Self Test Answers

Shriver & Atkins: Inorganic Chemistry 5e

Answers to self-tests and exercises CHAPTER 1

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 10 cm

Self-tests 1 0

S1.1

80 35

S1.2

d orbitals, 5 orbitals

S1.3

4

S1.4

3 p

S1.5

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

S1.6

Ni : [A [Ar]3d 4 s , Ni

S1.7

Period 4, Group 2, s 2, s block block

S1.8

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

7 −1 = R⎜ 2 − 2 ⎟ = 1.0288 X 10 10 m λ ⎝ 1 4 ⎠

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.

6 −1 = R⎜ 2 − 2 ⎟ = 9.7547 X 10 10 m λ ⎝ 1 3 ⎠

S1.10 S1.11

Br +

81 35

n

8

2

Br + γ

2+

8

: [A [Ar]3d

1.11

4

1 λ

1 ⎝ 12

= R⎜

1

1

1

−

-1

1 7 −1 = 1.0974 X 10 10 m ⎟ 2 ⎠

∞

1

1

1

⎛ 1

1

6 −1 = R⎜ 2 − 2 ⎟ = 8.2305 X 10 10 m λ ⎝ 1 2 ⎠ 1

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

1.12

0 up to n- X

Cs+

1.13

n2

1.14

Exercises 1.1

1.2 1.3

(a) 147 N+ 42 He→178 O+11 p + γ (b)

12 6

(c)

14 7

246 96

C+11 p→137 N + γ

N+ 01n→31 H+126 C

N

2

1

3

2

4 4

0 3

1 Cm+126C→257 112 Uub+ 0 n

The higher value of I I 2 for Cr relative to Mn is a consequence of the special stability of halffilled subshell configurations and the higher Zeff of a 3d 3d electron electron verses a 4 s electron.

l

l m

+1, 0, −1 +2, +1, …, −2 0 +3, +2, …, −3

Orbital designation 2 p

Number of orbitals 3

3d

5

4 s 4 f

1 7

1.15

n=5, l = 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

Ne + 42 He → 1225 Mg + 01 n

B:

σ = Z – Zeff ; σ = 5-2.42 = 2.58

Be+ 49 Be→126 C+ 42 He +201 n

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

1.4

22 10

1.5

9 4

1.6

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

1.17

1

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


2

ANSWERS TO SELF-TESTS AND EXERCISES

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. d shell.

1.22

(a) [He]2 s22 p2

1.30

(b) [He]2 s22 p5 (c) [Ar]4 s2 (d) [Ar]3d [Ar]3d 10 (e) [Xe]4 f 145d 106 s26 p3

1.23

CHAPTER 2

(f) [Xe]4 f 145d 106 s2

Self-tests

(a) [Ar]3d [Ar]3 d 14 s2

S2.1

2

(b) [Ar]3d [Ar]3d (c) [Ar]3d [Ar]3d 5

4 (d) [Ar]3d [Ar]3d

S2.2

(e) [Ar]3d [Ar]3d 6

(b) square planar

(f) [Ar]

S2.3

Linear

10 1 (g) [Ar]3d [Ar]3 d 4 s

S2.4

S2 : 1σg 2σu 3σg 1πu 2πg ;

(h) [Xe]4 f 7 1.24

(a) Angular

2–

2

2

2

4

4

Cl2 – : 1σg22σu23σg21πu42πg44σu1.

(a) Xe]4 f 145d 46 s2

S2.5

1σg22σu23σg21πu42πg4

(b) [Kr]4d [Kr]4d 6

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.

(f) [Kr]4d [Kr]4d

S2.9

–21 kJ mol –1

(a) S

S2.10

(a) +1/2

6

(c) [Xe]4 f

(d) [Xe]4 f 7 (e) [Ar] 2

1.25

(b) Sr

(b) +5

(c) V (d) Tc

Exercises

(e) In (f) Sm

2.1

(a) angular

1.26

See Figure 1.4.

(b) tetrahedral

1.27

(a) I 1 increases across the row except for a dip at S; (b) Ae tends to increase except for Mg

(c) tetrahedral

(filled subshell), P (half filled subshell), and at AR (filled shell).

1.28 1.29

2.2

(b) trigonal pyramidal

Radii of Period 4 and 5 d-metals d-metals are similar because of lanthanide contraction. 2 0 2 s and 2 p

(a) trigonal planar (c) square pyramidal

2.3

(a) T-shaped


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. d shell.

1.22

(a) [He]2 s22 p2

1.30

(b) [He]2 s22 p5 (c) [Ar]4 s2 (d) [Ar]3d [Ar]3d 10 (e) [Xe]4 f 145d 106 s26 p3

1.23

CHAPTER 2

(f) [Xe]4 f 145d 106 s2

Self-tests

(a) [Ar]3d [Ar]3 d 14 s2

S2.1

2

(b) [Ar]3d [Ar]3d (c) [Ar]3d [Ar]3d 5

4 (d) [Ar]3d [Ar]3d

S2.2

(e) [Ar]3d [Ar]3d 6

(b) square planar

(f) [Ar]

S2.3

Linear

10 1 (g) [Ar]3d [Ar]3 d 4 s

S2.4

S2 : 1σg 2σu 3σg 1πu 2πg ;

(h) [Xe]4 f 7 1.24

(a) Angular

2–

2

2

2

4

4

Cl2 – : 1σg22σu23σg21πu42πg44σu1.

(a) Xe]4 f 145d 46 s2

S2.5

1σg22σu23σg21πu42πg4

(b) [Kr]4d [Kr]4d 6

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.

(f) [Kr]4d [Kr]4d

S2.9

–21 kJ mol –1

(a) S

S2.10

(a) +1/2

6

(c) [Xe]4 f

(d) [Xe]4 f 7 (e) [Ar] 2

1.25

(b) Sr

(b) +5

(c) V (d) Tc

Exercises

(e) In (f) Sm

2.1

(a) angular

1.26

See Figure 1.4.

(b) tetrahedral

1.27

(a) I 1 increases across the row except for a dip at S; (b) Ae tends to increase except for Mg

(c) tetrahedral

(filled subshell), P (half filled subshell), and at AR (filled shell).

1.28 1.29

2.2

(b) trigonal pyramidal

Radii of Period 4 and 5 d-metals d-metals are similar because of lanthanide contraction. 2 0 2 s and 2 p

(a) trigonal planar (c) square pyramidal

2.3

(a) T-shaped



2.4

(b) square planar

(b) one

(c) linear

(c) none

(a)

(d) two 2.15

3

(a) 1σg22σu2 (b) 1σg22σu21πu2

Cl

Cl

(c) 1σg22σu21πu43σg1

I Cl

Cl

(d) 1σg22σu23σg21πu42πg3 2.16

The configuration for the neutral C2 would be 1σg21σu2 1πu4. The bond order would would be ½[2½[22+4] = 2.

2.17

(a) 1σg22σu23σg21πu42πg4

(b)

(b) 1

F

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

120 0

S 1800 F

2.5 2.6

2.18

(a) 2 (b) 1 (c) 2

F

2.19

(a) +0.5

(a) tetrahedral

(b) –0.5

(b) octahedral

(c) +0.5 2.20

(a) 176 pm (b) 217 pm (c) 221 pm

2.7

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

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

2.21

(a) 0

(a-c)

(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

2.13

(b) ionic

(d) Possibly stable in isolation (only bonding

(c) ionic

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

(a) sp2 (b) sp3 3

3

(c) sp d or spd 2 2

2

(d) p d or sp d. 2.14

(a) one

2.22

1

2.23

HOMO exclusively F; LUMO mainly S.

2.24

(a) electron deficient


4


(b) electron precise

CHAPTER 3 Self-tests S3.1 S3.2

See Fig. 3.7 and Fig. 3.32. See Figure 3.35.

S3.3

52%

S3.4

r h = ((3/2)1/2 – 1) r = r = 0.225 r

3.4 3.5

XA2

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

K 3C60

(b) CsCI

S3.5

409 pm

3.11

6

S3.6

401 pm

3.12

S3.7

FeCr 3

2B type and 4A type – in a distorted octahedral arrangement.

S3.8

X2A3.

3.13

S3.9

Ti CN = 6 (thought these are as two slightly different distances it is often described as 4 + 2) and O CN = 3.

(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.

S3.10

LaInO3

3.15

S3.11

2421 kJ mol

Lattice enthalpies for the di- and the trivalent ions. Also, the bond energy and third electron gain enthalpy for nitrogen will be large.

S3.12

Unlikely

3.16

4 four times the NaCl value or 3144 kJmol– .

S3.13

MgSO4 < CaSO4 < SrSO4 < BaSO4.

3.17

(a) 10906 kJmol –1

S3.14

NaClO4

S3.15

Schottky defects.

S3.16

Phosphorus and aluminium.

S3.17

The d x2-y2 x2-y2 and d z2 z2 have lobes pointing along the cell edges to th e nearest neighbor metals.

S3.18

–1

(a) n-type (b) p-type 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.

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 3.26

3.27

Yes. 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.

S4.9

Identify the acids and bases? (a) FeCl3 + Cl – –

[FeCl4] – , acid is FeCl3,

base is Cl .

(b) I – + I2 S4.10

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

I3 – , acid is I2, base is I – .

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

S4.11

CHAPTER 4

5

Draw the structure of BF3·OEt2?

Self-tests 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; 2H O, acid; hydroxide ion, conjugate base.

(c) NH3 + H2S

Exercises 4.1

NH4+ + HS –

Ammonia, base; NH4+, conjugate acid; hydrogen sulphide, acid; HS – , conjugate base.

S4.2

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.

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

S4.3

Sketch an outline of the s and p blocks of the periodic table, showing the elements that form acidic, basic, and amphoteric oxides?

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]+ 2+ 2+ < [Mn(H2O)6] < [Ni(H2O)6] < 3+ [Sc(H2O)6] .

S4.7

4.2

Predict p K a values? (a) H3PO4 p K a ≈ 3. The

of

the

2+

[Co(NH 3)5(OH)] .

(b) HSO4 – ? The conjugate base is SO4 – .

(b) H2PO4 p K a(2) ≈ 8. The actual value, –

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

given in Table 4.1, is 7.4.

(d) H2PO4 – ? The conjugate base is HPO42– .

(c) HPO42 – p K a(3) ≈ 13. The actual value,

(e)

given in Table 4.1, is 12.7.

Si(OH)4? The

conjugate

base

is

SiO(OH) 3 – .

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 TiO 2 to redissolve.

bases

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

actual value, given in Table 4.1, is 2.1.

S4.8

Identify the conjugate following acids?

(f) HS ? The conjugate base is S2– . 4.3

Identify the conjugate following bases?

acids

of

the


6

ANSWERS TO SELF-TESTS AND EXERCISES ClO4 – , cannot be studied in sulfuric acid.

(a) C5H5N (pyridine)? The conjugate acid is pyridinium ion, C5H6 N+.

(b)

HPO42 – ?

4.9 2–

The conjugate acid is H2PO4 .

(c) O2 – ? The conjugate acid is OH – .

electron withdrawing

4.11

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

4.12

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 K b of ethanoic acid?

4.13

K b = 5.6 × 10

What is the K a for C5H5NH+? –6

-

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

4.14

-

F will behave as a base in water.

O

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

Fe(III) complex, [Fe(OH 2)6] , is the stronger acid.

What are the structures and the p K a values of chloric (HClO3) and chlorous (HClO2) acid? O O Cl

Account for the trends in the p K a values of the conjugate acids of SiO 44 – , PO43 – , SO42 – , and ClO4 – ? The acidity of the four conjugate acids increases in the order HSiO43– < HPO42– < HSO4 – < HClO4.

K a = 5.6 × 10

4.8

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

–10

4.7

Is the p K a for HAsO42 – consistent with Pauling’s rules? No. Pauling’s rules are only approximate.

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

4.6

Is the – CN group electron donating or withdrawing?

O

the aluminum-containing species is more acidic.

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

Cl

H

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

H

Si(OH) 4, is more acidic.

O

(d) HClO3 or HClO4?

chloric acid

HClO4 is a stronger acid.

chlorous acid

(e) H2CrO4 or HMnO4?

Chloric acid, the predicted p K a = –2; actual value = –1.

HMnO4 is the stronger acid.

(f) H3PO4 or H2SO4?

Chlorous acid, the predicted p K a = 3; actual value = 2.

4.8

Which bases are too strong or too weak to be studied experimentally? (a) CO 32 – O2 – , ClO4 – , and NO3 – in water?

H2SO4 is a stronger acid.

4.15

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.

