of
CH3COOH and CH3COONa, CH3COOH reacts with added
base and CH3COO– combines with added acid, leaving [H +]
relatively unchanged. Although HCl and Cl - are a conjugate acidbase pair, Cl - has no tendency to combine with added acid to
form undissociated HCl. Any added acid simply increases [H +] in
an HCl-NaCl mixture. 15.14 (a) pH = 3.82 (b) pH = 3.96
15.16 (a) pH 4.60 (b) Na+(aq) CH3COO–(aq) H+(aq) Cl–(aq) ¡ CH3COOH(aq) Na+(aq) Cl–(aq)
(c) CH3COOH(aq) Na+(aq) OH–(aq) ¡
CH3COO–(aq) H2O(l) Na+(aq) 15.18 0.18 mol NaBrO
15.20 (a) pH 4.86 (b) pH 5.0 (c) pH 4.71
15.22 (a) [HCO3]\[H2CO3] 20 (b) [HCO3 -]> [H2CO3] = 10
15.24 360 cm3 of 0.10 M HCOO Na, 640 cm3 of 0.10 M HCOOH
15.26 (a) Curve B (b) pH at the approximate equivalence point
of curve A = 8.0, pH at the approximate equivalence point of
curve B = 7.0 15.28 (a) Above pH 7 (b) below pH 7 (c) at
pH 7 15.30 (a) 42.4 cm3 NaOH soln (b) 35.0 cm3 NaOH soln
(c) 29.8 cm3 NaOH soln 15.32 (a) pH 1.54 (b) pH 3.30
(c) pH 7.00 (d) pH 10.70 (e) pH 12.74
15.34 (a) pH = 7.00 (b) [HONH3+] 0.100 M, pH = 3.52
(c) [C3H5NH3+] 0.100 M, pH = 2.82 15.35 (a) All
equilibrium product expressions only contain reactants or
products whose concentration can change in the equilibrium.
The concentration of undissolved solid is constant, so does not
appear in the solubility product expression. (b) Ksp [Ag+][I–];
Ksp [SR2+][SO42–]; Ksp [Fe2+][OH–]2; Ksp [Hg22+][Br–]2;
15.37 (a) Ksp 7.63 10–9 (b) Ksp 2.7 10–9
(c) 5.3 * 10-4 mol Ba(IO3)2 dm–3 15.39 Ksp 2.3 10–9
15.41 (a) 7.1 * 10-7 mol AgBr dm–3
(b) 1.7 * 10-11 mol AgBr dm–3 (c) 5.0 * 10-12 mol AgBr dm–3
15.42 (a) 1.4 10–3 g dm–3 (b) 3.6 10–7 g dm–3 15.44 More
soluble in acid: (a) ZnCO3 (b) ZnS (d) AgCN (e) Ba3(PO4)2
15.46 [Cu2+] 2 10–12 M 15.48 K Ksp Kf 8 104
15.50 (a) Q 5 10–14, Q < Ksp; no Ca(OH)2 precipitates
(b) Q 9.4 10–6, Q < Ksp; no Ag2SO4 precipitates
15.51 pH 11.5 15.52 AgI will precipitate first, at
[I–] 4.2 10–13 M. 15.54 The first two experiments eliminate
group 1 and 2 ions (Figure 15.18). The absence of insoluble
carbonate precipitates in the filtrate from the third experiment
rules out group 4 ions. The ions that might be in the sample are
those from group 3, Al3+, Fe 3+, Zn2+, Cr 3+, Ni 2+, Co2+ or Mn2+,
and from group 5, NH 4 +, Na+ or K +. 15.56 (a) Make the
Answers to Selected Exercises
solution acidic with 0.5 M HCl; saturate with H 2S. CdS will
precipitate; ZnS will not. (b) Add excess base; Fe(OH)3(s)
precipitates, but Cr 3+ forms the soluble complex Cr(OH)4 -.
(c) Add (NH 4)2HPO 4 ;Mg 2+ precipitates as MgNH 4PO 4 ;K +
remains soluble. (d) Add 6 M HCl; precipitate Ag + as AgCl(s);
Mn2+ remains soluble. 15.58 (a) Base is required to increase
[PO4 3-] so that the solubility product of the metal phosphates
of interest is exceeded and the phosphate salts precipitate.
(b) Ksp for the cations in group 3 is much larger, and so to
exceed Ksp a higher [S 2-] is required. (c) They should all
redissolve in strongly acidic solution. 15.59 (a) pH = 3.025
(b) pH = 2.938 (c) pH = 12.862 15.61 (a) pH of buffer
A = pH of buffer B = 3.74. For buffers containing the same
conjugate acid and base components, the pH is determined by
the ratio of concentrations of conjugate acid and base. Buffers
A and B have the same ratio of concentrations, so their pH
values are equal. (b) Buffer capacity is determined by the
absolute amount of buffer components available to absorb
strong acid or strong base. Buffer A has the greater capacity
because it contains the greater absolute concentrations of
HCHO 2 and CHO 2-. (c) Buffer A: pH = 3.74, ¢pH = 0.00;
buffer B: pH = 3.66, ¢pH = -0.12 (d) Buffer A: pH = 3.74,
¢pH = 0.00; buffer B: pH = 2.74, ¢pH = -1.00 (e) The
results of parts (c) and (d) are quantitative confirmation that
buffer A has a significantly greater capacity than buffer B.
15.63 (a) molar mass 82.2 g mol–1 (b) Ka 3.8 10–7
15.65 (a) [OH–] 4.23 10–3 M, pH 11.63 (b) It will require
40.00 cm3 of 0.100 M HCl to reach the first equivalence point;
HCO3 - is the predominant species at this point. (c) An
additional 40.00 cm3 are required to form H2CO3, the
predominant species at the second equivalence point. (d) [H+]
1.2 10–4 M, pH 3.92 15.71 [KMnO4] [MnO4–] 0.11 M
15.74 [OH–] 1.7 10–11 M, pH of the buffer 3.22
15.77 (a) H+(aq) HCOO -(aq) Δ HCOOH(aq)
(b) K 5.56 103 (c) [Na+] [Cl–] 0.075 M, [H+] [HCOO–]
3.7 10–3 M, [HCOOH] 0.071 M
15.80 [Sr2+] [SO42–] 5.7 10–4 M, Ksp 3.2 10–7
Chapter 16
16.1 In a Brønsted-Lowry acid-base reaction, H + is transferred
from the acid to the base. In a redox reaction, one or more
electrons are transferred from the reductant to the oxidant.
° = 0.16 V
16.3 E cell
Switch
e
e
Voltmeter
Fe
anode
Fe(s)
NO3
Na
NO3
NO3
Fe2
NO3
NO3
Ni2
Fe2 (aq) 2 e Ni2 (aq) 2 e
Movement of cations
Movement of anions
Ni
cathode
Ni(s)
A-19
16.5 (a) A is reduced; B is oxidised. (b) Cathode:
A(aq) + 1e- ¡ A-(aq); anode: B(aq) ¡ B+(aq) + 1e (c) In a cell with a positive E° value, electrons flow spontaneously
from the anode to the cathode, so the anode half-reaction,
B(aq) ¡ B+(aq) + 1e -, is at higher potential energy.
16.7 Zinc, with a more negative E °red , is more easily oxidised
than iron. If conditions are favourable for oxidation, the zinc
coating will be preferentially oxidised, preventing iron from
corroding. 16.9 (a) Oxidation is the loss of electrons.
(b) Electrons appear on the products’ side (right side). (c) The
oxidant is the reactant that is reduced; it is the substance that
promotes the oxidation. (d) An alternative term for oxidant is
oxidising agent. 16.11 (a) True (b) false (c) true 16.12 (a) I,
+5 to 0; C, +2 to +4 (b) Hg, +2 to 0; N, -2 to 0 (c) N, +5 to
+2; S, -2 to 0 (d) Cl, +4 to +3; O, -1 to 0 16.13 (a) TiCl4(g)
+ 2 Mg(l) ¡ Ti(s) + 2 MgCl2(l) (b) Mg(l) is oxidised;
TiCl4(g) is reduced. (c) Mg(l) is the reductant; TiCl4(g) is the
oxidant. 16.15 (a) Sn2(aq) ¡ Sn4(aq) 2e, oxidation
(b) TiO2(s) + 4 H+(aq) + 2e- ¡ Ti2+(aq) + 2 H2O(l),
reduction (c) ClO3 -(aq) + 6 H+(aq) + 6e- ¡ Cl-(aq) +
3 H2O(l), reduction (d) 4 OH(aq) ¡ O2(g) 2 H2O(l) 4e, oxidation (e) SO32(aq) 2 OH(aq) ¡ SO42(aq) H2O(l) 2e, oxidation (f) N2(g) + 8 H+(aq) + reduction
(g) N2(g) + 6 H2O(l) + 2e- ¡ 2 NH3(g) + 6 OH-(aq),
reduction 16.16 (a) Cr2O7 2-(aq) + I-(aq) 8 H+(aq) ¡
2 Cr3+(aq) + IO3 -(aq) + 4 H2O(l); oxidising agent, Cr2O 7 2-;
reducing agent, I - (b) 4 MnO4 -(aq) + 5 CH3OH(aq) +
12 H+(aq) ¡ 4 Mn2 (aq) + 5 HCO2H(aq) + 11 H2O(l);
oxidising agent, MnO 4 - ; reducing agent, CH 3OH (c) I2(s) +
5 OCl-(aq) + H2O(l) ¡ 2 IO3 -(aq) + 5 Cl-(aq) + 2 H (aq);
oxidising agent, OCl -; reducing agent, I 2 (d) As2O3(s) +
2 NO3 -(aq) + 2 H2O(l) + 2 H+(aq) ¡ 2 H3AsO4(aq) +
N2O3(aq); oxidising agent, NO 3 -; reducing agent, As2O 3
(e) 2 MnO4 -(aq) + Br-(aq) + H2O(l) ¡ 2 MnO2(s) +
BrO3 -(aq) + 2 OH-(aq); oxidising agent, MnO 4 -; reducing
agent, Br - (f) Pb(OH)4 2-(aq) + CIO-(aq) ¡ PbO2(s) +
Cl-(aq) + 2 OH-(aq) + H2O(l); oxidising agent, ClO -; reducing
agent, Pb(OH)4 2- 16.17 (a) The reaction Cu2+(aq) +
Zn(s) ¡ Cu(s) + Zn2+(aq) is occurring in both figures. In
Figure 16.2 the reactants are in contact, while in Figure 16.3
the oxidation half-reaction and reduction half-reaction are
occurring in separate compartments. In Figure 16.2 the flow
of electrons cannot be isolated or utilised; in Figure 16.3
electrical current is isolated and flows through the voltmeter.
(b) Na + cations are drawn into the cathode compartment
to maintain charge balance as Cu2+ ions are removed.
16.18 (a) Fe(s) is oxidised, Ag+(aq) is reduced.
(b) Ag+(aq) + 1e- ¡ Ag(s); Fe(s) ¡ Fe2+(aq) + 2e (c) Fe(s) is the anode, Ag(s) is the cathode. (d) Fe(s) is
negative; Ag(s) is positive. (e) Electrons flow from the Fe
electrode (-) toward the Ag electrode (+). (f) Cations migrate
toward the Ag(s) cathode; anions migrate toward the Fe(s)
anode. 16.20 Electromotive force, emf, is the potential energy
difference between an electron at the anode and an electron at
the cathode of a voltaic cell. (b) One volt is the potential energy
difference required to impart 1 J of energy to a charge of
1 coulomb. (c) Cell potential, Ecell , is the emf of an
A-20
Answers to Selected Exercises
electrochemical cell. 16.22 (a) 2 H+(aq) + 2e- ¡ H2(g)
(b) A standard hydrogen electrode, SHE, has components that
are at standard conditions, 1 M H+(aq) and H2(g) at 1 bar.
(c) The platinum foil in a SHE serves as an inert electron carrier
and a solid reaction surface. 16.24 (a) A standard reduction
potential is the cell potential of a reduction half-reaction
measured against the standard hydrogen electrode at
standard conditions. (b) E °red = 0 (c) The reduction of Ag+(aq)
to Ag(s) is much more energetically favourable.
