Answers to self-tests and exercises

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 104 cm-1

Self-tests 1 0

+

S1.1

80 35 Br

S1.2

d orbitals, 5 orbitals

S1.3

4

S1.4

3p

S1.5

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

S1.6

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

S1.7

Period 4, Group 2, s block

S1.8

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

S1.9

81 35 Br

n

+ γ

1.11

1

λ 1

λ 1

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.

λ 1

λ

⎛1 1 ⎞ 7 −1 − ⎟ = 1.0974 X10 m 2 ⎝ 1 ∞2 ⎠

= R⎜

⎛1 1 ⎞ 7 −1 − ⎟ = 1.0288X10 m 2 2 ⎝1 ⎠ 4

= R⎜

⎛1 1⎞ − ⎟ = 9.7547X106 m −1 ⎝ 12 32 ⎠

= R⎜

⎛1 1⎞ − ⎟ = 8.2305X106 m −1 ⎝ 12 22 ⎠

= R⎜

S1.10

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

1.12

0 up to n-X

S1.11

Cs+

1.13

n2

1.14

Exercises 1.1

(a)

14 7

N+ 42 He→178 O+11p + γ

(b)

12 6

(c)

14 7

C+11p→137 N + γ 1 0

3 1

12 6

N+ n→ H+ C

1.2

246 96

1 Cm+126 C→257 112 Uub+ 0 n

1.3

The higher value of I2 for Cr relative to Mn is a consequence of the special stability of halffilled subshell configurations and the higher Zeff of a 3d electron verses a 4s electron. 4 2

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.

9 4

12 6

l

ml

Orbital designation

2

1

2p

3

2

3d

5

4 4

0 3

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

Number of orbitals 3

4s 4f

1 7

1.15

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

1.16

Li: σ = Z – Zeff ; σ = 3-1.28 = 1.72 Be: σ = Z – Zeff ; σ = 4-1.19 = 2.09

Ne + He → Mg + n

1.4

22 10

N

25 12

1 0

B: σ = Z – Zeff ; σ = 5-2.42 = 2.58

4 2

1 0

C: σ = Z – Zeff ; σ = 6-3.14 = 2.86

Be+ Be→ C+ He +2 n

N: σ = Z – Zeff ; σ = 7-3.83 = 3.17 O: σ = Z – Zeff ; σ = 8-4.45 = 3.55 F: σ = Z – Zeff ; σ = 9-5.10 = 3.90 1.17

1

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


ANSWERS TO SELF-TESTS AND EXERCISES

2 1.18

See Figs 1.11 through 1.16

1.19

See Table 1.6 and discussion.

1.20

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

1.21

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

1.22

(a) [He]2s22p2

1.30

(b) [He]2s22p5 (c) [Ar]4s2 (d) [Ar]3d10 (e) [Xe]4f145d106s26p3 (f) [Xe]4f145d106s2 1.23

(a) [Ar]3d14s2

CHAPTER 2 Self-tests S2.1

(b) [Ar]3d2 (c) [Ar]3d5 (d) [Ar]3d4 (e) [Ar]3d

S2.2

6

(b) square planar

(f) [Ar] 10

(g) [Ar]3d 4s

1

S2.3

Linear

S2.4

S22– : 1σg22σu23σg21πu42πg4 ;

(h) [Xe]4f 7 1.24

(a) Angular

Cl2– : 1σg22σu23σg21πu42πg44σu1.

(a) Xe]4f145d46s2

S2.5

1σg22σu23σg21πu42πg4

(b) [Kr]4d6

S2.6

½[2-2+4+2] = 3

S2.7

Bond order: C≡N, C=N, and C–N; Bond strength: C≡N > C=N > C–N.

S2.8

If it contains 4 or fewer electrons.

S2.9

–21 kJ mol–1

S2.10

(a) +1/2

6

(c) [Xe]4f

(d) [Xe]4f7 (e) [Ar] (f) [Kr]4d 1.25

2

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

(c) tetrahedral

1.28

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

1.29

2s2 and 2p0

2.2

(a) trigonal planar (b) trigonal pyramidal (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

(b)

(d) 1σg22σu23σg21πu42πg3 2.16

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

2.17

(a) 1σg22σu23σg21πu42πg4 (b) 1

F

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

F

S 1800 F 2.5 2.6

2.18

120 0

(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

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

2.8

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

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

2.21

(a) 0

(a-c)

-1

(b) 205 kJ mol 2.11

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

2.12

(a) covalent (b) ionic

(d) Possibly stable in isolation (only bonding and nonbonding orbitals are filled; not stable in solution because solvents would have higher proton affinity than He.

(c) ionic 2.13

(a) sp2 (b) sp3 (c) sp3d or spd3 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

3.4 3.5

XA2 K3C60

3.6

281 pm

3.7

429 pm

3.8

CuAu. Primitive 12 carat.

3.9

Zintl phase region

3.10

(a) 6:6 and 8:8

S3.1

See Fig. 3.7 and Fig. 3.32.

S3.2

See Figure 3.35.

S3.3

52%

S3.4

rh = ((3/2)1/2 – 1) r = 0.225 r

S3.5

409 pm

3.11

6

S3.6

401 pm

3.12

S3.7

FeCr3

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–1

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–1.

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

S3.18

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

Exercises 3.1

a ≠ b ≠ c and α = 90°, β = 90°, γ = 90°

3.2

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

3.3

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

(b) CsCI

(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. Vanandium carbide and manganese oxide.

3.24


ANSWERS TO SELF-TESTS AND EXERCISES 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. Ag2S and CuBr: p-type; VO2: n-type.

CHAPTER 4

S4.9

5

Identify the acids and bases? (a) FeCl3 + Cl– → [FeCl4]–, acid is FeCl3, base is Cl–. (b) I– + I2 → I3–, acid is I2, base is I–.

S4.10

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

S4.11

Draw the structure of BF3·OEt2?

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; H2O, acid; hydroxide ion, conjugate base.

Exercises 4.1

(c) NH3 + H2S → NH4+ + HS– Ammonia, base; NH4+, conjugate acid; hydrogen sulphide, acid; HS–, conjugate base. S4.2

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

S4.3

Calculate the pH of a 0.20 M tartaric acid solution?

Sketch an outline of the s and p blocks of the periodic table, showing the elements that form acidic, basic, and amphoteric oxides? The elements that form basic oxides are in plain type, those forming acidic oxides are in outline type, and those forming amphoteric oxides are in boldface type.

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

S4.7

4.2

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

of

the

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

(b) H2PO4 pKa(2) ≈ 8. The actual value, given in Table 4.1, is 7.4.

(c) CH3OH? The conjugate base is CH3O–. (d) H2PO4–? The conjugate base is HPO42–.

(c) HPO42– pKa(3) ≈ 13. The actual value, given in Table 4.1, is 12.7.

(e) Si(OH)4? 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 TiO2 to redissolve.

bases

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

–

S4.8

Identify the conjugate following acids?

(f) HS−? 4.3

The

conjugate

base

is

The conjugate base is S2–.

Identify the conjugate following bases?

acids

of

the



6

ClO4–, cannot be studied in sulfuric acid.

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

4.9

(b) HPO42–? The conjugate acid is H2PO42–. (c) O2–? The conjugate acid is OH–.

electron withdrawing

(d) CH3COOH? The conjugate acid is CH3C(OH)2+. (e) [Co(CO)4]–? The conjugate acid is HCo(CO)4. (f) CN–? 4.4

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

4.5

What is the Kb of ethanoic acid?

4.6

4.11

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

4.12

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

The conjugate acid is HCN.

Kb = 5.6 × 10

the order is HClO4 > HBrO3 > H2SO4 > HNO2. 4.13

–10

What is the Ka for C5H5NH+? -

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

4.14

-

F will behave as a base in water. 4.8

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

O

O

the aluminum-containing species is more acidic.

Cl

H O

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

H

Si(OH)4, is more acidic.

O

chloric acid

(d) HClO3 or HClO4? HClO4 is a stronger acid.

chlorous acid

(e) H2CrO4 or HMnO4?

Chloric acid, the predicted pKa = –2; actual value = –1.

HMnO4 is the stronger acid. (f) H3PO4 or H2SO4?

Chlorous acid, the predicted pKa = 3; actual value = 2. 4.8

Which bases are too strong or too weak to be studied experimentally? (a) CO32– 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.

–

strong

to

be

4.16

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

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

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

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

O

Cl

Account for the trends in the pKa values of the conjugate acids of SiO44–, PO43–, SO42–, and ClO4–? The acidity of the four conjugate acids increases in the order HSiO43– < HPO42– < HSO4– < HClO4.

Ka = 5.6 × 10–6 4.7

Is the –CN group electron donating or withdrawing?

4.17

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


ANSWERS TO SELF-TESTS AND EXERCISES 4.18

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

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

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 (oxide polyanions): Mo polyoxoanions: As, B, and Si. 4.19

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

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

–

4.25

–

2HCO 3-

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

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. 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.

The hard nitrogen atom will bond more strongly to the hard Lewis acid BF3. 4.29

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

4.30

Discuss relative basicities? (a) Acetone and DMSO? DMSO is the stronger base regardless of how hard or how soft the Lewis acid is. The ambiguity for DMSO is that both the oxygen atom and sulfur atom are potential basic sites.

Identifying acids and bases: (a) SO3 + H2O → HSO4– + H+? The acids in this reaction are the Lewis acids SO3 and H+ and the base is the Lewis base OH–. (b) Me[B12]– + Hg2+ → [B12] + MeHg+? The Lewis acid Hg2+ displaces the Lewis acid [B12] from the Lewis base CH3–. (c) KCl + SnCl2 → K+ + [SnCl3]–? The Lewis acid SnCl2 displaces the Lewis acid K+ from the Lewis base Cl–.

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

H2SO4 + HF ⇔ H3SO4+ + F-

4.24

<1

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

(b) CO2 and CaCO3?

4.23

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

(c) CH3HgI + HCl → CH3HgCl + HI?

H3PO4 + HPO42- ÅÆ 2H2PO4-

NH3 + HF

4-Me-py.

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

More balanced equations? (a) H3PO4 and Na2HPO4?

4.22

B(n-Bu)3.

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

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

Ca2+ +

Boron trichloride.

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

2[Fe(OH2)6]3+ →

CO 2 + CaCO 3 + H 2O

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

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– + 8H3O+ → P4O124– + 12H2O

4.21

7

(b) Me2S and DMSO? Depending on the EA and CA values for the Lewis acid, either base could be stronger. 4.31

Write a balanced equation dissolution of SiO2 by HF?

for

the



8

SiO2 + 6HF

2H2O + H2SiF6 or

SiO2 + 4HF

C6H5COOH + HF 2H2O + SiF4

Both a Brønsted acid–base reaction and a Lewis acid–base reaction. 4.32

(c) C6H5COOH?

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 Al2S3? The foul odor suggests H2S formation.

φ

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

acid-Cl- + I-

ΔfH = -20.0kJ/mol

Al2 O 3 + 3H 2 S

Al2 S 3 + 3H 2 O 4.33


acid-I- + Cl-

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

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

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

4.36

(a) CH3CH2OH? is:

CH 3CH 2OH + HF 3

+

HF

2MnO4− (aq) + 5Zn(s) + 16H+(aq) → 5Zn2+(aq) + 2Mn2+(aq) + 8H2O(l) S5.2

Does Cu metal dissolve in dilute HCl? No.