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

4.16

(b) HSO4 – , NO3 – , and ClO4 – , in H2SO4? HSO4 – , not too experimentally. –

strong

to

be

Arrange the following in order of increasing acidity? HSO4 , H3O+, H4SiO4, CH3GeH3, NH3, and HSO3F? increasing acidity is NH3 < CH3GeH3 < – + H4SiO4 < HSO4 < H3O < HSO3F.

studied

NO3 is of directly measurable base strength in liquid H2SO4.

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

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?

(d) AsF3(g) + SbF5(g)

polycations: Al, Cu, and Ti.

(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.

polyoxoanions: As, B, and Si.

The change in charge upon aqua ion polymerization?

4.26

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

4.20

+

8H3O 3+

2[Fe(OH 2)6]

→

P4O124–

+ 12H2O

4.21

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

4+

2-Me-py or 4-Me-py? 4.27

+

+ 2H3O

(a) H3PO4 and Na2HPO4?

4.28

NH2- + H2F+

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

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

4.29

4.30

are the Lewis acids SO3 and H+ and the base is the Lewis base OH – .

Depending on the E A and C A values for the Lewis acid, either base could be stronger.

–

Lewis acid [B12] from the Lewis base CH3 . Lewis acid SnCl2 displaces the Lewis acid K + from the Lewis base Cl – .

Discuss relative basicities? (a) Acetone and DMSO?

(b) Me2S and DMSO?

(b) Me[B12] – + Hg2+ [B12] + + 2+ MeHg ? The Lewis acid Hg displaces the K + + [SnCl3] – ? The

of line?

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.

Identifying acids and bases: (a) SO3 + H 2O HSO4 – + H+? The acids in this reaction

(c) KCl + SnCl2

Why is trimethylamine out

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

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

Choose between the two basic sites in Me2NPF2? The phosphorus atom in Me2 NPF2 is the softer of the two basic sites, so it will bond more strongly with the softer Lewis acid BH3

H2SO4 + HF ⇔ H3SO4+ + F4.23

[Ag (CN)2] –

2HCO 3-

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

NH3 + HF

Ph3P – SO2 +

CH3HgCl + HI? <1

(d) [AgCl2] – (aq) + 2 CN – (aq) (aq) + 2Cl – (aq)? >1

(b) CO2 and CaCO3?

4.22

Which of the following reactions have K eq > 1? (a) R 3P – BBr3 + R 3N – BF3 R 3P – BF3 + R 3N – BBr3? <1

(c) CH3HgI + HCl

H3PO4 + HPO42-  2H2PO4-

Ca 2+ +

4-Me-py.

(b) SO2 + Ph3P – HOCMe3 HOCMe3? > 1.

More balanced equations?

CO 2 + CaCO 3 + H 2O

B(n-Bu)3.

(b) More basic toward BMe 3: NMe3 or NEt3? NMe3.

→

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

Select the compound with the named characteristic? (a) Strongest Lewis acid: BF3, BCl3, or BBr3? BBr 3. BeCl2 or BCl3? Boron trichloride.

Write balanced equations for the formation of P4O124 – from PO43 – and for the formation of [(H2O) 4Fe(OH)2Fe(OH2)4]4+ from [Fe(OH2)6]3+? 4PO43– +

[AsF2][SbF6]?

The very strong Lewis acid SbF5 displaces the + – Lewis acid [AsF2] from the Lewis base F .

polyoxoanions (oxide polyanions): Mo

4.19

7

4.31

Write a balanced equation dissolution of SiO2 by HF?

for

the


8


SiO2 + 6HF

2H2O + H2SiF6

(c) C6H5COOH?

or SiO2 + 4HF

C6H5COOH + HF 2H2O + SiF4

Both a Brønsted acid–base reaction and a Lewis acid–base reaction.

4.32

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?

Write a balanced equation to explain the foul odor of damp Al 2S3? The foul odor

φ

Δf H

suggests H2S formation. Al 2 S 3 + 3H 2 O

4.33

decreases the activity of chloride relative to iodide, you can shift the following equilibrium to the right:

CHAPTER 5 Self-tests S5.1

acid-I - + Cl-

Half-reactions and balanced reaction for oxidation of zinc metal by permanganate ions? 2[MnO4− (aq) + 8H+ (aq) + 5e – + 4H2O(l)] reduction

(b) Favor basicity of R 3As over R 3N? Alcohols such as methanol or ethanol would be suitable.

5 [ Zn(s)

(c) Favor acidity of Ag+ over Al3+? An

→ Zn2+(aq) + 2e – ]

→

S5.2

Does Cu metal dissolve in dilute HCl? No.

acetonitrile, MeCN.

S5.3

Can Cr2O72– be used to oxidize Fe 2+, and would Cl – oxidation be a problem? Yes.

partially dehydroxylated one shown below, would provide Lewis acidic sites that could abstract Cl – :

-

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 2–

aqueous solution of SO4 precipitates as acid rain.

so is found in nature only combined with soft Lewis bases, the most common of which is S2– .

Write Brønsted acid – base reactions in liquid HF? (a) CH3CH2OH?

+

and H

ions

S5.6

Can Fe2+ disproportionate under standard conditions? No.

S5.7

bpy binding to Fe(III) or Fe(II)? Fe(II)

Why does Hg(II) occur only as HgS? Mercury(II) is a soft Lewis acid, and

4.36

oxidation

(d) Promote the reaction 2FeCl 3 + ZnCl2 Zn2+ + 2[FeCl4] ? A suitable solvent is Propose a mechanism for the acylation of benzene? An alumina surface, such as the

4.35

→ Mn2+(aq)

2MnO4− (aq) + 5Zn(s) + 16H+(aq) 5Zn2+(aq) + 2Mn2+(aq) + 8H2O(l)

example of a suitable solvent is diethyl ether. Another suitable solvent is H2O.

4.34

= -20.0kJ/mol

Al 2 O 3 + 3H 2 S

Describe solvent properties? (a) Favor displacement of Cl – by I – from an acid center? If you choose a solvent that

acid-Cl- + I-

C6H5COO - + H2F+

preferentially.

S5.8

Potential of AgCl/Ag,Cl – couple? E ox= – 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).

The balanced equation

is:

CH 3 CH 2 OH + HF N H

3

CH 3CH 2OH 2+ + F-

(b) NH3? The equation is N H + H F

4

+

+

F

-

S5.10

Frost diagram for thallium in aqueous acid?



5.3

9

Write balanced equations, if a reaction occurs, for the following species in aerated aqueous acid? (a) Cr2+? 4Cr (aq) + O2(g) + 4H (aq) → 4Cr (aq) E º = 1.65 V + 2H2O(l) 2+

+

3+

(b) Fe2+? 4Fe (aq) + O2(g) + 4H (aq) → 4Fe (aq) E º = 0.46 V + 2H2O(l) 2+

+

3+

(c) Cl – ? no reaction. (d) HOCl? No reaction. S5.11

The oxidation number of manganese?

(e) Zn(s)?

2+

Mn (aq)

S5.12

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

Compare the strength of NO3 – as an oxidizing agent in acidic and basic solution?

A competing reaction is: +

Zn(s) + 2H (aq) 0.763 V).

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

S5.13 S5.14

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

5.4

→ Zn2+(aq) + H2(g) ( E º =

Balanced equations for redox reactions? (a) Fe2+? (i) Fe2+ will not oxidize water.

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

(ii) Fe

2+

will not reduce water.

Exercises

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

5.1

Oxidation numbers?

(iv) disproportionation will not occur.

2 NO(g) + O2(g) → 2 NO2(g) +1 -2 0 +4 -2

(b) Ru2+? Ru2+ will not oxidize or reduce 2+

water. Ru will reduce O2 and in doing so 3+ 2+ will be oxidized to Ru . Ru will 3+ disproportionate in aqueous acid to Ru and metallic ruthenium.

2Mn3+(aq) + 2H2O → MnO2 + Mn2+ + 4H+(aq) +3 +1 -2 +4 -2 +2 +1

(c) HClO2? HClO2 will oxidize water, will

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

5.2

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.

→ CaH2(s) +2 -1

Suggest chemical reagents for redox transformations? (a) Oxidation of HCl to Cl2? S2O82– , H2O2, or α –PbO2 to oxidize Cl –

(d) Br2? Br 2 will not oxidize or reduce water. Br 2 will not reduce O2. Br 2 will not disproportionate in aqueous acid to Br – and HBrO.

to Cl2.

(b) Reducing Cr3+(aq) to Cr2+(aq)? metallic

5.5 Standard potentials vary with temperature in opposite directions? The

+

manganese, metallic zinc, or NH3OH .

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.

(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.

5.6

Balance redox reaction in acid solution: MnO4 – + H2SO3 → Mn2+ + HSO4 – ? pH dependence?


10


−

+

2MnO4 (aq) + 5H2SO3(aq) + H (aq) 2+ − 2Mn (aq) +5HSO3 (aq) + 3H2O(l)

→

5.13

38

K = 5.7 × 10

The potential decreases as the pH increases.

5.14 5.7

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

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

and

–

would again predominate. HSO4 predominant sulfur species at pH 6.

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

(b) The reduction of Fe 2O3(s)?

Frost diagram and standard potential for the HSO4 /S8(s) couple? 0.387 V

Using Frost diagrams? (a) What happens when Cl2 is dissolved in aqueous basic solution? Cl2 is thermodynamically

5.16

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

susceptible to disproportionation to Cl and ClO4 – when it is dissolved in aqueous base.

5.17

Reduction potential for MnO4 – to MnO2(s) at pH = 9.00? E = 0.98 V

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

5.18

Tendency of mercury species to act as an oxidizing agent, a reducing agent, or to undergo disproportionation? Hg2+ and

–

2+

Hg2 are both oxidizing agents. None of these species are likely to be good reducing 2+ agents. Hg2 is not likely to undergo disproportionation.

(b) What happens when Cl 2 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. Write equations for the following reactions: (a) N2O is bubbled into aqueous NaOH solution? –

5N2O(aq) + 2OH (aq) 4N2 (g) + H2O(l)

Zn(s) + I3 – (aq)

3I2(s) + 5ClO3 – (aq) + 3H2O(l) (aq) + 5Cl – (aq) + 6H+(aq)

→ 6IO3 –

(d) I2 → 2I – ? Acid or base, no difference. Determine the standard potential for the reduction of ClO4 – to Cl2? 1.392 V

will

undergo

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 Fe 2+ 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+].

Electrode potential for Ni /Ni couple at pH = 14? E =– 0.21 V Will acid or base most favour the following half-reactions? (a) Mn2+ → MnO4 – ? Base (c) H2O2 → O2? Base

5.12

5.20

2+

(b) ClO4 – → ClO3 – ? Acid

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

→ Zn2+(aq) + 3I – (aq)

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

5.11

5.19

→ 2NO3 – (aq) +

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

5.10

is the

5.15

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

5.9

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

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

5.8

Calculate the equilibrium constant for Au+(aq) + 2CN – (aq) → [Au(CN)2] – (aq)?

5.24

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


ANSWERS TO SELF-TESTS AND EXERCISES 11

CHAPTER 6

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

Self-tests S6.1

Γstr = A1g (Raman, polarized) + E g

Three S 4 axes.

S6.2

(Raman) + T 1u (IR).

Sketch the S 4 axis of an NH4+ ion. How many of these axes are there in the ion?

SF5Cl has C4v symmetry. stretching vibrations leads to:

Γstr = 3 A1 (IR and Raman, polarized)

(a) BF3 point group? D3h. (b) SO42 – point group? T d.

S6.3

Symmetry species of all five d orbitals of the central Xe atom in XeF 4 (D4h, Fig. 6.3)? dx2-y2 is B1g; dxy is B2g; dxz and dyz are Eg; dz2 is A1g.

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

S6.14 Symmetries of all the vibration modes of [PdCl4]2-? A1g + B1g + B2g + A2u + B2u + 2 E u S6.15

SALCs for sigma bonding in O? A1g + Eg

+T1u.

S6.4

What is the maximum possible degeneracy for an Oh molecule? 3.

Exercises

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.

6.1

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?

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 2C 4 C 2 ’ 2C 2 2C 2 i 2S 4 σh 2σv 2σd Γ3N 4 0 0 2 0 0 0 4 2 0 Reduces to A1g + B1g + E u A1g + B1g are Raman active. E u 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 SF 6?

Symmetry elements? (a) a C 3 axis and a

v

plane in the NH3 molecule?

N H

D2h character table, which is the Ag symmetry type.

S6.9

Analysis of the

N

H

H

H

H H

σ v

C 3

(b) a C 4 axis and a

h

plane in the square-

planar [PtCl4]2– ion?

