Inorganic Chemistry Solutions Manual [4 ed.] 0273742760, 9780273742760

Table of contents :
Chapter_01_X1a
Chapter_02_X1a
Chapter_03_X1a
Chapter_04_X1a
Chapter_05_X1a
Chapter_06_X1a
Chapter_07_X1a
Chapter_08_X1a
Chapter_09_X1a
Chapter_10_X1a
Chapter_11_X1a
Chapter_12_X1a
Chapter_13_X1a
Chapter_14_X1a
Chapter_15_X1a
Chapter_16_X1a
Chapter_17_X1a
Chapter_18_X1a
Chapter_19_X1a
Chapter_20_X1a
Chapter_21_X1a
Chapter_22_X1a
Chapter_23_X1a
Chapter_24_X1a
Chapter_25_X1a
Chapter_26_X1a
Chapter_27_X1a
Chapter_28_X1a
Chapter_29_X1a
FM_X1a
Ibc_X1a
Ifc_X1a

Citation preview

1

1 1.1

Basic concepts: atoms The notation: 50 Cr 24

shows that the atomic number, Z, is 24 and the mass number for the isotope is 50. Number of protons = Number of electrons = Z = 24 Number of neutrons = Mass number – Z = 50 – 24 = 26 For each isotope, Z = 24 and so there are 24 electrons and 24 protons. For mass numbers 52, 53 and 54, there are 28, 29 and 30 neutrons, respectively. 1.2 See Appendix 5 in H&S 1.3

Notation: 27 13 Al 79 81 35 Br 35 Br 54 56 26 Fe 26 Fe

57 58 26 Fe 26 Fe

1.4

‘Monotopic’ means that the element possesses only one isotope. Examples other than As include P, Na and Be. (a) Al is monotopic, i.e. there is only one naturally occurring isotope. Z = 13 Mass number = 27 Number of electrons = Number of protons = 13 Number of neutrons = 27 – 13 = 14 (b) Br (Z = 35) has 2 naturally occurring isotopes. Each isotope has 35 electrons and 35 protons. For the isotope with mass number 79: number of neutrons = 79 – 35 = 44 For the isotope with mass number 81: number of neutrons = 81 – 35 = 46 (c) Fe (Z = 26) has 4 naturally occurring isotopes. Each isotope has 26 electrons and 26 protons. For the isotope with mass number 54: number of neutrons = 54 – 26 = 28 For the isotope with mass number 56: number of neutrons = 56 – 26 = 30 For the isotope with mass number 57: number of neutrons = 57 – 26 = 31 For the isotope with mass number 58: number of neutrons = 58 – 26 = 32 Assume that 3H can be ignored since abundance is so low; error introduced by this assumption is negligible. The mass numbers of 1H and 2H are 1 and 2 respectively. Let % 1H = x, and % 2H = 100 – x Then: x × 1 (100 − x ) × 2 + 100 100 100.8 = x + 200 – 2 x x = 99.2

A r = 1.008 =

This result gives 99.2 % 1H and 0.8 % 2H. The values do not agree with those in Appendix 5 (99.985 % 1H and 0.015 % 2H) because we have used integral atomic masses for the isotopes. The accurate masses (5 sig. fig.) are 1.0078 and 2.0141, and if you work through the above calculation again, this gives 99.98 % 1H and 0.02 % 2H.

Basic concepts: atoms

2

S S

1.5

(a) Isotopic abundances: 32S 95.02 %, 33S 0.75 %, 34S 4.21 %, 36S 0.02 %. Relative intensities of peaks containing these isotopes must reflect their relative abundances. m/z = 256 is assigned to (32S)8 – the most abundant peak. m/z = 257 is assigned to (32S)7(33S). m/z = 258 is assigned to (32S)6(33S)2 and (32S)7(34S). m/z = 259 is assigned to (32S)6(33S)(34S). m/z = 260 is assigned to (32S)6(34S)2. (b) The structure of S8 is shown in 1.1; the parent ion arises from S8. Fragmentation by S–S bond cleavage produces S7, S6, S5, S4 ... and gives lower mass peaks.

1.6

(a)

S S

S S

S S

(1.1)

c = λν

λ= λ=

c

c in m s–1, λ in m, ν in Hz (s–1)

ν 2.997 ×108 12

3.0 ×10

= 1.0 ×10− 4 m

This lies in the far infrared region of the electromagnetic spectrum. See Appendix 4 in H&S

(b)

λ=

2.997 ×108 1.0 ×1018

= 3.0 ×10−10 m

This lies in the X-ray region of the electromagnetic spectrum. (c)

λ=

2.997 ×108 14

5.0 ×10

= 6.0 ×10−7 m

This electromagnetic radiation is in the visible region. 1.7

1.8

Refer to Fig. 1.3 in H&S and the accompanying discussion. Transitions to the level n = 1 belong to the Lyman series, therefore (a) and (e). Transitions to the level n = 2 belong to the Balmer series, therefore (b) and (d). Transitions to the level n = 2 belong to the Paschen series, therefore (c).

E = hν

ν=

c

λ

Units: λ in m

450 nm = 450 × 10–9 m

E = 4.41 × 10 −22 kJ

For the energy per mole, multiply by the Avogadro number: E = 4.41 × 10 −22 × 6.022 × 10 23 = 266 kJ mol −1

1.9

Equation 1.4 in H&S is: ⎛ 1

ν = R ⎜⎜

⎝2

2

−

1 ⎞ ⎟⎟ n2 ⎠

where R = 1.097 × 107 m–1

Basic concepts: atoms

3

For n = 3: ⎛1 ⎝4

1⎞ 9⎠

ν = 1.097 × 10 7 ⎜ − ⎟ = 1.524 × 10 6 m −1 Convert m–1 to m, and then to nm (1 nm = 10–9 m) to obtain the wavelength in nm:

ν =

λ=

1

λ

1 1.524 × 10 6

× 10 9 = 656.2 nm (to 4 sig. fig.)

For n = 4: ⎛1 ⎝4

ν = 1.097 × 10 7 ⎜ − λ=

1 6

1⎞ 6 −1 ⎟ = 2.057 × 10 m 16 ⎠

× 10 9 = 486.2 nm (to 4 sig. fig.)

2.057 × 10 Use the same method for n = 5, and n = 6, to obtain calculated values of λ of 434.1 nm and 410.2 nm respectively. Calculated values are consistent with observed lines in the Balmer series.

1.10

The radii, rn, are found using eq. 1.8 in H&S, with values of n = 2 and n = 3. rn =

ε 0h 2n 2 πme e 2

where: ε0 = permittivity of a vacuum h = Planck constant me = electron rest mass e = charge on an electron

For n = 2: rn =

8.854 ×10 −12 × (6.626 × 10 −34 ) 2 × 2 2 π × 9.109 × 10 −31 × (1.602 × 10 −19 ) 2

= 2.117 ×10 −10 m

or the radius may be quoted in pm: rn = 211.7 pm (1 pm = 10–12 m) For n = 3: 8.854 × 10 −12 × (6.626 × 10 −34 ) 2 × 32 rn = = 4.764 × 10 −10 m − 31 −19 2 π × 9.109 ×10 × (1.602 ×10 ) or:

rn = 476.4 pm

1.11

As the value of n increases, (a) the energy of an ns orbital increases, and (b) the size of the orbital increases.

1.12

(a) For the 6s orbital: n = 6, l = 0, ml = 0 (b) Each of the five 4d orbitals has a unique set of quantum numbers: n = 4, l = 2, ml = –2 n = 4, l = 2, ml = –1 n = 4, l = 2, ml = 0 n = 4, l = 2, ml = 1 n = 4, l = 2, ml = 2

1.13

(a) Principal quantum number = n. Each of the three 4p orbitals has a value of n = 4.

4

Basic concepts: atoms (b) For any np orbital, the orbital quantum number l = 1. Therefore the three 4p orbitals have the same value of l. (c) Each 4p orbital has a unique value of the magnetic quantum number, ml. For the three 4p orbitals, the sets of quantum numbers are: n = 4, l = 1, ml = –1 n = 4, l = 1, ml = 0 n = 4, l = 1, ml = 1 1.14

For information on radial nodes, see Figs. 1.7, 1.8 and 1.13 and accompanying discussion in H&S. Radial nodes follow the trend shown below: n

1

2

3

4

5

6

s p d f

0

1 0

2 1 0

3 2 1 0

4 3 2 1

5 4 3 2

(a) A 2s orbital has 1 radial node. (b) A 4s orbital has 3 radial nodes. (c) A 3p orbital has 1 radial node. (d) A 5d orbital has 2 radial nodes. (e) A 1s orbital has no radial nodes. (f) A 4p orbital has 2 radial nodes. 1.15

(a) Figure 1.5a in H&S shows a plot of R(r) against r for the 1s orbital of hydrogen. An s orbital has a finite value of R(r) at the nucleus. Figure 1.7 in H&S shows a plot of the radial distribution function, 4πr2R(r)2, against r for the hydrogen 1s orbital; at the nucleus, the function is zero. The maximum value of 4πr2R(r)2 for the 1s curve in Fig. 1.7 in H&S corresponds to the value of r at which the electron has the highest probability of being found. (b) For the 4s orbital of hydrogen, a plot of R(r) against r has both positive and negative values of R(r); it has a positive value at r = 0 (i.e. at the nucleus). The 4s orbital has 3 radial nodes; these are the three points at which the curve crosses the axis along which r is plotted. Values of 4πr2R(r)2 are always greater or equal to zero. A plot of 4πr2R(r)2 against r for the hydrogen 4s orbital has four points at which the function is zero: at r = 0 and at three points corresponding to the radial nodes. (c) Figure 1.6 in H&S shows a plot of R(r) against r for the hydrogen 3p orbital. At r = 0, R(r) = 0. The curve crosses the r axis once, and the function has both positive and negative values. Figure. 1.8 in H&S shows a plot of 4πr2R(r)2 against r for the 3p orbital of hydrogen. The curve shows that the orbital has one radial node. Values of 4πr2R(r)2 are always greater than or equal to zero.

1.16

Each orbital is defined by a set of three quantum numbers: n, l and ml. (a) 1s: n = 1 l = 0 ... (n – 1), so only l = 0 is possible; ml = +l, +(l – 1) ... 0 ... –(l – 1), –l , but since l = 0, only ml = 0 is possible. Therefore, the set of quantum numbers that defines the 1s orbital is: n = 1, l = 0, ml = 0.

Basic concepts: atoms

5

4πr2R(r)2

(b) 4s: n = 4 l = 0, 1, 2, 3, but for the s orbital, l = 0; as for (a) above, if l = 0, ml = 0. Therefore, the set of quantum numbers that defines the 4s orbital is: n = 4, l = 0, ml = 0. (c) 5s: n = 5 l = 0, 1, 2, 3, 4 but for the s orbital, l = 0; as above, if l = 0, ml = 0. Therefore, the set of quantum numbers that defines the 5s orbital is: n = 5, l = 0, ml = 0. 1.17

For a 3p atomic orbital: n = 3, l = 1, ml = +1, 0 or –1. Therefore, the three 3p orbitals are defined by the three sets of quantum numbers: n l ml 3 1 +1 3 1 0 3 1 –1

1.18

The combination of n = 4 and l = 3 refers to 4f atomic orbitals. For l = 3, the possible values of ml are +3, +2, +1, 0, –1, –2, –3, and so there are 7 atomic orbitals in the 4f set. The sets of 3 quantum numbers that uniquely define each 4f orbital are: n l ml n l ml 4 3 +3 4 3 –1 4 3 +2 4 3 –2 4 3 +1 4 3 –3 4 3 0

1.19

A hydrogen-like atom or hydrogen-like mononuclear ion contains one electron. To work out the answer to the problem, you need the atomic numbers (Z) of H (1), He (2) and Li (3). Thus, atomic H, He and Li contain 1, 2 and 3 electrons respectively. (a) H+ has no electrons, and is therefore not hydrogen-like. (b) He+ has 1 electron, and is therefore hydrogen-like. (c) He– has 3 electrons, and is therefore not hydrogen-like. (d) Li+ has 2 electrons, and is therefore not hydrogen-like. (e) Li2+ has 1 electron, and is therefore hydrogen-like.

1.20

(a) He+ is hydrogen-like, but it does not have the same nuclear charge as H. From footnote to Table 1.2 in H&S, radial part of the wavefunction R(r) is of the form:

He+

⎛ Z R (r ) = 2⎜⎜ ⎝ a0

H 0 0

1

2

3

Distance from nucleus, r / atomic units

Fig. 1.1 For answer 1.20b.

4

⎞ ⎟ ⎟ ⎠

3

2

⎛ Zr ⎞ ⎜ ⎟

e – ⎜⎝ a0 ⎟⎠

For H, Z = 1, but for He+, Z = 2. Thus, compared to the curve in Fig. 1.5a in H&S for the H atom, a similar curve for He+ will be of a similar shape, but shifted to the left and drawn in towards the nucleus. (b) The radial distribution functions for the 1s atomic orbitals in the H atom and He+ ion are shown in Fig. 1.1. Note how (compared to the H atom) the increased nuclear charge on He+ results in there being a greater probability of finding the electron closer to the nucleus.

Basic concepts: atoms 1.21

To find the energy of the 3s orbital of an H atom, use the equation: E=−

See eq. 1.16 in H&S E=−

k n2

1.312 ×10 3 32

= −146 kJ mol −1

In the H atom, orbitals with the same principal quantum number are degenerate, i.e. they have the same energy. Therefore the 3s and 3p atomic orbitals have the same energy. 1.22

0

Equation 1.16 in H&S is (with Z = 1): E=−

k n2

k = 1.312 × 103 kJ mol–1

For n = 2:

E=−

n=5 n=4 n=3 n=2

–500

For n = 1:

E=−

E / kJ mol–1

6

3

1.312 × 10 = −1312 kJ mol −1 1 1.312 ×103 = −328.0 kJ mol−1 4

–1000

n=1

For n = 3, E = –145.8 kJ mol–1, for n = 4, E = –82.00 kJ mol–1; for n = 5, E = –52.50 kJ mol–1.

–1500

Fig. 1.2 For answer 1.22.

The relative spacings of the energy levels are shown in Figure 1.2 and decrease as n increases. As one approaches the continuum (where n = ∞ ), the energy levels are extremely close together. 1.23

The six sets of quantum numbers that orbitals are: n = 5, l = 1, ml = –1, ms = +1/2 n = 5, l = 1, ml = –1, ms = –1/2 n = 5, l = 1, ml = 0, ms = +1/2 n = 5, l = 1, ml = 0, ms = –1/2 n = 5, l = 1, ml = 1, ms = +1/2 n = 5, l = 1, ml = 1, ms = –1/2

describe the six electrons in the three 5p

Electrons are spin-paired Electrons are spin-paired Electrons are spin-paired

1.24

The approximate order of orbital energies in increasing energy is: 1s < 2s < 3s < 3p < 3d < 4p < 6s < 6p

1.25

In the ground state, the 1s orbital of Li (Z = 3) is fully occupied. The third electron could occupy either a 2s or 2p orbital. This electron will experience the effective nuclear charge of the nucleus, being partially shielded from the nucleus by the 1s electrons. The 2s orbital penetrates the 1s orbital more than the 2p orbital does.

Basic concepts:atoms

7

Therefore, an electron in a 2p orbital is more shielded from the nucleus than an electron in a 2s orbital. Occupying the 2s orbital rather than the 2p orbital therefore leads to a lower energy system. Total number of electrons = atomic number, Z. In the notations below, the core electrons are written inside [ ]. The remaining electrons are the valence electrons. (a) Na Z = 11 Electronic configuration is [1s22s22p6 ] 3s1. (b) F Z = 9 Electronic configuration is [1s2] 2s22p5. (c) N Z = 7 Electronic configuration is [1s2] 2s22p3. (d) Sc Z = 21 Electronic configuration is [1s22s22p63s23p6] 4s23d1.

1.27

Approximate energy level diagrams for Na, F, N and Sc are as follows: 3s

Energy

Na

F

2p

Energy

1.26

2p 2s

2s 1s

2p 2s

Sc

Energy

N

Energy

1s

3d 4s 3p

1s

3s 2p 2s 1s

1.28

The ground state electronic configuration of B (Z = 5) is 1s22s22p1. For electrons in the 1s orbital: n = 1, l = 0, ml = 0; ms = +1/2 n = 1, l = 0, ml = 0; ms = –1/2 For electrons in the 2s orbital: n = 2, l = 0, ml = 0; ms = +1/2 n = 2, l = 0, ml = 0; ms = –1/2 For the electron in the 2p orbital: n = 2, l = 1, ml = 0 or +1 or –1; ms = +1/2 or –1/2

1.29

(a) Li: (b) O: (c) S: (d) Ca: (e) Ti:

Z=3 Ground state electronic configuration is 1s22s1. Z=8 Ground state electronic configuration is 1s22s22p4. Z = 16 Ground state electronic configuration is 1s22s22p63s23p4. Z = 20 Ground state electronic configuration is 1s22s22p63s23p64s2. Z = 22 Ground state electronic configuration is 1s22s22p63s23p64s23d2.

8

Basic concepts: atoms (f) Al:

1.30

Z = 13 Ground state electronic configuration is 1s22s22p63s23p1.

(a) F: Z=9 The ground state electronic configuration is:

1s2

2s22p5

Core Valence electrons electrons Energy

The energy level diagram for the valence electrons is:

2p

2s

Energy

(b) Al: Z = 13 The ground state electronic configuration is:

The energy level diagram for the valence electrons is:

1s22s22p6

3s23p1

Core electrons

Valence electrons

3p 3s

The energy level diagram for the valence electrons is:

Energy

(c) Mg: Z = 12 The ground state electronic configuration is:

1s22s22p6

3s2

Core electrons

Valence electrons

3p 3s

1.31 Ground state arrangement of 4 electrons in a set of p orbitals.

1.32

The arrangement of the np4 electrons is shown in the diagram in the margin. Hund’s (first) rule and the Pauli exclusion principle are used to determine this arrangement of electrons. See Section 1.9 in H&S for further details. (a) IE4 of Sn = 4th ionization energy. This is the energy required to remove the fourth electron from Sn in the gas phase. The equation that defines the process to which the value of IE4 of Sn refers is: Sn3+(g)

Sn4+(g) + e–

Energy is required, therefore the enthalpy change is positive. The reaction is endothermic. (b) (IE1 + IE2 + IE3) is the sum of the first three ionization energies of Al(g), and corresponds to the overall change: Al(g) Al3+(g) + 3e– . This can be summarized in a thermochemical cycle as shown alongside:

Al(g)

IE1 + IE2 + IE3

–3e–

IE1 –e–

Al+(g)

Al3+(g)

IE3 –e– IE2

–e–

Al2+(g)

1.33

Figure 1.3 shows the trend in the first four ionization energies of X. The value of IE1 is much lower than that of IE2, indicating that the first electron is removed relatively easily, but that the second and subsequent electrons require significantly more energy. This trend is characteristic of a group 1 element, so X is likely to be an alkali metal. (X is actually Rb.)

Ionization energy / kJ mol–1

Basic concepts:atoms

9

4000

2000

0 1

2

3

4

Electron removed

Fig. 1.3 For answer 1.33.

IE1 / kJ mol–1

600

400

200

0 Li Na K Rb Cs

1000

IE1 / kJ mol–1

IE1 / kJ mol–1

Fig. 1.4 For answer 1.34a.

(a) Descending group 1 (Fig. 1.4). All values are relatively low, i.e. easy ionization of the first electron: M(g) to M+(g). The general decrease in values down the group reflects the reduced attraction between the nucleus and the valence electron that is being ionized; screening effects increase as more shells of core electrons are added. (b) Descending group 13: see Fig. 1.5. A decrease from B to Al is expected because the core electrons of Al screen the valence 3p electron from the nuclear charge. The lack of a continued decreasing trend can be explained in terms of the failure of the d- and f-electrons (which have a low screening power) to compensate for the increase in nuclear charge. (c) Crossing the first row of the d-block: see Fig. 1.6. The overall trend is for a general increase in IE1, but differences between the values is small. For each metal, the ionization: M(g) M+(g) + e– corresponds to the removal of a 4s electron. (d) Crossing the row of elements from B to Ne: see Fig.1.7. General increase in IE1 is caused by an increase in Zeff. Nitrogen has a ground state electronic configuration 2s22p3 and the 2p level is half-occupied. There is a certain stability (see Box 1.7 in H&S) associated with this configuration and it is more difficult to ionize N than O. (e) Going from Xe to Cs: Xe (Z = 54) is a noble gas with ground state electronic configuration [Kr]5s24d105p6, so removal of an electron involves removal from a fully occupied quantum shell and requires a very large amount of energy. Cs (Z = 55) is an alkali metal with ground state electronic configuration [Xe]6s1. Removal

750

1000

IE1 / kJ mol–1

1.34

750

2500 2000

1500 500

500

1000 250

250

500

0

0 B Al Ga In Tl

Fig. 1.5 For answer 1.34b.

0 Sc Ti V Cr Mn Fe Co Ni Cu Zn

Fig. 1.6 For answer 1.34c.

B C N

O F Ne

Fig. 1.7 For answer 1.34d.

10

Basic concepts: atoms of the valence electron from a singly occupied orbital in Cs requires a much smaller (although still large) amount of energy than removing the first electron from the full shell in Xe (375.7 versus 1170 kJ mol–1). (f) Going from P to S: P (Z = 15) has a ground state configuration [Ne]3s23p3, while that of S (Z = 16) is [Ne]3s23p4. The values of IE1 are 1012 and 999.6 kJ mol–1, and the decrease is explained as for that from N to O in part (d) above.