S5.3

Can Cr2O72– be used to oxidize Fe2+, and would Cl– oxidation be a problem? Yes. Cl- oxidation is not a problem.

S5.4

Fuel cell emf with oxygen and hydrogen gases at 5.0 bar? E = 1.25 V.

S5.5

The fate of SO2 emitted into clouds? The aqueous solution of SO42– and H+ ions precipitates as acid rain.

S5.6

Can Fe2+ disproportionate under standard conditions? No.

S5.7

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

S5.8

Potential of AgCl/Ag,Cl– couple? Eox= – 1.38 V

S5.9

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

S5.10

Frost diagram for thallium in aqueous acid?

The balanced equation

CH 3CH 2OH 2+ + F-

(b) NH3? The equation is

NH

NH

oxidation

3+

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

4.35

Half-reactions and balanced reaction for oxidation of zinc metal by permanganate ions? 2[MnO4− (aq) + 8H+ (aq) + 5e– → Mn2+(aq) reduction + 4H2O(l)]

5 [ Zn(s) → Zn2+(aq) + 2e– ]

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

4.34

C6H5COO- + H2F+

4

+

+

F-


ANSWERS TO SELF-TESTS AND EXERCISES 5.3

9

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

4Cr2+(aq) + O2(g) + 4H+(aq) → 4Cr3+(aq) Eº = 1.65 V + 2H2O(l) (b) Fe2+?

4Fe2+(aq) + O2(g) + 4H+(aq) → 4Fe3+(aq) Eº = 0.46 V + 2H2O(l) (c) Cl–? no reaction. (d) HOCl? No reaction. S5.11

The oxidation number of manganese?

(e) Zn(s)?

Mn2+(aq) S5.12

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

–

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

A competing reaction is: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) (Eº = 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

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) Fe2+ will not reduce water.

Exercises

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

5.1

(iv) disproportionation will not occur.

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

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

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

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

LiCoO2(s) + C(s) → LiC(s) + CoO2(s) +1 +3 -2 0 +1-1 +4 -2 Ca(s) + H2(g) 0 0 5.2

→ CaH2(s) +2 -1

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

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

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

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

(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) → 2Mn2+(aq) +5HSO3− (aq) + 3H2O(l) The potential decreases as the pH increases. 5.7

5.13

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

5.14

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

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

(b) Mn? E = 0.55 V

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

(c) S? At pH 0, 0.387 V. At pH 14, SO42– would again predominate. HSO4– is the predominant sulfur species at pH 6.

and E = Eº – [(0.059V)/4][log(1/(p(O2)[H+]4)] (b) The reduction of Fe2O3(s)?

5.15

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 Cl2 is thermodynamically solution? susceptible to disproportionation to Cl– and ClO4– when it is dissolved in aqueous base.

5.16

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

5.17

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

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

5.18

Tendency of mercury species to act as an oxidizing agent, a reducing agent, or to undergo disproportionation? Hg2+ and Hg22+ are both oxidizing agents. None of these species are likely to be good reducing agents. Hg22+ is not likely to undergo disproportionation. Thermodynamic tendency of HO2 to undergo disproportionation? E = +1.275 V. will undergo (is positive), HO2 disproportionation. Dissolved carbon dioxide corrosive towards iron? Carbon dioxide and water generate carbonic acid which encourages the corrosion process by lowering solution pH. What is the maximum E for an anaerobic environment rich in Fe2+ and H2S? –0.1 V.

+ 6

Q = 1/[H ] and E = Eº – (RT/nF)(13.8 pH) 5.8

(b) What happens when Cl2 is dissolved in aqueous acid solution? Cl2 will not disproportionate. Cl2 is thermodynamically capable of oxidizing water. (c) Should HClO3 disproportionate in aqueous acid solution? Kinetic. 5.9

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

5N2O(aq) + 2OH–(aq) → 2NO3–(aq) + 4N2 (g) + H2O(l) (b) Zinc metal is added to aqueous acidic sodium triiodide? –

2+

–

Zn(s) + I3 (aq) → Zn (aq) + 3I (aq) (c) I2 is added to excess aqueous acidic HClO3?

3I2(s) + 5ClO3– (aq) + 3H2O(l) → 6IO3– (aq) + 5Cl– (aq) + 6H+(aq) 5.10

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

5.11

Will acid or base most favour the following half-reactions? (a) Mn2+ → MnO4–? Base (b) ClO4– → ClO3–? Acid (c) H2O2 → O2?

Base

(d) I2 → 2I–? Acid or base, no difference. 5.12

Determine the standard potential for the reduction of ClO4– to Cl2? 1.392 V

5.19

5.20

5.21 5.22

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

5.23

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

5.24

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


ANSWERS TO SELF-TESTS AND EXERCISES 11

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


S6.2

Γstr = A1g (Raman, polarized) + Eg

Sketch the S4 axis of an NH4+ ion. How many of these axes are there in the ion? Three S4 axes.

(Raman) + T1u (IR). SF5Cl has C4v symmetry. stretching vibrations leads to: Γstr = 3A1 (IR and Raman, polarized)

(a) BF3 point group? D3h. 2–

(b) SO4 point group? Td. S6.3

S6.4 S6.5

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

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

Is the skew form of H2O2 chiral? Yes.

S6.7

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

S6.8

Confirm that the symmetric mode is Ag? D2h character table, which is the Ag symmetry type. Show that the four CO displacements in the square-planar (D4h) [Pt(CO)4]2+ cation transform as A1g + B1g + Eu. How many bands would you expect in the IR and Raman spectra for the [Pt(CO)4]2+ cation? The reducible representation: D4h E 2C4 C2 2C2’ 2C2″ i 2S4 σh 2 σv 2 σd Γ3N 4 0 0 2 0 0 0 4 2 0 Reduces to A1g + B1g + Eu A1g + B1g are Raman active. Eu is IR active.

S6.10

Orbital symmetry for a tetrahedral array of H atoms in methane? A1

S6.11

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

S6.12

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

S6.13

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

Symmetry species of all five d orbitals of the central Xe atom in XeF4 (D4h, Fig. 6.3)? dx2-y2 is B1g; dxy is B2g; dxz and dyz are Eg; dz2 is A1g. What is the maximum possible degeneracy for an Oh molecule? 3.

S6.6

S6.9

Analysis of the

Predict how the IR and Raman spectra of SF5Cl differ from that of SF6?

S6.15 +T1u.

SALCs for sigma bonding in O? A1g + Eg

Exercises Symmetry elements? (a) a C3 axis and a σv plane in the NH3 molecule?

N H

N

H

H

H

H H

σv

C3

(b) a C4 axis and a σh plane in the squareplanar [PtCl4]2– ion?

Cl

Pt

Cl

Cl

Cl

Cl

Cl

C4 6.2

Pt

Cl

σh

S4 or i? (a) CO2? i (b) C2H2? i. (c) BF3? neither. (d) SO42–? three different S4.

6.3

Assigning point groups: (a) NH2Cl? Cs (b) CO32–? D3h (c) SiF4? Td (d) HCN? C∞v. (e) SiFClBrI? C1. (f) BrF4–? D4h.

Cl



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

6.12

and irreducible representations? A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu 6.13

(b) A p orbital? An infinite number of mirror planes that pass through both lobes and include the long axis of the orbital. In addition, the long axis is a Cn axis, where n can be any number from 1 to ∞. (c) A dxy orbital? Center of symmetry, three mutually perpendicular C2 axes, three mutually perpendicular mirror planes of symmetry, two planes that are rotated by 45º about the z axis from the xz plane and the yz plane. (d) A dz2 orbital? In addition to the symmetry elements possessed by a p orbital: (i) a center of symmetry, (ii) a mirror plane that is perpendicular to the C∞ axis, (iii) an infinite number of C2 axes that pass through the center of the orbital and are perpendicular to the C∞ axis, and (iv) an S∞ axis. 6.6

For the 90o-twisted form of B2F4 (D2d) The vibrations are: 3A1 (Raman, polarized) + B1 (Raman) + 2B2 (IR and Raman ) + 3E (IR and Raman). 6.14

(a) Take the 4 hydrogen 1s orbitals of CH4 and determine how they transform under Td. (b) Confirm that it is possible to reduce this representation to A1 + T2. (c) Which atomic orbitals on C can form MOs with H1s SALCs? Using symmetry Td, Γ3N reduces to: A1 + T2. The MOs would be constructed from SALCs with H1s and 2s and 2p atomic orbitals on C.

6.15

Use the projection operator method to construct the SALCs of A1 + T2 symmetry that derive from the four H1s orbitals in methane..

SO32– ion? (a) Point group? C3v (c) Which s and p orbitals have the maximum degeneracy? 3px and 3py orbitals are doubly degenerate. PF5? (a) Point group? D3h. (b) Degenerate MOs? 2. (c) Which p orbitals have the maximum degeneracy? 3px and 3py atomic orbitals are doubly degenerate

6.8 6.9

px = (1/2)(ϕ1 – ϕ2 + ϕ3 – ϕ3) (= T2)

trigonal bipyamidal geometry? No.

py = (1/2)(ϕ1 – ϕ2 – ϕ3 + ϕ3) (= T2)

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

pz = (1/2)(ϕ1 + ϕ2 – ϕ3 – ϕ3) (= T2)

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

6.11

s = (1/2)(ϕ1 + ϕ2 + ϕ3 + ϕ3) (= A1)

AsCl5 Raman spectrum consistent with a

(b) Perpendicular to the molecular plane? 1 6.10

IR and Raman to distinguish between: (a) planar and pyramidal forms of PF3, (b) planar and 90o-twisted forms of B2F4 (D2h and D2d respectively)? (a) Planar PF3, D3h, vibrations are: A1’ (Raman, polarized) + 2E’ (IR and Raman) + A2” (IR). Pyramidal PF3, C3v, vibrations are: 2A1 (IR and Raman, polarized) + 2E’ (IR and Raman)

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

(b) Degenerate MOs? 2

6.7

[AuCl4]− ion? Γ of all 3N displacements

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

SALCs for σ-bonds (a) BF3? (1/√3)(ϕ1 + ϕ2 + ϕ3) (= A1’) (1/√6)(2ϕ1 – ϕ2 – ϕ3) and (1/√2)(ϕ2 – ϕ3) (= E’) (b) PF5? (axial F atoms are ϕ4 + ϕ5) (1/√2)(ϕ4 + ϕ5) (= A1’) (1/√2)(ϕ4 − ϕ5) (= A2”) (1/√3)(ϕ1 + ϕ2 + ϕ3) (= A1’) (1/√6)(2ϕ1 – ϕ2 – ϕ3) and (1/√2)(ϕ2 – ϕ3) (= E’)



CHAPTER 7

Exercises 7.1

Self-tests S7.1

Give formulas corresponding to the following names? (a) Cisdiaquadichloroplatinum(II)? cis[PtCl2(OH2)2],

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

CO Ni

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

OC

(b) Diamminetetra(isothiocyanato) chromate(III)? [Cr(NCS)4(NH3)2] –. ,can exist as, cis[Cr(NCS)4(NH3)2] – or trans[Cr(NCS)4(NH3)2] –. (c) Tris(ethylenediamine)rhodium [Rh(en)3]3+.

