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Descriptive Inorganic Chemistry, Fifth Edition Answers to Odd-Numbered Questions 1
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Chapter 1
Exercises
1.1 (a) Surface where electron probability = 0.
(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
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.
1.11 (a) [Ne]3s1; (b) [Ar]4s23d8; (c) [Ar]4s13d10.
1.13 (a) [Ar]; (b) [Ar]; (c) [Ar]3d9.
1.15 1+ and 3+. Configuration of
[Xe]6s24f145d106p1.
1.17 1+. Configuration of [Kr]5s14d10.
1.19
1.21 E113: [Rn]7s25f1 46d1 07p 1 .E113 as +1 ion: [Rn]7s 25f1 46d1 0 .
E113 as +3 ion: [Rn]5f1 46d1 0 .
Beyond the Basics
1.23 9, 5, 121.
1.25 There are sevenforbitals. 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.
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 ofelements; but the table becomes very elongated.
2.7 The -iumending indicates a metal. The ending -on
has been used for non-metals.
2.9 With nuclei up to 26 protons, nuclear fusion is an
exothermic process.
2.11 (a) Lead (b) technetium (c) bromine.
2.13 Sodium, because it has an odd number of protons.
2.15 50.
2.17 (a) Several nonmetals have metallic luster.
(b) Diamond has the highest thermal conductivity ofall 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 4pelectrons willbe increased.
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4.3 High electrical conductivity, high thermal
conductivity, high reflectivity, and high boiling point.
4.5 3sand 3pband overlap means that electrons in the
full 3sband can spill over into the 3pband.
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 4/3r3, 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 gcm3.
4.21 145pm.
4.23 A suspension of gold nano-particles has a red color.
Chapter 5
Exercises5.1 (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 Hard and brittle crystals; high melting points;
electrically conducting in liquid phase and in aqueous
solution.
5.5 (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.
5.7 NaCl, because chloride is smaller than iodide; thecharge is more concentrated, and the ionic attraction
will be stronger.
5.9 Ag+, because it has the lowest charge density.
5.11 UCl3(837C); UCl4(590C); UCl5(327C); 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(2C) and WO3(1472C). The fluoride is
predominantly covalent while the oxide is ionic. Thefluoride 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
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 ofthe lead(IV) ion.
5.37 1.15(r++r).
5.39 164 pm
5.41 251 pm
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For barium sulfate: 1044 kJmol1
Yes.
6.39
For magnesium: H = +428 kJmol1. For lead: H =
+898 kJmol1The only significant difference between the two ions
is their hydration enthalpies.
6.41 Using the Kapustinskii equation: U = 622 kJmol1;
compared with 668 kJmol1experimentally and
636 kJmol1from the Born-Land equation.
6.43 H = +36 kJmol1
6.45 proton affinity = 1141 kJmol1
6.46 Step 1: G= +184 kJmol1
Step 2: G= 47 kJmol1
6.47 Ionic bond formation is best considered as acompetition for electrons.
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 tozero.
7.3 (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)
7.5 (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.
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) SO4
2(aq) + H3O+(aq)
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.
7.15 HF(H2SO4) + H2SO4(l) H2F+(H2SO4) +
HSO4(H2SO4)
HF is acting as a base and H2F+is the conjugate
acid,
7.17 HSeO4(base), H2SO4(conjugate acid); H2O
(acid), OH(conjugate base).
7.19 The HSe bond will be weaker than the HS bond.
Thus hydrogen selenide will be the stronger acid.
7.21 [Zn(OH2)6]2+
(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) +
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, Amust 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 Answers to Odd-Numbered Questions 7
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7.35 (a) No effect.
(b) Increasing pH. Se2(aq) + H2O(l)
HSe(aq) + OH(aq)
(c) Decreasing pH.
[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 G = 28 kJmol1. Magnesium oxide will be a
weaker base than calcium oxide (G = 59
kJmol1).
