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Descriptive Inorganic Chemistry, Fifth Edition
Chapter 1
Answers to Odd-Numbered Questions 1
Beyond the Basics
1.23
9, 5, 121.
Exercises
1.1
(a) Surface where electron probability = 0.
1.25
There are seven f orbitals. There are two separate
ways of depicting them and designating them: the
general set and the cubic set.
1.27
Hydrogen heads the alkali metal group. Helium
heads the alkaline earth metal group.
(b) No two electrons in an atom can have the same 4
quantum numbers.
(c) Attraction into a magnetic field by an unpaired
electron.
1.3
Chapter 2
Exercises
2.1
(a) Lanthanum through lutetium.
(b) Apparent radius of an atom in non-bonded contact
with another.
(c) Actual nuclear charge experienced by an electron.
2.3
Argon did not fit into any of the then-known groups.
Because the table was based on measured atomic
mass, argon should have been placed between
potassium and calcium.
2.5
The long form correctly depicts the order of
elements; but the table becomes very elongated.
1.5
5p.
1.7
Size of an orbital.
1.9
With parallel spins there is zero probability that the
electrons will occupy the same volume of space.
2.7
The -ium ending indicates a metal. The ending -on
has been used for non-metals.
1.11
(a) [Ne]3s1; (b) [Ar]4 s 2 3d 8 ; (c) [Ar]4s13d10.
2.9
With nuclei up to 26 protons, nuclear fusion is an
exothermic process.
1.13
(a) [Ar]; (b) [Ar]; (c) [Ar]3d 9 .
2.11
(a) Lead (b) technetium (c) bromine.
1.15
1+ and 3+. Configuration of
[Xe]6 s 2 4 f 14 5d 10 6p 1 .
2.13
Sodium, because it has an odd number of protons.
2.15
50.
1.17
1+. Configuration of [Kr]5s14d10.
2.17
(a) Several nonmetals have metallic luster.
(b) Diamond has the highest thermal conductivity of
all substances.
(c) Graphite is a good electrical conductor in two
dimensions.
2.19
Potassium. As the effective atomic charge on the
outermost electrons increases across a period.
2.21
The effective nuclear charge on the 4p electrons will
be increased.
1.19
1.21
E113: [Rn]7 s 2 5 f 1 4 6d 1 0 7p 1 .
E113 as +1 ion: [Rn]7s 2 5 f 1 4 6d 1 0 .
E113 as +3 ion: [Rn]5f 1 4 6d 1 0 .
2010 © W. H. Freeman and Company, All Rights Reserved
2
Answers to Odd-Numbered Questions
2.23
Descriptive Inorganic Chemistry, Fifth Edition
Using Slater’s rules.
3.9
The disadvantage of the Slater method is that it does
not distinguish between s and p electrons in terms of
shielding.
2.25
4s = 2.95; 3d = 5.60.
2.27
Phosphorus. Increasing nuclear charge across the
period, means increasing ionization energy.
2.29
Group 2. The size of the ionization energies
increases rapidly between the second and third
values.
2.31
First, magnesium. Second, sodium. Third,
magnesium.
2.33
Positive. Any gained electron would add to the 2s
orbital.
2.35
(a) 208; (b) 209; (c) 210.
3.11
Electron dot diagrams:
3.13
(a) Electron-dot diagram; (b) resonance structures;
(c) partial bond representation.
3.15
(a) Electron dot diagrams; (b) formal charge
structures; (c) partial bond representation
Beyond the Basics
2.37
Over time, the vast majority of these molecules have
escaped Earth’s gravitational field.
2.39
A member of the halogens.
2.41.
No specific answer.
Chapter 3
Exercises
3.1
(a) Linear combination of atomic orbitals.
(b) M.O. in which the increased electron density lies
between the two nuclei.
(c) Valence shell electron pair repulsion.
(d) Mixing atomic orbitals on a central atom.
(e) The axis containing the highest n-fold rotation
axis.
3.3
The bond order would be  and the ion would be
paramagnetic.
3.5
Bond order 2. Electron configuration:
(σ2s)2(σ*2s)2(π2p)4(σ2p)1.
3.7
3.
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Descriptive Inorganic Chemistry, Fifth Edition
Answers to Odd-Numbered Questions 3
d) Octahedral. Three C4 axes, four C3 axes, six C2
axes, nine σv planes, and a center of symmetry.
Beyond the Basics
3.41
The following show the overlap of an s orbital with a
typical d orbital to form a σ bond; overlap of a p
orbital with a typical d orbital to form a σ bond; and
overlap of a p orbital with a typical d orbital to form
a π bond.
3.17
3.19
(a) tetrahedral, V-shaped; (b) tetrahedral, trigonal
pyramidal;
(c) trigonal bipyramidal, linear; (d) octahedral, square
planar.
Linear: carbon disulfide and xenon difluoride; Vshaped: chlorine dioxide (< 109º), tin(II) chloride (<
120º), and nitrosyl chloride (< 120º).
3.21
Oxygen difluoride and phosphorus trichloride.
3.23
(a) sp3; (b) sp3; (c) sp3d; (d) sp 3 d 2 .
3.25
The hybrid orbitals used would be sp, accounting for
the linear arrangement.
3.27
Hydrogen selenide, because with more electrons, the
dispersion forces would be greater.
3.43
The NNO arrangement provides two possibilities
with only one formal charge per atom.
3.44
The formal charge electron arrangements for OCN−
have one single formal charge. For the isocyanate
ion, CNO−, at least two negative and one positive
formal charge exist.
The two feasible formal charge arrangements for the
CON− ion have five formal charges!
3.45
3.29
Polar: oxygen difluoride and phosphorus trichloride;
nonpolar: xenon difluoride and the tetrachloroiodate
ion.
3.31
Ammonia, because neighboring molecules will
hydrogen-bond to each other.
3.33
Hybridization: (a) sp; (b) sp2; (c) sp3; (d) sp3d; (e)
sp3d2.
3.35
9.
3.37
(a) One C3 axis, three C2 axes, three σv planes, one σh
plane, one improper S6 axis, and three S2 axes. D3h.
(b) One C4 axis and two σv planes. C4v.
(c) One C4 axis, four C2 axes, one σh plane, four σv
planes, a center of symmetry, one improper S4 axis,
and four S2 axes.
3.39
(a) Trigonal pyramidal. One C3 axis and three σv
planes. C3v.
(b) Trigonal planar. One C3 axis, three C2 axes, one
σh plane, and three σv planes. D3h.
(c) Tetrahedral. Four C3 axes, three C2 axes, and six
σv planes. Td.
3.47
NO+ and CN− are both triply bonded and will
therefore be energetically preferred.
3.49
There may be a significant ionic character to the
bonding in SbCl3 resulting in a higher boiling point
than otherwise expected.
3.51
One C5 axis, five C2 axes, one σh plane, and five σv
planes.
3.53
Ozone is infrared absorbing.
3.55
With low planetary mass, atmospheric gases on Mars
would be lost quickly. Volcanoes provided
replenishment of the atmospheric CO2. When the
core solidified, volcanic activity ceased and no more
of the ‘greenhouse gas’ was being pumped into the
atmosphere. The remaining atmosphere cooled.
Chapter 4
Exercises
4.1
(a) Metal consists of metal ions with free electrons.
(b) Smallest repeatable fragment of a crystal lattice
that.
(c) Combination of two or more solid metals.
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4
Answers to Odd-Numbered Questions
4.3
High electrical conductivity, high thermal
conductivity, high reflectivity, and high boiling point.
4.5
3s and 3p band overlap means that electrons in the
full 3s band can “spill over” into the 3p band.
4.7
For metallic behavior, the orbitals of the atoms must
overlap.
4.9
Cubic and hexagonal. Hexagonal.
4.11
Simple cubic unit cell contains 4 ×  atoms.
4.13
Same size atoms, adopt the same structure, must have
similar properties.
Beyond the Basics
4.15
The volume of the atom will be 4/3πr3, while the
volume of the cube will be (2r)3. The ratio of these
gives 0.52.
4.17
The length of the unit cell edge will be [4/(2)]r =
2.83r.
4.19
(a) 125 pm.
(b) 7.24 g·cm−3.
4.21
145pm.
4.23
A suspension of gold nano-particles has a red color.
Descriptive Inorganic Chemistry, Fifth Edition
5.11
UCl3 (837°C); UCl4 (590°C); UCl5 (327°C); UCl6
(179°). In this particular series, there does not appear
to be a clear divide between high (ionic) and low
(covalent) melting points.
5.13
WF6 (2°C) and WO3 (1472°C). The fluoride is
predominantly covalent while the oxide is ionic. The
fluoride is more polarizable than the oxide.
5.15
Tin(II) chloride has a higher melting point because
tin(II) has a fairly low charge density.
5.17
No, ionic compounds do not dissolve in nonpolar
solvents.
5.19
Magnesium chloride, because the dipositive smaller
magnesium ion has a significantly higher charge
density.
5.21
Lithium nitrate, because the lithium ion has a higher
charge density than the sodium ion.
5.23
The coordination number depends on the radius ratio.
5.25
The magnesium ion is smaller than the calcium ion.
5.27
Chapter 5
Exercises
5.1
(a) Distortion from a spherical shape.
(b) The holes between anions in the crystal packing.
(c) The diagram used to show the three bonding
categories: metallic, covalent, and ionic.
5.3
5.5
5.7
5.9
Hard and brittle crystals; high melting points;
electrically conducting in liquid phase and in aqueous
solution.
(a) K+, because the radius will be determined by the
inner orbitals.
(b) Ca2+, because the ions are isoelectronic but
calcium has one more proton.
(c) Rb+, because again the ions are isoelectronic, with
rubidium having two more protons than bromide.
NaCl, because chloride is smaller than iodide; the
charge is more concentrated, and the ionic attraction
will be stronger.
Ag+, because it has the lowest charge density.
5.29
(a) Metallic and a lesser contribution of ionic; (b)
covalent and a lesser contribution of ionic.
5.31
The choice would be metallic or covalent.
Beyond the Basics
5.33
Direct hybrid ionic-covalent bonding between pairs
of ions in the gas.
5.35
(a) Copper(II) chloride. The higher charge density
copper(II) ion.
(b) Lead(II) chloride. The very high charge density of
the lead(IV) ion.
5.37
1.15(r+ + r−).
5.39
164 pm
5.41
251 pm
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Descriptive Inorganic Chemistry, Fifth Edition
5.43
Answers to Odd-Numbered Questions 5
LiAt – wurtzite or sphalerite packing; NaAt and KAt,
sodium chloride packing; and RbAt and CsAt,
cesium chloride packing.
charges. The mathematical formula is known as
Coulomb’s law.
