Self Test Answers

Answers to self-tests and exercises

CHAPTER 1 Self-tests

S1.1 8035 Br + 1

0 n 8135 Br + γ

S1.2 d orbitals, 5 orbitals

S1.3 4

S1.4 3p

S1.5 The added p electron is in a different (p) orbital, so it is less shielded.

S1.6 8 2Ni :[Ar]3 4d s , 2 8Ni :[Ar]3d+

S1.7 Period 4, Group 2, s block

S1.8 Going down a group the atomic radius increases and the first ionization energy generally decreases.

S1.9 Group 16. The first four electrons are removed with gradually increasing values. Removing the fifth electron requires a large increase in energy, indicating breaking into a complete subshell.

S1.10 Adding another electron to C would result in the stable half filled p subshell.

S1.11 Cs+

Exercises 1.1 (a) 7

14 N+ 24He→ 8

17O+11p +γ

(b) 612C+1

1p→ 713N + γ

(c) 714 N+0

1n→13H+ 6

12C

1.2 96246Cm+ 6

12C→112257Uub+0

1n

1.3 The higher value of I2 for Cr relative to Mn is a consequence of the special stability of half-filled subshell configurations and the higher Zeff of a 3d electron verses a 4s electron.

1.4 22 4 25 110 2 12 0Ne He Mg n+ → +

1.5 49Be+4

9Be→ 612C+ 2

4He +201n

1.6 Since helium-4 is the basic building block, most additional fusion processes will produce nuclei with even atomic numbers.

1.7 Take the summation of the rest masses of all the nuclei of the products minus the masses of the nuclei of the reactants. If you get a negative number, energy will be released. But what you have calculated is the mass difference, which in the case of a nuclear reaction is converted to energy.

1.8 0.25

1.9 –13.2 eV

1.10 1524nm, 1.524 X 104 cm-1

1.11 1λ

= R1

12 −1

∞2⎛

⎝ ⎜

⎞

⎠ ⎟ = 1.0974 X107 m−1

1λ

= R1

12 −1

42⎛

⎝ ⎜

⎞

⎠ ⎟ = 1.0288X107 m−1

1λ

= R1

12 −1

32⎛

⎝ ⎜

⎞

⎠ ⎟ = 9.7547X106 m−1

1λ

= R1

12 −1

22⎛

⎝ ⎜

⎞

⎠ ⎟ = 8.2305X106 m−1

1.12 0 up to n-X

1.13 n2

1.14

N l ml Orbital designation

Number of orbitals

2 1 +1, 0, −1

2p 3

3 2 +2, +1, …, −2

3d 5

4 0 0 4s 1 4 3 +3, +2,

…, −3 4f 7

1.15 n=5, l = 3, and ml = -3,-2,-1,0,1,2,3

1.16 Li: σ = Z – Zeff ; σ = 3-1.28 = 1.72

Be: σ = Z – Zeff ; σ = 4-1.19 = 2.09

B: σ = Z – Zeff ; σ = 5-2.42 = 2.58

C: σ = Z – Zeff ; σ = 6-3.14 = 2.86

N: σ = Z – Zeff ; σ = 7-3.83 = 3.17

O: σ = Z – Zeff ; σ = 8-4.45 = 3.55

F: σ = Z – Zeff ; σ = 9-5.10 = 3.90

1.17 The 1s electrons shield the positive charge form the 2s electrons.

1

Shriver & Atkins: Inorganic Chemistry 5e

ANSWERS TO SELF-TESTS AND EXERCISES 2

1.18 See Figs 1.11 through 1.16

1.19 See Table 1.6 and discussion.

1.20 Table 1.6 shows SR > Ba < Ra. Ra is anomalous because of higher Zeff due to lanthanide contraction.

1.21 Anomalously high value for Cr is associated with the stability of a half filled d shell.

1.22 (a) [He]2s22p2

(b) [He]2s22p5

(c) [Ar]4s2

(d) [Ar]3d10

(e) [Xe]4f145d106s26p3

(f) [Xe]4f145d106s2

1.23 (a) [Ar]3d14s2

(b) [Ar]3d2

(c) [Ar]3d5

(d) [Ar]3d4

(e) [Ar]3d6

(f) [Ar]

(g) [Ar]3d104s1

(h) [Xe]4f 7

1.24 (a) Xe]4f145d46s2

(b) [Kr]4d6

(c) [Xe]4f6

(d) [Xe]4f7

(e) [Ar]

(f) [Kr]4d2

1.25 (a) S

(b) Sr

(c) V

(d) Tc

(e) In

(f) Sm

1.26 See Figure 1.4.

1.27 (a) I1 increases across the row except for a dip at S; (b) Ae tends to increase except for Mg (filled subshell), P (half filled subshell), and at AR (filled shell).

1.28 Radii of Period 4 and 5 d-metals are similar because of lanthanide contraction.

1.29 2s2 and 2p0

1.30


S2.1

S2.2 (a) Angular

(b) square planar

S2.3 Linear

S2.4 S22– : 1σg

22σu23σg

21πu42πg

4 ;

Cl2– : 1σg

22σu23σg

21πu42πg

44σu1.

S2.5 1σg22σu

23σg21πu

42πg4

S2.6 ½[2-2+4+2] = 3 S2.7 Bond order: C≡N, C=N, and C–N; Bond

strength: C≡N > C=N > C–N.

S2.8 If it contains 4 or fewer electrons.

S2.9 –21 kJ mol–1

S2.10 (a) +1/2

(b) +5

Exercises 2.1 (a) angular

(b) tetrahedral

(c) tetrahedral

2.2 (a) trigonal planar

(b) trigonal pyramidal

(c) square pyramidal

2.3 (a) T-shaped



(b) square planar

(c) linear

2.4 (a)

ICl Cl

Cl Cl

(b)

S

F

F

F

F

120 0

1800

2.5 (a) tetrahedral

(b) octahedral

2.6 (a) 176 pm

(b) 217 pm

(c) 221 pm

2.7 2(Si–O) = 932kJ > Si=O = 640kJ; therefore two Si–O are preferred and SiO2 should (and does) have four single Si–O bonds.

2.8 Multiple bonds are much stronger for period 2 elements than heavier elements

2.9 –483 kJ difference is smaller than expected because bond energies are not accurate. 2.10 (a) 0

(b) 205 kJ mol-1

2.11 Difference in electronegativities are AB 0.5, AD 2.5, BD 2.0, and AC 1.0. The increasing covalent character AD < BD < AC < AB.

2.12 (a) covalent

(b) ionic

(c) ionic

2.13 (a) sp2

(b) sp3

(c) sp3d or spd3

(d) p2d2 or sp2d.

2.14 (a) one

(b) one

(c) none

(d) two

2.15 (a) 1σg22σu

2

(b) 1σg22σu

21πu2

(c) 1σg22σu

21πu43σg

1

(d) 1σg22σu

23σg21πu

42πg3

2.16 The configuration for the neutral C2 would be 1σg

21σu2 1πu

4. The bond order would be ½[2-2+4] = 2.

2.17 (a) 1σg22σu

23σg21πu

42πg4

(b) 1

(c) There is no bond between the two atoms.

2.18 (a) 2

(b) 1

(c) 2

2.19 (a) +0.5

(b) –0.5

(c) +0.5

2.20

2.21 (a-c)

(d) Possibly stable in isolation (only bonding

and nonbonding orbitals are filled; not stable in solution because solvents would have higher proton affinity than He.

2.22 1

2.23 HOMO exclusively F; LUMO mainly S.

2.24 (a) electron deficient



(b) electron precise


S3.1 See Fig. 3.7 and Fig. 3.32. S3.2 See Figure 3.35. S3.3 52% S3.4 rh = ((3/2)1/2 – 1) r = 0.225 r

S3.5 409 pm S3.6 401 pm S3.7 FeCr3 S3.8 X2A3. S3.9 Ti CN = 6 (thought these are as two slightly

different distances it is often described as 4 + 2) and O CN = 3.

S3.10 LaInO3 S3.11 2421 kJ mol–1 S3.12 Unlikely

S3.13 MgSO4 < CaSO4 < SrSO4 < BaSO4.

S3.14 NaClO4 S3.15 Schottky defects. S3.16 Phosphorus and aluminium.

S3.17 The dx2-y2 and dz2 have lobes pointing along the cell edges to the nearest neighbor metals.

S3.18 (a) n-type

(b) p-type

Exercises 3.1 a ≠ b ≠ c and α = 90°, β = 90°, γ = 90° 3.2 Points on the cell corners at (0,0,0), (1,0,0),

(0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), and (1,1,1) and in the cell faces at (½,½,0), (½,1,½), (0,½,½) (½,½,1), (½,1,½), and (1,½,½).

3.3 (c) and (f) are not as they have neighbouring layers of the same position.

3.4 XA2 3.5 K3C60 3.6 281 pm 3.7 429 pm 3.8 CuAu. Primitive 12 carat. 3.9 Zintl phase region

3.10 (a) 6:6 and 8:8

(b) CsCI 3.11 6 3.12 2B type and 4A type – in a distorted

octahedral arrangement. 3.13 (a) ρ = 0.78, so fluorite

(b) FrI? ρ = 0.94, so CsCl (c) BeO? ρ = 0.19, so ZnS (d) InN? ρ = 0.46, so NaCl

3.14 CsCl. 3.15 Lattice enthalpies for the di- and the trivalent

ions. Also, the bond energy and third electron gain enthalpy for nitrogen will be large.

3.16 4 four times the NaCl value or 3144 kJmol–1.

3.17 (a) 10906 kJmol–1

(b) 1888 kJ mol–1

(c) 664 kJ mol–1 3.18 (a) MgSO4

(b) NaBF4

3.19 CsI < RbCl < LiF < CaO < NiO < AlN 3.20 Ba2+; solubilities decrease with increasing

radius of the cation.

3.21 (a) Schottky defects

(b) Frenkel defects

3.22 Solids have a greater number of defects as temperatures approaches their melting points.

3.23 The origin of the blue color involves electron

transfer from cationic centres. 3.24 Vanandium carbide and manganese oxide.



3.25 Yes. 3.26 A semiconductor is a substance with an

electrical conductivity that decreases with increasing temperature. It has a small, measurable band gap. A semimetal is a solid whose band structure has a zero density of states and no measurable band gap.

3.27 Ag2S and CuBr: p-type; VO2: n-type.


S4.1 (a) HNO3 + H2O → H3O+ + NO3–

HNO3, acid. Nitrate ion, conjugate base. H2O, base. H3O+, conjugate acid.

(b) CO32– + H2O → HCO3

– + OH–

carbonate ion, base; hydrogen carbonate, or bicarbonate, conjugate acid; H2O, acid; hydroxide ion, conjugate base.

(c) NH3 + H2S → NH4+ + HS–

Ammonia, base; NH4+, conjugate acid;

hydrogen sulphide, acid; HS–, conjugate base.

S4.2 What is the pH of a 0.10 M HF solution? pH= 2.24

S4.3 Calculate the pH of a 0.20 M tartaric acid solution?

pH=1.85

S4.4 Which solvent?

dimethylsulfoxide (DMSO) and ammonia.

S4.5 Is aKBrF4 an acid or a base in BrF3?

A base.

S4.6 Arrange in order of increasing acidity?

The order of increasing acidity is [Na(H2O)6]+ < [Mn(H2O)6]2+ < [Ni(H2O)6]2+ < [Sc(H2O)6]3+.

S4.7 Predict pKa values? (a) H3PO4 pKa ≈ 3. The actual value, given in Table 4.1, is 2.1.

(b) H2PO4– pKa(2) ≈ 8. The actual value,

given in Table 4.1, is 7.4.

(c) HPO42– pKa(3) ≈ 13. The actual value,

given in Table 4.1, is 12.7.

S4.8 What happens to Ti(IV) in aqueous solution as the pH is raised?

Treatment with ammonia causes the precipitation of TiO2. Further treatment with NaOH causes the TiO2 to redissolve.

S4.9 Identify the acids and bases?

(a) FeCl3 + Cl– → [FeCl4]–, acid is FeCl3, base is Cl–.

(b) I– + I2 → I3–, acid is I2, base is I–.

S4.10 The difference in structure between (H3Si)3N and (H3C)3N? The N atom of (H3Si)3N is trigonal planar, whereas the N atom of (H3C)3N is trigonal pyramidal.

S4.11 Draw the structure of BF3·OEt2?

Exercises 4.1 Sketch an outline of the s and p blocks of

the periodic table, showing the elements that form acidic, basic, and amphoteric oxides?

The elements that form basic oxides are in plain type, those forming acidic oxides are in outline type, and those forming amphoteric oxides are in boldface type.

4.2 Identify the conjugate bases of the following acids?

(a) [Co(NH3)5(OH2)]3+, conjugate base is [Co(NH3)5(OH)]2+.

(b) HSO4–? The conjugate base is SO4

–.

(c) CH3OH? The conjugate base is CH3O–.

(d) H2PO4–? The conjugate base is HPO4

2–.

(e) Si(OH)4? The conjugate base is SiO(OH)3

–.

(f) HS−? The conjugate base is S2–.

4.3 Identify the conjugate acids of the following bases?



(a) C5H5N (pyridine)? The conjugate acid is pyridinium ion, C5H6N+.

(b) HPO42–? The conjugate acid is H2PO4

2–.

(c) O2–? The conjugate acid is OH–.

(d) CH3COOH? The conjugate acid is CH3C(OH)2

+.

(e) [Co(CO)4]–? The conjugate acid is HCo(CO)4.

(f) CN–? The conjugate acid is HCN.

4.4 Calculate the [H3O+] and pH of a 0.10 M butanoic acid solution? pH=2.85

4.5 What is the Kb of ethanoic acid?

Kb = 5.6 × 10–10

4.6 What is the Ka for C5H5NH+?

Ka = 5.6 × 10–6

4.7 Predict if F- will behave as an acid or a base in water?

F- will behave as a base in water.

4.8 What are the structures and the pKa values of chloric (HClO3) and chlorous (HClO2) acid?

Cl

O

O OH Cl

O

OH

chloric acid chlorous acid Chloric acid, the predicted pKa = –2; actual

value = –1.

Chlorous acid, the predicted pKa = 3; actual value = 2.

4.8 Which bases are too strong or too weak to be studied experimentally? (a) CO3

2– O2–, ClO4

–, and NO3– in water?

CO32– is of directly measurable base strength.

O2–, is too strong to be studied experimentally in water.

ClO42– and NO3

– are too weak to be studied experimentally.

(b) HSO4–, NO3

–, and ClO4–, in H2SO4?

HSO4–, not too strong to be studied

experimentally.

NO3– is of directly measurable base strength

in liquid H2SO4.

ClO4–, cannot be studied in sulfuric acid.

4.9 Is the –CN group electron donating or withdrawing?

electron withdrawing

4.11 Is the pKa for HAsO42– consistent with

Pauling’s rules? No. Pauling’s rules are only approximate.

4.12 What is the order of increasing acid strength for HNO2, H2SO4, HBrO3, and HClO4?

the order is HClO4 > HBrO3 > H2SO4 > HNO2.

4.13 Account for the trends in the pKa values of the conjugate acids of SiO4

4–, PO43–, SO4

2–, and ClO4

–?

The acidity of the four conjugate acids increases in the order HSiO4

3– < HPO42– <

HSO4– < HClO4.

4.14 Which of the following is the stronger acid? (a) [Fe(OH2)6]3+ or [Fe(OH2)6]2+? The Fe(III) complex, [Fe(OH2)6]3+, is the stronger acid.

(b) [Al(OH2)6]3+ or [Ga(OH2)6]3+?

the aluminum-containing species is more acidic.

(c) Si(OH)4 or Ge(OH)4?

Si(OH)4, is more acidic.

(d) HClO3 or HClO4?

HClO4 is a stronger acid.

(e) H2CrO4 or HMnO4?

HMnO4 is the stronger acid.

(f) H3PO4 or H2SO4?

H2SO4 is a stronger acid.

4.15 Arrange the following oxides in order of increasing basicity? Al2O3, B2O3, BaO, CO2, Cl2O7, and SO3?

order of increasing basicity is Cl2O7 < SO3 < CO2 < B2O3 < Al2O3 < BaO.

4.16 Arrange the following in order of increasing acidity? HSO4

−, H3O+, H4SiO4, CH3GeH3, NH3, and HSO3F?

increasing acidity is NH3 < CH3GeH3 < H4SiO4 < HSO4

– < H3O+ < HSO3F.

4.17 Which aqua ion is the stronger acid, Na+ or Ag+? Ag+(aq).



4.18 Which of the following elements form oxide polyanions or polycations? Al, As, Cu, Mo, Si, B, Ti?

polycations: Al, Cu, and Ti.

polyoxoanions (oxide polyanions): Mo

polyoxoanions: As, B, and Si.

4.19 The change in charge upon aqua ion polymerization?

Polycation formation reduces the average positive charge per central M atom by +1 per M.

4.20 Write balanced equations for the formation of P4O12

4– from PO43– and for the formation

of [(H2O) 4Fe(OH)2Fe(OH2)4]4+ from [Fe(OH2)6]3+?

4PO43– + 8H3O+ → P4O12

4– + 12H2O

2[Fe(OH2)6]3+ →

[(H2O)4Fe(OH)2Fe(OH2)4+ + 2H3O+

4.21 More balanced equations?

(a) H3PO4 and Na2HPO4?

H3PO4 + HPO42- 2H2PO4

-

(b) CO2 and CaCO3?

CO2 + CaCO3 + H2O Ca2+ + 2HCO3-

4.22 Give the equations for HF in H2SO4 and HF in liquid NH3?

H2SO4 + HF ⇔ H3SO4+ + F-

NH3 + HF NH2- + H2F+

4.23 Why is H2Se a stronger acid than H2S? As you go down a family in the periodic chart, the acidy of the homologous hydrogen compounds increases.

4.24 Identifying elements that form Lewis acids? All of the p-block elements except nitrogen, oxygen, fluorine, and the lighter noble gases form Lewis acids in one of their oxidation states.

4.25 Identifying acids and bases: (a) SO3 + H2O → HSO4

– + H+? The acids in this reaction are the Lewis acids SO3 and H+ and the base is the Lewis base OH–.

(b) Me[B12]– + Hg2+ → [B12] + MeHg+? The Lewis acid Hg2+ displaces the Lewis acid [B12] from the Lewis base CH3

–.

(c) KCl + SnCl2 → K+ + [SnCl3]–? The Lewis acid SnCl2 displaces the Lewis acid K+ from the Lewis base Cl–.

(d) AsF3(g) + SbF5(g) → [AsF2][SbF6]? The very strong Lewis acid SbF5 displaces the Lewis acid [AsF2]+ from the Lewis base F–.

(e) EtOH readily dissolves in pyridine? A Lewis acid–base complex formation reaction between EtOH (the acid) and py (the base) produces the adduct EtOH–py.

4.26 Select the compound with the named characteristic? (a) Strongest Lewis acid: BF3, BCl3, or BBr3? BBr3.

BeCl2 or BCl3? Boron trichloride.

B(n-Bu)3 or B(t-Bu)3? B(n-Bu)3.

(b) More basic toward BMe3: NMe3 or NEt3? NMe3.

2-Me-py or 4-Me-py? 4-Me-py.

4.27 Which of the following reactions have Keq > 1? (a) R3P–BBr3 + R3N–BF3→ R3P–BF3 + R3N–BBr3? <1

(b) SO2 + Ph3P–HOCMe3 → Ph3P–SO2 + HOCMe3? > 1.

(c) CH3HgI + HCl → CH3HgCl + HI? <1

(d) [AgCl2]–(aq) + 2 CN–(aq) → [Ag (CN)2]–

(aq) + 2Cl–(aq)? >1

4.28 Choose between the two basic sites in Me2NPF2?

The phosphorus atom in Me2NPF2 is the softer of the two basic sites, so it will bond more strongly with the softer Lewis acid BH3

The hard nitrogen atom will bond more strongly to the hard Lewis acid BF3.

4.29 Why is trimethylamine out of line? Trimethyl amine is sterically large enough to fall out of line with the given enthalpies of reaction.

4.30 Discuss relative basicities? (a) Acetone and DMSO?

DMSO is the stronger base regardless of how hard or how soft the Lewis acid is. The ambiguity for DMSO is that both the oxygen atom and sulfur atom are potential basic sites.

(b) Me2S and DMSO? Depending on the EA and CA values for the Lewis acid, either base could be stronger.

4.31 Write a balanced equation for the dissolution of SiO2 by HF?



SiO + 6HF2 2H O + H SiF

SiO2 + 4HF

2 2 6

or

2H2O + SiF4 Both a Brønsted acid–base reaction and a Lewis acid–base reaction.

4.32 Write a balanced equation to explain the foul odor of damp Al2S3? The foul odor suggests H2S formation.

Al2S3 + 3H2O Al2O3 + 3H2S 4.33 Describe solvent properties? (a) Favor

displacement of Cl– by I– from an acid center? If you choose a solvent that decreases the activity of chloride relative to iodide, you can shift the following equilibrium to the right:

acid-Cl- + I- acid-I- + Cl- (b) Favor basicity of R3As over R3N? Alcohols such as methanol or ethanol would be suitable.

