Answers to Odd Numbered Questions

Descriptive Inorganic Chemistry, Fifth Edition Answers to Odd-Numbered Questions 1

2010 © W. H. Freeman and Company, All Rights Reserved

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]6s24 f145d106p1. 1.17 1+. Configuration of [Kr]5s14d10. 1.19

1.21 E113: [Rn]7s 25 f 1 46d 1 07p 1 .

E113 as +1 ion: [Rn]7s 25 f 1 46d 1 0 . E113 as +3 ion: [Rn]5 f 1 46d 1 0 .

Beyond the Basics 1.23 9, 5, 121. 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.

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. 2.7 The -ium ending 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 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.

2 Answers to Odd-Numbered Questions Descriptive Inorganic Chemistry, Fifth Edition


2.23 Using Slater’s rules.

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. 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.

3.9

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



3.17 (a) tetrahedral, V-shaped; (b) tetrahedral, trigonal

pyramidal; (c) trigonal bipyramidal, linear; (d) octahedral, square planar.

3.19 Linear: carbon disulfide and xenon difluoride; V-shaped: chlorine dioxide (< 109º), 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 3d 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. 3.29 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.

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.

3.43 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.

3.45 The two feasible formal charge arrangements for the CON− ion have five formal charges!

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.



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. 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 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; the

charge is more concentrated, and the ionic attraction will be stronger.

5.9 Ag+, because it has the lowest charge density.

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

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



5.43 LiAt – wurtzite or sphalerite packing; NaAt and KAt, sodium chloride packing; and RbAt and CsAt, cesium chloride packing.

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 ΔGº = −159 kJ·mol−1

6.9 N=N. 6.11 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 6.19

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

charges. The mathematical formula is known as Coulomb’s law.

6.27

6.29 Magnesium oxide will have a higher lattice energy. 6.31 ΔHf° = −913 kJ·mol−1; tabulated value is −933

kJ·mol−1.

6.33 250 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



For barium sulfate: −1044 kJ·mol−1 Yes.

6.39

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.

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.

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 H−Se bond will be weaker than the H−S 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, A− must be the stronger base.

Thus HB will be the stronger acid. 7.29 H3PO4(aq) + HPO4

2−(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).



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 ΔG° = −28 kJ·mol−1. Magnesium oxide will be a

weaker base than calcium oxide (ΔG° = −59 kJ·mol−1).

7.43 NO+ is a Lewis acid. Cl− is a Lewis base.

(NO)(AlCl4)(NOCl) + [(CH3)4N]Cl(NOCl) → [(CH3)4N](AlCl4)(NOCl) + NOCl(l)

7.45 (a) [NH4

+] = 3 × 10−17 mol·L−1 (b) [NH4

+] = 1×10−33 mol·L−1 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 of

both hard and soft bases. 7.59 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.

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−. 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+, 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.



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 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 Eº will increase to the point where insoluble, brown

manganese(III) oxide will be formed, thus discoloring 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 the element to its lower right in the periodic table.

9.3 The general formula: M+M3+(SO4

2−)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.

9.9 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) C≡O; (b) (C≡C)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 CN2

2− 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 10−6 of 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) 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

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 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.

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 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.

Hydrogen Chlorine Bond energy 432 kJ·mol−1 240 kJ·mol−1 Electron affinity −79 kJ·mol−1 −349 kJ·mol−1



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 metal ion range; all its common salts are colorless.

11.19 Potassium dioxide(1−) has a lower molar mass, and is

cheaper. 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 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.

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 the calcium carbonate to give a solution of calcium hydrogen carbonate.



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 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 12.25 ΔH° = +83 kJ·mol−1

ΔS° = 0.220 kJ·mol−1·K−1 T = 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 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 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. 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].



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)6

3+(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

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+][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 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 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. ΔH° = +624 kJ·mol−1

ΔS° = +0.354 kJ·mol−1·K−1 ΔG° = −181 kJ·mol−1

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.



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.

14.33

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 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) HCOOH(l) CO(g) + H2O(l) ⎯⎯ →⎯ 42SOH

CO2(g) + 2 Ca(s) → C(s) + 2 CaO(s) 2 CO(g) + O2(g) → 2 CO2(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) → SiF6

2−(aq) + 2 H+(aq) + 2 H2O(l)



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) ⎯→⎯Δ

2 SiO44−(aq) + 2 H+(aq) → Si2O7

6−(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) P3O9

3− (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

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 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:

Net energy change = 53 kJ . 15.29



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.

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; 120º; bond order would be 1.33 in

the 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 P4O10 are replaced by sulfur atoms.

15.49 Bonding between sodium and azide ions is likely to

be predominantly ionic whereas that in the heavy metal azides will be more covalent.

15.51 (a) K = 6 × 102 (b) K = 7×10−3

(c) Equilibrium is attained much more rapidly. 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.