7.43 NO+is a Lewis acid. Clis a Lewis base.
(NO)(AlCl4)(NOCl) + [(CH3)4N]Cl(NOCl)
[(CH3)4N](AlCl4)(NOCl) + NOCl(l)
7.45 (a) [NH4+] = 3 1017molL1
(b) [NH4+] = 11033molL1
7.47 (a) No. The reactants have the combinations
borderline-borderline and hard-hard.
(b) Yes. The products will be preferred where hard-
hard and soft-soft combinations result.
7.49 (a) Greater than 1.
(b) Less than 1.
7.51 (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.
7.53 (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 Zinc is a borderline acid, so it can be found as ores ofboth hard and soft bases.
7.59 H2CO3(aq) + MgSiO4(s) H2O(l) + SiO2(s) +MgCO3(s)
The atmospheric concentration of carbon dioxide hasdecreased, in part due to the formation of magnesium
carbonate minerals.
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 Fe
3+
(AlO6)
3
.
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.
8.3 (a) +3 (b) +5 (c) 3 (d) 3 (e) +5
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.
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
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)
6 HV2O73(aq) + 10 Cl(aq) + 6 H2O(l)
(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+, E = +0.762 V
8.25 (a) Au3+(aq) + 3 eAu(s) would be the strongeroxidizing agent.
(b) Al(s) Al3+(aq) + 3 ewould be the stronger
reducing agent.
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8.27 E = 0.267 V
8.29 E = +0.805 V
8.31 (a) Br2
(b) E = +0.52 V
(c) The Nernst expression does not have a pH-dependent term.
8.33 The most thermodynamically stable oxidation state is
+3.
8.35 At pH 0.00 because the reduction potential is lower.
Beyond the Basics8.37 The oxidation to carbon monoxide involves an
increase of entropy; thus the TS 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 = 493C
8.39 E will increase to the point where insoluble, brown
manganese(III) oxide will be formed, thusdiscoloring the toilet bowl.
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 theelement to its lower right in the periodic table.
9.3 The general formula: M+M3+(SO42)212H2O, whereM+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.
9.9 As the group is descended, the cation radii increase,
the ionic bond will weaken and the melting point will
be lower.
9.11 Scandium hydroxide, Sc(OH)3.
9.13 SO3(s) + H2O(l) H2SO4(aq)CrO3(s) + H2O(l) H2CrO4(aq)
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) CO; (b) (CC)2.
9.29 Yttrium.
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.
9.41
Group 15 Group 16 Group 17
Group 15 CN22 OCN
FCN
Group 16 OCN CO2 FCO+
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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 106of the signal itself.
10.7 Hydrogen rarely forms a negative ion.
10.9 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) HCCH(g) + 2 H2(g) H3CCH3(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 ammoniacompared to water can be explained in terms of the
much greater bond energy of dinitrogen (945
kJmol1) compared with that of dioxygen (498
kJmol1).
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
negative. Most covalent hydrides belong in the first
category.
10.17 KH; CaH2, GaH3, GeH4, AsH3, H2Se, HBr. The trend
is to increase by one H until germanium, then astepwise 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)
H = 525 kJ
Per mole, this is = 262 kJmol1, compared with
242 kJmol1for the combustion of pure
dihydrogen.
Chapter 11
Exercises
11.1 (a) 2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)
(b) Rb(s) + O2(g) RbO2(s)
(c) 2 KOH(s) + CO2(g) K2CO3(s) + H2O(l)
(d) 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 andelectricity. The alkali metals differ from typical
metals in that they are soft, extremely chemically
reactive, have low melting points and very lowdensities.
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 watermolecules.
11.9 Because the equilibrium of the synthesis reaction:
Na(l) + KCl(l) K(l) + NaCl(l) lies to the left.
Hydrogen Chlorine
Bond energy 432 kJmol1 240 kJmol1
Electron affinity 79 kJmol1 349 kJmol1
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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)
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 metalion range; all its common salts are colorless.