See Box 1.6 in H&S

1.35

(a) See Fig. 1.8. (b) Increase from H to He: removing an electron from a singly occupied 1s orbital in H, and from a fully occupied (spin-paired electrons) orbital in He. Decrease from He to Li: ground state electronic configuration of Li is 1s22s1, so the first electron to be ionized is from a half-occupied 2s orbital; much less energy is needed than for the removal of an electron from He (1s2). Increase from Li to Be: in Be, the first electron to be removed comes from a filled 2s orbital (spin-paired electrons) compared to the removal of the single electron from the same orbital in Li. Decrease from Be to B: Ground state electronic configuration of B is 1s22s22p1, so the first electron to be removed is from a singly occupied 2p orbital. This requires less energy than removal of the electron from the 1s22s2 configuration of Be. Trend from B to Ne: see answer 1.34d.

1.36

(a) For the reaction: O(g) + 2e– O2–(g) the energy change corresponds to the sum of the first two electron affinities. The use of energy terms labelled ‘electron affinities’ in tables of data requires particular care. Usually, enthalpy changes for the addition of an electron (ΔEAH) are required (e.g. in a thermochemical cycle) rather than the actual electron affinity (EA). These quantities are related by: ΔEAH(298 K) ≅ ΔEAU(0 K) = –EA Let the enthalpy change for the above reaction be ΔrH: ΔrH = ΔEAH1 + ΔEAH2 where ΔEAH1 and ΔEAH2 are the enthalpy changes for the processes: O(g) + e–

O–(g)

and

IE1 / kJ mol–1

O2–(g)

ΔrH = –141 + 798 = +657 kJ mol–1

From Table 1.5 in H&S:

Fig. 1.8 For answer 1.35.

O–(g) + e–

2500

He Ne

2000 F N

1500 1000

H

2

O

B

Li

500 1

C

Be

3

4

5

6

7

8 9 10 Atomic number, Z

Basic concepts: atoms

11

(b) The formation of O2–(g) from O(g) is highly endothermic, the contributing factor being the endothermic attachment of an electron to O–(g). This contrasts with exothermic attachment of an electron to neutral O(g). The difference arises because of the repulsive forces felt by the second electron as it approaches a negatively charged ion. The fact that many metal oxides composed of ions are thermodynamically stable means that there must be sufficient energy liberated as the Mn+(g) and O2–(g) ions come together (attractive forces) to more than offset the highly endothermic formation of O2–(g). The full thermochemical cycle (see Chapter 6) that must be considered is shown below for CaO: Ca(s) + 1/2O2(g)

See the discussion of lattice energies in Chapter 6.

ΔaHo(Ca, s) + ΔaHo(O, g)

ΔfHo(CaO, s)

CaO(s)

Ca(g) + O(g)

IE1 + IE2(Ca, g)

ΔlatticeHo(CaO, s) ∼ ΔU(0 Κ)

ΔEA(1)H(O, g) + ΔEA(2)H(O, g)

Ca2+(g) + O2–(g)

where: ΔaHo = enthalpy of atomization ΔlatticeHo = change in enthalpy associated with the formation of an ionic lattice ∼ lattice energy

In this thermochemical (Born-Haber) cycle, the atomization of Ca (from solid Ca to gaseous atoms) is endothermic, the atomization of O2 to gaseous O atoms is endothermic, the formation of Mn+(g) from M(g) is endothermic and the formation of O2– from O(g) is endothermic. Both the formation of CaO(s) from its elements in their standard states, and the formation of the ionic lattice from gaseous ions are exothermic. 1.37

The position of an element in the periodic table follows from the number of electrons. The mass number of an isotope is determined by the number of protons (equal to the number of electrons) and the number of neutrons. Each of Ar and K has three naturally occurring isotopes: 36Ar, 38Ar, 40Ar and 39K, 40K, 41K. The relative atomic mass depends on the percentage abundances of the isotopes. For K, the most abundant isotope (93.3 %) is the lightest (39K), while for Ar, the most abundant (99.6 %) is the heaviest (40Ar).

1.38

For evidence that the aufbau principle is only approximately true, inspect the electronic configurations in Table 1.3 in H&S; see the discussion accompanying this table.

1.39

List 1 S6 and S8 19F and 31P Isotope of hydrogen 12C and 13C Hydrogen ion Group 1 elements Same energy Negatively charged particle Spin-paired electrons Electron, proton, neutron Group 15 elements Cr, Mn, Fe

Correct match in list 2 Allotropes Monotopic elements Protium Isotopes of en element Proton Alkali metals Degenerate Electron ms = ±1/2 Fundamental particles Pnictogens d-Block elements

Comments See Box 1.1 in H&S See Appendix 5 in H&S See Table 10.1 in H&S

See Table 1.1 in H&S

Basic concepts: atoms 1.40

(a) Each noble gas has a complete shell of electrons, ns2np6. The first ionization energy corresponds to the removal of an electron from this filled level and is energetically unfavourable. See also answers 1.34 and 1.35 on p.9-10. (b) See answer 1.36, parts a and b. (c) In the ground state Li atom (Z = 3), the 1s orbital is fully occupied. The third electron could occupy either a 2s or 2p orbital. Think about which occupancy will give the lower energy configuration. An electron in a 2s or 2p orbital experiences the effective charge, Zeff, of a nucleus which is partly shielded by the 1s electrons. The 2p orbital penetrates the 1s orbital less than a 2s orbital does, and therefore a 2p electron is shielded from the nuclear charge more than a 2s electron. Thus, the electron occupies the 2s orbital in preference to the 2p since the former gives a lower energy system.

1.41

Be Compare Figure 1.9 with Fig. 1.16 in H&S; the latter shows the trend in values of 12000 IE1. The general appearance of the two graphs from Z = Mg 3 to 36 is similar, but is 8000 shifted by two elements, e.g. Ca maximum IE1 values correZn Kr spond to the noble gases, but 4000 B maximum IE3 values correAl Sc spond to group 2 metals. 0 Compare, for example, He 3 9 15 21 27 33 Z and Be. IE1 for He corresponds to the removal of one Fig. 1.9 Trend in values of IE3 for elements Li to Kr. electron from a 1s2 configuration. IE 3 for Be also corresponds to the removal of one electron from a 1s2 configuration, because the ground state configuration for a Be atom is 1s2 2s2.

1.42

Electron affinities are most often used in thermochemical cycles, e.g. Born-Haber cycle, and therefore it is the enthalpy change associated with the attachment of an electron to an atom or ion that is needed.

1.43

(a) The angular part of the wavefunction for the 1s and 2s orbitals is the same and does not depend on angles φ and θ :

See Table 1.2 in H&S

Third ionization energy / kJ mol–1

12

A(θ , φ ) =

16000

1 2 π

The orbital remains spherically symmetric. The angular parts of the wavefunction for the 2px and 3px orbitals are also the same, as are the angular parts of the wavefunction for the 2py and 3py orbitals, and for the 2pz and 3pz orbitals. The boundary surfaces for the angular parts of the wavefunctions of each pair of orbitals are therefore the same. (b) The function R(r) is the radial part of the wavefunction. The probability density is described by a plot of 4πr2R(r)2 (the radial distribution function) against r. For the 1s orbital of hydrogen, R(r) = 2e–r. When r = 0, R(r) = 2e0 = 2, i.e. R(r) has a finite value at the nucleus. The radial distribution function depends on r2, and is therefore zero when r = 0.

Basic concepts: atoms 1.44

13

The first plot below shows the abundances of the elements, and the second plot shows the same data on a log scale. 160000

Abundance

log Abundance

6

0

0 Ru

Os

Rh

Ir

Pd

Pt

Ru

Os

Rh

Ir

Pd

Pt

The wide range of abundances (from 11 for Os to 150000 for Pt) is clear in the direct plot on the left. The larger the range of values covered in a plot of abundance, the more useful the log plot becomes. A log plot gives information on the relative abundances and generates a more convenient scale. The right-hand plot allows you to see that the orders of magnitude of the abundances of Pd and Pt are similar, whereas the order of magnitude of Os is ≈ 4 times lower. 1.45

(a) Possible reasons why no other lines are visible in Fig. 1.18 in H&S: • no other lines fall in the visible region of the electromagnetic spectrum; • some emission lines are too faint to observe (intensity depends on probability of transition occurring); • only certain excited states are populated. (b) λ = 589 nm = 589 × 10–9 m c = 2.998 × 108 m s–1 = λν

ν =

2 .998 × 10 8 m s –1 589 × 10 – 9 m

= 5.09 × 1014 s–1 (to 3 sig. fig.)

14

2 2.1

Basic concepts: molecules (a) F2

(b) BF3 F

F

F

B F

(c) NH3

(d) H2Se H

H

N

H H

(e) H2O2

H

Se

Cl

Be

(f) BeCl2 O

H

H

(h) PF5

Si

F

F

H

F F

Allocate electrons to form N–H or N–F bonds, and then see how many unpaired electrons remain on each N atom: (a) N2H4 (b) N2F4 H H

N

F

N

H

N

F

H

N

F

F

N–N single bond X Isomers of N2F2 are possible: see answers 2.19f and 2.19g

F

P

H

H

2.2

Cl

O

(g) SiH4

H

F

(c) N2F2

N–N single bond (d) [N2H5]+

+

H

F N

N

H

N

N

H

H

F

N=N double bond

H

N–N single bond

Basic concepts: molecules 2.3

128 pm 117˚

218 pm

Figure 2.1 shows the experimental structure of O3. Resonance structures show possible covalent bonding descriptions. Overall bonding picture is a combination of resonance structures, although not all may contribute equally. Resonance structures do not exist as separate species. In O3, each O atom has 6 valence electrons. Possible resonance structures (ignoring those with unreasonably high charge separation, e.g. O2– and O2+ centres) are:

Fig. 2.1 Molecular structure of O3.

+ O

O O

2.4

15

O

O

–

–

O

(a) CO2

+ O

O

O

O

O

O

(b) SO2 O

C

O

O

C

O

S

O

O

S O

O

(c) OF2

(d) H2CO

H

O F

F

C

O

C

O

H H

O F

F H

2.5

Draw a Lewis structure for each molecule: (a) (b) N

O

O

(c) F

O N

F

The Lewis structures show that NO and NF2 are radicals (one unpaired electron on N). For O2, the Lewis structure shows that all valence electrons are paired. In fact, O2 is a diradical, and this can be rationalized by using MO theory. 2.6

(a) First write down the ground state electronic configuration to determine the valence electrons available. Then draw out a set of resonance structures. Li2 Li: Z=3 1s22s1 1 valence electron available The bonding in Li2 is described in a similar way to that in H2; resonance structures are: Li

B2 B:

Li

Z=5

Li+

1s22s22p1

Li–

Li–

Li+

3 valence electrons available

16

Basic concepts: molecules The bonding in B2 is described in terms of the following resonance structures (remember that in Lewis structures, odd electrons are paired wherever possible): B

B

B

B+

B

B–

B–

B+

C2 C: Z=6 1s22s22p2 4 valence electrons available The bonding in C2 is described in terms of the following resonance structures: C

+

–

C

C

+

C

C

–

–

C

C

+

C

+

C

–

C

(b) In each of the resonance structures drawn in part (a), all electrons are paired and so each molecule is predicted to be diamagnetic. There is therefore agreement with experiment for Li2 and C2, but not for B2. 2.7

(a) H2 H: Z=1 H

Na

(2.1)

S

(2.2)

N

N

(2.3)

N

N

(2.4)

Cl

1 valence electron available H+

H

H–

H–

H+

The left-hand resonance structure is consistent with an H–H single bond; bond order = 1. An ambiguity is not knowing to what extent the ionic resonance hybrids contribute. In the remaining parts of the question, only covalent resonance structures are considered; additional contributions may be made by ionic forms. (b) Na2 Na: Z = 11 1s22s22p63s1 1 valence electron available The bonding picture involves resonance structure 2.1 and this is consistent with a bond order of 1. (c) S2 S: Z = 16 1s22s22p63s23p4 6 valence electrons available The bonding picture will be analogous to that in O2 (see the end of Section 2.2 in H&S); it involves resonance structure 2.2, and is consistent with a bond order of 2. (d) N2 N: Z=7 1s22s22p3 5 valence electrons available Pairing of electrons, and obeying the octet rule, leads to 2.3 as the major contributing resonance structure. This is consistent with a bond order of 3. This conclusion assumes that resonance structure 2.4 (in which each N only has a sextet of valence electrons) does not contribute significantly. (e) Cl2 Cl: Z = 17 1s22s22p63s23p5 7 valence electrons available The bonding is described in terms of resonance structure 2.5, consistent with a bond order of 1.

Na

S

1s1

Cl

(2.5)

2.8

Ground state electronic configuration of He (Z = 2) is 1s2. Within VB theory, the resonance structures that could be drawn (remembering that electrons are paired so far as possible) are: He

He

He2+ He2–

He2– He2+

The double bond formation is not possible with only the 1s orbital per He atom, and the ionic form is unreasonable (look at ionization energies for He). It is

Basic concepts: molecules

17

concluded that He2 is not a viable species. (The real question is ‘What is the stability of He2 with respect to 2He?’ and VB theory does not give an answer to this.) (a) and (b) First, determine the available valence orbitals for each atom. He: σu*(1s) Z=2 The valence orbital is the 1s orbital. For the combination of two 1s atomic orbitals, the MO diagram is as shown on the right. Each He atom provides 2 electrons, so He2 1s 1s possesses 4 electrons which fully occupy both the σg(1s) and σu*(1s) MOs. In [He2]+, there are 3 electrons; these σg(1s) occupy the MOs shown in the diagram 2 1 2+ to give an electronic configuration σg(1s) σu*(1s) . In [He2] , there are 2 electrons and they occupy the MOs to give an electronic configuration of σg(1s)2. The bond order in [He2]+ is 1/2(2–1) = 1/2. The bond order in [He2]2+ is 1/2(2–0) = 1. Both [He2]+ and [He2]2+ are therefore viable species. Note that [He2]2+ is isoelectronic with H2.

2.10

(a) O: Z = 8 Ground state electronic configuration = 1s22s22p4. The 1s electrons are core electrons; the 2s and 2p orbitals and electrons are in the valence shell. Define the O–O internuclear axis to coincide with the z axis (diagram 2.6). For O2, the MO diagram is constructed by allowing: • 2s orbital of each O atom to interact together to give a σ-bonding combination and antibonding counterpart; • 2pz orbitals on the O atoms to interact to give σ-bonding MO and σ* MO; • 2px orbitals on the O atoms to interact to give π-bonding MO and π* MO; • 2py orbitals on the O atoms to interact to give π-bonding MO and π* MO. The π-bonding MOs are degenerate and so are the two π* MOs:

Energy

2.9

O

O

z

σu*(2pz)

Energy

(2.6)

πg*(2px) πg*(2py)

2px

2py

2pz

πu(2px) πu(2py)

2pz

2py

2px

σg(2pz) σu*(2s) 2s

–e–

O2 ⎯⎯→ [O2]+

⎯⎯→

+e–

2s

σg(2s) .

[O2]–

⎯⎯→

+e–

[O2]2–

(2.7)

Now put in the electrons. Each O atom has 6 valence electrons, so there are 12 electrons in O2 and these occupy the lowest lying MOs to give an electronic configuration of σg(2s)2σu*(2s)2σg(2pz)2πu(2px)2πu(2py)2πg*(2px)1πg*(2py)1. (b) Trend in O–O bond distances: O2, 121 pm; [O2]+, 112 pm; [O2]–, 134 pm; [O2]2–, 149 pm, i.e. [O2]2– > [O2]– > O2 > [O2]+. Using scheme 2.7, and assuming that the same qualitative MO diagram (above) is appropriate for each species, the electronic configurations of each species are:

18

Basic concepts: molecules [O2]2– [O2]– O2 [O2]+

σg(2s)2σu*(2s)2σg(2pz)2πu(2px)2πu(2py)2πg*(2px)2πg*(2py)2 σg(2s)2σu*(2s)2σg(2pz)2πu(2px)2πu(2py)2πg*(2px)2πg*(2py)1 σg(2s)2σu*(2s)2σg(2pz)2πu(2px)2πu(2py)2πg*(2px)1πg*(2py)1 σg(2s)2σu*(2s)2σg(2pz)2πu(2px)2πu(2py)2πg*(2px)1πg*(2py)0

The bond orders are therefore: 1/ (8 – 5) = 1.5 [O2]2– 1/2(8 – 6) = 1 [O2]– 2 + 1/ (8 – 3) = 2.5 1 O2 /2(8 – 4) = 2 [O2] 2 These bond orders are consistent with the sequence of bond lengths above, with the lowest bond order corresponding to the longest bond. Such a correlation can only be made along a series of closely related species. (c) A paramagnetic species contains one or more unpaired electrons: [O2]–, O2 and [O2]+ are paramagnetic, but [O2]2– is diamagnetic. 2.11

(a) CF4:

C (group 14) has 4 valence electrons F (group 17) has 7 valence electrons

F F

In CF4, C forms 4 single bonds and each atom has an octet of valence electrons. (b) O2:

C F

O (group 16) has 6 valence electrons

In O2, each O forms a double bond and each atom has an octet of valence electrons. (c) AsBr3:

O

As (group 15) has 5 valence electrons Br (group 17) has 7 valence electrons

Br

(c) SF2:

O

As

In AsBr3, As forms 3 single bonds and each atom has an octet of valence electrons.

Br

Br

S (group 16) has 6 valence electrons F (group 17) has 7 valence electrons

F

S

In SF2, S forms 2 single bonds and each atom has an octet of valence electrons. 2.12

F

F

P (group 15) has 5 valence electrons, and F (group 17) has 7 valence electrons. If all 5 valence electrons of the P atom are used to form P–F single bonds, the P atom has 10 electrons in its valence shell. A charge separated species with a P+ centre can be drawn so that each atom obeys the octet rule: F A set of 5 resonance structures must be drawn so as to make the 3 equatorial F P+ F bonds equivalent and the 2 axial bonds F– equivalent: F

F F

P+

F –

F

–

F

P+

F F

F

F–

P+

F F

F–

F F

F

P+

F F

F F

F

P+

F F

F F

–

F

Basic concepts: molecules 2.13

13

Group 15 16

14

Al

17

C

N

O

F

Si

P

S

Cl Br

δ– N

δ+ H

δ–

F

(2.8) –

δ C

+

δ H

δ P

(2.10)

Refer to the portion of the periodic table drawn in the margin. This shows the elements needed for the question. Remember that moving one place to the right in the table adds a valence electron, and on moving one place to the left, there is one less electron. (a) [NO2]+, CO2 and [N3]– are isoelectronic with each other. Each has 16 valence electrons. [NO2]– has 18 valence electrons. (b) [CN]–, N2, CO and [NO]+ are isoelectronic. Each has 10 valence electrons. [O2]2– has 14 valence electrons. (c) [SiF6]2–, [PF6]– and [AlF6]3– are isoelectronic. Each has 48 valence electrons. [BrF6]– has 50 valence electrons.

2.14

If two species are isoelectronic, they possess the same number of electrons. However, the term is often used to describe species which have the same number of valence electrons. F2 is isoelectronic with [O2]2–; it is also isoelectronic with Cl2 in terms of valence electrons. NH3 is isoelectronic with [H3O]+. [GaBr4]– is isoelectronic both with SiBr4 and [GaCl4]– in terms of valence electrons. [SH]– is isoelectronic with [OH]– in terms of valence electrons. [BH4]– is isoelectronic with [NH4]+. [AsF6]– is isoelectronic with SeF6. [PBr4]+ is isoelectronic with SiBr4; it is has the same number of valence electrons as [GaCl4]–. HF is isoelectronic with [OH]–.

2.15

When drawing a diagram to show a dipole moment, remember that by SI convention, the direction of the dipole moment is indicated by an arrow that points from the δ – to δ + end of the molecule. (a) N–H χP(N) = 3.0 χP(H) = 2.2 Polar: Nδ––Hδ+ See 2.8.

δ+

Br

(2.9) +

19

–

δ Cl

(2.11)

2.16

Group 13

14

15

16

17

B

C

N

O

F

Al

Si

P

S

Cl

(b) F–Br

χP(F) = 4.0

χP(Br) = 3.0 Polar: Fδ––Brδ+

See 2.9.

(c) C–H

χP(C) = 2.6

χP(H) = 2.2

See 2.10.

(d) P–Cl

χP(P) = 2.2

χP(Cl) = 3.2 Polar: Pδ+–Clδ–

(e) N–Br

χP(N) = 3.0

χP(Br) = 3.0 Non-polar

Slightly polar: Cδ––Hδ+

See 2.11.

Two species are isoelectronic if they possess the same total number of electrons.‘Isoelectronic’ is also used to mean ‘same number of valence electrons’, although strictly this usage should always be qualified (see also problem 2.14). In this answer, assume that ‘isoelectronic’ is strictly applied and that you are looking for pairs of species with the same total number of electrons. Rather than count the number of electrons, use the periodic table (see margin) to help you: moving one place to the right in the table means one more electron (e.g. going from O to F), and moving one place to the left means one less electron (e.g. going from S to P). A negative charge adds an electron; a positive charge takes one away, e.g. N+ is isoelectronic with C. The following species form isoelectronic pairs: HF and [OH]– NH3 and [H3O]+ CO2 and [NO2]+ SiCl4 and [AlCl4]– The remaining molecules do not have isoelectronic partners in the list.

20

Basic concepts: molecules 2.17 X Use Table 2.3 (‘Parent’ shapes) in H&S

Se

(a) H2Se Central atom is Se Se is in group 16, so number of valence electrons = 6 Number of bonding pairs (2 Se–H bonds) = 2 Number of lone pairs = 2 Total number of electron pairs = 4 = 2 bonding and 2 lone pairs ‘Parent’ shape = tetrahedral, see structure 2.12. Molecular shape = bent.

H H

(2.12)

–

H B H

H H

(2.13)

N F

F F

(2.14)

F F

Sb

F F

F

(2.15)

The VSEPR model is summarized in the following ‘rules’. • Each valence-shell electron pair of central atom E in molecule EXn with E–X single bonds is stereochemically significant; electron pair-electron pair repulsions determine the shape of EXn. • Electron-electron repulsions decrease in the order: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. • Where E–X bonds are of different bond orders, electron-electron repulsions decrease in the order: triple bond-single bond > double bond-single bond > single bond-single bond. • Electron-electron repulsions between bonding pairs in EXn depend on the difference between electronegativities of E and X; repulsions are less the more the E–X bonding electron density is drawn towards X.