6.27
Chapter 6
Exercises
6.1
(a) A reaction for which ΔG° is negative.
(b) A measure of disorder.
(c) A mole of a substance is formed from its
constituent elements in their standard phases at 298 K
and 100 kPa.
6.3
Negative. For a spontaneous reaction, the enthalpy
change must be negative.
6.5
ΔHf° = −286 kJ·mol−1;
ΔSf°
= −163 J·mol−1·K−1 = −0.163
kJ·mol−1·K−1
ΔGf°
= −237 kJ·mol−1
Spontaneous at SATP.
6.7
−1
ΔGº = −159 kJ·mol
6.9
6.11
N=N.
Approximate enthalpy of reaction = +112 kJ·mol−1
6.13
Sodium chloride, lithium fluoride, magnesium oxide.
6.15
[8/(3)½ − 6/2] = 1.62.
6.17
U = −2649 kJ·mol
−1
6.29
Magnesium oxide will have a higher lattice energy.
6.31
ΔHf° = −913 kJ·mol−1; tabulated value is −933
kJ·mol−1.
6.19
6.21
(EA H) = −454 kJ·mol−1
6.23
Because of the high charge density of the oxide, O2−.
Beyond the Basics
6.25
The term “the permittivity of free space” is a constant
that relates the attractive force between two point
6.33
250 kJ·mol−1.
6.35
The lattice energy of the calcium chloride will also be
much greater.
6.37
For calcium sulfate: −1021 kJ·mol−1
For strontium sulfate: −1132 kJ·mol−1
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6
Answers to Odd-Numbered Questions
For barium sulfate: −1044 kJ·mol−1
Yes.
Descriptive Inorganic Chemistry, Fifth Edition
7.7
(a) NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
(b) CN−(aq) + H2O(l) ⇌ HCN(aq) + OH−(aq)
(c) HSO4−(aq) + H2O(l) ⇌ SO42−(aq) + H3O+(aq)
6.39
7.9
ClNH2(aq) + H2O(l) ⇌ ClNH3+(aq) + OH–(aq)
7.11
H2SO4(l) + H2SO4(l) ⇌ H3SO4+(H2SO4) +
HSO4−(H2SO4)
7.13
(a) The ammonium ion, NH4+; (b) the amide ion,
NH2−.
For magnesium: ΔH = +428 kJ·mol−1. For lead: ΔH =
+898 kJ·mol−1
The only significant difference between the two ions
is their hydration enthalpies.
6.41
Using the Kapustinskii equation: U = −622 kJ·mol−1;
compared with −668 kJ·mol−1 experimentally and
−636 kJ·mol−1 from the Born-Landé equation.
6.43
ΔH = +36 kJ·mol−1
6.45
proton affinity = −1141 kJ·mol−1
6.46
Step 1: ΔG° = +184 kJ⋅mol−1
Step 2: ΔG° = −47 kJ⋅mol−1
6.47
Ionic bond formation is best considered as a
competition for electrons.
7.15
HSO4−(H2SO4)
HF is acting as a base and H2F+ is the conjugate
acid,
7.17
7.3
7.5
(a) H+(aq) + OH−(aq) → H2O(l)
(b) 2 HCO3−(aq) + Co2+(aq) → CoCO3(s) + H2O(l) +
CO2(g)
(c) OH−(aq) + CH3COOH(aq) → CH3COO−(aq) +
H2O(l)
(a) Pairs of species that differ in formula by one
ionizable hydrogen.
(b) Solvent that undergoes its own acid-base reaction.
(c) Ability of a substance to act as an acid or a base.
HSeO4− (base), H2SO4 (conjugate acid); H2O
(acid), OH− (conjugate base).
7.19
The H−Se bond will be weaker than the H−S bond.
Thus hydrogen selenide will be the stronger acid.
7.21
[Zn(OH2)6] (aq) + H2O(l) ⇌
[Zn(OH2)5(OH)]+(aq) + H3O+(aq)
7.23
The diprotic acid must be present in the least
proportion.
H2NNH2(aq) + H2O(l) ⇌ H2NNH3+(aq) + OH−(aq)
H2NNH3+(aq) + H2O(l) ⇌ +H3NNH3+(aq) +
Chapter 7
Exercises
7.1
Polar protic: solvents with a dielectric constant
between 50 and 100.
Dipolar aprotic: solvents with dielectric constant
between 20 and 50.
Nonpolar: solvents with dielectric constant close to
zero.
HF(H2SO4) + H2SO4(l) ⇌ H2F+(H2SO4) +
2+
OH−(aq)
7.25
(a) Acidic, because aluminum is a small high-charge
cation.
(b) Neutral, because the sodium ion will stay
unchanged.
7.27
With a smaller pKb, A− must be the stronger base.
Thus HB will be the stronger acid.
7.29
H3PO4(aq) + HPO42−(aq) ⇌ 2 H2PO4−(aq)
7.31
(a) N2O5; (b) CrO3; (c) I2O7.
7.33
(a) SiO2 (acid), Na2O (base); (b) NOF (acid), ClF3
(base); (c) Al2Cl6 (acid), PF3 (base).
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Descriptive Inorganic Chemistry, Fifth Edition
7.35
(a) No effect.
(b) Increasing pH.
HSe−(aq) + OH−(aq)
(c) Decreasing pH.
Se (aq) + H2O(l) ⇌
7.61
Dimethylsulfoxide must be a softer base than water.
7.63
In terms of the HSAB concept, the harder calcium
ion is likely to form a stronger bond to the water
molecules of hydration than the softer barium ion.
7.65
Fe3+(AlO6)3−.
2−
[Sc(OH2)6]3+(aq) + H2O(l) ⇌
[Sc(OH2)5(OH)]2+(aq) + H3O+(aq)
(d) Increasing pH. F−(aq) + H2O(l) ⇌ HF(aq) +
OH−(aq)
7.37
(a) Weakly basic; (b) neutral; (c) moderately basic;
(d) strongly basic.
7.39
(a) Strongly basic; (b) very strongly basic.
7.41
ΔG°
= −28 kJ·mol−1. Magnesium oxide will be a
weaker base than calcium oxide (ΔG° = −59
kJ·mol−1).
7.43
Answers to Odd-Numbered Questions 7
NO+ is a Lewis acid. Cl− is a Lewis base.
(NO)(AlCl4)(NOCl) + [(CH3)4N]Cl(NOCl) →
[(CH3)4N](AlCl4)(NOCl) + NOCl(l)
[NH4+] = 3 × 10−17 mol·L−1
[NH4+] = 1×10−33 mol·L−1
Chapter 8
Exercises
8.1
(a) A substance that will oxidize another.
(b) A two-dimensional plot of free energy against
temperature for series of reactions that involve
elements and their oxides, sulfides, or chlorides.
+3 (b) +5 (c) −3 (d) −3 (e) +5
8.3
(a)
8.5
(a) −2; (b) +2; (c) −1; (d) +6; (e) −2.
8.7
−1, +1, +3, +5, +7.
8.9
(a) +1; (b) +2; (c) +3; (d) +4; (e) +5. An increase by
units of +1 from Group 13 to Group 17.
8.11
(a) Nickel from +2 to 0, carbon from 0 to +2.
(b) Manganese from +7 to +2, sulfur from +4 to +6.
7.45
(a)
(b)
7.47
(a) No. The reactants have the combinations
borderline-borderline and hard-hard.
(b) Yes. The products will be preferred where hardhard and soft-soft combinations result.
8.13
NH4+(aq) + 3 H2O(l) → NO3−(aq) + 10 H+(aq) + 8 e−
8.15
N2H4(aq) + 4 OH−(aq) → N2(g) + 4 H2O(l) + 4 e−
(a) Greater than 1.
(b) Less than 1.
(a) Thallium(I), in analysis group I.
(b) Rubidium ion, in analysis group V.
(c) Radium ion, analysis group IV.
(d) Iron(III), in the analysis group III.
8.17
(a) 5 HBr(aq) + HBrO3(aq) → 3 Br2(aq) + 3 H2O(l)
(b) 2 HNO3(aq) + Cu(s) + 2 H+(aq) →
2 NO2(g) + Cu2+(aq) + 2 H2O(l)
8.19
(a) 12 V(s) + 10 ClO3−(aq) + 18 OH−(aq) →
7.49
7.51
7.53
6 HV2O73−(aq) + 10 Cl−(aq) + 6 H2O(l)
(a) MgSO4; (b) CoS.
Beyond the Basics
7.55
[S2-] = 1.1 x 10-22. Cadmium sulfide will precipitate.
Iron(II) sulfide will
not precipitate.
7.57
7.59
Zinc is a borderline acid, so it can be found as ores of
both hard and soft bases.
H2CO3(aq) + MgSiO4(s) → H2O(l) + SiO2(s) +
MgCO3(s)
The atmospheric concentration of carbon dioxide has
decreased, in part due to the formation of magnesium
carbonate minerals.
(b) 2 S2O42−(aq) + 3 O2(g) + 4 OH−(aq) →
4 SO42−(aq) + 2 H2O(l)
8.21
(a) Spontaneous.
(b) Non-spontaneous.
8.23
One example: Zn → Zn2+, Eº = +0.762 V
8.25
(a) Au3+(aq) + 3 e− → Au(s) would be the stronger
oxidizing agent.
(b) Al(s) → Al3+(aq) + 3 e− would be the stronger
reducing agent.
2010 © W. H. Freeman and Company, All Rights Reserved
8
Answers to Odd-Numbered Questions
8.27
E° = −0.267 V
8.29
E = +0.805 V
8.31
8.33
+3.
8.35
9.9
As the group is descended, the cation radii increase,
the ionic bond will weaken and the melting point will
be lower.
(a) Br2
(b) E° = +0.52 V
(c) The Nernst expression does not have a pHdependent term.
9.11
Scandium hydroxide, Sc(OH)3.
9.13
SO3(s) + H2O(l) → H2SO4(aq)
CrO3(s) + H2O(l) → H2CrO4(aq)
The most thermodynamically stable oxidation state is
9.15
(a) Al2O3, Sc2O3.
(b) P2O5, V2O5.
9.17
Tin.
9.19
N2, O2, F2; thus they have stronger dispersion
(London) forces.
9.21
Forming the Eu2+ ion would retain the half-filled d
orbital set.
9.23
(a) Indium(III) and bismuth(III); (b) cadmium(II)
and lead(II).