16.26 (a) Cr2+(aq) ¡ Cr3+(aq) + e -; Tl3+(aq) +
2e- ¡ Tl+(aq) (b) E °red = 0.78 V
(c)
Switch
e
e
Voltmeter
Inert (Pt)
anode
NO
O3
Na
Inert (Pt)
cathode
Movement of cations
Movement of anions
16.27 (a) E° = 0.823 V (b) E° = 1.89 V (c) E° = 1.211 V
(d) E° = 1.21 V 16.28 (a) 3 Ag(aq) Cr(s) ¡ 3 Ag(s)
+ Cr3(aq), E 1.54 V (b) Two of the combinations have
essentially equal E values: 2 Ag+(aq) + Cu(s) ¡ 2 Ag(s) +
Cu2+(aq), E° = 0.462 V; 3 Ni2+(aq) + 2 Cr(s) ¡ 3 Ni(s) +
2 Cr3+(aq), E° = 0.46 V 16.30 (a) Cl2(g) (b) Ni2+(aq)
(c) BrO3 -(aq) (d) O3(g) 16.32 (a) Cl2(aq), strong oxidant
(b) MnO4 -(aq), acidic, strong oxidant (c) Ba(s), strong
reductant (d) Zn(s), reductant 16.34 (a) Cu2+(aq) 6 O2(g) 6
Cr2O7 2-(aq) 6 Cl2(g) 6 H2O2(aq) (b) H2O2(aq) 6 I-(aq) 6
Sn2+(aq) 6 Zn(s) 6 Al(s) 16.35 Al and H 2C2O 4
16.36 (a) 2 Fe2+(aq) + S2O6 2-(aq) + 4 H+(aq) ¡ 2 Fe3+(aq) +
2 H2SO3(aq); 2 Fe2+(aq) + N2O(aq) + 2 H+(aq) ¡ 2 Fe3+(aq)
+ N2(g) + H2O(l); Fe2+(aq) + VO2 +(aq) + 2 H+(aq) ¡
Fe3+(aq) + VO2+(aq) + H2O(l) (b) E° = -0.17 V, ¢G° = 33 kJ;
E° = -2.54 V, ¢G° = 4.90 * 102 kJ; E° = 0.23 V,
¢G° = -22 kJ (c) K = 1.8 * 10-6; K = 1.2 * 10-86;
K = 7.8 * 103 16.37 ¢G° = 21.8 kJ, E °cell = -0.113 V
16.39 (a) E° = 0.16 V, K = 2.54 * 105 (b) E° = 0.277 V,
K = 2.3 * 109 (c) E° = 0.45 V, K = 1.5 * 1075
16.40 (a) K = 9.8 * 102 (b) K = 9.5 * 105 (c) K = 9.3 * 108
16.42 (a) The Nernst equation is applicable when the
components of an electrochemical cell are at non-standard
conditions. (b) Q = 1 (c) Q decreases and emf increases
16.44 (a) E decreases (b) E decreases (c) E decreases (d) no effect
16.46 (a) E° = 0.46 V (b) E = 0.29 V 16.48 (a) The
compartment with [Zn2+] 1.00 10–2 M is the anode.
(b) E° = 0 (c) E = 0.0668 V (d) In the anode compartment
[Zn2+] increases; in the cathode compartment [Zn2+] decreases
16.51 (a) The emf of a battery decreases as it is used. The
concentrations of products increase and the concentrations of
reactants decrease, causing Q to increase and Ecell to decrease.
(b) A D-size battery contains more reactants than a AA,
enabling the D to provide power for a longer time.
16.55 (a) The cell emf will have a smaller value. (b) NiMH
batteries use an alloy such as ZrNi 2 as the anode material. This
eliminates the use and disposal problems associated with Cd,
a toxic heavy metal. 16.56 The main advantage of a fuel cell
is that fuel is continuously supplied, so that it can produce
electrical current for a time limited only by the amount of
available fuel. For the hydrogen-oxygen fuel cell, this is also a
disadvantage because volatile and explosive hydrogen must be
acquired and stored. Alkaline batteries are convenient, but they
have a short lifetime, and the disposal of their zinc and
manganese solids is more problematic than disposal of
water produced by the hydrogen-oxygen fuel cell.
16.58 (a) anode: Fe(s) ¡ Fe2+(aq) + 2e -; cathode: O2(g) +
4 H+(aq) + 4e- ¡ 2 H2O(l) (b) 2 Fe2+(aq) + 6 H2O(l)
¡ Fe2O3 # 3 H2O(s) + 6 H+(aq) + 2e -; O2(g) + 4 H+(aq) +
4e- ¡ 2 H2O(l) 16.59 (a) Mg is called a ‘sacrificial anode’
because it has a more negative E °red than the pipe metal and is
preferentially oxidised when the two are coupled. It is sacrificed
to preserve the pipe. (b) E °red for Mg 2+ is -2.37 V, more
negative than most metals present in pipes, including Fe and
Zn. 16.62 (a) Electrolysis is an electrochemical process driven
by an outside energy source. (b) By definition, electrolysis
reactions are non-spontaneous. (c) 2 Cl-(l) ¡ Cl2(g) + 2e 16.64 (a) 236 g Cr(s) (b) 2.51 A 16.66 wmax 8.19 104
16.68 (a) 4.0 105 g Li (b) 0.24 kWh (mol Li)1
16.70 (a)2 Ni(aq) ¡ Ni(s) Ni2(aq) (b) 3 MnO4 2-(aq) +
4 H+(aq) ¡ 2 MnO4 -(aq) + MnO2(s) + 3 H2O(l)
(c) 3 H2SO3(aq) ¡ S(s) + 2 HSO4 -(aq) + 2 H+(aq) + H2O(l)
(d) Cl2(aq) + 2 OH-(aq) ¡ Cl-(aq) + ClO-(aq) + H2O(l)
16.71 (a) Fe(s) ¡ Fe2+(aq) + 2e -; 2 Ag+(aq) +
2e - ¡ Ag(s); Fe(s) + 2 Ag+(aq) ¡ Fe2(aq) + Ag(s)
Switch
e
e
Voltmeter
NO
O3
Fe
anode
Na
Ag
cathode
Movement of cations
Movement of anions
(b) Zn(s) ¡ Zn2+(aq) + 2e -; 2 H+(aq) + 2e- ¡ H2(g);
Zn(s) + 2 H+(aq) ¡ Zn2+(aq) + H2(g)
Switch
e
e
Voltmeter
Zn
anode
NO3
Na H2(g)
NO3 Zn2 NO3
NO3 H
Movement of cations
Movement of anions
Cathode
(standard
hydrogen
electrode)
Answers to Selected Exercises
(c) Cu ƒ Cu2+ ƒ ƒ ClO 3 -, Cl - ƒ Pt
(d) N2O 3 .
O
Switch
e
e
NO
O3
N
Na
Inert (Pt)
cathode
Movement of cations
Movement of anions
16.73 (a) The reduction potential for O2(g) in the presence of
acid is 1.23 V. O2(g) cannot oxidise Au(s) to Au+(aq), even in
the presence of acid. (b) Anything with a higher standard
reduction potential than 1.50 V, e.g., Co3+(aq), F2(g), H2O2(aq),
O3(g); marginally: BrO3 -(aq), Ce4+(aq), HClO(aq), MnO4 -(aq),
PbO2(s) (c) Au(s) is being oxidised; O2(g) is being reduced.
(d) 2 Na[Au(CN)2](aq) + Zn(s) ¡ 2 Au(s) + Zn2+(aq) +
2 Na+(aq) + 4 CN-(aq). Zn(s) is being oxidised; [Au(CN)2]-(aq)
is being reduced. 16.74 (a) E = 0.041 V (b) the inert electrode
where I is oxidised (c) No, at standard conditions, Cu(s) is
the anode. (d) 1.3 M I - 16.77 (a) In discharge, Cd(s) +
2 NiO(OH)(s) + 2 H2O(l) ¡ Cd(OH)2(s) + 2 Ni(OH)2(s).
In charging, the reverse reaction occurs. (b) E° = 1.25 V
(c) 1.25 V is the standard cell potential, E°. The concentrations
of reactants and products inside the battery are adjusted so
that the cell output is greater than E°. (d) K = 2 * 1042
16.79 (a) 6.4 * 106 C (b) 6.4 * 105 amp (c) 16 kWh
16.82 (a) H2(g) is being oxidised and N2(g) is being reduced.
(b) K = 6.9 * 105 (c) E° = 0.05755 V 16.87 E° = -0.928 V
Chapter 17
17.2 (a) Acid-base (Brønsted) (b) From left to right in
the reaction, the charges are: 0, 0, 1 +, 1 (c) NH3(aq) + H2O(l) Δ NH4 +(aq) + OH-(aq)
17.5 (a) N2O5 . Resonance structures other than the ones shown
are possible. Those with double bonds to the central oxygen do
not minimise formal charge and make a smaller contribution to
the net bonding model.
O
O
O
N
O
N
O
O
N
O
O
O
N
O
(b) N2O 4 . Resonance structures other than the ones shown are
possible.
O
O
O
N
N
O
O
N
O
O
N
O
(c) NO 2 . We place the odd electron on N because it is more
electronegative.
N
O
N
O
O
O
O
O
N
O
N
N
O
O
(e) NO. We place the odd electron on N because it is more
electronegative
N O
Voltmeter
Cu
anode
A-21
(f) N2O. The right-most structure does not minimise formal charge
and makes a smaller contribution to the net bonding model.
N N O
N N O
N N O
17.9 Metals: (b) Sr, (c) Mn, (e) Rh; nonmetals: (a) P, (d) Se,
(f) Kr; metalloids: none. 17.11 (a) O (b) Br (c) Ba (d) O
(e) Co 17.13 (a) N is too small a central atom to fit five
fluorine atoms, and it does not have available d orbitals, which
can help accommodate more than eight electrons. (b) Si does
not readily form p-bonds, which are necessary to satisfy the
octet rule for both atoms in the molecule. (c) As has a lower
electronegativity than N; that is, it more readily gives up
electrons to an acceptor and is more easily oxidised.
17.15 (a) Mg3N2(s) + 6 H2O(l) ¡ 2 NH3(aq) + 3 Mg(OH)2(s)
(b) 2 C3H7OH(l) + 9 O2(g) ¡ 6 CO2(g) + 8 H2O(l)
¢
(c) MnO2(s) + C(s)
¢
MnO2(s) + 2 C(s)
" CO(g) + MnO(s) or
" 2 CO(g) + Mn(s) or
¢
" CO (g) + Mn(s)
MnO2(s) + C(s)
2
(d) AlP(s) + 3 H2O(l) ¡ PH3(g) + Al(OH)3(s)
(e) Na2S(s) + 2 HCl(aq) ¡ H2S(g) + 2 NaCl(aq)
17.17 Like other elements in group 1, hydrogen has only one
valence electron and its most common oxidation number is +1.
17.19 (a) Mg(s) + 2 H+(aq) ¡ Mg2+(aq) + H2(g)
1000°C "
CO(g) + H (g)
(b) C(s) + H O(g)
2
2
1100°C
" CO(g) + 3 H (g)
(c) CH4(g) + H2O(g)
2
17.21 (a) NaH(s) + H2O(l) ¡ NaOH(aq) + H2(g)
(b) Fe(s) + H2SO4(aq) ¡ Fe2+(aq) + H2(g) + SO4 2-(aq)
(c) H2(g) + Br2(g) ¡ 2 HBr(g)
(d) 2 Na(l) + H2(g) ¡ 2 NaH(s)
¢ "
Pb(s) + H O(g)
(e) PbO(s) + H (g)
2
2
17.23 (a) Ionic (b) molecular (c) metallic 17.25 Xenon has a
lower ionisation energy than argon; because the valence
electrons are not as strongly attracted to the nucleus, they are
more readily promoted to a state in which the atom can form
bonds with fluorine. Also, Xe is larger and can more easily
accommodate an expanded octet of electrons.
17.26 (a) ClO3 -, +5 (b) HI, -1 (c) ICl3 ;I, +3, Cl, -1
(d) NaOCl, +1 (e) HClO 4 , +7 (f) XeF4 ;Xe, +4, F, -1
17.28 (a) iron(III) chlorate (b) chlorous acid (c) xenon
hexafluoride (d) bromine pentafluoride (e) xenon oxide
tetrafluoride (f) iodic acid 17.30 (a) Van der Waals
intermolecular attractive forces increase with increasing
number of electrons in the atoms. (b) F2 reacts with water.
F2(g) + H2O(l) ¡ 2 HF(g) + O2(g). That is, fluorine is too
strong an oxidising agent to exist in water. (c) HF has
extensive hydrogen bonding because fluorine is more
electronegative than the other halogens. (d) Oxidising power
is related to electronegativity. Electronegativity and oxidising
power decrease in the order given. 17.32 (a) As an oxidising
A-22
Answers to Selected Exercises
agent in steelmaking; to bleach pulp and paper; in
oxyacetylene torches; in medicine to assist in breathing
(b) synthesis of pharmaceuticals, lubricants and other
organic compounds where C ! C bonds are cleaved;
¢ "
2 Hg(l) + O2(g)
in water treatment. 17.34 (a) 2 HgO(s)
¢ "
(b) 2 Cu(NO ) (s)
2 CuO(s) + 4 NO (g) + O (g)
3 2
2
2
(c) PbS(s) + 4 O3(g) ¡ PbSO4(s) + 4 O2(g)
(d) 2 ZnS(s) + 3 O2(g) ¡ 2 ZnO(s) + 2 SO2(g)
(e) 2 K2O2(s) + 2 CO2(g) ¡ 2 K2CO3(s) + O2(g)
17.36 (a) acidic (b) acidic (c) basic (d) amphoteric
17.38 (a) H 2SeO3 , +4 (b) KHSO3 , +4 (c) H 2Te, -2
(d) CS 2 , -2 (e) CaSO 4 , +6
17.39 (a) 2 Fe3+(aq) + H2S(aq) ¡ 2 Fe2+(aq) + S(s) +
2 H+(aq)
(b) Br2(l) + H2S(aq) ¡ 2 Br-(aq) + S(s) + 2 H+(aq)
(c) 2 MnO4 -(aq) + 6 H+(aq) + 5 H2S(aq) ¡ 2 Mn2+(aq) +
5 S(s) + 8 H2O(l)
(d) 2 NO3 -(aq) + H2S(aq) + 2 H+(aq) ¡ 2 NO2(aq) + S(s) +
2 H2O(l)
17.41 (a)
2
(b)
S S
O Se O
Cl
Cl
O
Bent (free rotation
around S–S bond)
Trigonal pyramidal
(c)
O
O
S
Cl
O
H
O N O H
17.45 (a) O N O H
The molecule is bent around the central oxygen and nitrogen
atoms; the four atoms need not be coplanar. The right-most
form does not minimise formal charges and is less important in
the actual bonding model.
(b)
N
N
N
N
N
N
N
N
The molecule is linear.