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

2-

NC

(III)?

Ni

NC

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

CN CN

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

(e) Chlorotris(triphenylphosphine)rhodium (I)? [RhCl(PPh3)3]. S7.2

CO CO

2-

Cl

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

Co Cl

Cl Cl

S7.3

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

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

S7.4

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

H3N

NH3

H3N 7.2

Mn NH3

NH3 NH3

Write the formulas for the following complexes? (a) [CoCl(NH3)5]Cl2 (b) [Fe(OH2)6](NO3)3 (c) cis-[FeCl2(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. S7.6

(d) [Cr(NH3)5μ–OH–Cr(NH3)5]Cl5 7.3

Name the following complexes?

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

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

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

(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

M

M L

L

L

L

L

L

L

square planar

tetrahedral

(b) Isomers expected for MA2B2? tetrahedral complex, no isomers, for a squareplanar 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

Trigonal Bipyramidal

Square based pyramid

A = axial ligands

A = axial B = basal

E = equatorial ligands

7.6

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

[Mg(edta)(OH2)]2– 7.11

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

7.12

Which complexes have isomers?

[CoBrClI(OH2)] 7.13

Which complexes have isomers? (a) [Pt(ox)(NH3)2] no isomers (b) [PdBrCl(PEt3)2] has two isomers.

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

7.8 7.9

7.10

Explain the difference between monodentate, bidentate, and quadridentate? A monodentate ligand can bond to a metal atom only at a single atom, a bidentate ligand can bond through two atoms, a quadridentate ligand can bond through four atoms. 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. Draw structures of complexes that contain the ligands (a) en, (b) ox, (c) phen, and (d) edta? .

(c) [IrHCO(PR3)2] has two isomers. (d) [Pd(gly)2] has two isomers. 7.14

How many isomers are possible for the following complexes? (a) [FeCl(OH2)5]2+? None. (b) [IrCl3(PEt3)2]? 2 (c) [Ru(biby)3]2+? 2 (d) [CoCl2(en)(NH3)2]+? 4 (e) [W(CO)4(py)2] 2


ANSWERS TO SELF-TESTS AND EXERCISES 15 7.15

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

M

A A

B

D

E

E

D

E B

M C

B

B D

C

E C

A E

B

M

D

D

C

A

A

A

A

D E

M

A

C

C

D

S8.4

E A

D

S8.3

M

A

M

A

M

A

C

7.16

A C

A E

S8.2

A

D

A B

M

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

B

M

S8.5 A

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 planar).

S8.7

(c) cis-[RhCl2(NH3)4]+? not chiral. (d) [Ru(bipy)3]2+? chiral

Isomer shift for iron in Sr2FeO4? The Smaller and less positive. Why does the mass spectrum of ClBr3 have five peaks separated by 2 u? Halogen isomers.

Exercises

(e) fac-[Co(NO2)3(dien)]? Not chiral. (f) mer-[Co(NO2)3(dien)]? not planar).

TiO2 in sunscreens? Titania articles absorb this ultraviolet radiation Molecular shape and vibrational modes for XeF2? Trigonal bipyramidal, 4 total vibrational modes. (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?

Chiral (dien is

8.1

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

7.17

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

8.2

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

7.18

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

8.3

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

8.4

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

8.5

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

8.6

Wavenumber for O–O in O2+? In the region of 1800 cm 1.

8.7

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

8.8

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

N

N N

N

Ru N

N

N

N

Ru

N

N

N

N

Λ isomer

Δ isomer

7.19

Suggest a reason why Kf5 is so different? Because of a change in coordination.

7.20

Compare these values with those of ammonia given in exercise 7.19 and suggest why they are different? The chelate effect.


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



16

8.9

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

8.10

Form of 19F-NMR and 77Se-NMR spectra of 77SeF4? 19F NMR spectrum reveals two 1:3:3:1 quartets. The 77Se-NMR spectrum is a triplet of triplets. NMR spectral features for XeF5−? All 5 of the F atoms are chemically equivalent. Approximately 25% is present as 129Xe, I = 1/2, and in this case the 19F resonance is a doublet. The final result is a composite: two lines of 12.5% intensity from the 19F coupled to the 129Xe, and one remaining line of 75%.

8.11

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, gvalue anisotropy can be observed.

8.15

8.16 8.17

8.18

8.19

8.20

8.21 8.22

Isomer shift for iron in BaFe(VI)O4? A positive shift for Fe(VI) well below +0.2 mm s-1. Charge on Fe atoms in Fe4[Fe(CN)6]3? EPR and Mössbauer. No quadrupole splitting in Mössbauer spectrum of SbF5? The geometry must be close to cubic in the solid state. 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. Peaks in mass spectrum of Mo(C6H6)(CO)3? 258, 230, 200, 186, and 174. 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.

S9.1

Found in aluminosilicate minerals or sulfides? Cd and Pb will be found as sulfides. Rb and Sr can be found in aluminosilicate minerals. Cr and Pd can be found in both oxides and sulfides.

S9.2

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

S9.3

Shape of XeO4 and identify the Z + 8 compound with the same structure? A tetrahedral geometry. The same structure is SmO4.

S9.4

Comment on ΔfHө values? It is evident from the values that as we move down the group, steric crowding of the fluorines is minimized.

S9.5

Further data useful when drawing comparisons with the value for V2O5? We would have to know the products formed upon decomposition.

Exercises 9.1

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

9.2

Form saline hydrides, oxides and peroxides, and all the carbides react with water to liberate a hydrocarbon? the alkaline earth metals or Group 2 elements.

9.3

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

9.4

Born–Haber cycle for the formation of the hypothetical compound NaCl2? Which thermochemical step is responsible for the fact that NaCl2 does not exist? The second ionization energy of sodium is 4562 kJ mol-1 and is responsible for the fact that the compound does not exist.

9.5

Inert pair effect beyond Group 15? The relative stability of an oxidation state in which the oxidation number is 2 less than the group number is an example of the inert pair effect.

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.




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

10.3

(a) H2S? H = +1, S = –2. (b) KH? H = –1, K = +1. (c) [ReH9]2–?

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

9.7 9.8

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

9.9

Calculate ΔfHө for SeF6? ΔfHө = −1397 kJ mol-1.

Assign oxidation numbers to elements?

H = –1, Re = +7.

(d) H2SO4? H = +1, O = –2, S = +6. (e) H2PO(OH)? H bonded to an oxygen atom = +1. H bonded to the phosphorus atom? If they are assigned an oxidation number of +1, and O = –2, then P = +1. 10.4

Preparation of hydrogen gas?

(i) CH4(g) + H2O → CO(g) + 3H2(g) (1000°C) (ii) C(s) + H2O → CO(g) + H2(g) (1000°C) (iii) CO(g) + H2O → CO2(g) + H2 (g)


10.5

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

S10.1

(b) Trends in ΔfGº? See Table 10.1.

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

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

GeH4

Reactions of hydrogen compounds? (a) Ca(s) + H2(g) → CaH2(s).

10.6

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

10.7

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

10.8

Name and classify the following?

(b) NH3(g) + BF3(g) → H3N–BF3(g). (c) LiOH(s) + H2(g) → NR. S10.3

A procedure for making Et3MeSn?

2Et3SnH + 2Na → 2Na+Et3Sn– + H2 Na+Et3Sn– + CH3Br → Et3MeSn + NaBr

(a) BaH2? barium hydride.

Exercises

(b) SiH4?

silane.

10.1

(c) NH3?

ammonia,

Where does Hydrogen fit in the periodic chart? (a) Hydrogen in group 1? Hydrogen has one valence electron like the group 1 metals and is stable as H+, especially in aqueous media. (b) Hydrogen in group 17? Hydrogen can fill its 1s orbital and make a hydride H–. The halogens are diatomic gases just like hydrogen, but chemically it fits well in both group 1 and group 17.

(d) AsH3? Arsine. (e) PdH0.9? palladium hydride. (f) HI? hydrogen iodide. 10.9

(a) Hydridic character? Barium hydride (b) Brønsted acidity? Hydrogen iodide. (c) Variable composition? PdH0.9 .

(c) Hydrogen in group 14? There is no reason for hydrogen to be placed in this group. 10.2

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

Chemical characteristics of hydrides?

(d) Lewis basicity? Ammonia. 10.10

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

10.11

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



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.


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

S11.2

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

S11.3

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

(b) Increasing basicity toward a hard acid? H2Se < H2S < H2O. 10.15

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

10.16

Compare BH4–, AlH4–, and GaH4–? Since AlH4– is more “hydride-like,” it is the strongest reducing agent.

10.17

Compare period 2 and period 3 hydrogen compounds? Period 2 compounds: - except for B2H6, are all exoergic - tend to be weaker Brønsted acids and stronger Brønsted bases - bond angles in period 2 hydrogen compounds reflect a greater degree of sp3 hybridization

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

M2CO3(s) CO2(g)

2M+(g) + CO32−(g) O2−(g) + CO2(g) S11.5

- Several period 2 compounds exhibit strong hydrogen bonding.

of

LiNO3 decomposes in one step. LiNO3(s) → 1/2 Li2O(s) + NO2(g) + 1/4O2(g)

methylbismuthine, S11.6

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.

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

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

Suggest a method for the preparation of BiH3?

The redistribution BiH2Me.

2M+(g) +

KNO3(s) → KNO2(s) + 1/2O2(g)

- boiling points of HF, H2O, and NH3 are all higher than their respective period 3 homologues. 10.18

M2O(s) +

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

Exercises 11.1

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



11.2

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

CHAPTER 12

Trends of the fluorides and chlorides of the group 1 metals? Fluoride is a hard Lewis base and will form strong complexes with hard Lewis acids. The trends reverse for the chloride ion.

S12.1

11.3

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

11.4

Which is more likely to lead to the desired result? (a) Cs+ or Mg2+, form an acetate complex? Mg2+. (b) Be or Sr, dissolve ammonia? Strontium. +

in

S12.2

S12.3

↓

S12.4

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

11.7

Thermal stability of hydrides versus decompose to carbonates? Hydrides elements. Carbonates decompose to oxides.

11.8

The structures of CsCl and NaCl? 6-coordinate Na+, 8-coordinate Cs+.

11.9

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

Exercises 12.1

12.2

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

Because of a diagonal relationship between Be and Al. 12.3

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

M + H2O → M(OH)2; A = M(OH)2 M(OH)2 + CO2 → MCO3; B = MCO3 2MCO3 + 5C → 2MC2 + 3CO2; C = MC2 MC2 + 2H2O → M(OH)2 + C2H2 M(OH)2 + 2HCl → MCl2 + 2H2O; D = MCl2. 12.4

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

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

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

11.10 Predict the products of the following reactions?

(a) CH3Br + Li → Li(CH3) + LiBr

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

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

NaNH2 11.6

Use ionic radii to predict a structure type of BeSe?

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

Identify the compounds?

+ NH3

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.

+

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

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

BeCl2 is covalent; BaCl2 is ionic.

liquid

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

Self-tests

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?

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.

S13.2

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

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?

MgSeO4

Simple alkenes are inert towards LiBH4.