(c) Favor acidity of Ag+ over Al3+? An example of a suitable solvent is diethyl ether. Another suitable solvent is H2O.

(d) Promote the reaction 2FeCl3 + ZnCl2 →Zn2+ + 2[FeCl4]−? A suitable solvent is acetonitrile, MeCN.

4.34 Propose a mechanism for the acylation of benzene? An alumina surface, such as the partially dehydroxylated one shown below, would provide Lewis acidic sites that could abstract Cl–:

4.35 Why does Hg(II) occur only as HgS? Mercury(II) is a soft Lewis acid, and so is found in nature only combined with soft Lewis bases, the most common of which is S2–.

4.36 Write Brønsted acid–base reactions in liquid HF?

(a) CH3CH2OH? The balanced equation is:

CH3CH2OH + HF CH3CH2OH2+ + F-

(b) NH3? The equation is

(c) C6H5COOH?

C6H5COOH + HF C6H5COO- + H2F+

4.37 The dissolution of silicates by HF? both

4.38 Are the f-block elements hard? yes.

4.39 Calculate the enthalpy change for I2 with phenol?

ΔfH

φ

= -20.0kJ/mol


S5.1 Half-reactions and balanced reaction for oxidation of zinc metal by permanganate ions?

2[MnO4− (aq) + 8H+ (aq) + 5e– → Mn2+(aq)

+ 4H2O(l)] reduction 5 [ Zn(s) → Zn2+(aq) + 2e– ] oxidation 2MnO4

− (aq) + 5Zn(s) + 16H+(aq) → 5Zn2+(aq) + 2Mn2+(aq) + 8H2O(l)

S5.2 Does Cu metal dissolve in dilute HCl? No. S5.3 Can Cr2O7

2– be used to oxidize Fe2+, and would Cl– oxidation be a problem? Yes. Cl- oxidation is not a problem.

S5.4 Fuel cell emf with oxygen and hydrogen gases at 5.0 bar?

E = 1.25 V.

S5.5 The fate of SO2 emitted into clouds? The aqueous solution of SO4

2– and H+ ions precipitates as acid rain.

S5.6 Can Fe2+ disproportionate under standard conditions? No.

S5.7 bpy binding to Fe(III) or Fe(II)? Fe(II) preferentially.

S5.8 Potential of AgCl/Ag,Cl– couple? Eox= – 1.38 V S5.9 Latimer diagram for Pu? (a) Pu(IV)

disproportionates to Pu(III) and Pu(V) in aqueous solution; (b) Pu(V) does not disproportionate into Pu(VI) and Pu(IV).

S5.10 Frost diagram for thallium in aqueous acid?

N H 4+ + F -N H 3 + H F



S5.11 The oxidation number of manganese?

Mn2+(aq)

S5.12 Compare the strength of NO3– as an

oxidizing agent in acidic and basic solution?

Nitrate is a stronger oxidizing agent in acidic solution than in basic solution.

S5.13 The possibility of finding Fe(OH)3 in a waterlogged soil? Fe(OH)3 is not stable.

S5.14 The minimum temperature for reduction of MgO by carbon? 1800ºC or above.

Exercises 5.1 Oxidation numbers?

2 NO(g) + O2(g) → 2 NO2(g) +1 -2 0 +4 -2 2Mn3+(aq) + 2H2O → MnO2 + Mn2+ + 4H+(aq) +3 +1 -2 +4 -2 +2 +1

LiCoO2(s) + C(s) → LiC(s) + CoO2(s) +1 +3 -2 0 +1-1 +4 -2 Ca(s) + H2(g) → CaH2(s)

0 0 +2 -1

5.2 Suggest chemical reagents for redox transformations? (a) Oxidation of HCl to Cl2? S2O8

2–, H2O2, or α–PbO2 to oxidize Cl– to Cl2.

(b) Reducing Cr3+(aq) to Cr2+(aq)? metallic manganese, metallic zinc, or NH3OH+.

(c) Reducing Ag+(aq) to Ag(s)? The reduced form of any couple with a reduction potential less than 0.799 V.

(d) Reducing I2 to I–? The reduced form of any couple with a reduction potential less than 0.535 V.

5.3 Write balanced equations, if a reaction occurs, for the following species in aerated aqueous acid? (a) Cr2+?

4Cr2+(aq) + O2(g) + 4H+(aq) → 4Cr3+(aq) + 2H2O(l) Eº = 1.65 V

(b) Fe2+?

4Fe2+(aq) + O2(g) + 4H+(aq) → 4Fe3+(aq) + 2H2O(l) Eº = 0.46 V

(c) Cl–? no reaction.

(d) HOCl? No reaction.

(e) Zn(s)?

2Zn(s) + O2(g) + 4H+(aq) → 2Zn2+(aq) +2H2O(l) Eº = 1.99 V

A competing reaction is:

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) (Eº = 0.763 V).

5.4 Balanced equations for redox reactions?

(a) Fe2+?

(i) Fe2+ will not oxidize water.

(ii) Fe2+ will not reduce water.

(iii) Fe2+ will reduce O2 and in doing so will be oxidized to Fe3+.

(iv) disproportionation will not occur.

(b) Ru2+? Ru2+ will not oxidize or reduce water. Ru2+ will reduce O2 and in doing so will be oxidized to Ru3+. Ru2+ will disproportionate in aqueous acid to Ru3+ and metallic ruthenium.

(c) HClO2? HClO2 will oxidize water, will not reduce water. HClO2 will reduce O2 and in doing so will be oxidized to ClO3

–. HClO2 will disproportionate in aqueous acid to ClO3

– and HClO.

(d) Br2? Br2 will not oxidize or reduce water. Br2 will not reduce O2. Br2 will not disproportionate in aqueous acid to Br– and HBrO.

5.5 Standard potentials vary with temperature in opposite directions? The amino and cyano complexes must have different equilibrium shifts with respect to changes in temperature that results in the opposite directions of change for the cell potential.

5.6 Balance redox reaction in acid solution: MnO4

– + H2SO3 → Mn2+ + HSO4–? pH

dependence?



2MnO4− (aq) + 5H2SO3(aq) + H+(aq) →

2Mn2+(aq) +5HSO3− (aq) + 3H2O(l)

The potential decreases as the pH increases.

5.7 Write the Nernst equation for (a) The reduction of O2?

Q = 1/(p(O2)[H+]4)

and

E = Eº – [(0.059V)/4][log(1/(p(O2)[H+]4)]

(b) The reduction of Fe2O3(s)?

Q = 1/[H+]6 and E = Eº – (RT/nF)(13.8 pH)

5.8 Using Frost diagrams? (a) What happens when Cl2 is dissolved in aqueous basic solution? Cl2 is thermodynamically susceptible to disproportionation to Cl– and ClO4

– when it is dissolved in aqueous base.

The oxidation of ClO– is slow, so a solution of Cl– and ClO– is formed when Cl2 is dissolved in aqueous base.

(b) What happens when Cl2 is dissolved in aqueous acid solution? Cl2 will not disproportionate. Cl2 is thermodynamically capable of oxidizing water.

(c) Should HClO3 disproportionate in aqueous acid solution? Kinetic.

5.9 Write equations for the following reactions: (a) N2O is bubbled into aqueous NaOH solution?

5N2O(aq) + 2OH–(aq) → 2NO3–(aq) +

4N2 (g) + H2O(l)

(b) Zinc metal is added to aqueous acidic sodium triiodide?

Zn(s) + I3– (aq) → Zn2+(aq) + 3I–(aq)

(c) I2 is added to excess aqueous acidic HClO3?

3I2(s) + 5ClO3– (aq) + 3H2O(l) → 6IO3

–

(aq) + 5Cl– (aq) + 6H+(aq)

5.10 Electrode potential for Ni2+/Ni couple at pH = 14? E =– 0.21 V

5.11 Will acid or base most favour the following half-reactions? (a) Mn2+ → MnO4

–? Base

(b) ClO4– → ClO3

–? Acid

(c) H2O2 → O2? Base

(d) I2 → 2I–? Acid or base, no difference.

5.12 Determine the standard potential for the reduction of ClO4

– to Cl2? 1.392 V

5.13 Calculate the equilibrium constant for Au+(aq) + 2CN–(aq) → [Au(CN)2]–(aq)? K = 5.7 × 1038

5.14 Find the approximate potential of an aerated lake at pH = 6, and predict the predominant species? (a) Fe? 0.5 – 0.6 V

(b) Mn? E = 0.55 V

(c) S? At pH 0, 0.387 V. At pH 14, SO42–

would again predominate. HSO4– is the

predominant sulfur species at pH 6.

5.15 Frost diagram and standard potential for the HSO4

−/S8(s) couple? 0.387 V

5.16 Equilibrium constant for the reaction Pd2+(aq) + 4 Cl–(aq) ≡ [PdCl4]2–(aq) in 1 M HCl(aq)? K = 4.37 × 1010

5.17 Reduction potential for MnO4– to MnO2(s)

at pH = 9.00? E = 0.98 V 5.18 Tendency of mercury species to act as an

oxidizing agent, a reducing agent, or to undergo disproportionation? Hg2+ and Hg2

2+ are both oxidizing agents. None of these species are likely to be good reducing agents. Hg2

2+ is not likely to undergo disproportionation.

5.19 Thermodynamic tendency of HO2 to undergo disproportionation? E = +1.275 V. (is positive), HO2 will undergo disproportionation.

5.20 Dissolved carbon dioxide corrosive towards iron? Carbon dioxide and water generate carbonic acid which encourages the corrosion process by lowering solution pH.

5.21 What is the maximum E for an anaerobic environment rich in Fe2+ and H2S? –0.1 V.

5.22 How will edta4– complexation affect M2+ → M0 reductions? The reduction of a M(edta)2– complex will be more difficult than the reduction of the analogous M2+ aqua ion.

5.23 Which of the boundaries depend on the choice of [Fe2+]? Any boundary between a soluble species and an insoluble species will change as the concentration of the soluble species changes. The boundaries between the two soluble species, and between the two insoluble species, will not depend on the choice of [Fe2+].

5.24 Under what conditions will Al reduce MgO? Above about 1400ºC.




S6.1 Sketch the S4 axis of an NH4+ ion. How

many of these axes are there in the ion? Three S4 axes.

S6.2 (a) BF3 point group? D3h.

(b) SO42– point group? Td.

S6.3 Symmetry species of all five d orbitals of the central Xe atom in XeF4 (D4h, Fig. 6.3)?

dx2-y2 is B1g; dxy is B2g; dxz and dyz are Eg; dz2 is A1g. S6.4 What is the maximum possible degeneracy

for an Oh molecule? 3.

S6.5 A conformation of the ferrocene molecule that lies 4 kJ mol–1 above the lowest energy configuration is a pentagonal antiprism. Is it polar? No.

S6.6 Is the skew form of H2O2 chiral? Yes.

S6.7 Can the bending mode of N2O be Raman active? Yes.

S6.8 Confirm that the symmetric mode is Ag? D2h character table, which is the Ag symmetry type.

S6.9 Show that the four CO displacements in the square-planar (D4h) [Pt(CO)4]2+ cation transform as A1g + B1g + Eu. How many bands would you expect in the IR and Raman spectra for the [Pt(CO)4]2+ cation?

The reducible representation: D4h E 2C4 C2

2C2’

2C2″

i 2S4

σh 2σv 2σd

Γ3N 4 0 0 2 0 0 0 4 2 0

Reduces to A1g + B1g + Eu A1g + B1g are Raman active. Eu is IR active. S6.10 Orbital symmetry for a tetrahedral array

of H atoms in methane? A1

S6.11 Orbital symmetry for a square-planar array of H atoms? B2g.

S6.12 Which Pt atomic orbitals can combine with which of these SALCs? The atomic orbitals much have matching symmetries to generate SALCs. 5s and 4dz2 have A1g symmetry; the dx2-y2 has B1g symmetry; and 5px and 5py have Eu symmetry.

S6.13 Predict how the IR and Raman spectra of

SF5Cl differ from that of SF6?

SF6 has Oh symmetry. Analysis of the stretching vibrations leads to:

Γstr = A1g (Raman, polarized) + Eg

(Raman) + T1u (IR).

SF5Cl has C4v symmetry. Analysis of the

stretching vibrations leads to:

Γstr = 3A1 (IR and Raman, polarized)

+ 2B1 (Raman) + E (IR, Raman).

S6.14 Symmetries of all the vibration modes of [PdCl4]2-? A1g + B1g + B2g + A2u + B2u + 2Eu

S6.15 SALCs for sigma bonding in O? A1g + Eg +T1u.

Exercises 6.1 Symmetry elements? (a) a C3 axis and a σv

plane in the NH3 molecule?

NH H

HHN

HH

C3 σv (b) a C4 axis and a σh plane in the square-

planar [PtCl4]2– ion?

PtCl

Cl Cl

ClPt

Cl

Cl Cl

Cl

C4 σh 6.2 S4 or i? (a) CO2? i

(b) C2H2? i.

(c) BF3? neither.

(d) SO42–? three different S4.

6.3 Assigning point groups: (a) NH2Cl? Cs

(b) CO32–? D3h

(c) SiF4? Td

(d) HCN? C∞v.

(e) SiFClBrI? C1.

(f) BrF4–? D4h.



6.4 How many planes of symmetry does a benzene molecule possess? What chloro-substituted benzene has exactly four planes of symmetry? 7,and C6H3Cl3.

6.5 The symmetry elements of orbitals? (a) An s orbital? Infinite number of Cn axes, plus an infinite number of mirror planes of symmetry, plus center of inversion, i.

(b) A p orbital? An infinite number of mirror planes that pass through both lobes and include the long axis of the orbital. In addition, the long axis is a Cn axis, where n can be any number from 1 to ∞.

(c) A dxy orbital? Center of symmetry, three mutually perpendicular C2 axes, three mutually perpendicular mirror planes of symmetry, two planes that are rotated by 45º about the z axis from the xz plane and the yz plane.

(d) A dz2 orbital? In addition to the symmetry elements possessed by a p orbital: (i) a center of symmetry, (ii) a mirror plane that is perpendicular to the C∞ axis, (iii) an infinite number of C2 axes that pass through the center of the orbital and are perpendicular to the C∞ axis, and (iv) an S∞ axis.

6.6 SO32– ion? (a) Point group? C3v

(b) Degenerate MOs? 2

(c) Which s and p orbitals have the maximum degeneracy? 3px and 3py orbitals are doubly degenerate.

6.7 PF5? (a) Point group? D3h.

(b) Degenerate MOs? 2.

(c) Which p orbitals have the maximum degeneracy? 3px and 3py atomic orbitals are doubly degenerate

6.8 AsCl5 Raman spectrum consistent with a

trigonal bipyamidal geometry? No.

6.9 Vibrational modes of SO3? (a) In the plane of the nuclei? 5

(b) Perpendicular to the molecular plane? 1

6.10 Vibrations that are IR and Raman active? (a) SF6? None.

(b) BF3? The E′ modes are active in both IR and Raman.

6.11 Vibrations of a C6v molecule that are neither IR nor Raman active? Any A2, B1, or B2 vibrations of a C6v molecule will not be observed in either the IR spectrum or the Raman spectrum.

6.12 [AuCl4]− ion? Γ of all 3N displacements and irreducible representations? A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu

6.13 IR and Raman to distinguish between: (a)

planar and pyramidal forms of PF3, (b) planar and 90o-twisted forms of B2F4 (D2h and D2d respectively)?

(a) Planar PF3, D3h, vibrations are: A1’ (Raman, polarized) + 2E’ (IR and Raman) + A2” (IR). Pyramidal PF3, C3v, vibrations are: 2A1 (IR and Raman, polarized) + 2E’ (IR and Raman)

(b) For the planar form of B2F4 (D2h):

The vibrations are: 3Ag (Raman, polarized) + 2B2g (Raman) + B3g (Raman) + Au(inactive) + 2B1u (IR) + B2u (IR) + 2B3u (IR). For the 90o-twisted form of B2F4 (D2d) The vibrations are: 3A1 (Raman, polarized) + B1 (Raman) + 2B2 (IR and Raman ) + 3E (IR and Raman).

6.14 (a) Take the 4 hydrogen 1s orbitals of CH4

and determine how they transform under Td. (b) Confirm that it is possible to reduce this representation to A1 + T2. (c) Which atomic orbitals on C can form MOs with H1s SALCs?

Using symmetry Td, Γ3N reduces to: A1 + T2. The MOs would be constructed from SALCs

with H1s and 2s and 2p atomic orbitals on C. 6.15 Use the projection operator method to

construct the SALCs of A1 + T2 symmetry that derive from the four H1s orbitals in methane..

s = (1/2)(ϕ1 + ϕ2 + ϕ3 + ϕ3) (= A1)

px = (1/2)(ϕ1 – ϕ2 + ϕ3 – ϕ3) (= T2)

py = (1/2)(ϕ1 – ϕ2 – ϕ3 + ϕ3) (= T2)

pz = (1/2)(ϕ1 + ϕ2 – ϕ3 – ϕ3) (= T2) SALCs for σ-bonds (a) BF3? (1/√3)(ϕ1 + ϕ2 + ϕ3) (= A1’) (1/√6)(2ϕ1 – ϕ2 – ϕ3) and (1/√2)(ϕ2 – ϕ3) (=

E’) (b) PF5? (axial F atoms are ϕ4 + ϕ5) (1/√2)(ϕ4 + ϕ5) (= A1’) (1/√2)(ϕ4 − ϕ5) (= A2”) (1/√3)(ϕ1 + ϕ2 + ϕ3) (= A1’)

(1/√6)(2ϕ1 – ϕ2 – ϕ3) and (1/√2)(ϕ2 – ϕ3) (= E’)




S7.1 Give formulas corresponding to the following names? (a) Cis-diaquadichloroplatinum(II)? cis-[PtCl2(OH2)2],

trans-diaquadichloroplatinum(II), trans-[PtCl2(OH2)2].

(b) Diamminetetra(isothiocyanato) chromate(III)? [Cr(NCS)4(NH3)2] –. ,can exist as, cis-[Cr(NCS)4(NH3)2] – or trans-[Cr(NCS)4(NH3)2] –.

(c) Tris(ethylenediamine)rhodium (III)? [Rh(en)3]3+.

(d) Bromopentacarbonylmanganese (I)? [MnBr (CO)5].

(e) Chlorotris(triphenylphosphine)rhodium (I)? [RhCl(PPh3)3].

S7.2 What type of isomers are possible for [Cr(NO2)2•6H2O]? The hydrate isomers and linkage isomers of the NO2 group. Also, [Cr(ONO)(H2O)5]NO2 •H2O.

S7.3 Identifying isomers? Note that the two phosphine ligands in the trans isomer are related, therefore, they exhibit the same chemical shift.

S7.4 Sketches of the mer and fac isomers of [Co(gly)3]?

S7.5 Which of the following are chiral? (a) cis-[CrCl2 (ox)2]3–? Chiral .

(b) trans-[CrCl2(ox)2]3–? Not chiral.

(c) cis-[RhH(CO)(PR3)2]? Not chiral.

S7.6 Calculate all of the stepwise formation constants? Kf1 = 1 X 105. Kf2 will be 30% less or 30000, Kf3 = 9000, Kf4 = 2700, Kf5 = 810, and finally Kf6 = 243.

Exercises 7.1 Name and draw the structures of the

complexes? (a) [Ni(CO)4]? Nickel tetracarbonyl or tetracarbonyl nickel(0).

CO

NiOC CO

CO

(b) [Ni(CN)4]2–? Tetracyanonickelate (II).

2-

Ni

NC

NC CN

CN

(c) [CoCl4]2–? Tetrachlorocobaltate (II)

2-

Cl

CoCl Cl

Cl

(d) [Mn(NH3)6]2+? Hexaamminemanganesium (II)

2+

MnH3N

H3N NH3

NH3

NH3

NH3

7.2 Write the formulas for the following complexes?

(a) [CoCl(NH3)5]Cl2

(b) [Fe(OH2)6](NO3)3

(c) cis-[FeCl2(en)2]

(d) [Cr(NH3)5μ–OH–Cr(NH3)5]Cl5

7.3 Name the following complexes?

(a) cis-[CrCl2(NH3)4]+? cis-tetra(ammine)di(chloro)chromium(III)

(b) trans-[Cr(NCS)4(NH3)2]-? trans-

di(ammine)tetrakis(isothiocyanato)chromate (III)

(c) [Co(C2O4)(en)2]+? bis(ethylenediamine)oxalatocobalt(III).



7.4 Four-coordinate complexes? (a) Sketch the two observed structures?

(b) Isomers expected for MA2B2? tetrahedral complex, no isomers, for a square-planar complex, two isomers, cis and trans.

7.5 For five-coordinate complexes, sketch the two observed structures?

7.6 Six-coordinate complexes? (a) Sketch the two observed structures?

(b) Which one of these is rare? Trigonal prism.

7.7 Explain the difference between monodentate, bidentate, and quadridentate? A monodentate ligand can bond to a metal atom only at a single atom, a bidentate ligand can bond through two atoms, a quadridentate ligand can bond through four atoms.

7.8 What type of isomers do you get with

ambidentate ligands? linkage isomers.