(c)
H H
H
N
N
H
H
The geometry is tetrahedral around the left nitrogen, trigonal
pyramidal around the right.
(d)
O
O
N
3
2
(b) BaC2(s) + 2 H2O(l) ¡ Ba2+(aq) + 2 OH-(aq) + C2H2(g)
(c) 2 C2H2(g) + 5 O2(g) ¡ 4 CO2(g) + 2 H2O(g)
(d) CS2(g) + 3 O2(g) ¡ CO2(g) + 2 SO2(g)
(e) Ca(CN)2(s) + 2 HBr(aq) ¡ CaBr2(aq) + 2 HCN(aq)
17.55 (a)
800°C "
2 CH4(g) + 2 NH3(g) + 3 O2(g)
2 HCN(g) + 6 H2O(g)
cat
Tetrahedral
17.43 (a) NaNO 2 , +3 (b) NH 3 , -3 (c) N2O, +1 (d) NaCN, -3
(e) HNO 3 , +5 (f) NO 2 , +4
N
(d) NH3(aq) + H+(aq) ¡ NH4 +(aq)
(e) N2H4(l) + O2(g) ¡ N2(g) + 2 H2O(g)
17.47 (a) HNO2(aq) + H2O(l) ¡
NO3 -(aq) + 3 H+(aq) + 2e-, E°red = 0.94 V
(b) N2 (g) + H2O(l) ¡
N2O(aq) + 2 H+(aq) + 2e-, E°red = 1.77 V
17.49 (a) H 3PO3 , +3 (b) H3AsO3 , +3 (c) SbCl3 , +3
(d) Mg 3As2 , +5 (e) P2O 5 , +5 17.50 (a) Phosphorus is a larger
atom than nitrogen, and P has energetically available 3d
orbitals, which participate in the bonding, but nitrogen does
not. (b) Only one of the three hydrogens in H 3PO 2 is bonded
to oxygen. The other two are bonded directly to phosphorus
and are not easily ionised. (c) PH 3 is a weaker base than H 2O
so any attempt to add H + to PH 3 in the presence of H 2O causes
protonation of H 2O. (d) Because of the severely strained bond
angles in P4 molecules, white phosphorus is highly reactive.
17.52 (a) 2 Ca3PO4(s) + 6 SiO2(s) + 10 C(s) ¡
P4(g) + 6 CaSiO3(l) + 10 CO(g)
(b) PBr3(l) + 3 H2O(l) ¡ H3PO3(aq) + 3 HBr(aq)
(c) 4 PBr3(g) + 6 H2(g) ¡ P4(g) + 12 HBr(g)
¢ "
17.54 (a) ZnCO (s)
ZnO(s) + CO (g)
O
The ion is trigonal planar; it has three equivalent resonance
forms.
17.46 (a) Ba3N2(s) + 6 H2O(l) ¡ 3 Ba(OH)2(s) + 2 NH3(aq)
(b) 2 NO(g) + O2(g) ¡ 2 NO2(g)
(c) N2O5(g) + H2O(l) ¡ 2 H+(aq) + 2 NO3 -(aq)
(b) NaHCO3(s) + H+(aq) ¡ CO2(g) + H2O(l) + Na+(aq)
(c) 2 BaCO3(s) + O2(g) + 2 SO2(g) ¡ 2 BaSO4(s) + 2 CO2(g)
17.56 (a) H 3BO3 , +3 (b) SiBr4 , +4 (c) PbCl2 , +2
(d) Na 2B4O 7 # 10 H 2O, +3 (e) B2O 3 , +3 19.58 (a) Lead
(b) carbon, silicon and germanium (c) silicon
17.60 (a) Tetrahedral (b) Metasilicic acid will probably adopt
the single-strand silicate chain structure shown in
Figure 19.39(a). The Si to O ratio is correct and there are two
terminal O atoms per Si that can accommodate the two H
atoms associated with each Si atom of the acid.
17.62 (a) Diborane has bridging H atoms linking the two B
atoms. The structure of ethane has the C atoms bound directly,
with no bridging atoms. (b) B2H 6 is an electron-deficient
molecule. The 6 valence electron pairs are all involved in
B ¬ H sigma bonding, so the only way to satisfy the octet rule
at B is to have the bridging H atoms shown in Figure 19.41.
(c) The term ‘hydridic’ indicates that the H atoms in B2H 6
have a partial negative charge. 17.67 SiH 4 , CO, Mg. The
oxidation numbers of Si in SiO2, C in CO2 and Ca in CaO are at
their maximum and cannot increase by further combination
with O2. 17.69 (a) SO 3 (b) Cl2O 5 (c) N2O 3 (d) CO 2 (e) P2O 5
¢ "
17.75 GeO2(s) + C(s)
Ge(l) + CO2(g)
Ge(l) + 2 Cl2(g) ¡ GeCl4(l)
GeCl4(l) + 2 H2O(l) ¡ GeO2(s) + 4 HCl(g)
GeO2(s) + 2 H2(g) ¡ Ge(s) + 2 H2O(l)
17.77 The average Xe–F bond enthalpy in XeF2 and XeF4 is
134 kJ; in XeF6 it is 130 kJ. The decrease, as more F are packed
around the Xe probably indicates congestion leading to weaker
bonding. 17.79 [HClO] 0.036 M
Answers to Selected Exercises
Chapter 18
18.1 (a)
18.41
N
Cl
Pt
N
Cl
(b) coordination number = 4, coordination geometry = square
planar (c) oxidation number = +2 18.4 Structures (3) and (4)
are identical to (1); (2) and (5) are geometric isomers of (1).
18.8 Isolated atoms: (b) and (f); bulk metal: (a), (c), (d) and (e).
Although it may seem that atomic radius would be a property
of isolated atoms, recall that an isolated atom has no ‘edge’
and hence atomic radius only has meaning in a bulk sample.
18.9 Hf has a completed 4f subshell, while Zr does not. The
buildup in Z that accompanies the filling of the 4f orbitals,
along with additional shielding of valence electrons, offsets the
typical effect of the larger n value for the valence electrons of
Hf. Thus, the atomic radius of Hf is about the same as that of
Zr, the element above it in Group 14. 18.11 (a) ScF3 (b) CoF3
(c) ZnF2 (d) CrF6 18.13 Chromium, [Ar]4s 1 3d5, has six
valence electrons with similar ionisation energies, some or all
of which can be involved in bonding, leading to multiple stable
oxidation states. Al, [Ne]3s 2 3p1, has only three valence
electrons, which are all lost or shared during bonding,
producing the +3 state exclusively. 18.14 (a) Cr 3+, [Ar]3d 3
(b) Au3+, [Xe]4f 14 5d8 (c) Ru2+, [Kr]4d 6 (d) Cu+, [Ar]3d 10
(e) Mn4+, [Ar]3d 3 (f) Ir +, [Xe]4f 14 5d8 18.16 Ti 2+
18.18 Fe 2+ is a reducing agent that is readily oxidised to Fe 3+
in the presence of O2 from air. 18.20 The unpaired electrons in
a paramagnetic material cause it to be weakly attracted into a
magnetic field. A diamagnetic material, where all electrons are
paired, is very weakly repelled by a magnetic field. 18.22 E° will
become more negative as the stability (Kf value) of the complex
increases. 18.25 (a) A metal complex consists of a central metal
ion bonded to a number of surrounding molecules or ions. The
number of bonds formed by the central metal ion is the
coordination number. The surrounding molecules or ions are the
ligands. (b) Metal ions, by virtue of their positive charge and
empty d, s or p orbitals, act as electron-pair acceptors, or Lewis
acids. Ligands, which have at least one unshared electron pair,
act as electron-pair donors, or Lewis acids. 18.27 (a) +2
(b) 6 (c) 2 mol AgBr(s) will precipitate per mole of complex.
18.29 (a) Coordination number = 4, oxidation number = +2
(b) 5, +4 (c) 6, +3 (d) 5, +2 (e) 6, +3 (f) 4, +2 18.31 (a) 4 Cl (b) 4 Cl -, 1 O 2- (c) 4 N, 2 Cl - (d) 5 C (e) 6 O (f) 4 N
18.33 (a) A monodentate ligand binds to a metal via one atom,
a bidentate ligand through two atoms. (b) Three bidentate
ligands fill the coordination sphere of a six-coordinate
complex. (c) A tridentate ligand has at least three atoms
with unshared electron pairs in the correct orientation
to simultaneously bind one or more metal ions.
18.34 (a) Ortho-phenanthroline, o-phen, is bidentate
(b) oxalate, C2O4 2-, is bidentate (c) ethylenediaminetetraacetate,
EDTA, is pentadentate (d) ethylenediamine, en, is bidentate
18.37 (a) [Cr(NH 3)6](NO3)3 (b) [Co(NH 3)4CO3]2SO4
(c) [Pt(en)2Cl2]Br2 (d) K[V(H 2O)2Br4] (e) [Zn(en)2][HgI 4]
18.39 (a) tetraaminedichlororhodium(III) chloride
(b) potassium hexachlorotitanate(IV)
(c) tetrachlorooxomolybdenum(VI)
(d) tetraaquaoxalatoplatinum(IV) bromide
ONO
ONO
NH3
ONO
Pd
Pd
(a)
A-23
H3N
H3N
NH3
ONO
cis
trans
(b) [Pd(NH 3)2(ONO)2], [Pd(NH 3)2(NO 2)2]
(c)
N
N
N
N
N
V
V
N
N
Cl
Cl
Cl
Cl
N
(d) [Co(NH 3)4Br2]Cl, [Co(NH 3)4BrCl]Br 18.43 Yes. No
structural or stereoisomers are possible for a tetrahedral
complex of the form MA2B2 . The complex must be square
planar with cis and trans geometric isomers. 18.45 In the
trans isomer the Cl ¬ Co ¬ NH 3 bond angle is 180°; in the
cis isomer this bond angle is 90°. The cis isomer is chiral.
18.46 (a) One isomer (b) trans and cis isomers with 180° and
90° Cl ¬ Ir ¬ Cl angles, respectively (c) trans and cis isomers
with 180° and 90° Cl ¬ Fe ¬ Cl angles, respectively. The cis
isomer is optically active. 18.47 (a) Visible light has
wavelengths between 400 and 700 nm. (b) Complementary
colours are pairs of colours that in combination we perceive as
white light—so when one colour is absorbed by a material, we
see reflected or transmitted light from that material as the
complementary colour. (c) A coloured metal complex absorbs
visible light of its complementary colour. (d) 196 kJ mol1
18.49 Most of the attraction between a metal ion and a ligand
is electrostatic. Whether the interaction is ion-ion or ion-dipole,
the ligand is strongly attracted to the metal centre and can be
modelled as a point negative charge.
18.51 (a)
d 2 2, d 2
xy
z
dxy, dxz, dyz
(b) The magnitude of ¢ and the energy of the d-d transition for
a d 1 complex are equal. (c) ¢ = 203 kJ mol1 18.53 A yellow
colour is due to absorption of light around 400 to 430 nm, a
blue colour to absorption near 620 nm. The shorter wavelength
corresponds to a higher-energy electron transition and larger
¢ value. Cyanide is a stronger-field ligand, and its complexes
are expected to have larger ¢ values than aqua complexes.
18.55 All complexes in this exercise are six-coordinate octahedral.
(a)
(b)
d 4, high spin
(e)
(c)
d 5, high spin
d 5, low spin
18.56
(f)
d3
d 6, low spin
(d)
d8
high spin
18.58 [Pt(NH 3)6]Cl4 ;[Pt(NH 3)4Cl2]Cl2 ;[Pt(NH 3)3Cl3]Cl;
[Pt(NH 3)2Cl4]; K[Pt(NH 3)Cl5] 18.60 (a) In a square planar
complex, if one pair of ligands is trans, the remaining two
coordination sites are also trans to each other. The bidentate
ethylenediamine ligand is too short to occupy trans
A-24
Answers to Selected Exercises
coordination sites, so the trans isomer of [Pt(en)Cl2] is
unknown. (b) The minimum steric requirement for a transbidentate ligand is a medium-length chain between the two
coordinating atoms that will occupy the trans positions.
A polydentate ligand such as EDTA is much more likely to
occupy trans positions because it locks the metal ion in place
with multiple coordination sites and shields the metal ion from
competing ligands present in the solution.
18.61 (a) AgCl(s) + 2 NH3(aq) ¡ [Ag(NH3)2]+(aq) + Cl-(aq)
(b) [Cr(en)2Cl2]Cl(aq) + 2 H2O(l) ¡
[Cr(en)2(H2O)2]3+(aq) + 3 Cl-(aq);
+
3 Ag (aq) + 3 Cl (aq) ¡ 3 AgCl(s) (c)
Zn(OH3)2(aq) + 2 NaOH(aq) ¡ Zn(OH)2(s) + 2 NaNO3(aq);
Zn(OH)2(s) + 2 NaOH(aq) ¡ [Zn(OH)4]2-(aq) + 2 Na+(aq)
(d) Co2+(aq) + 4 Cl-(aq) ¡ [CoCl4]2-(aq)
18.62 (a)
d2
(b) Visible light with l = hc>¢ is absorbed by the complex,
promoting one of the d electrons into a higher-energy d orbital.
The remaining wavelengths are reflected or transmitted; the
combination of these wavelengths is the colour we see.
(c) [V(H 2O)6]3+ will absorb light with higher energy because it
has a larger ¢ than [VF6]3-. H 2O is in the middle of the
spectrochemical series and causes a larger ¢ than F -, a weakfield ligand. 18.64 [Co(NH 3)6]3+, yellow; [Co(H 2O)6]2+, pink;
[CoCl4]2-, blue 18.65 (a) [FeF6]4-. F - is a weak-field ligand
that imposes a smaller ¢ and longer l for the complex ion.