(b) [BH4]– Central atom is B B is in group 13, so number of valence electrons = 3 Add one extra electron for the negative charge Number of bonding pairs (4 B–H bonds) = 4 No lone pairs Total number of electron pairs = 4 = 4 bonding pairs ‘Parent’ shape = tetrahedral, see structure 2.13. Molecular shape = tetrahedral (c) NF3 Central atom is N N is in group 15, so number of valence electrons = 5 Number of bonding pairs (3 N–F bonds) = 3 Number of lone pairs = 1 Total number of electron pairs = 4 = 1 lone and 3 bonding pairs ‘Parent’ shape = tetrahedral, see structure 2.14. Molecular shape = trigonal pyramidal (d) SbF5 Central atom is Sb Sb is in group 15, so number of valence electrons = 5 Number of bonding pairs (5 Sb–F bonds) = 5 Number of lone pairs = 0 Total number of electron pairs = 5 = 5 bonding pairs ‘Parent’ shape = trigonal bipyramidal, see structure 2.15. Molecular shape = trigonal bipyramidal

Basic concepts: molecules (e) [H3O]+ Central atom is O O is in group 16, so number of valence electrons = 6 Subtract one electron to allow for the positive charge Number of bonding pairs (3 O–H bonds) = 3 Number of lone pairs = 1 Total number of electron pairs = 4 = 1 lone and 3 bonding pairs ‘Parent’ shape = tetrahedral, see structure 2.16. Molecular shape = trigonal pyramidal

+ O H

H H

(2.16)

(f) IF7 F F F

F

I

F

F F

(2.17)

I I

(2.18)

(h) [I3]+ Central atom is I I is in group 17, so number of valence electrons = 7 Subtract one extra electron to allow for the positive charge Number of bonding pairs (2 I–I bonds) = 2 Number of lone pairs = 2 Total number of electron pairs = 4 = 2 bonding and 2 lone pairs ‘Parent’ shape = tetrahedral, see structure 2.19. Molecular shape = bent

+ I

I I

(2.19)

(i) SO3 O S O

Central atom is I I is in group 17, so number of valence electrons = 7 Number of bonding pairs (7 I–F bonds) = 7 Number of lone pairs = 0 Total number of electron pairs = 7 = 7 bonding pairs ‘Parent’ shape = pentagonal bipyramidal, see structure 2.17. Molecular shape = pentagonal bipyramidal

(g) [I3]– Central atom is I I is in group 17, so number of valence electrons = 7 Add one extra electron for the negative charge Number of bonding pairs (2 I–I bonds) = 2 Number of lone pairs = 3 Total number of electron pairs = 5 = 2 bonding and 3 lone pairs ‘Parent’ shape = trigonal bipyramidal, see structure 2.18. Molecular shape = linear

–

I

21

O

(2.20)

2.18

Central atom is S S is in group 16, so number of valence electrons = 6 Each S–O bond is a double bond; number of S=O bonds = 3 For interelectronic repulsions, count each bond as 1 electron ‘pair’ Number of lone pairs = 0 Total number of electron ‘pairs’ = 3 = 3 bonding ‘pairs’ ‘Parent’ shape = molecular shape = trigonal planar, see structure 2.20.

Figure 2.2 shows the structure of SOF4. It is trigonal bipyramidal, and since S has 6 valence electrons, there are no stereochemically active lone pairs. Molecular and ‘parent’ shape = trigonal bipyramidal

Basic concepts: molecules

22

Valence electrons from S = 6, from O = 2, from each F = 1 Total valence electrons = 12 = 6 pairs available for bonding Total number of electron ‘pairs’ from Fig. 2.2 = 5 bonding ‘pairs’ Number of lone pairs = 0 One electron ‘pair’ in the VSEPR model arises from a formal double bond. The structure can be rationalized if one considers the bond orders (see Fig. 2.2). Since double bond–single bond repulsions > single bond–single bond repulsions, electronic repulsions are minimized if the O atom lies in the equatorial plane. Bond orders: (a) S–F single bonds; bond order 1; (b) S=O double bond; bond order 2.

F O

F

S

F F

Fig. 2.2 For answer 2.18.

2.19 X By SI convention, the arrow representing a dipole moment points from δ – to δ +

S

H H

(2.21)

O

C

O

O

C

(2.23)

(2.22)

S O

O

(2.24)

S

O

The shapes are found by using the VSEPR model – method as in answer 2.17. (a) H2S This is like H2Se (see answer 2.17a). First, consider each S–H bond: S–H χP(S) = 2.6 χP(H) = 2.2 Polar bond: Sδ––Hδ+ Now consider the molecule as a whole, taking into account lone pairs. The bond dipole moments reinforce each other, and the molecular dipole moment is enhanced by the 2 lone pairs. The direction of the resultant dipole moment is shown in 2.21. (b) CO2 The VSEPR model is consistent with a linear molecule (2 bonding ‘pairs’; C=O double bonds and no lone pairs), see structure 2.22. For the bond dipole moments: C–O χP(C) = 2.6 χP(O) = 3.4 Polar bond: Oδ––Cδ+ However, because the molecule is linear and symmetrical, the two bond dipole moments cancel one another (2.23), and the molecule is non-polar. (c) SO2 The VSEPR model is consistent with a bent molecule (2 bonding ‘pairs’: S=O double bonds and 1 lone pair), see structure 2.24. In order to work out if the molecule is polar, look first at each bond dipole moment: S–O χP(S) = 2.6 χP(O) = 3.4 Polar bond: Sδ+–Oδ– The bond vectors reinforce each other (2.24), but the resultant of these bond moments opposes effect of the lone pair. It is qualitatively difficult to assess direction of net molecular dipole moment. Experimental value for dipole moment in gas phase SO2 is 1.63 D; diagram 2.25 shows direction in which dipole moment acts. (d) BF3 The VSEPR model is consistent with trigonal planar BF3 (3 bonding and no lone pairs), see structure 2.26. Now consider each bond dipole moment: B–F χP(B) = 2.0 χP(F) = 4.0 Polar bond: Bδ+–Fδ– Each bond is polar, but the molecule is non-polar because the 3 bond dipole moments (which are vectors) cancel out. Show this by resolving the vectors into two opposing directions. Let each vector be V. The vectors act as shown below:

O

O

V

(2.25) V

F B F

F

(2.26)

60o 60o

V

Resolving vectors into opposing upward and downward directions: In an upward direction: total vector = V In a downward direction: total vector = 2 × V cos60 =2×V/2 =V

Therefore, the two equal and opposite vectors cancel out.

Basic concepts: molecules

F F

F

P

F F

(2.27) V1

N

N

N

F

F

N

F

F V2

(2.28)

(2.29)

F N

N

F

(2.30)

H

C

N

(2.31)

2.20 F

Cl

B

B

Cl

F

(i)

F

F

(ii)

(2.32)

F F

F F

P

F

P

F

Me

Me

F

(2.33) –

F Cl

F

P

Cl

F Cl

Cl

(2.34)

23

(e) PF5 The VSEPR model is consistent with a trigonal bipyramidal molecule (2.27). Determine whether each bond is polar by considering the electronegativity values: P–F χP(P) = 2.2 χP(F) = 4.0 Polar bond: Pδ+–Fδ– Each bond is polar, but the molecule is non-polar because: • the 3 bond dipole moments in the equatorial plane cancel out, and the reasoning for this is the same as for trigonal planar BF3 in part (d); • the 2 axial bond dipole moments are of equal magnitude but act in opposite directions, therefore the net dipole moment in the axial direction is zero. (f) cis-N2F2 Answer 2.2c showed a Lewis structure for N2F2 although we ignored the possibility of isomers. Starting from this Lewis structure, apply the VSEPR model to show that cis-N2F2 contains two trigonal planar N centres. The molecule is planar (2.28) because the N=N double bond constrains structure. Now consider each N–F bond: χP(N) = 3.0 χP(F) = 4.0 Polar bond: Nδ+–Fδ– Each bond is polar, but resultant moment of the 2 bonds opposes resultant moment due to the two lone pairs (2.29). It is impossible at a qualitative level to decide whether the vector V2 > V1, or V2 < V1. The experimental value of the molecular dipole moment is 0.16 D, and theoretical calculations indicate that V2 < V1. (g) trans-N2F2 As in part (f), start from the Lewis structure in answer 2.2c; use the VSEPR model to show that trans-N2F2 contains two trigonal planar N centres. The molecule is planar (2.30). Although each N–F bond is polar (as above), bond dipole moments cancel out, and dipole moments due to lone pairs cancel out. Therefore, trans-N2F2 is non-polar. (h) HCN Start from a Lewis structure and apply the VSEPR model. HCN is linear (2.31). Now consider electronegativities of the atoms: χP(H) = 2.2 χP(C) = 2.6 χP(N) = 3.0 + – δ δ Each bond is polar: H –C Cδ+–Nδ– The 2 bond dipole moments reinforce each other, and resultant molecular dipole moment acts in the direction shown in diagram 2.31. (a) BF2Cl The VSEPR model is consistent with a trigonal planar molecule (3 bonding pairs, no lone pairs). No stereoisomers are possible; any view that you draw can be converted into any other by a simple rotation, e.g. (i) to (ii) in structure 2.32. (b) POCl3 Starting from a Lewis structure, followed by application of the VSEPR model gives a tetrahedral structure. No isomers are possible. (c) MePF4 The structure of MePF4 is related to PF5 (trigonal bipyramidal, see answer 2.19e) but with one F atom replaced by a methyl group. The trigonal bipyramid contains 2 distinct sites: axial and equatorial. Two isomers are possible (2.33). (d) [PF2Cl4]– – F Start with a Lewis structure and apply the VSEPR model: Cl Cl [PF2Cl4]– is octahedral. Two F atoms can be mutually P adjacent or opposite giving cis- and trans-isomers Cl Cl respectively (2.34 and 2.35). F (2.35)

Basic concepts: molecules

24

2.21

(a) C

PPh3 Cl

Pt

C

O

O

(b) The remaining six MOs are represented schematically as follows: Cl

PPh3

C

(2.36)

O

C

1σ

O

C

2σ

C

O

π(2p)

O

π(2p)

Cl Cl

Pt

PPh3

C

PPh3

C

O

4σ

(2.37)

2.22

π*(2p)

(a) The structures of trans and cis-[PtCl2(PPh3)2] are shown in diagrams 2.36 and 2.37, respectively. On steric grounds, the trans-isomer should be favoured because the bulky PPh3 groups are further apart than in the cis isomer. (b) NSF3: S has 6 valence electrons. After the formation of the SN triple bond, S has 3 valence electrons which are involved in the formation of 3 S–F bonds. The VSEPR model is therefore consistent with a tetrahedal structure. SF4: S has 6 valence electrons. After the formation of 4 S–F bonds, a lone pair remains on the S atom. This lone pair is stereochemically active. Exercise: How are the structures below adapted to accommodate the octet rule? N

S

F F

F

2.23

(a) IF5 has a square-based pyramidal structure (2.39). There will be a net dipole moment acting along the I–F(axial) bond. Qualitatively, it is difficult to assess its direction. (b) Both Li and K are group 1 metals and for each, the first ionization involves the loss of the ns1 electron. For Li, n = 2, and for K, n = 4. The 4s electron in K is better shielded from the nuclear charge than the 2s electron in Li, and therefore the first ionization energy of K is lower than that of Li. (c) B is in group 13 and has 3 valence electrons. Once 3 B–I bonds are formed, all valence electrons have been used for bonding. P is in group 15 and has 5 valence electrons. The formation of 3 P–I bonds leaves a lone pair on the P atom. Using the VSEPR model, a trigonal planar geometry is expected for BI3, while the structure of PI3 is trigonal pyramidal, i.e. it is based on a tetrahedral arrangement of 1 lone and 3 bonding pairs of electrons.

2.24

(a) He has the electronic configuration 1s2. Removal of the first electron generates He+. Removal of the remaining 1s electron from this positively charged ion requires more energy than removal of the initial electron.

F

F

F

(c) Kr has 8 valence electrons and after the formation of 2 Kr–F bonds, 3 lone pairs remain. These preferentially occupy equatorial sites in a trigonal bipyramidal arrangement (structure 2.38).

(2.38)

I

S F

F

F

F

F

F Kr

O

F F

(2.39)

Basic concepts: molecules

25

(b) N2F2 has two isomers (look back at structures 2.28-2.29 on p. 23). The cisisomer is polar, whereas the trans-isomer is non-polar. Heating N2F2 at 373 K converts the trans-isomer to the cis-form. (c) S2 is (in terms of valence electrons) isoelectronic with O2. Figure 2.10 in H&S shows an energy level diagram for O2 and illustrates that the HOMO is degenerate and each orbital is singly occupied. Similarly, the HOMO of S2 is degenerate, i.e. πg*(3px)1π g*(3py)1. The unpaired electrons mean that S2 is paramagnetic. 2.25

X For details of the structure, see Fig. 17.5 in H&S

X Compare with the energy level diagram for F2 in Fig. 2.10 in H&S

2.26

(a) Bromine has two isotopes, 79Br and 81Br, each in approximately 50% abundance. Naturally occurring Br2 therefore exists as (79Br)2, (81Br)2 and (79Br81Br). In the mass spectrum of Br2, the envelope of peaks for the parent ion Br2+ consists of peaks corresponding to (79Br)2+, (81Br)2+ and (79Br81Br)+ at values of m/z = 158, 162 and 160 respectively. (b) Within a diatomic Br2 molecule, the intramolecular Br–Br distance in the solid state is 227 pm. The molecules pack in an ordered array and the distance of 331 pm corresponds to the shortest intermolecular separation. (c) Br2 has a single bond. This can be deduced from VB or MO theory. The ground state electronic configuration (valence electrons only) for Br2 is:

σg(4s)2σu*(4s)2σg(4pz)2πu(4px)2π u(4py)2πg*(4px)2π g*(4py)2 The bond order is therefore 1. On going from Br2 to Br2+, an electron is removed from an antibonding orbital, and the bond order increases. This is consistent with a shortening of the Br–Br bond. (a) Stereoisomers of [SiF3Me2]– : –

F F

Si

Me

F

Si

Me F

–

Me Me

F

Si

F F

–

Me F F Me

The isomer with the Me groups in the equatorial positions (2 F atoms axial) is preferred. (This preference is also seen in the isoelectronic PF3Me2.) Two approaches to rationalizing this: • on steric grounds, the smallest ligands tend to occupy the most crowded sites in the trigonal bipyramid, i.e. the axial sites with 90o angles to neighbouring ligands, as opposed to equatorial sites with 120o angles; • within the VSEPR model, the most electronegative ligands (i.e. those with the smallest bonding electron domains) occupy the more crowded axial sites. Both points are consistent with the observation that in [SiF3Me2]–, the Me groups occupy equatorial sites. (b) The complexes all contain Pt(II) and are square planar (see Chapter 20 in H&S for an explanation). For each of [PtCl4]2–, [PtCl3(PMe3)]– and [PtCl(PMe3)3]+, only one arrangement of ligands is possible, but for [PtCl2(PMe3)2], the ligands may be in cis- or trans-arrangements.

Basic concepts: molecules

26

2–

Cl Cl

Pt

–

Cl

Cl

Cl

Pt

Cl

Cl

Pt

Cl

Cl

Cl

P

Cl

Cl

(2.40)

N Cl

(2.41)

Cl Cl

PMe3

cis

(a) P is in group 15, and has 5 valence electrons. Therefore the ions present in [PCl4][PCl3F3] are [PCl4]+ and [PCl3F3]–. The shapes are determined by using the VSEPR model: [PCl4]+ Take the centre to be P+, with 4 valence electrons Number of bonding pairs (4 P–Cl bonds) = 4 Number of lone pairs = 0 Total number of electron pairs = 4 = 4 bonding pairs Molecular shape = tetrahedral [PCl3F3]– Take the centre to be P–, with 6 valence electrons Number of bonding pairs (6 P–X bonds) = 6 Number of lone pairs = 0 Total number of electron pairs = 6 = 6 bonding pairs Molecular shape = octahedral No stereoisomers are possible for [PCl4]+. For [PCl3F3]–, the Cl (or F) atoms may be in a mer- or fac-arrangement: –

F

Cl

Pt PMe3

trans

B

PMe3

Cl

PMe3

Cl

Pt PMe3

PMe3

2.27

Cl

PMe3

Cl

+

PMe3

–

Cl

F

Cl

Cl

Cl

P

F F

F

F

mer

fac

(b) BCl3 Central atom is B with 3 valence electrons (group 13) Number of bonding pairs (3 B–Cl bonds) = 3 Number of lone pairs = 0 Total number of electron pairs = 3 bonding pairs Molecular shape = trigonal planar (structure 2.40). NCl3 Central atom is N with 5 valence electrons (group 15) Number of bonding pairs (3 N–Cl bonds) = 3 Number of lone pairs = 1 Total number of electron pairs = 4 Molecular shape = trigonal pyramidal (structure 2.41).

Basic concepts: molecules

27

BCl3 is non-polar, whereas NCl3 is polar. In BCl3, the bond dipole moments cancel out (see answer 2.19d, page 22). In NCl3, the resultant of the bond dipole moments acts in a direction opposing the effect of the lone pair. The molecule is polar (dipole moment = 0.39 D) and the direction in which the dipole moment acts is shown in 2.42. Remember that by SI convention, the arrow representing a dipole moment points from δ – to δ +. N Cl

Cl Cl

(2.42)

2.28 2–

F F

F

Si

F

F F

(2.43)

F

(b) XeF4 Xe has 8 valence electrons (group 18) Number of bonding pairs (4 Xe–F bonds) = 4 Number of lone pairs = 2 Total number of electron pairs = 6 ‘Parent’ shape = octahedral Molecular shape = square planar (structure 2.44) All F atoms are equivalent.

F

Xe

F

F

(2.44)

+

F N F

F F

(2.45)

a

H F

F a

P Fb

(2.46)

a F F a

(a) [SiF6]2– Take the centre to be Si2–, with 6 valence electrons (Si is in group 14) Number of bonding pairs (6 Si–F bonds) = 6 Number of lone pairs = 0 Total number of electron pairs = 6 Molecular shape = octahedral All F atoms are equivalent (structure 2.43).

–

(c) [NF4]+ Take the central atom as N+, with 4 valence electrons (N is in group 15) Number of bonding pairs (4 N–F bonds) = 4 Number of lone pairs = 0 Total number of electron pairs = 4 Molecular shape = tetrahedral All F atoms are equivalent (structure 2.45). (d) [PHF5]– Take the central atom to be P–, with 6 valence electrons (P is in group 15) Number of bonding pairs (1 P–H and 5 P–F bonds) = 6 Number of lone pairs = 0 Total number of electron pairs = 6 Molecular shape = octahedral There are 2 F environments (structure 2.46), with the F atoms either trans or cis with respect to the H atom. (e) [SbF5]2– Take the centre to be Sb2–, with 7 valence electrons (Sb is in group 15) Number of bonding pairs (5 Sb–F bonds) = 5 Number of lone pairs = 1

Basic concepts: molecules

a

Fb F

Sb

F a

a F

2–

F a

Total number of electron pairs = 6 ‘Parent’ shape = octahedral Molecular shape = square-based pyramid There are 2 F environments (structure 2.47), with the F atoms in either basal (cis to the lone pair) or axial (trans to the lone pair) sites.

(2.47)

2.29

You can draw a Lewis structure for O2 so that each O atom obeys the octet rule. The ground state configuration of each oxygen atom is [He]2s 22p4 and there are two unpaired electrons per atom. In a Lewis structure of O 2, the electrons are paired up to give an O=O double bond:

O

O

or

O

O

Within valence bond theory, you can write resonance structures for O2 showing covalent and ionic contributions to the bonding, i.e. the bonding is described by a wavefunction, ψmolecule, with covalent and ionic contributions:

ψmolecule = N[ψ covalent + ψionic] It is expected that the covalent contribution will dominate, and therefore the bonding is realistically represented by a structure in which all electrons are paired and there is an O=O double bond. This predicts a diamagnetic molecule. However, experimental data show that O2 is a diradical with two unpaired electrons. Thus, valence bond theory is unable to account for the paramagnetism of O 2, although it is consistent with O2 possessing a short, relatively strong bond. The experimental bond length is 121 pm, and bond dissociation enthalpy is 498 kJ mol –1. Molecular orbital theory can be used to describe the bonding in terms of the MO diagram shown below. The construction of this diagram was described in answer 2.10 on p. 17. σu*(2p) πg*(2p)

Energy

28

πu(2p) 2px

2py

2pz

2pz

2py

σg(2p) σu*(2s) 2s

2s

σg(2s) O

O2

O

From the MO diagram, the following conclusions can be drawn: • O2 has a bond order of 2 (see p. 17 for working); • O2 has 2 unpaired electrons.

2px

Basic concepts: molecules

29

Thus, MO theory is more successful than VB theory in providing a model of the bonding in O2 that is consistent with the experimental facts. Historically, this result was one of the major achievements of early MO theory. 2.30

(a) Lewis structures for N2O: –

N

+ N

O

N

+ N

–

O

(b) SO2 Central atom is S S is in group 16, so number of valence electrons = 6 Each S–O bond is a double bond; number of S=O bonds = 2 Number of lone pairs = 1 Total number of electron ‘pairs’ = 3 (one lone pair and two double bonds) ‘Parent’ shape = trigonal planar Molecular shape is bent. NH3

N2O

CH4

CO2

Central atom is N N is in group 15, so number of valence electrons = 5 3 N–H single bonds use 3 valence electrons of N Number of lone pairs = 1 Total number of electron ‘pairs’ = 4 (one lone pair and three bonding pairs) ‘Parent’ shape = tetrahedral Molecular shape is trigonal pyramidal.