9.25
Thallium(I) bromide.
9.27
(a) C≡O; (b) (C≡C)2−.
9.29
Yttrium.
At pH 0.00 because the reduction potential is lower.
Beyond the Basics
8.37
The oxidation to carbon monoxide involves an
increase of entropy; thus the TΔS term will become
increasingly negative with increase in temperature.
The negative slope for this line will ultimately cross
the carbon dioxide line.
T = 766 K = 493ºC
8.39
Descriptive Inorganic Chemistry, Fifth Edition
Eº will increase to the point where insoluble, brown
manganese(III) oxide will be formed, thus
discoloring the toilet bowl.
Beyond the Basics
9.31
Because the synthetic route involves a negative free
energy change.
9.33
Add excess hydroxide ion.
9.35
(a) 12−, (b) 7.
9.37
Li (+1); Be (+2); B (+3); C (+4); N (+3); O (+2). For
Period 2, oxidation numbers reach a maximum at
carbon, then decrease.
Na (+1); Mg (+2); Al (+3); Si (+4); P (+5); S (+6); Cl
(+5). For Period 3, the oxidation number matches the
number of valence electrons except for chlorine.
9.39
Fe2O3 (+3); RuO4 (+8); OsO4 (+8). For ruthenium
and osmium, the oxidation number is the same as the
Group number.
Chapter 9
Exercises
9.1
(a) A pair of elements in a compound whose sum of
valence electrons adds up to eight.
(b) The relationship between an element and the
element to its lower right in the periodic table.
9.3
The general formula: M+M3+(SO42−)2·12H2O, where
M+ is potassium or ammonium and M3+ is aluminum,
chromium(III), or iron(III).
9.5
KF, CaF2, GaF3, GeF4, AsF5, SeF6, BrF5, KrF2.
The bonding in the potassium and calcium fluorides
is ionic, while that for the germanium, arsenic,
selenium, bromine, and krypton compounds is
covalent.
9.7
(a) Hydrogen gas, H2; (b) calcium metal.
9.41
Group 15
Group 16
Group 15
CN22−
OCN−
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Group 16
OCN−
CO2
Group 17
FCN
FCO+
Descriptive Inorganic Chemistry, Fifth Edition
Answers to Odd-Numbered Questions 9
negative. Most covalent hydrides belong in the first
category.
Chapter 10
Exercises
10.1
(a) A hydrogen atom bridging atoms in a covalent
bond in which the hydrogen is less electronegative.
(b) A hydrogen atom bridging atoms in a covalent
bond in which the hydrogen is more electronegative.
10.3
The ice cube consists of “heavy” water, deuterium
oxide.
10.5
The difference in absorption frequency is very small,
about 10−6 of the signal itself.
10.7
Hydrogen rarely forms a negative ion.
Bond energy
Electron affinity
10.9
Hydrogen
432 kJ·mol−1
−79 kJ·mol−1
Chlorine
240 kJ·mol−1
−349 kJ·mol−1
10.17
KH; CaH2, GaH3, GeH4, AsH3, H2Se, HBr. The trend
is to increase by one H until germanium, then a
stepwise decrease by one H to hydrogen bromide.
10.19
(a) Gas. It is a covalent hydride.
(b) Solid. This is an ionic hydride.
10.21
The closeness of the electronegativities of hydrogen
and carbon, and the ability to hydrogen bond.
Beyond the Basics
10.23 (a) Yes, liquid; (b) no, gas; (c) yes, liquid; (d) no,
gas.
10.25.
Looking at a generic Born-Haber cycle, where X = H
or Cl, we see that there are two features that differ.
10.27
Hydrogen and carbon monoxide.
H2O(l) + C(s) → H2(g) + CO(g)
The combustion reaction would therefore be:
H2(g) + CO(g) + O2(g) → H2O(g) + CO2(g)
ΔH = −525 kJ
Per mole, this is = −262 kJ·mol−1, compared with
−242 kJ·mol−1 for the combustion of pure
dihydrogen.
Enthalpy driven. The chemical equation is: N2(g) +
3 H2(g) → 2 NH3(g)
There is a decrease in the number of gas molecules,
hence a decrease in entropy.
10.11
(a) 2 KHCO3(s) → K2CO3(s) + H2O(g) + CO2(g)
(b) HC≡CH(g) + 2 H2(g) → H3C−CH3(g)
(c) PbO2(s) + 2 H2(g) → Pb(s) + 2 H2O(g)
(d) CaH2(s) + H2O(l) → Ca(OH)2(aq) + H2(g)
10.13
The much lesser enthalpy of formation of ammonia
compared to water can be explained in terms of the
much greater bond energy of dinitrogen (945
kJ·mol−1) compared with that of dioxygen (498
kJ·mol−1).
10.15
There are three categories of covalent hydrides: those
in which the hydrogen is nearly neutral; those in
which it is quite positive, and those in which it is
Chapter 11
Exercises
11.1
(a)
(b)
(c)
(d)
2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
Rb(s) + O2(g) → RbO2(s)
2 KOH(s) + CO2(g) → K2CO3(s) + H2O(l)
2 NaNO3(s) → 2 NaNO2(s) + O2(g)
11.3
They resemble “typical” metals in that they are shiny
and silvery and good conductors of heat and
electricity. The alkali metals differ from “typical”
metals in that they are soft, extremely chemically
reactive, have low melting points and very low
densities.
11.5
All common chemical compounds are water soluble.
They always form ions of +1 oxidation state.
Their compounds are almost always ionic.
11.7
The most likely argument is that the hydroxide ion
can hydrogen bond with the surrounding water
molecules.
11.9
Because the equilibrium of the synthesis reaction:
Na(l) + KCl(l) → K(l) + NaCl(l) lies to the left.
2010 © W. H. Freeman and Company, All Rights Reserved
10
Answers to Odd-Numbered Questions
11.11
(a) Sodium hydroxide; (b) anhydrous sodium
carbonate; (c) sodium carbonate decahydrate.
11.13
(a) Loss of water by a hydrated salt in a low-humidity
environment.
(b) Chemical similarities of one element and the
element to its lower right in the periodic table.
11.15
CO2(g) + NH3(aq) + H2O(l) → NH4+(aq) +
HCO3−(aq)
HCO3−(aq) + Na+(aq) → NaHCO3(s)
2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g)
CaCO3(s) → CaO(s) + CO2(g)
Descriptive Inorganic Chemistry, Fifth Edition
2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g)
2 KOH(aq) + CO2(g) → K2CO3(aq) + H2O(l)
K2CO3(aq) + CO2(g) + H2O(l) → 2 KHCO3(aq)
2 KO2(s) + CO2(g) → K2CO3(s) + 2 O2(g)
3 K+(aq) + [Co(NO2)6]3−(aq) → K3[Co(NO2)6](s)
Beyond the Basics
11.25 Current = 6.94 x 104 A
11.27
In the series LiF to CsF, there is an increasing
mismatch in ion sizes. For the series LiI to CsI, there
is a decreasing mismatch in ion sizes.
11.29
NaBF4. The hydration energy will more probably
exceed the (lower) lattice energy, making the
compound more soluble.
11.31
Either: that there is appreciable covalent bonding in
the lithium hydride, or that the lithium ion is so small
that the lattice consists of touching hydride ions with
lithium ions “rattling around” in the lattice holes..
11.33
LiF and KI.
11.35
Calcium-40 is a “doubly magic” nucleus with filled
shells of protons and neutrons.
CaO(s) + H2O(l) → Ca(OH)2(s)
2 NH4+(aq) + 2 Cl−(aq) + Ca(OH)2(s)
→ 2 NH3(aq) + CaCl2(aq) + 2 H2O(l)
The problems: disposal of waste calcium chloride,
and the high energy requirements.
11.17
The ammonium ion is monopositive; its salts are all
soluble; its size is about the middle of the alkali metal
ion range; all its common salts are colorless.
11.19
Potassium dioxide(1−) has a lower molar mass, and is
cheaper.
Chapter 12
11.21
The ammonium ion is large.
Exercises
12.1
(a)
11.23
Lithium:
6 Li(s) + N2(g) → 2 Li3N(s)
2 Li(s) + Cl2(g) → 2 LiCl(s)
Li(s) + C4H9Cl(solv) → LiC4H9(solv) + LiCl(s)
4 Li(s) + O2(g) → 2 Li2O(s)
2 Li(s) + H2O(l) → 2 LiOH(aq) + H2(g)
Li2O(s) + H2O(l) → 2 LiOH(aq)
2 LiOH(aq) + CO2(g) → Li2CO3(aq) + H2O(l)
Li2O(s) + CO2(g) → Li2CO3(s)
Sodium:
2 Na(s) + Cl2(g) → 2 NaCl(s)
2 Na(s) + H2O(l) → 2 NaOH(aq) + H2(g)
2 Na(s) + O2(g) → Na2O2(s)
Na2O2(g) + H2O(l) → 2 NaOH(aq) + H2O2(aq)
2 NaOH(aq) + CO2(g) → Na2CO3(aq) + H2O(l)
Na2CO3(aq) + CO2(g) + H2O(l) → 2 NaHCO3(aq)
2 Na(s) + 2 NH3(l) → 2 NaNH2(NH3) + H2(g)
Na2O2(s) + CO2(g) → Na2CO3(s) + O2(g)
Potassium
Na(l) + KCl(l) → K(g) + NaCl(l)
2 K(s) + Cl2(g) → 2 KCl(s)
K(s) + O2(g) → KO2(s)
2 KO2(s) + 2 H2O(l) → 2 KOH(aq) + H2O2(aq) +
O2(g)
(b)
2 Ca(s) + O2(g) → 2 CaO(s)
CaCO3(s) → CaO(s) + CO2(g)
(c)
Ca(HCO3)2(aq) → CaCO3(s) + H2O(l) +
CO2(g)
(d)
CaO(s) + 3 C(s) → CaC2(s) + CO(g)
12.3
(a) Barium; (b) barium.
12.5
The higher charge density magnesium ion will cause
the water molecules surrounding it during the
hydration step to become much more ordered than
with the lower charge density sodium ion.
12.7
They form 2+ ions exclusively and their salts tend to
be highly hydrated.
12.9
Steric hindrance.
12.11
Rainwater, an aqueous solution of carbon dioxide,
percolates into limestone deposits, reacting with the
calcium carbonate to give a solution of calcium
hydrogen carbonate.