Group 1 hydroxides are more soluble than group 2 hydroxides, and therefore have higher OH− concentrations. 12.7

12.8

4

4

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

Anhydrous Mg, and Ca sulphates are preferred as drying agents, because of the higher affinity of Mg and Ca sulphates for water. 12.9

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

LiBH4 + NH4Cl S13.3

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 . 2

BCl3(g) + py(l) → Cl3B − py(s)

2

(c) BBr3 and F3BN(CH3)3?

BBr3(l) + F3BN(CH3)3(s) → BF3(g) + Br3BN(CH3)3(s) S13.4

BeBr2 should be covalent. MgBr2 should be ionic. BaBr2 should be ionic 12.12

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) BCl3 and ethanol, (b) BCl3 and pyridine in hydrocarbon solution, (c) BBr3 and F3BN(CH3)3?

(b) BCl3 and pyridine in hydrocarbon solution?

Predict structures for BeTe and BaTe.

2

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

5

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

BH3NH3 + LiCl + H2

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

BeTe, close to ZnS-like structure. BaTe, close to CsCl-like structure. 12.11

THF

(a) BCl3 and ethanol?

Salts of divalent ions have low solubility. 12.10

11

S13.1

10

7, arachno species. The structure of B4H10:

Predict the products of the following reactions? (a) MgCl2 + 2LiC2H5 → 2LiCl + Mg(C2H5)2 (b) Mg + (C2H5)2Hg → Mg(C2H5)2 + Hg (c) Mg + C2H5HgCl → C2H5MgCl + Hg S13.6


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

–

[B10H11 (AlCH3)]

13

2

3

6


ANSWERS TO SELF-TESTS AND EXERCISES 21 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

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

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?

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 + GaBr3 → Me2SGaBr3 + AlCl3. (b) TlCl and formaldehyde (HCHO) in acidic aqueous solution? No reaction.

Exercises 13.1

Give a balanced chemical equation and conditions for the recovery of boron? B2O3 + 3Mg → 2B + 3MgO ΔH < 0

13.2

Describe the bonding in (a) BF3? Covalent. (b) AlCl3? In the solid state, a layered structure. At melting point, dimers. (c) B2H6? Electron-deficient dimer.

13.3

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

13.4

BCl3 > BF3 > AlCl3 (a) BF3N(CH3)3 + BCl3 BCl3N(CH3)3 + BF3 (BCl3 > BF3) (b) BH3CO + BBr3 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?

13.9

Diborane is extremely toxic, and the boron containing product of combustion is a solid, B2O3. 13.10

BF3

13.11

A

(b) 13.12

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

heat

B

Given NaBH4, a hydrocarbon of your choice, and appropriate ancillary reagents and solvents, give formulas and conditions for the synthesis of (a) B(C2H5)3, (b) Et3NBH3? (a) BCl3 + 3C2H5MgCl B(C2H5)3 + 3MgCl2

H2O

CaF2

Using BCl3 as a starting material and other reagents of your choice, devise a synthesis for the Lewis acid chelating agent, F2B– C2H4–BF2?

2BCl3 + 2Hg Æ B2Cl4 + 2HgCl2 B2Cl4 + 4AgF Æ B2F4 + 4AgCl B2F4 + C2H4 Æ F2CH2CH2BF2

Identify compounds A, B, and C?

LiAlH4

(a) BH3 + (CH3)2C=CH2 B[CH2-CH(CH3)2]3 (b) BH3 + CH CH B(CH=CH2)3 Diborane has been used as a rocket propellant. Calculate the energy released from 1.00 kg of diborane given the following values of ΔfHө/kJ mol-1: B2H6 = 31, H2O = -242, B2O3 = -1264. The combustion reaction is B2H6 (g) + 3 O2(g) → 3 H2O (g) + B2O3 (s). What would be the problem with diborane as a fuel? -73,172 kJ.

TlBr3 + NaBr → NaTlBr4 13.5

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

Heat, Ether

[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

+ C2H2 (2) B10H12(SEt2)2 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

B H

13.13

H

Which boron hydride would you expect to be more thermally stable, B6H10 or B6H12? Give a generalization by which the thermal stability of a borane can be judged?

B6H10 13.14 13.15

(a) Give a balanced chemical equation (including the state of each reactant and product) for the air oxidation of pentaborane(9). (b) Describe the probable disadvantages, other than cost, for the use of pentaborane as a fuel for an internal combustion engine? (a) 2B5H9 (l) + 12O2 (g) 5B2O3 (s) + 9H2O (l)

13.16

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

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

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

B

H H

H H

C C B

H

H

B

B Fe B H

B B B

H H

B H

B H

13.18

Heat

(a) nido . (b) 12. (c) The total number of valence elections is (10x3)+(14x1)=44; the number of cluster valence is the remainder of 44-20=24. 13.17

B H

How many skeletal electrons are present in B5H9? 14

H

H

H

H

B

B B

B C

C

H

H

B

(a) What are the similarities and differences in structure of layered BN and graphite (Section 13.9)? (b) Contrast their reactivity with Na and Br2. (c) Suggest a rationalization for the differences in structure and reactivity. (a) Their structures? Both of these substances have layered structures. (b) Their reactivity with Na and Br2? Graphite reacts, boron nitride is unreactive.

(c) Explain the differences? The large HOMO–LUMO gap in BN means it is more difficult to remove an electron from it than from the HOMO of graphite. 13.19

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

3 PhNH3+Cl− + 3 BCl3 → Ph3N3B3Cl3 + 9 HCl (b) Me3N3B3H3?

3 MeNH3+Cl− + 3 BCl3 → Me3N3B3Cl3 + 9 HCl Me3N3B3Cl3 + 3 LiH → Me3N3B3H3 + 3 LiCl The structures:



Exercises

13.20

Silicon forms the chlorofluorides SiCl3F, SiCl2F2, and SiClF3. Sketch the structures of these molecules?

14.2

Explain why CH4 burns in air whereas CF4 does not. The enthalpy of combustion of CH4 is −888 kJ mol−1 and the C–H and C–F bond enthalpies are −413 and −489 kJ mol−1 respectively? The bond enthalpy of a C–F bond is higher than the bond enthalpy of a C–H bond.

14.3

SiF4 reacts with (CH3)4NF to form [(CH3)4N][SiF5]. (a) Use the VSEPR rules to determine the shape of the cation and anion in the product; (b) Account for the fact that the 19F NMR spectrum shows two fluorine environments? (a) The cation is [(CH3)4N]+

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

14.1

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

+

Self-tests S14.1

CH3

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

N C H3

(a) With potassium? Potassium results in a material with a higher conductivity. (b) With bromine? Bromine can remove electrons from the π-symmetry HOMOs of graphite. This also results in a material with a higher conductivity. S14.2

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

The anion is SiF5–

S14.3 13

13

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

13

-

F F Si

CH4: ΔfH= –61kJ mol–1 SiH4: ΔfH= +39 kJ mol–1 Propose a synthesis of D13CO2– starting from 13CO?

CH3

H3C

F

F F (b) There are two different fluorine environments.

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



24 8-

14.11

O

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

Sn-NMR

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

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

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

Charge = –8. 14.5

14.12

O

Predict the appearance of the spectrum of Sn(CH3)4? Doublet

119

14.7

Use the data in Table 14.2 and the additional bond enthalpy data given here to calculate the enthalpy of hydrolysis of CCl4 and CBr4. Bond enthalpies/(kJ mol–1): O–H = 463, H–Cl = 431, H–Br = 366? ΔhH˚CCl4 = 110 kJmol–1 ΔhH˚CBr4 = 86 kJmol–1

14.8

Identify the compounds A to F:? (A) SiCl4 (B) SiRCl3 (C) RSi(OH)3. (D) RSiOSiR +H2O (E) SiR4 (F)? SiO2

YYYYY14.9 (a) Summarize the trends in relative stabilities of the oxidation states of the elements of Group 14, and indicate the elements that display the inert pair effect. (b) With this information in mind, write balanced chemical reactions or NR (for no reaction) for the following combinations, and explain how the answer fits the trends. (i) Sn2+(aq) + PbO2(s) (excess) → (air excluded) (ii) Sn2+(aq) + O2(air) →? (a) +4 is the most stable oxidation state for the lighter elements, but +2 is the most stable oxidation state of Pb. Pb therefore displays the inert-pair effect. (b) (i) Sn2+ + PbO2 + 4 H+ → Sn4+ + Pb2+ + 2H2O, (ii) 2Sn2+ + O2 + 4H+ → 2Sn4+ + 2H2O. 14.10

Use data from Resource section 3 to determine the standard potential for each of the reactions in Exercise 14.5 (b). In each case, comment on the agreement or disagreement with the qualitative assessment you gave for the reactions? (i) V = +1.31 V. (ii) V= 1.08 V. Both reactions agree with the predictions made in Excercise 14.5.

14.13

Preferably without consulting reference material, draw a periodic table and indicate the elements that form saline, metallic, and metalloid carbides? Ionic(silane) Metallic Metalloid carbides carbides carbides Group I Li, Na, K, elements Rb, Cs Group II Be, Mg, Ca, elements Sr, Ba Group 13 Al B elements Group 14 Si elements 3d-Block Sc, Ti, V, elements Cr, Mn, Fe, Co, Ni 4d-Block Zr, Nb, Mo, elements Tc, Ru 5d-Block La, Hf, Ta, elements W, Re, Os 6d-Block Ac elements Lanthanides Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu

14.14 Describe the preparation, structure and classification of (a) KC8, (b) CaC2, (c) K3C60? (a) KC8? Formed by heating graphite with potassium vapor or by treating graphite with a solution of potassium in liquid ammonia. There is a layered structure of alternating sp2 carbon atoms and potassium ions, a saline carbide. (b) CaC2?

Ca(l) + 2C(s) → CaC2(s) or

CaO(s) + 3C(s) → CaC2(s) + CO(g)



Calcium carbide contains discrete C22– ions. (c) K3C60? A solution of C60 can be treated with elemental potassium. It is ionic. 14.15

Write balanced chemical equations for the reactions of K2CO3 with HCl(aq) and of Na4SiO4 with aqueous acid? K2CO3(aq) + 2HCl(aq) → 2KCl(aq) + CO2(g) + H2O(l)

NH3 + ClO− + H+ → [H3N—Cl-O]− + H+ → H2NCl + H2O H2NCl + NH3 → H2NNH2 + HCl Both can be thought of as redox reactions. S15.5

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

Na4SiO4(aq) + 4HCl(aq) → 4NaCl(aq) + SiO2(s) + 2H2O(l) 14.16

14.17

Describe in general terms the nature of the [SiO3]n2ν 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 SiO4 tetrahedra. The two-dimensional aluminosilicate layers in kaolinite represent another way of connecting SiO4 tetrahedra.

Exercises 15.1

P As Sb Bi O S Se Te

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

Self-tests S15.1

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

S15.2

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

Dimehylhydrazine ignites spontaneously, and produces less CO2. S15.3

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?

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?

N

(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.

CHAPTER 15

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

15.2

Type of element Nonmetal Nonmetal nonmetal metalloid metalloid nonmetal nonmetal nonmetal nonmetal

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 → 2Li3N

4NH3(aq) + 7O2(g) → 6H2O(g) + 4NO2(g)

2Li3N + 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? The second process is considerably cheaper than the first, because lithium is very expensive. 15.4

NO2−(aq) + NO3−(aq) + H2O(l) (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)

NH4NO3(s) + H2O Æ NH4+ + NO3-(aq)

(d) N3–?