7.9 Which ligand could act like a chelating ligand? (a) Triphenylphosphite, no/ (b) Bis(dimethyl)phosphino ethane (dmpe)

yes. (c) Bipyridine (bipy), yes. (d) Pyrazine, no.

7.10 Draw structures of complexes that contain the ligands (a) en, (b) ox, (c) phen, and (d) edta? .

ML L

L L

M

LL

L

L

tetrahedral square plana

[Mg(edta)(OH2)]2–

7.11 What types of isomers are [RuBr(NH3)5]Cl and [RuCl(NH3)5]Br? Ionization isomers.

7.12 Which complexes have isomers?

[CoBrClI(OH2)]

7.13 Which complexes have isomers?

(a) [Pt(ox)(NH3)2] no isomers

(b) [PdBrCl(PEt3)2] has two isomers.

(c) [IrHCO(PR3)2] has two isomers.

(d) [Pd(gly)2] has two isomers.

7.14 How many isomers are possible for the following complexes?

(a) [FeCl(OH2)5]2+? None.

(b) [IrCl3(PEt3)2]? 2

(c) [Ru(biby)3]2+? 2

(d) [CoCl2(en)(NH3)2]+? 4

(e) [W(CO)4(py)2] 2

r

M

A

A

E

EE

Trigonal Bipyramidal

A = axial ligands

E = equatorial ligands

M

A

BB

B B

Square based pyramid

A = axialB = basal



7.15 Draw all possible isomers for [MA2BCDE]? Including optical isomers, 15 isomers are possible! .

M

A

EC D

ABM

A

DE C

ABM

A

CD E

AB

M

A

AD C

EBM

A

AC D

EBM

A

AD E

CB

M

A

CB D

AEM

A

BE C

ADM

A

ED B

AC

7.16 Which of the following complexes are chiral? (a) [Cr(ox)3]3–? Chiral

(b) cis-[PtCl2(en)]? Chiral (The en is not planar).

(c) cis-[RhCl2(NH3)4]+? not chiral.

(d) [Ru(bipy)3]2+? chiral

(e) fac-[Co(NO2)3(dien)]? Not chiral.

(f) mer-[Co(NO2)3(dien)]? Chiral (dien is not planar).

7.17 Which isomer, Λ or Δ, is the complex Mn(acac)3, shown in the exercise? The Λ isomer.

7.18 Draw both isomers, Λ or Δ, of the complex [Ru(en)3]+2 ?

7.19 Suggest a reason why Kf5 is so different?

Because of a change in coordination. 7.20 Compare these values with those of

ammonia given in exercise 7.19 and suggest why they are different? The chelate effect.


S8.1 Main features of the CrO2 powder XRD pattern? XRD pattern for CrO2 will show

identical reflections to those of rutile TiO2 but shifted to slightly higher diffraction angles.

S8.2 TiO2 in sunscreens? Titania articles absorb this ultraviolet

radiation S8.3 Molecular shape and vibrational modes for

XeF2? Trigonal bipyramidal, 4 total vibrational modes.

S8.4 (a) 77Se-NMR spectrum consists of a triplet of triplets? The triplet of triplets.

(b) Proton resonance of the hydrido ligand consist of eight equal intensity lines? yes.

S8.5 EPR signal of new material arises from W sites?

14% of naturally occurring tungsten is 183W, which has I = ½. Thus, the signal is split into 2 lines.

S8.6 Isomer shift for iron in Sr2FeO4? The Smaller and less positive.

S8.7 Why does the mass spectrum of ClBr3 have five peaks separated by 2 u? Halogen isomers.

Exercises 8.1 How would you determine crystalline

components in mineral sample? Powder X-ray diffraction.

8.2 Why are there no diffraction maxima in borosilicate glass? Glass has no long-range periodicity or order.

8.3 What is the minimum size of a cubic crystal that can be studied? 0.5 μm by 0.5 μm by 0.5 μm.

Ru

N

N

N

N

N

N

Ru

N

N

N

N

N

N

Δ isomerΛ isomer

8.4 Wavelength of neutron at 2.20 km/s? 1.80 × 10–12 m or 180 pm.

8.5 Order of stretching frequencies? The smaller effective mass of the oscillator for CN− causes the molecule to have the higher stretching frequency. The bond order for NO is 2.5, and N is heavier than C, hence CO has a higher stretching frequency than NO.

8.6 Wavenumber for O–O in O2+? In the region

of 1800 cm 1.

8.7 UV photoelectron spectrum of NH3? The band at 11 eV is due to the lone pair and the pyramidal angle. The ionised molecule has greater planarity, thus the long progression.

8.8 Raman bands assignments? N(SiH3)3 is planar. N(CH3) 3 is pyramidal.



8.9 Single 13C peak in NMR? Chemically distinct carbonyls are exchanging position sufficiently quickly.

8.10 Form of 19F-NMR and 77Se-NMR spectra of 77SeF4? 19F NMR spectrum reveals two 1:3:3:1 quartets. The 77Se-NMR spectrum is a triplet of triplets.

8.11 NMR spectral features for XeF5−? All 5 of

the F atoms are chemically equivalent. Approximately 25% is present as 129Xe, I = 1/2, and in this case the 19F resonance is a doublet. The final result is a composite: two lines of 12.5% intensity from the 19F coupled to the 129Xe, and one remaining line of 75%.

8.12 g-values? 1.94, 1.74, and 1.56.

8.13 Slower process, NMR or EPR? NMR.

8.14 Differences in EPR spectrum for d-metal with one electron in solution versus frozen? In aqueous solution at room temperature, molecular tumbling removes the effect of the g-value anisotropy. In frozen solution, g-value anisotropy can be observed.

8.15 Isomer shift for iron in BaFe(VI)O4? A positive shift for Fe(VI) well below +0.2 mm s-1.

8.16 Charge on Fe atoms in Fe4[Fe(CN)6]3? EPR and Mössbauer.

8.17 No quadrupole splitting in Mössbauer spectrum of SbF5? The geometry must be close to cubic in the solid state.

8.18 No peak in the mass spectrum of Ag at 108 u? Two isotopes, 107Ag (51.82%) and 109Ag (48.18%). Compounds that contain silver will have two mass peaks.

8.19 Peaks in mass spectrum of Mo(C6H6)(CO)3? 258, 230, 200, 186, and 174.

8.20 Cyclic voltammogram of Fe(III) complex? The complex undergoes a reversible one-electron reduction with a reduction potential of 0.21 V. Above 0.720 V the complex is oxidized.

8.21 Zeolite of composition CaAl2Si6O16.nH2O, determine n.? n=7.2 As an integer, n = 7.

8.22 Ratio of cobalt to acetylacetonate in the product? The ratio is 3:1.


S9.1 Found in aluminosilicate minerals or sulfides?

Cd and Pb will be found as sulfides. Rb and Sr can be found in aluminosilicate

minerals. Cr and Pd can be found in both oxides and

sulfides. S9.2 Sulfur forms catenated polysulfides

whereas polyoxygen anions are unknown? Owing to a strong tendency to form strong double bonds, it is more likely that polyoxygen anions will form pi bonds that limit extended bonding owing to restrictions on pi orbital overlap through multiple bridging centres.

S9.3 Shape of XeO4 and identify the Z + 8

compound with the same structure? A tetrahedral geometry. The same structure is

SmO4. S9.4 Comment on ΔfHө values? It is evident from

the values that as we move down the group, steric crowding of the fluorines is minimized.

S9.5 Further data useful when drawing

comparisons with the value for V2O5? We would have to know the products formed upon decomposition.

Exercises 9.1 Maximum stable oxidation state? (a) Ba;

+2, (b) As; +5, (c) P; +5, (d) Cl; +7. 9.2 Form saline hydrides, oxides and

peroxides, and all the carbides react with water to liberate a hydrocarbon?

the alkaline earth metals or Group 2 elements. 9.3 Elements vary from metals through

metalloids to non-metals; form halides in oxidation states +5 and +3 and toxic gaseous hydrides? Elements in Group 15.

9.4 Born–Haber cycle for the formation of the

hypothetical compound NaCl2? Which thermochemical step is responsible for the fact that NaCl2 does not exist?

The second ionization energy of sodium is 4562 kJ mol-1 and is responsible for the fact that the compound does not exist.

9.5 Inert pair effect beyond Group 15? The

relative stability of an oxidation state in which the oxidation number is 2 less than the group number is an example of the inert pair effect.



9.6 Ionic radii, ionization energy, and metallic character? Metallic character, ionic radii decrease across a period and down a group. Ionization energy increases across a period and decreases down a group.

9.7 Names of ores? (a) Mg; MgCO3 magnesite,

(b) Al; Al2O3 bauxite, and (c) Pb; PbS galena. 9.8 Identify the Z + 8 element for P.

Similarities? V (vanadium). 9.9 Calculate ΔfHө for SeF6? ΔfHө = −1397 kJ mol-1.


S10.1 Which of the following CH4, SiH4,, or GeH4

would best H+ or H- donor?

CH4, the strongest Bronsted acid. GeH4 would be the best hydride donor.

S10.2 Reactions of hydrogen compounds?

(a) Ca(s) + H2(g) → CaH2(s).

(b) NH3(g) + BF3(g) → H3N–BF3(g).

(c) LiOH(s) + H2(g) → NR.

S10.3 A procedure for making Et3MeSn?

2Et3SnH + 2Na → 2Na+Et3Sn– + H2

Na+Et3Sn– + CH3Br → Et3MeSn + NaBr

Exercises 10.1 Where does Hydrogen fit in the periodic

chart? (a) Hydrogen in group 1? Hydrogen has one valence electron like the group 1 metals and is stable as H+, especially in aqueous media.

(b) Hydrogen in group 17? Hydrogen can fill its 1s orbital and make a hydride H–. The halogens are diatomic gases just like hydrogen, but chemically it fits well in both group 1 and group 17.

(c) Hydrogen in group 14? There is no reason for hydrogen to be placed in this group.

10.2 Low reactivity of hydrogen? Hydrogen exists as a diatomic molecule (H2). It has a high bond enthalpy. It also only has two electrons shared between two protons.

10.3 Assign oxidation numbers to elements?

(a) H2S? H = +1, S = –2.

(b) KH? H = –1, K = +1.

(c) [ReH9]2–? H = –1, Re = +7.

(d) H2SO4? H = +1, O = –2, S = +6.

(e) H2PO(OH)? H bonded to an oxygen atom = +1. H bonded to the phosphorus atom? If they are assigned an oxidation number of +1, and O = –2, then P = +1.

10.4 Preparation of hydrogen gas?

(i) CH4(g) + H2O → CO(g) + 3H2(g) (1000°C)

(ii) C(s) + H2O → CO(g) + H2(g) (1000°C)

(iii) CO(g) + H2O → CO2(g) + H2 (g)

10.5 Properties of hydrides of the elements? (a) Position in the periodic table? See Figure 10.2.

(b) Trends in ΔfGº? See Table 10.1.

(c) Different molecular hydrides? Molecular hydrides are found in groups 13/III through 17/VII.

10.6 What are the physical properties of water without hydrogen bonding? It most likely would be a gas at room temperature; ice would be denser than water.

10.7 Which molecule has the stronger hydrogen bonds? S–H···O has a weaker hydrogen bond than O–H···S.

10.8 Name and classify the following?

(a) BaH2? barium hydride.

(b) SiH4? silane.

(c) NH3? ammonia,

(d) AsH3? Arsine.

(e) PdH0.9? palladium hydride.

(f) HI? hydrogen iodide.

10.9 Chemical characteristics of hydrides?

(a) Hydridic character? Barium hydride

(b) Brønsted acidity? Hydrogen iodide.

(c) Variable composition? PdH0.9 .

(d) Lewis basicity? Ammonia.

10.10 Phases of hydrides of the elements? BaH2 and PdH0.9 are solids, none is a liquid, and SiH4, NH3, AsH3, and HI are gases.

10.11 Structures of H2Se, P2H4, and H3O+? The Lewis structures of these three species are:



H3O+ should be trigonal pyramidal.

10.12 The reaction that will give the highest proportion of HD? Reaction (b) will produce 100% HD and no H2 or D2.

10.13 Most likely to undergo radical reactions? (CH3)3SnH, the tin compound is the most likely to undergo radical reactions with alkyl halides.

10.14 Arrange H2O, H2S, and H2Se in order?

(a) Increasing acidity? H2O < H2S < H2Se.

(b) Increasing basicity toward a hard acid? H2Se < H2S < H2O.

10.15 The synthesis of binary hydrogen compounds? (i) direct combination of the elements, (ii) protonation of a Brønsted base, and (iii) metathesis using a compound such as LiH, NaBH4, or LiAlH4.

10.16 Compare BH4–, AlH4

–, and GaH4–? Since

AlH4– is more “hydride-like,” it is the

strongest reducing agent.

10.17 Compare period 2 and period 3 hydrogen compounds? Period 2 compounds:

- except for B2H6, are all exoergic

- tend to be weaker Brønsted acids and stronger Brønsted bases

- bond angles in period 2 hydrogen compounds reflect a greater degree of sp3 hybridization

- Several period 2 compounds exhibit strong hydrogen bonding.

- boiling points of HF, H2O, and NH3 are all higher than their respective period 3 homologues.

10.18 Suggest a method for the preparation of BiH3?

The redistribution of methylbismuthine, BiH2Me.

3BiH2Me → 2BiH3 + BiMe3

10.19 Describe the compound formed between water and Kr? A clathrate hydrate.

10.20 Potential energy surfaces for hydrogen bonds? (See Figure 10.9) The surface for the H2O, Cl– system has a double minimum, while the surface for the bifluoride ion has a single minimum.

10.21.1 Dihydrogen as an oxidizing agent? It’s reaction with an active s-block metal such as sodium.


S11.1 Change in cell parameter for CsCl? At 445 °C the CsCl structure changes to rock-salt and assumes the face centered cubic.

S11.2 Lattice enthalpies of formation? LiF is 625 kJ mol−1 and for NaF is 535 kJ mol−1.

S11.3 Trend is stability of Group 1 ozonides? Group 1 ozonides are less stable compared to the superoxides.

S11.4 Sketch the thermodynamic cycle of Group 1 carbonate. M2CO3(s) M2O(s) + CO2(g) 2M+(g) + CO3

2−(g) 2M+(g) + O2−(g) + CO2(g)

S11.5 Explain the differences in temperature of decomposition of LiNO3 and KNO3? KNO3 decomposes in two steps at two different temperatures.

KNO3(s) → KNO2(s) + 1/2O2(g)

2KNO2(s) → K2O(s) + 2NO2(g) + 1/2O2(g)

LiNO3 decomposes in one step.

LiNO3(s) → 1/2 Li2O(s) + NO2(g) + 1/4O2(g)

S11.6 Predicted 7Li NMR of Li3N? Two peaks in the NMR spectrum at low temperature. Only one resonance in the NMR at high temperature.

Exercises

11.1 (a) Why are group 1 metals good reducing agents? They have one valence electron in the ns1 subshell, and relatively low first ionization energies.



(b) Why are group 1 metals poor complexing agents? They are large, electropositive metals and have little tendency to act as Lewis acids.

11.2 Trends of the fluorides and chlorides of the group 1 metals?

Fluoride is a hard Lewis base and will form strong complexes with hard Lewis acids. The trends reverse for the chloride ion.

11.3 Synthesis of group 1 alkyls? Most alkyl lithiums are made using elemental lithium with the corresponding alkyl chlorides.

11.4 Which is more likely to lead to the desired result? (a) Cs+ or Mg2+, form an acetate complex? Mg2+.

(b) Be or Sr, dissolve in liquid ammonia? Strontium.

(c) Li+ or K+, form a complex with C2.2.2? Potassium ion.

11.5 Identify the compounds?

NaOH ← H2O + Sodium metal + O2 → Na2O2 + heat → Na2O

+ NH3

↓

NaNH2

11.6 Trends in solubility? Higher for LiF and CsI, lower for CsF and LiI.

11.7 Thermal stability of hydrides versus

carbonates? Hydrides decompose to elements. Carbonates decompose to oxides.

11.8 The structures of CsCl and NaCl? 6-coordinate Na+, 8-coordinate Cs+. 11.9 The effect of the alkyl group on the

structure of lithium alkyls? Whether a molecule is monomeric or polymeric is based on the streric size of the alkyl group – less bulky alkyl groups lead to polymerization.

11.10 Predict the products of the following reactions?

(a) CH3Br + Li → Li(CH3) + LiBr (b) MgCl2 + LiC2H5 → Mg(C2H5)Br + LiBr (c) C2H5Li + C6H6 → LiC6H5 + C2H6


S12.1 Predict whether (a) BeCl2 and (b) BaCl2 are predominantly ionic or covalent?

BeCl2 is covalent; BaCl2 is ionic. S12.2 Calculate the lattice enthalpies for CaO

and CaO2 and check that the above trend is confirmed?

Calcium oxide and calcium peroxide are 3465 kJ mol–1 and 3040 kJ mol–1.

S12.3 Use ionic radii to predict a structure type

of BeSe?

According to Table 3.6 should be close to ZnS-like structure.

S12.4 Calculate the lattice enthalpy of MgF2 and

comment on how it will affect the solubility compared to MgCl2?

MgF2 is 2991 kJ mol–1, will reduce solubility compared to MgCl2.

Exercises 12.1 Why are compounds of beryllium covalent

whereas those of the other group 2 elements are predominantly ionic?

Be has large polarizing power and a high charge density due.

12.2 Why are the properties of beryllium more similar to aluminium and zinc than to magnesium?

Because of a diagonal relationship between Be and Al. 12.3 Identify the compounds A, B, C, and D of

the group 2 element M?

M + H2O → M(OH)2; A = M(OH)2 M(OH)2 + CO2 → MCO3; B = MCO3 2MCO3 + 5C → 2MC2 + 3CO2; C = MC2 MC2 + 2H2O → M(OH)2 + C2H2 M(OH)2 + 2HCl → MCl2 + 2H2O; D = MCl2.

12.4 Why does beryllium fluoride form a glass when cooled from a melt?

BeF2 adopts SiO2 like arrangement. 12.5 Why is magnesium hydroxide a much more

effective antacid than calcium or barium hydroxide? Mg(OH)2 is sparingly soluble and mildly basic.



12.6 Explain why Group 1 hydroxides are much more corrosive to metals than Group 2 hydroxides?

Group 1 hydroxides are more soluble than group 2 hydroxides, and therefore have higher OH− concentrations.

12.7 Which of the salts MgSeO4 or BaSeO4 would be expected to be more soluble in water?

MgSeO4

12.8 Which Group 2 salts are used as drying agents and why?

Anhydrous Mg, and Ca sulphates are preferred as drying agents, because of the higher affinity of Mg and Ca sulphates for water.

12.9 How do group 2 salts give rise to scaling

from hard water?

Salts of divalent ions have low solubility. 12.10 Predict structures for BeTe and BaTe.

BeTe, close to ZnS-like structure. BaTe, close to CsCl-like structure.

12.11 Use the data in Table 1.7 and the Ketelaar triangle in Fig. 2.38 to predict the nature of the bonding in BeBr2, MgBr2, and BaBr2.

BeBr2 should be covalent. MgBr2 should be ionic. BaBr2 should be ionic

12.12 The two Grignard compounds C2H5MgBr

and 2,4,6-(CH3)3C6H2MgBr dissolve in THF. What differences would be expected in the structures of the species formed in these solutions?

C2H5MgBr will be tetrahedral with two molecules of solvent coordinated to the magnesium. The bulky organic group in 2,4,6-(CH3)3C6 H2MgBr leads to a coordination number of two.

12.13 Predict the products of the following

reactions?

(a) MgCl2 + 2LiC2H5 → 2LiCl + Mg(C2H5)2 (b) Mg + (C2H5)2Hg → Mg(C2H5)2 + Hg (c) Mg + C2H5HgCl → C2H5MgCl + Hg


S13.1 11B nuclei have I = 3/2. Predict the number of lines and their relative intensities in the 1H-NMR spectrum of BH4

–? 4. Relative intensity ratio is 1:3:3:1.

S13.2 Write an equation for the reaction of LiBH4 with propene in ether solvent and a 1:1 stoichiometry and another equation for its reaction with ammonium chloride in THF with the same stoichiometry?

Simple alkenes are inert towards LiBH4.

LiBH4 + NH4Cl THF

BH3NH3 + LiCl + H2

S13.3 Write and justify balanced equations for plausible reactions between (a) BCl3 and ethanol, (b) BCl3 and pyridine in hydrocarbon solution, (c) BBr3 and F3BN(CH3)3?

(a) BCl3 and ethanol?

BCl3(g) + 3 EtOH(l) → B(OEt)3(l) + 3 HCl(g)

(b) BCl3 and pyridine in hydrocarbon solution?

BCl3(g) + py(l) → Cl3B − py(s)

(c) BBr3 and F3BN(CH3)3?

BBr3(l) + F3BN(CH3)3(s) → BF3(g) + Br3BN(CH3)3(s)

S13.4 Suggest a reaction or series of reactions for the preparation of N, N’, N’’-trimethyl-B,B’,B’’-trimethylborazine starting with methylamine and boron trichloride?

Cl3BNCH3 + 3CH3MgBr (CH3)3B3N3(CH3)3 + 3Mg(Br, Cl)2

S13.5 How many framework electron pairs are

present in B4H10 and to what structural category does it belong? Sketch its structure?