Cl

Cl

Cl

Pt

Pt Cl

Cl

Cl

Cl

C 4

6.2

Cl

σ h

S 4 or i ? (a) CO2? i

(b) C2H2? i. (c) BF3? neither. (d) SO42– ? three different S 4. 6.3

Assigning point groups: (a) NH2Cl? C s (b) CO32– ? D3h (c) SiF4? T d (d) HCN? C ∞v . (e) SiFClBrI? C 1. (f) BrF4 – ? D4h.


12

6.4

6.5


How many planes of symmetry does a benzene molecule possess? What chlorosubstituted benzene has exactly four planes of symmetry? 7,and C6H3Cl3. The symmetry elements of orbitals? (a) An s orbital? Infinite number of C n axes, plus

6.12

6.13

planes that pass through both lobes and include the long axis of the orbital. In addition, the long axis is a C n axis, where n can be any number from 1 to ∞.

(c) A d xy orbital? Center of symmetry, three

(b) For the planar form of B2F4 ( D2h): The vibrations are: 3 Ag (Raman, polarized) + 2 B2g (Raman) + B3g (Raman) + Au(inactive) + 2 B1u (IR) + B2u (IR) + 2 B3u (IR).

mutually perpendicular C 2 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.

For the 90o-twisted form of B2F4 ( D2d) The vibrations are: 3 A1 (Raman, polarized) + B1 (Raman) + 2 B2 (IR and Raman ) + 3 E (IR and Raman).

(d) A d z2 orbital?

6.14

SO32– ion? (a) Point group? C 3v (b) Degenerate MOs? 2

are doubly degenerate.

PF5? (a) Point group? D3h. (b) Degenerate MOs? 2. (c) Which p orbitals have the maximum degeneracy? 3 px and 3 py atomic orbitals are

6.8 6.9

Use the projection operator method to construct the SALCs of A 1 + T2 symmetry that derive from the four H1s orbitals in methane.. 1

s = ( / 2)(ϕ 1 + ϕ 2 + ϕ 3 + ϕ 3) (= A1)

AsCl5 Raman spectrum consistent with a

px = (1 / 2)(ϕ 1 – ϕ 2 + ϕ 3 – ϕ 3) (= T 2)

trigonal bipyamidal geometry? No.

py = ( / 2)(ϕ 1 – ϕ 2 – ϕ 3 + ϕ 3) (= T 2)

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

pz = (1 / 2)(ϕ 1 + ϕ 2 – ϕ 3 – ϕ 3) (= T 2)

1

Vibrations that are IR and Raman active? (a) SF6? None. (b) BF3? The E ′ modes are active in both IR and Raman.

6.11

6.15

doubly degenerate

(b) Perpendicular to the molecular plane? 6.10

(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 T d, Γ3N reduces to: A1 + T 2. The MOs would be constructed from SALCs with H1s and 2s and 2p atomic orbitals on C.

(c) Which s and p orbitals have the maximum degeneracy? 3 px and 3 py orbitals 6.7

IR and Raman to distinguish between: (a) planar and pyramidal forms of PF 3, (b) planar and 90o-twisted forms of B2F4 ( D2h and D2d respectively)? (a) Planar PF3, D3h, vibrations are: A1’ (Raman, polarized) + 2 E ’ (IR and Raman) + A2” (IR). Pyramidal PF3, C 3v, vibrations are: 2 A1 (IR and Raman, polarized) + 2 E ’ (IR and Raman)

(b) A p orbital? An infinite number of mirror

6.6

of all 3 N displacements and irreducible representations?

A1g + A2g + B1g + B2g + E g + 2 A2u + B2u + 3 E u

an infinite number of mirror planes of symmetry, plus center of inversion, i.

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 C 2 axes that pass through the center of the orbital and are perpendicular to the C ∞ axis, and (iv) an S ∞ axis.

[AuCl4] ion?

Vibrations of a C 6v molecule that are neither IR nor Raman active? Any A2, B 1, or B 2 vibrations of a C 6v molecule will not be observed in either the IR spectrum or the Raman spectrum.

1

SALCs for -bonds (a) BF3? 1

( / √3)(ϕ 1 + ϕ 2 + ϕ 3) (= A1’) (1 / √6)(2ϕ 1 – ϕ 2 – ϕ 3) and (1 / √2)(ϕ 2 – ϕ 3) (= E ’)

(b) PF5? (axial F atoms are ϕ 4 + ϕ 5) (1 / √2)(ϕ 4 + ϕ 5) (= A1’) (1 / √2)(ϕ 4 − ϕ 5) (= A2”) (1 / √3)(ϕ 1 + ϕ 2 + ϕ 3) (= A1’) 1 1 ( / √6)(2ϕ 1 – ϕ 2 – ϕ 3) and ( / √2)(ϕ 2 – ϕ 3) (= E ’)



CHAPTER 7

Exercises 7.1

Self-tests

Name and draw the structures of the complexes? (a) [Ni(CO)4]? Nickel tetracarbonyl or tetracarbonyl nickel(0).

S7.1

Give formulas corresponding to the following names? (a) Cisdiaquadichloroplatinum(II)? cis-

CO

[PtCl2(OH2)2],

Ni

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

OC

(b) Diamminetetra(isothiocyanato) chromate(III)?

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

[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 [Rh(en)3]3+.

2-

NC (III)?

CN

2-

Cl

What type of isomers are possible for [Cr(NO2)2 6H2O]? The hydrate isomers

Co

and linkage isomers of the NO2 group. Also, [Cr(ONO)(H 2O)5]NO2 H2O.

Cl

Identifying isomers? Note that the two

Cl Cl

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

phosphine ligands in the trans isomer are related, therefore, they exhibit the same chemical shift.

S7.4

CN

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

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

S7.3

Ni

NC

(d) Bromo pentacarbonylmanganese (I)? [MnBr (CO)5].

S7.2

CO CO

Hexaamminemanganesium (II) 2

NH3

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

H3N

Mn

H3N

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-[FeCl 2(en)2]

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)(PR 3)2]? Not chiral. S7.6

Calculate all of the stepwise formation constants? K f1 = 1 X 105. K f2 will be 30% less or 30000, K f3 = 9000, K f4 = 2700, K f5 = 810, and finally K f6 = 243.

(d) [Cr(NH3)5μ –OH–Cr(NH 3)5]Cl5 7.3

Name the following complexes? (a) cis-[CrCl2(NH3)4]+? cistetra(ammine)di(chloro)chromium(III) -

(b) trans-[Cr(NCS)4(NH3)2] ? transdi(ammine)tetrakis(isothiocyanato)chromate (III)

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


14


7.4

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

L

L

M

M L

L

L

L

L

tetrahedral

(b)

Isomers

expected

square planar

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? A

A

B

B

E

M

M

E B

E

B

A

7.6

Trigonal Bipyramidal

Square based pyramid

A = axial ligands

A = axial

E = equatorial ligands

B = basal

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

[Mg(edta)(OH 2)]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.

(b) Which one of these is rare? Trigonal 7.7

prism.

(c) [IrHCO(PR 3)2] has two isomers.

Explain the difference between monodentate, bidentate, and quadridentate? A monodentate ligand can

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

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 7.9

What type of isomers do you get with ambidentate ligands? linkage isomers. 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? .

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



7.15

Draw all possible isomers for [MA2BCDE]?

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

Including optical isomers, 15 isomers are possible! . A

A A

B C

A

B

M D

E

M D

C

E

D

C

TiO2 in sunscreens? Titania articles absorb this ultraviolet radiation

A

B

M

E

S8.2

A

S8.3

Molecular shape and vibrational modes for XeF2? Trigonal bipyramidal, 4 total vibrational modes.

A B

A E

B

C

C

M D

E

B

D

D

M

C E

A

A

A

A

A

A

B

A

D

M D

E

S8.5 A

C

M

C

S8.4

M

A

E

7.16

A

M D

C B

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

B E

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

S8.6

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

S8.7

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

Exercises

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

8.1

Chiral (dien is

not planar).

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

8.2

or , of the complex

N

N

8.5

Order of stretching frequencies?

8.6

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

Δ isome

Suggest a reason why K f5 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.

Self-tests Main features of the CrO2 powder XRD pattern? XRD pattern for CrO2 will show

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. of 1800 cm 1.

8.7

UV photoelectron spectrum of NH 3? 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.

CHAPTER 8

S8.1

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

N

N

N

7.19

8.4

Ru

N

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

μm by 0.5 μm. N

N

Ru

Λ isome

8.3

N

N

N

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

Draw both isomers, [Ru(en)3]+2 ?

N

How would you determine crystalline components in mineral sample? Powder Xray diffraction.

isomer.

7.18

Why does the mass spectrum of ClBr 3 have five peaks separated by 2 u? Halogen isomers.

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

7.17

Isomer shift for iron in Sr 2FeO4? The Smaller and less positive.

planar).

(f) mer -[Co(NO2)3(dien)]?

(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. EPR signal of new material arises from W sites?

8.8

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


16

8.9


Single

13

C peak in NMR? Chemically

S9.1

distinct carbonyls are exchanging position sufficiently quickly.

8.10

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.

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 XeF 5 ? All 5 of

S9.2

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

Found in aluminosilicate minerals or sulfides?

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 Δf Hө 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 V 2O5? We would have to know the products formed upon decomposition.

Charge on Fe atoms in Fe 4[Fe(CN)6]3? EPR and Mössbauer.

8.17

No quadrupole splitting in Mössbauer spectrum of SbF 5? 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

8.22

9.1 9.2

Self-tests

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 NaCl 2? Which thermochemical step is responsible for the fact that NaCl2 does not exist?

Zeolite of composition CaAl2Si6O16.nH2O, determine n.? n=7.2 As an integer, n = 7. Ratio of cobalt to acetylacetonate in the product? The ratio is 3:1.

CHAPTER 9

Maximum stable oxidation state? (a) Ba; +2, (b) As; +5, (c) P; +5, (d) Cl; +7.

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

8.21

Exercises

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

10.3

(a) H2S? H = +1, S = –2.

decrease across a period and down a group. Ionization energy increases across a period and decreases down a group.

9.7

Assign oxidation numbers to elements?

(b) KH? H = –1, K = +1. (c) [ReH9]2 – ? H = –1, Re = +7.

Names of ores? (a) Mg; MgCO3 magnesite, (b) Al; Al2O3 bauxite, and (c) Pb; PbS galena.

(d) H2SO4? H = +1, O = –2, S = +6.

9.8

Identify the Z + 8 element for P. Similarities? V (vanadium).

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

9.9

ө Calculate H for SeF6? f

(e) H2PO(OH)? H bonded to an oxygen

10.4

ө Δ H = −1397 kJ mol-1. f

Preparation of hydrogen gas?

→ CO(g) + 3H2(g) (1000°C)

(i) CH4(g) + H2O (ii) C(s) + H2O

→ CO(g) + H2(g) (1000°C)

(iii) CO(g) + H2O

CHAPTER 10 10.5

→ CO2(g) + H2 (g)

Self-tests

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

S10.1

(b) Trends in

10.2.

Which of the following CH4, SiH4,, or GeH4 would best H+ or H donor? CH4, the strongest Bronsted acid. would be the best hydride donor.

S10.2

(a) Ca(s) + H2(g) → CaH2(s).

groups 13/III through 17/VII.

10.6

S10.3

A procedure for making Et3MeSn? 2Et3SnH + 2Na

10.7

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

→ 2Na+Et3Sn – + H2

Na+Et3Sn – + CH3Br → Et3MeSn + NaBr

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.

(b) NH3(g) + BF3(g) → H3 N–BF3(g). (c) LiOH(s) + H2(g) → NR.

See Table 10.1.

(c) Different molecular hydrides? Molecular hydrides are found in

GeH4

Reactions of hydrogen compounds?

f G º?

10.8

Name and classify the following? (a) BaH2? barium hydride.

Exercises

(b) SiH4? silane.

10.1

(c) NH3?

Where does Hydrogen fit in the periodic chart? (a) Hydrogen in group 1? Hydrogen

(d) AsH3? Arsine.

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.

10.2

ammonia,

(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) Hydrogen in group 14? There is no

(c) Variable composition? PdH0.9 .

reason for hydrogen to be placed in this group.

(d) Lewis basicity? Ammonia.

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.

10.10

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:


18


10.21.1 Dihydrogen as an oxidizing agent?

It’s reaction with an active s-block metal such as sodium.

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.

10.15

Self-tests 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

S11.3

Trend is stability of Group 1 ozonides?

kJ mol−1 and for NaF is 535 kJ mol−1.