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

3NaNH2(l) + NaNO3 →

4

15.5

2

3

at elevated temperatures:

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) PCl4+, (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

Eº = 0.839 V.

HPO32– and H2PO2– ions will be much better reducing agents than oxidizing agents. 15.16

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

LiAlH4

15.12

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

AS

Cl2

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 = AsR3 (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) + 8H2O(l) → 2H3PO4(aq) + 10HCl(aq) (c) AlCl3?

tetrahedral.

PCl5 + AlCl3 → [PCl4]+[AlCl4]− (d) NH4Cl? Cyclic molecules or linear chain polymers nPCl5 + nNH4Cl → −[(N = P(Cl)2)n]− + 4nHCl

15.18

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, N2O, 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

16.6

In Presence of Cl–

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

16.7

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

2

5

3

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

16.9

2

2+

2

O2

-0.13

1.51

HO2-

2

H2O2

16.10

in acidic solution, calculate ΔrG 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

16.11

(a) The disproportionation of H2O2 and HO2? 1.068 V. (b) Catalysis by Cr2+? of decomposing H2O2.

Predict whether any of the following will be reduced by thiosulfate ions,S2O32–, in acidic conditions: VO2+, Fe3+, Cu+, Co3+?

SF reacts with BF to form [SF ][BF ]. Use VSEPR theory to predict the shapes of the cation and anion? 4

3

3

4

SF4+ trigonal pyramidal, BF4− tetrahedral.

Cr2+ is not capable

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

S2O62−and S2O32−. Use the standard potential data in Resource section 3 to predict whether SeO32– is more stable in acidic or basic solution?

Fe3+ and Co3+ will be reduced.

2

2

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

The SeO32– is marginally more stable in acid solutions.

ө

2–

(a) Give the formula for Te(VI) in acidic aqueous solution and contrast it with the formula for S(VI). (b) Offer a plausible explanation for this difference? (b) An explanation? 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 K2O are acidic. 16.2

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

(a) Formulas? H5TeO6– and HSO4–

State whether the following oxides are acidic, basic, neutral, or amphoteric: CO , P O , SO , MgO, K O, Al O , CO? 2

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

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

Exercises 16.1

3

S2O82– > SO42– > SO32–

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

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


ANSWERS TO SELF-TESTS AND EXERCISES 29 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? 19

16.13

(a) SF4 + (CH3)4NF → [(CH3)4N]+[SF5]−; (b) square pyramidal structure; (c) two F environments. Identify the sulfur-containing compounds A, B, C, D, E, and F?

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

B r2

liqu id

lower

soft er

I2

soli d

lowest

soft est

H e

gas

N e A r K r X e

gas gas gas gas

gree n dark redbro 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–, IBr2–, 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:

Exercises 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

gas

lower

Har dnes s of hali de ion hard est soft er

Col or

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?

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 diffusion 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

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

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

3

3

a). b) IF3:

X = IF4N(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

17.6

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

2

Cl2 and F2. 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? 3

3

(a) Difference in volatility? Ammonia exhibits hydrogen bonding. (b) Explain the difference in basicity? The strong electron-withdrawing effect in NF3 reduces the basicity.

c).IF3, two. IF4–, one. 17.9

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

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 trimethylsilyl cyanide? Trimethylsilyl cyanide contains an Si–CN single bond. 17.8

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

6

SbCl5 is trigonal bipyramidal, FClO3 is pyramidal, and ClF6 is octahedral. 17.10

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

+

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

17.7

3

2

3

3

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



(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

IF6+,

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

(a) The structures of [IF6]+ and IF7? octahedral; IF7, pentagonal bipyramid.

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

(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, Cs symmetry. 2

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

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

(a) SbF5? increases the acidity of BrF3. (b) SF6? No effect on the acidity or basicity of BrF3.

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

(c) CsF? Increases the basicity of BrF3. 17.16

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

17.22

+

5

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

2

(b) CH3OH? Methanol, being an organic compound, is readily oxidized by strong oxidants. (c) F2? No.

(b) E at pH 0 and pH 7 for ClO4–? Eº = 1.201 V (see Appendix 2).

At pH 7, V = 0.788 V 17.23

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

4

3

2

Br3− + I2 → 2 IBr + Br−

2

2

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

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

2

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

17.18

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

17.24

Which oxidizing agent reacts more readily in dilute aqueous solution, perchloric acid or periodic acid? Give a mechanistic explanation for the difference?

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 standard 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 atmosphere? 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

2

least Argon.

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:

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 excess of F2:

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

(c) XeO3?

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) XeOF4? (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. 18.5


(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 −

−

4−

2 HXeO4 (aq) + 2 OH (aq) → XeO6 (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 with other xenon fluoride anions? Pentagonal bipyramid.



Use molecular orbital theory to calculate the bond order of the diatomic species E2+ with E=He and Ne? He22+ = 0.5, Ne22+ = 0.5.

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

the group oxidation number is not achieved by N?

19.2

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.

19.3

State the trend in the stability of the group oxidation state on descending a group of metallic elements in the d block. Illustrate the trend using standard potentials in acidic solution for Groups 5 and 6. The group oxidation number increases in stability as you descend a group.

19.4

For each part, give balanced chemical equations or NR (for no reaction) and rationalize your answer in terms of trends in oxidation states.

Predict the appearance of the 129Xe-NMR spectrum of XeOF3+. 1:3:3:1 quartet

18.10 Predict the appearance of the 19F-NMR spectrum of XeOF4. Strong central line, two other lines symmetrically distributed around the central line.


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?

(a) Cr2+ (aq) + Fe3+ (aq) →?

Cr2+(aq) + Fe3+(aq) → Cr3+(aq) + Fe2+(aq) (b) CrO42– (aq) + MoO2 (s) →?

2 CrO42− + 3 MoO2(s) + 10 H+ →

+

+4; VO2 S19.2

S19.3

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

2 Cr3+ + 3 H2MoO4 + 2 H2O (c) MnO4– (aq) + Cr3+ (aq) →?

6 MnO4− + 10Cr3+ + 11H2O → 6Mn2+ + 5Cr2O72− + 22 H+ 19.5

Describe the probable structure of the compound formed when Re3Cl9 is dissolved in a solvent containing PPh3?

2+

More likely for Ni2+ to form a sulphide. Hardness decreases from left to right in the d block.

Exercises 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 oxidation number can be reached but is a powerful oxidizing agent by O, and those for which

2+

2

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

19.1

(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 across Period 4. (c) Give a balanced chemical equation for the reaction.

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 rutile or fluorite structures, and (c) indicate the region 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.

19.7

Write a balanced chemical equation for the reaction that occurs when cis[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. II

+

19.9

(a) [Re(O)2 (py)4]+, (b) [V (O)2 (ox)2]3–, (c) [Mo(O)2(CN)4]4–, (d) [VOCl4]2–? The first three are dioxo complexes and will have cisdioxometal structural units. (d) VOCl42–, has a square-pyramidal structure with an apical oxo ligand. 19.10

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) NiI2? ionic compound with significant degree of covalent character 2

4

2

2

(b) NbCl4? significantly covalent. (c) FeF2? ionic.

−

cis-[Ru LCl(OH2)] + Ox + OH → cis[RuIIILCl(OH)]+ + Red + H2O

(d) PtS ? significant amount of covalent character.

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

(e) WCl2 ? metal–metal bonding.

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

19.11

As oxidation state of the metal increases, ability to accept electron density from an OH– or O2– ligand increases. 19.8

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

(a) [Mo2(O2CCH3)4]? σ2π4δ2 configuration, molybdenum-molybdenum quadruple bond. (b) [Cr2(O2CC2H5)4]? σ2π4δ2,chromium– chromium quadruple bond.

Give plausible balanced chemical reactions (or NR for no reaction) for the following combinations, and state the basis for your answer: (a) MoO42– (aq) plus Fe2+ (aq) in acidic solution? No reaction. (b) The preparation of [Mo6O19]2−(aq) from K2MoO4(s)? 6 MoO42−(aq) + 10 H+(aq) → [Mo6O19]2−(aq) + 5 H2O(l)

(c) [Cu2(O2CCH3)4]? σ2π4δ2δ*2π*4σ*2 configuration, no metal–metal bond in this molecule. 19.12

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

19.13

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

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

5 ReCl5(s) + 2 MnO4−(aq) + 12 H2O(l) → 5 ReO4−(aq) + 2 Mn2+(aq) + 25 Cl−(aq) + 24 H+(aq) (d) MoCl2(s) plus warm HBr(aq)? 6 MoCl2(s) + 2 Br−(aq) → [Mo6Cl12]2−(aq) + Br2(aq) (e) TiO(s) with HCl(aq) under an inert 2TiO (s) + 6H+ (aq) Æ atmosphere? 3+ 2Ti + H2(g) + 2H2O(l) E˚ = +0.37 V.

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

Hg2+ (aq) + Cd(s) Æ Hg(l) + Cd2+ (aq)

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



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

O O O

Cr Cr O

19.14

O

O

S20.5

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

S20.6

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

S20.7

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

S20.8

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. 17500 cm–1 and 22400 cm–1.

S20.9

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

O O

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

[Ru2Cl2(ClCO2)4]- : 2.5 +, mixed valence at the Ru center, in line with the energy level scheme [0.5 (8-3)]. [Ru2 (CH3COCH3)2(ClCO2)4]: 2+, the bond order of 2.0 should also be expected [0.5x(84)].

CHAPTER 20

Oh 1 D are are

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 d7 configurations? A high-spin d7 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 23,800 cm–1 are of this type.

S20.2 The magnetic moment of the complex [Mn(NCS)6]4– is 6.06μB. What is its electron 3 2 configuration? t2g eg S20.3

S20.4

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). Suggest an interpretation of the photoelectron spectra of [Fe(C5H5)2] and [Mg(C5H5)2] shown in Fig. 20.18? In the spectrum of [Mo(CO)6], the ionization energy around 8 eV was attributed to the t2g electrons that are largely metal-based. The

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



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

(c) [Fe(CN)6]3–

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

(e) The Co2+ complex.

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

(d) The ruthenium complex. 20.6

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

.20.7

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

20.8

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

20.9

The spectrum of d1 Ti3+(aq) is attributed to a single electronic transition eg ← t2g. The band shown in Fig. 20.3 is not symmetrical and suggests that more than one state is involved. Suggest how to explain this observation using the Jahn-Teller theorem? The electronic excited state of Ti(OH2)63+ has the configuration t2g0eg1, 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

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

(f) Tetrahedral [FeCl4]2−? d6, high spin, four unpaired electrons, LFSE = 0.6 ΔT. (g) Tetrahedral [Ni(CO)4]? d10, 0 unpaired electrons, LFSE = 0. 20.2

Both H- and P(C6H5)3 are ligands of similar field strength, high in the spectrochemical series. Recalling that phosphines act as π acceptors, is π-acceptor character required for strong-field behaviour? What orbital factors account for the strength of each ligand? No.