(b) [V(H 2O)6]2+. V 2+ has a lower charge, so the interaction
with the ligand will produce a weaker field, a smaller Δ
and a longer absorbed wavelength. (c) [CoCl4]2-. Cl is a weak-field ligand that imposes a smaller ¢ and a
longer l for the complex ion. 18.67 In carbonic anhydrase,
Zn2 (Lewis acid) withdraws electrons from H2O (Lewis base).
The O—H bond is polarised and H becomes more ionisable
and thus more acidic than in the bulk solvent.
18.69 K 4[Mn(ox)2Br2] 18.70 47.3 mg Mg2+ dm3,
53.4 mg Ca2+ dm3 18.71 ¢E = 3.02 * 10-19 J photon-1,
l = 657 nm. The complex will absorb in the visible around
660 nm and appear blue-green.
Chapter 19
19.1 (a) 24Ne: 10 p, 14 n, just above the belt of stability. Reduce
the neutron-to-proton ratio via b-decay. (b) 32Cl: 17 p, 15 n,
just below the belt of stability. Increase the neutron-to-proton
ratio via positron emission or orbital electron capture.
(c) 108Sn: 50 p, 58 n, just below the belt of stability. Increase the
neutron-to-proton ratio via positron emission or orbital
electron capture. (d) 216Po: 84 p, 132 n, just beyond the belt
of stability. Nuclei with atomic numbers 84 tend to decay
via alpha emission, which decreases both protons and
neutrons. 19.4 (a) 7 minutes. (b) 0.1 min1
(c) The fraction of 88Mo remaining after 12 min is 0.3/1.0 0.3.
In(Nt/No) kt (0.099)(12) 1.188; Nt/No e1.188 88
0
55
0.30. (d) 88
42Mo ¡ 41Nb 1e 19.7 (a) 25Mn:
25 p, 30 n (b) 201Hg: 80 p, 121 n (c) 39K: 19 p, 20 n 19.9 (a) 10n
90
0
(b) 42He or 42a (c) 00γ or γ 19.11 (a) 90
37Rb ¡ 38Sr 1e
72
0
72
(b) 34Se 1e (orbital electron) ¡ 33As
76
0
226
222
4
(c) 76
36Kr ¡ 35Br 1e (d) 88Ra ¡ 86Rn 2He
211
211
0
50
0
19.13 (a) 82Pb ¡ 83Bi 1b (b) 25Mn ¡ 50
24Cr 1e
179
0
179
230
226
4
(c) 74W 1e ¡ 73Ta (d) 90Th ¡ 88Ra 2He
19.14 There are 7 a particle emissions. The change in atomic
number in the series is 10. Each a particle results in an atomic
number lower by two. The 7 a particle emissions alone would
cause a decrease of 14 in atomic number. Each b particle
emission raises the atomic number by one. To obtain the
observed lowering of 10 in the series, there must be 4 b
emissions. 19.16 (a) 85B—low neutron/proton ratio,
positron emission (for low atomic numbers, positron
emission is more common than orbital electron capture)
(b) 68
29Cu—high neutron/proton ratio, beta emission
32
(c) 15P—slightly high neutron/proton ratio, beta emission
(d) 39
17Cl—high neutron/proton ratio, beta emission
19.18 (a) Stable: 39
19K—odd proton, even neutron more
abundant than odd proton, odd neutron; 20 neutrons is a
magic number. (b) Stable: 209
83Bi—odd proton, even neutron
more abundant than odd proton, odd neutron; 126 neutrons is
a magic number. (c) Stable: 58
28Ni—even proton, even neutron
more likely to be stable than even proton, odd neutron;
65
19.20 (a) 42He, both
28Ni has high neutron/proton ratio
18
(b) 8O has a magic number of protons, but not neutrons
66
208
(c) 40
20Ca, both (d) 30Zn has neither (e) 82Pb, both
19.22 Radioactive: 58
29Cu—odd proton, odd neutron, low
neutron/proton ratio. 206Po—high atomic number.
Stable: 62
28Ni—even proton, even neutron, stable
neutron/proton ratio. 108
47Ag—stable neutron/proton ratio, (one
of 5 stable odd proton/odd neutron nuclides).
184
W—even proton, even neutron, stable neutron/proton ratio
19.23 Protons and alpha particles are positively charged and
must be moving very fast to overcome electrostatic forces
which would repel them from the target nucleus. Neutrons
are electrically neutral and not repelled by the nucleus.
1
1
32
7
0
19.24 (a) 32
16S 0n ¡ 1p 15P (b) 4Be 1e (orbital
7
187
187
0
electron) ¡ 3Li (c) 75Re ¡ 76Os 1e
2
1
99
235
1
135
(d) 98
42Mo 1H ¡ 0n 43Tc (e) 92U 0n ¡ 54Xe 99
1
238
4
241
1
38Sr 2 0n 19.26 (a) 92U 2He ¡ 94Pu 0n
14
4
17
1
56
4
(b) 7N 2He ¡ 8O 1H (c) 26Fe 2He ¡
60
0
29Cu 1e 19.28 Chemical reactions do not affect the
character of atomic nuclei. The energy changes involved in
chemical reactions are much too small to allow us to alter
nuclear properties via chemical processes. Therefore, the nuclei
that are formed in a nuclear reaction will continue to emit
radioactivity regardless of any chemical changes we bring to
bear. However, we can hope to use chemical means to separate
radioactive substances, or remove them from foods or a
portion of the environment. 19.30 28.1 mg; 0.201 mg
19.33 (a) ~ 1011 a particles emitted in 5.0 min (b) ~ 10m Ci
19.35 2.20 103 yr 19.37 4.378 109 g 19.39 2.98912 1011 J/23Na nucleus required; 1.80009 1013 J/mol23Na
Answers to Selected Exercises
1
0n
144
94
1
¡ 58Ce 36Kr 2 0n 19.53 The extremely
high temperature is required to overcome the electrostatic
charge repulsions between the nuclei so that they come
together to react. 19.55 Hydroxyl radical is more toxic to
living systems, because it produces other radicals when it
reacts with molecules in the organism. This often starts a
disruptive chain of reactions, each producing a different free
radical. Hydroxide ion, OH, on the other hand, will be readily
neutralised in the buffered cell environment. Its most common
reaction is ubiquitous and innocuous: H OH ¡ H2O.
The acid-base reactions of OH are usually much less
disruptive to the organism than the chain of redox reactions
initiated by •OH radical. 19.57 (a) 3.2 108 dis/s (b) 1.5 103 Gy (c) 1.5 102 Sv 19.58 The stable nucleus is 210
82Pb.
19.60 The most massive radionuclides will have the highest
neutron/proton ratios. Thus, they are most likely to decay by a
process that lowers this ratio, beta emission. The least massive
nuclides, on the other hand, will decay by a process that
increases the neutron/proton ratio, positron emission or
orbital electron capture.
19.62
180
Nt (dis/min)
150
120
90
60
30
0
0
2
4
6 8 10 12 14 16 18
Time (hrs)
5.0
4.5
l n (Nt)
19.41 (a) 1.2305 1012 J/nucleon (b) 1.37312 1012
J/nucleon (c) 1.37533 1012 J/nucleon 19.43 (a) 1.71 105 kg/d (b) 2.1 108 g 235U (about 230 tons of 235U per day)
19.45 We can use Figure 19.9 to see that the binding energy per
nucleon (which gives rise to the mass defect) is greatest for
nuclei of mass numbers around 50. Thus (a) 59
27Co should
possess the greatest mass defect per nucleon. 19.47 The 59Fe
would be incorporated into the diet component, which in turn
is fed to the rabbits. After a time blood samples could be
removed from the animals, the red blood cells separated,
and the radioactivity of the sample measured. If the iron in
the dietary compound has been incorporated into blood
haemoglobin, the blood cell sample should show beta
emission. Samples could be taken at various times to
determine the rate of iron uptake, rate of loss of the iron
from the blood, and so forth. 19.49 (a) Control rods
control neutron flux so that there are enough neutrons to
sustain the chain reaction but not so many that the core
overheats. (b) A moderator slows neutrons so that they are
more easily captured by fissioning nuclei.
1
160
72
1
239
19.51 (a) 235
92U 0n ¡ 62Sm 30Zn 4 0n (b) 94Pu A-25
4.0
3.5
3.0
0
2
4
6 8 10 12 14 16 18
Time (hrs)
19.63 7.2 109 g Sr 19.66 10Be is slightly higher.
19.71 2.425 103 nm; it lies at the short wavelength end of
the range of observed gamma ray wavelengths.
Chapter 20
20.1 (a) A greater volume than 22.7 dm3 (b) The gas will occupy
more volume at 85 km than at 50 km. 20.3 Ozone concentration
varies with altitude because conditions favourable to ozone
formation and unfavourable to its decomposition vary with
altitude. Above 60 km, there are too few O 2 molecules. Below
30 km, there are too few O atoms. Between 30 km and 60 km, O 3
concentration varies depending on the concentrations of O, O 2
and M*. 20.4 CO2(g) dissolves in seawater to form H2CO3(aq),
which dissociates to form HCO3 and then CO32, as well as
H(aq) ions, decreasing the pH of the ocean. Carbon is removed
from the ocean as CaCO3(s) in the form of sea shells, coral and
other carbonates. 20.8 (a) Its temperature profile
(b) troposphere, 0 to 12 km; stratosphere, 12 to 50 km;
mesosphere, 50 to 85 km; thermosphere, 85 to 110 km
20.11 8.3 * 1016 CO molecules 20.13 570 nm
20.15 (a) Photodissociation is cleavage of a bond such that two
neutral species are produced. Photoionisation is absorption of a
photon with sufficient energy to eject an electron, producing an ion
and the ejected electron. (b) Photoionisation of O2 requires
1205 kJ mol1. Photodissociation requires only 495 kJ mol1.
At lower elevations, high-energy short-wavelength solar
radiation has already been absorbed. Below 90 km, the increased
concentration of O 2 and the availability of longer wavelength
radiation cause the photodissociation process to dominate.
20.17 A hydrofluorocarbon is a compound that contains
hydrogen, fluorine and carbon; it contains hydrogen in place
of chlorine. HFCs are potentially less harmful than CFCs
because photodissociation does not produce Cl atoms, which
catalyse the destruction of ozone. 20.19 (a) The C ¬ F bond
requires more energy for dissociation than the C ¬ Cl bond and
is not readily cleaved by the available wavelengths of UV light.
(b) Chlorine is present as chlorine atoms and chlorine oxide
molecules, Cl and ClO, respectively. 20.22 (a) H2SO4(aq) +
CaCO3(s) ¡ CaSO4(s) + H2O(l) + CO2(g) (b) The CaSO4(s)
would be much less reactive with acidic solution, since it would
require a strongly acidic solution to shift the relevant equilibrium
to the right. CaSO4(s) + 2 H+(aq) Δ Ca2+(aq) + 2 HSO4 -(aq)
20.24 (a) Ultraviolet (b) 357 kJ mol1 (c) The average C ¬ H bond
energy from Table 7.4 is 413 kJ mol1. The C ¬ H bond energy in
CH 2O, 357 kJ>mol, is less than the ‘average’ C ¬ H bond energy.
A-26
(d)
Answers to Selected Exercises
O
H
H hn
C
H
O
Chapter 21
C H
21.1 (c) and (d) are the same
21.5
CH3
20.26 0.099 M Na+ 20.28 4.36 * 105 g CaO 20.29 The
minimum pressure required to initiate reverse osmosis is
greater than 5.1 atm. 20.31 (a) CO2(g), HCO 3 -,
H2O(l), SO 4 2-, NO 3 -, HPO 4 2-, H 2PO 4 - (b) CH4(g), H2S(g),
NH3(g), PH3(g) 20.33 2.5 g O 2 20.35 Mg2+(aq) +
Ca(OH)2(s) ¡ Mg(OH)2(s) + Ca2+(aq) 20.38 4 FeSO4(aq) +
O2(aq) + 2 H2O(l) ¡ 4 Fe3+(aq) + 4 OH - + 4 SO42-(aq);
Fe3+(aq) + 3 HCO3-(aq) ¡ Fe(OH)3(s) + 3 CO2(g)
20.42 The averate molar mass is 34.77 g mol1. If all O2 has
photodissociated then the average molar mass is 37.0 g mol1.
20.45 (a) Photodissociation of CBrF3 to form Br atoms requires
less energy than the production of Cl atoms and should occur
hn "
CF3(g) + Br(g);
readily in the stratosphere. (b) CBrF3(g)
Br(g) + O3(g) ¡ BrO(g) + O2(g) 20.47 The formation of
NO(g) is endothermic, so K increases with increasing
temperature. The oxidation of NO(g) to NO2(g) is exothermic,
so the value of K decreases with increasing temperature.
20.49 (a) CH4(g) + 2 O2(g) ¡ CO2(g) + 2 H2O(g)
(b) 2 CH4(g) + 3 O2(g) ¡ 2 CO(g) + 4 H2O(g) (c) 3.0 dm3
oxygen, 14.0 dm3 dry air 20.51 Most of the 390 watts m2
radiated from the Earth’s surface is in the infrared region of the
spectrum. Greenhouse gases absorb much of this radiation,
which warms the atmosphere close to the Earth’s surface and
makes the planet liveable. 20.53 (a) NO(g) + hn ¡ N(g) +
O(g) (b) NO(g) + hn ¡ NO+(g) + e - (c) NO(g) +
O3(g) ¡ NO2(g) + O2(g) (d) 3 NO2(g) + H2O(l) ¡
2 HNO3(aq) + NO(g) 20.55 Because NO has an odd electron,
like Cl(g), it could act as a catalyst for decomposition of
stratospheric ozone. This would lead to an increase in the
amount of damaging ultraviolet radiation that reaches the
troposphere. In the troposphere NO is oxidised to NO 2 by O 2 .