11.19 Potassium dioxide(1) has a lower molar mass, and ischeaper.
11.21 The ammonium ion is large.
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)
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 Basics11.25 Current = 6.94 x 104A
11.27 In the series LiF to CsF, there is an increasing
mismatch in ion sizes. For the series LiI to CsI, thereis a decreasing mismatch in ion sizes.
11.29 NaBF4. The hydration energy will more probably
exceed the (lower) lattice energy, making thecompound more soluble.
11.31 Either: that there is appreciable covalent bonding inthe 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.
Chapter 12
Exercises
12.1 (a) 2 Ca(s) + O2(g) 2 CaO(s)
(b) 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 thecalcium carbonate to give a solution of calcium
hydrogen carbonate.
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12.13 Ca(OH)2(aq) + Mg2+(aq) Mg(OH)2(s) +
Ca2+(aq)
Mg(OH)2(s) + 2 HCl(aq) MgCl2(aq) + 2 H2O(l)
Mg2+(MgCl2) + 2 eMg(l)
2 Cl-(MgCl2) Cl2(g) + 2 e
12.15 (a) Ca(OH)2(hydrated lime) or CaO (quicklime); (b)
Mg(OH)2;
(c) MgSO47 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 C4ion.
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 MgCl26
H2O(s).
(b) Mg(s) + Cl2(g) MgCl2(s)
Beyond the Basics
12.25 H = +83 kJmol1
S = 0.220 kJmol1K1T = 377 K
12.27 The formula is actually [Mg(OH2)6]2+
[SO4H2O]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 G = 92 kJmol1. Less favorable, for at a higher
temperature, the low- melting magnesium will be aliquid. The reason for synthesizing at a higher
temperature is the greatly increased rate of reaction.
12.35 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)
(d) 2 B4H10(g) + 11 O2(g) 4 B2O3(s) + 10
H2O(g)
13.3 The bridging oxygen atoms have an oxidation
number of 1:
13.5 An arachno-cluster.
13.7 -1,042kJ. The major factors are the weak fluorine-
fluorine bond, and the exceedingly strong boron-
fluorine 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 Bcations 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.
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 thesmallest because all the valence electrons have been
lost.
13.25 Cl3Al[O(C2H5)2].
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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 eachside 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 delectrons, behavesmore like a Group 3 element than a Group 13
element.
13.39
13.41 208 kJmol1. B.O. = 1.
13.43 Hf(B2O3) = 1271 kJmol1
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+][B12
4].
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)
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 atomswith organic side groups.
14.5 Diamond is a very hard, transparent, colorless solid
that is a good conductor of heat but a non-conductorof electricity. Graphite is a soft, slippery, black solid
that is a poor conductor of heat but a good conductor
of electricity. C60is 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 discretemolecules, such as C60. These individual nonpolarunits can become solvated by nonpolar or low-
polarity 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.
H = +624 kJmol1
S = +0.354 kJmol1K1
G = 181 kJmol1
14.13 It is the lower bond energy of the C=S bond
compared to the C=O bond that makes such a large
difference.
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14.15 sphybrid orbitals are formed.
14.17 Silicon in silane molecule has empty 3dorbitals 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 adsorptionagents; 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.
14.33
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 CNand 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)
HCOOH(l) CO(g) + H2O(l) 42SOH
CO2(g) + 2 Ca(s) C(s) + 2 CaO(s)
2 CO(g) + O2(g) 2 CO
2(g)
CO(g) + 2 H2(g) CH3OH(l) catalyst
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)
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SiO2(s) + 2 NaOH(l) Na2SiO3(s) + H2O(g)
SiO2(s) + 3 C(s) SiC(s) + 2 CO(g)
SiO2(s) + 2 Na2CO3(l) Na4SiO4(s) + 2 CO2(g)
2 SiO44(aq) + 2 H+(aq) Si2O7
6(aq) + H2O(l)
Beyond the Basics
14.43 Its a calcium ion mimic.
14.45 Sodium and calcium ions can leach out.
14.47 (a) A six-membered ring structure, Si3O3, withalternating 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 G = +48 kJmol1, a positive value indicates
decomposition will be favored.