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

Group 1 ozonides are less stable compared to the superoxides.

The synthesis of binary hydrogen compounds? (i) direct combination of the

S11.4 Sketch the thermodynamic cycle of Group 1 carbonate.

elements, (ii) protonation of a Brønsted base, and (iii) metathesis using a compound such as LiH, NaBH4, or LiAlH4.

10.16

CHAPTER 11

M2CO3(s) CO2(g)

M2O(s) +

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

2M+(g) + CO32−(g) 2 O −(g) + CO2(g)

S11.5

2M+(g) +

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

- Several period 2 compounds exhibit strong hydrogen bonding.

KNO3(s)

→ KNO2(s) + 1/2O2(g)

- boiling points of HF, H2O, and NH3 are all higher than homologues.

10.18

their

respective

period

2KNO2(s)

3

LiNO3 decomposes in one step.

Suggest a method for the preparation of BiH3? The redistribution BiH2Me.

of

→ K 2O(s) + 2NO2(g) + 1/2O2(g)

LiNO3(s)

→ 1/2 Li2O(s) + NO2(g) + 1/4O2(g)

methylbismuthine,

S11.6

the NMR spectrum at low temperature. Only one resonance in the NMR at high temperature.

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.

Predicted 7Li NMR of Li3N? Two peaks in

Exercises 11.1

(a) Why are group 1 metals good reducing agents? They have one valence electron in the n s1 subshell, and relatively low first ionization energies.



11.2

(b)

Why are group 1 metals poor complexing agents? They are large,

CHAPTER 12

electropositive metals and have little tendency to act as Lewis acids.

Self-tests

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

S12.1

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?

11.4

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

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

liquid

(c) Li+ or K +, form a complex with C2.2.2? Potassium ion. 11.5

BeCl2 is covalent; BaCl2 is ionic.

Most alkyl lithiums are made using elemental lithium with the corresponding alkyl chlorides.

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

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

Use ionic radii to predict a structure type of BeSe? According to Table 3.6 should be close to ZnS-like structure.

S12.4

Identify the compounds?

Calculate the lattice enthalpy of MgF2 and comment on how it will affect the solubility compared to MgCl2? –1

NaOH ← H2O + Sodium metal + O2 → Na2O2 + heat → Na2O + NH3

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

Exercises 12.1

↓ NaNH2 11.6

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

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

11.7

11.8

decompose to elements. Carbonates decompose to oxides.

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

The structures of CsCl and NaCl?

Because of a diagonal relationship between Be and Al.

Thermal stability of carbonates? Hydrides

hydrides versus

12.2

6-coordinate Na+, 8-coordinate Cs+.

11.9

Why are compounds of beryllium covalent whereas those of the other group 2 elements are predominantly ionic?

The effect of the alkyl group on the structure of lithium alkyls? Whether a

12.3

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.

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

12.4

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

Identify the compounds A, B, C, and D of the group 2 element M?

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.


20

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

12.8

Which of the salts MgSeO or BaSeO would be expected to be more soluble in water? MgSeO4

Simple alkenes are inert towards LiBH4.

4

S13.2

4

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

How do group 2 salts give rise to scaling from hard water?

LiBH4 + NH4Cl

S13.3

2

2

2

The two Grignard compounds C H MgBr and 2,4,6-(CH ) C H MgBr dissolve in THF. What differences would be expected in the structures of the species formed in these solutions? 2

3

3

6

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

(b) BCl3 and pyridine in hydrocarbon solution? BCl3(g) + py(l) → Cl3B − py(s)

(c) BBr3 and F3BN(CH3)3? BBr 3(l) + F3BN(CH3)3(s) Br 3BN(CH3)3(s)

S13.4

BeBr 2 should be covalent. MgBr 2 should be ionic. BaBr 2 should be ionic

12.12

5

→

BF3(g)

+

Suggest a reaction or series of reactions for the preparation of N, N’, N’’-trimethylB,B’,B’’-trimethylborazine starting with methylamine and boron trichloride? Cl3BNCH3 + 3CH3MgBr (CH3)3B3 N3(CH3)3 + 3Mg(Br, Cl)2

2

S13.5

How many framework electron pairs are present in B H and to what structural category does it belong? Sketch its structure? 4

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

BH3 NH3 + LiCl + H2

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

Predict structures for BeTe and BaTe.

Use the data in Table 1.7 and the Ketelaar triangle in Fig. 2.38 to predict the nature of the bonding in BeBr , MgBr , and BaBr .

THF

(a) BCl3 and ethanol?

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

12.11

B nuclei have I = 3/2. Predict the number of lines and their relative intensities in the 1 H-NMR spectrum of BH4 – ? 4. Relative intensity ratio is 1:3:3:1.

Salts of divalent ions have low solubility.

12.10

11

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?

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

12.9

S13.1

10

7, arachno species. The structure of B4H10:

Predict the products of the following reactions? MgCl2 + 2LiC2H5 → 2LiCl + Mg(C2H5)2 (b) Mg + (C2H5)2Hg → Mg(C2H5)2 + Hg (c) Mg + C2H5HgCl → C2H5MgCl + Hg

(a)

S13.6

CHAPTER 13 Self-tests

Propose a plausible product for the reaction between Li[B H ] and Al (CH ) ? 10

[B10H11 (AlCH3)] –

13

2

3

6



S13.7

Propose a synthesis for the polymer precursor 1,7-B C H (Si(CH ) Cl) from 1,2B C H and other reagents of your choice? 10

10

2

2

10

3

2

2

13.6

12

Δ

1,2 - B10C2H12 ⎯ ⎯→ 1,7-B10C2H12 (90%) + 1,12B10C2H12 (10%) 1,7-B10C2H10Li2 + 2Si(CH3)2Cl2 B10C2H10(Si(CH3)2 + 2LiCl

S13.8

→

1,7 -

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

No, it explodes in air. B2H6 + 3 O2 → B2O3 + 3 H2O2

13.7

Predict how many different boron environments would be present in the proton-decoupled 11B-NMR of a) B5H11, b) B4H10? a) 3 b) 2.

13.8

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

(Me)2SalCl3 + GaBr 3 → Me2SGaBr 3 + AlCl3.

(a) BH3 + (CH3)2C=CH2 B[CH2-CH(CH3)2]3

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

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

Exercises 13.1

Give a balanced chemical equation and conditions for the recovery of boron? B2O3 + 3Mg → 2B + 3MgO

13.2

13.9

Δ H < 0

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) BF3 N(CH3)3 + BCl3 N(CH3)3 + BF3 (b) BH3CO + BBr 3

13.4

13.10

Diborane has been used as a rocket propellant. Calculate the energy released from 1.00 kg of diborane given the following values of Δf Hө/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?

Using BCl3 as a starting material and other reagents of your choice, devise a synthesis for the Lewis acid chelating agent, F 2B– C2H4 –BF2? 2BCl3 + 2Hg  B2Cl4 + 2HgCl2 B2Cl4 + 4AgF  B2F4 + 4AgCl B2F4 + C2H4  F2CH2CH2BF2

13.11

Identify compounds A, B, and C? LiAlH4 BF3

CH

Diborane is extremely toxic, and the boron containing product of combustion is a solid, B2O3.

BCl3 (BCl3 > BF3) NR

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?

CH

-73,172 kJ.

TlBr 3 + NaBr → NaTlBr 4

13.5

(c) C = B2O3 Does B2H6 survive in air? If not, write the equation for the reaction?

Given NaBH4, a hydrocarbon of your choice, and appropriate ancillary reagents and solvents, give formulas and conditions for the synthesis of (a) B(C 2H5)3, (b) Et3NBH3? Heat, Ether

A

(a) BCl3 + 3C2H5MgCl B(C2H5)3 + 3MgCl2

(b)

H2O

CaF2

13.12 C

B heat

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

[HN(C2H5)3]Cl + NaBH4 H2 + H3BN(C2H5)3 + NaCl

Draw the B12 unit that is a common motif of boron structures; take a viewpoint along a C2 axis?


22


(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 THF

(5) 2Na2[B9C2H11] + FeCl2 2NaCl + Na2[Fe(B9C2H11)2] 2-

H

B H

H

13.13

Which boron hydride would you expect to be more thermally stable, B 6H10 or B6H12? Give a generalization by which the thermal stability of a borane can be judged?

13.15

(b)

product

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

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?

B Fe B H

H

C

B

B B

H H

C

H

B

B

B

H

H H

B H

B H

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.

of

(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)?

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

B

H

Heat

(a) nido . (b) 12. (c) The total number of valence elections is

13.17

B

(g)

The boron containing combustion is a solid, B2O3.

13.16

C H

H

(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? 2B5H9 (l) + 12O2 5B2O3 (s) + 9H2O (l)

B

B

How many skeletal electrons are present in B5H9? 14

(a)

B

H

B6H10

13.14

H

C

H

H

B

B H

H

(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 → Ph3 N3B3Cl3 + 9 HCl

(b) Me3N3B3H3? 3 MeNH3+Cl− + 3 BCl3 → Me3 N3B3Cl3 + 9 HCl Me3 N3B3Cl3 + 3 LiH → Me3 N3B3H3 + 3 LiCl The structures:



Exercises

13.20

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 CF 4 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?

Give the structural type and describe the structures of B4H10, B5H9, and 1,2B10C2H12? 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.

CHAPTER 14

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)4 N]+ +

Self-tests S14.1

CH3

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

N

(a) With potassium? Potassium results in a

C H3

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 CH 4 and SiH4?

The anion is SiF5 –

F Si

Propose a synthesis of D from 13CO?

13

13

CO2 –

13

starting

CO(g) + 2MnO2(s) → CO2(g) + Mn2O3(s) 2Li(s) + D2 → 2LiD(s) CO2(g) + LiD(et) → Li+D13CO2−(et)

13

F

F

–1

S14.3

-

F

CH4: Δf H= –61kJ mol –1 SiH4: Δf H= +39 kJ mol

CH3

H3C

F

(b) There are two different fluorine environments.

14.4 Draw the structure and determine the charge on the cyclic anion [Si 4O12] ?


24


14.11

8O

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

Si O O

O

Si Si O

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

O

O Si

14.12

(a) Describe the trend in band gap energy, E g, for the elements carbon (diamond) to

Sn-NMR

tin (grey). (b) Does the electrical conductivity of silicon increase or decrease when its temperature is changed from 20 C to 40 C?

14.6 Predict the appearance of the H-NMR spectrum of Sn(CH 3)4? Doublet

(a) There is a decrease in band gap energy from carbon (diamond) to grey tin. (b) Increase.

O

Charge = –8.

14.5

14.7

Predict the appearance of the spectrum of Sn(CH 3)4? Doublet

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 – ): O–H = 463, H–Cl = 431, H–Br = 366? ΔhH ΔhH

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

˚

˚

14.13

Preferably without consulting reference material, draw a periodic table and indicate the elements that form saline, metallic, and metalloid carbides?

˚

˚

14.8

Identify the compounds A to F:? (A) SiCl4 (B) SiRCl3 (C) RSi(OH) 3. (D) RSiOSiR +H2O (E) SiR 4 (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) Sn (aq) + PbO2(s) (excess) → (air excluded) (ii) Sn (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.

Group I elements Group II elements Group 13 elements Group 14 elements 3d-Block elements 4d-Block elements 5d-Block elements 6d-Block elements Lanthanides

Ionic(silane) carbides Li, Na, K, Rb, Cs Be, Mg, Ca, Sr, Ba Al

Metallic carbides

Metalloid carbides

B Si Sc, Ti, V, Cr, Mn, Fe, Co, Ni Zr, Nb, Mo, Tc, Ru La, Hf, Ta, W, Re, Os Ac 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) K 3C60? (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)


ANSWERS TO SELF-TESTS AND EXERCISES 25 Calcium carbide contains discrete C22– ions.

(c) K 3C60? A solution of C60 can be treated

NH3 + ClO− + H → [H3 N—Cl-O] − + H → H2 NCl + H2O +

with elemental potassium. It is ionic.

+

H2 NCl + NH3 → H2 NNH2 + HCl Both can be thought of as redox reactions.

14.15

Write balanced chemical equations for the reactions of K 2CO3 with HCl(aq) and of Na4SiO4 with aqueous acid?

S15.5

K 2CO3(aq) + 2HCl(aq) → 2KCl(aq) + CO2(g) + H2O(l)

3

Describe in general terms the nature of the [SiO3]n ion in jadeite and the silicaalumina framework in kaolinite? The structures of jadeite and kaolinite consist of extended one- and two-dimensional structures, respectively. The [SiO32– ]n ions in jadeite are a linear polymer of SiO 4 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.