Central atom is taken as N+ (see part (a)) N+ has 4 valence electrons All valence electrons of the central N+ are used in bonding (see part (a)) ‘Parent’ shape = molecular shape = linear

Central atom is C C is in group 14, so number of valence electrons = 4 4 C–H single bonds and no lone pairs Total number of electron ‘pairs’ = 4 ‘Parent’ shape = molecular shape = tetrahedral

Central atom is C C is in group 14, so number of valence electrons = 4 2 C=O bonds and no lone pairs Total number of electron ‘pairs’ = 2 ‘Parent’ shape = molecular shape = linear

(c) The only radical species are O2 and NO. Although O2 has an even number of electrons, the highest occupied MOs are degenerate and singly occupied, making O2 a diradical (see the MO diagram in answer 2.29). NO is a radical because it contains an odd number of electrons (11 valence electrons) and so the HOMO is singly occupied.

Basic concepts: molecules (d) O3 absorbs in the UV region. Its presence in the Earth’s upper atmosphere is essential for protecting the planet’s surface from over-exposure to UV radiation from the Sun (see Box 14.6 in H&S). (e) MO diagram for N2 from two N atoms: σu*(2p) Energy

30

πg*(2p)

2py

2px

πu(2p)

2pz

2pz

2py

2px

σg(2p) σu*(2s) 2s

2s

σg(2s) N

N

N2

The bond order in N2 is 3. The very strong bond (945 kJ mol–1, see Appendix 12 in H&S) makes N2 chemically inert. (f) The monoatomic gases (He, Ne, Ar, Kr, Xe) are the group 18 (noble) gases. 2.31

(a) O

C

C

O

The left-hand resonance structure shows that O has an octet of electrons, but C has a sextet. In the right-hand structure, both atoms possesses an octet. (b) Compare the right-hand resonance structure in part (a) with a Lewis structure of N2: N

N

N2 and CO is isoelectronic; N, C– and O+ are all isoelectronic. (c) You should construct an MO diagram like that in Fig. 2.15a in H&S. The HOMO is essentially a carbon-centred MO (3σ MO) and the orbital character points outwards and corresponds to a lone pair. See the diagram of the 3σ MO drawn at the right-hand side of Fig. 2.15b in H&S. (d) CO competes with O2 for the haemoglobin binding sites and without treatment, O2 does not displace CO. Severe CO poisoning is therefore fatal. In a hyperbaric chamber, pure O2 at a pressure of 1.4 bar displaces haemoglobin-bound CO. 2.32

(a) S

S H

H

O

+

–

O

–

O

S

+ O

Basic concepts: molecules

31

(b) Both are bent molecules and are polar. By SI convention, the direction of the dipole moment is indicated by an arrow that points from the δ – to δ + end of the molecule: You have to take into account both the effects of the lone pairs and the differences in S S H electronegativities of the O O H atoms.

Energy

(c) MO diagram for HO• :

2px

2py

2pz

1s 2σ

1σ

2s

O

HO

H

Let the H–O bond lie along the z-axis. HO• contains 7 valence electrons and is a radical. To a first approximation (not allowing for any mixing of orbital character), the 1σ MO is localized on O and is non-bonding. The 2σ MO arises from overlap of the O 2pz and H 1s orbitals. The 2px and 2py orbitals are non-bonding, and so the 2σ MO is the only bonding orbital. The bond order is 1.

32

3 3.1 Pauling electronegativity values are listed in Table 2.2 of H&S

Cl

S

B Cl

O

Cl

(3.1)

O

Introduction to molecular symmetry Use the VSEPR model to obtain the molecular shapes. The method of working is as in answer 2.17. After you have the shape of the molecule, if necessary, use Pauling electronegativity values to obtain the bond dipole moments and then consider the molecular dipole moment. A detailed method of working was given in answer 2.19. (a) BCl3 B (group 13), 3 valence electrons; 3 bonding pairs and no lone pairs; trigonal planar molecule (3.1). Each B–Cl bond is polar: χP(B) = 2.0, χP(Cl) = 3.2, but the molecule is non-polar. See answer 2.19d for resolution of vectors in the related BF3.

(3.2)

(b) SO2 Bent molecule (3.2); see answer 2.19c for a detailed answer.

P

P

Br Br Br

Br

(3.3)

Br Br

(3.4)

S

C

S

(d) CS2 C (group 16), 4 valence electrons; forms two C=S double bonds; there are no lone pairs on the C atom, so the molecule is linear (3.5). Each bond is approximately non-polar: χP(C) = χP(S) = 2.6. The molecule is non-polar. See also answer 2.19b for a detailed answer to CO2, which is related to CS2.

(3.5)

H

H

C F

F F

(3.6)

C F

(c) PBr3 P (group 15), 5 valence electrons; 3 bonding pairs and 1 lone pair; trigonal pyramidal molecule (3.3). Each P–Br bond is polar: χP(P) = 2.2, χP(Br) = 3.0. The net size and direction of the molecular dipole moment depends on the extent to which the lone pair offsets (or not) the resultant moment due to the bond dipoles. The result is shown in structure 3.4. By SI convention, the dipole-arrow points from δ– to δ+.

F F

(3.7)

Questions 3.2-3.13: general notes

(e) CHF3 The molecule is tetrahedral (like CH4 or CF4). Two types of bond must be considered: C–H: χP(C) = 2.6, χP(H) = 2.2, so slightly polar bond (see 3.6). C–F: χP(C) = 2.6, χP(F) = 4.0, so polar bond (see 3.6). The resultant molecular dipole moment is as shown in diagram 3.7.

A symmetry operation is an operation performed on an object which leaves it in a configuration which is indistinguishable from, and superimposable on, the original configuration. A symmetry operation is carried out with respect to points, lines or planes, the latter being the symmetry elements.

Introduction to molecular symmetry 3.2 C2

σv

σv′

Fig. 3.1 Principal axis of rotation, and the two mirror planes in H2O.

3.3

33

(a) E is the identity operator. It identifies the molecular configuration. The operator E leaves the molecule unchanged. All objects can be operated upon by the identity operator E. (b) A plane of symmetry (mirror plane) is denoted by σ. (c) The symmetry operation of rotation about an n-fold axis (the symmetry element) is denoted by Cn, in which the angle of rotation is 360o/n, where n = 1, 2, 3, 4 ... (d) An Sn axis is an n-fold improper rotation axis: rotation through 360o/n about the axis is followed by reflection through a plane perpendicular to the same axis. If the mirror plane lies perpendicular to the principal axis, it is denoted σh. If the plane contains the principal axis, it is denoted σv. To see the difference between σv and σv′, it is best to use an example: H2O is a simple example. Figure 3.1 shows the principal (C2) axis in a molecule of H2O, and the two mirror planes, both of which contain the principal rotation axis. The plane which bisects the molecule is labelled σv, and the plane in which the molecule lies is labelled σv′. A σd plane contains the principal rotation axis, and also bisects the angle between two adjacent 2-fold axes. (a) An 8-vertex star. The symmetry of the star is such that rotation through (360/8)o = 45o gives another star superimposable on the first one; the * is used in the diagram to clarify the rotation that has occurred. This operation is repeated 7 more times to get back to the first orientation. The highestorder axis is an 8-fold axis (C8) running through the centre of the star, perpendicular to the plane of the paper. (b) An ellipse. The symmetry is such that rotation through (360/2)o = 180o as shown in the diagram gives another ellipse superimposable on the first one. This operation is repeated once more to get back to the first orientation. The highest-order rotation axis of the ellipse is a 2-fold axis (C2). There are 2 other C2 axes: both lie in the plane of the paper, one horizontal, and one vertical with respect to the ellipse drawn. (c) A pentagon. The symmetry is such that rotation through (360/5) o = 72 o gives another pentagon superimposable on the first one. This operation is repeated 4 more times to get back to the first orientation. The principal axis is a 5-fold axis (C5). (d) The symmetry of this shape is such that rotation through (360/3)o = 120o gives a shape superimposable on the first one. This operation is repeated twice more to get back to the first orientation. The principal axis is a 3-fold axis (C3).

*

*

Rotate through 45o

Rotate through 180o

*

Rotate through 72o

*

*

*

*

Rotate through 120o

*

34

Introduction to molecular symmetry 3.4

SO2 is a bent molecule (see answer 2.4b). It must possess an E operator – all molecules do. The other symmetry operators are: A σv plane:

A C2 axis:

C2

A σv′ plane:

3.5

(3.8)

The symmetry operators are E, C2, σv and σv′.

Fig. 2.1 in H&S shows one view of the H2O2 molecule. Another two views are shown here in structures 3.8 and 3.9. In 3.9, the molecule is viewed along the O–O bond. The symmetry operator that H2O2 possesses is a C2 axis running through the midpoint of the O–O bond in the direction shown in the left-hand diagram below. Its operation is shown in the right-hand diagram below; look carefully at the perspective in the diagrams.

(3.9) H(1)

C2 3.6

F

H(2)

The diagrams below summarize the answer.

σv C2

B

F

F

F

F

(3.10)

C2

(3.11)

C3

σh

C2

σv 3.7 Cl B F

Br

(3.12)

H(1)

C2

Cl

B

H(2)

Each σv plane contains one B–F bond.

σv

(a) On going from BF3 to BClF2 (3.10 to 3.11): the C3 axis is lost, two C2 axes are lost, and two σv planes are lost. (Colour the left-hand F atom in the diagrams above to signify a change to Cl, and confirm which of the symmetry operators are no longer valid). (b) On going from BClF2 to BBrClF (3.11 to 3.12): C2 axis is lost, and the σv plane is lost. (Use the diagrams above again to help you confirm these facts). (c) Each molecule has a σh plane – the plane containing the atoms.

Introduction to molecular symmetry 3.8

First, draw the structures of the molecules and ions: N H

See Chapters 15 and 16 for representations of the bonding in SO3, [SO4]2– and [NO3]– so that each atom obeys the octet rule

3.9

H

O

H P

Br

S Br Br

O

Cl

F S

F F

F

(3.13)

(3.14)

Cl

N

O

O

–

O

O

O

O

The structures of [SO4]2– and [NO3]– are drawn so as to emphasize the equivalence of the bonds (delocalized bonding) and the symmetry; see margin note. Note: (i) each possesses a C3 axis; (ii) only the planar molecules possess a σh plane. The answers are therefore: (a) C3 axis but no σh plane: NH3, PBr3, [SO4]2–. (b) C3 axis and a σh plane: SO3, AlCl3, [NO3]–. First, draw out the structures of the molecules and ions. Use the VSEPR model to show why [ICl4]– and XeF4 are square planar (2 lone pairs and 4 bonding pairs): Cl

3.10

S

Al

O

2–

O

Cl

Cl

As above, the structure drawn for [SO4]2– emphasizes its symmetry and does not illustrate a bonding picture

35

C Cl

Cl Cl

Cl

I

Cl Cl

2–

O

–

F F

S O

O O

Si F

F

F

Xe

F F

F

The only two species with a C4 axis and a σh plane are the square planar ones: [ICl4]– and XeF4. For each, first draw out the structure (using VSEPR to help you) and then work out the number of mirror planes. (a) SF4: the structure is disphenoidal (3.13). SF4 contains 2 mirror planes as shown in 3.14: one plane contains the S and 2 F(axial) atoms, and the other contains the S and 2 F(equatorial) atoms. (b) H2S: the bent structure is analogous to that of H2O (although the bond angle is different) and so the 2 mirror planes are as shown in Fig. 3.1. (c) SF6: octahedral structure (6 bonding pairs, 0 lone pairs). There are 9 mirror planes. The 3 σh planes are shown in 3.15, and each contains the S and 4 F atoms. The 6 σd planes can be considered in 3 sets of 2. One set is shown in 3.16. Each plane contains the S and two opposite F atoms (i.e. it contains a C4 axis), and bisects the other two F–S–F axes. The other 2 sets of σd planes are similarly constructed, starting with a different C4 axis (i.e. F–S–F axis).

(3.15)

(3.16)

(d) SOF4 has a see-saw geometry (Fig. 2.2, p. 22). SOF4 contains 2 mirror planes, analogous to those in SF4 (3.14).

Introduction to molecular symmetry

36

(e) SO2: this has a bent structure and so has the same symmetry as H2S in part (b). It has 2 mirror planes. (f) SO3: the structure is trigonal planar (see answer 2.17i, p. 21). There are 4 mirror planes: one σh and 3 σv planes. Their positions are analogous to those in BF3, answer 3.6.

Questions 3.11-3.12: general notes

A centre of inversion: If reflection of all parts of a molecule through the centre of the molecule produces an indistinguishable configuration, the centre is a centre of inversion (centre of symmetry), designated by the symbol i.

3.11

(a) Si is in the same group as C, so Si2H6 is expected to have the same structure as ethane. (b) A staggered conformer (3.17) will be the most favoured in terms of steric energy. (c) The midpoint of the Si–Si bond is an inversion centre. This point is shown in structure 3.17. To confirm that this is a centre of inversion, reflect each point of the molecule through centre i and show that an identical point is generated. For example, take H atom a ... reflect through i ... you end up at H atom a′ which is indistinguishable from atom a. (d) The eclipsed conformer 3.18 is the least favoured in terms of steric energy. (e) This conformer does not contain a centre of inversion. To confirm this, try taking the midpoint of the Si–Si bond again as in part (c). Reflection of all parts of the molecule through this point does not lead to an indistinguishable configuration.

3.12

In each part of this answer, if you are unsure of the reason why a molecule does or does not have an inversion centre, take the central atom as a trial inversion centre, and try inverting parts of the molecule through this point. (a) BF3: B, group 13; 3 valence electrons; trigonal planar; structure 3.19. No inversion centre. (b) SiF4: Si, group 14; 4 valence electrons; tetrahedral; structure 3.20. No inversion centre. (c) XeF4: Xe, group 18; 8 valence electrons; square planar (4 bonding and 2 lone pairs); structure 3.21. The Xe centre lies on an inversion centre. (d) PF5: P, group 15; 5 valence electrons; trigonal bipyramidal; structure 3.22. No inversion centre. (e) [XeF5]– : Xe, group 18; 8 valence electrons; pentagonal planar (5 bonding and 2 lone pairs); structure 3.23. No inversion centre. (f) SF6: S, group 16; 6 valence electrons; octahedral; structure 3.24. The S centre lies on an inversion centre. (g) C2F4: planar structure 3.25; this is related to ethene. There is an inversion centre at the midpoint of the C=C bond.

a

i a′

(3.17)

(3.18)

F B F

F

(3.19) F Si F

F F

(3.20)

F F

Xe

F

F

P

F

F

(3.21)

F F

F

(3.22)

–

F F

F

Xe F

F F

S

F

F F

(3.23)

F F

F

(3.24)

F C

C

F

F

(3.25)

Introduction to molecular symmetry H C H

C

C

H H

37

(h) The C=C=C unit is linear, but the presence of two adjacent π-bonds places a restriction on the orientation of the two CH2 groups: they are orthogonal (3.26). Therefore, there is no inversion centre.

(3.26)

3.13

A linear molecule possesses an ∞-fold axis of rotation. This applies to both symmetrical (e.g. F2, 3.27) and asymmetrical (e.g. HCN, 3.28) molecules. C∞ (3.27)

Questions 3.14-3.21: general notes

3.14 N F

(3.28)

Before attempting questions 3.14-3.21, make sure that you have studied worked examples 3.4-3.7 in H&S, in which point group assignments are made with accompanying explanations. When reading through answers 3.14-3.21, ensure that you have Fig. 3.10 from H&S available for reference. This gives a flowchart for assigning point groups. NF3: trigonal pyramidal structure 3.29. To determine the point group, apply the strategy shown in Fig. 3.10 in H&S:

START

F

C∞

F

(3.29)

Is the molecule linear?

No

Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Are there 3 C2 axes perpendicular to the principal axis? Is there a σh plane? Are there n (i.e. 3) σv planes containing

No Yes: C3 axis

the Cn axis?

No No Yes

STOP

Conclusion: the point group is C3v.

F F

S

F

F F

Cl

C4 F S

F

F F

Cl

A member of the D∞h point group must contain a C∞ axis and, therefore, the species is linear. See answer 3.13.

3.16

SF5Cl: structure 3.30. This is an example of a molecule that we loosely call ‘octahedral’, but which does not possess octahedral symmetry, i.e. it does not belong to the Oh point group. To determine the point group, apply the strategy in Fig. 3.10 in H&S:

START

(3.30)

F

3.15

Is the molecule linear?

No

Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Are there 4 C2 axes perpendicular to the principal axis? Is there a σh plane? Are there n (i.e. 4) σv planes containing

No Yes: C4 (see 3.31)

the Cn axis? Conclusion: the point group is C4v.

(3.31)

No No Yes

STOP

38

Introduction to molecular symmetry 3.17 C2

σv

σv′ Fig. 3.2 Principal axis of rotation, and the two mirror planes in BrF3.

3.18 C2

If BrF3 were trigonal planar or trigonal pyramidal, the molecule would have a C3 axis. The fact that it belongs to the C2v point group means that the principal axis is a C2 axis. The other 3-coordinate structure possibility is T-shaped, so the next step in the answer is to work out the symmetry properties of a T-shaped BrF3 molecule. Apart from the E operator, T-shaped BrF3 contains a C2 axis, a σv plane, and a σv′ plane as shown in Fig. 3.2. These are consistent with the C2v point group. Now see if this agrees with the VSEPR model: Central atom is Br Br is in group 17; number of valence electrons = 7 Number of bonding pairs (3 Br–F bonds) = 3 Number of lone pairs = 2 Total number of electron pairs = 5 ‘Parent’ shape = trigonal bipyramidal Molecular shape = T-shaped, see structure 3.32

3.19

F C Cl

(3.33)

Cl Cl Cl

C F

F F

(3.34)

F

Br F

(3.32)

The structure of [XeF5]– is shown in Fig. 3.3. To determine the point group, apply the strategy shown in Fig. 3.10 in H&S:

START

Fig. 3.3 Pentagonal planar structure of [XeF5]–. One of the 5 C2 axes is shown; each of the other 4 C2 axes coincides with a different Xe–F bond vector.

F

Is the molecule linear?

No

Does it have Td, Oh or Ih symmetry? Is there a Cn axis?

No Yes: C5 (perpendicular to plane of paper)

Are there 5 C2 axes perpendicular to the principal axis? Is there a σh plane perpendicular to the principal axis?

Yes (Fig. 3.3) Yes

STOP

Conclusion: the point group is D5h. This question is best done by first grouping compounds as follows, taking into account the number of different substituents: (a) CCl4 and (e) CF4 (b) CCl3F and (d) CClF3 (c) CCl2F2 Take (a) and (e) together: each is regular tetrahedral and belongs to the Td point group. Take (b) and (d) together: molecules 3.33 and 3.34 belong to the same point group, which is assigned as follows:

START

Is the molecule linear?

No

Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Are there 3 C2 axes perpendicular to the principal axis? Is there a σh plane? Are there n (i.e. 3) σv planes containing

No Yes: C3 axis

the Cn axis? Conclusion: the point group is C3v.

No No Yes

STOP

Introduction to molecular symmetry

(c) CCl2F2: see structure 3.35 in which the orientation has been chosen so as to help you in finding the symmetry elements. To assign the point group:

F Cl Cl

39

START

C

Is the molecule linear?

No

No Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Yes: C2 axis (horizontal through C in 3.35) Are there 2 C2 axes perpendicular to the principal axis? No Is there a σh plane? No Are there n (i.e. 2) σv planes containing

F

(3.35)

the C2 axis?

Yes

STOP

Conclusion: the point group is C2v. 3.20 F

F S

F

O

(a) To assign a point group to SF4, first draw the structure (see answer 3.24) which is shown in 3.36.

START F

S

F

F F

F

(3.36)

(3.37)

The σv planes are shown in structure 3.14

3.21

Is the molecule linear?

No

No Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Yes: C2 axis (horizontal through S in 3.36) Are there 2 C2 axes perpendicular to the principal axis? No Is there a σh plane? No Are there n (i.e. 2) σv planes containing the C2 axis?

Yes

STOP

Conclusion: the point group is C2v. (b) The structures of SOF4 and SF4 are related – compare 3.37 with 3.36. The presence of the O atom does not change the symmetry. The O atom lies on the C2 axis. The point group is still C2v. The tetrahedron, octahedron and icosahedron are drawn below:

Tetrahedron

Octahedron

Icosahedron

The highest symmetry is possessed by the icosahedron; the Ih point group possesses the highest number of symmetry elements. Questions 3.22-3.23: general notes

For a molecule containing n atoms, the number of degrees of vibrational freedom can be determined by using the equations: Linear molecule: Number of degrees of vibrational freedom = 3n – 5 Non-linear molecule: Number of degrees of vibrational freedom = 3n – 6 For a vibrational mode to be IR active, the vibration must give rise to a change in the molecular dipole moment.