7, arachno species. The structure of B4H10:

S13.6 Propose a plausible product for the

reaction between Li[B10H13] and Al2(CH3)6? [B10H11 (AlCH3)]–



S13.7 Propose a synthesis for the polymer precursor 1,7-B10C2H10(Si(CH3)2Cl)2 from 1,2-B10C2H12 and other reagents of your choice?

1,2 - B10C2H12 1,7-B10C2H12 (90%) + 1,12-B10C2H12 (10%)

⎯→⎯Δ

1,7-B10C2H10Li2 + 2Si(CH3)2Cl2 → 1,7 - B10C2H10(Si(CH3)2 + 2LiCl

S13.8 Propose, with reasons, the chemical equation (or indicate no reaction) for reactions between (a) (CH3)2SAlCl3 and GaBr3?

(Me)2SalCl3 + GaBr3 → Me2SGaBr3 + AlCl3.

(b) TlCl and formaldehyde (HCHO) in acidic aqueous solution? No reaction.

Exercises 13.1 Give a balanced chemical equation and

conditions for the recovery of boron? B2O3 + 3Mg → 2B + 3MgO ΔH < 0

13.2 Describe the bonding in (a) BF3? Covalent.

(b) AlCl3? In the solid state, a layered structure. At melting point, dimers. (c) B2H6? Electron-deficient dimer.

13.3 Arrange the following in order of increasing Lewis acidity: BF3, BCl3, AlCl3. In the light of this order, write balanced chemical reactions (or no reaction) for (a) BF3N(CH3)3 + BCl3 →, (b) BH3CO + BBr3→?

BCl3 > BF3 > AlCl3 (a) BF3N(CH3)3 + BCl3

BCl3N(CH3)3 + BF3 (BCl3 > BF3) (b) BH3CO + BBr3 NR

13.4 Thallium tribromide (1.11 g) reacts quantitatively with 0.257 g of NaBr to form a product A. Deduce the formula of A. Identify the cation and anion?

TlBr3 + NaBr → NaTlBr4

13.5 Identify compounds A, B, and C?

ABF3

H2O

LiAlH4

BCheat

CaF2

(a) A = B2H6 (b) B = B(OH)3

rvive in air? If not, write the

2O3 + 3 H2O2

3.7 Predict how many different boron

13.8 products from the

(c) C = B2O3

13.6 Does B2H6 suequation for the reaction?

No, it explodes in air. B2H6 + 3 O2 → B

1environments would be present in the proton-decoupled 11B-NMR of a) B5H11, b) B4H10? a) 3 b) 2.

Predict the hydroboration of (a) (CH3)2C=CH2, (b) CH CH?

(a) H3 + B (CH )2C=CH2 3 B[CH2-CH(CH3)2]3

(b) BH3 + CH CH B(CH=CH2)3 Diborane ha

13.9 s been used as a rocket

extremely toxic, and the boron

3.10 Using BCl3 as a starting material and other

B2Cl4 + 2HgCl2

3.11 Given NaBH4, a hydrocarbon of your

(a) BCl3 + 3C2H5MgCl

propellant. Calculate the energy released from 1.00 kg of diborane given the following values of ΔfHө/kJ mol-1: B2H6 = 31, H2O = -242, B2O3 = -1264. The combustion reaction is B2H6 (g) + 3 O2(g) → 3 H2O (g) + B2O3 (s). What would be the problem with diborane as a fuel?

-73,172 kJ.

Diborane iscontaining product of combustion is a solid, B2O3.

1reagents of your choice, devise a synthesis for the Lewis acid chelating agent, F2B–C2H4–BF2?

2BCl3 + 2Hg B2Cl4 + 4AgF B2F4 + 4AgCl B2F4 + C2H4 F2CH2CH2BF2 1

choice, and appropriate ancillary reagents and solvents, give formulas and conditions for the synthesis of (a) B(C2H5)3, (b) Et3NBH3?

Heat, Ether

B(C2H5)3 + 3MgCl2

(b) [HN(C2H5)3]Cl + NaBH4

H2 + H3BN(C2H5)3 + NaCl

3.12 Draw the B12 unit that is a common motif

1of boron structures; take a viewpoint along a C2 axis?



13.13 Which boron hydride would you expect to

be more thermally stable, B6H10 or B6H12? Give a generalization by which the thermal stability of a borane can be judged?

B6H10

13.14 How many skeletal electrons are present in B5H9? 14

13.15 (a) Give a balanced chemical equation (including the state of each reactant and product) for the air oxidation of pentaborane(9). (b) Describe the probable disadvantages, other than cost, for the use of pentaborane as a fuel for an internal combustion engine?

(a) 2B5H9 (l) + 12O2 (g) Heat

5B2O3 (s) + 9H2O (l)

(b) The boron containing product of combustion is a solid, B2O3.

13.16 (a) From its formula, classify B10H14 as closo, nido, or arachno. (b) Use Wade’s rules to determine the number of framework electron pairs for decaborane(14). (c) Verify by detailed accounting of valence electrons that the number of cluster valence electrons of B10H14 is the same as that determined in (b)?

(a) nido . (b) 12.

(c) The total number of valence elections is (10x3)+(14x1)=44; the number of cluster valence is the remainder of 44-20=24.

13.17 Starting with B10H14 and other reagents of your choice, give the equations for the synthesis of [Fe(nido-B9C2H11)2]2-, and sketch the structure of this species?

(1) B10H14 + 2SEt2 B10H12(SEt2)2 + H2

(2) B10H12(SEt2)2 + C2H2 B10C2H12 + 2SEt2 + H2 (3) 2 B10C2H12 + 2EtO– + 4EtOH

2 B9C2H12– + 2B(OEt)3 + 2H2

(4) Na[B9C2H12] + NaH Na2[B9C2H11] + H2

(5) 2Na2[B9C2H11] + FeCl2 THF

2NaCl + Na2[Fe(B9C2H11)2]

BB

B

B

CC

BB

B

B

B

FeH

H

H

HH

H

H

H

H

H

H

H

H

H

HH

H

H

H

H

H

H

BB

B

B

CC

BB

B

B

B

2-

13.18 (a) What are the similarities and

differences in structure of layered BN and graphite (Section 13.9)? (b) Contrast their reactivity with Na and Br2. (c) Suggest a rationalization for the differences in structure and reactivity. (a) Their structures? Both of these substances have layered structures.

(b) Their reactivity with Na and Br2? Graphite reacts, boron nitride is unreactive.

(c) Explain the differences? The large HOMO–LUMO gap in BN means it is more difficult to remove an electron from it than from the HOMO of graphite.

13.19 Devise a synthesis for the borazines (a) Ph3N3B3Cl3 and (b) Me3N3B3H3, starting with BCl3 and other reagents of your choice. Draw the structures of the products? (a) Ph3N3B3Cl3?

3 PhNH3+Cl− + 3 BCl3 → Ph3N3B3Cl3 + 9 HCl

(b) Me3N3B3H3?

3 MeNH3+Cl− + 3 BCl3 → Me3N3B3Cl3 + 9 HCl

Me3N3B3Cl3 + 3 LiH → Me3N3B3H3 + 3 LiCl

The structures:



13.20 Give the structural type and describe the

structures of B4H10, B5H9, and 1,2-B10C2H12?

B4H10, is an arachno borane. B5H9, is a nido borane. 1,2-B10C2H12 is a closo carborane

13.21 Arrange the following boron hydrides in

order of increasing Brønsted acidity, and draw a structure for the probable structure of the deprotonated form of one of them: B2H6, B10H14, B5H9? In a series of boranes, the acidity increases as the size of the borane increases.


S14.1 Describe how the electronic structure of graphite is altered when it reacts with (a) potassium, (b) bromine?

(a) With potassium? Potassium results in a material with a higher conductivity.

(b) With bromine? Bromine can remove electrons from the π-symmetry HOMOs of graphite. This also results in a material with a higher conductivity.

S14.2 Use the bond enthalpy data in Table 14.2 and above to calculate the standard enthalpy of formation of CH4 and SiH4?

CH4: ΔfH= –61kJ mol–1

SiH4: ΔfH= +39 kJ mol–1 S14.3 Propose a synthesis of D13CO2

– starting from 13CO?

13CO(g) + 2MnO2(s) → 13CO2(g) + Mn2O3(s)

2Li(s) + D2 → 2LiD(s) 13CO2(g) + LiD(et) → Li+D13CO2

−(et)

Exercises 14.1 Silicon forms the chlorofluorides SiCl3F,

SiCl2F2, and SiClF3. Sketch the structures of these molecules?

14.2 Explain why CH4 burns in air whereas CF4

does not. The enthalpy of combustion of CH4 is −888 kJ mol−1 and the C–H and C–F bond enthalpies are −413 and −489 kJ mol−1 respectively?

The bond enthalpy of a C–F bond is higher than the bond enthalpy of a C–H bond.

14.3 SiF4 reacts with (CH3)4NF to form

[(CH3)4N][SiF5]. (a) Use the VSEPR rules to determine the shape of the cation and anion in the product; (b) Account for the fact that the 19F NMR spectrum shows two fluorine environments? (a) The cation is [(CH3)4N]+

N

CH3

CH3H3C

+

CH3

The anion is SiF5

–

Si F

F

F

F

F

-

(b) There are two different fluorine environments.

14.4 Draw the structure and determine the charge on the cyclic anion [Si4O12]ν ?



O

O O

O

OSi

Si

Si

Si

O

O

O

8-

Charge = –8.

14.5 Predict the appearance of the 119Sn-NMR spectrum of Sn(CH3)4? Doublet 14.6 Predict the appearance of the 1H-NMR spectrum of Sn(CH3)4? Doublet

14.7 Use the data in Table 14.2 and the

additional bond enthalpy data given here to calculate the enthalpy of hydrolysis of CCl4 and CBr4. Bond enthalpies/(kJ mol–1): O–H = 463, H–Cl = 431, H–Br = 366?

ΔhH˚CCl4 = 110 kJmol–1 ΔhH˚CBr4 = 86 kJmol–1

14.8 Identify the compounds A to F:? (A) SiCl4 (B) SiRCl3 (C) RSi(OH)3. (D)

RSiOSiR +H2O (E) SiR4 (F)? SiO2 YYYYY14.9 (a) Summarize the trends in

relative stabilities of the oxidation states of the elements of Group 14, and indicate the elements that display the inert pair effect. (b) With this information in mind, write balanced chemical reactions or NR (for no reaction) for the following combinations, and explain how the answer fits the trends. (i) Sn2+(aq) + PbO2(s) (excess) → (air excluded) (ii) Sn2+(aq) + O2(air) →? (a) +4 is the most stable oxidation state for the lighter elements, but +2 is the most stable oxidation state of Pb. Pb therefore displays the inert-pair effect. (b) (i) Sn2+ + PbO2 + 4 H+ → Sn4+ + Pb2+ + 2H2O, (ii) 2Sn2+ + O2 + 4H+ → 2Sn4+ + 2H2O.

14.10 Use data from Resource section 3 to determine the standard potential for each of the reactions in Exercise 14.5 (b). In each case, comment on the agreement or disagreement with the qualitative assessment you gave for the reactions?

(i) V = +1.31 V. (ii) V= 1.08 V. Both reactions agree with the predictions made in Excercise 14.5.

14.11 Give balanced chemical equations and conditions for the recovery of silicon and germanium from their ores?

SiO2(s) + C(s) → Si(s) + CO2(g) ΔH < 0 GeO2(s) + 2H2(g) → Ge(s) + 2H2O(g)

ΔH < 0 14.12 (a) Describe the trend in band gap energy,

Eg, for the elements carbon (diamond) to tin (grey). (b) Does the electrical conductivity of silicon increase or decrease when its temperature is changed from 20˚C to 40˚C?

(a) There is a decrease in band gap energy

from carbon (diamond) to grey tin. (b) Increase.

14.13 Preferably without consulting reference

material, draw a periodic table and indicate the elements that form saline, metallic, and metalloid carbides?

Ionic(silane) carbides

Metallic carbides

Metalloid carbides

Group I elements

Li, Na, K, Rb, Cs

Group II elements

Be, Mg, Ca, Sr, Ba

Group 13 elements

Al B

Group 14 elements

Si

3d-Block elements

Sc, Ti, V, Cr, Mn, Fe, Co, Ni

4d-Block elements

Zr, Nb, Mo, Tc, Ru

5d-Block elements

La, Hf, Ta, W, Re, Os

6d-Block elements

Ac

Lanthanides Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu

14.14 Describe the preparation, structure and classification of (a) KC8, (b) CaC2, (c) K3C60? (a) KC8? Formed by heating graphite with

potassium vapor or by treating graphite with a solution of potassium in liquid ammonia. There is a layered structure of alternating sp2 carbon atoms and potassium ions, a saline carbide.

(b) CaC2? Ca(l) + 2C(s) → CaC2(s) or

CaO(s) + 3C(s) → CaC2(s) + CO(g)



Calcium carbide contains discrete C22– ions.

(c) K3C60? A solution of C60 can be treated with elemental potassium. It is ionic.

14.15 Write balanced chemical equations for the

reactions of K2CO3 with HCl(aq) and of Na4SiO4 with aqueous acid?

K2CO3(aq) + 2HCl(aq) → 2KCl(aq) + CO2(g) + H2O(l)

Na4SiO4(aq) + 4HCl(aq) → 4NaCl(aq) + SiO2(s) + 2H2O(l)

14.16 Describe in general terms the nature of the [SiO3]n

2ν ion in jadeite and the silica-alumina framework in kaolinite?

The structures of jadeite and kaolinite consist of extended one- and two-dimensional structures, respectively. The [SiO3

2–]n ions in jadeite are a linear polymer of SiO4 tetrahedra. The two-dimensional aluminosilicate layers in kaolinite represent another way of connecting SiO4 tetrahedra.

14.17 (a) How many bridging O atoms are in the

framework of a single sodalite cage? (b) Describe the (supercage) polyhedron at the centre of the Zeolite A structure in Fig. 14.3? (a) 48.

(b) Eight sodalite cages are linked together to form the large cage of zeolite A.


S15.1 Consider the Lewis structure of a segment of the structure of bismuth shown in Fig. 15.2. Is this puckered structure consistent with the VSEPR model? Yes.

S15.2 Refined hydrocarbons and liquid hydrogen are also used as rocket fuel. What are the advantages of dimethylhydrazine over these fuels?

Dimehylhydrazine ignites spontaneously, and produces less CO2.

S15.3 From trends in the periodic table, decide whether phosphorus or sulphur is likely to be the stronger oxidizing agent? Sulphur.

S15.4 Summarize the reactions that are used for the synthesis of hydrazine and hydroxylamine. Are these reactions best described as electron-transfer processes or nucleophilic displacements?

NH3 + ClO− + H+ → [H3N—Cl-O]− + H+ → H2NCl + H2O

H2NCl + NH3 → H2NNH2 + HCl

Both can be thought of as redox reactions.

S15.5 When titrated against base a sample of polyphosphate gave end points at 30.4 and 45.6 cm3. What is the chain length?

There are (30.4 cm3)/(7.6 cm3) = 4 strongly acidic OH groups per molecule. A molecule with 2 terminal OH groups and four further OH groups is a tetrapolyphosphate.

Exercises 15.1 List the elements in Groups 15 and indicate

the ones that are (a) diatomic gases, (b) nonmetals, (c) metalloids, (d) true metals. Indicate those elements that display the inert-pair effect?

Type of element

Diatomic gas?

Inert pair effect?

N Non-metal

yes No

P Non-metal

no No

As nonmetal no No Sb metalloid no No Bi metalloid no Yes O nonmetal yes No S nonmetal no No Se nonmetal no No Te nonmetal no No

15.2 (a) Give complete and balanced chemical equations for each step in the synthesis of H3PO4 from hydroxyapatite to yield (a) high-purity phosphoric acid and (b) fertilizer-grade phosphoric acid. (c) Account for the large difference in costs between these two methods?

(a) high-purity phosphoric acid?

2Ca3(PO4)2 + 10C + 6SiO2 →

P4− + 10CO + 6CaSiO3

P4 (pure) + 5O2 → P4O10

P4O10 + 6H2O → 4H3PO4 (pure)

(b) Fertilizer grade H3PO4?

Ca5(PO4)3OH + 5H2SO4 →

3H3PO4 (impure) + 5CaSO4 + H2O

(c) Account for the difference in cost?



Fertilizer-grade phosphoric acid involves a single synthetic step for a product that requires little or no purification.

15.3 Ammonia can be prepared by (a) the hydrolysis of Li3N or (b) the high-temperature, high-pressure reduction of N2

by H2. Give balanced chemical equations for each method starting with N2, Li, and H2, as appropriate. (c) Account for the lower cost of the second method?

(a) Hydrolysis of Li3N?

6Li + N2 → 2Li3N

2Li3N + 3H2O → 2NH3 + 3Li2O

(b) Reduction of N2 by H2?

N2 + 3H2 → 2NH3

(c) Account for the difference in cost? The second process is considerably cheaper than the first, because lithium is very expensive.

15.4 Show with an equation why aqueous solutions of NH4NO3 are acidic?

NH4NO3(s) + H2O NH4+ + NO3

-(aq)

15.5 Carbon monoxide is a good ligand and is toxic. Why is the isoelectronic N2 molecule not toxic?

N2 itself, with a triple bond between the two atoms, is strikingly unreactive.

15.6 Compare and contrast the formulas and stabilities of the oxidation states of the common nitrogen chlorides with the phosphorus chlorides?

The only isolable nitrogen chloride is NCl3, and it is thermodynamically unstable. Both PCl3 and PCl5 are stable.

15.7 Use the VSEPR model to predict the probable shapes of (a) PCl4

+, (b) PCl4–, (c)

AsCl5?

(a) PCl4+? tetrahedron.

(b) PCl4–? A see-saw.

(c) AsCl5? A trigonal bipyramid.

15.8 Give balanced chemical equations for each of the following reactions: (a) oxidation of P4 with excess oxygen, (b) reaction of the product from part (a) with excess water, (c) reaction of the product from part (b) with a solution of CaCl2 and name the product?

(a) Oxidation of P4 with excess O2?

P4 + 5O2 → P4O10

(b) Reaction of the product from part (a) with excess H2O?

P4O10 + 6H2O → 4H3PO4

(c) Reaction of the product from part (b) with CaCl2?

2H3PO4(l) + 3CaCl2(aq) →

Ca3(PO4)2(s) + 6HCl(aq)

15.9 Starting with NH3(g) and other reagents of your choice, give the chemical equations and conditions for the synthesis of (a) HNO3, (b) NO2

– , (c) NH2OH, (d) N3

–?

(a) HNO3?

4NH3(aq) + 7O2(g) → 6H2O(g) + 4NO2(g)

High temperature

(b) NO2–?

2NO2(aq) + 2OH−(aq) →

NO2−(aq) + NO3

−(aq) + H2O(l)

(c) NH2OH? cold aqueous acidic solution

NO2−(aq) + 2HSO3

−(aq) + H2O(l) → NH3OH+(aq) + 2SO4

2−(aq)

(d) N3–? at elevated temperatures:

3NaNH2(l) + NaNO3 →

NaN3 + 3NaOH + NH3(g)

2NaNH2(l) + N2O → NaN3 + NaOH + NH3

15.10 Write the balanced chemical equation corresponding to the standard enthalpy of formation of P4O10(s). Specify the structure, physical state (s, l, or g), and allotrope of the reactants. Do either of the reactants differ from the usual practice of taking as reference state the most stable form of an element?

P4(s) + 5O2(g) → P4O10(s)

15.11 Without reference to the text, sketch the general form of the Frost diagrams for phosphorus (oxidation states 0 to +5) and bismuth (0 to +5) in acidic solution and discuss the relative stabilities of the +3 and +5 oxidation states of both elements?

(i) Bi(III) is much more stable than Bi(V), and (ii) P(III) and P(V) are both about equally stable.



15.12 Are reactions of NO2–

as an oxidizing agent generally faster or slower when pH is lowered? Give a mechanistic explanation for the pH dependence of NO2

– oxidations?

The rates of reactions in which nitrite ion is reduced are increased as the pH is lowered.

15.13 When equal volumes of nitric oxide (NO) and air are mixed at atmospheric pressure a rapid reaction occurs, to form NO2 and N2O4. However, nitric oxide from an automobile exhaust, which is present in the parts per million concentration range, reacts slowly with air. Give an explanation for this observation in terms of the rate law and the probable mechanism?

The rate law must be more than first order in NO concentration.

15.14 Give balanced chemical equations for the reactions of the following reagents with PCl5 and indicate the structures of the products: (a) water (1:1), (b) water in excess, (c) AlCl3, (d) NH4Cl?

(a) H2O? tetrahedral POCl3.

PCl5 + H2O → POCl3 + 2HCl

(b) H2O in excess?

2PCl5(g) + 8H2O(l) → 2H3PO4(aq) + 10HCl(aq)

(c) AlCl3? tetrahedral.