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) + 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 AsO4

2−(aq) + 4 H+(aq) → 2 AsH3(g) + 8 H2O(l)

2 AsH3(g) 2 As(s) + 3 H2(g) ⎯→⎯Δ



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 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:

16.19 The structure is probably based on the S8 ring.

16.21

16.23 Ca(OH)2(s) + CO2(g) → CaCO3(s) + 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)

(b) Ba2+(aq) + SO42−(aq) → BaSO4(s) 16.37 There is a very high activation energy barrier to the

reaction SO2 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.



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) 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 of sulfur cannot exceed 6.

16.59 SO3

2−(aq) + S2O82−(aq) + H2O(l) → 3 SO4

2−(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 S2O3

2−(aq) + I2(aq) → S4O62−(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 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) (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; 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 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.



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)

(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 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.

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) → SiF6

2−(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 S2O3

2−(aq) → 2 I−(aq) + S4O62−(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 be

decomposing.


and Company, All Rights Reserved 2010 © W. H. Freeman

17.47

17.49 Dichlorine heptaoxide. It is the oxide in the higher

oxidation state. 17.51 The bond angles will be approximately 109½º.

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 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.

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 ½. 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 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)


mpany, All Rights Reserved

2010 © W. H. Freeman and Co

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) XeO6

4−(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 Ar−F bond energy = 77.5 kJ·mol-1. Chapter 19

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 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]Cl·2 H2O, (d) K3[Mo(CN)8].

19.13 (a) The d6 configuration in an octahedral field:

(b) The d6 configuration in a tetrahedral field:

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: d0, −0.0 Δtet, ascending to d2,

−1.2 Δtet, descending to d5, −0.0 Δtet, repeating to

d10, −0.0 Δtet.

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



19.29 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.

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, the

maximum 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 FeO4

2−(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 a

water molecule. 20.23 According to Fajan’s 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, Al2Cl6 and 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)

⎯→⎯Δ

TiCl4(g) + O2(g) TiO2(s) + 2 Cl2(g) ⎯→⎯Δ

TiCl4(g) + 2 Mg(l) Ti(s) + 2 MgCl2(l) ⎯→⎯Δ

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: CrO4

2−(aq) + 2 Ag+(aq) → Ag2CrO4(s) CrO4

2–(aq) + H2O(l) ↔ HCrO4–(aq) + OH–(aq)

2 CrO42−(aq) + 2 H+(aq) → Cr2O7

2−(aq) + H2O(l) Cr2O7

2−(aq) + 2 NH4+(aq) → (NH4)2Cr2O7(s)

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4 H2O(l) Cr2O7

2−(aq) + 14 H+(aq) + 6 e− → 2 Cr3+(aq) + 7 H2O(l) Cr2O7

2−(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) → CrO4

2−(aq) + 2 Cl−(aq) + 2 H2O(l) Cr2O7

2–(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)



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 S2O3

2−(aq) → [Fe(S2O3)2]−(aq) [Fe(S2O3)2]−(aq) + Fe3+(aq) → 2 Fe2+(aq) + S4O6

2−(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)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 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)

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 Cr2O7

2−(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) 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.

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 π-bonding 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.



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) 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 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 PdF3 has the formulation of: (Pd2+)[PdF6]2–. 21.18 It has a stable, water-soluble, species at near-neutral

pH 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 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

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 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

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, cadmium plating has been discontinued.

22.11 Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)

Zn(OH2)62+(aq) + 4 NH3(aq) → Zn(NH3)4

2+(aq) + 6 H2O(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)



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)4

2+(NH3) (b) Zn(NH2)2(NH3) + 2 NH2

–(NH3) → Zn(NH2)42–

(NH3) 22.21 Zinc oxide. 22.23 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.

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)4 postassium 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 or bis(pentahaptocyclopentadienyl)iron(II) (c) hexahaptobenzenetricarbonylchromium(0) (d) pentahaptocyclopentadienyltricarbonyltungsten(I) (e) bromopentacarbonylmanganese(I)

23.13 Cr(CO)6 Fe(CO)5 Ni(CO)4

MnCO

CO

COOC

OC Mn

COCO

COOC

CO

23.15 V(CO)6 is a seventeen electron complex. 23.17 (a) 1 Mn–Mn bond

(b) 2 Mn–Mn bonds

Mn Mn

CO

OC CO

OC

(c) 1 Fe–Fe bond

Fe Fe

COCOOC

COCO

(d) no Mn-Mn bonds

Mn Mn

Br

Br

COCO

COCO

OC

OC

OCOC

23.19 (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 Co

(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] + Br2 → 2 [Mn(CO)5Br] (j) [HMn(CO)5] + CO2 → [(CO)5MnCOOH]

23.21 (a) +3

(b) +1 Beyond the Basics 23.23

(η5-C5H5)2Ni + Ni(CO)4 Ni Ni

OC

CO

2

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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