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

S15.4

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

3

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

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

14.16

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

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 N Nonmetal P Nonmetal As nonmetal Sb metalloid Bi metalloid O nonmetal S nonmetal Se nonmetal Te nonmetal

15.2

Diatomic gas? yes

Inert pair effect? No

no

No

no no no yes no no no

No No Yes No No No No

(a) Give complete and balanced chemical equations for each step in the synthesis of H PO 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? 3

4

(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?


26


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 Li N or (b) the hightemperature, high-pressure reduction of N by H . Give balanced chemical equations for each method starting with N , Li, and H , as appropriate. (c) Account for the lower cost of the second method?

P4O10 + 6H2O → 4H3PO4

(c) Reaction of the product from part (b) with CaCl2? 2H3PO4(l) + 3CaCl2(aq) →

3

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

2

2

15.9

2

2

Starting with NH3(g) and other reagents of your choice, give the chemical equations and conditions for the synthesis of (a) HNO , (b) NO2 – , (c) NH OH, (d) N3 – ? 3

(a) Hydrolysis of Li3N?

(a) HNO3?

6Li + N2 → 2Li3 N

4NH3(aq) + 7O2(g) → 6H2O(g) + 4NO2(g)

2Li3 N + 3H2O → 2NH3 + 3Li2O

High temperature

(b) Reduction of N2 by H2?

(b) NO2 – ?

N2 + 3H2 → 2NH3

2NO2(aq) + 2OH−(aq) →

(c) Account for the difference in cost?

15.4

The second process is considerably cheaper than the first, because lithium is very expensive.

(c) NH2OH? cold aqueous acidic solution

Show with an equation why aqueous solutions of NH NO are acidic?

NO2−(aq) + 2HSO3−(aq) + H2O(l) → NH3OH+(aq) + 2SO42−(aq)

NH4 NO3(s) + H2O  NH4+ + NO3-(aq)

(d) N3 – ? at elevated temperatures:

4

15.5

2

NO2−(aq) + NO3−(aq) + H2O(l)

3

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

3NaNH2(l) + NaNO3 →

2

NaN3 + 3NaOH + NH3(g) 2NaNH2(l) + N2O → NaN3 + NaOH + NH3

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?

15.10

4

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) PCl 4+, (b) PCl4 – , (c) AsCl ? 5

(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 P 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 CaCl and name the product? 4

2

(a) Oxidation of P4 with excess O2? P4 + 5O2 → P4O10

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

Write the balanced chemical equation corresponding to the standard enthalpy of formation of P O (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? 10

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.


ANSWERS TO SELF-TESTS AND EXERCISES 27 and H2PO22− useful reducing agents?

as

oxidizing

or

E º = 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? D

LiAlH4

15.12

Cl2

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 NO 2 – oxidations?

AS

When equal volumes of nitric oxide (NO) and air are mixed at atmospheric pressure a rapid reaction occurs, to form NO and N O . 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?

C

(a) A = AsCl3 (b) B = AsCl5 (c) C = AsR 3 (d) D = AsH3

2

2

4

15.17

Sketch the two possible geometric isomers of the octahedral [AsF Cl ] – and explain how they could be distinguished by 19F– NMR? 4

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

15.14

B

3RMgBr

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

15.13

Cl2/hv

A

2

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

Give balanced chemical equations for the reactions of the following reagents with PCl and indicate the structures of the products: (a) water (1:1), (b) water in excess, (c) AlCl , (d) NH Cl? 5

3

4

(a) H2O? tetrahedral POCl3. PCl5 + H2O → POCl3 + 2HCl

(b) H2O in excess? 2PCl5(g) + 10HCl(aq)

8H2O(l)

→

2H3PO4(aq)

+

(c) AlCl3? tetrahedral. PCl5 + AlCl3 → [PCl4]+[AlCl4]−

(d) NH4Cl? Cyclic molecules or linear chain polymers nPCl5 + n NH4Cl →

−[(N = P(Cl)2)n]− + 4nHCl

15.18

Identify the nitrogen compounds A, B, C, D, and E? (a) A = NO2 (b) B = HNO3; C = NO

15.15

Use standard potentials (Resource section 3) to calculate the standard potential of the reaction of H3PO2 with Cu2+. Are HPO22−

(c) D = N2O4 (d) E= NO2 (e) F = NH4+


28

15.19


Use the Latimer diagrams in Resource 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, N 2O, NH3OH+, H4P2O6, and P.

16.3

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

16.4

Which of the solvents ethylenediamine (which is basic and reducing) or SO (which is acidic and oxidizing) might not react with (a) Na S , (b) K Te ? 2

2

CHAPTER 16 16.5

Determine whether the decomposition of H O is spontaneous in the presence of either Br or Cl ? 2

2

2

The decomposition of H2O2 is thermodynamically favored in presence of Cl – .

16.6

16.7

Probable structures of SO2F and (CH3) 3NSO2, and predict reactions with OH-. Both are trigonal pyramidal. OH- displaces F— or (CH3) 3 N—.

2

2

2

16.8

3

(a) Use standard potentials (Resource section 3) to calculate the standard potential of the disproportionation of H O in acid solution. (b) Is Cr a likely catalyst for the disproportionation of H O ? (c) Given the Latimer diagram 2

-0.13

1.51

HO2-

O2

16.9

2

2

H2O2

in acidic solution, calculate ΔrG ө for the disproportionation of hydrogen superoxide (HO ) into O and H O , and compare the result with its value for the disproportionation of H O ? 2–

2

2

Use the standard potential data in Resource section 3 to predict which oxoanions of sulfur w ill disproportionate in acidic conditions? 2 2 S2O6 −and S2O3 −.

2+

2

(a) Give the formula for Te(VI) in acidic aqueous solution and contrast it with the formula for S(VI). (b) Offer a plau sible explanation for this difference?

Tellurium is a larger element than sulfur and can increase its coordination number.

CO2, SO3, P2O5, and Al2O3 are amphoteric. CO is neutral; MgO and K 2O are acidic.

16.2

Predict which oxidation states of Mn will be reduced by sulfite ions in basi c conditions?

(b) An explanation?

State whether the following oxides are acidic, basic, neutral, or amphoteric: CO , P O , SO , MgO, K O, Al O , CO? 3

2–

(a) Formulas? H5TeO6 – and HSO4 –

Exercises

5

2–

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

-

2

Rank the following species from the strongest reducing agent to the strongest oxidizing agent: SO42–,SO32–,O3SO2SO32–? 2–

In Presence of Cl –

16.1

3

S2O8 > SO4 > SO3

2

The decomposition of H2O2 is – thermodynamically favored in presence of Br

S16.2

2

ethylenediamine is a better solvent than sulfur dioxide.

Self-tests S16.1

4

Use the standard potential data in Resource section 3 to predict whether SeO32– is more stable in acidic or basic solution? The SeO32– is marginally more stable in acid solutions.

16.10

Predict whether any of the following will be reduced by thiosulfate ions,S 2O32– , in acidic conditions: VO2+, Fe3+, Cu+, Co3+? Fe3+ and Co3+ will be reduced.

2

(a) The disproportionation of H2O2 and HO2? 1.068 V.

SF reacts with BF to form [SF ][BF ]. Use VSEPR theory to predict the s hapes of the cation and anion?

(b) Catalysis by Cr2+?

SF4+ trigonal pyramidal, BF4− tetrahedral.

2

2

2+

Cr

16.11

is not capable

4

3

3

4

of decomposing H2O2.

(c) The disproportionation of HO2? Δr G° = 157 kJ. For the disproportionation of H2O2 (part (a)), Δr G° = 103 kJ.

16.12

Tetramethylammonium fluoride (0.70 g) reacts with SF (0.81 g) to form an ionic product. (a) Write a balanced equation for 4



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

B r 2

liqu id

lower

soft er

I2

soli d

lowest

soft est

H e

gas

N e A r K r X e

gas

19

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

16.13

Identify the sulfur-containing compounds A, B, C, D, E, and F? A = S2Cl2, B = S4 N4, C = S2 N2, D = K 2S2O3, E = S2O62−, F = SO2.


One source of iodine is sodium iodate, NaIO . Which of the reducing agents SO(aq) or Sn (aq) would seem practical from the standpoints of thermodynamic feasibility and plausible judgements about cost? Standard potentials are given in Resource section 3?

gas gas gas

gree n dark red bro wn dark viol et Col orle ss colo rless colo rless colo rless colo rless

3

2+

2

17.2

Both. SO2 is cheaper .

S17.2

Predict the 19F-NMR pattern for IF7?

For F:

2 resonances.

S17.3

F2 + H2SO4 → CaSO4 + 2 HF

From the perspective of structure and bonding, indicate several polyhalides that are analogous to [py–I–py]+, and describe their bonding?

2 HF + 2 KF → 2 K +HF2− 2 K +HF2− + electricity → F2 + H2 + 2 KF For Cl:

Examples include I3 – , IBr 2 – , ICl2 – , and IF2 – . The three centers contribute four electrons to three molecular orbitals, one bonding, one nonbonding, and one antibonding.

2 Cl− + 2 H2O + electricity → Cl2 + H2 + 2 OH− For Br and I:

Exerci ses 17.1

Preferably without consulting reference material, write out the halogens and noble gases as they appear in the periodic table, and 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?

F

Phy sica l stat e

Electron egativit y

gas

highest (4.0)

2

C l2

Describe how the halogens are recovered from their naturally occurring halides and rationalize the approach in terms of standard potentials. Give balanced chemical equations and conditions where appropriate?

gas

lower

Har dnes s of hali de ion hard est soft er

Col or

2 X− + Cl2 → X2 + 2 Cl− I−)

17.3

(X− = Br −,

Sketch a choralkali cell. Show the half-cell reactions and indicate the direction of diff usion of the ions. Give the chemical equation 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) → Cl2(g) + 2 e− cathode: 2 H2O(l) + 2 e− → 2 OH−(aq) + H2(g)

light yell ow yell ow-

-

unwanted reaction: 2 OH−(aq) + Cl2(aq) → ClO−(aq) + Cl−(aq) + H2O(l)


30

17.4


Sketch the form of the vacant s* orbital of a dihalogen molecule and desc ribe its role in the Lewis acidity of the dihalogens?

shapes of IF and the cation and anion in X, (c) predict how many 19F-NMR signals would be observed in IF and X? 3

3

a). b) IF3:

X = IF4 N(CH3)4 Different possible arrangements of

Since the 2σu* antibonding orbital is the LUMO for a X2 molecule, it is the orbital that accepts the pair of electrons from a Lewis base.

17.5

Which dihalogens are thermodynamically capable of oxidizing H O to O ? 2

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

2

Cl2 and F2.

17.6

Nitrogen trifluoride, NF , boils at –129 C and is devoid of Lewis basicity. By contrast, the lower molar mass compound NH 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?

17.9

(a) Difference in volatility? Ammonia

17.10

3

3

˚

c).IF3, two. IF4 – , one.

˚

exhibits hydrogen bonding.

Use the VSEPR model to predict the shapes of SbCl , FClO , and [ClF ]+? 5

Indicate the product of the reaction between ClF and SbF ? 5

The strong electron-withdrawing effect in NF3 reduces the basicity.

Based on the analogy between halogens and pseudohalogens write: (a) the balanced equation for the probable reaction of cyanogen, (CN) , with aqueous sodium hydroxide, (b) the equation for the probable reaction of excess thiocyanate with the oxidizing agent MnO (s) in acidic aqueous solution, (c) a plausible structure for trimethylsilyl cyanide? 2

2

(a) The reaction of NCCN with NaOH? NCCN(aq) + 2 OH−(aq) → CN−(aq) + NCO−(aq) + H2O(l)

(b) The reaction of SCN – with MnO2 in aqueous acid? 2SCN−(aq) + MnO2(s) + 4 H+(aq) → (SCN)2(aq) + Mn2+(aq) + 2 H2O(l)

(c) The structure of trimethylsily l cyanide? Trimethylsilyl cyanide contains an Si–CN single bond.

Given that 1.84 g of IF reacts with 0.93 g of [(CH ) N]F to form a product X, (a) identify X, (b) use the VSEPR model to predict the 3

3

4

5

[ClF4]+[SbF6] – .

17.11

Sketch all the isomers of the complexes MCl F and MCl F . Indicate how many fluorine environments would be indicated in the 19F-NMR spectrum of each isomer? 4

17.8

6

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

(b) Explain the difference in basicity?