2010 © W. H. Freeman and Company, All Rights Reserved
Descriptive Inorganic Chemistry, Fifth Edition
12.13
Mg2+(MgCl2) + 2 e− → Mg(l)
2 Cl-(MgCl2) → Cl2(g) + 2 e−
12.15
(a) Ca(OH)2 (hydrated lime) or CaO (quicklime); (b)
Mg(OH)2;
(c) MgSO4·7 H2O.
12.17
Lead is used because it has the highest atomic
number of the common, non-radioactive elements.
12.19
Both form tough oxide coatings over their surface;
they are amphoteric, forming beryllate and aluminate
anions; they form carbides containing the C4− ion.
12.21
Magnesium ion is a key component of chlorophyll.
12.23
(a)
Mg(s) + HCl(aq) → MgCl2(aq) + H2(g)
then evaporate to crystallize MgCl2·6
H2O(s).
(b)
Mg(s) + Cl2(g) → MgCl2(s)
Beyond the Basics
ΔH°
ΔS°
T
= +83 kJ·mol−1
= 0.220 kJ·mol−1·K−1
= 377 K
12.27
The formula is actually [Mg(OH2)6]2+[SO4·H2O]2−.
12.29
BeH+. This ion would possess a single bond.
12.31
Ca3N2(s) + 4 NH3(l) → 3 Ca(NH2)2(NH3)
12.33
ΔGº = −92 kJ·mol−1. Less favorable, for at a higher
temperature, the low- melting magnesium will be a
liquid. The reason for synthesizing at a higher
temperature is the greatly increased rate of reaction.
12.35
(d) 2 B4H10(g) + 11 O2(g) → 4 B2O3(s) + 10
H2O(g)
Ca(OH)2(aq) + Mg2+(aq) → Mg(OH)2(s) +
Ca2+(aq)
Mg(OH)2(s) + 2 HCl(aq) → MgCl2(aq) + 2 H2O(l)
12.25
Answers to Odd-Numbered Questions 11
13.3
number of –1:
13.5
An arachno-cluster.
13.7
-1,042kJ. The major factors are the weak fluorinefluorine bond, and the exceedingly strong boronfluorine bond.
13.9
Al3+ is surrounded by the partially negative oxygen
atoms of the six water molecules.
13.11
The hydrated aluminum ion acts as a Bronsted-Lowry
acid.
13.13
The potential environmental hazards are “red mud;”
hydrogen fluoride gas; the carbon oxides; and
fluorocarbon compounds produced.
13.15
Aluminum fluoride is a typical ionic compound.
Both aluminum bromide and aluminum iodide are
covalently bonded dimers. Aluminum chloride is a
borderline case.
13.17
A spinel has the formula AB2X4, where A is a
dipositive metal ion, B is a tripositive metal ion, and
X is a dinegative ion. In the reverse spinel, the A
cations occupy octahedral sites while half of the B
cations occupy the tetrahedral sites.
13.19
Gallium(III) fluoride must consist of an ionic lattice
of gallium(3+) and chloride(1−) ions.
13.21
In acid conditions, the soluble Al(OH2)63+ is
produced. The aluminum ion is very toxic to fish.
The species is probably Na2BeCl4,.
Chapter 13
Exercises
13.1
(a) 3 K(l) + AlCl3(s) → Al(s) + 3 KCl(s)
(b) B2O3(s) + 2 NH3(g) → 2 BN(s) + 3 H2O(g)
(c) 2 Al(s) + 2 OH−(aq) + 6 H2O(l) → 2
[Al(OH)4]−(aq) + 3 H2(g)
The bridging oxygen atoms have an oxidation
Beyond the Basics
13.23 The metallic radius is a measure of the atomic size.
The covalent radius will be smaller because there is
orbital overlap. The ionic radius is by far the
smallest because all the valence electrons have been
lost.
13.25
Cl3Al[O(C2H5)2].
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12
Answers to Odd-Numbered Questions
13.27
The beryllium ion will resemble the aluminum ion.
[Be(OH2)4]2+(aq) + H2O(l) → [Be(OH2)3(OH)]+(aq)
+ H3O+(aq)
13.29
Let number of ions of magnesium = x, then: x = +3.
13.31
3 GaCl(s) → GaCl3(s) + 2 Ga(s)
There are equal moles (in the same phase) on each
side of the equation.
13.33
4 AlCl3(s) + CH3CN(l) → [Al(CH3CN)6]3+(CH3CN)
+ 3 [AlCl4]−(CH3CN)
13.35
Ga(OH2)63+(aq) → GaO(OH)(s) + H2O(l) + 3
H3O+(aq)
Addition of acid will shift the equilibrium to the left.
13.37
Aluminum, lacking any “inner” d electrons, behaves
more like a Group 3 element than a Group 13
element.
13.39
Descriptive Inorganic Chemistry, Fifth Edition
14.3
(a) An element forming chains of its atoms.
(b) Low density silicates with numerous cavities in
the structure.
(c) Non-metallic inorganic compounds.
(d) Chains of alternating silicon and oxygen atoms
with organic side groups.
14.5
Diamond is a very hard, transparent, colorless solid
that is a good conductor of heat but a non-conductor
of electricity. Graphite is a soft, slippery, black solid
that is a poor conductor of heat but a good conductor
of electricity. C60 is black and a nonconductor of
heat and electricity.
14.7
Diamond and graphite both have network covalent
bonded structures. The solvation process cannot
provide the energy necessary to break nonpolar
covalent bonds. The fullerenes consist of discrete
molecules, such as C60. These individual nonpolar
units can become solvated by nonpolar or lowpolarity solvent molecules and hence dissolve.
14.9
The three classes are ionic, covalent, and metallic.
14.11
SiO2(s) + 3 C(s) → SiC(s) + 2 CO(g), entropy
driven.
ΔH°
= +624 kJ·mol−1
ΔS°
ΔG°
13.41
208 kJ·mol−1. B.O. = 1.
13.43
ΔHf(B2O3) = −1271 kJ·mol−1
13.45
Using the atomic radius of zirconium would give a
ratio of sizes of close to unity: not an NaCl packing
pattern. The structure must be [Zr4+][B124−].
14.13
= +0.354 kJ·mol−1·K−1
= −181 kJ·mol−1
It is the lower bond energy of the C=S bond
compared to the C=O bond that makes such a large
difference.
Chapter 14
Exercises
14.1
(a) Li2C2(s) + 2 H2O(l) → 2 LiOH(aq) + C2H2(g)
(b) SiO2(s) + 2 C(s) → Si(l) + 2 CO(g)
(c) CuO(s) + CO(g) → Cu(s) + CO2(g)
(d) Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l)
CaCO3(s) + CO2(g) + H2O(l) → Ca(HCO3)2(aq)
(e) CH4(g) + 4 S(l) → CS2(g) + 2 H2S(g)
(f) SiO2(s) + 2 Na2CO3(l) → Na4SiO4(s) + 2
CO2(g)
(g) PbO2(s) + 4 HCl(aq) → PbCl4(aq) + 2 H2O(l)
PbCl4(aq) → PbCl2(s) + Cl2(g)
2010 © W. H. Freeman and Company, All Rights Reserved
Descriptive Inorganic Chemistry, Fifth Edition
Answers to Odd-Numbered Questions 13
14.33
14.15
sp hybrid orbitals are formed.
14.17
Silicon in silane molecule has empty 3d orbitals that
can be involved in the oxidation process.
14.19
The synthesis of HFC-134a requires a complex,
expensive multistep procedure.
14.21
It absorbs wavelengths in the infrared region that are
currently transparent.
14.23
14.25
Trigonal planar.
14.27
There are three Fe2+ ions and two Fe3+ ions per
formula.
14.29
Zeolites are used as ion exchangers; as adsorption
agents; for gas separation; and as specialized
catalysts.
14.31
Any polymer molecules that leak in breast implants
cannot be broken down by normal bodily processes.
14.35
PbO(s) + H2O(l) → PbO2(s) + 2 H+(aq) + 2 e−
PbO(s) + 2 H+(aq) + 2 e− → Pb(s) + H2O(l)
14.37
CN− and CO.
14.39
The lack of the range of synthetic pathways.
14.41
Carbon:
4 CO(g) + Ni(s) → Ni(CO)4(g)
CO(g) + Cl2(g) → COCl2(g)
CO(g) + S(s) → COS(g)
H 2SO 4
HCOOH(l) ⎯⎯
⎯→ CO(g) + H2O(l)
CO2(g) + 2 Ca(s) → C(s) + 2 CaO(s)
2 CO(g) + O2(g) → 2 CO2(g)
catalyst
CO(g) + 2 H2(g) ⎯⎯⎯→ CH3OH(l)
2 C(s) + O2(g) → 2 CO(g)
C(s) + O2(g) → CO2(g)
Na2C2(s) + 2 H2O(l) → 2 NaOH(aq) + C2H2(g)
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
Al4C3(s) + H2O(l) → 3 CH4(g) + 4 Al(OH)3(s)
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
CH4(g) + 4 S(l) → CS2(g) + 2 H2S(g)
CS2(g) + 3 Cl2(g) → CCl4(g) + S2Cl2(l)
CS2(g) + S2Cl2(l) → CCl4(g) + 6 S(s)
CH4(g) + NH3(g) → HCN(g) + 3 H2(g)
HCN(aq) + H2O(l) → H3O+(aq) + CN−(aq)
Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l)
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) +
CO2(g)
CaCO3(s) + H2O(l) + CO2(g) → Ca(HCO3)2(aq)
Silicon:
Si(s) + HCl(g) → SiHCl3(g) + H2(g)
2 CH3Cl(g) + Si(s) → (CH3)2SiCl2(l)
SiO2(s) + 2 C(s) → Si(s) + 2 CO(g)
SiO2(s) + 6 HF(aq) → SiF62−(aq) + 2 H+(aq) + 2
H2O(l)
2010 © W. H. Freeman and Company, All Rights Reserved
14
Answers to Odd-Numbered Questions
SiO2(s) + 2 NaOH(l) → Na2SiO3(s) + H2O(g)
SiO2(s) + 3 C(s) → SiC(s) + 2 CO(g)
Δ
SiO2(s) + 2 Na2CO3(l) ⎯
Na4SiO4(s) + 2 CO2(g)
⎯→
2 SiO44−(aq) + 2 H+(aq) → Si2O76−(aq) + H2O(l)
Beyond the Basics
14.43 It’s a calcium ion mimic.
14.45
Sodium and calcium ions can leach out.
14.47
(a) A six-membered ring structure, Si3O3, with
alternating silicon and oxygen atoms.
(b) P3O93−
(c) S3O9.