PCl5 + AlCl3 → [PCl4]+[AlCl4]−

(d) NH4Cl? Cyclic molecules or linear chain polymers

nPCl5 + nNH4Cl →

−[(N = P(Cl)2)n]− + 4nHCl

15.15 Use standard potentials (Resource section 3) to calculate the standard potential of the reaction of H3PO2 with Cu2+. Are HPO2

2−

and H2PO22− useful as oxidizing or

reducing agents?

Eº = 0.839 V.

HPO32– and H2PO2

– ions will be much better reducing agents than oxidizing agents.

15.16 Identify the compounds A, B, C, and D?

AAS B

C

D

Cl2

LiAlH4

Cl2/hv

3RMgBr

(a) A = AsCl3

(b) B = AsCl5 (c) C = AsR3 (d) D = AsH3

15.17 Sketch the two possible geometric isomers of the octahedral [AsF4Cl2]–

and explain how they could be distinguished by 19F–NMR?

The cis isomer gives two 19F signals and the trans isomer gives one signal.

15.18 Identify the nitrogen compounds A, B, C, D, and E?

(a) A = NO2

(b) B = HNO3; C = NO

(c) D = N2O4

(d) E= NO2

(e) F = NH4+



1 Use the Latimer diagr5.19 ams in Resource

R 16

ermine whether the decomposition of

ition of H2O2 is

ion of H2O2 is –

S16.2 obable structures of SO2F and with

xercise whether the following oxides are

mphoteric.

16.2 urce

section 3 to determine which species of N and P disproportionate in acid conditions?

The species of N and P that disproportionate+are N2O4, NO, N2O, NH3OH , H4P2O6, and P.

CHAPTE

elf-tests S

S16.1 DetH2O2 is spontaneous in the presence of either Br2 or Cl2?

The decomposthermodynamically favored in presence of Br–

In Presence of Cl–

The decompositthermodynamically favored in presence of Cl.

Pr -

(CH3) 3NSO2, and predict reactions-. OH

Both are trigonal pyramidal. OH- displaces F— or (CH3) 3N—.

es E16.1 Stat

acidic, basic, neutral, or amphoteric: CO2, P2O5, SO3, MgO, K2O, Al2O3, CO?

CO2, SO3, P2O5, and Al2O3 are aCO is neutral; MgO and K2O are acidic.

(a) Use standard potentials (Resosection 3) to calculate the standard potential of the disproportionation of H2O2

in acid solution. (b) Is Cr2+ a likely catalyst for the disproportionation of H2O2? (c) Given the Latimer diagram

O2 HO2--0.13 1.51

H2O2 in acidic solution, calculate Δr

ө for the

of H2O2 and

Cr ? Cr is not capable

tion of HO2? ΔrG° = 157 kJ. For the disproportionation of H2O2 (part (a)), ΔrG° = 103 kJ.

16.3

tronger.

or SO2 (which not react

16.5 pecies from the

agent: SO4 ,SO3 ,O3SO2SO3 ?

16.6 c

16.7 formula for Te(VI) in acidic

sible

ncrease its

16.8

ill disproportionate in

16.9

3 to predict whether ble in acidic or basic

16.10 ether any of the following will be

: VO , Fe , Cu , Co ?

16.11 F4]. Use hapes of the

cation and anion?

6.12 g) nic

product. (a) Write a balanced equation for

Gdisproportionation of hydrogen superoxide (HO2–

) into O2 and H2O2, and compare the result with its value for the disproportionation of H2O2?

(a) The disproportionation HO2? 1.068 V.

(b) Catalysis by 2+ 2+

of decomposing H2O2.

(c) The disproportiona

Which hydrogen bond would be stronger: S—H . . . O or O—H . . . S?

O–H hydrogen bonds are s

16.4 Which of the solvents ethylenediamine (which is basic and reducing) is acidic and oxidizing) might with (a) Na2S4, (b) K2Te3?

ethylenediamine is a better solvent than sulfur dioxide.

Rank the following sstrongest reducing agent to the strongest oxidizing 2– 2– 2–

S2O82– > SO4

2– > SO32–

Predict which oxidation states of Mn will be reduced by sulfite ions in basiconditions?

Mn (+VII, +VI, +V, +IV, +III) will be reduced by sulfite ions in basic solution.

(a) Give theaqueous solution and contrast it with the formula for S(VI). (b) Offer a plauexplanation for this difference?

(a) Formulas? H5TeO6– and HSO4

–

(b) An explanation? Tellurium is a larger element than sulfur and can icoordination number.

Use the standard potential data in Resource section 3 to predict which oxoanions of sulfur wacidic conditions?

S2O62−and S2O3

2−. Use the standard potential data in Resource section

2–SeO3 is more stasolut on? i

The SeO32– is marginally more stable in acid

solutions.

Predict whreduced by thiosulfate ions,S2O3

2–, in acidic conditions 2+ 3+ + 3+

Fe3+ and Co3+ will be reduced.

SF4 reacts with BF3 to form [SF3][BVSEPR theory to predict the s

SF4+ trigonal pyramidal, BF4

− tetrahedral.

1 Tetramethylammonium fluoride (0.70reacts with SF4 (0.81 g) to form an io



the reaction and (b) sketch the structure of the anion. (c) How many lines would be observed in the 19F-NMR spectrum of the anion?

(a) SF4 + (CH3)4NF → [(CH3)4N]+[SF5]−; (b) square pyramidal structure; (c) two F environm

ents. .13

of iodine is sodium iodate, O3. Which of the reducing agents

SO (aq) or Sn2+(aq) would seem practical

S17.2 r IF7?

and several polyhalides that

to [py–I–py]+, and describe

orbitals, one bonding, one

Exerci17.1 reference

erial, write out the halogens and noble y appear in the periodic table,

16 Identify the sulfur-containing compounds A, B, C, D, E, and F?

A = S2Cl2, B = S4N4, C = S2N2, D = K2S2O3,

E = S2O62−, F = SO2.

CHAPTER 17

Self-tests

S17.1 One source NaI

2

from the standpoints of thermodynamic feasibility and plausible judgements about cost? Standard potentials are given in Resource section 3?

Both. SO2 is cheaper.

Predict the 19F-NMR pattern fo

2 resonances.

S17.3 From the perspective of structure bonding, indicateare analogoustheir bonding?

Examples include I3–, IBr2

–, ICl2–, and IF2

–. The three centers contribute four electrons to three molecular nonbonding, and one antibonding.

ses Preferably without consulting matgases as theand indicate the trends in (a) physical state (s, l, or g) at room temperature and pressure, (b) electronegativity, (c) hardness of the halide ion, (d) color?

Physical

Electronegativity

Hardness of

Color

state

halide ion

F2

hest (4.0)

light yellow

gas hig hardest

Cl2

gas lower softer

yellow-

green

Br2

liquid

lower softer

darkred-brown

I2 solid

lowest softest

darkviolet

He

gas Colorless

Ne

gas colorless

A gas or

colrless

Kr

gas colorless

Xe

gas colorless

17.2 Describe how the halogens are recovered from their naturally occurring halides and rationalize the approach in terms of

CaSO4 + 2 HF

2 KF → 2 K+HF2−

2 + 2 KF

Cl2 → X2 + 2 Cl− (X− = Br−,

17.3 ow thhe dir

usion of the ions. Give the chemical

) → Cl2(g) + 2 e

cathod

H2O(l)

standard potentials. Give balanced chemical equations and conditions where appropriate?

For F:

F2 + H2SO4 →

2 HF +

2 K+HF2− + electricity → F2 + H

For Cl:

2 Cl− + 2 H2O + electricity → Cl2 + H2 + 2 OH−

For Br and I:

2 X− + I−)

Sketch a choralkali cell. Sh e half-cell reactions and indicate t ection of diffequation for the unwanted reaction that would occur if OH– migrated through the membrane and into the anode compartment?

A drawing of the cell is shown in Figure 17.3.

anode: 2 Cl−(aq −

e: 2 H2O(l) + 2 e− → 2 OH−(aq) + H2(g)

- unwanted reaction:

2 OH−(aq) + Cl2(aq) → ClO−(aq) + Cl−(aq) +



17.4 Skea dihalogen ribe its role

tch the form of the vacant s* orbital of molecule and desc

in the Lewis acidity of the dihalogens?

Since the 2σu* antibonding orbital is the LUMO for a X2 molecule, it is the orbital that

17.5 le of oxidizing H2O to O2?

17.6 at –129˚C void of Lewis basicity. By

.

effect in NF3

17.7 ogy between halogens and pseudohalogens write: (a) the balanced

→ CN−(aq) +

(b aqueous aci

Mn (aq) + 2 H2O(l)

(c) l cyan ins

17.8 3 reacts with 0.93 g of [(CH3)4N]F to form a product X, (a) identify X, (b) use the VSEPR model to predict the

accepts the pair of electrons from a Lewis base.

Which dihalogens are thermodynamically capab

Cl2 and F2. Nitrogen trifluoride, NF3, boilsand is decontrast, the lower molar mass compound NH3 boils at –33˚C and is well known as a Lewis base. (a) Describe the origins of this very large difference in volatility. (b) Describe the probable origins of the difference in basicity?

(a) Difference in volatility? Ammonia exhibits hydrogen bonding

(b) Explain the difference in basicity? The strong electron-withdrawingreduces the basicity.

Based on the anal

equation for the probable reaction of cyanogen, (CN)2, with aqueous sodium hydroxide, (b) the equation for the probable reaction of excess thiocyanate with the oxidizing agent MnO2(s) in acidic aqueous solution, (c) a plausible structure for trimethylsilyl cyanide?

(a) The reaction of NCCN with NaOH?

NCCN(aq) + 2 OH−(aq) NCO−(aq) + H2O(l)

) The reaction of SCN– with MnO2 ind?

2SCN−(aq) + MnO2(s) + 4 H+(aq) → (SCN)2(aq) + 2+

The structure of trimethylsilyide? Trimethylsilyl cyanide conta

an Si–CN single bond.

Given that 1.84 g of IF

shapes of IF3 and the cation and anion in X, (c) predict how many 19F-NMR signals would be observed in IF3 and X?

a). X = IF4N(CH3)4 b) Different possible arrangements of IF3:

The shape of anion IF4 is Square planar. The shape of cation (CH3)4N+ is Tetrahedral.

c)

7.9 l to predict the shapes 5 3 ClF6]+?

7.10

17.11 of the complexes l3F3. Indicate how many

ents would be indicated

–

.IF3, two. IF4–, one.

1 Use the VSEPR modeof SbCl , FClO , and [

SbCl5 is trigonal bipyramidal, FClO3 is pyramidal, and ClF6 is octahedral.

1 Indicate the product of the reaction between ClF5 and SbF5?

[ClF4]+[SbF6] –.

Sketch all the isomersMCl4F2 and MCfluorine environmin the 19F-NMR spectrum of each isomer?

MCl4F2, the cis isomer, 1; the trans isomer, 2. MCl3F3, the fac isomer, 1 the mer isomer, 2.



17.12 (a) Use the VSEPR model to predict the probable shapes of [IF6]+ and IF7. (b) Give a plausible chemical equation for the preparation of [IF6][SbF6]?

(a) The structures of [IF6]+ and IF7? IF6+,

octahedral; IF7, pentagonal bipyramid.

(b) The preparation of [IF6] [SbF6]?

IF7 + SbF5 → [IF6][SbF6]

7.14 al, Cs

17.15 the following s likely to make liquid BrF3 a

a stronger Lewis

he acidity or basicity 3

17.16 rf IF5

+? Two resonances.

17.17

xplain ) F2,

Since SbF5 cannot be oxidized, it 3.

reasoning?

2 IBr + Br−

17.19 not?

arge cations stabilize large, unstable anions.

17.13 Predict the shape of the doubly chlorine–bridged I2Cl6 molecule by using the VSEPR model and assign the point group?

A planar dimer (I2Cl6).

1 Predict the structure and identify the point group of ClO F? trigonal pyramid2

symmetry.

Predict whether each ofsolutes istronger Lewis acid orbase: (a) SbF5, (b) SF6, (c) CsF?

(a) SbF5? increases the acidity of BrF3.

(b) SF6? No effect on tof BrF .

(c) CsF? Increases the basicity of BrF3.

Predict the appea ance of the 19F-NMR spectrum o

. Predict whether each of the following compounds is likely to be dangerously explosive in contact with BrF3 and eyour answer: (a) SbF5, (b) CH3OH, (c(d) S2Cl2?

(a) SbF5?will not form an explosive mixture with BrF

(b) CH3OH? Methanol, being an organic compound, is readily oxidized by strong oxidants.

(c) F2? No.

(d) S2Cl2? S2Cl2 will be oxidized to higher valent sulfur fluorides.

17.18 The formation of Br3–from a

tetraalkylammonium bromide and Br2 is only slightly exoergic. Write an equation (or NR for no reaction) for the interaction of [NR4][Br3] with I2 in CH2Cl2 solution and give your

Br3− + I2 →

Explain why CsI3(s) is stable with respect to the elements but NaI3(s) is

L

17.20 Write plausible Lewis structures for (a) ClO2 and (b) I2O6 and predict their shapes and the associated point group?

(a).

(b) I2O5?

17.21 (a) Give the formulas and the probable relative acidities of perbromic acid and periodic acid. (b) Which is the more stable?

rmulas are HBrO4 and H5IO6. The in iodine’s ability to expand its

coordination shell.

(b) Relative stabilities? Periodic acid is thermodynamically more stable.

17.22 (a) Describe the expected trend in the standard potential of an oxoanion in a

17.23

cission.

17.24 readily chloric acid

a mechanistic erence?

(a) The fodifference lies

solution with decreasing pH. (b) Demonstrate this phenomenon by calculating the reduction potential of ClO4

–

at pH = 7 and comparing it with the tabulated value at pH = 0?

(a) The expected trend? E decreases as the pH increases.

(b) E at pH 0 and pH 7 for ClO4–?

Eº = 1.201 V (see Appendix 2).

At pH 7, V = 0.788 V

With regard to the general influence of pH on the standard potentials of oxoanions, explain why the disproportionation of an oxoanion is often promoted by low pH?

Low pH results in a kinetic promotion: protonation of an oxo group aids oxygen–halogen bond s

Which oxidizing agent reacts morein dilute aqueous solution, peror periodic acid? Giveexplanation for the diff

Periodic acid.



17.25 lly –

dard potentials.) (b)

7.26 following compounds present

4, (

losion hazard.

)

CHAP

f-tests

S18.1

e production of perxenate

O q OH−(aq) → XeO64−(aq) + Xe(g)

+ O2(g) + 2 H2O(l)

lain why helium is present in low concentration in the atmosphere even

tive decay.

8.2 Which of the noble gases would you choose (a) The lowest-temperature

igerant? Helium.

) XeF2? Xe and F2

ut have a large

) → XeF6(s)

18.4 (b) XeO2F2?

(a) For which of the following anions is disproportionation thermodynamicafavourable in acidic solution: OCl2, ClO2 , ClO2

– , and ClO4

– ? (If you do not know the

properties of these ions, determine them from a table of stanFor which of the favourable cases is the reaction very slow at room temperature?

The rates of disproportionation are probably HClO > HClO2 > ClO3

–.

1 Which of the an explosion hazard? (a) NH4ClO4, (b) Mg(ClO4)2, (c) NaClO d) [Fe(H2O)6][ClO4]2. Explain your reasoning?

(a) NH4ClO4? Ammonium perchlorate is a dangerous compound, since the N atom of the NH4

+ ion is in its lowest oxidation state and can be oxidized.

(b) Mg(ClO4)2? Not an explosion hazard.

(c) NaClO4? Not an exp

(d) [Fe(H2O)6] [ClO4]2? An explosion hazard, since Fe(II) can be oxidized to Fe(III).

17.27 Use standard potentials to predict which of the following will be oxidized by ClO– ions in acidic conditions: (a) Cr3+, (b) V3+, (c) Fe2+, (d) Co2+?

(a) Cr3+? No. (b) V3+? Yes. (c) Fe2+? Yes. (dCo2+? No.

TER 18 Sel

Write a balanced equation for the decomposition of xenate ions in basic solution for thions, xenon, and oxygen

4−(a ) + 2 2 HXe

xercises E

18.1 Exp

though it is the second most abundant element in the universe?

Helium present in today’s atmosphere is the product of ongoing radioac

1as refr

(b) An electric discharge light source requiring a safe gas with the lowest ionization energy? Xenon.

(c) The least expensive inert atmosphere? Argon.

18.3 By means of balanced chemical equations and a statement of conditions, describe a suitable synthesis of (aat 400ºC, or photolyze Xe and F2 in glass:

Xe(g) + F2(g) → XeF2(s)

(b) XeF6? High temperature, bexcess of F2:

Xe(g) + 3 F2(g

(c) XeO3?

XeF6(s) + 3 H2O(l) → XeO3(s) + 6 HF(g)

Draw the Lewis structures of (a) XeOF4? (c) XeO6

4−?

18.5 ive the formula and describe the structure of a noble gas species that is isostructural with (a) ICl4

−? XeF4

(b) IBr2–? Linear geometry. Isostructural

with XeF2.

(c) BrO3–? Trigonal pyramidal geometry.

Isostructural with XeO3.

(d) ClF? Isostructural with the cation XeF+.

G

18.6 (a) Give a Lewis structure for XeF7–?

ith

bipyramid.

(b) Speculate on its possible structures by using the VSEPR model and analogy wother xenon fluoride anions? Pentagonal



18.7 Use molecular orbital theory to calculate the bond order of the diatomic species E2

+ with E=He and Ne? He2

2+ = 0.5, Ne22+ = 0.5.

18.8 Identify the xenon compounds A, , C, D, and E?

4

other lines b ted around the central

CHAP

19.1 Refer to the appropriate Latimer diagram the

e and formula of the species

n?

19.2 Suggest a use for molybdenum(IV) sulfide makes use of its solid-state structure. onalize your suggestion? MoS2 used as

S19.3

ntaining PPh3?

olecular species such as Re3Cl123–

Exerci19.1

es of the d block,

tion

ose for which

oup oxidation number is not achieved by N?

B

A = XeF2(g) B = [XeF]+[MeBF3]

− C = XeF6 D = XeO3 E = XeF (g).

18.9 Predict the appearance of the 129Xe-NMR spectrum of XeOF3

+. 1:3:3:1 quartet

18.10 Predict the appearance of the 19F-NMR spectrum of XeOF4.

Strong c nte ral line, two symmetrically distri uline.

TER 19 Self-tests

Sin Resource section 3 and identify oxidation statthat is thermodynamically favoured when an acidic aqueous solution of V2+ is exposed to oxyge

+4; VO2+

SthatRatia lubricant. The slipperiness of MoS2 is because of the ease with which one layer can glide over another.

Describe the probable structure of the compound formed when Re3Cl9 is dissolved in a solvent co

Discrete mor Re3Cl9(PPh3)3 are formed.

ses Without reference to a periodic table, sketch the first seriincluding the symbols of the elements. Indicate those elements for which the group oxidation number is common by C, those for which the group oxidanumber can be reached but is a powerful oxidizing agent by O, and th

the gr

19.2

As atoms become heavier, more energy is needed to vaporize them.

19.3 State the trend in the stability of the group oxidation state on descending a group of metallic elements in the d block. Illustrate the trend using standard potentials in

in as you descend a group.

19.4 action) and

Fe2+(aq)

(aq) →? − 3+ 1H2O →

19.5 q), is more presence of

ith the tren s Pe al

More

Hardnes in the d block.

19.6

e or fluorite n of

example?

Explain why the enthalpy of sublimation of Re(s) is significantly greater than that of Mn(s)?

acidic solution for Groups 5 and 6. The group oxidation number increases stability

For each part, give balanced chemical quations or NR (for no ree

rationalize your answer in terms of trends in oxidation states.

(a) Cr2+ (aq) + Fe3+ (aq) →?

Cr2+(aq) + Fe3+(aq) → Cr3+(aq) +

(b) CrO42– (aq) + MoO2 (s) →?

2 CrO42− + 3 MoO2(s) + 10 H+ →

2 Cr3+ + 3 H2MoO4 + 2 H2O

(c) MnO4– (aq) + Cr3+

6 MnO4 + 10Cr + 1

6Mn2+ + 5Cr2O72− + 22 H+

(a) Which ion, Ni2+(aq) or Mn2+(alikely to form a sulfide in the H2S? (b) Rationalize your answer w

ds in hard and soft character acrosriod 4. (c) Give a balanced chemic

equation for the reaction.

likely for Ni2+ to form a sulphide.

s decreases from left to right

Ni2+(aq) + H2S(aq) → NiS(s) + 2 H+(aq)

Preferably without reference to the text (a) write out the d block of the periodic table, (b) indicate the metals that form difluorides with the rutilstructures, and (c) indicate the regiothe periodic table in which metal-metal bonded halide compounds are formed, giving one



Metal–metal bonded halide compounds are found within bold border. Example: Sc5Cl6.