On dissolving in water, NO 2 disproportionates into NO3-(aq)
and NO(g). Over time the NO in the troposphere will be
converted into NO3–, which is in turn incorporated into soils.
20.56 (a) PNO2 1.9 10–6 kPa (b) 2.2 * 1019 NO2 molecules
20.60 9 days. 20.63 (a) 17e -, 8.5e - pairs
O
N
O
O
N
O
(b) The fact that NO 2 is an electron-deficient molecule indicates
that it will be highly reactive. Dimerisation results in the
formation of a N ¬ N single bond which completes the octet
of both N atoms. In a closed system, NO 2 and N2O 4 exist in
equilibrium. In an urban environment, NO 2 is produced from
hot car combustion. At these temperatures, equilibrium favours
the monomer because the reaction is exothermic. (c) NO 2 is an
oxidising agent and CO is a reducing agent, so we expect
products to contain N in a more reduced form, NO or N2 , and C
in a more oxidised form, CO 2 . (d) No. Because NO 2 is a very
reactive odd-electron molecule, we expect it to react or
photodisssociate before it can migrate to the stratosphere.
H3C¬CH¬CH2¬ CH2¬CH¬CH2¬ CH3
CH2
CH3
Br
21.7 CH3CH2CH2CHCH3 21.8 (a) CH3Cl Cl° ¡
° Cl HCl, °CH Cl Cl° ¡ CH Cl (b) Free-radical
CH
2
2
2 2
halogenations are indiscriminant because of the reactivity of
the free radical and the chain reaction profile. (c) An excess
of chlorine is best though 2 equivalents will suffice.
21.9 Carbon, hydrogen, oxygen, nitrogen, sulfur, phosphorus,
fluorine and chlorine. Oxygen, nitrogen, fluorine and chlorine
are more electronegative than carbon; sulfur has the same
electronegativity as carbon. 21.11 An empirical formula
shows the simplest mole ratio of elements in a compound. For
example, an empirical formula of CH2 implies that the
molecule contains only carbon and hydrogen and that they are
present in a 1:2 ratio. A molecular formula shows the exact
number and kinds of atoms in a molecule. For example, C6H12
((CH2)6) tells us the number of carbons and hydrogens actually
found within the molecule. A structural formula shows the
bonding of atoms within a molecule. In this instance, it shows
that the hydrocarbon is cyclic.
H
H
H
C
H
C
H C
H
H C
C H
C
H
H
H H
21.13 (a) A straight-chain hydrocarbon has all carbon atoms
connected in a continuous chain. A branched-chain
hydrocarbon has a branch; at least one carbon atom is bound
to three or more carbon atoms. (b) An alkane is a complete
molecule composed of carbon and hydrogen in which all
bonds are s-bonds. An alkyl group is a substituent formed by
removing a hydrogen atom from an alkane. (c) Alkanes are
said to be saturated because they cannot undergo addition
reactions, such as those characteristic of carbon-carbon double
bonds. 21.15 (a) 109.5° (b) sp3 hybrid orbitals
21.17 Constitutional isomers have the same molecular formula
but a different connectivity for the atoms. Constitutional
isomers are not interconvertable without the breaking or
formation of a C – C bond. For example, butane and 2methylpropane are constitutional isomers. Conformational
isomers are isomers that differ only in their three-dimensional
arrangement in space by rotation about a single C – C bond;
for example, the eclipsed and staggered forms of butane.
21.19 65. 21.21 (a) A compound that contains hydrogen and
carbon atoms (b) yes (c) CH3CH2CH3 (d) butane:
CH3CH2CH2CH3, molecular formula C4H10, empirical
formula C2H5.
A-27
Answers to Selected Exercises
21.23 (a)
H
H
H
we would expect polyethylene to be more soluble in organic
solvents such as hexane.
21.40
H
C
C
H C
H
H C
C H
C
H
H
H H
(b) No. The difference lies in one extra C – C bond and two
less C – H bonds. (c) No, its molecular formula is C6H10.
Cyclohexane has a molecular formula of C6H12.
H H
H
H
C
C
H C
H C
C
H
C
H
H H
21.25 Butane and 2-methylpropane are constitutional isomers.
Butane molecules are able to maximise their London dispersive
forces when compared with 2-methylpropane which is more
globular. The energy associated with this stabilising interaction
needs to be overcome for butane to boil. The higher boiling
point of butane equates to this extra energy.
21.27 (a) This molecule exists in the form that minimises
1,3-diaxial repulsions. To do this, the larger group (C(CH3)3)
is equatorial (as shown). The Br atom is much smaller and
hence occupies an axial position (as shown). (b) Br axil,
C(CH3)3 eq
21.30
(a)
(b)
CH3
CH3
H
H
Br
H
H
CH3
H
H
21.32
(a)
CH3
CH3
cyclobutane
methylcyclopropane
21.42 (a) 2-methylhexane (b) 4-ethyl-2,4-dimethyldecane
CH3
(c) CH3
CH2
CH2
CH
CH2
CH3
CH3
CH2
(d) CH3
CH2
CH2
CH2
CH
CH3
CH2
C
CH3
CH3
CH2
CH2
CH3
CH
(e)
CH2
CH2
CH2
21.44 (a) 2,3-dimethylheptane (b) methylcyclobutane
(c) 1,4-dichlorocyclohexane 21.46 Ethane, ethanol and
ethylene all have the prefix eth which indicates they all contain
two carbons in their molecular formulae. 21.48 The
compound (shown right) would be lowest in energy because
the two methyl groups are furthest apart. 21.50 (a) C8H16,
ethylcyclohexane (b) C4H10, 2-methylpropane (c) C6H12,
1,2,3-trimethylcyclopropane (d) C6H14, 3-methylpentane.
Constitutional isomers:
(a)
(b)
(b)
CH2CH3
Br
H
H
H
H
H
H
H
H
Br
CH2CH3
21.34 (a) Because they are a carbon network and do not
contain C – H bonds. They do not follow the general
molecular formula CnH2n+2. (b) Not really; the bonding
geometry of carbon in a nanotube is very different from that in
cyclohexane. There are also problems similar to 1,3-diaxial
repulsions with trying to include hydrogens on the inner
surface of the tube. 21.36 Cycloalkanes are able to have
geometric isomerism due to the restricted (in fact, impossible)
rotation about any C – C bond within the cyclic framework.
Alkanes are able to rotate freely about C – C bonds and so
are able to interconvert freely. Such isomers are known as
conformational isomers. 21.38 The polymer shown here is
an alkane. Since molecules are more soluble in ‘like’ solvents,
(c)
(d)
21.52 (a) In a substitution reaction one atom or group of atoms
replaces another atom. An oxidation reaction typically occurs
with the loss of electrons within a chemical reaction. In terms
of alkanes, combustion is classed as an oxidation reaction.
(b) The reaction is:
C
H
+ Br2
light
¡
C
Br
This substitution reaction is classed as regiospecific because it
occurs preferentially at one carbon, namely the tertiary carbon.
(c) The reaction is:
C5H12 + 8O2 ¡ 5CO2 + 6H2O
2,2-dimethylpropane
A-28
Answers to Selected Exercises
21.54 ¢Hcomb>mol CH 2 for cyclopropane = 696.3 kJ, for
cyclopentane = 663.4 kJ. ¢Hcomb>CH 2 group for cyclopropane
is greater because C3H 6 contains a strained ring. When
combustion occurs, the strain is relieved and the stored energy
is released. 21.56 Since all the carbon and hydrogen within
the alkane is expressed as CO2 and H2O in the product, we can
use the mass of each produced to calculate the molar ratios of
C : H. Number moles CO2 ⫽ 4.800 g/42 g mol–1 ⫽
0.11 moles ⫽ number moles of C. Number moles
H2O ⫽ 2.475 g/18 g mol–1 ⫽ 0.1375 moles. Number moles
H ⫽ 2 ⫻ 0.1375 ⫽ 0.275 moles. Ratio C : H ⫽ 0.11 : 0.275 ⫽
1 : 2.5 or 2 : 5. Since the general molecular formula for an
aliphatic alkane is CnH2n+2, the smallest possible molecular
formula must be C4H10.
21.58
(a)
(b)
(compounds II, V, VI), the product of highest yield will be
isomer IV, in which substitution at the tertiary carbon has
taken place. 21.66 (a) Constitutional isomers (b) Geometric
isomers; mirror images (c) At 3° carbons (where methyl groups
bond to the rings).
Chapter 22
22.2 DNA, hands, gloves, feet, socks, sea shells, twins and can
openers, to name a few. 22.3 (a) and (c) 22.5 (a) Mirror
images (b) Stereoisomerism; they are enantiomers.
22.6 Structures (3) and (4) are identical to (1); (2) and (5) are
geometric isomers of (1). 22.7 Chiral centres are marked with
an asterisk.
H
OH
(a)
(b)
N * * S
CH3
RO
*C
H3C
(c)
(d)
Cl
H C H
H
21.62
(a)
(c)
H
H
H C C C H
H HH
CH3
CH3
>
most stable
least stable
(CH3)2CCH2CH3 > (CH3)2CHCH2CH2 > CH3
most stable
least stable
OH
HO
*
H
N
O
*
CH3
HO
*
>
CH3
*
penicillin V
(d)
H3C
N
COOH
lactic acid
(b)
O
R = C6H5
(c)
21.60 (a)
(b)
O
H
COOH
camphor
epinephrine
22.8 (a) No. The name only indicates which way this
enantiomer rotates plane-polarised light. (b) and (c):
*
*
* = chiral centre
21.64
Br
Br
I
II
Br
Br
Br
III
IV
V
Br
(d) (R)-limonene 22.11 (a) An organic molecule must rotate
plane-polarised light. (b) A racemic mixture contains an equal
amount of enantiomers. The effect of one enantiomer on planepolarised light is cancelled by its enantiomer, which rotates the
plane-polarised light by an equal amount in the other
direction. 22.13 (a) –Br > –OH > –CH2CH3 > –CH3
(b) –OCH3 > –COOH > –CH(CH3)2 > –H
(c) –NH2 > –COOH > –CH2CH2CH2COOH > –H
(d) –OH > –COOCH3 > –COOH > –CH2OH
22.15 (a) They are constitutional isomers. (b) They are
geometric isomers that are nonsuperimposable mirror images.
They are enantiomers. 22.16 (a) cis-1,2-dimethylcyclopentane
(left) ⫽ (1S, 2R)-dimethylcyclopentane. trans-1,2dimethylcyclopentane (right) ⫽ (1S,2R)-dimethylcyclopentane
(b)
CH both substituents are either axial or equational.
3
Br
VI
VII
Despite the fact that there are nine primary positions
(compounds I, III, VII) and six secondary positions
CH3
.
(c) (1S,2S)-dimethylcyclohexane.
Answers to Selected Exercises
22.18 Chiral centres are marked with an asterisk.
22.32
(a)
H3C
*
HO
CH3
*
*
O
*
HO
CH3
H
H
HO
H
H
H
Cl
Cl
O
N(CH3)2
*
22.34
22.20 (a) Stereocentre is marked with an asterisk.
NH3
ONO
Pd
Pd
H3N
NH3
ONO
cis
trans
(b) [Pd(NH 3)2(ONO)2], [Pd(NH 3)2(NO2)2]
H = HO
OH
H
H
H
OH HO
COOH
COOH
OH
H
H
H
OH
OH
COOH
COOH
enantiomers
meso compound
22.28
(a)
(b)
CH3
COOH
CH3
H
OH
H
OH
COOH
(d)
CH3
H
Cl
H
Cl
Br
O
Br
CH3
22.30
H
H3C
CHO
COOH
(1R, 2R)
N
N
OHC
N
⫹
N
N
V
V
N
N
Cl
Cl
Cl
Cl
N
(d) [Co(NH 3)4Br2]Cl, [Co(NH 3)4BrCl]Br 22.36 Yes. No
structural or stereoisomers are possible for a tetrahedral
complex of the form MA2B2 . The complex must be square
planar with cis and trans geometric isomers. 22.38 In the
trans isomer the Cl ¬ Co ¬ NH 3 bond angle is 180°; in the
cis isomer this bond angle is 90°. The cis isomer is chiral.
22.39 (a) One isomer (b) trans and cis isomers with 180° and
90° Cl ¬ Ir ¬ Cl angles, respectively (c) trans and cis isomers
with 180° and 90° Cl ¬ Fe ¬ Cl angles, respectively. The cis
isomer is optically active. 22.41 (a) In a square planar
complex, if one pair of ligands is trans, the remaining two
coordination sites are also trans to each other. The bidentate
ethylenediamine ligand is too short to occupy trans
coordination sites, so the trans isomer of [Pt(en)Cl2] is
unknown. (b) The minimum steric requirement for a transbidentate ligand is a medium-length chain between the two
coordinating atoms that will occupy the trans positions.
A polydentate ligand such as EDTA is much more likely to
occupy trans positions because it locks the metal ion in place
with multiple coordination sites and shields the metal ion from
competing ligands present in the solution.