40

Introduction to molecular symmetry 3.22

(a) SO2 is non-linear (see answer 2.19c, p. 22). Number of degrees of vibrational freedom = 3n – 6 = (3 × 3) – 6 = 3 (b) SiH4 is non-linear (tetrahedral, like CH4). Number of degrees of vibrational freedom = 3n – 6 = (3 × 5) – 6 = 9

H

C

(c) HCN is linear (see 3.38). Number of degrees of vibrational freedom = 3n – 5 = (3 × 3) – 5 = 4

N

(3.38)

(d) H2O is non-linear. Number of degrees of vibrational freedom = 3n – 6 = (3 × 3) – 6 = 3 (e) BF3 is non-linear (trigonal planar, see 2.26, p. 22) Number of degrees of vibrational freedom = 3n – 6 = (3 × 4) – 6 = 6 3.23

First, draw out the structure of the molecule and determine if there is a molecular dipole moment (μ). A vibrational mode is only IR active if it gives rise to a change in molecular dipole moment (Δμ > 0). For a non-linear molecule, total number of modes of vibrational freedom = 3n – 6. (a) H2O is polar (3.39). Now consider the 3 modes of vibrational freedom:

(3.39)

Symmetric stretch: Δμ > 0 IR active

Asymmetric stretch: Δμ > 0 IR active

Deformation: Δμ > 0 IR active

(b) SiH4 is tetrahedral and non-polar (like CH4); Td symmetry. Vibrational modes are shown in Fig. 3.16 in H&S: 4 modes of vibrational freedom (2 degenerate pairs) are IR active and give rise to 2 bands in the IR spectrum. (c) PCl3 has a trigonal pyramidal structure and is polar (3.40 shows two views of PCl3). The vibrational modes are:

non-planar

(3.40)

–

–

Symmetric stretch: Δμ > 0

IR active

Symmetric deformation: Δμ > 0

IR active

+ –

planar

(3.41)

Doubly degenerate (asymmetric) stretch: Δμ > 0 IR active Doubly degenerate deformation: Δμ > 0 IR active There are 6 modes of vibrational freedom giving rise to 4 bands in the IR spectrum. (d) AlCl3 has a trigonal planar structure and is non-polar (3.41 shows two views of AlCl3). The modes of vibrational freedom can be represented in a similar way to those for PCl3, but the molecular framework of AlCl3 is planar. As a consequence of this, the symmetric stretch in AlCl3 is IR inactive because Δμ = 0. The IR active

Introduction to molecular symmetry

41

modes for AlCl3 are the symmetric deformation, doubly degenerate stretch, and doubly degenerate deformation, i.e. 5 modes of vibrational freedom giving rise to 3 bands in the IR spectrum. (e) CS2 is linear and is non-polar. Now consider the vibrational modes:

Asymmetric stretch: Δμ > 0 IR active

Doubly degenerate deformation: Δμ > 0 IR active The deformation is a bending mode and is doubly degenerate (see Fig. 3.11 in H&S for more detail). Two fundamental bands are therefore seen in the IR spectrum due to the asymmetric stretch and the deformation. Symmetric stretch: Δμ = 0 IR inactive

(f) HCN is a linear molecule and is polar. Now consider the vibrational modes: like CS2, there will be 3 vibrational modes, but an important point about a linear XYZ molecule is that the modes can essentially be assigned to an X–Y stretch, a Y–Z stretch, and an XYZ deformation (double degenerate). This arises because each of the symmetric and asymmetric stretches is dominated by the stretch of one or other of the bonds. Thus, in HCN we can represent the vibrational modes as:

CN bond stretch: Δμ > 0 IR active

CH bond stretch: Δμ > 0 IR active

Doubly degenerate deformation: Δμ > 0 IR active

Three fundamental absorptions are seen in the IR spectrum. 3.24 O D

D

(3.42)

The D2O molecule is bent (3.42) and has C2v symmetry. The C2v character table (Appendix 3 of H&S) is: C2v

E

C2

σv(xz)

σv'(yz)

A1

1

1

1

1

z

x2, y2, z2

A2

1

1

–1

–1

Rz

xy

B1

1

–1

1

–1

x, Ry

xz

B2

1

–1

–1

1

y, Rx

yz

The method of working is as for SO2 in Section 3.7 in H&S. Work out the reducible representation shown below by considering the effects on the bonds of the molecule of the symmetry operations in the character table. How many bonds are unchanged? See Fig. 3.3 in H&S for definition of planes in H2O

E

C2

σv(xz)

σv'(yz)

2

0

0

2

This row of characters equals the sum of the rows of characters for the A1 and B2 representations. Therefore, the D2O molecule has two non-degenerate stretching modes. Both are IR active, because, in the right-hand column of the character table, there is a z entry for the A1 mode and a y entry for the B2 mode.

42

Introduction to molecular symmetry 3.25

+ _

Deformation (bending or scissoring) modes result in changes to the bond angles of a molecule. Figure 3.14 in H&S shows the vibrational modes of SO3 (a planar molecule). The symmetric deformation is illustrated in structure 3.43. Consider the effect of each symmetry operation of the D3h point group on this deformation mode. Using the notation that a ‘1’ means ‘no change’, and a ‘–1’ means a ‘reversal of the direction of the vector’ gives:

+

+

E

C3

C2

σh

S3

σv

1

1

–1

–1

–1

1

(3.43)

Compare this line of characters with those in the D3h character table in Appendix 3 in H&S. A match is found for the A2'' representation and therefore the symmetric deformation has A2'' symmetry. 3.26

CBr4 is tetrahedral (3.44) and belongs to the Td point group. From the Td character table (Appendix 3 in H&S), the symmetry operations for the Td point group are E, C3, C2, S4 and σd. To construct a reducible representation for the vibrational stretching modes, consider the effects of these operators on the bonds in the CBr4 molecule. Enter into a table the number of bonds that are left unchanged when each symmetry operation is applied in turn:

(3.44)

E

C3

C2

S4

σd

4

1

0

0

2

3

0

–1

–1

1

T2

1

1

1

1

1

A1

This reduces to:

i.e. the sums of these rows of characters give the row in the reducible representation. 3.27

The six IR active modes of vibration of a molecule with Td symmetry form two triply degenerate sets (see Fig. 3.16 in H&S). Therefore, only two absorptions appear in the IR spectrum.

3.28

(a) Al2Cl6 (3.45) is a non-linear molecule. Therefore, for n atoms: Number of degrees of vibrational freedom = 3n – 6 = (3 × 8) – 6 = 18

y z x

(3.45)

(b) Look up the D2h character table in Appendix 3 of H&S. The question asks about stretching modes of vibration. Therefore, construct a reducible representation by considering the effects of the symmetry operations of the D2h point group on the different types (terminal and bridging) of bonds of Al2Cl6 : E

C2(z)

C2(y)

C2(x)

i

σ(xy)

σ(xz)

σ(yz)

Terminal

4

0

0

0

0

0

4

0

Bridge

4

0

0

0

0

0

0

4

Introduction to molecular symmetry

43

These representations reduce to Ag + B2g + B1u + B3u and Ag + B3g + B1u + B2u, respectively, showing that the terminal Al–Cl stretching modes are of Ag, B2g, B1u and B3u symmetries, and the stretching modes for the bridges are of Ag, B3g, B1u and B2u symmetries. In order to see if these are IR active, look at the right-hand columns in the D2h character table. If the symmetry label is associated with an x, y or z in the table, then the mode is IR active. Therefore, the B1u, B2u and B3u modes are IR active. 3.29

The octahedral (Oh) structure of [AlF6]3– is shown below along with one of each type of rotation axis. The octahedron also contains an inversion centre, and σh and σd planes. C4, S4, C2

C3, S6

C2

Look up the Oh character table in Appendix 3 of H&S. Consider the stretching modes for [AlF6]3–. Construct a reducible representation by working out how many Al–F bonds are left unchanged by each symmetry operation of the Oh point group: E

C3

C2

C4

6

0

0

2

C2 (= C24) i

S4

S6

σh

σd

0

0

0

4

0

2

This reduces to A1g + Eg + T1u (show that the sum of the rows of characters for these representations matches the reducible representation). To see which modes are IR active, check for the occurrence of an x, y or z in the right-hand part of the character table. Only the T1u mode is IR active (see Fig. 3.18 in H&S). 3.30 C4, C2 X OC

M

OC

CO

C2''

CO X C2'

(3.46)

trans-M(CO)4X2 belongs to the D4h point group. Rotational axes are shown in structure 3.46 (see also Fig. 3.4 in H&S). The question is concerned with stretching the C–O bonds. Therefore, construct a reducible representation by considering the effects of the symmetry operations of the D4h point group on the C–O bonds. How many C–O bonds are left unchanged by each symmetry operation? E

C4

C2

C 2'

C2''

i

S4

σh

σv

σd

4

0

0

2

0

0

0

4

2

0

This reduces to A1g + B1g + Eu (sum the rows of characters for these representations in the D4h character table, and check that this matches the reducible representation). The symmetries of the carbonyl stretching modes of trans-M(CO)4X2 are therefore A1g, B1g and Eu. In the D4h character table, only the Eu mode is associated with an x, y or z in the right-hand part of the table, and therefore, only the Eu mode is IR active.

44

Introduction to molecular symmetry 3.31

OC

CO

Pt

2+

CO

OC

Firstly, determine the point group to which [Pt(CO)4]2+ belongs, assuming a square planar structure (3.47). Using Fig. 3.10 in H&S, you can show that the point group is D4h. Hence, the method of working for the first part of the problem is as for question 3.30. There are three CO stretching modes: A1g, B1g and Eu. Now look at the right-hand columns of the D4h character table in Appendix 3 in H&S. Part of the table is reproduced below:

D4h

D4h

(3.47)

A1g

x2 + y2, z2

B1g

x2 – y2 (x,y)

Eu

IR active

Raman active

The appearance of the (x,y) term in the row for the Eu representation shows that this mode is IR active. The remaining stretching modes are IR inactive. Each of the A1g and B1g modes (but not the Eu mode) is Raman active because there is a product term (e.g. z2) in the right-hand column. Thus, the appearance of the band at 2235 cm–1 in the IR spectrum that is absent from the Raman spectrum, and the appearance of two bands (2257 and 2281 cm–1) in the Raman spectrum that are missing from the IR spectrum, are consistent with [Pt(CO)4]2+ having D4h symmetry. 3.32 OC

X

M

OC

C2

X C2v

(3.48)

cis-M(CO)2X2 and trans-M(CO)2X2 are 4-coordinate complexes, and must be square planar molecules (a tetrahedral complex would not possess cis- and trans-isomers). First determine the point groups. cis-M(CO)2X2 (3.48) has one C2 axis, while transM(CO)2X2 (3.49) has three. cis-M(CO)2X2 is non-centrosymmetric (no inversion centre), but trans-M(CO)2X2 is centrosymmetric. Use Fig. 3.10 in H&S to confirm that the point groups are C2v and D2h, respectively. Now consider the CO stretching modes of vibration. Use the C2v and D2h character tables (Appendix 3 in H&S) to construct reducible representations. Then reduce them to determine the number and symmetries of the CO stretching modes of each isomer. cis-M(CO)2X2 (C2v):

C2(z) OC

M

X

X

C2(y)

CO C2(x) D2h

(3.49)

CO

E

C2

σv(xz)

σv'(yz)

2

0

0

2

(see Fig. 3.3 in H&S for definition of planes)

This reduces to A1 + B2. (This is same as for SO2, described in detail in Section 3.7 in H&S.) From the right-hand column in the C2v character table, both the A1 and B2 stretching modes are IR active, and two absorptions are expected in the region close to 2000 cm–1 in the IR spectrum of cis-M(CO)2X2. trans-M(CO)2X2 (D2h): E CO

2

C2(z) 0

C2(y) 0

C2(x)

i

2

0

σ(xy) 2

σ(xz)

σ(yz)

2

0

This reduces to Ag + B3u. From the right-hand column in the D2h character table, the B3u stretching mode is IR active, and the Ag mode is Raman active. Therefore, one

Introduction to molecular symmetry

45

absorption is expected close to 2000 cm–1 in the IR spectrum of trans-M(CO)2X2. Therefore, the cis- and trans-isomers can be distinguished by the number of observed IR spectroscopic absorptions arising from the CO stretches. 3.33

D3h

(a) Point group of PF5 (e.g. of XY5, 3.50) is D3h. See worked example 3.5 in H&S. To determine the number and symmetries of the stretching modes of vibration of PF5, first work out a reducible representation by working out how many P–F bonds are left unchanged by each symmetry operation of the D3h character table: E

C3

C2

σh

S3

σv

5

2

1

3

0

3

This reduces to 2A1' + E' + A2'' (check that the sums of the rows of characters for these representations in the D3h character table match the reducible represenation above). There are therefore four modes of vibration for PF5, two with A1' symmetry, one with A1'', and one doubly degenerate E' mode. (b) Part (a) showed that there are four stretching modes. To determine which of these are IR active, check the right-hand side of the D3h character table and look for x, y or z labels. Only the E' and A2'' modes are IR active, and this is consistent with the IR spectrum of gaseous PF5 exhibiting two absorptions. The right-hand columns in the D3h character table show that a vibrational mode with A1' symmetry is Raman active. PF5 has two stretching modes with A1' symmetry, and therefore two bands are observed in the Raman spectrum.

(3.50)

3.34

(a) A chiral molecule is one that lacks an improper (Sn) axis of symmetry. It is also possible to identify a chiral molecule by its lack of an inversion centre, i, and plane of symmetry. However, there are a few species that are achiral even though they lack an inversion centre and plane of symmetry, i.e. species which possess Sn symmetry where n is an even number. (b) A pair of enantiomers consists of two molecular species which are mirror images of each other and are non-superposable. (c) A helical chain can be exemplified by catena-sulfur, Sn, or catena-selenium, Sen. The chains may possess a left- or right-handed twist, and are therefore chiral.

3.35

(a) Consider the diagram on the right. Inversion through point i takes you from point A to A'. Alternatively, rotation through 180o followed by reflection through a plane perpendicular to the axis of rotation also takes you from A to A':

A

i

A'

C2 A

Rotation Reflection

σh A'

(b) An S1 improper rotation consists of a 360o rotation followed by reflection through a plane perpendicular to the axis of rotation. Since the rotation takes you back to where you started, the net result is simply reflection.

46

Introduction to molecular symmetry 3.36-3.41

These are Web-based problems with mainly interactive question and answer sessions.

3.42

(a) The IR absorption band at 2143 cm–1 arises from the stretching mode of the CO molecule:

O

C

C

O

(b) The asymmetric stretch of the CO2 molecule (see Fig. 3.11 in H&S) comes at 2349 cm–1 and may interfere with the band for CO if the absorptions are broadened. (c) The fundamental stretching mode for N2 at 2359 cm–1 is IR inactive (no change in dipole moment) and so no band appears in the IR spectrum. (d) The toxicity of CO arises from the fact that it binds to Fe(II) in haemoglobin and competes for the O2-binding sites in blood. Under physiological conditions. CO binds irreversibly to haem and prevents O2 from being carried in the bloodstream. A pressure of O2 of ≈ 1.4 bar is needed to displace CO from haemoglobin. (e) First convert the units, then find the molecular mass of CO and the number of moles: 0.20 mg m–3 = 0.20 × 10–3 mg dm–3 = 0.20 × 10–6 g dm–3 Mr CO = 12.01 + 16.00 = 28.01 g mol–1 Moles of CO = (0.20 × 10–6 g dm–3) / (28.01 g mol–1) = 7.1 × 10–9 mol dm–3 Number of molecules of CO = (7.1 × 10–9) × (6.02 ×1023) = 4.3 ×1023 (2 sig. fig.) 3.43

O2 is a non-polar diatomic molecule, and the fundamental stretching mode does not lead to a change in dipole moment. Therefore the vibrational mode is IR inactive. Once bound to Fe, stretching the coordinated O2 molecule leads to a change in dipole moment because the molecule is no longer in a symmetrical environment: N(His) (Porph)N

N(Porph) Fe(III)

O

O

(Porph)N

N(Porph) O

O

Thus, the stretching mode of the metal-bound molecule is IR active.

47

4 Experimental techniques 4.1

Gas chromatography (GC) is a separation technique in which the mobile phase is a gas and so is suitable for the detection and quantification of H2.

4.2

The two components on the spot TLC plate (Fig. 4.1) have Rf values of 0.52 and 0.15 when the eluting solvent mixture is CH3CN : H2O in a ratio 1 : 3 (see margin). Critical factors to replicate separation when changing from TLC to column LC: (i) The stationary phase must be the same (e.g. alumina or silica). However, stationary phases for column chromatography are available commercially in different particle and pore sizes. Differences between the properties of the stationary phase particles on the TLC plate and in the column will affect separation. (ii) The mobile phase must remain the same. (iii) When the sample is introduced on to the top of the stationary phase, it needs to be as a concentrated solution which is allowed to form a tight band prior to eluting with the solvent. The order of elution of the components should be replicated in the column with respect to the plate; the difference in Rf values of 0.52 and 0.15 is large. Spot TLC is much faster than a column and uses far less solvent. Therefore it is used to establish the best solvent system for preparative column LC or preparative plate chromatography.

Rf = 0.52 Rf = 0.15

Fig. 4.1 Diagram of TLC plate for answer 4.2.

4.3

Near UV: 250-400nm; visible 400-750 nm. Ru3(CO)12 absorbs at 392 nm which falls in the near UV region. However the absorption is broad and tails into the visible. The compound is orange (see Table 24.3 in H&S), consistent with absorption of violet light (see the colour wheel and data in Table 19.2 in H&S). Because Ru3(CO)12 is coloured, visual detection is possible during column chromatographic separation of the compound.

4.4

CHN analysis gives the % of each of C, H and N in a compound. Consider the monomer AlMe3 and dimer Al2Me6. Each contains 49.99% C and 12.58% H. To distinguish between monomer and dimer, mass spectrometry is a suitable technique.

4.5

Reaction of NbCl4(THF)2 with py under reducing conditions gives NbClx(py)y which contains 50.02% C, 4.20% H and 11.67% N. You start with Nb(IV) and therefore the product contains either Nb(III) or Nb(II). Assume that 6-coordinate Nb is retained. Possible compounds could be: NbCl3(py)3

Mr = 436.56

%C = {(15 × 12.01) / 436.56} × 100 = 41.27% NbCl2(py)4

Mr = 480.21

%C = {(20 × 12.01) / 480.21} × 100 = 50.02% Check N and H: %N = {(4 × 14.01) / 480.21} × 100 = 11.67%

Experimental techniques %H = {(20 × 1.008) / 480.21} × 100 = 4.20% Therefore 4.6 N

x = 2 and y = 4.

The structure of bpy is shown in the margin. Each N atom has a lone pair and protonation will give [Hbpy]+. Common mineral acids are HCl, HNO3 and H2SO4. Try HCl: bpy + HCl J [Hbpy]Cl

N

bpy

For [Hbpy]Cl, Mr = 192.64 %C = {(10 × 12.01) / 192.64} × 100 = 62.34% %H = {(9 × 1.008) / 192.64} × 100 = 4.71% %N = {(2 × 14.01) / 192.64} × 100 = 14.55% This gives a good fit to the experimental data of C 62.35, H 4.71, N 14.54 %, and therefore the mineral acid is HCl and the isolated compound is the ionic salt [Hbpy]+Cl–. 4.7

C60.xCHBr3 loses solvent and the final weight loss is 41%. Mr = (60 × 12.01) + x{12.01 + 1.008 + 3(79.90)} = 720.6 + 252.72x Assume that all solvent is lost. Therefore, 252.72x is 41% of Mr: 252.72x = 0.41{720.6 + 252.72x} which solves to give x = 1.98 ≈ 2, consistent with the two-step process described in the question.

4.8

CaSO4.2H2O J CaSO4.1/2H2O Mr = 172.18 145.16 g mol–1 Change in mass = 172.18 – 145.16 = 27.02 g % weight loss = (27.02 / 172.18) × 100 = 15.69% At 433K:

CaSO4.2H2O J γ-CaSO4 Mr = 145.16 136.15 Change in mass = 145.16 – 136.15 = 9.01 g % weight loss with respect to the original sample = (9.01 / 172.18) × 100 = 5.23%

At 463K:

A rough sketch of the TGA curve is shown on the right.

100 %mass

48

100 - 15.69 = 84.31% 84.31 - 5.23 = 79.08%

0 300

433 400

463

500 Temperature / K

Experimental techniques

49

4.9

(a) Treatment of acid is required for digestion of the sample. (b) The detailed method that you should outline is analogous to that described in worked example 4.1. For a detailed account, see: S. Ching, R. P. Neupane and T. P. Gray (2006) J. Chem. Educ., vol. 83, p. 1674. (c) Water content is determined using TGA. Water will be lost before there is any thermal decomposition of the anhydrous mineral because the H2O molecules are sited between the layers of the inorganic structure and are held by hydrogen bonds.

4.10

Cr(CO)6 is neutral and volatile, and EI MS can be used for analysis. The spacings of the peak envelopes are all equal to 28 mass units, corresponding to sequential loss of CO. The ions are: [Cr(CO)6]+ J [Cr(CO)5]+ J [Cr(CO)4]+ J [Cr(CO)3]+ J [Cr(CO)2]+ J [Cr(CO)]+ J [Cr]+ Relative intensities of different peaks indicate the relative stabilities of the ions.

4.11

Interpretation of the FAB mass spectrum of [Pd(PPh3)4] with NOBA matrix: The base peak comes at m/z 279.1 and the isotope pattern shows that the ion does not contain Pd. Fragmentation of [Pd(PPh3)4] is most likely to involve loss of PPh3 groups, and [PPh3 + H]+ would give a peak at m/z 263.1. The difference between the observed m/z 279.1 and m/z 263.1 is 16.0, suggesting an O atom. Phosphanes easily oxidize in the air (see Chapter 15 in H&S) and so the base peak at m/z 279.1 is assigned to [Ph3PO + H]+. In FAB MS, an organic matrix is usually used and gives rise to peaks in the mass spectrum. NOBA matrix (4.1) gives a peak at m/z 154.0 assigned to [NOBA + H]+, and therefore this peak can be ignored.