14.59 Energy released = 394 kJmol1
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)
(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.
15.7 (a) Nitrogen has a very strong nitrogen-nitrogen triplebond.
(b) Kinetic factors can lead to other products.
15.9 Air. Cool the mixture and have the argon condenseout.
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 isa red powdery solid that consists of long polymer
chains.
15.25 Ammonia must be the stronger base.
15.27 NO bond order is 2:
Net energy change = 53 kJ .
15.29
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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 H2S2O7and H6Si2O7.
15.41 (a) Rapid algae growth leading to a depletion ofdissolved 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.
Beyond the Basics
15.43 PH4+and Cl, then BCl4
.
PH3(g) + HCl(l) PH4+(HCl) + Cl(HCl)
Cl(HCl) + BCl3(HCl) BCl4(HCl)
15.45 Trigonal planar; 120; bond order would be 1.33 inthe first case and 1.17 in the other.
15.47 The most obvious structure would be that in which
the four terminal oxygen atoms in P4O10are replacedby sulfur atoms.
15.49 Bonding between sodium and azide ions is likely to
be predominantly ionic whereas that in the heavymetal azides will be more covalent.
15.51 (a) K = 6 102
(b) K = 7103
(c) Equilibrium is attained much more rapidly.
15.53 H = 57 kJmol1
15.55 Mass Na2HPO4= 4.0 g, mass NaH2PO4= 8.6 g
15.57 Assuming that the PCl bond has about the same
energy in PCl5and PCl3, the dissociation energy is =
412 kJmol1,
For the decomposition of PF5, the energy change will
be = 825 kJmol1.
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 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.
15.71 NO2+
and CNO
15.73 A very large low-charge anion might stabilize the
pentanitrogen cation.
15.75 (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) + 6H2O(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 largevolume 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)
2 AsH3(g) 2 As(s) + 3 H2(g)
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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)
(f) Na2SO3(aq) + H2SO4(aq) Na2SO4(aq) +
SO2(g) + H2O(l)
(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 increasesthe 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 ofeach 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:
16.19 The structure is probably based on the S8ring.
16.21
16.23 Ca(OH)2(s) + CO2(g) CaCO3(s) + H2O(l)
16.25 At higher temperatures, S8rings 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 pureporbitals in its bonding.
16.29 Sulfuric acid can act as an acid; as a dehydratingagent, 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)
(b) Ba2+(aq) + SO42(aq) BaSO4(s)
16.37 There is a very high activation energy barrier to the
reaction SO2SO3.
16.39 The large tetramethylammonium cation will stabilizethe 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 ahealthy existence.
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Beyond the Basics
16.45 The value of 668 kJ is much less than the 1209 kJ
for sulfur hexafluoride. This difference is accountedfor 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 theconjugate base of a weak acid.
16.49 Concentration in ppb = 2 105ppb
16.51 (a) Length of side = 400 pm
(b) Thus length of side = 339 pm.
16.53
16.55 mass = 94 tonne
16.57 Apparent oxidation number S: [S] = +8, an
impossible value because the oxidation number ofsulfur cannot exceed 6.
16.59 SO32(aq) + S2O8
2(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) S4O6
2(aq) + 2 I(aq)
S2O32(aq) + 2 H+(aq) H2S2O3(aq)
H2S2O3(aq) H2O(l) + S(s) + SO2(g)
16.65 The triple-bond structure is more likely.
16.67 Rubidium or cesium. A large low-charge cation is
necessary.
16.69
16.71. The species would be isoelectronic and isostructuralwith 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) + 5NaCl(s) + 3 H2O(l)
(e) I2(s) + 5 F2(g) 2 IF5(s)
(f) BrCl3(l) + 2 H2O(l) 3 HCl(aq) + HBrO2(aq)
17.3 Fluorine has a very weak fluorine-fluorine bond; itscompounds 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 theother halides.
17.5 The reaction with nonmetals is strongly enthalpy-
driven.