Thus there are two ways for a complex to develop a large value of Δ0, by possessing ligands that are π-acids or by possessing ligands that are strong σ-bases (or both). 20.3

20.4

20.5

Estimate the spin-only contribution to the magnetic moment for each complex in Exercise 20.1. Spin-only contributions are: complex N μSO = [(N)(N + 2)]1/2 3+ [Co(NH3)6] 0 0 [Fe(OH2)6]2+ 4 4.9 [Fe(CN)6]3– 1 1.7 [Cr(NH3)6]3+ 3 3.9 [W(CO)6] 0 0 [FeCl4]2– 4 4.9 [Ni(CO)4] 0 0 Solutions of the complexes [Co(NH3)6]2+, [Co(OH2)6]2+ (both Oh), and [CoCl4]2- are colored. One is pink, another is yellow, and the third is blue. Considering the spectrochemical series and the relative magnitudes of ΔT and ΔO, assign each color to one of the complexes. [CoCl4]2– is blue, [Co(NH3)6]2+ is yellow, [Co(OH2)6]2+ is pink. 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. 3+

(b) The Fe complex.

(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



(b) 3p2? 20.13

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

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

D, 3P, 1S. The ground term is 3P.

The gas-phase ion V3+ has a 3F ground term. The 1D and 3P terms lie, respectively, 10 642 and 12 920 cm–1 above it. The energies of the terms are given in terms of Racah parameters as E(3F) = A - 8B, E(3P) = A + 7B, E(1D) = A - 3B + 2C. Calculate the values of B and C for V3+. B = (12920 cm– 1 )/(15) = 861.33 cm–1 and C = 3167.7 cm–1.

20.20

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

20.21

Solutions of [Cr(OH2)6]3+ ions are pale blue– green but the chromate ion, CrO42–, is an intense yellow. Characterize the origins of the transitions and explain the relative intensities. The blue-green color of the Cr3+ ions in [Cr(H2O)6]3+ is caused by 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.

20.22

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

20.14 Write the d-orbital configurations and use the Tanabe–Sugano diagrams (Resource section 6) to identify the ground term of (a) Low-spin [Rh(NH3)6]3+? 1A1g. (b) [Ti(H2O)6]3+? 2T2g. (c) High-spin [Fe(H2O)6]3+? 6A1g. 20.15

Using the Tanabe-Sugano diagrams in Resource section 6, estimate ΔO and B for (a) [Ni(H2O)6]2+ (absorptions at 8500, 15400 and 26000 cm-1) ? Δ0 = 8500 cm–1 and B ≈ 770 cm–1. (b) [Ni(NH3)6]2+ (absorptions at 10750, 17500 and 28200 cm-1)? Δ0 = 10,750 cm–1 and B ≈ 720 cm–1.

20.16

The spectrum of [Co(NH3)6]3+ has a very weak band in the red and two moderate intensity bands in the visible to near-UV. How should these transitions be assigned? The first two transitions listed above correspond to two low-spin bands. The very weak band in the red corresponds to a 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 d6 Co3+ ion in [CoF6]3– is also high spin, but in this case a single spin-allowed transition makes the complex colored and gives it a oneband 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



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 MnO4ions 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 d0. Therefore, no ligand field transitions are possible.

The difference in energy between the two transitions, E(t2) – E(e) = 13700 cm–1, is just equal to ΔT. 20.24

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

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

Exercises 21.1

The rate constants for the formation of [CoX(NH3)5]2+ from [Co(NH3)5OH2]3+ for X = Cl2, Br2, N3–, and SCN¯ differ by no more than a factor of two. What is the mechanism of the substitution? Dissociative.

21.2

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

21.3

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

21.4

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

rate = (kKE[Mn(OH2)62+][X−])/(1 + KE[X−]) 21.5


Calculate the second-order rate constant for the reaction of trans-[PtCl(CH3) (PEt3)2] with NO2– in MeOH, for which npt = 3.22 ? k2(NO2−) = 100.71 = 5.1 M−1s−1

S21.2

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

S21.3

3

3

4

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-

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

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.

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 is mechanism? The dissociative.

21.8

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



21.9

21.10

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

(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) 2phen]. There is a kinetic pathway with activation parameters Δ‡ H = 1101 kJ mol–1 and Δ ‡S = 142 J K –1 mol –1. Propose a mechanism.

A possible mechanism is loss of one dimethyl sulfide ligand, followed by the coordination of 1,10-phenanthroline. 21.11

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

Predict the products of the following reactions: (a) [Pt(PR3)4]2+ + 2Cl–? cis[PtCl2(PR3)2]. (b) [PtCl4]2– + 2PR3? trans-[PtCl2(PR3)2]. (c) cis-[Pt(NH3)2(py)2]2+ + 2Cl–? trans[PtCl2(NH3)(py)].

21.15

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+?

The order of increasing rate is [Ir(NH3)6]3+ < [Rh(NH3)6]3+ < [Co(NH3)6]3+ < [Ni(OH2)6]2+ < [Mn(OH2)6]2+. 21.16

State the effect on the rate of dissociatively activated reactions of Rh(III) complexes of each of (a) an increase in the positive charge on the complex? decreased rate. (b) changing the leaving group from NO3– to Cl–? Decreased rate. (c) changing the entering group from Cl– to I–? This change will have little or no effect on the rate.

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(N3)(NH3)5]2+, [V(OH2)6]2+} {[Co(N3)(NH3)5]2+, [V(OH2)6]2+} → {[Co(N3)(NH3)5]2+, [V(OH2)5]2+, H2O} {[Co(N3)(NH3)5]2+, [V(OH2)5]2+, H2O} → [(NH3)5Co–N=N=N–V(OH2)5]4+ [(NH3)5Co–N=N=N–V(OH2)5]4+ → [(NH3)5Co– N=N=N–V(OH2)5]4+ [(NH3)5Co–N=N=N–V(OH2)5]4+ → [Co(OH2)6]2+ + [V(N3)(OH2)5]2+ The outer sphere pathway: [Co(N3)(NH3)5]2+ + [V(OH2)6]2+ → {[Co(N3)(NH3)5]2+ [V(OH2)6]2+}



40

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

Using these values gives K12 = 1.81 × 1028. Substitution of these values in the MarcusCross relationship gives k12 = 8.51 × 1015 The dm3mol−1s−1.21.21 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 The product will be complex? [W(CO)5(NEt3)], and the quantum yield will be 0.4.

{[Co(N3)(NH3)5] +,[V(OH2)6]3+} → {[Co(OH2)6]2+ , [V(OH2)6]3+ 21.18

The compound [Fe(SCN)(OH2)5]2+ can be detected in the reaction of [Co(NCS)(NH3)5]2+ with Fe2+(aq) to give Fe3+(aq) and Co2+(aq). What does this observation suggest about the mechanism? Appears to be an inner-sphere electron transfer reaction.

21.19

Calculate the rate constants for electron transfer in the oxidation of [V(OH2)6]2+ (Eσ (V3+/V2+) = –0.255 V) and the oxidants (a) [Ru(NH3)6]3+ (EO(Ru3+/Ru2+) = + 0.07 V), (b) [Co(NH3)6]3+ (EO(Co3+/Co2+) = +0.10 V). Comment on the relative sizes of the rate constants. (a) [Ru(NH3)6]3+ k = 4.53 × 103 dm3mol−1s−1 (b) [Co(NH3)6]

3+

−2

3

−1 −1

k = 1.41 × 10 dm mol s

Relative sizes? The reduction of the Ru complex is more thermodynamically favoured and faster. 21.20

Calculate the rate constants for electron transfer in the oxidation of [Cr(OH2)6]2+ (EO–(Cr3+/Cr2+) = –0.41 V) and each of the oxidants [Ru(NH3)6]3+ (EO(Ru3+/Ru2+) = +0.07 V), [Fe(OH2)6]3+ (EO(Fe3+/Fe2+) = +0.77 V) and [Ru(bpy)3]3+ (EO(Ru3+/Ru2+) = +1.26 V). Comment on the relative sizes of the rate constants (a) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1; k22 (Ru3+/Ru2+ for the hexamine complex) = 6.6 × o 103 dm3mol−1s−1; f12 = 1; K12 = e [nF ε /RT] where εo = 0.07 V – (–0.41V) = 0.48 V; n = 1; F = 96485 C; R = 8.31 Jmol−1K−1 and T = 298 K. Using these values gives K12 = 1.32 × 108. Substitution of these values in the MarcusCross relationship gives k12 = 2.95 × 103 dm3mol−1s−1. (b) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1; k22 (Fe3+/Fe2+ for the aqua complex) = 1.1 o dm3mol−1s−1; f12 = 1; K12 = 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. Using these values we get K12 = 9.26 × 1019. Substitution of these values in the MarcusCross relationship gives k12 = 3.19 × 107 dm3mol−1s−1. (c) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1; k22 (Ru3+/Ru2+ for the bipy complex) = 4 × 108 o dm3mol−1s−1; f12 = 1; K12 = 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.

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.

CHAPTER 22 Self-tests No.

S22.1

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. Electron count, 16; oxidation number, +4. 2

2

S22.3

What is the formal name of [Ir(Br)2(CH3)(CO)(PPh3)2]? Dibromocarbonylmethylbis(triphenylphosphine)iridium(III).

S22.4

Which of the two iron compounds Fe(CO)5 and [Fe(CO)4(PEt3)] will have the higher CO stretching frequency? Which will have the longer M–C bond? Fe(CO)5

S22.5

Show that both are 18-electron species. (a) [(η6-C7H8)Mo(CO)3] (49)? η6-C7H8 = 6 electrons, Mo atom = 6. carbonyl = 2 electrons, total = 18. (b) [(η7-C7H7)Mo(CO)3]+(51)? η7-C7H7+ = 6 electrons, Mo atom = 6, 3 COs = 6, total = 18.

S22.6

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

(e) La(η5-Cp*)3? tris(pentamethylcyclopentadienyl)lanthanum(I II), 18e–; (f) Fe(η3-allyl)(CO)3Cl? 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 tetrahedron with 4 terminal Cp rings and four capping COs.

(g) Fe(CO)4(PEt3)? tetracarbonyltriethylphosphineiron(0); (h) Rh(CO)2(Me)(PPh3)? dicarbonylmethyltriphenylphosphinerhodium( I),16e(i) Pd(Cl)(Me)(PPh3)2? chloromethylbis(triphenylphosphine) palladium(II), 16e–;

S22.10 If Mo(CO)3L3 is desired, which of the ligands P(CH3)3 or P(t-Bu)3 would be preferred? Give reasons for your choice. PMe3 would be preferred.

(j) Co(η5-C5H5)(η4-C4Ph4)? cyclopentadienyltetraphenylcylcobutadinecob alt(I), 18e–; (k) [Fe(η5-C5H5)(CO2)]–? dicarbonylcyclopentadienylferrate(0), 18e–;

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

(l) Cr(η6-C6H6)(η6-C7H8)? benzenecycloheptatrienechromium(0), 18e–;

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

(m) Ta(η5-C5H5)2Cl3? trichlorobiscyclopentadineyltanatalum(V), 18e–;

Cl Ph3P

Cl

Cl

Ph

PPh3

Ph3P

(n) Ni(η5-C5H5)NO? cyclopentadineylnitrosylnickel(0),18e–.

PPh3

Pd

Pd

+

PhCl

Cl

Cl

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

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?

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(PEt3)2 (Me)(Cl)] The ethyl group in does not? [Pt(PEt3)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) C6H6? η6, η4, and η2.

(b) Co(η4-C4H4)(η5-C5H5)? 18.