Chapter 23
23.1 (c), (d) 23.3 Molecule (c)
23.5
O
O
C
H
2
1
⫹
(c)
H
CH3
H
H3C
ONO
H3N
(b) Because nature is very specific in its use of chirality for a
particular function, we would expect the enantiomer not to
have the same effect, or a drastically lowered effect at best.
22.22 (a) S (b) R (c) S 22.24 [a]D ⫽ ⫹65, l ⫽ 5 cm ⫽ 0.5 dm,
c ⫽ 8/25 g/mL. Substituting into Equation 21.11 gives
a ⫽ ⫹10.4°. Sucrose is dextrorotatory.
22.26
COOH
COOH
COOH
COOH
(c)
CBr3
ONO
(a)
*
NH
HO
H
Cl
*
*
*
SH
NCH3
* H *
(b)
H
N
O
A-29
CH3
COOH
O
C
O
O
O
23.6 (a) The molecule contains a double bond and so shows
some trigonal planar geometry. (b) The molecule contains a
triple bond and so some portion of the molecule has a linear
geometry. 23.8 Cyclic alkane: C6H12, saturated; cyclic alkene:
C6H10, unsaturated; linear alkyne: C6H10, unsaturated;
aromatic hydrocarbon: C6H6, unsaturated. 23.10 CnH2n–2
A-30
Answers to Selected Exercises
23.28 (a)
23.12
H
H
C
1-hexene
C
H
2-methyl-1-pentene
2-hexene
2,3-dimethyl-1-butene
H
C
C
H
H
All bond angles are 120˚
(b) The very small difference in the C – C single bond length is
an effect of conjugation.
23.30 (a)
(b)
s-bond
2-methyl-2-pentene
3-hexene
(c)
2,3-dimethyl-2-butene
3-methyl-1-pentene
p-bond
4-methyl-2-pentene
3-methyl-2-pentene
2-ethyl-1-butane
4-methyl-1-pentene
3
23.14 (a) These are sp hybridised. Their lobes point to the
corners of a tetrahedron. (b) These are sp2 hybridised and form
a trigonal planar geometry. The unhybridised p orbital is
used for p-bonding. (c) These are sp hybridised and form
a linear geometry. The two unhybridised p orbitals are
used for p-bonding. 23.16 (a) cis-6-methyl-3-octene
(b) 4,4-dimethyl-1-hexyne (c) 3-chloro-1-propyne
23.18 Geometric isomerism in alkenes is the result of restricted
rotation about the double bond. In alkanes bonding sites are
interchangeable by free rotation about the C – C single bonds.
In alkynes there is only one additional bonding site on a triply
bound carbon, so no isomerism results.
23.20
H
HC
CH CH
CH CH
2
C
H3C
3
3
2
C
C
H
E-2-pentene
H
(d) A s-bond is stronger since it relies on substantial overlap of
the two orbitals. p-Bonds are generally weaker because the
p-orbitals overlap in a less complete fashion. This overlap is
dictated by the s-bond length. 23.32 (a) 18 valence electrons
(b) 16 valence electrons form s-bonds (c) 2 valence electrons
form p-bonds (d) No valence electrons are nonbonding
(e) The left and central C atoms are sp2 hybridised; the right C
atom is sp3 hybridised.
23.33 (a)
O
O
C
H
Z-2-pentene
Cyclopentene cannot exhibit E,Z isomerism. The E isomer is
too strained to be feasible. 23.22 (a) (Z)-3-chloro-4-methyl3-heptene (b) (E)-2-chloro-2-pentene (c) (E)-1-bromo-1-chloro2-methyl-1-butene (d) (E)-1-amino-3-methyl-2-hexene
23.24 (a) 2 unhybridised orbitals, 2 p-bonds (b) 1 s-bond and
2 p-bonds (c) The p-bonds impart restricted rotation about a
C – C bond, thereby reducing the number of possible
conformers a molecule can possess. 23.26 Geometrically, this
is difficult. The C – C bond lengths associated with the ring
are not long enough to accommodate a triple bond, despite its
smaller bond length. The resulting molecule would be highly
reactive.
O
H3C
O
(b) Yes, see part (a) (c) The existence of more than one
resonance structure is a good indication of delocalisation.
23.35 (a)
H
H H
H
C
H
H
H
methane
3
C
C
H3C
C
C
H
H
H
ethane
(b) sp3 (c) The carbons in ethane are able to exist as sp2 hybrid
orbitals, which would lead to multiple bonding. Methane
cannot have multiple bonds between the carbon and hydrogen
atoms. 23.37 (a) An addition reaction is the addition of some
reagent to the two atoms that form a multiple bond. In a
substitution reaction one atom or group of atoms replaces
another atom. Alkenes typically undergo addition, while
aromatic hydrocarbons usually undergo substitution.
(b)
CH3
CH3 H
CH3
CH
C
C
CH3 + Br2
CH3
CH3 H
CH3
CH
C
C
Br Br
(b)
23.39 (a)
Cl2
Cl2
NH4F
CH3
Answers to Selected Exercises
23.41 (a) One
(b)
(c)
Br
(d)
OH
A-31
OCH3
Br
CHCH3
HBr
(c) The hydrogen atom in HBr adds to the carbon atom of the
double bond bearing the most hydrogen atoms bonded
directly to it. (d) Yes!
23.42
H
H
H
H
H
HBr
Br
H
H
H
Br
23.53 Polyethylene has many CH2 groups (>100) and is formed
by the polymerisation of ethylene. Decane is not classed as a
polymer because it doesn’t have enough repeat units, nor is it
formed in any appreciable quantity by the polymerisation of
ethylene, the smallest alkene.
23.55 (a)
(b)
(c)
Ph
Cl
23.57 Addition or free-radical polymerisation
intermediate
F F F F F F
23.44 (a) H2/Pd (b) H2O/H2SO4 (c) H2/Pd (1 equiv.) (d) Br2
(e) Br2/CH3OH
23.45
(a)
(b)
CH2
H
H
C
C
C
H
CH3
H
(c)
CH3
F F F F F F F F
23.59 HDPE has a higher average molecular mass and less
chain branching which leads to a higher crystallinity.
23.61 Is the neoprene biocompatible? Does it provoke
inflammatory reactions? Does neoprene meet the physical
requirements of a flexible lead? Will it remain resistant to
degradation and maintain elasticity? Can neoprene be
prepared in sufficiently pure form so that it can be classified
as medical grade?
23.63
O
O
N
23.47
(a)
O
N
O
N
N
(b)
CH3
CH2
(c)
Chapter 24
24.1 (a) alkene, ether (b) carboxylic acid, carboxylic ester
(c) ketone, alkene and alcohol (d) phenol, alcohol, ether,
amine, alkene (e) alcohol, ether, haloalkane (hemiacetal).
24.3 Tertiary carbocation is the most easily formed because it
is the most stable.
(d)
CH3
HO
OH
23.49 (a)
H2/Pd
24.6 Five
24.7
(b) To put two D atoms on a single carbon, it is necessary that
one of the already existing C ¬ H bonds in ethylene be broken
while the molecule is adsorbed, so that the H atom moves off
as an adsorbed atom and is replaced by a D atom. This requires
a larger activation energy than simply adsorbing C2H4 and
adding one D atom to each carbon.
23.52
OH
(b)
(a)
OH
CH2CH3
Br
Cl
OH
OH
OH
HO
OH
24.9 (a) The OH group allows for hydrogen bonding. This
interaction is associated with an amount of energy that needs
to be overcome in order for the liquid to boil. The methyl
group does not hydrogen bond. (b) Two ether oxygens; a
much longer ether that will have more significant London
A-32
Answers to Selected Exercises
dispersion forces. 24.11 Despite the ability of both alcohols to
hydrogen bond with water, 1-hexanol is more alkane-like
because of its longer alkyl chain. This portion of the molecule
does not interact strongly with water. 24.13 Alcohols are far
more reactive, while ethers are inert. Alcohols are useful
solvents in reactions that also use it as a reagent, for example,
esterification. 24.15 (a) diethyl ether, or ethoxyethane
(b) propoxymethane (c) 3-methyltetrahydrofuran
(d) 2,6-dimethyltetrahydropyran.
24.17
OH
(a) CH3CH2CHCH3
most stable
least stable
24.23 2-methyl-2-propanol is a 3° alcohol, which are not
oxidised. 1-propanol is a 1° alcohol and so is easily oxidised.
24.24 Dioxane can act as a hydrogen bond acceptor, so it
will be more soluble than cyclohexane in water.
24.26 one 24.27 (a) 5-bromo-2-methylcyclohexanol;
(b) 6-chloro-1-methylcyclohexene 24.28 the C – Cl bond is
polar with d at carbon and d at chlorine due to difference
in electronegativity.
24.29 (a)
(b)
Cl
I
CH3
(c)
I
CN
24.40 (a) 1-bromopropane and 2-methyl-1-propanol
(b) chloroethane and 2,2-dimethyl-1-propanol
(c) chloroethane and cyclopentanol
24.41
OCH2CH3
1. NaOH/CH3CH2I
HO
24.43
24.34 React ethanol with sodium. Once the alkoxide is formed,
add in bromoethane.
24.33 (a) HClaq (b) H2/Pd, NaOH (c) H2SO4 (d) SOCl2,
NaOCH(CH3)2
24.36
Cl
OCH CH
Cl
NaOCH2CH3
b-elimination
OH
2-methyl-pentan-2-ol
24.44 (a)
(b) II and III
I
II
III
24.46 This is a difficult one! The norbornane ring system is
heavily strained as a result of the bridging CH2 group. Having
a double bond at the base of the bridge (Zaitsev product most substituted double bond) would increase the strain
further. The geometry at the carbon found at the base of the
bridge would not be trigonal planar.
24.47
(b)
(a) H3C
(c)
3
CH3
(d)
CH3
24.49
2
OH
2. H2O/H
CH3
24.30 (a) NaOH (b) H2O/H2SO4 (c) H2SO4 (d) SOCl2/Cr
24.32 (a)
(b)
O Na
substitution
(d) CH3CH2NH2CH3
(e)
H3C
NaOCH2CH3
OCH3
OH
(note inversion of
stereochemistry)
CH3S
(b)
(b) HOCH 2CH 2OH
(c) CH 3CH 2OCH 2CH 3
24.19 (a) 4-ethyl-2-methyl-3-hexanol (b) butane-2,3-diol
(c) 2,4-dimethyl-3-pentanol (d) cyclohexane-1,2,3-triol.
24.21
(c)
24.38
(a)
2-methylpropene
24.50 (a) Step 1: HCl. Step 2: KOtBu/tBuOH. Alternatively:
Step 1: H2O/H2SO4. Step 2: H2SO4 (b) Step 1: HCL.
Step 2: CH3ONa/CH3OH. (c) Step 1: H2O/H2SO4.
Step 2: H2SO4. Product: 2-chloro-2-methylcyclohexane
(d) Propane
Answers to Selected Exercises
24.52
24.79 (a)
OH
H2O
H2SO4
1. NaH (base)
Cl
2.
HCl
Cl
CH3
CH3
CH3
O
H
H
H
CH3
(b)
24.53 (a) iv (b) H2O < CH3CO2 < HO
24.55 (a) CH3CH2OH, (b) CH2 ! CH2
24.57 The tertiary haloalkane. This alkyl chloride is much
more hindered. 24.59 The rate doubles.
24.61
NaCN
CH3
H
This intermediate is favoured because it generates a 3°
carbocation.
There is an inversion of stereochemistry leading to the R
configuration 24.63 (a) Methoxide is more nucleophilic
because the nucleophilicity of tert-butoxide is diminished by
steric effects (b) CH3I KOCH(CH3)2 24.64 (a) Doubles the
rate; (b) No change as this is not the rate-determining step.
24.66 (a) No, the rate-determining step is the loss of the
leaving group to form the intermediate carbocation,
(b) Rate k[haloalkane], (c) polar solvents
24.67 (a) H C
3
Br
(b) Because this is a secondary alkyl bromide, the reaction
occurs through both an SN1 and SN2 mechanism, leading to
a mix of diastereoisomers with slightly more inversion
than retention at the center undergoing substitution.
24.69 The newly formed double bond in this compound is
conjugated with the phenyl ring. 24.71 (a) B, (b) A
24.73 (a) (i)
(ii) CH3CH2CHBrCH2CH3 or
Br
CH3CHBrCH2CH2CH3,
(iii) CH3CH2CH(CH3)CH2Br
24.75 E1
24.77
CH3
CH3
(a)
(b)
SN2
SN1
CN
CH3
CH2
CN
I
CH2
CH3
(c)
H2O
OH2
CH3
(d)
CH3
CH3
O
H
CH3
H 2O
OH
H
CH3
H3O
CH3
This reaction is catalytic in H3O.