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 half-equation, the reduction potential will not be pH
sensitive.
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17.11 The HF bond is particularly strong..
17.13 Mass of calcium sulfate = 4.1 1012g = 4.1 106
tonne
17.15 Zero.
17.17 (a) 2 Cr(s) + 3 Cl2(g) 2 CrCl3(s)
(b) Cr(s) + 2 ICl(l) CrCl2(s) + I2(s)
17.19 Iron(III) iodide will not be stable because iodide ionis a reducing agent.
17.21 H = 7838 kJ. It would be a good propellantbecause 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 S2O3
2(aq) 2 I(aq) + S4O62(aq)
17.25 Steric hindrance.
17.27
17.29 It would start to show some metallic properties; thediatomic element might be a significant electrical
conductor; common oxidation state of 1; form an
insoluble compound with silver ion. Astatine shouldform interhalogen compounds.
17.31 Structure (c), with the charge on the sulfur atom,
must be the major contributor.
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)
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) + S4O6
2(aq)
2 I(aq) + Cl2(g) I2(aq) + 2 Cl(aq)
I(aq) + I2(aq) I3
(aq)
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 bedecomposing.
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17.47
17.49 Dichlorine heptaoxide. It is the oxide in the higher
oxidation state.
17.51 The bond angles will be approximately 109.
17.53 2 NH4ClO4(s) N2(g) + Cl2(g) + 2 O2(g) + 4 H2O(g)
17.55
17.57 Tl+(I3). Iodide is a reducing agent.
17.59 (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.
17.61 (a) ClF3(l) + BF3(g) ClF2+(ClF3) + BF4
(ClF3)
(b) ClF3(l) + KF(s) K+(ClF3) + ClF4
(ClF3)
(c) In (a), the BF bond is much stronger than the Cl
F bond. In (b), the ClF bond strength must begreater than the energy needed to extract a
fluoride ion from the potassium fluoride lattice.
Chapter 18
Exercises
18.1 (a) Xe(g) + 2 F2(g) XeF4(s)
(b) XeF4(s) + 2 PF3(g) 2 PF5(g) + Xe(g)
18.3 Descending, the melting and boiling points increase,
as do the densities.
18.5 Helium cannot be solidified under normal pressure;
when cooled close to absolute zero, liquid helium
becomes an incredible thermal conductor.18.7 The bond order must be .
18.9 The weakness of the fluorine-fluorine bond that has
to be broken, and the comparative strength of the
xenon-fluorine bond.
18.11
18.13 The double-bonded structure probably makes a majorcontribution 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 KrF22 (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)
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XeOF4(l) + 2 H2O(l) XeO3(s) + 4 HF(l)
XeO3(s) + OH(aq) HXeO4
(aq)
2 HXeO4(aq) + 2 OH
(aq) XeO6
4(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)
Beyond the Basics
18.23 XeF2(SbF5) + SbF5(l) XeF+(SbF5) + SbF6
(SbF5)
18.25 The ArF bond energy = 77.5 kJmol-1.
Chapter 19
Exercises
19.1 (a) Element belonging to the d-block.
(b) Molecules or ions covalently bonded to a centralmetal ion.
(c) Energy separation between different members ofthe metals d-orbital set.
19.3 The cyanide ligand stabilizes low oxidation states and
stabilizes normal ones.
19.5 [Pt(NH3)4]2+[PtCl4]
2
19.7 The geometric isomers are:
There are two optical (chiral) isomers.
19.9 (a) Ammonium pentachlorocuprate(II); (b)pentaammineaquacobalt(III) bromide; (c) potassium
tetracarbonylchromate(-III); (d) potassium
hexafluoronickelate(IV); (e) tetraamminecopper(II)
perchlorate.
19.11 (a) [Mn(OH2)6](NO3)2, (b) Pd[PdF6], (c)
[CrCl2(OH2)4]Cl2 H2O, (d) K3[Mo(CN)8].