4.12

(a) Pb has four isotopes and this will dominate the pattern of each peak envelope:

HO

NO2 NOBA m/z 153

(4.1)

100%

206 208

204

204 1.4% 206 24.1% 207 22.1% 208 52.5% m/z

(b) Lead(II) acetate is Pb(O2CMe)2. EI MS gives radical cations, so no addition of H+ or Na+.The lowest mass peak envelope at m/z 208.0 arises from [Pb]+. The difference between m/z 224.0 and 208.0 is 16 mass units, so m/z 224.0 may be assigned to [Pb + O]+. The peaks at m/z 326.0 and 267.0 are assigned to [M]+ (parent ion) and [M – MeCO2]+. 4.13

Combinations of Ag+ with PPh3 and MeCN lead to singly charged ions. If two Ag+ are present, addition of nitrate ion keeps the overall charge +1. The observed ions are assigned as follows: m/z 369 = [Ag(PPh3)]+ m/z 410 = [Ag(PPh3) + MeCN]+ m/z 633 = [Ag(PPh3)2]+ m/z 802 = [Ag2(PPh3)2 + NO3]+ + m/z 893 = [Ag(PPh3)3] m/z 1064 = [Ag2(PPh3)3 + NO3]+

50

Experimental techniques For a 1 : 1 ratio of Ag+ : PPh3 in MeCN, the base peak is [Ag(PPh3) + MeCN]+ with [Ag(PPh3)]+ being the next most abundant peak. Even when the ligand is not in excess, the ions [Ag(PPh3)2]+ and [Ag2(PPh3)2 + NO3]+ are observed. Once PPh3 is present in excess (1 : 2, 1 : 3, or 1 : 4), the base peak is [Ag(PPh3)2]+, indicating that this is a particularly stable ion in the mass spectrometer. For 1 : 3 and 1 : 4 ratios, MeCN is not observed binding to Ag+, and [Ag(PPh3)3]+ becomes an important species. 4.14

Using 12C, 1H, 14N and 16O:

O

Mr = 298

O N

N H

N H N

C16H18N4O2 (a) The base peak is the most intense peak in the mass spectrum and the intensity is set arbitrarily to 100%. (b) The molecule is neutral and combines with H+, Na+ or K+ in the ESI mass spectrometer. The peaks at m/z 299.2 and 321.1 arise from [M + H]+ and [M + Na]+, respectively. 4.15

(a) From the structure shown, note that the ligand is [L]– (deprotonated phenolic OH) and so with the coordinated Cl– ion, the oxidation state of copper is +2.

100%

H

I

N N

Cu

527.9 528.9 529.9 530.9 531.9

Cl

m/z

4.16

121 122 123 124

100%

m/z

O I

(b) The complex is neutral and possibilities for a positively charged ion are [M + H]+, [M + Na]+ or [M – Cl]+. A peak envelope around m/z 529 fits [M – Cl]+. For the isotope pattern (see margin), look at Appendix 5 in H&S. Cu has two isotopes, 63Cu (69.2%) and 65Cu (30.8%). Note these are two mass units apart. Iodine is monotopic, and C, H, N and O are essentially monotopic. The isotope distribution in the peak around m/z 529 is therefore dominated by the isotope pattern of Cu. (c) For the minor peak envelope, the charge on the ion is +1, because the spacings between the peaks in the envelope are one mass unit. If the ligand in the complex is represented by L–, then the ion with m/z 994 is assigned to [M – Cl + L + H]+, i.e. [CuL2 + H]+. [Me4Sb][Ph2SbCl4] is a salt containing [Me4Sb]+ and [Ph2SbCl4]– ions. Both ions contain Sb(V). Positive mode ESI MS detects the cation and negative mode, the anion. Data needed: see Appendix 5 in H&S. (a) m/z 181 (100 %), 182 (4.5 %), 183 (74.6 %) and 184 (3.4 %) arise from [Me4Sb]+ (from the positive mode ESI MS). Sb has two isotopes, two mass units apart (see margin); this pattern dominates the peak envelope (m/z 181 contains 121Sb and m/z 183 contains 123Sb). In the negative mode MS, the peak envelope with m/z 415 (48.8 %), 416 (6.4 %), 417 (100 %), 418 (13.1 %), 419 (78.6 %), 420 (10.2 %), 421 (30.1 %), 422 (3.9 %) and 423 (5.7 %) arises from [Ph2SbCl4]–. The isotope pattern is influenced mainly by the four Cl atoms (two isotopes, 35Cl and 37Cl in a 3 : 1 ratio) and the Sb atom

Experimental techniques

51

(two isotopes as above). Since [Ph2SbCl4]– contains 12 C atoms, the presence of contributes significantly to the isotope pattern. (b) Use VSEPR (Sb, group 15, 5 valence electrons). [Me4Sb]+ : Sb+ (4 valence electrons) forms 4 Sb–C bonds and there are no lone pairs on the Sb centre. [Ph2SbCl4]– : Sb– (6 valence electrons) forms 2 Sb–C and 4 Sb–Cl bonds and there are no lone pairs on the Sb centre. 13C

+

Me Sb Me

Cl Me Me

–

Ph Sb

Cl

–

Ph

Cl

Cl

Cl

Cl

Sb

Ph Cl

Ph

Cl

trans-isomer

cis-isomer

Tetrahedral, no isomers

Octahedral

Mass spectrometric data provide no information about isomers. 4.17

For the ligand, L, m/z = 301.1. MALDI-TOF technique is used, therefore H+, Na+ or K+ can be added to neutral fragments. Peak envelopes for the product of the reaction of PtCl2 with L (1 : 2 ratio): m/z 302.1 [L + H]+ m/z 833.2 [PtL2Cl]+ m/z 869.1 [PtL2Cl2+ H]+ m/z 891.1 [PtL2Cl2+ Na]+

Cl N NH

N

N

N

L

The complex is [PtL2Cl2]. Two isomers (cis and trans) are possible. When a matrix is present in MS, it is useful to check the m/z of the matrix (αcyano-4-hydroxycinnamic acid, m/z = 189.0) in case the matrix is associated with the observed ions. 4.18

MALDI-TOF technique is used, and so H+, Na+ or K+ can be added to neutral fragments. The observed peaks are m/z 615.7 (base peak) and 637.7, and are assigned to: m/z 615.7 [L+ H]+ m/z 637.7 [L+ Na]+

H2N

Me

OH

OH

(4.2)

N

N

N O

NH

NH2

O

N

N

N

O

HN

NH2

The proposed formula is [Cu4L2(O2CMe)2(py)4(MeOH)2]. Let this be represented by M. The structure of H3L is shown . O in 4.2. The starting material is Cu(O2CMe) 2 H2O and 2+ contains Cu . Make a note of the charges on metal ions and ligands: four Cu2+, two L3–, two [MeCO2]– and neutral OH MeOH and py ligands. MALDI-TOF is used and so

Me

O

HN

O

H2N

4.19

NH

52

Experimental techniques H+, Na+ or K+ add to neutral fragments. The most intense peaks in the peak envelopes come at m/z 977 and 611 and are assigned as follows: m/z 977 m/z 611

[M – 2py – MeOH + H]+ [M – L – 2Cu – 2py – 2MeOH + 2H]+

4.20

For centrosymmetric molecules, the rule of mutual exclusion states that vibrations that are IR active are Raman inactive, and vibrations that are IR inactive are Raman active. For examples, choose homonuclear diatomic molecules, e.g. N2 and O2.

4.21

ν3 corresponds to the asymmetric stretch. The wavenumber is given by the equation: 1 2πc

ν =

k

μ

where: 1

μ

=

m + m2 1 = 1 m1 + m2 m1m2

Boron is common to [BF4]–, [BCl4]–, [BBr4]– and [BI4]– ; let mass of B be m1. m2 follows the order I > Br > Cl > F. As the terms 1/μ and (1/μ)1/2 become smaller along the series F to Cl to Br to I, so the wavenumber also becomes smaller. 4.22

The [N3]– ion is linear: –

–

+ N

N

N

The vibrational modes are: –

+ N

N

–

–

N

N

Symmetric stretch ν1

+ N

–

–

N

N

Asymmetric stretch ν3

+ N

–

N

Deformation (degenerate) ν2

There must be a change in dipole moment for the vibrational mode to be IR active. Only ν2 and ν3 are IR active. Assignments: 2041 cm–1 (ν3), 645 cm–1 (ν2) and 1344 cm–1 (ν1). See Fig. 3.11 in H&S for CO2, isoelectronic with [N3]–. 4.23

When urea bonds through the O atom, the contribution from resonance form A decreases. This means that there is less C=O character and more C–O character. Therefore νCO is lower. But the observation is that νCO shifts from 1683 cm–1 in the free ligand to 1725cm–1 in [Pt(urea)6]Cl2. This suggests that there is a greater contribution from resonance structure A, and therefore urea binds through the N atom. This can be checked by looking at what happens to ν(CN): it moves from 1471 cm–1 in the free ligand to1395 cm–1 in the complex, consistent with more C–N character and less C=N character, i.e. greater contribution from resonance structure A. H2N

C

NH2

+ H2N

NH2

C

O

O

A

B

–

H2N

+ NH2

C O C

–

Experimental techniques

53

4.24

Li3[PO4] contains the [PO4]3– ion. If [PO4]3– has Td symmetry (tetrahedral), there will be two IR active modes whereas for D4h symmetry (square planar), three vibrational modes are IR active (see Table 4.2 and Figs. 3.16 and 3.17 in H&S). Two bands at 1034 and 591 cm–1 are observed, consistent with Td symmetry.

4.25

(a) Of the observed absorptions at λmax = 245, 276, 284, 324 and 569 nm , the band at 569 nm lies in the visible region. This wavelength corresponds to absorption of green light (see Fig 4.15) and the complementary (observed) colour is purple/red (see Table 19.2 in H&S). (b) The lowest energy absorption has the longest wavelength: E = hν =

hc

λ

and is therefore that at 569 nm. (c) Use the Beer-Lambert law: A = εmax × c × l For λmax = 245 nm, εmax = 48200 dm3 mol–1 cm–1. The concentration of the solution is 2.0 × 10–5 mol dm–3, and the cuvette holding the solution has a path length of 1 cm: A = 48200 × 2.0 × 10–5 × 1 = 0.96 (2 sig. fig., no units) 4.26

(a) π* I π transitions are ligand-based, and involve MOs with pyridine and/or alkyne π-bonding (occupied MO) and π*-antibonding (vacant MO) character. In the compounds in the question, the alkyne and bpy units are directly bonded and are conjugated: N

R3PAu

AuPR3

N

(b) There is no absorption in the visible region, and therefore the compound is not coloured. (c) When comparing the UV-VIS spectra of a series of compounds, you should use plots of ε against λ rather than A against λ. This is because ε is a property of the compound, whereas A depends on ε and the concentration (Beer-Lambert law). 4.27

A blue (or hypsochromic) shift in wavelength is towards the blue-end of the spectrum, i.e. towards shorter wavelength (higher energy).

4.28

Measurement in Hz gives a value of the coupling constant J which is independent of the magnetic field. If it were recorded as a chemical shift difference (δ1 – δ2) ppm, the value would depend on the field strength. For example, consider a coupling constant JHH = 10 Hz. In a 100 MHz 1H NMR spectrum, this corresponds to a chemical shift difference (δ1 – δ2) = 0.1 ppm, but in a 250 MHz spectrum, JHH = 10 Hz corresponds to a value of (δ1 – δ2) = 0.04 ppm. Reporting coupling constants in Hz makes it possible to compare spectra recorded at different field strengths.

Experimental techniques

54

F

F a C

Cb

F

O

4.29

Values of spin-spin coupling constants decrease with the bond separation of the nuclei that are spin-coupled. The fact that it is possible to observe long range couplings between 31P and 19F, and between 31P and 1H, but not between nonequivalent and remote 1H nuclei, suggests that for a pair of directly attached nuclei, values of JPF and JPH are much greater than JHH. Typical values for directly attached nuclei are: JPF ≤ 1500 Hz, JPH ≤ 800 Hz, JHH ≤ 10 Hz.

4.30

See structure 4.3. There are 2 13C environments, labelled a and b in 4.3. Both 1H and 19F are NMR active (1H: 100%, I = 1/2; 19F: 100%, I = 1/2). Nucleus 13C(a) couples to 3 adjacent 19F to give a binomial quartet. Because J(13C-19F) values are large for directly attached nuclei, long range coupling is also seen. Therefore, 13C(b) couples to the 3 19F nuclei (a 2-bond coupling) to give another binomial quartet. The quartet with J = 284 Hz is assigned to 13C(a), and that with J = 44 Hz to 13C(b). No coupling of 13C(b) to the 1H nucleus is observed (they are not directly attached).

4.31

Consider structure 4.4. In the 31P NMR spectrum, Ph2PH exhibits a doublet with a large JPH (directly attached 1H and 31P nuclei). In the spectrum of Ph3P, a singlet is expected. (This ignores any small couplings to the ortho-H atoms of the phenyl rings which might be resolved). It would be instructive to run the proton-decoupled 31P NMR spectrum (31P{1H} NMR spectrum). This instrumentally removes 31P-1H coupling and in the spectrum of Ph2PH, the doublet will collapse to a singlet on going from the 31P NMR to 31P{1H} NMR spectrum.

4.32

(a) The binomial decet (10 line pattern) arises from coupling of the 31P nucleus to 9 equivalent 1H nuclei (4.5), each with I = 1/2. There is free rotation about each P–C and C–H single bond. The value of 2.7 Hz corresponds to JPH. (b) In the 1H NMR spectrum, one signal is observed (all protons equivalent) and it is a doublet due to coupling to the 31P nucleus. The magnitude of JPH = 2.7 Hz must be the same as is observed in the 31P NMR spectrum.

4.33

(a) See structure 4.6. The 29Si nucleus couples to two directly attached, equivalent 1H nuclei, H(c), to give a triplet, J SiH = 194 Hz. 1 (b) In the H NMR spectrum, there are 3 proton environments (a, b and c in 4.6), but the question only asks about protons H(c). Although 29Si is spin active with I = 1/ , it is only present in 4.7% abundance, and so 95.3% of protons attached to Si do 2 not couple to 29Si; these protons give a singlet in the 1H NMR spectrum. 4.7% of protons H(c) couple to 29Si, and these protons give a Singlet for the protons not doublet, the centre of which coupling to 29Si coincides with the singlet arising from 95.3% of the One half of doublet assigned protons. The spectrum is to the protons coupling to 29Si JSiH= 194 Hz shown in Fig. 4.2, which also shows where the coupling constant (194 Hz) is measured. Fig. 4.2 Simulation of part of the 1H NMR

O

H

(4.3)

Trigonal pyramidal

R P

R = H or Ph

(4.4)

P

CH3 CH3

H 3C

(4.5)

c

H

aH

H

c Ha

Si C

H a

C H

C

bH

a

Hb

(4.6)

spectrum of compound 4.6.

4.34

Figure 4.3 shows the 11B NMR spectra of THF.BH3 and PhMe2P.BH3, and the structures of these adducts are shown in 4.7 and 4.8. (a) The 11B nucleus couples to 3 equivalent 1H nuclei to give a binomial quartet. This corresponds to the observed

Experimental techniques

55

Quartet due to 11B-1H coupling

Each line of the quartet is split into a doublet by 11B-31P coupling; the two quartets are shown in solid and hashed lines respectively.

H O

B H

H

JBH

(4.7)

Me

(a)

Me

JBH

(b)

H

P Ph

B

JBP H

H

(4.8)

Fig. 4.3 Simulations of the 11B NMR spectra of (a) THF.BH3 and (b) PhMe2P.BH3.

spectrum in Fig. 4.3a, and a value of JBH can be measured between any pair of adjacent lines, e.g. as shown in the figure. (b) The 11B nucleus couples to 3 equivalent 1H nuclei to give a binomial quartet, and couples to the 31P nucleus to give a doublet. The overall signal is a doublet of quartets, which can be represented in two stages as follows. Comparison of the ‘stick’ spectrum with Fig. 4.3b allows you to see where to measure values of JBH and JBP. (Exercise: The coupling constants can also be measured between other pairs of lines in the spectrum. Where are these pairs of lines in Fig. 4.3b?) 4.35

298 K

175 K

Fig. 4.4 Simulations of 19F NMR spectra of SF4 at 298 K and 175 K.

(a) SF4

F

Axial

S (group 16), 6 valence electrons F Number of S–F single bonds = 4 S Number of lone pairs = 1 Equatorial F Total number of electron pairs = 5 F ‘Parent’ shape = trigonal bipyramidal (4.9) (4.9) Molecular shape = see-saw (b) A temperature dependent 19F NMR spectrum shows SF4 is stereochemically non-rigid; more thermal energy at 298 K than at 175 K. At 298 K, a singlet is observed (Fig. 4.4, top), therefore all F atoms must be equivalent, and the axial and equatorial F atoms are exchanging positions. At 175 K, two equal intensity triplets are observed (Fig. 4.4, bottom): there are two F environments (axial and equatorial, 4.9). Each axial F couples to 2 equatorial F’s to give a binomial triplet. Each

Experimental techniques

56

equatorial F couples to 2 axial F’s to give a binomial triplet. The two triplets occur at different chemical shifts. Because the two-site exchange involves equally populated sites (i.e. 2 F’s in each environment), the chemical shift of the singlet at 298 K lies in the midpoint of the chemical shifts of the two triplets. The two spectra in Fig. 4.4 are aligned to show the relative positions of the signals. (For unequally populated sites, the limiting high temperature peak lies at a weighted average shift value.) 4.36 F

F Si F

F

F

F

F

P

F

F

F

(4.10)

(4.11)

F

O

F

S

S

F

F

F

F

F

(4.12)

(4.13)

F C F

F F

(4.14)

4.37

(a) SiF4 Si (group 14) forms 4 Si–F single bonds; no lone pairs ‘Parent’ shape = molecular shape = tetrahedral; see 4.10 All F atoms are equivalent, so a singlet in the 19F NMR spectrum is consistent with a static molecular structure. (b) PF5 P (group 15) forms 5 P–F single bonds; no lone pairs ‘Parent’ shape = molecular shape = trigonal bipyramidal; see 4.11 Two F environments (axial and equatorial), so a singlet in the 19F NMR spectrum is not consistent with a static molecular structure. (c) SF6 S (group 16) forms 6 S–F single bonds; no lone pairs ‘Parent’ shape = molecular shape = octahedral; see 4.12 All F atoms are equivalent, so a singlet in the 19F NMR spectrum is consistent with a static molecular structure. (d) SOF2 S (group 16) forms 2 S–F single bonds and 1 S=O double bond; 1 lone pair ‘Parent’ shape = tetrahedral Molecular shape = trigonal pyramidal; see 4.13 The F atoms are equivalent, so a singlet in the 19F NMR spectrum is consistent with a static molecular structure. (e) CF4 C (group 14) forms 4 C–F single bonds; no lone pairs ‘Parent’ shape = molecular shape = tetrahedral; see 4.14 All F atoms are equivalent, so a singlet in the 19F NMR spectrum is consistent with a static molecular structure. Because of the periodic relationship between C and Si, answer (e) is effectively the same as part (a). Berry pseudo-rotation (Fig. 4.5) is a dynamic process usually invoked to explain the exchange of axial and equatorial atoms in a trigonal bipyramidal species (or axial and basal atoms in a square-based pyramidal species). Only a small amount of energy is needed to interconvert trigonal bipyramidal and square-based pyramidal structures. No bonds are broken and the interconversion involves small

Fig. 4.5 Schematic representation of the Berry pseudorotational process for a 5-coordinate species, starting on the left-hand side with a trigonal bipyramidal structure.

2

2

2 5

5 4

1

5 4

1

1 4

3

3

3

Experimental techniques

57

changes of the bond angles subtended at the central atom. Sequential bond angle changes in the directions shown in Fig. 4.5 result in atoms 4 and 5 moving from equatorial to axial sites, and atoms 2 and 3 moving from axial to equatorial sites. The intermediate structure in Fig. 4.5 is a square-based pyramid, atom 1 being axial. Repetition of these exchanges means that every substituent ‘visits’ both equatorial and axial sites in the trigonal bipyramidal structure. 5-Coordinate species undergoing Berry pseudo-rotation include PF5 and Fe(CO)5. 4.38

H

P

Me Me

C H H

(4.15) O– P+ Me

In PPh3, the 3 phenyl rings adopt a paddlewheel arrangement (Fig. 4.6). Any rotation about the P–C bond is hindered; compared to PMe3, the barrier to rotation is higher. In tetrahedral SiMe4, there is free rotation about the Si–C and C–H bonds. In interpreting 1H NMR spectra, assume that the H atoms are equivalent because there is rotation about the Si–C bond on the NMR spectroscopic timescale.

Me Me

(4.16)

c

c

a P

P

The 5 singlets in the 29Si NMR spectrum arise from SiCl4, SiBr4 and 3 species formed by Cl/Br exchange: SiCl3Br, SiCl2Br2 and SiClBr3.

4.40

Structure 4.17 shows that [P 5Br 2]+ contains 3 P environments in a 2 : 1 : 2 ratio. The doublet of triplets with J values of 321 and 149 Hz is assigned to P(b); 31P–31P coupling is between adjacent phosphorus nuclei. The triplet of triplets is assigned to P(a). The larger J value is due to coupling between 31P(a) and 31P(b); the smaller J value (26 Hz) is for the longer range coupling between 31P(a) and 31P(c). The triplet of doublets is assigned to P(c). The larger J value arises from coupling between 31P(c) and 31P(b); the J value of 26 Hz arises from the longer range coupling between 31P(c) and 31P(a).

4.41

The CO ligands in W(CO)6 (4.18) are equivalent and one signal is observed in the 13C NMR spectrum. Since the natural abundance of 13C is 1.1%, enriching samples with 13C is often used. Assume complete enrichment. There are 5 naturally occurring isotopes of W, but only 183W is NMR active (14.3%, I = 1/2). Therefore, 14.3% of the 13C nuclei are attached to 183W and appear in the 13C NMR spectrum as a doublet. 13C nuclei bonded to non-spin active W give a singlet. Thus, the 13C NMR spectrum of W(CO)6 appears as a doublet (i.e. satellites) superimposed on a singlet.

Br Br

P b

(4.17)

O C OC

W

CO CO

OC C O

(4.18)

Fig. 4.6 A space-filling model of PPh3 showing the paddle-wheel arrangement of the Ph groups; P atom is shown in black.

4.39

b P

P

‘Static solution structure’ is somewhat misleading, because low energy dynamic processes, in particular free rotation about single bonds, still occur on the NMR spectroscopic timescale. Take the examples given in the question to illustrate the answer. PMe3: see structure 4.15 which shows one CH3 group in full. In solution, rotations about the single C–H and P–C bonds occur, and the P–C bond rotation makes all 9 H atoms equivalent in solution, although they may not necessarily be equivalent in the solid state. In OPMe3 (4.16), rotations about the C–P and C–H bonds again occur. Rotations about P–C bonds mean that the 9 H atoms are equivalent. The discussion above for PMe3 applies to OPMe3.