14.49
Tin(II) chloride is the Lewis acid, while the chloride
ion, the Lewis base.
14.51
Mg2SiO4(s) + 2 “H2CO3(aq)” → 2 MgCO3(s) +
SiO2(s) + 2 H2O(l)
14.53
A = CH4; B = S; C = CS2; D = H2S; E = Cl2; F = CCl4
CH4(g) + 4 S(s) → CS2(g) + 2 H2S(g)
CS2(g) + 2 Cl2(g) → CCl4(g) + 2 S(s)
CH4(g) + 4 Cl2(g) → CCl4(g) + 4 HCl(g)
14.55
Y is Sn(C2H5)4, Z is SnCl(C2H5)3
3 Sn(C2H5)4(l) + SnCl4(l) → 4 SnCl(C2H5)3(l)
14.57
ΔGº = +48 kJ·mol−1, a positive value indicates
decomposition will be favored.
14.59
Energy released = 394 kJ·mol–1
Descriptive Inorganic Chemistry, Fifth Edition
15.7
(a) Nitrogen has a very strong nitrogen-nitrogen triple
bond.
(b) Kinetic factors can lead to other products.
15.9
Air. Cool the mixture and have the argon condense
out.
15.11
A solution of the ion is acidic, not neutral, and its
compounds are all very thermally unstable.
15.13
15.15
Volume of gas = 2.8 L
15.17
Hydrogen bonding in ammonia molecules.
15.19
The shapes are:
15.21
High pressure favors the reaction direction that will
result in the lesser moles of gas.
15.23
White phosphorus is a very reactive, white, waxy
substance that consists of P4, while red phosphorus is
a red powdery solid that consists of long polymer
chains.
15.25
Ammonia must be the stronger base.
15.27
N−O bond order is 2:
Chapter 15
Exercises
15.1
(a) AsCl3(l) + 3 H2O(l) → H3AsO3(aq) + 3 HCl(g)
(b) 3 Mg(s) + N2(g) → Mg3N2(s)
(c) NH3(g) + 3 Cl2(g) → NCl3(l) + 3 HCl(g)
(d) CH4(g) + H2O(g) → CO(g) + 3 H2(g)
(e) N2H4(l) + O2(g) → N2(g) + 2 H2O(g)
(f) NH4NO3(aq) → N2O(g) + 2 H2O(l)
(g) 2 NaOH(aq) + N2O3(aq) → 2 NaNO2(aq) + H2O(l)
(h) 2 NaNO3(s) → 2 NaNO2(s) + O2(g)
Net energy change = 53 kJ
15.29
(i) P4O10(g) + C(s) → P4(g) + 10 CO(g)
15.3
Arsenic has both metallic and nonmetallic allotropes.
15.5
Difference in boiling points; different acid-base
properties; difference in their combustions.
2010 © W. H. Freeman and Company, All Rights Reserved
.
Descriptive Inorganic Chemistry, Fifth Edition
Answers to Odd-Numbered Questions 15
This much higher value results from fluorine bonds
to other elements being stronger than those of
chlorine to the same element..
15.59
[NF4]+F−
15.61
[A] Red phosphorus; [B] white phosphorus; [C]
tetraphosphorus decaoxide;
[D] phosphoric acid; [E] phosphorus trichloride; [F]
phosphorus pentachloride; [G]
phosphorous/phosphonic acid.
4 P(s) → P4(s)
P4(s) + 5 O2(g) → P4O10(s)
P4O10(s) + 6 H2O(l) → 4 H3PO4(aq)
P4(s) + 6 Cl2(g) → 4 PCl3(l)
PCl3(l) + Cl2(g) → PCl5(s)
PCl5(s) + 4 H2O(l) → H3PO4(aq) + 5 HCl(g)
PCl3(l) + 3 H2O(l) → H3PO3(aq) + 3 HCl(g)
15.63
Li3N(s) + 3 H2O(l) → 3 LiOH(aq) + NH3(g)
This would be uneconomical.
15.65
HONH2 (or NH2OH, hydroxylamine); H2NNO2;
(NH2)2CO (urea).
15.67
Beyond the Basics
15.43 PH4+ and Cl−, then BCl4−.
PH3(g) + HCl(l) → PH4+(HCl) + Cl−(HCl)
Cl−(HCl) + BCl3(HCl) → BCl4−(HCl)
2 NCl3(g) → N2(g) + 3 Cl2(g)
The reaction is highly exothermic due primarily to
the strength of the nitrogen-nitrogen triple bond.
15.69
Only two hydrogen atoms are replaced because the
structure contains only two hydroxyl groups.
15.45
Trigonal planar; 120º; bond order would be 1.33 in
the first case and 1.17 in the other.
15.71
NO2+ and CNO−
15.73
15.47
The most obvious structure would be that in which
the four terminal oxygen atoms in P4O10 are replaced
by sulfur atoms.
A very large low-charge anion might stabilize the
pentanitrogen cation.
15.75
Bonding between sodium and azide ions is likely to
be predominantly ionic whereas that in the heavy
metal azides will be more covalent.
(a) Silver(I) or lead(II) or mercury(I).
(b) N3−(aq) + H2O(l) → HN3(aq) + OH−(aq)
(c) The azide ion will decompose on heating.
15.77
3 (NH4)[N(NO2)2](s) + 4 Al(s) → 2 Al2O3(s) + 6
H2O(g) + 6 N2(g)
Reasons for its exothermicity: (a) the formation of
dinitrogen; (b) the formation of water; (c) the
formation of aluminum oxide.
It would be a good propellant because of the large
volume of gas produced per mole of ADN.
15.79
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
9 H2(g) + 2 AsO42−(aq) + 4 H+(aq) → 2 AsH3(g) + 8
H2O(l)
15.31
Steric hindrance by bromine.
15.33
In the azide ion, the double-double nitrogen-nitrogen
bond is strongly preferred.
15.35
NH2OH(aq) + BrO3−(aq) → NO3−(aq) + Br−(aq) +
H3O+(aq)
15.37
NOF(g) + SbF5(l) → NO+(SbF5) + SbF6−(SbF5)
15.39
H2S2O7 and H6Si2O7.
15.41
(a) Rapid algae growth leading to a depletion of
dissolved dioxygen.
(b) Mutually beneficial relationship between two
organisms.
(c) Use of a chemical compound to combat disease.
(d) Calcium hydroxide phosphate that is the bone
material.
15.49
K = 6 × 102
K = 7×10−3
Equilibrium is attained much more rapidly.
15.51
(a)
(b)
(c)
15.53
ΔHº = −57 kJ·mol−1
15.55
Mass Na2HPO4 = 4.0 g, mass NaH2PO4 = 8.6 g
15.57
Assuming that the P−Cl bond has about the same
energy in PCl5 and PCl3, the dissociation energy is =
412 kJ·mol−1,
For the decomposition of PF5, the energy change will
be = 825 kJ·mol−1.
Δ
2 AsH3(g) ⎯
⎯→ 2 As(s) + 3 H2(g)
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16
Answers to Odd-Numbered Questions
Descriptive Inorganic Chemistry, Fifth Edition
16.21
Chapter 16
Exercises
16.1
(a) 2 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
(b) BaS(s) + 4 O3(s) → BaSO4(s) + 4 O2(g)
(c) BaO2(s) + 2 H2O(l) → Ba(OH)2(aq) + H2O2(aq)
(d) 2 KOH(aq) + CO2(g) → K2CO3(aq) + H2O(l)
K2CO3(aq) + CO2(g) + H2O(l) → 2 KHCO3(aq)
(e) Na2S(aq) + H2SO4(aq) → Na2SO4(aq) + H2S(g)
16.23
Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(l)
(f) Na2SO3(aq) + H2SO4(aq) → Na2SO4(aq) +
SO2(g) + H2O(l)
16.25
At higher temperatures, S8 rings break into S2
molecules analogous to O2.
16.27
The closeness of the bond angle in H2Te to 90°
suggests that the central tellurium atom is using pure
p orbitals in its bonding.
16.29
Sulfuric acid can act as an acid; as a dehydrating
agent, as an oxidizing agent, as a sulfonating agent,
and as a base with stronger acids.
16.31
Sulfur trioxide.
16.33
The formal charge representations are:
16.35
(a) H2S(g) + Pb(CH3COO)2(aq) → PbS(s) + 2
CH3COOH(aq)
(g) 8 Na2SO3(aq) + S8(s) → 8 Na2S2O3(aq)
16.3
Its electrical resistivity is low enough to be
considered metallic.
16.5
(a) Finely divided metals that are spontaneously
flammable in air.
(b) Different crystal forms of an element.
(c) Unusual type of equilibria found with hemoglobin
in which addition of one oxygen molecule increases
the ease of addition of subsequent oxygen molecules.
16.7
Photosynthesis has resulted in the conversion to
dioxygen of most of the carbon dioxide.
16.9
Bond order, about 1.
16.11
Larger. Because of steric crowding.
16.13
The oxidation number of +1 for oxygen is a result of
each atom being sandwiched between a more
electronegative fluorine atom.
16.15
Among the Group 16 elements, it is only sulfur that
readily catenates.
16.17
The structures are:
(b) Ba2+(aq) + SO42−(aq) → BaSO4(s)
16.19
16.37
There is a very high activation energy barrier to the
reaction SO2 Æ SO3.
16.39
The large tetramethylammonium cation will stabilize
the large, low-charge ozonide ion.
16.41
The NS2+ ion is isoelectronic and isostructural with
carbon disulfide, CS2.
16.43
We require only small quantities of selenium for a
healthy existence.
The structure is probably based on the S8 ring.
2010 © W. H. Freeman and Company, All Rights Reserved
Descriptive Inorganic Chemistry, Fifth Edition
Beyond the Basics
16.45 The value of −668 kJ is much less than the −1209 kJ
for sulfur hexafluoride. This difference is accounted
for by the chlorine-chlorine bond being stronger.
16.47
The ammonium salt will be less basic than the
calcium salt because the ammonium ion is the
conjugate base of a weak acid.
16.49
Concentration in ppb = 2 × 10–5 ppb
16.51
(a)
(b)
Length of side = 400 pm
Thus length of side = 339 pm.
Answers to Odd-Numbered Questions 17
16.65
The triple-bond structure is more likely.
16.67
Rubidium or cesium. A large low-charge cation is
necessary.
16.69
16.53
16.71.
The species would be isoelectronic and isostructural
with the carbonate ion and the nitrate ion.