17.7

3

2

3

3

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 [IF ] and IF . (b) Give a plausible chemical equation for the preparation of [IF ][SbF ]?

17.19

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

17.20

6

6

+

7

Explain why CsI (s) is stable with respect to the elements but NaI (s) is not? 3

3

Large cations stabilize large, unstable anions.

6

IF6+,

Write plausible Lewis structures for (a) ClO and (b) I O and predict their shapes and the associated point group? 2

octahedral; IF7, pentagonal bipyramid.

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

2

6

(a).

IF7 + SbF5 → [IF6][SbF6]

17.13

Predict the shape of the doubly chlorine– bridged I Cl molecule by using the VSEPR model and assign the point group? 2

(b) I2O5?

6

A planar dimer (I2Cl6).

17.14

Predict the structure and identify the point group of ClO F? trigonal pyramidal, C s 2

symmetry.

17.15

Predict whether each of the following solutes is likely to make liquid BrF a stronger Lewis acid or a stronger Lewis base: (a) SbF , (b) SF , (c) CsF?

17.21

3

5

6

(b) SF6? No effect on the acidity or basicity

(a) The formulas are HBrO4 and H5IO6. The difference lies in iodine’s ability to expand its coordination shell.

of BrF3.

(b) Relative stabilities? Periodic acid is

(c) CsF? Increases the basicity of BrF3.

thermodynamically more stable.

(a) SbF5? increases the acidity of BrF3.

17.16

Predict the appearance of the F-NMR spectrum of IF ? Two resonances. 19

5

17.22

+

.

17.17

Predict whether each of the following compounds is likely to be dangerously explosive in contact with BrF and explain your answer: (a) SbF , (b) CH OH, (c) F , (d) S Cl ? 3

5

2

3

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

2

(a) The expected trend? E decreases as the

2

pH increases.

(a) SbF5?

Since SbF5 cannot be oxidized, it will not form an explosive mixture with BrF3.

(b) E at pH 0 and pH 7 for ClO4 – ?

(b) CH3OH? Methanol, being an organic

E º = 1.201 V (see Appendix 2).

compound, is readily oxidized by strong oxidants.

At pH 7, V = 0.788 V

17.23

(c) F2? No. (d) S2Cl2? S2Cl2 will be oxidized to higher valent sulfur fluorides.

17.18

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

The formation of Br from a tetraalkylammonium bromide and Br is only slightly exoergic. Write an equation (or NR for no reaction) for the interaction of [NR ][Br ] with I in CH Cl solution and give your reasoning?

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 scission.

– 3

2

4

3

2

Br 3− + I2 → 2 IBr + Br −

2

2

17.24

Which oxidizing agent reacts more readily in dilute aqueous solution, per chloric acid or periodic acid? Give a mechanistic explanation for the diff erence? Periodic acid.


32

17.25


(a) For which of the following anions is disproportionation thermodynamically favourable in acidic solution: OCl , ClO2 – , ClO2 – , and ClO4 – ? (If you do not know the properties of these ions, determine them from a table of stan dard potentials.) (b) For which of the favourable cases is the reaction very slow at room temperature?

18.2

2

The rates of disproportionation are probably HClO > HClO2 > ClO3 – .

17.26

(b) An electric discharge light source requiring a safe gas with the lowest ionization energy? Xenon. (c) The least atmosphere? Argon. 18.3

Which of the following compounds present an explosion hazard? (a) NH ClO , (b) Mg(ClO ) , (c) NaClO , ( d) [Fe(H O) ][ClO ] . Explain your reasoning? 4

4

2

4

2

4

6

4

expensive

inert

By means of balanced chemical equations and a statement of conditions, describe a suitable synthesis of (a ) XeF2? Xe and F2 at 400ºC, or photolyze Xe and F2 in glass:

2

Xe(g) + F2(g) → XeF2(s)

(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) XeF6? High temperature, but have a large

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

(c) XeO3?

excess of F2: Xe(g) + 3 F2(g) → XeF6(s)

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

(c) NaClO4? Not an explosion hazard. (d)

[Fe(H2O)6]

[ClO4]2? An explosion

hazard, since Fe(II) can be oxidized to Fe(III).

17.27

Which of the noble gases would you choose as (a) The lowest-temperature refrigerant? Helium.

18.4 Draw the Lewis structures of (a) XeOF 4? (b) XeO2F2? (c) XeO64 ?

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. (d) Co2+? No.

CHAPTER 18

18.5

Self-tests S18.1

(b) IBr2 – ? Linear geometry. Isostructural with XeF2.

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

2 HXeO4−(aq) + 2 OH−(aq) → XeO64−(aq) + Xe(g) + O2(g) + 2 H2O(l)

Give the formula and describe the structure of a noble gas species that is isostructural with (a) ICl4−? XeF4

(c) BrO3 – ? Trigonal pyramidal geometry. Isostructural with XeO3.

(d) ClF? Isostructural with the cation XeF+. 18.6

(a) Give a Lewis structure for XeF7 – ?

Exercises 18.1

Explain why helium is present in low concentration in the atmosphere even though it is the second most abundant element in the universe? Helium present in today’s atmosphere is the product of ongoing radioactive decay.

(b) Speculate on its possible structures by using the VSEPR model and analogy w ith other xenon fluoride anions ? Pentagonal bipyramid.



18.7

Use molecular orbital theory to calculate the bond order of the diatomic species E 2+ with E=He and Ne? He22+ = 0.5, Ne22+ = 0.5.

the group oxidation number is not achieved by N?

18.8 Identify the xenon compounds A, B, C, D, and E? A = XeF2(g) +

B = [XeF] [MeBF3] C = XeF6 D = XeO3 E = XeF4 (g).

18.9

−

Predict the appearance of the spectrum of XeOF3+.

19.2 129

Xe-NMR

Explain why the enthalpy of sublimation of Re(s) is significantly greater than that of Mn(s)? As atoms become heavier, more energy is needed to vaporize them.

1:3:3:1 quartet

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

19.3

Strong central line, two other lines symmetrically distri bu ted around the central line.

CHAPTER 19

The group oxidation number increases in stability as you descend a group.

19.4

Self-tests S19.1

Refer to the appropriate Latimer diagram in Resource section 3 and identify the oxidation state and formula of the species that is thermodynamically favoured when an acidic aqueous solution of V2+ is exposed to oxygen? +4; VO2

S19.2

S19.3

Cr 2+(aq) + Fe3+(aq) → Cr 3+(aq) + Fe2+(aq)

(b) CrO42– (aq) + MoO2 (s) →? 2 + 2 CrO4 − + 3 MoO2(s) + 10 H →

2 Cr 3+ + 3 H2MoO4 + 2 H2O

Suggest a use for molybdenum(IV) sulfide that makes use of its solid-state structure. Rationalize your suggestion? MoS2 used as a lubricant. The slipperiness of MoS2 is

(c) MnO4 – (aq) + Cr3+ (aq) →? 6 MnO4− + 10Cr

3+

19.5

Describe the probable structure of the compound formed when Re 3Cl9 is dissolved in a solvent containing PPh3?

(a) Which ion, Ni (aq) or Mn (aq), is more likely to form a sulfide in the presence of H S? (b) Rationalize your answer with the trends in hard and soft character acros s Period 4. (c) Give a balanced chemic al equation for the reaction. 2+

2+

2

More likely for Ni

2+

to form a sulphide.

Hardness decreases from left to right in the d block.

Exerci ses Without reference to a periodic table, sketch the first series of the d block, including the symbols of the elements. Indicate those elements for which the group oxidation number is common by C, those for which the group oxida tion number can be reached but is a powerful oxidizing agent by O, and th ose for which

+ 11H2O →

6Mn2+ + 5Cr 2O72− + 22 H+

Discrete molecular species such as Re3Cl123– or Re3Cl9(PPh3)3 are formed.

19.1

For each part, give balanced chemical equations or NR (for no re action) and rationalize your answer in terms of trends in oxidation states. (a) Cr2+ (aq) + Fe3+ (aq) →?

+

because of the ease with which one layer can glide over another.

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 acidic solution for Groups 5 and 6.

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

19.6

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 rutil e or fluorite structures, and (c) indicate the regio n of the periodic table in which metal-metal bonded halide compounds are formed, giving one example?


34


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

19.9

Speculate on the structures of the following species and present bonding models to justify your answers: (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

19.7

Write a balanced chemical equation for the cisreaction that occurs when [RuLCl(OH2)]+ (see Fig. 19.9) in acidic 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 . Give other examples and a reason for the redox state of the metal center affecting the extent of protonation of coordinated oxygen.

three are dioxo complexes and will have cisdioxometal structural units. (d) VOCl42– , has a square-pyramidal structure with an apical oxo ligand.

19.10

2

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

(b) NbCl4? significantly covalent. (c) FeF2? ionic. (d) PtS ? significant amount of covalent character.

(e) WCl2 ? metal–metal bonding. 19.11

As oxidation state of the metal increases, – ability to accept electron density from an OH 2– or O ligand increases.

19.8

molybdenum-molybdenum quadruple bond.

(b) (c)

2+

5 ReO4 (aq) + 2 Mn (aq) + 25 Cl (aq) + 24 H+(aq)

warm

HBr(aq)?

6 MoCl2(s) + 2 Br −(aq) → [Mo6Cl12]2−(aq) + Br 2(aq)

(e) TiO(s) with HCl(aq) under an inert atmosphere? 2TiO (s) + 6H+ (aq)  2Ti3+ + H2(g) + 2H2O(l) E = +0.37 V. ˚

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

[Cu2(O2CCH3)4]?

σ2π4δ2δ*2π*4σ*2

19.12

Explain the differences in the following redox couples, measured at 25 oC? Higher oxidation states become more stable on descending a group.

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

plus

σ2π4δ2,chromium–

configuration, no metal–metal bond in this molecule.

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

MoCl2(s)

[Cr2(O2CC2H5)4]?

chromium quadruple bond.

[Mo6O19]2−(aq) + 5 H2O(l)

(d)

Indicate the probable occupancy of s, p, and d bonding and antibonding orbitals, and the bond order for the followi ng tetragonal prismatic complexes? (a) [Mo2(O2CCH3)4]? σ2π4δ2 configuration,

Give plausible balanced chemical reactions (or NR for no reaction) for the following combinations, and state the basi s for your answer: (a) MoO42 – (aq) plus Fe2+ (aq) in acidic solution? No reaction. (b) The preparation of [Mo6O19]2 (aq) from K 2MoO4(s)? 6 MoO42−(aq) + 10 H+(aq) →

2

compound with significant degree of covalent character

+

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

4

2

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

Which of the following are likely to have structures that are typical of (a) predominantly ionic, (b) significantly covalent, (c) metal-metal bonded compounds: NiI , NbCl , FeF , PtS, and WCl ? Rationalize the differences and speculate on the structures? (a) NiI 2? ionic

Hg2+ (aq) + Cd(s)  Hg(l) + Cd2+ (aq)

19.13

Addition of sodium ethan oate to aqueous solutions of Cr(II) gives a red diamagnetic product. Draw the structure of the product, noting any features of interest?



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

What terms arise from a p1d 1 configuration? 1F and 3F terms are possible.

S20.6

Identifying ground terms (Hint: Because d 9 is one electron short of a closed shell wit h L = 0 and S = 0, treat it on the same footing as a d1 configuration.) (a) 2 p2? 3P (called a

O

O O

Cr O

S20.5

Cr

O O

“triplet P” term). (b) 3d 9? 2D (called a “doublet D” term).

O O

S20.7

What terms in a d 2 complex of Oh symmetry correlate with the 3F and 1D terms of the free atom? An F terms are 3

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.

S20.8

-

[Ru2Cl2(ClCO2)4] : 2.5 +, mixed valence at the Ru center, in line with the energy level scheme [0.5 (8-3)].

3

3

T1g, T2g, and A2g. Similarly, D terms are 1 T2g and 1Eg.

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

–1

17500 cm and 22400 cm .

S20.9

[Ru2 (CH3COCH3)2(ClCO2)4]: 2+, the bond order of 2.0 should also be expected [0.5x(84)].

CHAPTER 20

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 ε max= 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 lowlying π* orbitals.)

Self-tests S20.1

(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.

What is the LFSE for both high- and lowspin d 7 configurations? A high-spin d 7 configuration is t25geg2. High spin, 0.8 Δo.

(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 –1 23,800 cm are of this type.