24.81 Bromination using Br2/CCl4; followed by two
dehydrohalogenations using a strong base such as NaOEt or
NaNH2
24.83
HO
OH
OH2
O
H
H
O
H
H
H2O
H3O
O
H 2O
O
H
24.85
OH
H2SO4
H3O
heat
heat
OH
OEt
(c)
Ph
(d)
E1
H3C
E2
Ph
(e)
CH3
H
O
H3C
Cl
SN1
A-33
Cl
NaOH
O
A-34
Answers to Selected Exercises
Chapter 25
(b)
25.1
(a)
(b)
OH
C
CH3CH2CH2CH2CH2CH2OH
H
1-hexanol
Conditions: PCC
cyclohexanol
Conditions: K2Cr2O7/H2SO4
HO
(c)
OH
(c)
(d)
CH2OH
CH2
25.9
(a)
OH
(b)
H
COOH
O
2,6-dimethylheptan-1-ol
Conditions: CrO3/H2SO4/H2O
25.4
phenylmethanol
Conditions: K2Cr2O7/H2SO4
HN
NH
OH
pyranose
OH
OH
OHC
O
O
CH3
OH
25.12 (a) ester (b) acetal (c) alcohol (d) hemiacetal (e) acetal
(f) ether 25.14 (a) alkyne, aldehyde (b) alkene, carboxylic acid,
aromatic (c) alkyl halide, ketone (d) alkene, ester (e) amine,
amide 25.16 (a) 2-pentanone (b) cyclopentanone (c) 3-bromo2-methylcyclopentanone (d) propanal (e) 3-nitrobenzaldehyde
(f) 4-methyl-2-pentanone
25.18
O
O
furanose
H
25.20 (a)
O
OH
CH3HN
CH2OH
O
O
C
pyranose
H3C
OH
25.5 (a)
(b)
O
CHO
+ CH3CHO
CH3
(b) polar (c) weak electrostatic interactions e.g. dipole-dipole
(d) 1-propanol has stronger electrostatic interactions in the
form of hydrogen bonding. 25.21 (a) Step 1: H2/Pd. Step 2:
NaOH (b) K2Cr2O7/H2SO4 (c) H2SO4
25.23
O
OH
CH3
(d)
(c)
O
O
product 2
product 1
25.25
O
C
(e)
CH3
OHC
CHO
+ CH3CH2MgBr
O
25.6
C
CH2CH3
O
25.8 (a) K2Cr2O7/H2SO4 (b) NaBH4/CH3OH
25.10 (a)
OH
+ CH3MgBr
C
H3C
CH2CH3
+ PhMgBr
Answers to Selected Exercises
25.27
(a)
25.41 (a)
(b)
OH
(b)
NC OH
OH
HOOC
OH
(c)
A-35
(c)
CN
(d)
OH
HO
25.29
(a)
(b) HN
NH2
(e)
OH
N
OH
HOOC
COOH
25.43 (a and b)
N
(c)
O
O
O
N
25.31
(a) H3CH2CO
OCH2CH3
(b)
O
O
(c)
O
keto
keto
enolate
(c) As a result of these canonical forms, the methyl protons are
more acidic than they would be in pentane.
(d)
O
This enol is conjugated into the
aromatic ring.
25.45
O
25.33 (a) Nucleophilic addition. (b) acid chloride > aldehyde >
ketone > ester > amide 25.35 Step 1: HBr; Step 2: Mg, ether;
Step 3: HCHO; Step 4: H3O
25.36 (a)
NHCH2CH3
OH
25.47 Br2/H, H2O; NaCN
25.49
O
O
Br
Step 1:
Step 2:
(b)
N
H
(c)
NHCH3
25.38
O
O
25.50 (a) any molecule with molecular formula C(H2O)n,
where n is an integer (b) a simple sugar (c) two
monosaccharides covalently bonded to each other (d) a
stereoisomer about C1 (e) a monosaccharide containing an OR
group (R ≠ H) at position C-1 25.52 (a) In the linear form of
galactose, the aldehydic carbon is C1. Carbon atoms 2, 3, 4 and
5 are chiral because they each carry four different groups.
(b) Both the b form (shown here) and the a form (OH on
carbon 1 on same side of ring as OH on carbon 2) are possible.
CH2OH
5
HO
25.40
O
C4
H
C
O
H
OH
H
C
3
OH
1
C
2
H
OH
Galactose
C
H
A-36
Answers to Selected Exercises
25.54 The empirical formula of glycogen is C6H10O5. The
six-membered ring form of glucose is the unit that forms the
basis of glycogen. The monomeric glucose units are joined by
a linkages. 25.56 (a) C6H12O6 (b) C6(H2O)6
(c)
(d) b-anomer
OH
H
HO
HO
OH
HO
H OH
H
H
25.58 Galactose is an aldohexose; ribose is an aldopentose.
25.60 (a) mutarotation (b) 72% 25.62 (a) disaccharide
(b) C1 on both saccharide units (c) The glycosidic bond is
at C1 of the saccharide unit on LHS; it’s a 1,4-glycosidic bond
(d) b-anomer 25.64 (a) D-glucosamine
(b)
OH
H
HO
HO
OH
HO
H NH2
H
H
25.66 (a)
CHO
H
HO
(b)
OH
O
C
HO
CH2OH
H
H
CH2OH
CH2OH
CHO
(c)
(d)
HO
H
HO
H
CH2OH
O
C
H
HO
H
CH2OH
OH
H
OH
CHO
25.68 (a) two (b) 22 4 (c) only one expected to be found
because of the specificity of nature and its enzymes.
(c)
(d)
OH
2,6-dimethylheptan-1-ol
Conditions: CrO3/H2SO4/H2O
26.1 (a) aromatic, amide, alkyl halide and alcohol, nitro group
(b) aromatic, carboxylic acid, ester (c) phenol, amide
(d) ether, aromatic, alkane, amine, alcohol, ester.
26.3
(a)
(b)
OH
CH3CH2CH2CH2CH2CH2OH
OH
cyclohexan-1,2-diol
1-hexanol
Conditions: PCC
phenylmethanol
Conditions: K2Cr2O7/H2SO4
26.4 Wax 2 will have the lower melting point. As a result of the
kink produced by the double bond, this wax cannot pack and
interact as well as wax 1.
26.6 (a)
O
O
H
H
N
N C
C
n
repeat unit
(b) The amide functionality when combined with the
inflexible benzene ring linkers would make this polymer
more suitable as a high-strength, high-temperature
polymer. 26.7 (a) Methanoic acid (b) butanoic acid
(c) 3-methylpentanoic acid 26.8 The presence of both
¬ OH and ¬ C “ O groups in pure acetic acid leads us to
conclude that it will be a strongly hydrogen-bonded substance.
That the melting and boiling points of pure acetic acid are
both higher than those of water, a substance we know to
be strongly hydrogen-bonded, supports this conclusion.
26.10 Carboxylic acids are more acidic because the negative
charge of the conjugate base is stabilised by delocalisation.
Alcohols, with the exception of phenols, are unable to
delocalise the negative charge of their conjugate bases.
26.11 The carboxylic acid will react with the bicarbonate
solution, resulting in the evolution of CO2 (bubbles). The ester
will not. 26.13 Acid A is most acidic. The trifluoromethane
group is inductive, making the O – H bond weaker and
compound A more acidic than compound B, for instance.
26.15 (a) K2Cr2O7/H2SO4 (b) KMnO4
26.17
O
O
H
H
Chapter 26
CH2OH
CH2
Cl
C
C
O
H
vs
Cl
H
pH 3.37, 2.43, 1.74, 1.35
respectively
C
C
O
H
Cl
inductive nature of Cl
weakens O—H bond
by withdrawing
electron density
26.18 (a) ester (b) acetal (c) carboxylic acid (d) hemi-acetal.
26.20
O
C
Ph
OCH3 CH3MgBr,
O
C
CH3
OCH3
PhMgBr
Answers to Selected Exercises
26.21
(a)
O
26.29 (a)
O
(b)
O
CH2
HC
O
O
O
(a) CH3CH2O
CH2
C
(b) CH3N
O
O
CCH3
(c)
O
N-methylethanamide or
N-methylacetamide
26.24
CCH3
Phenylacetate
O
(a) CH3CH2C
O
O
O
R
R
R, R, R = long alkyl chains
Ethylbenzoate
H
R
O
Ph
26.22
A-37
CH3 NaOH
O
CH3CH2C
(b) A saturated fatty acid contains alkyl groups only. They are
normally solids or semisolids. An unsaturated fatty acid
contains double bonds which cause ‘kinks’ in the structure. As
a result, unsaturated fatty acids are more likely to be liquids at
room temperature. (c) The saturated triglyceride will have a
higher melting point.
26.30 (a) polar head group, hydrophobic chain
(b)
O
Na CH3OH
O
(b) CH3C
O
NaOH
O
CH3C
hydrophilic
Na
O
OH
hydrophobic
26.26
(a)
(b)
26.31
(a)
O
COOCH2CH3
COOH
(c)
CH3
(d)
HO
OH
O
O
(b)
26.27 The chiral centre is marked with an asterisk.
O
CH2
HC *
CH2
O
O
O
O
O
C15H31
COO(CH2)4CH3
(c)
PhH2COOC
C18H37
C20H41
COOCH2Ph
(d)
O
O
26.33 (a) SOCl2/pyridine (b) CH3NH2 (2 equivalents)
(c) CH3CH2CH2OH 26.34 Reagent A: K2Cr2O7/H2SO4.
Reagent B: SOCl2. Reagent C: NH3.
A-38
Answers to Selected Exercises
26.36
OH
26.50
OH
O
O
O
O
+ HNO3
NO2
O
O
O
O
n
H2/Pd
OH
OH
CH3OH/H
Ac2O
O
O
n
NHCCH3
CH3
NH2
H3C
O
O
OH
n
HO
O
26.52
26.38
HOOC
H2N
COOH
HO
O
OH
O
NH2
HO
HO
OH
O
H
N
N
H
O
O
n
26.40
O
HOOC
COOH
H2N
NH2
O
O
O
and
26.42
R
H
N
R
O
O
R
O
(CH2)3C
O
O
O
n H 2N
O
H
N
N
H
26.44
O
O
H H
n
O
O
H
O
N
(CH2)3C
OH
O
n
O
NH
(CH2)3C
O
NH
(CH2)3C
n
26.46 Poly(ethylene adipate) would be more suitable. It is more
hydrophilic and flexible. 26.48 Current vascular graft
materials cannot be lined with cells similar to those in the native
artery. The body detects that the graft is ‘foreign’, and platelets
attach to the inside surfaces, causing blood clots. The inside
surfaces of future vascular implants need to accommodate a
lining of cells that do not attract or attach to platelets.
Its strength and rigidity are a result of the cross-linking nature
of the polymer. The addition of aromatic rings also imparts
rigidity and strength. 26.53 The fewer steps in a process, the
less waste is generated. Processes with fewer steps require less
energy at the site of the process, and for subsequent cleanup or
disposal of waste. 26.55 (a) H2O (b) It is better to prevent
waste than to treat it. Produce as little nontoxic waste as
possible. Chemical processes should be efficient. Raw materials
should be renewable.
A-39
Answers to Selected Exercises
Chapter 27
27.20 (a)
27.1 (a)
(b)
(b)
27.2 (a) and (d)
27.4
O
CH2
27.3 (ii)
CH2
CH2
27.6 Benzene. The carbon hybridization of benzene is sp2
while that of cyclohexane is sp3. The more s character in
benzene causes its hydrogens to be more acidic than those of
cyclohexane. 27.8 Conjugation, cyclic planar structures
containing [4n 2} p electrons. 27.11 (a) 2-nitrotoluene or
o-nitrotoluene (b) 1,4-dibromobenzene or p-dibromobenzene
(c) 3-methyphenol (d) 1-chloro-2-methyl-2-phenylbutane
(e) 3-hydroxybenzaldehyde. 27.12 (a) Having two
contributors to the structure of benzene which differ by the
alternation of double and single bonds, on average each bond
length is neither single nor double but somewhere in between.
(b) answer (a) explains why.
27.13
H
H
H
H
(a)
H
CH2
etc.
O
H
CH2
p-bond
C
C
C
C
C
C
C
C
C
C
H
H
H
H
C
C
C
C
C
C
C
C
C
C
H
H
H
H
H
H
(b) This is because of the influence of the two cannonical
structures. Each C – C bond is neither a single nor a double
bond, but is best represented as something in between; for
example, as a ‘1.5’ bond.
has no contributing structures
27.22 The benzylic alcohol will undergo dehydration more
rapidly since it yields a secondary benzylic carbocation, which
is more stable than secondary carbocation of the aliphatic
alcohol.
27.23
OH
OH
OH
OH
Cl
Cl
H3C
CH3
27.24 They must have a planar, cyclic structure and contain an
unhybridized p orbital in each atom in the ring. Finally, the
cycle must contain [4n 2] p electrons 27.27 (a) An addition
reaction is the addition of some reagent to the two atoms that
form a multiple bond. In a substitution reaction one atom or
group of atoms replaces another atom. Alkenes typically
undergo addition, while aromatic hydrocarbons usually
undergo substitution.
(b)
CH3
CH3 H
CH3
CH
C
C
CH3 + Br2
CH3
CH3 H
CH3
CH
C
C
Br Br
(c)
Cl
Cl
H
27.15 The change in phase, from milky to clear, upon repeated
heating and cooling. 27.18 (a) Thermodynamics associated
with the hydrogenation of benzene, cyclohexene and 1,4cyclohexadiene. The difference in reactivity of cyclohexene
and benzene to bromination. (b) Aromatic systems have a
delocalised p-bonding system in which each p-orbital connects
to form a ‘super p-way’. This is explained in the following
diagram:
C
H
C
C
H
H
+ Cl2
H
H
FeCl3
H
+ HCl
H
Cl
Cl
27.29
Cl
H2C “ CH2
HBr
CH3CH2Br
CH2CH3
CH3CH2Br
AlBr3
C
C
H
(c)
H
H
H
CH3
27.31
(a)
(b)
CH3
CH3
NO2
C
H
Br
Cl
A-40
Answers to Selected Exercises
(c)
27.48 The deactivating nature of the NO2 group will preclude
it from undergoing reaction under Friedel-Crafts conditions.