19.7

ic solution at +0.2 V is made strongly basic at the same potential. Write a balanced equation for each of the successive reactions when this same complex at pH = 6 and +0.2 V is exposed to progressively more oxidizing environments up to +1.0 V

–

tions llowing

s for your q) in

er an inert O + (a

(Hg2+ (aq) + Cd(s) Cd2+ (aq)

19.9

r answers:

]

e cis-o 4 , has a

an apical oxo

(b) significantly

19.11

ng

3)4]? σ2π4δ2 configuration,

2CC2H5)4]? σ2π4δ2,chromium–

19.12 following

19.13 oate to aqueous

Write a balanced chemical equation for the reaction that occurs when cis-[RuLCl(OH2)]+ (see Fig. 19.9) in acid

. Give other examples and a reason for the redox state of the metal center affecting the extent of protonation of coordinated oxygen.

cis-[RuIILCl(OH2)]+ + Ox + OH− → cis-[RuIIILCl(OH)]+ + Red + H2O

cis-[RuIILCl(OH2)]+ + H2O → cis-[RuIIILCl(OH)]+ + H3O+ + e−

cis-[RuIIILCl(OH2)]+ + H2O → cis-[RuIVLCl(OH)]+ + H3O+ + e−

As oxidation state of the metal increases, ability to accept electron density from an OHor O2– ligand increases.

19.8 Give plausible balanced chemical reac(or NR for no reaction) for the focombinations, and state the basianswer: (a) MoO4

2– (aq) plus Fe2+ (aacidic solution? No reaction.

(b) The preparation of [Mo6O19]2−(aq) from K2MoO4(s)? 6 MoO4

2−(aq) + 10 H+(aq) → [Mo6O19]2−(aq) + 5 H2O(l)

(c) ReCl5 (s) plus KMnO4(aq)?

5 ReCl5(s) + 2 MnO4−(aq) + 12 H2O(l) →

5 ReO4−(aq) + 2 Mn2+(aq) + 25 Cl−(aq) +

24 H+(aq)

(d) MoCl2(s) plus warm HBr(aq)? 6 MoCl2(s) + 2 Br−(aq) → [Mo6Cl12]2−(aq) + Br2(aq)

(e) TiO(s) with HCl(aq) undatmosphere? 2Ti (s) + 6H q) 2Ti3+ + H2(g) + 2H2O(l) E˚ = +0.37 V.

f) Cd(s) added to Hg2+(aq)? Hg(l) +

Speculate on the structures of the following species and present bonding models to justify you

(a) [Re(O)2 (py)4]+, (b) [V (O)2 (ox)2]3–, (c) [Mo(O)2(CN)4

4–, (d) [VOCl4]2–? The first three are dioxo complexes and will hav

metal structural units. (d) VOCl 2–dioxsquare-pyramidal structure withligand.

19.10 Which of the following are likely to have structures that are typical of (a) predominantly ionic,covalent, (c) metal-metal bonded compounds: NiI2, NbCl4, FeF2, PtS, and WCl2? Rationalize the differences and speculate on the structures? (a) NiI2? ionic compound with significant degree of covalent character

(b) NbCl4? significantly covalent.

(c) FeF2? ionic.

(d) PtS ? significant amount of covalent character.

(e) WCl2 ? metal–metal bonding.

Indicate the probable occupancy of s, p, and d bonding and antibonding orbitals, and the bond order for the followitetragonal prismatic complexes?

(a) [Mo2(O2CCHmolybdenum-molybdenum quadruple bond.

(b) [Cr2(Ochromium quadruple bond.

(c) [Cu2(O2CCH3)4]? σ2π4δ2δ*2π*4σ*2 configuration, no metal–metal bond in this molecule.

Explain the differences in the redox couples, measured at 25oC?

Higher oxidation states become more stable . on descending a group

Addition of sodium ethansolutions of Cr(II) gives a red diamagnetic product. Draw the structure of the product, noting any features of interest?



Cr OO

OO O

OO

CrO

19.14 Consider the two ruthenium complexes in

Table 19.8. Using the bonding scheme depicted in Figure 19.19, confirm the bonding orders and electron configurations given in the table.

[Ru2Cl2(ClCO2)4]- :

2.5 +, mixed valence at the Ru center, in line with the energy level scheme [0.5 (8-3)].

[Ru2 (CH3COCH3)2(ClCO2)4]: 2+, the bond order of 2.0 should also be expected [0.5x(8-4)].


S20.1 What is the LFSE for both high- and low-spin d7 configurations? A high-spin d7 configuration is t2

5geg

2. High spin, 0.8 Δo.

S20.2 The magnetic moment of the complex [Mn(NCS)6]4– is 6.06μB. What is its electron configuration? t2g

3eg2

S20.3 Account for the variation in lattice

enthalpy of the solid fluorides in which each metal ion is surrounded by an octahedral array of F– ions: MnF– (2780 kJ mol–1), FeF2 (2926 kJ mol–1), CoF2 (2976 kJ mol–1), NiF2 (3060 kJ mol–1), and ZnF2 (2985 kJ mol–1).

If it were not for ligand field stabilization energy (LFSE), MF2 lattice enthalpies would increase from Mn(II) to Zn(II).

S20.4 Suggest an interpretation of the photoelectron spectra of [Fe(C5H5)2] and [Mg(C5H5)2] shown in Fig. 20.18?

In the spectrum of [Mo(CO)6], the ionization energy around 8 eV was attributed to the t2g electrons that are largely metal-based. The

differences in the 6–8 eV region can be attributed to the lack of d electrons for Mg(II).

S20.5 What terms arise from a p1d1 configuration? 1F and 3F terms are possible.

S20.6 Identifying ground terms (Hint: Because d9 is one electron short of a closed shell with L = 0 and S = 0, treat it on the same footing as a d1 configuration.) (a) 2p2? 3P (called a “triplet P” term). (b) 3d9? 2D (called a “doublet D” term).

S20.7 What terms in a d2 complex of Oh symmetry correlate with the 3F and 1D terms of the free atom? An F terms are 3T1g, 3T2g, and 3A2g. Similarly, D terms are 1T2g and 1Eg.

S20.8 Use the same Tanabe-Sugano diagram to predict the energy of the first two spin-allowed quartet bands in the spectrum of [Cr(OH2)6]3+ for which Δo = 17 600 cm-1 and B = 700 cm-1.

17500 cm–1 and 22400 cm–1.

S20.9 The spectrum of [Cr(NCS)6]3- has a very weak band near 16 000 cm-1, a band at 17 700 cm-1 with εmax= 5160 dm3 mol-1 cm-1, a band at 23 800 cm-1 with εmax= 5130 dm3 mol-1 cm-1, and a very strong band at 32 400 cm-1. Assign these transitions using the d3 Tanabe-Sugano diagram and selection rule considerations. (Hint: NCS- has low-lying π* orbitals.)

(i) The very low intensity of the band at 16,000 cm–1 is a clue that it is a spin-forbidden transition, probably 2Eg ← 4A2g.

(ii) Spin-allowed but Laporte-forbidden bands typically have ε ~ 100 M–1 cm–1, so it is likely that the bands at 17,700 cm–1 and 23,800 cm–1 are of this type.

(iii) The band at 32,400 cm–1 is probably a charge transfer band, since its intensity is too high to be a ligand field (d–d) band.

Exercises 20.1 Determine the configuration (in the form

t2g x egy or ext2

y, as appropriate), the number of unpaired electrons, and the ligand-field stabilization energy in terms of ΔO or ΔT and P for each of the following complexes using the spectrochemical series to decide, where relevant, which are likely to be high-spin and which low-spin.

(a) [Co(NH3)6]3+? d6, low spin, no unpaired electrons, LFSE = 2.4Δ0. (b) [Fe(OH2)6]2+? d6, high spin, four unpaired electrons, LFSE = 0.4Δ0.



(c) [Fe(CN)6]3–? d5, low spin, one unpaired electron. LFSE is 2.0Δ0.

(d) [Cr(NH3)6]3+? d3, three unpaired electrons, low spin, LFSE = 1.2Δ0.

(e) [W(CO)6]? d6, low spin, no unpaired electrons, LFSE = 2.4Δ0.

(f) Tetrahedral [FeCl4]2−? d6, high spin, four unpaired electrons, LFSE = 0.6 ΔT.

(g) Tetrahedral [Ni(CO)4]? d10, 0 unpaired electrons, LFSE = 0.

20.2 Both H- and P(C6H5)3 are ligands of similar field strength, high in the spectrochemical series. Recalling that phosphines act as π acceptors, is π-acceptor character required for strong-field behaviour? What orbital factors account for the strength of each ligand?

No.

Thus there are two ways for a complex to develop a large value of Δ0, by possessing ligands that are π-acids or by possessing ligands that are strong σ-bases (or both).

20.3 Estimate the spin-only contribution to the magnetic moment for each complex in Exercise 20.1. Spin-only contributions are:

complex N μSO = [(N)(N + 2)]1/2 [Co(NH3)6]3+ 0 0 [Fe(OH2)6]2+ 4 4.9 [Fe(CN)6]3– 1 1.7 [Cr(NH3)6]3+ 3 3.9 [W(CO)6] 0 0 [FeCl4]2– 4 4.9 [Ni(CO)4] 0 0

20.4 Solutions of the complexes [Co(NH3)6]2+,

[Co(OH2)6]2+ (both Oh), and [CoCl4]2- are colored. One is pink, another is yellow, and the third is blue. Considering the spectrochemical series and the relative magnitudes of ΔT and ΔO, assign each color to one of the complexes. [CoCl4]2– is blue, [Co(NH3)6]2+ is yellow, [Co(OH2)6]2+ is pink.

20.5 For each of the following pairs of complexes, identify the one that has the larger LFSE:

(a) [Cr(OH2)6]2+ or [Mn(OH2)6]2+ (b) [Mn(OH2)6]2+ or [Fe(OH2)6]3+ (c) [Fe(OH2)6]3+ or [Fe(CN)6]3- (d) [Fe(CN)6]3– or [Ru(CN)6] (e) tetrahedral [FeCl4]2–or tetrahedral [CoCl4]2–

(a) The chromium complex.

(b) The Fe3+ complex.

(c) [Fe(CN)6]3–

(d) The ruthenium complex.

(e) The Co2+ complex.

20.6 Interpret the variation, including the overall trend across the 3d series, of the following values of oxide lattice enthalpies (in kJ mol–1). All the compounds have the rock-salt structure: CaO (3460), TiO (3878), VO (3913), MnO (3810), FeO (3921), CoO (3988), NiO (4071)? There are two factors that lead to the values: (i) decreasing ionic radius from left to right across the d block, and (ii) LFSE.

.20.7 A neutral macrocyclic ligand with four donor atoms produces a red diamagnetic low-spin d8 complex of Ni(II) if the anion is the weakly coordinating perchlorate ion. When perchlorate is replaced by two thiocyanate ions, SCN–, the complex turns violet and is high-spin with two unpaired electrons. Interpret the change in terms of structure? Shift from square planar to tetragonal complex.

20.8 Bearing in mind the Jahn-Teller theorem, Predict the structure of [Cr(OH2)6]2+? Elongated octahedron.

20.9 The spectrum of d1 Ti3+(aq) is attributed to a single electronic transition eg ← t2g. The band shown in Fig. 20.3 is not symmetrical and suggests that more than one state is involved. Suggest how to explain this observation using the Jahn-Teller theorem? The electronic excited state of Ti(OH2)6

3+ has the configuration t2g0eg

1, and so the excited state possesses eg degeneracy. Therefore, the “single” electronic transition is really the superposition of two transitions, one from an Oh ground-state ion to an Oh excited-state ion, and a lower energy transition from an Oh ground-state ion to a lower energy distorted excited-state ion.

20.10 Write the Russell–Saunders term symbols for states with the angular momentum quantum numbers (L,S):

(a) L = 0, S = 5/2? 6S

(b) L = 3, S = 3/2? 4F

(c) L = 2, S = 1/2? 2D

(d) L = 1, S = 1? 3P

20.11 Identify the ground term from each set of terms: (a) 3F, 3P, 1P, 1G? 3F.

(b) 5D, 3H, 3P, 1G, 1I? 5D

(c) 6S, 4G, 4P, 2I? 6S



20.12 Give the Russell-Saunders terms of the configurations and identify the ground term? (a) 4 s1? 2S.

(b) 3p2? 1D, 3P, 1S. The ground term is 3P.

20.13 The gas-phase ion V3+ has a 3F ground term. The 1D and 3P terms lie, respectively, 10 642 and 12 920 cm–1 above it. The energies of the terms are given in terms of Racah parameters as E(3F) = A - 8B, E(3P) = A + 7B, E(1D) = A - 3B + 2C. Calculate the values of B and C for V3+. B = (12920 cm–

1)/(15) = 861.33 cm–1 and C = 3167.7 cm–1.

20.14 Write the d-orbital configurations and use the Tanabe–Sugano diagrams (Resource section 6) to identify the ground term of (a) Low-spin [Rh(NH3)6]3+? 1A1g.

(b) [Ti(H2O)6]3+? 2T2g.

(c) High-spin [Fe(H2O)6]3+? 6A1g.

20.15 Using the Tanabe-Sugano diagrams in Resource section 6, estimate ΔO and B for (a) [Ni(H2O)6]2+ (absorptions at 8500, 15400 and 26000 cm-1) ? Δ0 = 8500 cm–1 and B ≈ 770 cm–1.

(b) [Ni(NH3)6]2+ (absorptions at 10750, 17500 and 28200 cm-1)? Δ0 = 10,750 cm–1 and B ≈ 720 cm–1.

20.16 The spectrum of [Co(NH3)6]3+ has a very weak band in the red and two moderate intensity bands in the visible to near-UV. How should these transitions be assigned? The first two transitions listed above correspond to two low-spin bands. The very weak band in the red corresponds to a spin-forbidden transition.

20.17 Explain why [FeF6]3- is colorless whereas [CoF6]3– is colored but exhibits only a single band in the visible. No spin-allowed transitions are possible for the Fe3+; the complex is expected to be colorless. The d6 Co3+ ion in [CoF6]3– is also high spin, but in this case a single spin-allowed transition makes the complex colored and gives it a one-band spectrum.

20.18 The Racah parameter B is 460 cm–1 in [Co(CN)6]3- and 615 cm-1 in [Co(NH3)6]3+. Consider the nature of bonding with the two ligands and explain the difference in nephelauxetic effect? Ammonia and cyanide ion are both σ-bases, but cyanide is also a π-acid.

20.19 An approximately ‘octahedral’ complex of Co(III) with ammine and chloro ligands gives two bands with εmax between 60 and 80 dm3 mol–1 cm–1, one weak peak with

εmax52 dm3 mol-1 cm-1, and a strong band at higher energy with εmax= 2 x 104 dm3 mol–1 cm–1. What do you suggest for the origins of these transitions? First, the intense band at relatively high energy is undoubtedly a spin-allowed charge-transfer transition. The two bands with εmax = 60 and 80 M–1 cm–1 are probably spin-allowed ligand field transitions. The very weak peak is most likely a spin-forbidden ligand field transition.

20.20 Ordinary bottle glass appears nearly colorless when viewed through the wall of the bottle but green when viewed from the end so that the light has a long path through the glass. The color is associated with the presence of Fe3+ in the silicate matrix. Suggest which transitions are responsible for the color? The faint green color, which is only observed when looking through a long pathlength of bottle glass, is caused by spin-forbidden ligand field transitions.

20.21 Solutions of [Cr(OH2)6]3+ ions are pale blue–green but the chromate ion, CrO4

2–, is an intense yellow. Characterize the origins of the transitions and explain the relative intensities. The blue-green color of the Cr3+ ions in [Cr(H2O)6]3+ is caused by spin-allowed but Laporte-forbidden ligand field transitions. The relatively low molar absorption coefficient is the reason that the intensity of the color is weak. The oxidation state of chromium in dichromate dianion is Cr(VI); the intense yellow color is due to LMCT transitions.

20.22 Classify the symmetry type of the d orbital in a tetragonal C4v symmetry complex, such as [CoCl(NH3)5] –, where the Cl lies on the z-axis. (a) Which orbitals will be displaced from their position in the octahedral molecular orbital diagram by π interactions with the lone pairs of the Cl– ligand? (b) Which orbital will move because the Cl– ligand is not as strong a base as NH3? (c) Sketch the qualitative molecular orbital diagram for the C4v complex. The Cl atom lone pairs of electrons can form π molecular orbitals with dxz and dyz. These metal atomic orbitals are π-antibonding MOs, and so they will be raised in energy.



20.23 Consider the molecular orbital diagram for a tetrahedral complex (based on Fig. 20.7) and the relevant d-orbital configuration and show that the purple color of MnO4

- ions cannot arise from a ligand-field transition. Given that the wavenumbers of the two transitions in MnO4

- are 18 500 and 32 200 cm–1, explain how to estimate ΔT from an assignment of the two charge-transfer transitions, even though ΔT cannot be observed directly. The oxidation state of manganese in permanganate anion is Mn(VII), which is d0. Therefore, no ligand field transitions are possible.

The difference in energy between the two transitions, E(t2) – E(e) = 13700 cm–1, is just equal to ΔT.

20.24 The lowest energy band in the spectrum of [Fe(OH2)6]3+ (in 1M HClO4) occurs at lower energy than the equivalent transition in the spectrum of [Mn(OH2)6]2+. Explain why this is.

The extra charge of the iron complexes keeps the eg and t2g levels close


S21.1 Calculate the second-order rate constant

for the reaction of trans-[PtCl(CH3) (PEt3)2] with NO2

– in MeOH, for which npt = 3.22 ? k2(NO2

−) = 100.71 = 5.1 M−1s−1

S21.2 Given the reactants PPh3, NH3, and [PtCl4]2-, Propose efficient routes to both cis- and trans- [PtCl2(NH3)(PPh3)]?

S21.3 Use the data in Table 21.8 to estimate an appropriate value for KE and calculate kr2 for the reactions of V(II) with Cl- if the observed second-order rate constant is 1.2x102 dm3 mol–1 s–1. KE = 1 M–1, k = 1.2 × 102 s–1.

Exercises 21.1 The rate constants for the formation of

[CoX(NH3)5]2+ from [Co(NH3)5OH2]3+ for X = Cl2, Br2, N3

–, and SCN¯ differ by no more than a factor of two. What is the mechanism of the substitution? Dissociative.

21.2 If a substitution process is associative, why may it be difficult to characterize an aqua ion as labile or inert? The rate of an associative process depends on the identity of the entering ligand and, therefore, it is not an inherent property of [M(OH2)6]n+.

21.3 The reactions of Ni(CO)4 in which phosphines or phosphites replace CO to give Ni(CO)3L all occur at the same rate regardless of which phosphine or phosphite is being used. Is the reaction d or a? d.

21.4 Write the rate law for formation of [MnX(OH2)5]1 from the aqua ion and X–. How would you undertake to determine if the reaction is d or a?

rate = (kKE[Mn(OH2)62+][X−])/(1 + KE[X−])

21.5 Octahedral complexes of metal centers with high oxidation numbers or of d metals of the second and third series are less labile than those of low oxidation number and d metals of the first series of the block. Account for this observation on the basis of a dissociative rate-determining step.

Metal centers with high oxidation numbers have stronger bonds to ligands than metal centers with low oxidation numbers. Furthermore, period 5 and 6 d-block metals have stronger metal ligand bonds.

21.6 A Pt(II) complex of tetramethyldiethylenetriamine is attacked by Cl- 105 times less rapidly than the diethylenetriamine analogue. Explain this observation in terms of an associative rate-determining step. The greater steric hindrance.

21.7 The rate of loss of chlorobenzene, PhCl, from [W(CO)4L(PhCl)] increases with increase in the cone angle of L. What does this observation suggest about the mechanism? The mechanism is dissociative.

21.8 The pressure dependence of the replacement of chlorobenzene (PhCl) by piperidine in the complex [W(CO)4(PPh3)(PhCl)] has been studied. The volume of activation is found to be 111.3 cm3 mol–1. What does this value suggest about the mechanism? Mechanism must be dissociative.



21.9 Does the fact that [Ni(CN)5]3– can be

isolated help to explain why substitution reactions of [Ni(CN)4]2– are very rapid? For a detectable amount of [Ni(CN)5]3– to build up in solution, the forward rate constant kf must be numerically close to or greater than the reverse rate constant kr.

21.10 Reactions of [Pt(Ph)2(SMe2)2] with the bidentate ligand 1,10-phenanthroline (phen) give [Pt(Ph)2phen]. There is a kinetic pathway with activation parameters Δ‡H = 1101 kJ mol–1 and Δ‡S = 142 J K–1 mol–1. Propose a mechanism.

A possible mechanism is loss of one dimethyl sulfide ligand, followed by the coordination of 1,10-phenanthroline.

21.11 A two-step synthesis for cis- and trans-[PtCl2(NO2) (NH3)] – start with [PtCl4]2–?

21.12 How does each of the following affect the rate of square-planar substitution reactions? (a) Changing a trans ligand from H to Cl? Rate decreases.

(b) Changing the leaving group from Cl– to I–? The rate decreases.

(c) Adding a bulky substituent to a cis ligand? The rate decreases.

(d) Increasing the positive charge on the complex? Rate increase.

21.13 The rate of attack on Co(III) by an entering group Y is nearly independent of Y with the spectacular exception of the rapid reaction with OH–. Explain the anomaly. What is the implication of your explanation for the behaviour of a complex lacking Brønsted acidity on the ligands? (i) The general trend: octahedral Co(III) complexes undergo dissociatively activated ligand substitution.