19.13 (a) The d6configuration in an octahedral field:
(b) The d6configuration in a tetrahedral field:
19.15 The largest value of 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: d0, 0.0 tet, ascending to d2,
1.2 tet, descending to d5, 0.0 tet, repeating to
d10, 0.0 tet.
19.21 Normal spinel, because the Cr3+ion will have a
greater CFSE than that of the Ni2+ion.
19.23 [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
M3+(aq)
(b) pH = 3.38
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19.29 For zinc, with its filled d10orbitals, there is no CFSE.
For nickel, a square-planar geometry will maximize
CFSE and it will enable some degree of bonding tooccur.
19.31 (a) [Cr(OH2)6]3+
3Cl, hexaaquachromium(III)
chloride;(b) [Cr(OH2)5Cl]2+2Cl
,
pentaaquachlorochromium(III) chloride;
(c) [Cr(OH2)4Cl2]+Cl,
tetraaquadichlorochromium(III) chloride.
19.33 Fluoride is a weaker field ligand than chloride.
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 For the earlier part of the Period 4 elements, themaximum oxidation number is the same as the group
number. For the later members, the oxidation state of
+2 predominates.
20.5 Titanium(IV) chloride vaporizes readily.
20.7 (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)
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 awater molecule.
20.23 According to Fajans Rules, cations with non-noble-gas 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, Al2Cl6and Fe2Cl6. On the other hand,iron(III) oxide is basic, while the oxide of aluminum
is amphoteric.
20.29 Titanium:
TiO2(s) + 2 C(s) + 2 Cl2(g) TiCl4(g) + 2
CO(g)
TiCl4(g) + O2(g) TiO2(s) + 2 Cl2(g)
TiCl4(g) + 2 Mg(l) Ti(s) + 2 MgCl2(l)
Vanadium:
[H2VO4](aq) + 4 H+(aq) + eVO2+(aq) + 3
H2O(l)
VO2+(aq) + 2 H+(aq) + eV3+(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) Cr2O7
2(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 e2 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 e2 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) + eMnO4
2(aq)
MnO42
(aq) + 2 H2O(l) + 2 eMnO2(s) + 4
OH(aq)
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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) + 6H2O(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) + Fe
3+(aq) 2 Fe
2+(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) + eFe2+(aq) + 2 eFe(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)4
2(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 e4 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) + 6H2O(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)
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 cyanideion.
20.37 (a) The high-charge cation (Fe3+) will have asomewhat low lattice energy when
combined with a low-charge anion (ClO4).
(b) Either: Ammonia and water are quite high inthe 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.
20.39 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 This corresponds to a full neutron shell.
20.47 3+, as the shared oxygen would have an oxidation
state of 2. The linear shape suggests there is a -
bonding CrOCr 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, theabsorption of light will be more and more in the
visible part of the spectrum.
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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 -bonding through the dorbitals.
20.57 (a) Fe(s) + O2(g) Fe2O3(s)(b) Sodium silicate prevents the continuation of theoxidation.
(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) 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 5dfluorides, 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 40C, and
is very soluble in low-polarity, organic solvents.
21.9 Ruthenium, rhodium, palladium, osmium, iridium,and platinum.
21.11 The 3dtransition metals tend to have lower oxidationstates than those of the 4dand 5dseries. The smaller
3dions cannot accommodate as many ligands.
21.13 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.
21.15 End-on overlap of a pair of orbitals; diagonal
overlap of a pair of orbitals; and the side-to-side
overlap of a pair of orbitals.
21.17 PdF3has the formulation of: (Pd2+)[PdF6]
2.
21.18 It has a stable, water-soluble, species at near-neutralpH making it transportable by biological fluids.
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-
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 groupnumber.
21.26 In the complex shown, each iodide bridges threeniobium atoms. [Nb6I8]
3+
21.28 The Re3Cl9structure involves a central triangle of
rhenium atoms with bridging and terminal chlorineatoms in a polymeric structure.