22.8 Provide plausible reasons for the differences in IR wavenumbers between each of the following pairs: (a) Mo(PF3)3(CO)3 2040, 1991 cm–1 versus Mo(PMe3)3(CO)3 1945, 1851 cm–1? CO bands of the trimethylphosphine complex are 100 cm–1 or more lower in frequency. PMe3 is primarily a σ-donor ligand. PF3 is primarily a π-acid ligand. (b) MnCp(CO)3 2023, 1939 cm–1 vs. MnCp*(CO)3 2017, 1928 cm–1? CO bands of the Cp* complex are lower in frequency than the corresponding bands of the Cp complex. Cp* is a stronger donor ligand than Cp.

(c) Co(η3-C3H5)(CO)3? 18.

22.9

(d) Cyclooctadiene? η2 and η4 . (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. Cs symmetry complex, 3

Draw plausible structures and give the electron count of (a) Ni(η3-C3H5)2 (b) (c) Co(η3Co(η4-C4H4)(η5-C5H5) C3H5)(CO)3. If the electron count deviates from 18, is the deviation explicable in terms of periodic terms. (a) Ni(η3-C3H5)2 16, very common for group 9 and group 10 elements.

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.

22.5

State the two common methods for the preparation of simple metal carbonyls and illustrate your answer with chemical equations. Is the selection of method based on thermodynamic or kinetic considerations? (1) Mo(s) + 6 CO(g) → Mo(CO)6(s) (high temperature and pressure required)

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

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

→

The reason that the second method is preferred is kinetic. 22.6

Suggest a sequence of reactions for the preparation of Fe(CO)3(dppe), given iron metal, CO, dppe (Ph2PCH2CH2PPh2), and other reagents of your choice. Fe(s) + 5 CO(g) → Fe(CO)5(l) (high temperature and pressure required)

Fe(CO)5(l) + diphos(s) → Fe(CO)3(diphos)(s) + 2 CO(g) 22.7

Suppose that you are given a series of metal tricarbonyl compounds having the respective symmetries C2v, D3h, and Cs. Without consulting reference material, which of these should display the greatest number of CO stretching bands in the IR

22.10

Decide which of the two complexes W(CO)6 or IrCl(CO)(PPh3)2 should undergo the faster exchange with 13CO. Justify your answer. The Ir complex. It has an A mechanism.

22.11

Which metal carbonyl in each of (a) [Fe(CO)4]2– or [Co(CO)4]– (b) [Mn(CO)5]– or [Re(CO)5]–should be the most basic toward a proton? What are the trends on which your answer is based ? (a) [Fe(CO)4]2– The trend involved is the greater affinity for a cation that a species with a higher negative charge has.



(b) The rhenium complex. The trend involved is the greater M–H bond enthalpy for a period 6 metal ion relative to a period 4 metal ion in the same group. 22.12

Using the 18-electron rule as a guide, indicate the probable number of carbonyl ligands in (a) W(η6-C6H6)(CO)n, (b) Rh(η5C5H5)(CO)n, and (c) Ru3(CO)n. (a) 3 (b) 2 (c) 12

22.13

Propose two syntheses for MnMe(CO)5, both starting with Mn2(CO)10, with one using Na and one using Br2? You may use other reagents of your choice. (i) Reduce Mn2(CO)10 with Na to give Mn(CO)5–; react with MeI to give MnMe(CO)5. (ii), Oxidize with Br2. to give MnBr(CO)5; displace the bromide with MeLi to give MnMe(CO)5.

22.14

Give the probable structure of the product obtained when Mo(CO)6 is allowed to react first with LiPh and then with the strong carbocation reagent, CH3OSO2CF3. Mo(CO)5(C(OCH3)Ph)

22.15

Na[W(η5-C5H5)(CO)3] reacts with 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.

22.17

(a) Reflux Mo(CO)6 with cycloheptatriene to give [Mo(η6-C7H8)(CO)3]; treat with the trityl tetrafluoroborate to give [Mo(η7C7H7)(CO)3]BF4. (b) React [IrCl(CO)(PPh3)2] with MeCl, then expose to CO atmosphere to give [IrCl2(COMe)(CO)(PPh3)2]. 22.18

W

CO CO CO

η1 22.16

W CO CO C:

B: η3

-2 C O

-C O

-H

Fe OC

Fe

CO CO

A

OC

H CO

B

[π-C5H5Fe(CO)2]2 C

The compound B shows two 1H NMR resonances due to Fe-H proton and the aromatic Cp ring. C shows a single 1H NMR resonance because of aromatic Cp ring. 22.19

W CO CO CO

η2

Treatment of TiCl4 at low temperature with EtMgBr gives a compound that is unstable above 270ºC. However, treatment of TiCl4 at low temperature with MeLi or LiCH2SiMe3 gives compounds that are stable at room temperature. Rationalize these observations. The compounds formed are TiR4; ethyl has the low energy β-hydride elimination decomposition.

When Fe(CO)5 is refluxed with cyclopentadiene compound A is formed which has the empirical formula C8H6O3Fe and a complicated 1H NMR spectrum. Compound A readily loses CO to give compound B with two 1H-NMR resonances, one at negative chemical shift (relative intensity one) and one at around 5ppm (relative intensity 5). Subsequent heating of B results in the loss of H2 and the formation of compound C. Compound C has a single 1H-NMR resonance and the empirical formula C7H5O2Fe. Compounds A, B, and C all have 18 valence electrons: identify them and explain the observed spectroscopic data. + F e (C O ) 5

+ A:

Suggest syntheses of (a) [Mo(η7C7H7)(CO)3]BF4 from Mo(CO)6 and (b) from [IrCl2(COMe)(CO)(PPh3)2] [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)3Na. 22.20

Which compound would you expect to be more stable,RhCp2 or RuCp2? Give a plausible explanation for the difference in terms of simple bonding concepts



44

RuCp2 has 18 electrons. Give the equation for a workable reaction for the conversion of Fe(η5-C5H5)2 to Fe(η5C5H5) (η5-C5H4COCH3) and (b) Fe(η5C5H5) (η5-C5H4CO2H)

22.21

(b) Can these CVE values be derived from the 18-electron rule? No. (c) Determine the probable geometry of and [Co6(C)(CO)15]2[Fe6(C)(CO)16]2– ? The iron complex probably contains an octahedral Fe6 array. The cobalt complex probably contains a trigonal-prismatic Co6 array.

(a) Fe(η5 – C5H5)2 + CH3COCl → 5 Fe(η – C5H5)(η5 – C5H4COCH3) + HCl 22.27

(b) Cl

Fe

22.22

22.23

22.24

+

O

O Cl

A lC l 3 C H 2C l2

Cl

Fe

H 2 O , t-B u O K DME

C O 2H

Fe

Sketch the a1’ symmetry-adapted orbitals for the two eclipsed C5H5 ligands stacked together with D5h symmetry. Identify the s, p, and d orbitals of a metal atom lying between the rings that may have nonzero overlap, and state how many a1’ molecular orbitals may be formed. The symmetry-adapted orbitals of the two eclipsed C5H5 rings in a metallocene are shown in Resource Section 5, the dz2 orbital on the metal has a1′ symmetry, as does s. Three a1′ MOs will be formed.

22.28 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.

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.


Derive the ground state of the Tm3+ ion 3 H6.

S23.2

The product of the reaction above is in fact a hydride bridged dimer (9). Suggest a strategy to ensure that the hydride is monomeric. A simple approach would be to use the Cp rings substituted by bulky groups and examine rate of formation.

S23.3

Use the Frost diagrams and data in Resource section 2 to determine the most stable uranium ion in acid aqueous solution in the presence of air and give its formula. UO22+ if sufficient oxygen is present.

Write a plausible mechanism, giving your reasoning, for the reactions: (a) [Mn(CO)5(CF2)]+ + H2O → [Mn(CO)6]+ + 2HF?

(i) The F atoms render the C atom subject to nucleophilic attack (ii)two equivalents of HF are eliminated (b) Rh(C2H5) (CO) RhH(CO)(PR3)2 + C2H4? elimination reaction. 22.25

22.26

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.

(PR3)2 → A β-hydrogen

Given mechanism of CO insertion, what rate constant can be extracted from rate data? Rate = ka[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


ANSWERS TO SELF-TESTS AND EXERCISES 45 state and correlate this deviation with electronic structure. ? (a) Balanced equation 2 Ln(s) + 6 H3O+(aq) →

2 Ln3+(aq) + 3 H2(g) + 6 H2O(l) (b) Redox potentials The potentials for the Ln0/Ln3+ oxidations in acid solution range from a low of 1.99 V for europium to 2.38 V.

23.10

Predict a structure type for BkN based on the ionic radii r(Bk3+) = 96 pm and r(N3−) = 146 pm. The rock-salt structure.

23.11

Describe the general nature of the distribution of the elements formed in the thermal neutron fission of 235U, and decide which of the following highly radioactive nuclides are likely to present the greatest radiation hazard in the spent fuel from nuclear power reactors: (a) 39Ar, (b) 228Th, (c) 90Sr, (d) 144Ce. 90Sr and 144Ce.

(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.

CHAPTER 24

S24.1

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

How would you expect the first and second ionization energies of the lanthanoids to vary across the series? Sketch the graph that you would get if you plotted the third ionization energy of the lanthanoids versus atomic number? Identify elements at any peaks or troughs and suggest a reason for their occurrence? First and second IEs would show general increase across the lanthanoids. With the third, anomalies arise.

S24.2

Why does increased pressure reduce the conductivity of K+ more than that of Na+ in β-alumina? Because K+ is larger than Na+.

S24.3

Rationalize the observation that FeCr2O4 is a normal spinel? The A2+ ions (Fe2+ in this example) occupy tetrahedral sites and the B3+ ions (Cr3+) occupy octahedral sites.

23.4

7

23.5

Derive the ground state of the Tb3+ ion. F6

23.6

Predict the magnetic moment of a compound containing the Tb3+ ion. μ= 9.72 μB

23.7

Explain why stable and readily isolable carbonyl complexes are unknown for the lanthanoids? Carbonyl compounds need back-bonding from metal orbitals of the appropriate symmetry.

23.8

Suggest a synthesis of neptunocene from NpCl4? Np(COT)2 was prepared by the reaction of K2COT with NpCl4 under inert atmosphere.

23.9

Account for the similar electronic spectra of Eu3+ complexes with various ligands and the variation of the electronic spectra of Am3+ complexes as the ligand is varied. The 5f orbitals of the actinide ions interact strongly with ligand orbitals, and the splitting of the 5f subshell, as well as the color of the complex, varies as a function of ligand.

Self-tests

Exercises 24.1

NiO doped with Li2O? The electronic conductivity of the solid increases owing to formation of Ni1–xLix)O.

24.2

What is a crystallographic shear plane? Both.

24.3

How might you distinguish between a solid solution and a series of discrete crystallographic shear plane structures? A solid solution would contain a random collection of crystallographic shear planes.

24.4

Wurtzite crystal structure and bottleneck? The wurtzite structure is shown in Figure 3.35. The normal sites for cations in this structure are the tetrahedral holes. The bottleneck involves the space formed by three close packed anions.