Chapter 17
Exercises
17.1
(a) UO2(s) + 4 HF(g) → UF4(s) + 2 H2O(l)
(b) CaF2(s) + H2SO4(l) → 2 HF(g) + CaSO4(s)
(c) SCl4(l) + 2 H2O(l) → SO2(g) + 4 HCl(g)
(d) 3 Cl2(aq) + 6 NaOH(aq) → NaClO3(aq) + 5
NaCl(s) + 3 H2O(l)
16.55
mass = 94 tonne
16.57
Apparent oxidation number S: [S] = +8, an
impossible value because the oxidation number of
sulfur cannot exceed 6.
(e) I2(s) + 5 F2(g) → 2 IF5(s)
16.59
SO32−(aq) + S2O82−(aq) + H2O(l) → 3 SO42−(aq) + 2
H+(aq)
16.61
E = −1.48 V
16.63
[A] Sulfur dioxide; [B] potassium hydroxide; [C]
potassium sulfite; [D] sulfur; [E] thiosulfate ion; [F]
tetrathionate ion; [G] thiosulfuric acid.
SO2(g) + 2 KOH(aq) → K2SO3(aq)
K+(aq) + [B(C6H5)4]−(aq) → K[B(C6H5)4](s)
K2SO3(aq) + S(s) → K2S2O3(aq)
2 S2O32−(aq) + I2(aq) → S4O62−(aq) + 2 I−(aq)
S2O32−(aq) + 2 H+(aq) → H2S2O3(aq)
H2S2O3(aq) → H2O(l) + S(s) + SO2(g)
(f) BrCl3(l) + 2 H2O(l) → 3 HCl(aq) + HBrO2(aq)
17.3
Fluorine has a very weak fluorine-fluorine bond; its
compounds with metals are often ionic when those of
the comparable chlorides are covalent; it forms the
strongest hydrogen bonds known; it tends to stabilize
high oxidation states; the solubility of its metal
compounds is often quite different than those of the
other halides.
17.5
The reaction with nonmetals is strongly enthalpydriven.
17.7
I2(s) + 7 F2(g) → 2 IF7(s) There is a decrease of
seven moles of gas in this reaction..
17.9
Because hydrogen ion does not appear in the halfequation, the reduction potential will not be pH
sensitive.
2010 © W. H. Freeman and Company, All Rights Reserved
18
Answers to Odd-Numbered Questions
17.11
The H−F bond is particularly strong..
17.13
Mass of calcium sulfate = 4.1 × 1012 g = 4.1 × 106
tonne
17.15
Zero.
17.17
(a) 2 Cr(s) + 3 Cl2(g) → 2 CrCl3(s)
Descriptive Inorganic Chemistry, Fifth Edition
2 F−(KH2F3) → F2(g) + 2 e−
HF(aq) + OH−(aq) → H2O(l) + F−(aq)
HF(aq) + F−(aq) → HF2−(aq)
6 HF(aq) + SiO2(s) → SiF62−(aq) + 2 H+(aq) + 2
H2O(l)
4 HF(g) + UO2(s) → UF4(s) + 2 H2O(g)
UF4(s) + F2(g) → UF6(g)
Chlorine:
P4(s) + 10 Cl2(g) → 4 PCl5(s)
2 Fe(s) + 3 Cl2(g) → 2 FeCl3(s)
3 Cl2(g) + NH3(g) → NCl3(l) + 3 HCl(g)
Cl2(aq) + 2 OH−(aq) → Cl−(aq) + ClO−(aq) + H2O(l)
ClO−(aq) + H+(aq) → HClO(aq)
2 ClO−(aq) + Ca2+(aq) → Ca(ClO)2(s)
Cl2(g) + H2(g) → 2 HCl(g)
2 HCl(g) + Fe(s) → FeCl2(s) + H2(g)
3 Cl2(aq) + 6 OH−(aq) → ClO3−(aq) + 5 Cl−(aq) + 3
H2O(l)
ClO3−(aq) + H2O(l) → ClO4−(aq) + 2 H+(aq) + 2 e−
2 ClO3−(aq) + 4 H+(aq) + 2 Cl−(aq) → 2 ClO2(aq) +
Cl2(g) + 2 H2O(l)
Iodine:
I2(s) + Cl2(g) → 2 ICl(s)
I2(s) + 2 S2O32−(aq) → 2 I−(aq) + S4O62−(aq)
2 I−(aq) + Cl2(g) → I2(aq) + 2 Cl−(aq)
I−(aq) + I2(aq) → I3−(aq)
(b) Cr(s) + 2 ICl(l) → CrCl2(s) + I2(s)
17.19
Iron(III) iodide will not be stable because iodide ion
is a reducing agent.
17.21
ΔH
= −7838 kJ. It would be a good propellant
because it produces a large number of small gas
molecules.
17.23
10 H2S(g) + 6 I2O5(s) → 10 SO2(g) + 6 I2(s) + 10
H2O(l)
I2(s) + 2 S2O32−(aq) → 2 I−(aq) + S4O62−(aq)
17.25
Steric hindrance.
17.27
17.29
17.31
It would start to show some metallic properties; the
diatomic element might be a significant electrical
conductor; common oxidation state of −1; form an
insoluble compound with silver ion. Astatine should
form interhalogen compounds.
Structure (c), with the charge on the sulfur atom,
must be the major contributor.
17.35
Chlorine oxidation state = +1, oxygen = –1.
17.37
The iodide anion will stabilize the large low-charge
cation.
17.39
BrF would be an analog of Cl2.
17.41
(a) (CN)2; (b) AgCN, or Pb(CN)2, or Hg2(CN)2.
17.43
P(CN)3
Beyond the Basics
17.45 The ammonium hydrogen fluoride may be
decomposing.
17.33
Fluorine:
Cl2(g) + 3 F2(g) → 2 ClF3(g)
S(s) + 3 F2(g) → SF6(g)
BrO3−(aq) + F2(g) + 2 OH−(aq) → BrO4−(aq) + 2
F−(aq) + H2O(l)
2 Fe(s) + 3 F2(g) → 2 FeF3(s)
H2(g) + F2(g) → 2 HF(g)
2010 © W. H. Freeman and Company, All Rights Reserved
Descriptive Inorganic Chemistry, Fifth Edition
Answers to Odd-Numbered Questions 19
(b) ClF3(l) + KF(s) → K+(ClF3) + ClF4–(ClF3)
(c) In (a), the B–F bond is much stronger than the Cl
F bond. In (b), the Cl–F bond strength must be
greater than the energy needed to extract a
fluoride ion from the potassium fluoride lattice.
17.47
Chapter 18
Exercises
18.1
(a) Xe(g) + 2 F2(g) → XeF4(s)
(b) XeF4(s) + 2 PF3(g) → 2 PF5(g) + Xe(g)
17.49
Dichlorine heptaoxide. It is the oxide in the higher
oxidation state.
18.3
Descending, the melting and boiling points increase,
as do the densities.
17.51
The bond angles will be approximately 109½º.
18.5
Helium cannot be solidified under normal pressure;
when cooled close to absolute zero, liquid helium
becomes an incredible thermal conductor.
The bond order must be ½.
18.7
18.9
17.53
2 NH4ClO4(s) → N2(g) + Cl2(g) + 2 O2(g) + 4 H2O(g)
The weakness of the fluorine-fluorine bond that has
to be broken, and the comparative strength of the
xenon-fluorine bond.
18.11
17.55
17.57
Tl+(I3)–. Iodide is a reducing agent.
17.59
–
17.61
(a) The azide (N3 ) ion, acts as a pseudohalide ion.
Thus it can form a pseudo-interhalide ion.
(b) Higher.
(c) There will be a trigonal bipyramid electron-pair
arrangement.
(d) By a large cation.
(a) ClF3(l) + BF3(g) → ClF2+(ClF3) + BF4–(ClF3)
18.13
The double-bonded structure probably makes a major
contribution to the bonding.
18.15
Using the calculation method:
(a)
+4
(b)
+6
(c)
+8
18.17
Rubidium or cesium.
18.19
2 Au + 7 KrF2 → 2 (KrF)+(AuF6–) + 5 Kr
18.21
Xe(g) + F2(g) → XeF2(s)
2 XeF2(s) + 2 H2O(l) → 2 Xe(g) + O2(g) + 4 HF(l)
Xe(g) + 2 F2(g) → XeF4(s)
Xe(g) + 3 F2(g) → XeF6(s)
XeF6(s) + H2O(l) → XeOF4(l) + 2 HF(l)
2010 © W. H. Freeman and Company, All Rights Reserved
20
Answers to Odd-Numbered Questions
XeOF4(l) + 2 H2O(l) → XeO3(s) + 4 HF(l)
XeO3(s) + OH−(aq) → HXeO4−(aq)
2 HXeO4−(aq) + 2 OH−(aq) → XeO64−(aq) + Xe(g) +
O2(g) + H2O(l)
XeO64−(aq) + 2 Ba2+(aq) → Ba2XeO6(s)
Ba2XeO6(s) + 2 H2SO4(aq) → 2 BaSO4(s) + XeO4(g)
+ 2 H2O(l)
Descriptive Inorganic Chemistry, Fifth Edition
19.13
(a) The d6 configuration in an octahedral field:
Beyond the Basics
18.23 XeF2(SbF5) + SbF5(l) → XeF+(SbF5) + SbF6–(SbF5)
18.25
The Ar−F bond energy = 77.5 kJ·mol-1.
Chapter 19
(b) The d6 configuration in a tetrahedral field:
Exercises
19.1
(a) Element belonging to the d-block.
(b) Molecules or ions covalently bonded to a central
metal ion.
(c) Energy separation between different members of
the metal’s d-orbital set.
19.3
19.5
The cyanide ligand stabilizes low oxidation states and
stabilizes normal ones.
[Pt(NH3)4]2+[PtCl4]2−
19.7
The geometric isomers are:
19.15
The largest value of Δ is for the cobalt(III) complex,
the others being cobalt(II) because the splitting
increases with increase in oxidation state.
19.17
(a)
[ReF6]2–, the heavier metal has greater
crystal field splitting.
(b)
[Fe(CN)6]3–, the higher charge has greater
crystal field splitting.
19.19
ConfigurationCFSE: d 0 , −0.0 Δtet, ascending to d 2 ,
−1.2 Δtet, descending to d 5 , −0.0 Δtet, repeating to
d10, −0.0 Δtet.
Normal spinel, because the Cr3+ ion will have a
greater CFSE than that of the Ni2+ ion.
There are two optical (chiral) isomers.