S20.2 The magnetic moment of the complex [Mn(NCS)6]4– is 6.06 μB. What is its electron 3 2 configuration? t 2g eg 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 t 2g electrons that are largely metal-based. The

(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 ext2y, 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+? d 6, low spin, no unpaired electrons, LFSE = 2.4Δ0. (b) [Fe(OH2)6]2+? d 6, high spin, unpaired electrons, LFSE = 0.4Δ0.

four


36

ANSWERS TO SELF-TESTS AND EXERCISES (c) [Fe(CN)6]3 – ? d 5, low spin, one unpaired

(c) [Fe(CN) 6]3–

electron. LFSE is 2.0Δ0. 3+

(d) [Cr(NH3)6] ?

(d) The ruthenium complex. 3

d , three electrons, low spin, LFSE = 1.2Δ0.

unpaired

(e) [W(CO)6]? d 6, low spin, no unpaired

(e) The Co2+ complex. 20.6

electrons, LFSE = 2.4Δ0.

(f) Tetrahedral [FeCl4]2 ? d 6, high spin, four unpaired electrons, LFSE = 0.6 ΔT.

(g) Tetrahedral [Ni(CO)4]? d 10, 0 unpaired electrons, LFSE = 0.

20.2

that lead to the values: (i) decreasing ionic radius from left to right across the d block, and (ii) LFSE.

-

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?

.20.7

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 [Co(NH 3)6]3+ [Fe(OH2)6]2+ 3– [Fe(CN) 6] [Cr(NH3)6]3+ [W(CO)6] [FeCl4]2– [Ni(CO) 4]

20.4

N 0 4 1 3 0 4 0

μSO = [( N )( N + 2)]1/2

20.8

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

20.9

The spectrum of d1 Ti3+(aq) is attributed to a single electronic transition eg ← t 2g. 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)63+ has the configuration t 2g0eg1, 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 excitedstate ion, and a lower energy transition from an Oh ground-state ion to a lower energy distorted excited-state ion.

20.10

[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.

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.

0 4.9 1.7 3.9 0 4.9 0

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,

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 rocksalt structure: CaO (3460), TiO (3878), VO (3913), MnO (3810), FeO (3921), CoO (3988), NiO (4071)? There are two factors

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

(b) 3 p2? 20.13

εmax52 dm3 mol-1 cm-1, and a strong band at higher energy with εmax= 2 x 10 4 dm3 mol –1

Give the Russell-Saunders terms of the configurations and identify the ground term? (a) 4 s1? 2S. 1

3

1

cm –1. What do you suggest for the origins of these transitions? First, the intense band

3

D, P, S. The ground term is P. 3+

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 spinforbidden ligand field transition.

3

The gas-phase ion V has a F 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 - 8 B, E (3P) = A + 7 B, E (1D) = A - 3 B + 2C . Calculate the values of B and C for V3+. B = (12920 cm –

20.20

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. 3+

color, which is only observed when looking through a long pathlength of bottle glass, is caused by spin-forbidden ligand field transitions.

6

(c) High-spin [Fe(H2O)6] ? A1g. 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 ≈

20.21

770 cm –1.

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

3+

and B ≈ 720 cm .

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 spinforbidden 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 d 6 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 πacid.

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

Solutions of [Cr(OH2)6]3+ ions are pale blue– green but the chromate ion, CrO42– , is an intense yellow. Characterize the origins of the transitions and explain the relative intensities. The blue-green color of the Cr 3+ ions in [Cr(H2O)6] is caused by spinallowed 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.

–1

20.16

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 Fe 3+ in the silicate matrix. Suggest which transitions are responsible for the color? The faint green

20.22

Classify the symmetry type of the d orbital in a tetragonal C 4v 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 C 4v complex. The Cl atom lone pairs of electrons can form π molecular orbitals with d xz and d yz. These metal atomic orbitals are π-antibonding MOs, and so they will be raised in energy.


38

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 MnO 4ions 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 chargetransfer transitions, even though ΔT cannot be observed directly . The oxidation state of manganese in permanganate anion is Mn(VII), which is d 0. Therefore, no ligand field transitions are possible. The difference in energy between the two transitions, E (t 2) – 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(OH 2)6]2+. Explain why this is. The extra charge of the iron complexes keeps the eg and t2g levels close

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 = (kK E[Mn(OH2)62+][X−])/(1 + K E[X−])

21.5


Calculate the second-order rate constant for the reaction of trans-[PtCl(CH3) (PEt3)2] with NO2 – in MeOH, for which npt = 3.22 ? k 2(NO2 ) = 100.71 = 5.1 M 1s 1

S21.2

Given the reactants PPh , NH , and [PtCl ] , Propose efficient routes to both cis- and trans- [PtCl2(NH3)(PPh3)]? 3

3

4

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.

2-

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 ratedetermining 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 S21.3

Use the data in Table 21.8 to estimate an appropriate value for K E and calculate k r2 for the reactions of V(II) with Cl- if the observed second-order rate constant is 1.2x102 dm3 mol –1 s –1. K E = 1 M –1, k = 1.2 × 102 s –1.

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

For a detectable amount of [Ni(CN)5]3– to build up in solution, the forward rate constant k f must be numerically close to or greater than the reverse rate constant k r.

21.10

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

21.14

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

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

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?

Predict the products of the following reactions: (a) [Pt(PR 3)4]2+ + 2Cl – ? cis[PtCl2(PR 3)2].

(b) [PtCl4]2 – + 2PR 3? trans-[PtCl2(PR 3)2]. (c) cis-[Pt(NH3)2(py)2]2+ + 2Cl – ? trans[PtCl2(NH3)(py)].

21.15

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

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+? 3+

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

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.

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.

(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(N 3)(NH3)5]2+, [V(OH2)6]2+} →

{[Co(N3)(NH3)5]2+, [V(OH2)6]2+} → 2+ 2+ {[Co(N3)(NH3)5] , [V(OH2)5] , H2O} 2+ 2+ {[Co(N3)(NH3)5] , [V(OH2)5] , H2O} → [(NH3)5Co–N=N=N–V(OH 2)5]4+

[(NH3)5Co–N=N=N–V(OH 2)5]4+ → [(NH3)5Co– N=N=N–V(OH 2)5]4+ 4+

[(NH3)5Co–N=N=N–V(OH 2)5] [V(N3)(OH2)5]2+

2+

→ [Co(OH2)6]

The outer sphere pathway: [Co(N3)(NH3)5]2+ + [V(OH2)6]2+ → {[Co(N3)(NH3)5]2+ [V(OH 2)6]2+}

+


40

ANSWERS TO SELF-TESTS AND EXERCISES 2+

2+

+

Using these values gives K 12 = 1.81 × 10 . Substitution of these values in the Marcus15 Cross relationship gives k 12 = 8.51 × 10 The dm3mol−1s−1.21.21 28

{[Co(N3)(NH3)5] ,[V(OH 2)6] } → {[Co(N3)(NH3)5] , [V(OH2)6]3+} {[Co(N3)(NH3)5] +,[V(OH 2)6]3+} → {[Co(OH2)6]2+ , 3+ [V(OH2)6]

21.18

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

The compound [Fe(SCN)(OH2)5]2+ can be detected in the reaction of 2+ 2+ [Co(NCS)(NH3)5] with Fe (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+ ( E σ (V3+/V2+) = –0.255 V) and the oxidants (a) [Ru(NH3)6]3+ ( E O(Ru3+/Ru2+) = + 0.07 V), (b) [Co(NH3)6]3+ ( E O(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+ ( E O –(Cr3+/Cr2+) = –0.41 V) and each of the oxidants [Ru(NH3)6]3+ ( E O(Ru3+/Ru2+) = +0.07 V), [Fe(OH2)6]3+ ( E O(Fe3+/Fe2+) = +0.77 V) and [Ru(bpy)3]3+ ( E O(Ru3+/Ru2+) = +1.26 V). Comment on the relative sizes of the rate constants

[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.


Is Mo(CO)7 likely to be stable?

S22.2

What is the electron count for and oxidation number of platinum in the anion of Zeise’s salt, [PtCl3(C2H4)] – ? Treat CH =CH as a neutral two-electron donor.

(a) k 11 (Cr 3+/Cr 2+) = 1 × 10−5 dm3mol−1s−1; k 22

(Ru3+/Ru2+ for the hexamine complex) = 6.6 ×

2

2

Electron count, 16; oxidation number, +4.

o

103 dm3mol−1s−1; f 12 = 1; K 12 = e [nF ε /RT] o where ε = 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 K 12 = 1.32 × 108. Substitution of these values in the MarcusCross relationship gives k 12 = 2.95 × 103 dm3mol−1s−1. (b) k 11 (Cr 3+/Cr 2+) = 1 × 10−5 dm3mol−1s−1; k 22 (Fe3+/Fe2+ for the aqua complex) = 1.1

No.

S22.3

What is the formal [Ir(Br)2(CH3)(CO)(PPh3)2]?

name

of

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.

o

dm3mol−1s−1; f 12 = 1; K 12 = e[nF ε /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. 19 Using these values we get K 12 = 9.26 × 10 . Substitution of these values in the Marcus7 Cross relationship gives k 12 = 3.19 × 10 dm3mol−1s−1. (c) k 11 (Cr 3+/Cr 2+) = 1 × 10−5 dm3mol−1s−1; k 22 (Ru3+/Ru2+ for the bipy complex) = 4 × 108 o

dm3mol−1s−1; f 12 = 1; K 12 = e[nF ε /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.

(a) [( 6-C7H8)Mo(CO)3] (49)?

η6-C7H8 = 6 electrons, Mo atom = 6. carbonyl = 2 electrons, total = 18.

(b) [( 7-C7H7)Mo(CO)3]+(51)? η7-C7H7+ = 6 electrons, Mo atom = 6, 3 COs = 6, total = 18.

S22.6

Propose a synthesis Mn(CO)4(PPh3)(COCH3) starting [Mn2(CO)10], PPh3, Na and CH3I.

for with



Mn2(CO)10 + 2 Na → 2Na[Mn(CO) 5]

Exercises

Na[Mn(CO) 5] + CH3I → Mn(CH3)(CO) 5 + NaI

22.1

Mn(CH3)(CO)5 + PPh3 → Mn(CO)4(PPh3)(COCH3)

S22.7

S22.8

S 22.9

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

(a) Fe(CO)5? Pentacarbonyliron(0), 18e – ; (b) Mn2(CO)10?

(c) V(CO)6? hexcarbonylvanadium(0), 17e – (d) [Fe(CO)4]2– ? tetracarbonylferrate(–2), 18e – ;

(e) La( 5-Cp*)3? tris(pentamethylcyclopentadienyl)lanthanum(I – II), 18e ;

(f) Fe( 3-allyl)(CO)3Cl? (g) Fe(CO)4(PEt3)?

tetracarbonyltriethylphosphineiron(0);

(h) Rh(CO)2(Me)(PPh3)? dicarbonylmethyltriphenylphosphinerhodium( I),16e

(i) Pd(Cl)(Me)(PPh3)2?

tetrahedron with 4 terminal Cp rings and four capping COs.

chloromethylbis(triphenylphosphine) palladium(II), 16e – ;

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.

(j) Co( 5-C5H5)( 4-C4Ph4)? cyclopentadienyltetraphenylcylcobutadinecob alt(I), 18e – ;

PMe3 would be preferred.

(k) [Fe( 5-C5H5)(CO2)] – ?

S22.11 Assess the relative substitutional reactivities of indenyl and fluorenyl (86) compounds? Fluorenyl compounds are more

dicarbonylcyclopentadienylferrate(0), 18e – ;

(l) Cr( 6-C6H6)( 6-C7H8)?

reactive than indenyl.

benzenecycloheptatrienechromium(0), 18e – ;

S22.12 Show that the reaction is an example of reductive elimination?

(m) Ta( 5-C5H5)2Cl3? trichlorobiscyclopentadineyltanatalum(V), 18e – ;

Cl Cl

PPh3

Ph3P

(n) Ni( 5-C5H5)NO?

PPh3

Pd

Pd Ph

–

allyltricarbonylchloroiron(II), 18e ;

The compound [Fe4(Cp)4(CO)4] is a darkgreen 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

Cl

–

decacarbonyldimanganese(0), 18e ;

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.

Ph3P

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?

+

PhCl

Cl

The decrease in both coordination number and oxidation number by 2.

–

cyclopentadineylnitrosylnickel(0),18e .

Cl

22.2

Sketch an η2 the interactions of 1,4butadiene 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

S22.13 Explain why [Pt(PEt3)2(Et)(Cl)] readily decomposes, whereas [Pt(PEt 3)2 (Me)(Cl)] does not? The ethyl group in [Pt(PEt 3)2(Et)(Cl)] is prone to β -hydride elimination.