Chapter 22
Exercises
22.1 (a) Zn(s) + Br2(l) ZnBr2(s)
(b) ZnCO3(s) ZnO(s) + CO2(g)
22.3 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 withboth acids and bases, and they are both strong Lewis
acids.
22.7 Cd(OH)2(s) + 2 eCd(s) + 2 OH(aq)
2 Ni(OH)2(s) + 2 OH(aq) 2 NiO(OH)(s) + 2
H2O(l) + 2 e
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, cadmiumplating has been discontinued.
22.11 Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g)
Zn(OH2)62+(aq) + 4 NH3(aq) Zn(NH3)42+(aq) + 6H2O(l)
Zn2+
(aq) + 2 OH(aq) Zn(OH)2(s)
Zn(OH)2(s) + 2 OH(aq) Zn(OH)4
2(aq)
Zn(OH)2(s) ZnO(s) + H2O(l)
ZnO(s) + 2 H+(aq) Zn2+(aq) + H2O(l)
ZnCO3(s) ZnO(s) + CO2(g)
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Beyond the Basics
22.13 Mercury(I) undergoes a disproportionation
equilibrium.
22.15 The metals are very different in size.
22.17 Sulfur. Mercury(II) is a soft acid. Sulfur is a soft
base.
22.19 (a) Zn(NH2)2(NH3) + 2 NH4+(NH3)
Zn(NH3)42+(NH3)
(b) Zn(NH2)2(NH3) + 2 NH2(NH3) Zn(NH2)4
2
(NH3)
22.21 Zinc oxide.
22.23 Hydrogen sulfide is in a two-step equilibrium withthe sulfide ion. When acidified, the increased
hydronium-ion concentration will drive the
equilibria to the left.
Chapter 23
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
23.3 (a) Bi(CH3)5
(b) Si(C6H5)4 tetraphenyl silane
(c) KB(C6H5)4postassium teraphenylborane(d) Li4(CH3)4(e) (C2H5)MgCl
23.5 C2H5MgBr will be tetrahedral with two molecules of
solvent coordinated to the magnesium.
23.7 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 orbis(pentahaptocyclopentadienyl)iron(II)
(c)hexahaptobenzenetricarbonylchromium(0)
(d)pentahaptocyclopentadienyltricarbonyltungsten(I)
(e)bromopentacarbonylmanganese(I)
23.13 Cr(CO)6
Fe(CO)5
Ni(CO)4
Mn
COCO
COOC
OC Mn
COCO
COOC
CO
23.15 V(CO)6is a seventeen electron complex.
23.17 (a) 1 MnMn bond
(b) 2 MnMn bonds
Mn Mn
CO
OCCO
OC
(c) 1 FeFe bond
Fe Fe
COCOOCCO
CO
(d) no Mn-Mn bonds
Mn Mn
Br
Br
CO
CO
COCO
OC
OC
OCOC
23.19 (a) [Cr(CO)6] + 3 CH3CN [Cr(CO)3(CH3CN)3] + 3 CO
(b) [Mn2(CO)10] + H22 [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
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(e) 23.29
(f)
(g)
[PtCl2(PMe3)2] + LiCH2CH2CH2CH2Li 2 LiCl+ (PMe3)2Pt
(h) [Ni(CO)4]+ PF3 [Ni(CO)3PF3] + CO
(i)
[Mn2(CO)10] + Br22 [Mn(CO)5Br]
(j) [HMn(CO)5] + CO2[(CO)5MnCOOH]
23.21 (a) +3
(b)+1
Beyond the Basics
23.23
(5-C5H5)2Ni + Ni(CO)4 Ni Ni
OC
CO
2
23.25
A = tricarbonyl(5-cyclopentadienyl)(1-propenyl)tungsten(II)
B = dicarbonyl(5-cyclopentadienyl)(3-propenyl)tungsten(II)
C = tricarbonyl(5-cyclopentadienyl)(2-propenyl)tungsten(II)
hexafluorophosphate
Evolved gas = propene
23.27 Ti(S2CEt2)4.
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