19.21
19.23
19.9
19.11
(a) Ammonium pentachlorocuprate(II); (b)
pentaammineaquacobalt(III) bromide; (c) potassium
tetracarbonylchromate(-III); (d) potassium
hexafluoronickelate(IV); (e) tetraamminecopper(II)
perchlorate.
(a) [Mn(OH2)6](NO3)2, (b) Pd[PdF6], (c)
[CrCl2(OH2)4]Cl·2 H2O, (d) K3[Mo(CN)8].
[Ni(OH2)6]2+(aq) + 2 det(aq) → [Ni(det)2]2+(aq) +
6 H2O(l)
The chelate effect.
Beyond the Basics
19.25 The ligand is probably too large to fit in addition to
the three chloro-ligands.
19.27
(a) M2+ should disproportionate as the sum of the
potentials is positive.
3 M2+(aq) → M(s) + 2
3+
M (aq)
(b) pH = 3.38
2010 © W. H. Freeman and Company, All Rights Reserved
Descriptive Inorganic Chemistry, Fifth Edition
19.29
19.31
19.33
For zinc, with its filled d10 orbitals, there is no CFSE.
For nickel, a square-planar geometry will maximize
CFSE and it will enable some degree of π bonding to
occur.
3+
−
(a) [Cr(OH2)6] ·3Cl , hexaaquachromium(III)
chloride;
(b) [Cr(OH2)5Cl]2+·2Cl−,
pentaaquachlorochromium(III) chloride;
(c) [Cr(OH2)4Cl2]+·Cl−,
tetraaquadichlorochromium(III) chloride.
Fluoride is a weaker field ligand than chloride.
Answers to Odd-Numbered Questions 21
20.19
Chromium(VI) oxide. The very high charge density
of the chromium metal ion will result in covalent
bond formation.
20.21
Chromium(III) ion will lose a hydrogen ion to a
water molecule.
20.23
According to Fajan’s Rules, cations with non-noblegas configurations are likely to have a more covalent
character.
20.25
(a) FeO(OH), (b) Fe3+, (c) Fe2+.
20.27
They both form anhydrous chlorides that react with
water. In the gas phase, their chlorides exist as
dimers, Al2Cl6 and Fe2Cl6. On the other hand,
iron(III) oxide is basic, while the oxide of aluminum
is amphoteric.
20.29
Titanium:
Chapter 20
Exercises
20.1
(a) TiCl4(l) + O2(g) → TiO2(s) + 2 Cl2(g)
(b) Na2Cr2O7(s) + S(l) → Cr2O3(s) + Na2SO4(s)
(c) Cu(OH)2(s) → CuO(s) + H2O(l)
20.3
20.5
20.7
For the earlier part of the Period 4 elements, the
maximum oxidation number is the same as the group
number. For the later members, the oxidation state of
+2 predominates.
Titanium(IV) chloride vaporizes readily.
(a)
MnO4−(aq) + 8 H+(aq) + 5 e− →
Mn2+(aq) + 4 H2O(l)
(b)
MnO4−(aq) + 2 H2O(l) + 3 e− → MnO2(s)
+ 4 OH−(aq)
20.9
Fe(s) + 2 HCl(g) → FeCl2(s) + H2(g)
2 Fe(s) + 3 Cl2(g) → 2 FeCl3(s)
20.11
(a) Cobalt, (b) Copper, (c) Chromium.
20.13
The two reactants are the hexaaquairon(III) ion and
thiosulfate ion:
Fe3+(aq) + 2 S2O32−(aq) →
[Fe(S2O3)2]−(aq)
[Fe(S2O3)2]−(aq) + Fe3+(aq) → 2
Fe2+(aq) + S4O62−(aq)
20.15
(a) Fluoride stabilizes high oxidation states.
(b) Low spin.
20.17
2 FeO42−(aq) + 2 NH3(aq) + 10 H+(aq) → 2 Fe3+(aq)
+ N2(g) + 8 H2O(l)
Δ
TiO2(s) + 2 C(s) + 2 Cl2(g) ⎯
⎯→ TiCl4(g) + 2
CO(g)
Δ
TiCl4(g) + O2(g) ⎯
⎯→ TiO2(s) + 2 Cl2(g)
Δ
TiCl4(g) + 2 Mg(l) ⎯
⎯→ Ti(s) + 2 MgCl2(l)
Vanadium:
[H2VO4]–(aq) + 4 H+(aq) + e– → VO2+(aq) + 3
H2O(l)
VO2+(aq) + 2 H+(aq) + e– → V3+(aq) + H2O(l)
[V(OH2)6]3+(aq) + e– → [V(OH2)6]2+(aq)
Chromium:
CrO42−(aq) + 2 Ag+(aq) → Ag2CrO4(s)
CrO42–(aq) + H2O(l) ↔ HCrO4–(aq) + OH–(aq)
2 CrO42−(aq) + 2 H+(aq) → Cr2O72−(aq) + H2O(l)
Cr2O72−(aq) + 2 NH4+(aq) → (NH4)2Cr2O7(s)
(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4 H2O(l)
Cr2O72−(aq) + 14 H+(aq) + 6 e− → 2 Cr3+(aq) + 7
H2O(l)
Cr2O72−(aq) + 2 K+(aq) → K2Cr2O7(s)
K2Cr2O7(s) + H2SO4(aq) → 2 CrO3(s) + K2SO4(aq) +
H2O(l)
K2Cr2O7(s) + 4 NaCl(s) + 6 H2SO4(l) → 2 CrO2Cl2(l)
+ 2 KHSO4(s)
+ NaHSO4(s) + 3 H2O(l)
CrO2Cl2(l) + 4 OH−(aq) → CrO42−(aq) + 2 Cl−(aq) +
2 H2O(l)
Cr2O72–(aq) + 14 H+(aq) + 6 e– → 2 Cr3+(aq) + 7
H2O(l)
2 Cr3+(aq) + Zn(s) → 2 Cr2+(aq) + Zn2+(aq)
2 Cr2+(aq) + 4 CH3COO-(aq) + 2 H2O(l) →
Cr2(CH3COO)4(OH2)2(s)
Manganese:
MnO4−(aq) + e− → MnO42−(aq)
MnO42−(aq) + 2 H2O(l) + 2 e− → MnO2(s) + 4
OH−(aq)
2010 © W. H. Freeman and Company, All Rights Reserved
22
Answers to Odd-Numbered Questions
MnO4−(aq) + 2 H2O(l) + 3 e− → MnO2(s) + 4
OH−(aq)
MnO4−(aq) + 8 H+(aq) + 5 e− → Mn2+(aq) + 4 H2O(l)
Mn2+(aq) + 2 OH−(aq) → Mn(OH)2(s)
Mn(OH)2(s) + OH−(aq) → MnO(OH)(s) + H2O(l) +
e−
Iron:
[Fe(OH2)6]3+(aq) + SCN–(aq) →
[Fe(SCN)(OH2)5]2+(aq) + H2O(l)
[Fe(OH2)6]3+(aq) + 4 Cl–(aq) ↔ [FeCl4]–(aq) + 6
H2O(l)
Fe3+(aq) + 3 OH−(aq) → FeO(OH)(s) + H2O(l)
[Fe(OH2)6]3+(aq) + e− → [Fe(OH2)6]2+(aq)
Fe3+(aq) + 2 S2O32−(aq) → [Fe(S2O3)2]−(aq)
[Fe(S2O3)2]−(aq) + Fe3+(aq) → 2 Fe2+(aq) +
S4O62−(aq)
Fe2+(aq) + 2 OH−(aq) → Fe(OH)2(s)
[Fe(OH2)6]2+(aq) + NO(aq) → [Fe(NO)(OH2)5]2+(aq)
+ H2O(l)
Fe(OH)2(s) + OH−(aq) → FeO(OH)(s) + H2O(l) + e−
Fe2+(aq) + 2 e− → Fe(s)
2 Fe(s) + 3 Cl2(g) → 2 FeCl3(s)
Fe(s) + 2 HCl(g) → FeCl2(s) + H2(g)
Cobalt:
[Co(OH2)6]3+(aq) + e− → [Co(OH2)6]2+(aq)
[Co(OH2)6]2+(aq) + 4 Cl−(aq) → [CoCl4]2−(aq) + 6
H2O(l)
Co2+(aq) + 2 OH−(aq) → Co(OH)2(s)
Co(OH)2(s) + 2 OH−(aq) → Co(OH)42−(aq)
Co(OH)2(s) + OH−(aq) → CoO(OH)(s) + H2O(l) + e−
[Co(OH2)6]2+(aq) + 6 NH3(aq) → [Co(NH3)6]2+(aq) +
6 H2O(l)
[Co(NH3)6]2+(aq) → [Co(NH3)6]3+(aq) + e−
O2(g) + 2 H2O(l) + 4 e− → 4 OH−(aq)
Nickel:
Ni(CO)4(g) → Ni(s) + 4 CO(g)
Ni(s) → Ni2+(aq) + 2 e−
[Ni(OH2)6]2+(aq) + 4 Cl−(aq) → [NiCl4]2−(aq) + 6
H2O(l)
Ni2+(aq) + 2 OH−(aq) → Ni(OH)2(s)
[Ni(OH2)6]2+(aq) + 6 NH3(aq) → [Ni(NH3)6]2+(aq) +
6 H2O(l)
Copper:
2 Cu(s) + 2 H+(aq) + 4 Cl−(aq) → 2 [CuCl2]−(aq) +
H2(g)
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
Cu(s) → Cu2+(aq) + 2 e−
[Cu(OH2)6]2+(aq) + 4 NH3(aq) → [Cu(NH3)4]2+(aq) +
6 H2O(l)
[Cu(OH2)6]2+(aq) + 4 Cl−(aq) → [CuCl4]2−(aq) + 6
H2O(l)
Cu2+(aq) + 2 OH−(aq) → Cu(OH)2(s)
Cu(OH)2(s) + 2 OH−(aq) → [Cu(OH)4]2−(aq)
Cu(OH)2(s) → CuO(s) + H2O(l)
Descriptive Inorganic Chemistry, Fifth Edition
Beyond the Basics
20.31 Addition of an anhydrous calcium compound will
result in formation of the hexaaquacalcium ion.
Addition of an anhydrous zinc compound results in
the formation of the competing complexation.
20.33
Cr2O72−(aq) + H2O(l) → 2 CrO42−(aq) + 2 H+(aq)
Pb2+(aq) + CrO42−(aq) → PbCrO4(s)
20.35
(a) Nickel(II) hydroxide.
(b) This should be the square planar
tetracyanonickelate(II) ion.