42


with a single d-block metal atom such as cobalt? (a) C2H4? η2 (b) Cyclopentadienyl? Can be η5, η3, or η1.

C s symmetry complex, 3

(c) C6H6? η6, η4, and η2. 2

4

(d) Cyclooctadiene? η and η . (e) Cyclooctatetraene?η8, η6, η4, η2. 22.4

spectrum? Check your answer and give the number of expected bands for each by consulting Table 22.7.

Draw plausible structures and give the electron count of (a) Ni( 3-C3H5)2 (b) Co( 4-C4H4)( 5-C5H5) (c) Co( 3C3H5)(CO)3. If the electron count deviates from 18, is the deviation explicable in terms of periodic terms.

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 –1 100 cm or more lower in frequency. PMe3 is primarily a a

(b) Co( 4-C4H4)( 5-C5H5)? 18. 22.9

(2) 2 CoCO3(s) + 2 H2(g) + 8 CO(g) Co2(CO)8(s) + 2 CO2 + 2 H2O

(b) Does the electron count for each metal in your structure agree with the 18electron 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

→

The reason that the second method is preferred is kinetic.

Suggest a sequence of reactions for t he preparation of Fe(CO)3(dppe), given iron metal, CO, dppe (Ph2PCH2CH2PPh2), and other reagents of your choice.

rule are common for cyclopentadienyl complexes to the right of the d block.

22.10

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

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.

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)

22.6

π-acid ligand.

of the Cp* complex are lower in frequency than the corresponding bands of the Cp complex. Cp* is a stronger donor ligand than Cp.

9 and group 10 elements.

22.5

PF3 is primarily

(b) MnCp(CO)3 2023, 1939 cm –1 vs. MnCp*(CO)3 2017, 1928 cm –1? CO bands

(a) Ni( 3-C3H5)2 16, very common for group

(c) Co( 3-C3H5)(CO)3? 18.

σ-donor ligand.

Suppose that you are given a series of metal tricarbonyl compounds having the respective symmetries C 2v, D3h, and C s. Without consulting reference material, which of these should display the greatest number of CO stretching bands in the IR

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

22.13

22.17

Using the 18-electron rule as a guide, indicate the probable number of carbonyl ligands in (a) W(η6-C6H6)(CO)n, (b) Rh(η5C5H5)(CO)n, and (c) Ru3(CO)n. (a) 3 (b) 2 (c) 12 Propose two syntheses for MnMe(CO)5, both starting with Mn2(CO)10, with one using Na and one using Br 2? You may use other reagents of your choice. (i) Reduce Mn(CO)5 – ;

(a) Reflux Mo(CO)6 with cycloheptatriene to give [Mo(η6-C7H8)(CO) 3]; treat with the trityl 7 tetrafluoroborate to give [Mo(η C7H7)(CO)3]BF4. (b) React [IrCl(CO)(PPh 3)2] with MeCl, then expose to CO atmosphere to give [IrCl2(COMe)(CO)(PPh 3)2].

22.18

Mn2(CO)10 with Na to give react with MeI to give MnMe(CO)5. (ii), Oxidize with Br 2. 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 3chloroprop-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.

A:

CO CO CO

η1 22.16

W

B:

η3

CO CO

C:

-2CO

OC

-H

Fe

CO CO

A

OC

H CO

B

[π-C5H5Fe(CO)2]2 C

1

The compound B shows two H NMR resonances due to Fe-H proton and the 1 aromatic Cp ring. C shows a single H NMR resonance because of aromatic Cp ring.

22.19

CO CO CO

η2

The compounds formed are TiR 4; ethyl has the low energy β-hydride elimination decomposition.

-C O Fe

W

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.

When Fe(CO)5 is refluxed with cyclopentadiene compound A is formed which has the empirical formula C 8H6O3Fe 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 H 2 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. + F e ( C O )5

+ W

Suggest syntheses of (a) [Mo( 7C7H7)(CO)3]BF4 from Mo(CO)6 and (b) [IrCl2(COMe)(CO)(PPh3)2] from [IrCl(CO)(PPh3)2]?

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)3 Na.

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


44


(b) Can these CVE values be derived from the 18-electron rule? No.

RuCp2 has 18 electrons.

22.21

Give the equation for a workable reaction for the conversion of Fe( 5-C5H5)2 to Fe( 5C5H5) ( 5-C5H4COCH3) and (b) Fe( 5C5H5) ( 5-C5H4CO2H) (a) Fe(η5 – C5H5)2 + CH3COCl 5 Fe(η – C5H5)(η5 – C5H4COCH3) + HCl

(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

(b) Cl

Fe

22.22

+

O

O Cl

A l C l3 C H 2 C l2

Fe

Cl

CO 2 H H2O, t-BuOK DME

Fe

Sketch the a1’ symmetry-adapted orbitals for the two eclipsed C 5H5 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 a 1’ 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 d z2 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( η5C5H5)(η4-C5H6)]+ whereas Fe(η5-C5H5)2 reacts with strong acid to yield [Fe( η5C5H5)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

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.


Derive the ground state of the Tm3+ ion 3

S23.2

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

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.

H6.

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

(ii)two equivalents of HF are eliminated

(b) Rh(C2H5) (CO) (PR 3)2 RhH(CO)(PR 3)2 + C2H4? A β-hydrogen

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. UO22+ if sufficient oxygen is present.

elimination reaction.

22.25

22.26

Given mechanism of CO insertion, what rate constant can be extracted from rate data? Rate = k a[RMn(CO) 5]

(a) What cluster valence electron (CVE) count is characteristic of octahedral and trigonal prismatic complexes? octahedral M6, 86; trigonal prismatic M6, 90.

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) →

23.10

The rock-salt structure.

3+

2 Ln (aq) + 3 H2(g) + 6 H2O(l)

(b) Redox potentials The potentials for the

23.11

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

Self-tests S24.1

Synthesis for Sr2MoO4? Sealed tube, high temperature, 6 SrO(s) + Mo(s) + 2 MoO3(s)  3 Sr 2MoO4.

S24.2

Why does increased pressure reduce the conductivity of K + more than that of Na + in -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+ 3+ ions (Cr ) occupy octahedral sites.

7

23.5

Derive the ground state of the Tb 3+ ion. F6

23.6

Predict the magnetic moment of a compound containing the Tb 3+ ion.

Exercises 24.1

μ= 9.72 μB

24.2 Explain why stable and readily isolable carbonyl complexes are unknown for the lanthanoids? Carbonyl compounds need

Suggest a synthesis of neptunocene from NpCl4?

24.3

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 5 f orbitals of the actinide ions interact strongly with ligand orbitals, and the splitting of the 5 f subshell, as well as the color of the complex, varies as a function of ligand.

The electronic conductivity of the solid increases owing to formation of Ni1–xLix)O.

What is a crystallographic shear plane? 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.

Np(COT) 2 was prepared by the reaction of K 2COT with NpCl4 under inert atmosphere.

23.9

NiO doped with Li2O?

Both.

back-bonding from metal orbitals of the appropriate symmetry.

23.8

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.

CHAPTER 24

would show general increase across the lanthanoids. With the third, anomalies arise.

23.7

Predict a structure type for BkN based on the ionic radii r (Bk 3+) = 96 pm and r (N3−) = 146 pm.

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) Sr 2WMnO6.


46

24.7

ANSWERS TO SELF-TESTS AND EXERCISES Where might intercalated Na+ ions reside in the ReO3 structure? The unit cell for

24.21

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

AlP > InSb.

24.22


Classify oxides as glass-forming or nonglass-forming? BeO, B2O3, and to some

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?

24.15

diameter is 8.42 nm.

S25.3

Host for QDs? MCM-41.

Exercises

Li2 NiF4

Synthesis of LiTaS2? Direct reaction of TaS2

properties different from those of a molecule or an extended solid.

25.1

electrochemical

25.2

24.16

Oxotetrahedral species in 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

framework

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.

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.

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

candidate to consider.

24.20

Hemispherical imprint for 2 nm nanoparticle? The radius is 4.21 nm and the

(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

with BuLi, or through insertion of Li ions.

24.19

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.

High-temperature superconductors? All except Gd2Ba2Ti2Cu2O11, which contains only Cu2+, are superconductors.

extent GeO2, glass forming. Transition metal and rare earth oxides are typically nonglassforming.

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.

S25.2 24.12

Order of band gaps? BN > C(diamond) >

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 “topdown” 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? Topdown methods allow for precise control over the spatial relationships. Bottom-up methods



allow for the precise spatial control over atoms and molecules.

25.6

25.12

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

25.14

(a) Steps in solutions synthesis of nanoparticles? (i) solvation; (ii) stable nuclei

(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.

(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.

(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.

of nanometer dimensions formed; (iii) growth of particles to desired size occurs.

25.9

PVD versus CVD? In CVD, the atomic

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) 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.

(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

25.11

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.

Vapor-phase versus solution-based techniques? (a) Vapor-phase, large sizes. (b) Vapor-phase, more agglomeration.

25.17 25.10

(a) What is a core-shell nanoparticle?

(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 (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.

Common features of self-assembly?

Static versus dynamic self-assembly? Static self-assembly is when a system selfassembles 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.


48


25.19

Compare SAMs and cell membranes? SAM

25.20

can structurally resemble a phospholipid bilayer. What is morphosynthesis? Control of nanoarchitectures in inorganic materials through changes in synthesis parameters.

25.21

(a) What are the two classes of inorganicorganic 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

25.23

25.24.

25.25

(a) Why is dispersion important in nanocomposites? They lead to increased exposed surface areas. (b) Why is dispersion difficult? The often nonpolar organic

S26.4

Without R groups attached to the Zr center there is no preference for specific binding during polymerization.

Exercises 26.1

(b) H2 plus O2 plus an electrical arc? Not catalysis..

(c) The production of Li 3N and its reaction with H2O? Not catalysis. 26.2

per unit time per unit amount of catalyst.

A bionanomaterial and its application?

product is formed relative to by-products.

PPF/PPF-DA is an injectable bionanomaterial used for bone-tissue engineering.

(c) Catalyst? A substance that increases the

(a) Biomimetics? Designing nanomaterials that mimic biological systems. (b) example of biomimetics? Cellulose fibers in paper have

(d) Catalytic cycle? Sequence of chemical

been used to template the growth of titanium oxide nanotubes.

(e) Catalyst support?

Bionanocomposites and improved mechanical strength? Biomimetics need

(b) Selectivity? How much of the desired

rate of a reaction but is not itself consumed. reactions involving the catalyst that transform the reactants into products. 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

Self-tests

S26.3

Define the following terms? (a) Turnover frequency? The amount of product formed

polymers do not have strong interactions with the polar or ionic inorganic components.

CHAPTER 26

S26.2

Which of the following constitute catalysis? (a) H2 and C2H4 in contact with Pt? An example of genuine catalysis.

trabecular and cortical bone tissues, which combine compressibility and tensile strength. Hybrid alumoxane nanoparticles dispersed in PPF/PPF-DA shows enhanced strength.

S26.1

Demonstate that the polymerization of propene with a simple Cp2ZrCl2 catalyst would give rise to atactic polypropene?

Which of the following processes would be worth investigating? (a) The splitting of H2O into H2 and O2? Not worthwhile.

The effect of added phosphine on the catalytic activity of RhH(CO)(PPh3)3?

(b) The decomposition of CO2 into C and O2? A waste of time.

Added phosphine will result in a lower concentration of the catalytically active 16electron complex.

(c) The combination of N2 with H2 to produce NH3? Very worthwhile reaction to

-Alumina heated to 900ºC, cooled, and exposed to pyridine vapor? complete

(d) The hydrogenation of double bonds in vegetable oil? The process can be readily

dehydroxylation.

set up with existing technology.

Would a pure silica analog of ZSM-5 be an active catalyst for benzene alkylation (see Figure 26.24)? No.

try to catalyze efficiently at 80ºC.

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

exchange of the methyl group chemisorbed –CHR(CH3) group 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.

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.

in the (R =

CHAPTER 27 Self-tests

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

S27.1 S27.2

3D structure can place any particular atom in a suitable position for axial coordination.

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

S27.3

26.11

(a)

Attack

by

dissolved

S27.4

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. (a) Enhanced acidity? When Al3+ replaces Si4+ on lattice site charge is balanced by H3O+.

S27.5

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

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

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

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

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.

(b) Three other ions? Ga3+, Co3+, and Fe3+. 26.12

Why does saline contain NaCl? Osmotic balance.

the loss of ethane.

26.10

Is Iron (II) expected to be present in the cell as uncomplexed ions? No. Unusual coordination of Mg? The protein’s

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.

Self Test Answers

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