(ii) The anomalously high rate of substitution by OH– signals an alternate path, that of base hydrolysis.

(iii) The implication is that a complex without protic ligands will not undergo anomalously fast OH– ion substitution.

21.14 Predict the products of the following reactions: (a) [Pt(PR3)4]2+ + 2Cl–? cis-[PtCl2(PR3)2].

(b) [PtCl4]2– + 2PR3? trans-[PtCl2(PR3)2].

(c) cis-[Pt(NH3)2(py)2]2+ + 2Cl–? trans-[PtCl2(NH3)(py)].

21.15 Put in order of increasing rate of substitution by H2O the complexes: [Co(NH3)6]3+, [Rh(NH3)6]3+, [Ir(NH3)6]3+, [Mn(OH2)6]2+, [Ni(OH2)6]2+?

The order of increasing rate is [Ir(NH3)6]3+ < [Rh(NH3)6]3+ < [Co(NH3)6]3+ < [Ni(OH2)6]2+ < [Mn(OH2)6]2+.

21.16 State the effect on the rate of dissociatively activated reactions of Rh(III) complexes of each of (a) an increase in the positive charge on the complex? decreased rate.

(b) changing the leaving group from NO3–

to Cl–? Decreased rate.

(c) changing the entering group from Cl– to I–? This change will have little or no effect on the rate.

(d) changing the cis ligands from NH3 to H2O? Decreased rate.

21.17 Write out the inner- and outer-sphere pathways for reduction of azidopentaamminecobalt(III) ion with V2+(aq). What experimental data might be used to distinguish between the two pathways?

The inner-sphere pathway:

[Co(N3)(NH3)5]2+ + [V(OH2)6]2+ → [[Co(N3)(NH3)5]2+, [V(OH2)6]2+}

{[Co(N3)(NH3)5]2+, [V(OH2)6]2+} → {[Co(N3)(NH3)5]2+, [V(OH2)5]2+, H2O}

{[Co(N3)(NH3)5]2+, [V(OH2)5]2+, H2O} → [(NH3)5Co–N=N=N–V(OH2)5]4+

[(NH3)5Co–N=N=N–V(OH2)5]4+ → [(NH3)5Co–N=N=N–V(OH2)5]4+

[(NH3)5Co–N=N=N–V(OH2)5]4+ → [Co(OH2)6]2+ + [V(N3)(OH2)5]2+

The outer sphere pathway:

[Co(N3)(NH3)5]2+ + [V(OH2)6]2+ → {[Co(N3)(NH3)5]2+ [V(OH2)6]2+}



{[Co(N3)(NH3)5]2+,[V(OH2)6]2+} → {[Co(N3)(NH3)5]+ , [V(OH2)6]3+}

{[Co(N3)(NH3)5] +,[V(OH2)6]3+} → {[Co(OH2)6]2+ , [V(OH2)6]3+

21.18 The compound [Fe(SCN)(OH2)5]2+ can be detected in the reaction of [Co(NCS)(NH3)5]2+ with Fe2+(aq) to give Fe3+(aq) and Co2+(aq). What does this observation suggest about the mechanism? Appears to be an inner-sphere electron transfer reaction.

21.19 Calculate the rate constants for electron transfer in the oxidation of [V(OH2)6]2+ (Eσ (V3+/V2+) = –0.255 V) and the oxidants (a) [Ru(NH3)6]3+ (EO(Ru3+/Ru2+) = + 0.07 V), (b) [Co(NH3)6]3+ (EO(Co3+/Co2+) = +0.10 V). Comment on the relative sizes of the rate constants.

(a) [Ru(NH3)6]3+ k = 4.53 × 103 dm3mol−1s−1

(b) [Co(NH3)6]3+ k = 1.41 × 10−2 dm3mol−1s−1

Relative sizes? The reduction of the Ru complex is more thermodynamically favoured and faster.

21.20 Calculate the rate constants for electron transfer in the oxidation of [Cr(OH2)6]2+ (EO–(Cr3+/Cr2+) = –0.41 V) and each of the oxidants [Ru(NH3)6]3+ (EO(Ru3+/Ru2+) = +0.07 V), [Fe(OH2)6]3+ (EO(Fe3+/Fe2+) = +0.77 V) and [Ru(bpy)3]3+ (EO(Ru3+/Ru2+) = +1.26 V). Comment on the relative sizes of the rate constants

(a) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1; k22 (Ru3+/Ru2+ for the hexamine complex) = 6.6 × 103 dm3mol−1s−1; f12 = 1; K12 = e [nF εo/RT] where εo = 0.07 V – (–0.41V) = 0.48 V; n = 1; F = 96485 C; R = 8.31 Jmol−1K−1 and T = 298 K. Using these values gives K12 = 1.32 × 108. Substitution of these values in the Marcus-Cross relationship gives k12 = 2.95 × 103 dm3mol−1s−1.

(b) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1; k22 (Fe3+/Fe2+ for the aqua complex) = 1.1 dm3mol−1s−1; f12 = 1; K12 = e[nF εo/RT] where εo = 0.77 V – (−0.41V) = 1.18 V; n = 1; F = 96485 C; R = 8.31 Jmol−1K−1 and T = 298 K. Using these values we get K12 = 9.26 × 1019. Substitution of these values in the Marcus-Cross relationship gives k12 = 3.19 × 107 dm3mol−1s−1.

(c) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1; k22 (Ru3+/Ru2+ for the bipy complex) = 4 × 108 dm3mol−1s−1; f12 = 1; K12 = e[nF εo/RT] where εo = 1.26 V – (–0.41V) = 1.67 V; n = 1; F = 96485 C; R = 8.31 Jmol−1K−1 and T = 298 K.

Using these values gives K12 = 1.81 × 1028. Substitution of these values in the Marcus-Cross relationship gives k12 = 8.51 × 1015 dm3mol−1s−1.21.21 The photochemical substitution of [W(CO)5(py)] (py = pyridine) with triphenylphosphine gives W(CO)5(P(C6H5)3). In the presence of excess phosphine, the quantum yield is approximately 0.4. A flash photolysis study reveals a spectrum that can be assigned to the intermediate W(CO)5. What product and quantum yield do you predict for substitution of [W(CO)5(py)] in the presence of excess triethylamine? Is this reaction expected to be initiated from the ligand field or MLCT excited state of the complex? The product will be [W(CO)5(NEt3)], and the quantum yield will be 0.4.

21.22 From the spectrum of [CrCl(NH3)s]2+ shown in Fig. 20.32, propose a wavelength for photoinitiation of reduction of Cr(III) to Cr(II) accompanied by oxidation of a ligand. ~250 nm.


S22.1 Is Mo(CO)7 likely to be stable? No.

S22.2 What is the electron count for and oxidation number of platinum in the anion of Zeise’s salt, [PtCl3(C2H4)]–? Treat CH2=CH2 as a neutral two-electron donor. Electron count, 16; oxidation number, +4.

S22.3 What is the formal name of [Ir(Br)2(CH3)(CO)(PPh3)2]? Dibromocarbonylmethylbis(triphenylphos-

phine)iridium(III).

S22.4 Which of the two iron compounds Fe(CO)5 and [Fe(CO)4(PEt3)] will have the higher CO stretching frequency? Which will have the longer M–C bond? Fe(CO)5

S22.5 Show that both are 18-electron species.

(a) [(η6-C7H8)Mo(CO)3] (49)?

η6-C7H8 = 6 electrons, Mo atom = 6. carbonyl = 2 electrons, total = 18.

(b) [(η7-C7H7)Mo(CO)3]+(51)? η7-C7H7+ = 6

electrons, Mo atom = 6, 3 COs = 6, total = 18.

S22.6 Propose a synthesis for Mn(CO)4(PPh3)(COCH3) starting with [Mn2(CO)10], PPh3, Na and CH3I.



Mn2(CO)10 + 2 Na → 2Na[Mn(CO)5]

Na[Mn(CO)5] + CH3I → Mn(CH3)(CO)5 + NaI

Mn(CH3)(CO)5 + PPh3 → Mn(CO)4(PPh3)(COCH3)

S22.7 The IR spectrum of [Ni2(η5-Cp)2(CO)2] has a pair of CO stretching bands at 1857 cm–1 (strong) and 1897 cm–1 (weak). Does this complex contain bridging or terminal CO ligands, or both? (Substitution of η5-C5H5 ligands for CO ligands leads to small shifts in the CO stretching frequencies for a terminal CO ligand.) Bridging

S22.8 By using the same molecular orbital diagram, comment on whether the removal of an electron from [Fe(η5-Cp)2] to produce [Fe(η5-Cp)2]+ should produce a substantial change in M–C bond length relative to neutral ferrocene. It will not.

S 22.9 The compound [Fe4(Cp)4(CO)4] is a dark-

green solid. Its IR spectrum shows a single CO stretch at 1640 cm–1. The 1H NMR spectrum is a single line even at low temperatures. From this spectroscopic information and the CVE, propose a structure for [Fe4(Cp)4(CO)4]. TA tetrahedron with 4 terminal Cp rings and four capping COs.

S22.10 If Mo(CO)3L3 is desired, which of the ligands P(CH3)3 or P(t-Bu)3 would be preferred? Give reasons for your choice. PMe3 would be preferred.

S22.11 Assess the relative substitutional reactivities of indenyl and fluorenyl (86) compounds? Fluorenyl compounds are more reactive than indenyl.

S22.12 Show that the reaction is an example of reductive elimination?

PdPh PPh3

Ph3P Cl

Cl

Cl

PdPh3P Cl

Cl PPh3

+ PhCl

The decrease in both coordination number and

oxidation number by 2.

S22.13 Explain why [Pt(PEt3)2(Et)(Cl)] readily decomposes, whereas [Pt(PEt3)2 (Me)(Cl)] does not? The ethyl group in [Pt(PEt3)2(Et)(Cl)] is prone to β-hydride elimination.

Exercises 22.1 Name the species, draw the structures of,

and give valence electron counts to the metal atoms? Do any of the complexes deviate from the 18-electron rule? If so, how is this reflected in their structure or chemical properties?

(a) Fe(CO)5? Pentacarbonyliron(0), 18e–;

(b) Mn2(CO)10? decacarbonyldimanganese(0), 18e–;

(c) V(CO)6? hexcarbonylvanadium(0), 17e–

(d) [Fe(CO)4]2–? tetracarbonylferrate(–2), 18e–;

(e) La(η5-Cp*)3? tris(pentamethylcyclopentadienyl)lanthanum(III), 18e–;

(f) Fe(η3-allyl)(CO)3Cl? allyltricarbonylchloroiron(II), 18e–;

(g) Fe(CO)4(PEt3)? tetracarbonyltriethylphosphineiron(0);

(h) Rh(CO)2(Me)(PPh3)? dicarbonylmethyltriphenylphosphinerhodium(I),16e-

(i) Pd(Cl)(Me)(PPh3)2? chloromethylbis(triphenylphosphine) palladium(II), 16e–;

(j) Co(η5-C5H5)(η4-C4Ph4)? cyclopentadienyltetraphenylcylcobutadinecobalt(I), 18e–;

(k) [Fe(η5-C5H5)(CO2)]–? dicarbonylcyclopentadienylferrate(0), 18e–;

(l) Cr(η6-C6H6)(η6-C7H8)? benzenecycloheptatrienechromium(0), 18e–;

(m) Ta(η5-C5H5)2Cl3? trichlorobiscyclopentadineyltanatalum(V), 18e–;

(n) Ni(η5-C5H5)NO? cyclopentadineylnitrosylnickel(0),18e–.

22.2 Sketch an η2 the interactions of 1,4-butadiene with a metal atom and (b) do the same for an η4 interaction.

22.3 What hapticities are possible for the

interaction of each of the following ligands



with a single d-block metal atom such as cobalt? (a) C2H4? η2 (b) Cyclopentadienyl? Can be η5, η3, or η1.

(c) C6H6? η6, η4, and η2.

(d) Cyclooctadiene? η2 and η4 .

(e) Cyclooctatetraene?η8, η6, η4, η2.

22.4 Draw plausible structures and give the electron count of (a) Ni(η3-C3H5)2 (b) Co(η4-C4H4)(η5-C5H5) (c) Co(η3-C3H5)(CO)3. If the electron count deviates from 18, is the deviation explicable in terms of periodic terms.

(a) Ni(η3-C3H5)2 16, very common for group 9 and group 10 elements.

(b) Co(η4-C4H4)(η5-C5H5)? 18.

(c) Co(η3-C3H5)(CO)3? 18.

22.5 State the two common methods for the preparation of simple metal carbonyls and illustrate your answer with chemical equations. Is the selection of method based on thermodynamic or kinetic considerations?

(1) Mo(s) + 6 CO(g) → Mo(CO)6(s) (high temperature and pressure required)

(2) 2 CoCO3(s) + 2 H2(g) + 8 CO(g) → Co2(CO)8(s) + 2 CO2 + 2 H2O

The reason that the second method is preferred is kinetic.

22.6 Suggest a sequence of reactions for the preparation of Fe(CO)3(dppe), given iron metal, CO, dppe (Ph2PCH2CH2PPh2), and other reagents of your choice.

Fe(s) + 5 CO(g) → Fe(CO)5(l) (high temperature and pressure required)

Fe(CO)5(l) + diphos(s) → Fe(CO)3(diphos)(s) + 2 CO(g)

22.7 Suppose that you are given a series of metal tricarbonyl compounds having the respective symmetries C2v, D3h, and Cs. Without consulting reference material, which of these should display the greatest number of CO stretching bands in the IR

spectrum? Check your answer and give the number of expected bands for each by consulting Table 22.7.

Cs symmetry complex, 3

22.8 Provide plausible reasons for the differences in IR wavenumbers between each of the following pairs: (a) Mo(PF3)3(CO)3 2040, 1991 cm–1 versus

Mo(PMe3)3(CO)3 1945, 1851 cm–1? CO bands of the trimethylphosphine complex are 100 cm–1 or more lower in frequency. PMe3 is primarily a σ-donor ligand. PF3 is primarily a π-acid ligand. (b) MnCp(CO)3 2023, 1939 cm–1 vs. MnCp*(CO)3 2017, 1928 cm–1? CO bands of the Cp* complex are lower in frequency than the corresponding bands of the Cp complex. Cp* is a stronger donor ligand than Cp.

22.9 The compound Ni3(C5H5)3(CO)2 has a single CO stretching absorption at 1761 cm–1. The IR data indicate that all C5H5 ligands are pentahapto and probably in identical environments. (a) On the basis of these data, propose a structure.

(b) Does the electron count for each metal in your structure agree with the 18-electron rule? If not, is nickel in a region of the periodic table where deviations from the 18-electron rule are common? No. Deviations from the rule are common for cyclopentadienyl complexes to the right of the d block.

22.10 Decide which of the two complexes W(CO)6 or IrCl(CO)(PPh3)2 should undergo the faster exchange with 13CO. Justify your answer. The Ir complex. It has an A mechanism.

22.11 Which metal carbonyl in each of (a) [Fe(CO)4]2– or [Co(CO)4]– (b) [Mn(CO)5]– or [Re(CO)5]–should be the most basic toward a proton? What are the trends on which your answer is based ? (a) [Fe(CO)4]2– The trend involved is the greater affinity for a cation that a species with a higher negative charge has.



(b) The rhenium complex. The trend involved is the greater M–H bond enthalpy for a period 6 metal ion relative to a period 4 metal ion in the same group.

22.12 Using the 18-electron rule as a guide, indicate the probable number of carbonyl ligands in (a) W(η6-C6H6)(CO)n, (b) Rh(η5-C5H5)(CO)n, and (c) Ru3(CO)n.

(a) 3 (b) 2 (c) 12 22.13 Propose two syntheses for MnMe(CO)5,

both starting with Mn2(CO)10, with one using Na and one using Br2? You may use other reagents of your choice. (i) Reduce Mn2(CO)10 with Na to give Mn(CO)5

–; react with MeI to give MnMe(CO)5. (ii), Oxidize with Br2. to give MnBr(CO)5; displace the bromide with MeLi to give MnMe(CO)5.

22.14 Give the probable structure of the product obtained when Mo(CO)6 is allowed to react first with LiPh and then with the strong carbocation reagent, CH3OSO2CF3.

Mo(CO)5(C(OCH3)Ph)

22.15 Na[W(η5-C5H5)(CO)3] reacts with 3-chloroprop-1-ene to give a solid, A, which has the molecular formula W(C3H5)(C5H5)(CO)3. Compound A loses carbon monoxide on exposure to light and forms compound B, which has the formula W(C3H5)(C5H5)(CO)2. Treating compound A with hydrogen chloride and then potassium hexafluorophosphate, K+PF6–, results in the formation of a salt, C. Compound C has the molecular formula [W(C3H6)(C5H5)(CO)3]PF6. Use this information and the 18-electron rule to identify the compounds A, B, and C. Sketch a structure for each, paying particular attention to the hapticity of the hydrocarbon.

WCO

CO

COA:W

COCOB:

WCO

CO

COC:

η3η2η1

+

22.16 Treatment of TiCl4 at low temperature

with EtMgBr gives a compound that is unstable above 270ºC. However, treatment of TiCl4 at low temperature with MeLi or LiCH2SiMe3 gives compounds that are stable at room temperature. Rationalize these observations.

The compounds formed are TiR4; ethyl has the low energy β-hydride elimination decomposition.

22.17 Suggest syntheses of (a) [Mo(η7-

C7H7)(CO)3]BF4 from Mo(CO)6 and (b) [IrCl2(COMe)(CO)(PPh3)2] from [IrCl(CO)(PPh3)2]?

(a) Reflux Mo(CO)6 with cycloheptatriene to give [Mo(η6-C7H8)(CO)3]; treat with the trityl tetrafluoroborate to give [Mo(η7-C7H7)(CO)3]BF4.

(b) React [IrCl(CO)(PPh3)2] with MeCl, then expose to CO atmosphere to give [IrCl2(COMe)(CO)(PPh3)2].

22.18 When Fe(CO)5 is refluxed with cyclopentadiene compound A is formed which has the empirical formula C8H6O3Fe and a complicated 1H NMR spectrum. Compound A readily loses CO to give compound B with two 1H-NMR resonances, one at negative chemical shift (relative intensity one) and one at around 5ppm (relative intensity 5). Subsequent heating of B results in the loss of H2 and the formation of compound C. Compound C has a single 1H-NMR resonance and the empirical formula C7H5O2Fe. Compounds A, B, and C all have 18 valence electrons: identify them and explain the observed spectroscopic data.

+ Fe(C O )5

FeO C C O

C O

-2 C O -C O

F e

O CC O

H

-H

A B [π-C5H5Fe(CO)2]2 C The compound B shows two 1H NMR

resonances due to Fe-H proton and the aromatic Cp ring. C shows a single 1H NMR resonance because of aromatic Cp ring.

22.19 When Mo(CO)6 is refluxed with

cyclopentadiene compound D is formed which has the empirical formula C8H5O3Mo and an absorption in the IR spectrum at 1960 cm–1. Compound D can be treated with bromine to yield E or with Na/Hg to give compound F. There are absorptions in the IR spectra of E and F at 2090 and 1860 cm–1, respectively. Compounds D, E, and F all have 18 valence electrons: identify them and explain the observed spectroscopic data.

D = C5H5Mo(CO)3. E = C5H5Mo(CO)3Br. F = C5H5Mo(CO)3Na.

22.20 Which compound would you expect to be more stable,RhCp2 or RuCp2? Give a plausible explanation for the difference in terms of simple bonding concepts



RuCp2 has 18 electrons. 22.21 Give the equation for a workable reaction

for the conversion of Fe(η5-C5H5)2 to Fe(η5-C5H5) (η5-C5H4COCH3) and (b) Fe(η5-C5H5) (η5-C5H4CO2H)

(a) Fe(η5 – C5H5)2 + CH3COCl → Fe(η5 – C5H5)(η5 – C5H4COCH3) + HCl

(b)

F e +

C l O

C l A lC l3C H 2C l2

Fe

O C l

H 2O , t-B uO K

D M EFe

C O 2H

22.22 Sketch the a1’ symmetry-adapted orbitals for the two eclipsed C5H5 ligands stacked together with D5h symmetry. Identify the s, p, and d orbitals of a metal atom lying between the rings that may have nonzero overlap, and state how many a1’ molecular orbitals may be formed.

The symmetry-adapted orbitals of the two eclipsed C5H5 rings in a metallocene are shown in Resource Section 5, the dz2 orbital on the metal has a1′ symmetry, as does s. Three a1′ MOs will be formed.

22.23 The compound Ni(η5-C5H5)2 readily adds one molecule of HF to yield [Ni(η5-C5H5)(η4-C5H6)]+ whereas Fe(η5-C5H5)2 reacts with strong acid to yield [Fe(η5-C5H5)2H]+. In the latter compound the H atom is attached to the Fe atom. Provide a reasonable explanation for this difference.

Protonation of FeCp2 at iron does not change its number of valence electrons.

22.24 Write a plausible mechanism, giving your reasoning, for the reactions: (a) [Mn(CO)5(CF2)]+ + H2O → [Mn(CO)6]+ + 2HF?

(i) The F atoms render the C atom subject to nucleophilic attack

(ii)two equivalents of HF are eliminated

(b) Rh(C2H5) (CO) (PR3)2 → RhH(CO)(PR3)2 + C2H4? A β-hydrogen elimination reaction.