58

Experimental techniques 4.42 Se

S

Se

Se

Se

a Se

S

b Se

S

S

S

S

S

Se

S

Se

S

S

S

S

S

S

S

S

S

S

S

SeS7

S

S

a Se

S

S

1,3-Se2S6

1,2-Se2S6 a Se

b Se

a Se

b Se

b Se

S

S

S

S

S

Se b

S

Se c

S

S

S

Se a

S

S

Se c

S

1,2,4-Se3S5

S

1,2,3-Se3S5

S

1,2,5-Se3S5

S

1,2,3,4-Se4S4

In the 77Se NMR spectrum of the above compounds, there will be 1 signal for each of SeS7, 1,2-Se2S6 and 1,3-Se2S6, 2 signals for 1,2,3-Se3S5 and 1,2,3,4-Se4S4 (see a and b labels above), and 3 signals for 1,2,4-Se3S5 and 1,2,5-Se3S5 (see a to c labels). 4.43

11B

is 80% abundant (I = 3/2). Coupling between 19F and 11B in BFCl2 gives rise to a 1 : 1 : 1 : 1 quartet (11B has 4 spin states) in the 19F NMR spectrum. The 19F nuclei in BF2Cl are equivalent and give one signal (1 : 1 : 1 : 1 quartet). The chemical shifts of the 19F NMR signals for BFCl2 and BF2Cl are expected to be different.

4.44

SbMe5 is expected to have a trigonal bipyramidal structure, and therefore there are two Me environments (axial and equatorial) in the static structure. 5-Coordinate molecules are often fluxional on the NMR spectroscopic timescale, the energy barrier to axial-equatorial exchange typically being low. Exchange of the Me groups in SbMe5 on the NMR timescale leads to the observation of only one signal in the 1H NMR spectrum.

4.45

The structures to be considered are shown below with the predicted number of 19F NMR signals and coupling patterns arising from 19F-19F coupling: F F

1 19F environment gives rise to a singlet

F

Nb

F

F F

b

b

F

a

F aF

Nb Cl

Cl

a F

F

Fa

F

2 19F environments; a doublet assigned to F(a), a quintet assigned to F(b)

Nb

F

b

F

F

F

aF

Nb

a F

Cl

Cl

Cl

In the trans-isomer, 1 19F environment; a singlet

In the cis-isomer, 2 19 F environments; two triplets

Experimental techniques

Cl Cl a

Nb

F Cl

a F

F b

In the mer-isomer, 2 19F environments; a doublet for F(a), a triplet for F(b)

F F

Nb

Cl

Pt

In the fac-isomer, 1 19F environment, giving a singlet

Au

Au

P

P

Ph

Ph

Ph H

H

(4.20)

4.48 a

F a F

Me P Fb

(4.22)

a F a F

–

Cl

Cl

Cl

Cl

Nb

Cl F

F

F

In the trans-isomer, 1 19F environment; a singlet

In the cis-isomer, 1 19F environment; a singlet

(a) Structure 4.20 shows the compound for the question. The signal at δ 3.60 ppm is assigned to the CH2 group and the triplet arises from coupling to 2 equivalent 31P (I = 1/ , 100%) nuclei. 31P{1H} means that the 31P NMR spectrum is proton2 decoupled and therefore the observed signal is a singlet. (b) Structure 4.21 shows a schematic representation R of the P-containing cation shown in Fig. 4.38 in H&S. Me The 4-coordinate P atom can formally be assigned a P positive charge. Each of the other P atoms carries a P P+ + lone pair of electrons. The P centre has two different organic substituents. Of the two PR groups on the R R left of structure 4.21, one has the R group on the R = Me same side of the ring as an Me substituent, while the (4.21) other has the R group on the opposite side of the ring. Thus, the P atoms are non-equivalent, and 3 signals are expected in the 31P NMR spectrum.

(4.19)

Ph

Cl

4.47

PEt3

Cl

Cl Cl

(a) [PF6]– is octahedral with 6 equivalent F atoms. There is one signal in the 19F NMR spectrum, but 31P is spin active (100% I = 1/2), and so the signal is a doublet. (b) In trans-[PtI2(PEt3)2] (4.19), the 2 P atoms are equivalent and give one signal in the 31P NMR spectrum. The notation 31P{1H} means that the spectrum is protondecoupled and therefore the observed signal pattern is not due to 31P-1H coupling. 195Pt is spin active (I = 1/ ), but is only present in 33.8% abundance. Approximately 2 one third of the 31P nuclei couple to 195Pt giving a doublet. Two thirds of the 31P nuclei do not couple to 195Pt and therefore give a singlet. The relative integrals of the superimposed singlet and doublet are in a ratio 2 : 1, and the observed threeline pattern appears as a 1 : 4 : 1 signal.

I

Cl

Cl

4.46 PEt3 I

Nb

F

F Cl

59

The [PF5Me]– ion (4.22) is octahedral. One F atom is trans to the Me group, and 4 F atoms are cis to the Me substituent. Thus, 2 signals are observed in the 19F NMR spectrum. The signal for F(b) appears as a doublet of quintets due to coupling to 31P and 19F(a). The signal for F(a) is a doublet of doublets because there is to coupling to 31P and 19F(b). Therefore, 3 coupling constants can be measured: JPF(a), JPF(b) and JF(a)F(b).

60

Experimental techniques

b PPh3 Ph3P a

Rh

4.49

The square planar Rh(PPh3)3Cl (4.23) has 2 P environments. With only 31P-31P coupling, the 31P NMR spectrum would exhibit a doublet for P(a) and a triplet for P(b). The value of JPP must be the same for each signal. Therefore, from the data in the question, JP(a)P(b) = 38 Hz. Additional coupling arises because 103Rh is spin active (100%, I = 1/2). The doublet of doublets is assigned to P(a) with JP(a)P(b) = 38 Hz and JRhP(a) = 145 Hz. The doublet of triplets is assigned to P(b) with JP(a)P(b) = 38 Hz and JRhP(b) = 190 Hz.

4.50

[BH4]– is tetrahedral and all H atoms are equivalent, and so there is one signal in the 1H NMR spectrum. The 1H nuclei couple to 11B (I = 3/ 2, 80.1%) to give a 1 : 1 : 1 : 1 four line pattern since there are four spin states for 11B. For naturally occurring boron, 19.9% is 10B (I = 3, 7 spin states) and coupling of the protons to this nucleus gives rise to a 1 : 1 : 1 : 1 : 1 : 1 : 1 seven line pattern. The multiplets are superimposed on one another (centred at δ –3.60 ppm) and are both observed only in well resolved spectra. A figure of the spectrum is given in M. Zanger et al. (2005) J. Chem. Educ., vol. 82, p. 1390. The coupling constants are 80.5 Hz for J B H and 27.1 Hz for J B H. On a 400 MHz NMR spectrometer, 1 ppm = 400 Hz, and therefore the 80.5 Hz coupling will correspond to ≈ 0.2 ppm. On a 100 MHz NMR spectrometer, 1 ppm = 100 Hz, and therefore the 80.5 Hz coupling will correspond to ≈ 0.8 ppm. Thus, on going from a 400 to 100 MHz instrument, the spectrum will be ‘spread out’. The literature reference given above compares spectra recorded at 400 and 90 MHz.

PPh3 a

Cl

(4.23)

11

4.51 H315N

15

NH3

Pt Cl

Cl

1

1

(a) The NH3 groups in square planar cis-[Pt(15NH3)2Cl2] are equivalent (4.24), and the 15N NMR spectrum shows one signal. 15N couples to 3 equivalent 1H to give a 1 : 3 : 3 : 1 binomial quartet. 195Pt is spin active (I = 1/2, 33.8%), and therefore the signal has satellites arising from additional coupling to 195Pt; only 33.8% of the 15N is affected. The final signal is a doublet of quartets superimposed on a quartet. (b) The observed coupling constants are J N H = 74 Hz and J N Pt = 303 Hz. Note that 303 ≈ 74 × 4. The left-hand diagram below shows the 1 : 3 : 3 : 1 quartet for the 15N-1H coupling in the 15N NMR spectrum, and below it, the doublet of quartets for the 15N-195Pt and 15N-1H coupling. Together, these result in the apparent octet shown in the right-hand diagram. 15

(4.24)

10

1

15

195

JNH 74 Hz

JNPt 303 Hz

4.52

(a) Ground state electronic configuration of Na is [Ne]3s1. NaCl is ionic and contains Na+ and Cl–. The electronic configuration of Na+ is [Ne]. There are no unpaired electrons (Cl– also has a closed shell configuration) and so NaCl is EPR silent.

Experimental techniques

61

(b) 14N I = 1 gives rise to a 3-line pattern: 2nI + 1 = 2(1)(1) + 1 = 3 The EPR spectrum should appear as shown: B/G

4.53 2+

O H2 O

V

H 2O

OH2

[VO(OH2)5]2+ is vanadium(IV), therefore d1. Isotopes of V: 50V, 0.25% and 51V, 99.75%. The EPR spectrum of [VO(OH2)5]2+ shows an 8-line pattern, consistent with the pattern being dominated by coupling of the odd electron to 51V (I = 7/2). 2nI + 1 = 2(1)(7/2) + 1 = 8

OH2 OH2

g-values determined were gzz =1.932 and gxx = gyy =1.979, and therefore the system must be anisotropic with axial symmetry (x and y axes are equivalent but are different from the principal axis, z). Structure 4.25 shows [VO(OH2)5]2+; the principal axis contains the V=O unit.

(4.25)

4.54

Isotopes of Cu: 63Cu, 69.2% and 65Cu, 30.8%, both I = 3/2. Cu2+ is d9, so one unpaired electron. (a) For each isotope: 2nI + 1 = 2(1)(3/2) + 1 = 4 Therefore, 4 peaks are expected for each isotope. (b) Because of coupling to both 63Cu and 65Cu, there are two superimposed EPR spectra. The 2 pairs of outer peaks in Fig. 4.39 in H&S belong to the separate spectra arising from 63Cu and 65Cu. The two inner peaks are superimpositions of the EPR signals of the two spectra, thus the greater intensities of the central peaks. (c) Look at Fig. 4.28 in H&S. Since the hyperfine coupling constant A(65Cu) is 1.07 × A(63Cu), the outer lines belong to the spectrum arising from coupling to 65Cu, and the inner spectrum is from coupling of the unpaired electron to 63Cu. Values of A (in G) can be determined from Fig. 4.39 in H&S from the spacings of the peaks (see Fig. 4.28 in H&S). From Fig. 4.39 in H&S, estimates can be made of A(63Cu) = 140 G and A(65Cu) = 150 G. (d) The gyromagnetic factor, g, is given by: g sample =

hν μ B × Bsample

As you collect the data together for the calculation, ensure the units are consistent. Read the value of B corresponding to the centre of the spectrum in Fig. 4.39 in H&S. This gives Bsample = 3360 G = 3360 × 10–4 T. The EPR spectrum was measured as 9.75 GHz. Thus, ν = 9.75 GHz = 9.75 × 109 Hz = 9.75 × 109 s–1

μB = Bohr magneton = 9.274 × 10–24 J T–1 h = Planck constant = 6.626 × 10–34 J s

62

Experimental techniques

g sample =

(6.626 × 10 −34 J s)(9.75 × 10 9 s −1 ) (9.274 × 10 − 24 J T −1 )(3360 ×10 − 4 T)

= 2.07 4.55

(a) The method is as for problem 4.54d. From Fig. 4.40a in H&S, read the value of B corresponding to the centre of the spectrum: Bsample = 3475 G = 3475 × 10–4 T. Then: g sample =

(6.626 ×10 −34 J s)(9.75 × 10 9 s −1 ) (9.274 × 10 − 24 J T −1 )(3475 × 10 − 4 T )

= 2.00 (b) For 2 equivalent 14N with I = 1: Multiplicity = 2nI + 1 = 2(2)(1) + 1 = 5 This corresponds to Fig. 4.40a in which there are 5 peaks in the spectrum. (c) Fig. 4.40a: A is found from the separation of the peaks: A1 = A2 = 30 G In Fig. 4.40b: the spectrum can be analysed as two sets of 3 signals. A1 is the smaller of the two splittings, and A2 is measured from the larger separation: A1 = 10 G

A2 = 30 G

4.56

For 14N, I = 1 and for 1H, I = 1/2. The hyperfine interactions of the unpaired electron with 14N and 1H nuclei can be considered as the sum of these interactions. (a) When A(14N) = A(1H) = 30 G: (b) A(14N) = 30G, A(1H) = 10 G: A1 = A2 = 30 G

A1 = 10 G

A2 = 30 G

4.57

59Co,

100%, I = 7/2. Therefore, for each of the different Co2+ sites: Multiplicity = 2nI + 1 = 2(1)(7/2) + 1 = 8

Experimental techniques

63

Look at the EPR spectrum in Fig. 4.41 in H&S. There are 15 peaks, although you expect 2 × 8 = 16 peaks. Therefore two peaks must overlap in the middle of the spectrum. (a) To calculate g-values for the two sites, measure Bsample for each of the two spectra. Locate the middle of each of the overlapping 8-line spectra in Fig. 4.41 in H&S. Site 1: Bsample = 1550 G. Site 2: Bsample = 1920 G. The method is as for problem 4.54d. g1 =

( 6 .626 × 10 − 34 J s )( 9 .214 × 10 9 s − 1 ) ( 9 .274 × 10 − 24 J T −1 )(1550 × 10 − 4 T )

= 4.25 g2 =

( 6 .626 × 10 − 34 J s )( 9 .214 × 10 9 s − 1 ) (9 .274 × 10 − 24 J T −1 )(1920 × 10 − 4 T )

= 3.43 (b) The overlapping of the two EPR spectra causes the central peak to be more intense. (c) If the g-values were closer together, the EPR spectra would overlap even more. The appearance of the overall spectrum, and the number of lines observed, depends upon the difference in the values of g1 and g2. 4.58

The technique must (i) detect Fe, (ii) be sensitive to oxidation state, and (iii) be sensitive to geometrical environment. Mössbauer spectroscopy matches all three criteria. The spectrum gives values of isomer shift (measured in mm s–1) which gives a measure of the electron density on the 57Fe atom. The oxidation state and coordination environment of Fe in clays can be found from a combination of the isomer shift and quadrupole splitting data.

4.59

(a) X-rays are scattered by the electrons surrounding the nuclei in atoms in the solid. An H atom has only one electron which is used for covalent bond formation. The electron density close to the H nucleus in a compound is therefore very small (H has no core electrons). In a covalent X–H bond, the bonding electrons are shared between X and H, and so the location of the H nucleus is based on the location of the bonding electrons (i.e. not the nucleus). As a result, an X–H bond distance determined by X-ray diffraction is typically shorter than the same separation determined by neutron diffraction (see below). (b) Neutron diffraction is used to accurately locate H atoms because neutrons are diffracted by atomic nuclei. Thus direct location of the position of the H atom (nucleus) is possible. Use of the technique is limited by the number of available neutron sources.

4.60

An amorphous material has no long-range ordering of its atoms or molecules. Polycrystalline materials (powders) consist of huge numbers of microcrystals, each of which possesses long-range ordering of the constituent atoms or molecules. In a single crystal, the crystal lattice extends continuously through the solid, and the diffraction pattern consists of well defined spots.

64

Experimental techniques 4.61

(a) PhB(OH)2 contains a trigonal planar B atom. The two OH groups can form hydrogen bonds to adjacent molecules in the solid state, forming a network: H O

H

O

H

O

B

H

O

B

B O

H

O H

(b) H2SO4 is tetrahedral with two OH groups which can form hydrogen bonds to O atoms in neighbouring molecules in the solid state. This results in an infinite network of interconnected molecules: O

O S

O

O O

H

O

H

H

O O

O

O S

S H

H

O

O

O O

H

S O O

H

H

(c) Carboxylic acids with one functional group per molecule typically form hydrogen-bonded dimers in the solid state: O

O

H

O

H

O

4.62

Use the data from a single crystal structure to calculate a powder pattern. An experimental powder pattern of the bulk sample is recorded and compared to the calculated pattern. A match means that the single crystal is representative of the bulk sample. Problems arising when comparing single crystal versus bulk sample include polymorphs and solvates. Elemental analysis or TGA can detect the presence of solvent molecules in bulk samples (see answer 4.64a).

4.63

(a) A thermal ellipsoid indicates the volume in space in which there is a probability of finding a particular atom and shows the thermal motion of the atom. The motion depends upon the amount of energy that the molecule possesses, and so can be minimized by collecting X-ray diffraction data at low temperatures. The representation of atoms by thermal ellipsoids is an anisotropic representation. The IUPAC definition of anistropy is ‘the property of molecules and materials to exhibit variations in physical properties along different molecular axes of the substance’. (b) Usually, H atoms are represented isotropically because they are not directly located; their positions are fixed in chemically sensible sites with respect to the atoms to which they are bonded, the latter having been directly located.

Experimental techniques

65

4.64

(a) A single crystal is found to be a dihydrate but the bulk sample can be shown to be anhydrous by using TGA. If the bulk sample were also a dihydrate, the water of crystallization should be released at temperatures lower than those needed to decompose the sample (see worked example 4.2 in H&S). (b) Use the data from a single crystal structure to calculate a powder pattern. When more bulk material has been prepared, record the powder pattern by X-ray diffraction and ensure that there is a good fit with that calculated from the single crystal structure. (c) Elemental analysis should confirm that the samples of compound X have the same composition. The fact that the samples are polymorphs can be confirmed using single crystal X-ray diffraction (for full structural characterization) or powder diffraction (for screening samples).

4.65

(a) The wavenumber of the IR absorption is related to the reduced mass, m, by the equation:

1 2πc

ν =

k

μ

where k is the force constant of the bond. Naturally occurring CO is predominantly 12C16O (see Appendix 5 in H&S). On going from 12C16O to 13C16O, assume that the force constant of the bond remains the same, and so only the reduced mass changes. The reduced mass, μ, is given by: 1

μ

=

m + m2 1 = 1 m1 + m2 m1m2

where m1 and m2 are the isotopic masses in kg. However, it is not necessary to convert to kg if you are dealing with a ratio as the conversion factors will cancel:

ν (12 C16 O) μ (13 C16 O) = ν (13 C16 O) μ (12 C16 O) 1 12

16

μ ( C O)

=

12 + 16 12 × 16

1 13

16

μ ( C O)

=

13 + 16 13 × 16

⎛ 12 + 16 ⎞⎛ 13 × 16 ⎞ = ⎜ ⎟⎜ ⎟ ν ( C O) ⎝ 12 × 16 ⎠⎝ 13 + 16 ⎠ 2170

13

16

ν (13 C16 O) = 2130 cm −1 Complete labelling of CO with 13C causes a shift in the absorption to lower wavenumber (lower energy) by 40 cm–1. (b) Free CO gives rise to an IR absorption at 2170 cm–1. For both [Fe(CO)4]2– and [Fe(CO)6]2+, the observed absorptions are for asymmetric stretching modes of the ion (see Figs. 3.16 and 3.18 in H&S). The values of 1788 cm–1 in [Fe(CO)4]2– and 2204 cm–1 in [Fe(CO)6]2+ are consistent with a weakening of the CO bond on going from CO to [Fe(CO)4]2– and a strengthening on going to [Fe(CO)6]2+ (see Chapter 24 in H&S).

66

Experimental techniques (c) Td symmetry for [Fe(CO)4]2– means a tetrahedral anion with four equivalent 13C atoms. Therefore in the 13C NMR spectrum, one signal is expected. D3h symmetry for Fe(CO)5 means a trigonal bipyramidal molecule with three equatorial and two axial 13C atoms. A static structure in solution would give two signals (ratio 3 : 2) in the 13C NMR spectrum, but 5-coordinate structures are often dynamic in solution (Berry pseudo-rotation). Thus the axial and equatorial sites exchange on the NMR timescale and one signal is observed. 4.66

The complex you have made is [Ru(py)6][BF4]2. (a) Elemental analysis confirms the composition of the compound, but gives no structural information. (b) In the absence of crystallographic data, the presence of the [BF4]– ion could be confirmed using 19F and/or 11B NMR spectroscopy. e.g. in the 19F NMR spectrum, a 1 : 1 : 1 : 1 signal (coupling to 11B, 80%, I = 3/2) would be observed. The chemical shift should be compared to that of an authentic sample, e.g. of NaBF4. (c) Solution 1H and/or 13C NMR spectra would confirm the equivalence of the pyridine ligands. The py ligand (4.26) coordinates through the N atom. Three signals are expected in either the 1H or 13C NMR spectrum. (d) Single crystal X-ray diffraction is used to confirm the solid state structure. (e) The ESI mass spectrum could be measured in either positive or negative mode. Negative mode should show [BF4]– at m/z = 87; the isotope pattern of the peak envelope is determined by the two isotopes of B (11B, 80% and 10B, 20%). The positive mode shows peak envelopes arising from [M – BF4]+ and [M – 2BF4]2+ at m/z = 663.2 and 288.1, respectively. Isotope patterns dominated by Ru, and by B and Ru for m/z = 663.2.

4.67

(a) The elemental composition (usually %C, H, N) is the same for X and X2. Therefore, elemental analysis does not distinguish between them. (b) 1H NMR spectroscopy is only useful if the environment of the organic group, R, is affected by dimer formation, either because of symmetry or the role of the R group in dimer formation. For example, Ph3Al dimerizes as follows:

4 3 2 N

(4.26)

2

Ph Ph

Al

Ph

Ph Ph

Al

Al

Ph Ph

In the monomer, there is one Ph environment, but in the dimer there are two (terminal and bridging, see Chapter 23 in H&S). However, the dimer is dynamic in solution at 310 K and only one Ph environment is apparent unless the spectrum is recorded at low temperatures. Thus, NMR spectroscopy is not necessarily helpful. (c) Provided the parent ion is observed in the mass spectrum, you can use this technique to distinguish between monomer and dimer. However, be aware that molecular aggregation can occur in the mass spectrometer. (d) In solution, monomer and dimer may be in equilibrium. Crystallization may favour the formation of a dimer. Thus, solid state data (irrespective of temperature) which confirms the presence of a dimer tells you nothing about solution speciation or whether a dimer will dissociate into monomers at higher temperatures.