(c) This must involve the addition of a fifth cyanide
ion.
20.37
(a)
20.39
The high-charge cation (Fe3+) will have a
somewhat low lattice energy when
combined with a low-charge anion (ClO4–).
(b)
Either: Ammonia and water are quite high in
the spectrochemical series. Or: Ammonia
and water are hard bases.
(c)
Bromide is more easily reduced than
chloride; thus the charge transfer takes place
at a lower energy.
Ni = +2, S = −1.
20.41
[A] Nickel(II) sulfide; [B] hydrogen sulfide; [C]
hexaaquanickel(II) ion;
[D] sulfur dioxide; [E] sulfur; [F] and [G] disulfur
dichloride and sulfur dichloride; [H]
hexaamminenickel(II) ion; [I] nickel(II) hydroxide;
[J] nickel metal; [K] tetracarbonylnickel(0).
NiS(s) + 2 H+(aq) → Ni2+(aq) + H2S(g)
2 H2S(g) + 3 O2(g) → 2 H2O(l) + 2 SO2(g)
2 H2S(g) + SO2(g) → 2 H2O(l) + 3 S(s)
2 S(s) + Cl2(g) → S2Cl2(l)
S(s) + Cl2(g) → SCl2(l)
[Ni(OH2)6]2+(aq) + 6 NH3(aq) → [Ni(NH3)6]2+(aq) +
6 H2O(l)
Ni2+(aq) + 2 OH−(aq) → Ni(OH)2(s)
Ni2+(aq) + Zn(s) → Ni(s) + Zn2+(aq)
Ni(s) + 4 CO(g) → Ni(CO)4(l)
20.43
Vanadium.
20.45
20.47
This corresponds to a full neutron shell.
3+, as the shared oxygen would have an oxidation
state of –2. The linear shape suggests there is a πbonding Cr−O−Cr system.
20.49
Presumably the chloride ligand has preferentially
stabilized the 3+ oxidation state of the iron.
20.51
As the halide ion is more readily oxidized, the
absorption of light will be more and more in the
visible part of the spectrum.
2010 © W. H. Freeman and Company, All Rights Reserved
Descriptive Inorganic Chemistry, Fifth Edition
20.53
Calcium will replace the Mn2+. Iron would most
likely replace the Mn3+. Titanium would replace the
silicon. Aluminum could replace the Mn3+.
20.55
Some form of π-bonding through the d orbitals.
20.57
(a) Fe(s) + O2(g) → Fe2O3(s)
(b) Sodium silicate prevents the continuation of the
oxidation.
(c) The red-hot iron would have reacted with water to
give hydrogen gas. The explosion would have
resulted from a hydrogen/oxygen mixture.
Chapter 21
Exercises
21.1
(a)
density cations result in low lattice energies and such
salts should be water soluble.
21.24
Though thorium is an actinoid, the early actinoids
favor oxidation states matching their analogous group
number.
21.26
In the complex shown, each iodide bridges three
niobium atoms. [Nb6I8]3+
21.28
The Re3Cl9 structure involves a central triangle of
rhenium atoms with bridging and terminal chlorine
atoms in a polymeric structure.
Chapter 22
2 [Ag(CN)2]−(aq) + Zn(s) → 2 Ag(s) +
[Zn(CN)4]2−(aq)
(b)
2 Au(s) + 3 Cl2(g) → 2 AuCl3(s)
21.3
Discussing the 5d fluorides, the oxidation number
seems to ‘plateau’ at seven.
21.5
(a) automobile engine lubricant; (b) antibacterial.
21.7
Osmium(VIII) oxide has a melting point of 40°C, and
is very soluble in low-polarity, organic solvents.
21.9
Ruthenium, rhodium, palladium, osmium, iridium,
and platinum.
21.11
The 3d transition metals tend to have lower oxidation
states than those of the 4d and 5d series. The smaller
3d ions cannot accommodate as many ligands.
21.13
Answers to Odd-Numbered Questions 23
Exercises
22.1
(a)
(b)
22.3
Zn(s) + Br2(l) → ZnBr2(s)
ZnCO3(s) → ZnO(s) + CO2(g)
Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)
Zn2+(aq) + CO32−(aq) → ZnCO3(s)
22.5
(a) Zinc and magnesium have the following
similarities: their cations are 2+ ions of similar size,
they are colorless, and they both form hexahydrates.
Both elements form soluble chlorides and sulfates,
and insoluble carbonates.
(b) The only two common features are that both zinc
and aluminum are amphoteric metals, reacting with
both acids and bases, and they are both strong Lewis
acids.
22.7
Cd(OH)2(s) + 2 e− → Cd(s) + 2 OH−(aq)
2 Ni(OH)2(s) + 2 OH−(aq) → 2 NiO(OH)(s) + 2
For Pd: +2 and +3. For Pt: +2, +4, and +6. Square
planar is common for the lower oxidation states,
octahedral geometry for the +6.
H2O(l) + 2 e−
21.15
orbitals; diagonal
End-on overlap of a pair of
overlap of a pair of
orbitals; and the side-to-side
overlap of a pair of
orbitals.
22.9
Cadmium metal was used as a coating for paper clips
primarily because it was a ‘sacrificial anode.’ As
cadmium compounds are highly toxic, cadmium
plating has been discontinued.
21.17
21.18
PdF3 has the formulation of: (Pd2+)[PdF6]2–.
It has a stable, water-soluble, species at near-neutral
pH making it transportable by biological fluids.
22.11
Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)
Zn(OH2)62+(aq) + 4 NH3(aq) → Zn(NH3)42+(aq) + 6
H2O(l)
Zn2+(aq) + 2 OH−(aq) → Zn(OH)2(s)
Zn(OH)2(s) + 2 OH−(aq) → Zn(OH)42−(aq)
Zn(OH)2(s) → ZnO(s) + H2O(l)
ZnO(s) + 2 H+(aq) → Zn2+(aq) + H2O(l)
ZnCO3(s) → ZnO(s) + CO2(g)
Beyond the Basics
21.20 Fluorine tends to promote metals to their highest
oxidation states. WF6.
21.22
The potassium halides are all water-soluble while all
of the silver halides are insoluble. Low-charge-
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24
Answers to Odd-Numbered Questions
Beyond the Basics
22.13 Mercury(I) undergoes a disproportionation
equilibrium.
22.15
22.17
22.19
22.21
22.23
Descriptive Inorganic Chemistry, Fifth Edition
23.13
Cr(CO)6
Fe(CO)5
Ni(CO)4
CO
The metals are very different in size.
OC
CO OC
(a) Zn(NH2)2(NH3) + 2 NH4 (NH3) →
Zn(NH3)42+(NH3)
(b) Zn(NH2)2(NH3) + 2 NH2–(NH3) → Zn(NH2)42–
(NH3)
Zinc oxide.
V(CO)6 is a seventeen electron complex.
23.17
(a)
1 Mn–Mn bond
(b)
2 Mn–Mn bonds
CO
OC
Mn
(c)
Hg(CH3)2 + 2 Na → 2 NaCH3 + Hg
23.9
(a) LiCH3 + LiBr
(b) 2 LiCl + Mg(C2H5)2
(c) Mg(C2H5)2 + Hg
(d) Li(C6H5) + C2H6
(e) C2H5MgCl + Hg
(f) B(CH2CH2CH3)3
(g) Sn(C2H5)4 + 4 MgCl2
23.11
(a) hexacarbonylchromium(0)
(b) ferrocene or
bis(pentahaptocyclopentadienyl)iron(II)
(c) hexahaptobenzenetricarbonylchromium(0)
(d) pentahaptocyclopentadienyltricarbonyltungsten(I)
(e) bromopentacarbonylmanganese(I)
CO
1 Fe–Fe bond
OC
COCO CO
CO
Fe
(a) Bi(CH3)5
(b) Si(C6H5)4 tetraphenyl silane
(c) KB(C6H5)4 postassium teraphenylborane
(d) Li4(CH3)4
(e) (C2H5)MgCl
23.7
Mn
OC
Exercises
23.1
(a) organometallic
(b) not organometallic as the bond B-O not B-C
(c) organometallic
(d) not organometallic as nitrogen is not metallic
(e) not organometallic as there is no Na-C bond
(f) organometallic
(g) organometallic
C2H5MgBr will be tetrahedral with two molecules of
solvent coordinated to the magnesium.
CO
23.15
Hydrogen sulfide is in a two-step equilibrium with
the sulfide ion. When acidified, the increased
hydronium-ion concentration will “drive” the
equilibria to the left.
23.5
CO
CO
+
Chapter 23
23.3
Mn
Mn
OC
Sulfur. Mercury(II) is a soft acid. Sulfur is a soft
base.
CO
CO
(d)
no Mn-Mn bonds
OC
OC
Br
Mn
OC
OC
23.19
Fe
CO
Mn
Br
CO
CO
CO
(a) [Cr(CO)6] + 3 CH3CN → [Cr(CO)3(CH3CN)3] + 3 CO
(b) [Mn2(CO)10] + H2 → 2 [HMn(CO)5]
(c) [Mo(CO)6] +
(CH3)2PCH2CH2P(Ph)CH2CH2P(CH3)2 →
[Mo(CO)3((CH3)2PCH2CH2P(Ph)CH2CH2P(CH3)2)] +
3 CO
(d) [Fe(CO)5] + 1,3-cyclohexadiene → 2 CO +
(CO)3Fe
2010 © W. H. Freeman and Company, All Rights Reserved
Descriptive Inorganic Chemistry, Fifth Edition
Answers to Odd-Numbered Questions 25
(e)
23.29
(f)
(g)
2 LiCl + (PMe3)2Pt
[PtCl2(PMe3)2] + LiCH2CH2CH2CH2Li
(h) [Ni(CO)4]+ PF3 → [Ni(CO)3PF3] + CO
(i) [Mn2(CO)10] + Br2 → 2 [Mn(CO)5Br]
(j) [HMn(CO)5] + CO2 → [(CO)5MnCOOH]
23.21
(a) +3
(b) +1
Beyond the Basics
23.23
OC
(η5-C5H5)2Ni + Ni(CO)4
2
Ni
Ni
CO
23.25
A = tricarbonyl(η5-cyclopentadienyl)(η1-propenyl)tungsten(II)
B = dicarbonyl(η5-cyclopentadienyl)(η3-propenyl)tungsten(II)
C = tricarbonyl(η5-cyclopentadienyl)(η2-propenyl)tungsten(II)
hexafluorophosphate
Evolved gas = propene
23.27
Ti(S2CEt2)4.
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