22.25 Given mechanism of CO insertion, what rate constant can be extracted from rate data? Rate = ka[RMn(CO)5]

22.26 (a) What cluster valence electron (CVE)

count is characteristic of octahedral and trigonal prismatic complexes?

octahedral M6, 86; trigonal prismatic M6, 90.

(b) Can these CVE values be derived from the 18-electron rule? No.

(c) Determine the probable geometry of [Fe6(C)(CO)16]2– and [Co6(C)(CO)15]2-

? The iron complex probably contains an octahedral Fe6 array. The cobalt complex probably contains a trigonal-prismatic Co6 array.

22.27 Based on isolobal analogies, choose the groups that might replace the group in boldface in (a) Co3(CO)9CH→ OCH3, N(CH3)2, or SiCH3? SiCH3.

(b) (OC)5MnMn(CO)5 → I, CH2, or CCH3? I.

22.28 Ligand substitution reactions on metal clusters are often found to occur by associative mechanisms, and it is postulated that these occur by initial breaking of an M-M bond, thereby providing an open coordination site for the incoming ligand. If the proposed mechanism is applicable, which would you expect to undergo the fastest exchange with added13CO, Co4(CO)12 or Ir4(CO)12? Suggest an explanation. Co. This is because metal–metal bond strengths increase down a group in the d block.


S23.1 Derive the ground state of the Tm3+ ion 3H6. S23.2 The product of the reaction above is in fact

a hydride bridged dimer (9). Suggest a strategy to ensure that the hydride is monomeric. A simple approach would be to use the Cp rings substituted by bulky groups and examine rate of formation.

S23.3 Use the Frost diagrams and data in Resource section 2 to determine the most stable uranium ion in acid aqueous solution in the presence of air and give its formula. UO2

2+ if sufficient oxygen is present.

Exercises 23.1 (a) Give a balanced equation for the

reaction of any of the lanthanoids with aqueous acid. (b) Justify your answer with redox potentials and with a generalization on the most stable positive oxidation states for the lanthanoids. (c) Name the two lanthanoids that have the greatest tendency to deviate from the usual positive oxidation



state and correlate this deviation with electronic structure. ? (a) Balanced equation 2 Ln(s) + 6 H3O+(aq) →

2 Ln3+(aq) + 3 H2(g) + 6 H2O(l)

(b) Redox potentials The potentials for the Ln0/Ln3+ oxidations in acid solution range from a low of 1.99 V for europium to 2.38 V.

(c) Two unusual lanthanides Ce4+, Eu2+.

23.2 Explain the variation in the ionic radii between La3+ and Lu3+.

The lanthanide contraction.

23.3 From a knowledge of their chemical properties, speculate on why Ce and Eu were the easiest lanthanoids to isolate before the development of ion-exchange chromatography. Ce4+ and Eu2+, unusual oxidation states were used in separation procedures.

23.4 How would you expect the first and second ionization energies of the lanthanoids to vary across the series? Sketch the graph that you would get if you plotted the third ionization energy of the lanthanoids versus atomic number? Identify elements at any peaks or troughs and suggest a reason for their occurrence? First and second IEs would show general increase across the lanthanoids. With the third, anomalies arise.

23.5 Derive the ground state of the Tb3+ ion. 7F6 23.6 Predict the magnetic moment of a compound containing the Tb3+ ion. μ= 9.72 μB

23.7 Explain why stable and readily isolable

carbonyl complexes are unknown for the lanthanoids? Carbonyl compounds need back-bonding from metal orbitals of the appropriate symmetry.

23.8 Suggest a synthesis of neptunocene from NpCl4?

Np(COT)2 was prepared by the reaction of K2COT with NpCl4 under inert atmosphere.

23.9 Account for the similar electronic spectra of Eu3+ complexes with various ligands and the variation of the electronic spectra of Am3+ complexes as the ligand is varied. The 5f orbitals of the actinide ions interact strongly with ligand orbitals, and the splitting of the 5f subshell, as well as the color of the complex, varies as a function of ligand.

23.10 Predict a structure type for BkN based on the ionic radii r(Bk3+) = 96 pm and r(N3−) = 146 pm. The rock-salt structure.

23.11 Describe the general nature of the distribution of the elements formed in the thermal neutron fission of 235U, and decide which of the following highly radioactive nuclides are likely to present the greatest radiation hazard in the spent fuel from nuclear power reactors: (a) 39Ar, (b) 228Th, (c) 90Sr, (d) 144Ce. 90Sr and 144Ce.


S24.1 Synthesis for Sr2MoO4? Sealed tube, high temperature, 6 SrO(s) + Mo(s) + 2 MoO3(s)

3 Sr2MoO4.

S24.2 Why does increased pressure reduce the conductivity of K+ more than that of Na+ in β-alumina? Because K+ is larger than Na+.

S24.3 Rationalize the observation that FeCr2O4 is a normal spinel? The A2+ ions (Fe2+ in this example) occupy tetrahedral sites and the B3+ ions (Cr3+) occupy octahedral sites.

Exercises 24.1 NiO doped with Li2O? The electronic

conductivity of the solid increases owing to formation of Ni1–xLix)O.

24.2 What is a crystallographic shear plane? Both.

24.3 How might you distinguish between a solid solution and a series of discrete crystallographic shear plane structures? A solid solution would contain a random collection of crystallographic shear planes.

24.4 Wurtzite crystal structure and bottleneck? The wurtzite structure is shown in Figure 3.35. The normal sites for cations in this structure are the tetrahedral holes. The bottleneck involves the space formed by three close packed anions.

24.5 Synthesis of? (a) MgCr2O4 - heat (NH4)2Mg(CrO4)2 gradually to 1100-1200°C; (b) SrFeO3Cl - heat SrO + SrCl2 + Fe2O3 in a sealed tube; (c) Ta3N5 - heat Ta2O5 under NH3 at 700°C.

24.6 Products of reactions? (a) LiNiO2, (b) Sr2WMnO6.



24.7 Where might intercalated Na+ ions reside in the ReO3 structure? The unit cell for ReO3 is shown in Figure 24.16(a), the structure is very open.

24.8 Antiferromagnet ordering? In an antiferromagnetic substance, the spins on different metal centers are coupled into an antiparallel alignment. As the temperature approaches 0 K, the net magnetic moment goes to zero.

24.9 Magnetic measurements on ferrite? inverse spinel.

24.10 Using LFSEs determine site preference for

A= Ni(II) and B = Fe(III)? Better ligand field stabilization and a strong preference for inverse spinel.

24.11 High-temperature superconductors? All

except Gd2Ba2Ti2Cu2O11, which contains only Cu2+, are superconductors.

24.12 Classify oxides as glass-forming or

nonglass-forming? BeO, B2O3, and to some extent GeO2, glass forming. Transition metal and rare earth oxides are typically nonglass-forming.

24.13 Which metal sulfides might be glass

forming? Metalloid and nonmetal sulfides. 24.14 Examples of spinels containing? (a)

sulfide? Zn(II)Cr(III)2S4 (b) fluoride? Li2NiF4

24.15 Synthesis of LiTaS2? Direct reaction of TaS2

with BuLi, or through electrochemical insertion of Li ions.

24.16 Oxotetrahedral species in framework

structures? Be, Ga, Zn, and P. 24.17 Formulas for structures isomorphous with

SiO2 containing Al, P, B, and Zn replacing Si? (AlP)O4, (BP)O4, (ZnP2)O6.

24.18 Mass percent of hydrogen in NaBH4 and hydrogen storage? 10.7 %, it is a good candidate to consider.

24.19 Formula for this lithium aluminium magnesium dihydride and structures?

Mg1-x(M)xH2y (M=Al, Li), The Li and Al will be incorporated as metal hydride solid solutions.

24.20 Color intensity differences in Egyptian blue pale versue blue-green spinel? The Cu site in Egyptian blue is square planar. In copper aluminate spinel blue, the site is tetrahedral.

24.21 Order of band gaps? BN > C(diamond) >

AlP > InSb. 24.22 Fulleride structures? In Na2C60, all of the

tetrahedral holes are filled with sodium cations. In Na3C60, all of the tetrahedral holes and all of the octahedral holes are filled with sodium cations.


S25.1 Synthesis of core-shell nanoparticles? The thermodynamic driving force is adjusted to a level that allows for heterogeneous nucleation of the shell material on the core but prevents homogeneous nucleation of the shell material.

S25.2 Hemispherical imprint for 2 nm

nanoparticle? The radius is 4.21 nm and the diameter is 8.42 nm.

S25.3 Host for QDs? MCM-41. Exercises 25.1 (a) Surface areas? 3.14 × 102

nm2 versus 3.14 × 106 nm2 (a factor of 104). (b) Nanoparticles based on size? The 10 nm particle. (c) Nanoparticles based on properties? A nanoparticle should exhibit properties different from those of a molecule or an extended solid.

25.2 Electron length and quantum

confinement? A characteristic length is the exciton Bohr radius.

25.3 Why are QDs better for bioimaging? One

light source can be used to excite different quantum dots.

25.4 Band energies for QD versus bulk

semiconductor? The energies of the band edges for a QD nanocrystal are more widely separated.

25.5 (a) Top-down versus bottom-up? The “top-

down” approach requires one to “carve out” nanoscale features from a larger object. The “bottom-up” approach requires one to “build up” nanoscale features from smaller entities.

(b) Advangtages and disadvantages? Top-down methods allow for precise control over the spatial relationships. Bottom-up methods



allow for the precise spatial control over atoms and molecules.

25.6 (a) What are SPMs? A method to image the

microscale features by scanning a very small probe over the surface and measuring some physical interaction between the tip and the material. (b) SPM and a specific nanomaterial? Local magnetic domains of magnetic nanomaterials, such as nanorods of iron oxide, can be imaged using magnetic force microscopy.

25.7 SEM versus TEM? SEM, an electron beam

is scanned over a material, and an image is generated by recording the intensity of secondary or back-scattered electrons. TEM, an electron beam is transmitted through the materials, and the image is the spatial variation in the number of transmitted electrons.

SEM sample preparation, ensure the material is conductive. In TEM, the sample needs to be made transparent to the electron beam..

25.8 (a) Steps in solutions synthesis of

nanoparticles? (i) solvation; (ii) stable nuclei of nanometer dimensions formed; (iii) growth of particles to desired size occurs.

(b) Why should the last two steps occur independently? So that nucleation fixes the total number of particles and growth leads to controlled size and narrow size distribution. (c) Stabilizers? Prevent surface oxidation and aggregation, and they limit traps for the holes and electrons, and improve quantum yields and luminescence.

25.9 Vapor-phase versus solution-based

techniques? (a) Vapor-phase, large sizes. (b) Vapor-phase, more agglomeration.

25.10 (a) What is a core-shell nanoparticle? (b) How are they made? Nucleation in one

solution, then grow the particle in another. (c) Purpose? In biosensing, the dielectric property of the shell can control the surface plasmon of the core. In drug delivery, the shell could react with a specific location and the core could be used as a treatment.

25.11 (a) Homogeneous versus heterogeneous? Homogeneous nucleation leads to solid formation throughout the vapour phase. (b) thin film? Heterogeneous nucleation is preferred in thin-film growth. (c) nanoparticles? Homogeneous nucleation is generally preferred for nanoparticle synthesis.

25.12 PVD versus CVD? In CVD, the atomic

species of interest are bound chemically to other species. Also, their thermal energies are typically rather low. In PVD, the atomic species of interest are typically atomic.

25.13 (a) Purpose of QD layers? Multiple layers of quantum dots can increase the intensity of any optical absorption or emission. They can also be used to form quantum cascade lasers. (b) Limitations? The limitations come from the requirements on how coherent the interface between the two materials must be.

25.14 (a) Applications of quatum wells? Lasers

and optical sensors. (b) Why are they used over other materials? They exhibit properties that are not observed in molecular or traditional solid state materials. (c) How are they made? Molecular beam epitaxy.

25.15 Superlattices and improved properties? AlN and TiN - different elastic constants,

large number of interfaces spaced on the nanoscale.

Results are much improved hardness values. 25.16 (a) Self-assembly? (a) Offers methods to

bridge bottom-up and top-down approaches to synthesis. (b) In nanotechnology? Offers a route to assemble nanosized particles into macroscopic structures.

25.17 Common features of self-assembly? (i) molecular or nanoscale subunits; (ii)

spontaneous assembly of the subunits; (iii) noncovalent interactions between the assembled subunits; (iv) longer-range structures arising from the assembly process.

Shell

Core 25.18 Static versus dynamic self-assembly? Static

self-assembly is when a system self-assembles to a stable state, eg. a liquid crystal. Dynamic self-assembly is when the system is oscillating between states and is dissipating energy in the process, eg. an oscillating chemical reaction.



25.19 Compare SAMs and cell membranes? SAM can structurally resemble a phospholipid bilayer.

25.20 What is morphosynthesis? Control of nanoarchitectures in inorganic materials through changes in synthesis parameters.

25.21 (a) What are the two classes of inorganic-

organic nanocomposites? Class I, hybrid materials where no covalent or ionic bonds are present. Class II, at least some of the components are linked through chemical bonds. (b) Examples? Class I, block copolymers. Class II, polymer/clay nanocomposites.

25.22 (a) Why is dispersion important in

nanocomposites? They lead to increased exposed surface areas. (b) Why is dispersion difficult? The often nonpolar organic polymers do not have strong interactions with the polar or ionic inorganic components.

25.23 A bionanomaterial and its application?

PPF/PPF-DA is an injectable bionanomaterial used for bone-tissue engineering.

25.24. (a) Biomimetics? Designing nanomaterials

that mimic biological systems. (b) example of biomimetics? Cellulose fibers in paper have been used to template the growth of titanium oxide nanotubes.

25.25 Bionanocomposites and improved

mechanical strength? Biomimetics need trabecular and cortical bone tissues, which combine compressibility and tensile strength. Hybrid alumoxane nanoparticles dispersed in PPF/PPF-DA shows enhanced strength.


S26.1 The effect of added phosphine on the catalytic activity of RhH(CO)(PPh3)3?

Added phosphine will result in a lower concentration of the catalytically active 16-electron complex.

S26.2 γ-Alumina heated to 900ºC, cooled, and exposed to pyridine vapor? complete dehydroxylation.

S26.3 Would a pure silica analog of ZSM-5 be an active catalyst for benzene alkylation (see Figure 26.24)? No.

S26.4 Demonstate that the polymerization of propene with a simple Cp2ZrCl2 catalyst would give rise to atactic polypropene?

Without R groups attached to the Zr center there is no preference for specific binding during polymerization.

Exercises 26.1 Which of the following constitute catalysis?

(a) H2 and C2H4 in contact with Pt? An example of genuine catalysis.

(b) H2 plus O2 plus an electrical arc? Not catalysis..

(c) The production of Li3N and its reaction with H2O? Not catalysis.

26.2 Define the following terms? (a) Turnover frequency? The amount of product formed per unit time per unit amount of catalyst.

(b) Selectivity? How much of the desired product is formed relative to by-products.

(c) Catalyst? A substance that increases the rate of a reaction but is not itself consumed.

(d) Catalytic cycle? Sequence of chemical reactions involving the catalyst that transform the reactants into products.

(e) Catalyst support? Generally a ceramic like γ-alumina or silica gel.

26.3 Classify the following as homogeneous or heterogeneous catalysis? (a) The increased rate of SO2 oxidation in the presence of NO? Homogeneous.

(b) The hydrogenation of oil using a finely divided Ni catalyst? Heterogeneous.

(c) The conversion of D-glucose to a D,L mixture by HCl? Homogeneous.

26.4 Which of the following processes would be worth investigating? (a) The splitting of H2O into H2 and O2? Not worthwhile.

(b) The decomposition of CO2 into C and O2? A waste of time.

(c) The combination of N2 with H2 to produce NH3? Very worthwhile reaction to try to catalyze efficiently at 80ºC.

(d) The hydrogenation of double bonds in vegetable oil? The process can be readily set up with existing technology.

26.5 Why does the addition of PPh3 to RhCl(PPh3)3 reduce the hydrogenation



turnover frequency? The catalytic species that enters the cycle is RhCl(PPh3)2(Sol) (Sol = a solvent molecule).

26.6 Explain the trend in rates of H2 absorption by various olefins catalyzed by RhCl(PPh3)3? In both cases, the alkene that is hydrogenated more slowly has a greater degree of substitution and so is sterically more demanding.

26.7 Hydroformylation catalysis with and without added P(n-Bu)3? The transformation of (E) into CoH(CO)4 must be the rate-determining step in the absence of added P(n-Bu)3. In the presence of added P(n-Bu)3, the formation of either (A) or (E) is the rate-determining step.

26.8 How does starting with MeCOOMe instead of MeOH lead to ethanoic anhydride instead of ethanoic acid using the Monsanto acetic acid process? The reaction of the ethanoate ion with the acetyl iodide leads to ethanoic anhydride.

26.9 Suggest a reason why? (a) Ring opening alkene metathesis polymerization (ROMP) proceeds? ROMP can result in reduced steric strain. (b) Ring-closing metathesis (RCM) reaction proceeds? RCM results in the loss of ethane.

26.10 (a) Attack by dissolved hydroxide? Structure C in Figure 26.11 (b) Attack by coordinated hydroxide? Structure E given in Figure 26.11 (c) Can one differentiate the stereochemistry? Yes.

26.11 (a) Enhanced acidity? When Al3+ replaces Si4+ on lattice site charge is balanced by H3O+.

(b) Three other ions? Ga3+, Co3+, and Fe3+. 26.12 Why is the platinum-rhodium in

automobile catalytic converters dispersed on the surface of a ceramic rather than used in the form of thin foil? A thin foil of platinum-rhodium will not have as much surface area as an equal amount of small particles finely dispersed on the surface of a ceramic support.

26.13 Devise a plausible mechanism to explain the deuteration of 3,3-dimethylpentane? (i) The mechanism of deuterium exchange is probably related to the reverse of the last two reactions in Figure 26.20.

(ii) The second observation can be explained by invoking a mechanism for rapid deuterium

exchange of the methyl group in the chemisorbed –CHR(CH3) group (R = C(CH3)2(C2H5)).

26.14 Why does CO decrease the effectiveness of Pt in catalyzing the reaction 2H+(aq) + 2e– → H2(g)? Platinum has a strong tendency to chemisorb CO.

26.15 Describe the role of an electrocatalyst? Platinum is the most efficient electrocatalyst for accelerating oxygen reduction at the fuel cell cathode, but is expensive.


S27.1 Is Iron (II) expected to be present in the cell as uncomplexed ions? No.

S27.2 Unusual coordination of Mg? The protein’s 3D structure can place any particular atom in a suitable position for axial coordination.

S27.3 Why does saline contain NaCl? Osmotic balance.

S27.4 Explain the significance of the Calcium ion pumps activation by calmodulin? The binding of calmodulin is a signal informing the pump that the cytoplasmic Ca2+ level has risen above a certain level.

S27.5 Why are iron-porphyrin complexes unable

to bind O2 reversibly? The Fe(II) gets oxidized to Fe(III), yielding an oxo-Iron(III) porphyrin complex.

S27.6 What is the nature of binding at Cu blue centers as indicated by the EPR spectrum? There is greater covalence in blue Cu centers than in simple Cu(II) compounds.

S27.7 What is the nature of an active site with Copper (III)? Diamagnetic with square-planar geometry.

S27.8 Why mercury is so toxic because of the action of enzymes containing cobalamin? Cobalamins are very active methyl transfer reactions, which can methylate anything in the cell.

S27.9 Suggest experiments that could establish

the structure of the MoFe cofactor? EPR, single-crystal X-ray diffraction, and EXAFS.

S27.10 Why might Cu sensors be designed to bind Cu(I) rather than Cu(II)? Cu(I) has an almost unique ability to undergo linear coordination by sulphur-containing ligands.



Exercises 27.1 Lanthanides versus calcium? They are hard

Lewis acids and prefer coordination by hard bases. Gd3+ has excellent fluorescence.

27.2 Substituting Co2+ for Zn2+? Co(II) commonly adopts distorted tetrahedral and five-coordinate geometries typical of Zn(II) in enzymes. Co(II) is paramagnetic, enabling Zn enzymes to be studied by EPR.

27.3 Compare the acid/base catalytic activities of Zn(II), Fe(III), and Mg(II)? Acid strengths: Fe(III) > Zn(II) > Mg(II). Ligand binding rates are Mg(II) > Zn(II) > Fe(III).

27.4 Propose a physical method for the determination of Fe(V)? EPR or Mossbauer.

27.5 Interpret the Mossbauer spectra of ferredoxin? In the spectrum, the oxidized spectrum is consistent with the iron atoms having the same valence, intermediate between the 3+ and the 2+ states, with one pair spin-up and the other pair spin-down.

27.6 Explain the differences in the structures of the oxidized and reduced forms of the P-cluster in nitrogenase? The change in structure suggest that the coupling of proton and electron transfer can also occur at the P-cluster, by controlling protonation of the exchangeable ligands.

27.7 What metals are involved in the synthesis of acetyl groups? Co(II), Fe(II) and Cu(I).


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