24.5

Synthesis of? (a) MgCr2O4 - heat (NH4)2Mg(CrO4)2 gradually to 1100-1200°C; (b) SrFeO3Cl - heat SrO + SrCl2 + Fe2O3 in a sealed tube; (c) Ta3N5 - heat Ta2O5 under NH3 at 700°C.

24.6

Products of reactions? (a) LiNiO2, (b) Sr2WMnO6.



46 24.7

Where might intercalated Na+ ions reside in the ReO3 structure? The unit cell for ReO3 is shown in Figure 24.16(a), the structure is very open.

24.8

Antiferromagnet ordering? In an antiferromagnetic substance, the spins on different metal centers are coupled into an antiparallel alignment. As the temperature approaches 0 K, the net magnetic moment goes to zero.

24.9

Magnetic measurements on ferrite? inverse spinel.

24.10

Using LFSEs determine site preference for A= Ni(II) and B = Fe(III)? Better ligand field stabilization and a strong preference for inverse spinel.

24.11

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

24.12

Classify oxides as glass-forming or nonglass-forming? BeO, B2O3, and to some extent GeO2, glass forming. Transition metal and rare earth oxides are typically nonglassforming.

24.13

Which metal sulfides might be glass forming? Metalloid and nonmetal sulfides.

24.14

Examples of spinels containing? (a) (b) fluoride? sulfide? Zn(II)Cr(III)2S4 Li2NiF4

24.15

Synthesis of LiTaS2? Direct reaction of TaS2 with BuLi, or through electrochemical insertion of Li ions.

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 candidate to consider.

24.19

Formula for this lithium aluminium magnesium dihydride and structures?

Order of band gaps? BN > C(diamond) > AlP > InSb.

24.22

Fulleride structures? In Na2C60, all of the tetrahedral holes are filled with sodium cations. In Na3C60, all of the tetrahedral holes and all of the octahedral holes are filled with sodium cations.


Synthesis of core-shell nanoparticles? The thermodynamic driving force is adjusted to a level that allows for heterogeneous nucleation of the shell material on the core but prevents homogeneous nucleation of the shell material.

S25.2

Hemispherical imprint for 2 nm nanoparticle? The radius is 4.21 nm and the diameter is 8.42 nm.

S25.3

Host for QDs? MCM-41.

Exercises 25.1

(a) Surface areas? 3.14 × 102 nm2 versus 3.14 × 106 nm2 (a factor of 104). (b) Nanoparticles based on size? The 10 nm particle. (c) Nanoparticles based on properties? A nanoparticle should exhibit properties different from those of a molecule or an extended solid.

25.2

Electron length and quantum confinement? A characteristic length is the exciton Bohr radius.

25.3

Why are QDs better for bioimaging? One light source can be used to excite different quantum dots.

25.4

Band energies for QD versus bulk semiconductor? The energies of the band edges for a QD nanocrystal are more widely separated.

25.5

(a) Top-down versus bottom-up? The “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

framework

Mg1-x(M)xH2y (M=Al, Li), The Li and Al will be incorporated as metal hydride solid solutions. 24.20

24.21

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.



allow for the precise spatial control over atoms and molecules. 25.6

25.7

25.8

(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. 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.. (a) Steps in solutions synthesis of nanoparticles? (i) solvation; (ii) stable nuclei of nanometer dimensions formed; (iii) growth of particles to desired size occurs. (b) Why should the last two steps occur independently? So that nucleation fixes the total number of particles and growth leads to controlled size and narrow size distribution. (c) Stabilizers? Prevent surface oxidation and aggregation, and they limit traps for the holes and electrons, and improve quantum yields and luminescence.

25.9

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

25.10

(a) What is a core-shell nanoparticle?

25.11

(a) Homogeneous versus heterogeneous? Homogeneous nucleation leads to solid formation throughout the vapour phase. (b) thin film? Heterogeneous nucleation is (c) preferred in thin-film growth. nanoparticles? Homogeneous nucleation is generally preferred for nanoparticle synthesis.

25.12

PVD versus CVD? In CVD, the atomic species of interest are bound chemically to other species. Also, their thermal energies are typically rather low. In PVD, the atomic species of interest are typically atomic.

25.13

(a) Purpose of QD layers? Multiple layers of quantum dots can increase the intensity of any optical absorption or emission. They can also be used to form quantum cascade lasers. (b) Limitations? The limitations come from the requirements on how coherent the interface between the two materials must be.

25.14

(a) Applications of quatum wells? Lasers and optical sensors. (b) Why are they used over other materials? They exhibit properties that are not observed in molecular or traditional solid state materials. (c) How are they made? Molecular beam epitaxy.

25.15

Superlattices and improved properties? AlN and TiN - different elastic constants, large number of interfaces spaced on the nanoscale. Results are much improved hardness values.

25.16

(a) Self-assembly? (a) Offers methods to bridge bottom-up and top-down approaches to synthesis. (b) In nanotechnology? Offers a route to assemble nanosized particles into macroscopic structures.

25.17

Common features of self-assembly? (i) molecular or nanoscale subunits; (ii) spontaneous assembly of the subunits; (iii) noncovalent interactions between the assembled subunits; (iv) longer-range structures arising from the assembly process.

25.18

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.

Shell Core

(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.



48

25.19 25.20

25.21

25.22

25.23

25.24.

25.25

Compare SAMs and cell membranes? SAM can structurally resemble a phospholipid bilayer. What is morphosynthesis? Control of nanoarchitectures in inorganic materials through changes in synthesis parameters. (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. (a) Why is dispersion important in nanocomposites? They lead to increased exposed surface areas. (b) Why is dispersion difficult? The often nonpolar organic polymers do not have strong interactions with the polar or ionic inorganic components.

S26.4

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

Exercises 26.1

(c) The production of Li3N and its reaction with H2O? Not catalysis. 26.2

Define the following terms? (a) Turnover frequency? The amount of product formed per unit time per unit amount of catalyst. (b) Selectivity? How much of the desired product is formed relative to by-products. (c) Catalyst? A substance that increases the rate of a reaction but is not itself consumed. (d) Catalytic cycle? Sequence of chemical reactions involving the catalyst that transform the reactants into products.

(a) Biomimetics? Designing nanomaterials that mimic biological systems. (b) example of biomimetics? Cellulose fibers in paper have been used to template the growth of titanium oxide nanotubes.

(e) Catalyst support? Generally a ceramic like γ-alumina or silica gel. 26.3

Classify the following as homogeneous or heterogeneous catalysis? (a) The increased rate of SO2 oxidation in the presence of NO? Homogeneous. (b) The hydrogenation of oil using a finely divided Ni catalyst? Heterogeneous. (c) The conversion of D-glucose to a mixture by HCl? Homogeneous.

CHAPTER 26 26.4

Self-tests S26.1

Which of the following constitute catalysis? (a) H2 and C2H4 in contact with Pt? An example of genuine catalysis. (b) H2 plus O2 plus an electrical arc? Not catalysis..

A bionanomaterial and its application? PPF/PPF-DA is an injectable bionanomaterial used for bone-tissue engineering.

Bionanocomposites and improved mechanical strength? Biomimetics need trabecular and cortical bone tissues, which combine compressibility and tensile strength. Hybrid alumoxane nanoparticles dispersed in PPF/PPF-DA shows enhanced strength.

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

D,L

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 try to catalyze efficiently at 80ºC.

S26.2

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

(d) The hydrogenation of double bonds in vegetable oil? The process can be readily set up with existing technology.

S26.3

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

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

26.7

26.8

26.9

26.10

26.11

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. 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. How does starting with MeCOOMe instead of MeOH lead to ethanoic anhydride instead of ethanoic acid using the Monsanto acetic acid process? The reaction of the ethanoate ion with the acetyl iodide leads to ethanoic anhydride.

exchange of the methyl group in the chemisorbed –CHR(CH3) group (R = C(CH3)2(C2H5)). 26.14

Why does CO decrease the effectiveness of Pt in catalyzing the reaction 2H+(aq) + 2e– → H2(g)? Platinum has a strong tendency to chemisorb CO.

26.15

Describe the role of an electrocatalyst? Platinum is the most efficient electrocatalyst for accelerating oxygen reduction at the fuel cell cathode, but is expensive.

CHAPTER 27 Self-tests S27.1 S27.2

S27.3

Why does saline contain NaCl? Osmotic balance.

S27.4

Explain the significance of the Calcium ion pumps activation by calmodulin? The binding of calmodulin is a signal informing the pump that the cytoplasmic Ca2+ level has risen above a certain level.

S27.5

Why are iron-porphyrin complexes unable to bind O2 reversibly? The Fe(II) gets oxidized to Fe(III), yielding an oxo-Iron(III) porphyrin complex.

S27.6

What is the nature of binding at Cu blue centers as indicated by the EPR spectrum? There is greater covalence in blue Cu centers than in simple Cu(II) compounds.

Why is the platinum-rhodium in automobile catalytic converters dispersed on the surface of a ceramic rather than used in the form of thin foil? A thin foil of platinum-rhodium will not have as much surface area as an equal amount of small particles finely dispersed on the surface of a ceramic support.

S27.7

What is the nature of an active site with Copper (III)? Diamagnetic with squareplanar geometry. 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.

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.

S27.9

Suggest a reason why? (a) Ring opening alkene metathesis polymerization (ROMP) proceeds? ROMP can result in reduced steric strain. (b) Ring-closing metathesis (RCM) reaction proceeds? RCM results in the loss of ethane. (a) Attack by dissolved hydroxide? Structure C in Figure 26.11 (b) Attack by coordinated hydroxide? Structure E given in Figure 26.11 (c) Can one differentiate the stereochemistry? Yes. (a) Enhanced acidity? When Al3+ replaces Si4+ on lattice site charge is balanced by H3O+. (b) Three other ions? Ga3+, Co3+, and Fe3+.

26.12

26.13

Is Iron (II) expected to be present in the cell as uncomplexed ions? No. Unusual coordination of Mg? The protein’s 3D structure can place any particular atom in a suitable position for axial coordination.

(ii) The second observation can be explained by invoking a mechanism for rapid deuterium

S27.8

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.



50

Exercises 27.1

Lanthanides versus calcium? They are hard Lewis acids and prefer coordination by hard bases. Gd3+ has excellent fluorescence.

27.2

Substituting Co2+ for Zn2+? Co(II) commonly adopts distorted tetrahedral and five-coordinate geometries typical of Zn(II) in enzymes. Co(II) is paramagnetic, enabling Zn enzymes to be studied by EPR.

27.3

Compare the acid/base catalytic activities Acid of Zn(II), Fe(III), and Mg(II)? strengths: Fe(III) > Zn(II) > Mg(II). Ligand binding rates are Mg(II) > Zn(II) > Fe(III). Propose a physical method for the EPR or determination of Fe(V)? Mossbauer.

27.4

27.5

Interpret the Mossbauer spectra of ferredoxin? In the spectrum, the oxidized spectrum is consistent with the iron atoms having the same valence, intermediate between the 3+ and the 2+ states, with one pair spin-up and the other pair spin-down.

27.6

Explain the differences in the structures of the oxidized and reduced forms of the Pcluster in nitrogenase? The change in structure suggest that the coupling of proton and electron transfer can also occur at the Pcluster, by controlling protonation of the exchangeable ligands. What metals are involved in the synthesis of acetyl groups? Co(II), Fe(II) and Cu(I).

27.7

Answers to self-tests and exercises

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