27.50 CH3Cl/AlCl3; KMnO4/H, heat; HNO3/H2SO4
27.52 HNO3, H2SO4; Br2, FeBr3; KMnO4, H
27.54
Br
(d)
CH3
CH3 O
CH2CH3
CH3
H3C
27.33 (a) CH3Cl/AlCl3 (b) (CH3)2CHCl/AlCl3 (c) Cl2/FeCl3
(d) CH3COCl/AlCl3.
27.34
(a)
(b)
H
H
H
C
C
(d)
CH2
CH2
CH3
27.36 When benzene undergoes electrophilic substitution, the
product retains the stability associated with an aromatic
system. This stability would be lost in electrophilic addition
because the product is not aromatic.
27.37 (a)
H
H
H
E
E
E
(b) No, the carbocation is not aromatic since the ring contains
an sp3 centre.
27.39
NH2
NH2
NH2
H
Br
H
Br
27.40 (a)
(b)
O
27.41 (a)
(b)
(b) terephthalic acid
COOH
28.1 (a) molecule (i), disaccharide (b) molecule (iii), amino
acid (c) molecule (ii), organic base 28.3 Ala-His-Tyr-Ser,
AHYS 28.6 (a) alkaloid (b) carbohydrate (c) alkaloid
(d) nucleoside (e) amino acid (f) nucleoside
28.8 (a) cyclohexylamine (b) 1-amino-2-phenylethane
(c) N-methylethanamine (d) 3-pentanamine
(e) N,N-diethylcyclobutylamine (f) 5-bromo-2-ethyl-aniline
28.10
(a)
(b)
NO2
NH2
N
(c)
NH2
(d)
N
COOH
(e)
NH2
H
Br
27.56 (a)
Chapter 28
H
(c)
NO2
HOOC
H
NO2
CH3
C
H
H3CO
H
Br
NH2
COOH
28.12 (a) secondary amine (b) tertiary amine (c) primary amine
28.14 Aniline is a weaker base because the lone pair on nitrogen
interacts with the p-bonds of the phenyl ring. This means that
cyclohexylamine is a stronger nucleophile than aniline.
28.16
O
(a)
N
H2N
H
O
Cl
27.42 benzene
27.46
CH3
H
NO2
Answers to Selected Exercises
28.28 (a)
Cl
(b)
H
C
HO
COOH
O 2N
Cl
O
N
H
O
N
O2N
(b) two, cis 2 trans (c) trans more stable, reduces steric
interaction. 28.29 (a) A compound containing both amine
and carboxylic acid groups connected to the same carbon atom.
(b) By a condensation reaction of the amino group of one
amino acid with the carboxylic acid of another amino acid.
(c) F phenylalanine, Q glutamine, Pro proline,
Tyr tyrosine. 28.31 (a) g (b) b (c) a (d) b (e) d
28.33 (a)
O
O
O
H3N ¬CH ¬C ¬ OH
Pd 2
O
NH2
H2/Pd
NO2
NH2
2
HO
CH3
HO
H3C
(b) condensation reaction. 28.26 Only the major part of the
mechanism is shown. Proton transfers have not been shown.
O
O
(a)
H3C
(b)
H2N
O
C
O
H3N ¬CH ¬C ¬ O
H3N ¬CH ¬C ¬O
CH2
CH2
CH ¬CH3
CH 2
CH3
CH2
in diethyl ether
in diethyl ether in aqueous layer
28.20 (a) N-methylbutanamide (b) N,N-dimethylacetamide
(c) 2-methylpropanamide (d) N-cyclohexylpropanamide
28.22 (a) tertiary (b) secondary (c) primary (d) secondary
(e) tertiary (f) tertiary
28.24 (a)
CH3O
O
NH
Cl
C
C
high pH
(b) The first pKa value represents the point of deprotonation
of the carboxylic acid group, while the second pKa value
represents the point of deprotonation of the ammonium group.
28.35 (a) This is the point within a titration curve in which the
zwitterion is the predominant form in solution. (b) 6.01
28.37 (a) 0.13 (b) 0 28.39 The result is because of the
inductive effect of the ammonium group.
28.41 (a)
O
O
NH3 Cl
add
2 M HCl
CH3
low pH
N
H2N
NO2
CH2
NH2
Leu
Lys
(b) An aqueous solution containing both Lys and Leu with
pH 8 would mean that Lys is positively charged while Leu
would be negatively charged. Since electrophoresis separates
based on charge, Lys would migrate one way and Leu the
other.
28.43
(a)
Br
NH2
xs NH3
Ph
O
C
(d)
O
O
COO
H3O
Cl
NH3
Ph
O
C
Ph
COOH
NH2CH3
(c)
H2N ¬CH ¬C¬O
CH3
O
(b)
N
O
O2N
28.18
(a) NO2
COO
H3O
O
NH2Ph
NH
A-41
(b)
O
NH2
NaCN
Ph
H
NH4Cl
Ph
CN
NH4
A-42
Answers to Selected Exercises
28.45 (a)
O
O
O
H2N ¬CH ¬C ¬ NH¬ CH ¬C ¬OH
CH2
CH2
OH
CH 2
O
O
H2N ¬CH ¬C ¬ NH¬ CH¬C¬ NH¬ CH ¬C ¬OH
CH2
CH 2
CH 3
OH
CH2
CH2
Phe-Ser-Ala
(b) Val-Gly-Asp is a hydrophilic protein with acidic properties.
Phe-Ser-Ala would be more hydrophobic and neutral.
28.53
O
O
O
NH2
O
O
H2N ¬CH ¬C ¬ NH¬ CH¬C¬ NH¬ CH ¬C ¬OH
H2N ¬CH ¬C ¬ NH¬ CH ¬C ¬OH
CH2
CH2
CH 2
OH
CH3
CH 2
OH
28.55 Pro-Phe-Arg-Pro-Pro-Gly-Phe-Ser-Pro
28.57 (a) 6 M HCl/heat (b) Trypsin hydrolysis would aid in
sequencing around lysine. Chymotripsin would cleave about
Phe. 28.59 Cys-Gly-Phe-Lys-Cys 28.61 ΔG° 13 kJ
28.63 The presence of ionisable or polar groups make the
vitamin more hydrophilic.
CH2
CH2
NH2
(b) 4
28.47 (a)
28.65
O
O
O
CH3
O
O
OH
H2N¬ CH¬ C ¬ NH¬ CH ¬ C¬ NH¬ CH ¬ C ¬ OH
CH¬ CH3
H
Cl
SOCl2
N
CH 2
N
1. NH3
2. aqueous base
SH
CH2
CH3
O
(b) 27: Ser-Ser-Ser, Phe-Phe-Phe, Thr-Thr-Thr, Ser-Thr-Thr,
Thr-Ser-Thr, Thr-Thr-Ser, Thr-Phe-Phe, Phe-Thr-Phe,
Phe-Phe-Thr, Phe-Thr-Thr, Thr-Phe-Thr, Thr-Thr-Phe,
Phe-Phe-Ser, Ser-Phe-Phe, Phe-Ser-Phe, Phe-Ser-Ser,
Ser-Ser-Thr, Ser-Thr-Ser, Thr-Ser-Ser, Ser-Ser-Phe, Ser-Phe-Ser,
Ser-Thr-Phe, Ser-Phe-Thr, Phe-Thr-Ser, Thr-Ser-Phe,
Phe-Ser-Thr, Thr-Phe-Ser. 28.49 The primary structure of
a protein relates to the arrangement of amino acids along its
chain, e.g. Phe-Lys-Ala. The secondary structure corresponds
to an intramolecular region of structural regularity as found
in a-helices and b-sheets. The tertiary structure of a protein
relates to its overall shape. In the case of enzymes, the tertiary
structure leads to the formation of pockets (active sites) in
which chemistry takes place.
28.51 (a)
O
O
O
NH2
N
28.68 A nucleotide consists of a nitrogen-containing aromatic
compound, a sugar in the furanose (five-membered) ring form
and a phosphoric acid group. The structure of deoxycytidine
monophosphate is
NH2
O
H
CH 2
C
CH3
Val-Gly-Asp
OH
P
O
H2N¬ CH¬ C ¬ NH¬ CH ¬ C¬ NH¬ CH ¬ C ¬ OH
CH¬ CH3
N
O
O
CH2
O
C H
H C
H C
C H
OH
O
28.70
N
O
H
C4H7O3CH2OH HPO42
C4H7O3CH2¬O ¬PO32 H2O
28.72 In the helical structure for DNA, the strands of the
polynucleotides are held together by hydrogen-bonding
interactions between particular pairs of bases. Adenine and
Answers to Selected Exercises
thymine are one base pair, and guanine and cytosine are
another. For each adenine in one strand, there is a cytosine in
the other, so that total adenine equals total thymine, and total
guanine equals total cytosine. 28.74 (a) CCATGA (b) DNA
based since it contains T instead of U (c) This notation is a way
of expressing the linkage and sequence in an unambiguous
way.
28.76 (a)
O
O
O
O ¬P ¬ O¬P ¬ O ¬ P¬ O¬Adenosine
O
O
O
O
O
H 2O
H2PO4
O
(b) negative (c) AMP (adenosine monophosphate) and
inorganic phosphate 28.78 (a) Since purine bases pair with
pyrimidine bases, there must be equal amounts of both in
double-stranded DNA. (b) This does not necessarily apply to
single-stranded DNA where pairing does not occur.
28.80 (a) phenylalanine (b) polystyrene
28.82
O
Br
NH
N
HO
O
O
H
H
H
H
OH
(c) 1 signal (d) 3 signals in ratio 3:2:2 (e) 1 signal (f) 3 signals
in ratio 2:2:1 (g) 3 signals in ratio 3:2:3 29.16 (a) 420 Hz
(b) 2.1 ppm (c) 840 Hz 29.18 The real structure of BHT would
have four signals in its 1H NMR spectrum in the ratio 3:2:18:1.
The proposed structure would have six signals in the ratio
3:1:1:1:9:9 29.20 (a) doublet, quartet (b) doublet and septet
(c) singlet
29.22
spin diagram
without influence of Hb
with influence of Hb
O ¬ P¬O ¬ P¬ O¬ Adenosine
O
A-43
OH
28.84 Phe-Pro-Val-Asp-Pro-Ile
Chapter 29
29.2 (d) is the most likely due to the position of the C ! O
stretch and minimal C – H stretching band.
29.4 (a) Number of absorptions in 1H and 13C spectra: 2 vs 4.
(b) Position of CH2 signal in 1H NMR spectrum. For the first
molecule, this would occur around 4–4.5 ppm, the second
around 2–2.5 ppm. (c) Number of absorptions in 1H spectra:
3 vs 2. 29.6 M+ 192. m/z 161 [M – OCH3]+;
105 [M – C6H5CH2CH2]+; 77 [C6H5]+. DNP test (ketones)
and hydroxylamine tests (carboxylic acids) would yield
positive results. 29.7 (a) OH (b) C ! O (c) C ! C
(d) COOH (e) amide 29.9 (a) OH stretch present for
CH3CH2CH2CH2OH (b) OH stretch present for
CH2 ! CHCH2OH (c) C ! O stretch present for pyridone
(d) C ! C stretch present for CH3CH ! CHCH2CH3
(e) N – H stretch present for CH3CH2NHCH3 29.11 If picric
acid were a carboxylic acid it would contain stretches at
around 1700 cm–1 and 3500 cm–1. As a phenol, it contains an
N, it is a strong and
absorption at 3500 cm–1. 29.13 (a) C
distinctive signal in a unique part of the IR spectrum.
(b) alkyne however weak, especially if symmetrical.
29.14 (a) 4 signals in ratio 3:2:6:3 (b) 3 signals in ratio 3:2:1
29.24 The two products differ by the number of signals present
in the 1H NMR spectrum of each. The major product is the one
that has two signals in the ratio 3:1. The minor product
contains five signals. 29.26 0.12 J mol–1 29.28 (a) 2, no
splitting (b) 3, singlet, doublet, septet (c) 2, triplet, quartet
(d) 4, 2 singlet, 2 triplet. 29.30 (a) formic acid
(b) ethyl formate, methyl acetate 29.32 (a) 13C is
approximately 1% abundant, 1H is 100% abundant. (b) These
small signals are due to 13C – H coupling. (c) Because
deuterium, which gives rise to the multiplet, has a frequency
range outside that irradiated for the decoupling experiment.
29.34 9.5 ppm CHO, 2.2 ppm CH2 and 1.0 ppm CH3.
29.36
Isotope(s) of CBr4
m/z
(12C)(79Br)4
(13C)(79Br)4
(12C)(79Br)3(81Br)
(13C)(79Br)3(81Br)
(12C)(79Br)2(81Br)2
(13C)(79Br)2(81Br)2
(12C)(79Br)(81Br)3
(13C)(79Br)(81Br)3
(12C)(81Br)4
(13C)(81Br)4
328
329
330
331
332
333
334
335
336
337
29.38 C4H8O, IDH 1 29.40 (a) The magnet is used to
deflect charged particles thereby separating fragments by
mass. (b) The mass of the two isotopes of chlorine are 35
and 37. The atomic weight takes into account the relative
distributions of these two isotopes in nature.
29.42 (a) x m/z, y % relative abundance or signal intensity
(b) Charged species are influenced by a magnetic field more
strongly than uncharged species. The effect of the magnetic
field on charged moving particles is the basis of their
separation by mass. 29.44 (a) 5 (b) 1 (c) 6 (d) 7 (e) 3 (f) 7
29.46 Iodobutane 29.48 Bottle A: Compound 5, Bottle B:
Compound 3. 29.50 3-chloropropanoic acid
29.52 HOCH2CH2CN 29.54 HOCH2CH2CH2NH2
29.56
O
O
O