Experimental techniques

67

4.68

Solution NMR spectroscopy (as opposed to solid state studies) is usually carried out, therefore the first criterion is to find a suitable solvent. A range of NMR active nuclei is available, 1H and 13C being the most routinely used. Information obtained from the spectra: • number of signals tells you how many environments of the observed nucleus in a molecule; • you gain information about symmetry; • coupling between nuclei helps you to build up a picture of the connectivity of the atoms which in turn gives structural information; • line widths of the signals and variable temperature NMR spectroscopy gives information about dynamic processes. For a complete answer, it is useful to give some examples.

4.69

(a) Ar Ni = 58.69 Ar Zn = 65.41 The differences above result in a difference in elemental analyses: [Zn(en)3]Cl2 %C 22.76, H 7.64, N 26.54 %C 23.25, H 7.81, N 27.12 [Ni(en)3]Cl2 Unfortunately, the experimental data (%C 23.00, H 7.71, N 26.92) lie in between these two sets of values and so do not distinguish between the complexes. (b) In, e.g. ESI MS, loss of Cl– is expected to occur to give [Ni(en)3Cl]+ or [Zn(en)3Cl]+. Both the m/z values and isotope patterns distinguish the ions: 100%

[Ni(en)3Cl]+

m/z

279 280 281 282 283 284 285 286

278 279

273 274 275 276 277

100%

m/z

[Zn(en)3Cl]+

(c) Zn2+ has a d10 electronic configuration while Ni2+ is d8. The en ligand is bidentate and [Zn(en)3]2+ and [Ni(en)3]2+ are octahedral complexes. While [Zn(en)3]2+ is diamagnetic, [Ni(en)3]2+ is paramagnetic. The solution 1H NMR spectrum of [Zn(en)3]Cl2 should be well resolved, but that of the nickel(II) complexes may be broad, and if resolved, will be paramagnetically shifted. (d) Because the atomic numbers of Zn and Ni are close (28 and 30), and both are in +2 oxidation state, it would be difficult to unambiguously distinguish between them by X-ray diffraction alone. The M–N bond lengths may differ slightly on going from Ni to Zn. (e) Atomic absorption spectroscopy (AAS) detects metals by observing the diagnostic absorption spectrum of gaseous atoms of the metal. The absorption spectrum of each metal is characteristic. 4.70

(a) Since the peak at m/z 64 appears after the volcanic eruption, it is likely to arise from SO2. The increase in intensity of the peak at m/z 34 is likely to caused by an increase in the amount of H2S.

68

Experimental techniques (b) Earth’s atmosphere (Fig. 15.1 in H&S) contains N2 (78%), O2 (21%), Ar (0.92%), CO2 (0.04%) and very small amounts of other gases. From Appendix 5 in H&S: N 14N 99.63% 15N 0.37% O 16O 99.76% 17O 0.04% 18O 0.20% Ar 36Ar 0.34% 38Ar 0.06% 40Ar 99.6% Therefore, the gas most likely to interfere with detection of an ion at m/z 34 is O2; m/z 32 for [(16O)2]+ and 34 for [(16O)(18O)]+ will be the main peaks. (c) [Ar]+ gives m/z 40; [O2]+ m/z 32, and [N2]+ m/z 28. (d) m/z 18 arises from H2O. Water vapour is produced in large amounts after a volcanic eruption. (e) Helium is a monatomic gas and is monotopic: 4He. The ion is [He]+ at m/z 4. 4.71

(a) S is in group 16 and has 6 valence electrons. [S2]– has a total of 13 valence electrons:

–

[S3]– has a total of 19 valence electrons and is bent:

–

S

S

S

S S

(b) Both ions are radicals and EPR spectroscopy detects paramagnetic species. Rotation of the ions generates approximately spherical species and therefore they behave isotropically. (c) [S3]– absorbs at 595 nm which corresponds to absorption of green light (see Fig 4.15 in H&S) and the complementary (observed) colour is violet (see Table 19.2 in H&S). (d) 370 nm is in the near UV close to the high energy end of the visible range. The absorption is broad and tails into the visible region, allowing it to contribute to the colour of the pigment. (e) Absorption by [S2]– alone (blue end of the visible) leads to a yellow colour (see Table 19.2 in H&S), while absorption of [S3]– alone (green light absorbed) gives a violet colour. By varying the ratios of the two, a range of observed colours can be produced.

69

5 Bonding in polyatomic molecules 5.1

X Normalization: see Box 1.3 and discussion with eq. 2.2 in H&S

(a), (b) Hybrid orbitals are generated by mixing atomic orbitals which are close in energy. A set of spatially-directed orbitals, for application within VB theory, is derived. The character of a hybrid orbital depends on the atomic orbitals involved and their percentage contributions, e.g. an sp2 hybrid comprises 1/3 s and 2/3p orbital character. Each hybrid orbital points along an internuclear vector or towards a lone pair within a molecule; e.g. in CH4, sp3 hybridization is used to obtain 4 equivalent hybrid orbitals, each pointing along a different C–H internuclear vector. Hybridization provides a convenient way to develop a bonding picture using localized σ-bonds; unused orbitals such as p atomic orbitals can be used to form πbonds, e.g. in CO2 (see answer 5.9). (c) Eqs 5.1 and 5.2 in H&S are:

ψ sp hybrid =

1

ψ sp hybrid =

1

2 2

(ψ 2 s + ψ 2 p x ) (ψ 2 s −ψ 2 p x )

The equations refer to two sp hybrid orbitals. For normalized wavefunctions, the sum of the squares of the normalization factors must equal unity. Check this is true. For the first wavefunction: 1 1 ψ sp hybrid = ψ 2s + ψ 2 px 2 2 2

2

⎛ 1 ⎞ ⎛ 1 ⎞ 1 1 ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ = + = 1 2 2 2 2 ⎝ ⎠ ⎝ ⎠ Similarly for the second wavefunction.

∴ Normalized

Questions 5.2-5.4: general notes

In hybridization schemes: • mixing orbitals of the same phase corresponds to constructive interference of waves; • mixing orbitals of opposite phases corresponds to destructive interference of waves; • changing the sign of a wavefunction changes its phase.

5.2

(a) The information in Fig. 5.4 in H&S only allows a qualitative answer. It is easily seen that the first (additive) combination yields the hybrid orbital shown: x

x

+

x

2px

2s

For the second hybrid, take the combinations of atomic orbitals in 2 stages. Step 1: X Watch signs! Think about phases!

x

x

x

2px

2s

Now add in the 2py character:

y

x

x +

2py

Bonding in polyatomic molecules

70

For the last hybrid, again take the orbital combinations in 2 stages. Step 1: x

x

2px

2s

X Watch signs! Think about phases!

x

Now add in the 2py character: y x

x

2py

(b) The method is as in answer 5.1b. Take each wavefunction in turn:

1

ψ sp 2 hybrid =

3

ψ 2s +

2 ψ 2 px 3

Sum of the squares of the normalization factors is: 1

ψ sp 2 hybrid =

3

ψ 2s −

1 6

ψ 2 px +

1 2

1 3

ψ 2s −

1 6

ψ 2 px −

1 2

H

H

H

1 1 1 + + = 1 ∴ Normalized 3 6 2

The method of working is as answer 5.2a, but now you must work in three dimensions; use Fig. 5.1 to help you. Equation 5.6 in H&S is:

ψ sp 3 hybrid =

C

1 1 1 + + = 1 ∴ Normalized 3 6 2

ψ 2 py

Sum of the squares of the normalization factors is: 5.3

∴ Normalized

ψ 2 py

Sum of the squares of the normalization factors is:

ψ sp 2 hybrid =

1 2 + =1 3 3

(

1 ψ 2s +ψ 2 px +ψ 2 p y +ψ 2 pz 2

)

Take the combinations below in a stepwise manner as in answer 5.2a. Because of the vector properties of the 2p orbitals, as each 2p contribution is added in, the resultant hybrid orbital changes direction:

H

z

+

z

+

y

+

x

Equation 5.7 in H&S is: y x

Fig. 5.1 The relationship between a cube and a tetrahedron (for answer 5.3). The edges of the cube coincide with a set of Cartesian axes.

ψ sp 3 hybrid =

(

1 ψ 2 s + ψ 2 p x −ψ 2 p y −ψ 2 p z 2

)

Again, consider the contributions from the atomic orbitals in a stepwise manner and watch the signs: z

+

–

y

–

x

Equations 5.8 and 5.9 in H&S can be correlated to the last two diagrams in Fig. 5.6a in H&S in a similar way to the worked answers above.

Bonding in polyatomic molecules

71

(a) Take the shaded lobes of the px and dx2–y2 orbital to point along the +x axis, and the shaded lobe of the py orbital to point along the +y axis. In the xy plane, the orbital combinations to give 4 sp2d hybrid orbitals are:

5.4

y

X Watch signs! Think about phases!

y

y x

x +

x

x +

y

y

y x

x

x

x

+

y

y

y

y x

x

x

+

y

y

y x

y

x

x

(b) Available for hybridization are one s, two p, and one d orbital. Each hybrid orbital must contain the same amount of s character; since there are 4 hybrid orbitals, each contains 25% s character. Each hybrid orbital must contain the same amount of p character, i.e. 50% p character. Each hybrid contains 25% d character. 5.5

(a) SiF4: Si, group 14, 4 valence electrons, see structure 5.1. The molecular structure is tetrahedral; 4 substituents and no lone pairs, therefore sp3 hybridization. (b) [PdCl4]2– : Square planar Pd(II) complex 5.2, therefore sp2d is appropriate. (c) NF3: N, group 15, 5 valence electrons, see structure 5.3. The molecular structure is trigonal pyramidal; 3 bonding pairs and 1 lone pair, therefore sp3 hybridization. (d) F2O: O, group 16, 6 valence electrons, see structure 5.4. The molecular structure is bent; 2 bonding pairs and 2 lone pairs, therefore sp3 hybridization. (e) [CoH5]4– is a 5-coordinate Co(I) complex. Whether this is trigonal bipyramidal or square-based pyramidal (5.5, see Fig. 10.14 in H&S), sp3d hybridization is appropriate. (f) [FeH6]4– is an octahedral Fe(II) complex (structure 5.6, see Fig. 10.14 in H&S); sp3d2 hybridization. (g) CS2: C, group 14, 4 valence electrons, see structure 5.7. The molecule is linear and the C atom is sp hybridized. (h) BF3: B, group 13, 3 valence electrons, see structure 5.8. The molecular structure is trigonal planar; 3 bonding pairs and no lone pairs, therefore sp2 hybridization.

F Si F

F F

(5.1) 2– Cl Pd

Cl Cl

Cl

(5.2) 4–

H N F

F

F F

F

(5.3)

H

O

(5.4)

Co

H

H H

(5.5)

4–

H H

Fe

H S

C

S

F

B

H

H

F

H

(5.6)

F

(5.7)

(5.8)

72

Bonding in polyatomic molecules 5.6

(a) See 2.28 and 2.29 (p. 23) for structures of cis- and trans-N2F2. Both isomers contain N in a non-linear environment with 1 lone pair, so sp2 hybridization. (b) Fig. 2.1 in H&S shows that each O atom in H2O2 is in a non-linear environment with 2 lone pairs. Hence, sp3 hybridization is appropriate.

5.7

(a) Regular geometries for a 5-coordinate species are trigonal bipyramidal and square-based pyramidal (5.9). C3v symmetry for PF5 is consistent with a trigonal bipyramidal structure (5.10). (b) VB theory describes the bonding in PF5 in terms of a set of resonance structures in which the P atom retains an octet of electrons: F

(5.9) F

P

P

F

F

F F

P

F–

F F

F F–

P

F

F

F

F

F–

F

F

F–

F F

F

(5.10)

F

F–

P

F

P

F

F

F

5.8

F

F

(a) Trigonal planar (isoelectronic with [NO3]–, see worked example 5.2 in H&S). (b) The resonance structures which contribute the most are: O O

C

O O

O

C

O

O O

(c) The bonding description is like that for and can be summarized as follows:

C O

[NO3]–

in worked example 5.2 in H&S

O O

O

C

C

O O

O

X Check that the number of electrons used in the scheme = number of valence electrons available (24)

C–O σ-bond

sp2 hybrid orbital on oxygen, occupied by a lone pair of electrons

One pair of electrons in each of two oxygen 2p orbitals

One pair of electrons for C–O π-bond

This scheme corresponds to one resonance structure (1 C=O double and 2 C–O single bonds). Each resonance structure may be similarly described. 5.9

(a) CO2 is linear (see answer 2.19b, p.22). (b) For a linear triatomic molecule, the central atom can be considered to be sp hybridized.

Bonding in polyatomic molecules

73

(c) An appropriate bonding scheme is summarized below: Carbon sp hybrid orbitals

Oxygen 2p orbital

Oxygen sp2 hybrid orbital

π-Bond

O

C

Oxygen sp2 hybrid orbitals each containing 2 electrons

O

σ-Bond σ-Bond C

O

O

(5.11)

X

H

X

X

π-Bond Carbon 2p orbitals

O

(d) The scheme shows the formation of two C=O double bonds. (e) Lewis structures are shown in 5.11. These are consistent with the bonding schemes developed using hybridized atomic orbitals in the VB model. 5.10

Read through the first part of Section 5.4 in H&S. Your answer to this problem should be constructed from the information in Section 5.4 in H&S, and should include a simple example such as an MO diagram for linear XH2.

5.11

In VB theory, localized bonds arise because a wavefunction is set up to describe each X–H interaction. Using a hybrid orbital approach for linear XH2, the X atom can be considered to be sp hybridized and each X–H interaction described as in Fig. 5.2. In MO theory, molecular orbitals are constructed using contributions from atomic orbitals of all the atoms (where this is allowed by symmetry). For linear XH2, the interactions between the atomic orbitals of X and the ligand group orbitals (LGOs) of the H----H fragment are considered (Fig. 5.3). Each of the bonding MOs possesses bonding character delocalized over all 3 atoms. Assuming X has at least 2 valence electrons (e.g. Be), then the bonding MOs are filled. The πu MOs are non-bonding, and the σg* and σu* MOs are antibonding.

H

or

H

C

H

Fig. 5.2 In linear XH2, each X–H bond can be described by the overlap of an sp hybrid orbital and H 1s orbital. Each bond comprises a localized 2c-2e interaction.

σu*

Energy

O

Oxygen 2p orbital

σg* 2py,2py (πu)

Out-of-phase 1s orbitals = LGO(2), σu

πu

In-phase 1s orbitals = LGO(1), σg

2pz (σu)

σu (H–X–H bonding)

2s (σg)

σg (H–X–H bonding)

Fig. 5.3 An MO diagram for the formation of linear XH2; the atoms are defined to lie along the z axis.

X

XH2

H X H

H----H

z

74

Bonding in polyatomic molecules 5.12 z y

O H

(a) First draw the H2O molecular framework with respect to the axis set specified in the question: see 5.12. The left-hand column in Table 5.6 in H&S lists the atomic or ligand group orbitals, and reading down each of the next 6 columns gives the % composition of each MO along with the eigenvector (i.e. sign of each contributing wavefunction). MOs can be constructed as follows, with the relative sizes of the lobes reflecting the % contributions:

H

(5.12) x

ψ1

ψ2

ψ3

ψ4

ψ5

ψ6

(b) The number of valence electrons available = 6 (from O) + 2 (from 2H) = 8. Hence MOs ψ1, ψ2, ψ3 and ψ4 are occupied. ψ1 and ψ2 provide the O–H bonding character, and ψ3 and ψ4 correspond to the lone pairs. 5.13

(a) The BH3 molecule is defined as lying in the xy plane. The H atoms lie in the nodal plane of the B 2pz orbital. There is no net overlap between the H 1s and B 2pz orbitals, so the B 2pz orbital becomes a non-bonding MO in BH3. (b) Schematic representations are:

a1 '

e'

5.14

a1 '*

e'* Antibonding MOs

Bonding MOs

Schematic representations are:

X The second e orbital is shown in Fig. 5.19 in H&S e

a1 *

e*

5.15

[NH4]+ is isoelectronic with CH4. Thus, the description of the bonding in [NH4]+ is essentially the same as in CH4. Refer to the discussion in Section 5.5 in H&S and to Figs. 5.20 and 5.21 in H&S. In each, replace C by N+ (isoelectronic species).

5.16

I – I + (a) The Lewis structures give 2c-2e localized I–I I I I I bonds in each case. This I does not explain the variI ation in bond lengths. (b) The bonding in I2 can be described as in F2 (Fig. 2.8 in H&S, replacing 2s and 2p orbitals by 5s and 5p). This gives a bond order of 1. For bent [I3]+, an MO diagram can be constructed using that for H2O as a basis (Fig. 5.15 in H&S) because I+ and O are isoelectronic in terms of valence electrons. Each I–I bond order is 1. For linear [I3]–, an MO diagram can be constructed using that for XeF2 as a basis

Bonding in polyatomic molecules

75

(Fig. 5.30 in H&S) because I– and Xe are isoelectronic. This leads to one occupied bonding MO which is delocalized over all three atoms, and therefore an I–I bond order of 1/2. The conclusion from MO theory is: • I–I bond orders: I2 ≈ [I3]+ > [I3]– • Expected trend in I–I bond lengths: I2 ≈ [I3]+ < [I3]– This agrees with the experimental values of 267 pm in I2, 268 pm in [I3]+, and 290 pm in [I3]–. 5.17

(a) For BCl3 to have D3h symmetry, it must contain a σh mirror plane, a C3 principal axis, three C2 axes and three σv mirror planes. The molecule is trigonal planar: Cl

Each of the three Cl–B–Cl bond angles is 120o.

B Cl

For NCl3 to possess C3v symmetry, it must contain a C3 principal axis and three σv mirror planes. Its structure is therefore trigonal pyramidal. Each Cl–N–Cl bond angle is Ka(2) > Ka(3) ... because it is more difficult to remove H+ from an anion than from a neutral species. (b) In structure 7.28, the [DTPA]5– ligand is drawn showing only the 8 donor atoms. A likely coordination environment for the 9-coordinate Gd3+ ion in [Gd(DTPA)(OH2)]2– is a tricapped trigonal prism or a distorted variation of this structure. This polyhedron is illustrated on the left below, and a proposed representation of the complex is shown on the right. 2–

O

N O

N

Ga O

H2O

N

O

O

The proposed structure that you draw does not have to be identical to the one illustrated above, but it must have the donor atoms in sites that are realistic in terms of the ligand structure and the coordination sphere. See Box 4.3 in H&S for an illustration of the structure of a related 9-coordinate Gd(III) complex. (c) Your answer should include comments on the following points. • The largest value of log K corresponds to the most thermodynamically stable complex. • The [DTPA]5– ligand contains N- and O-donors, both of which are hard donor atoms (hard Lewis bases). The principle of ‘hard and soft acids and bases’ (HSAB) is useful for rationalizing the observed trends in complex stabilities. In aqueous solution, complexes formed between class (a), or hard, metal ions (metal ions are Lewis acids) and ligands containing donor atoms from groups 15 and 16 show the following trends in stabilities: N >> P > As > Sb O >> S > Se > Te Class (b), or soft, metal ions prefer soft donor atoms, e.g. S rather than O, and P rather than N. Of the metal ions given in the question, Gd3+ and Fe3+ are hard acids while Ag+ is a soft metal ion. The pattern in values of log K is consistent with a favourable hard metal ion-hard ligand match, but a less favourable soft metal ion-hard donor atom combination. • [DTPA]5– is a chelating ligand, and thus is able to form particuarly stable complexes with metal ions. The values of log K of 22.5 and 27.3 are large, and illustrate the ‘chelate effect’.

114

Acids, bases and ions in aqueous solution 7.33

(a) Use of βn corresponds to an overall stability constant, in constrast to Kn which refers to an indiviual step. This, log β4 for [Pd(CN)4]2– refers to the process: Pd2+(aq) + 4[CN]–(aq)

[Pd(CN)4]2–(aq)

(b) Data available: Pd2+(aq) + 4[CN]–(aq)

[Pd(CN)4]2–(aq)

Pd(CN)2(s) + 2[CN]–(aq)

[Pd(CN)4]2–(aq)

log β4 = 62.3 log K = 20.8

The equilibrium that corresponds to Ksp is: Pd(CN)2(s)

Pd2+(aq) + 2[CN]–(aq)

Construct a Hess-type cycle that combines these reactions. In the cycle shown below, single arrows rather than equilibrium signs are used to give the cycle a sense of ‘direction’:

Pd2+(aq) + 4[CN]–(aq)

ΔGo1

[Pd(CN)4]2–(aq) . ΔGo2

ΔGo3

Pd(CN)2(s) + 2[CN]–(aq)

ΔGo3 = ΔGo2 – ΔGo1 –RT ln K3 = –RT ln K2 + RT ln K1 ln K3 = ln K2 – ln K1 log K3 = log K2 – log K1 log K3 = 20.8 – 62.3 = –41.5 7.34

K3 = Ksp = 10–41.5

(a) The pH of a 0.10 mol dm–3 aqueous CuSO4 solution is 4.17. This solution contains [Cu(OH2)6]2+ ions and the equilibrium for its acid dissociation is: [Cu(OH2)6]2+(aq) + H2O(l)

[H3O]+(aq) + [Cu(OH2)5(OH)]+(aq)

The concentration of H+ ions is determined as follows: pH = –log [H]+ = 4.17 [H]+ = 6.76 × 10–5 mol dm–3

Acids, bases and ions in aqueous solution

Ka =

115

[H 3O + ][Cu(OH 2 ) 5 (OH) + ] [Cu(OH 2 ) 6 2 + ]

=

[ H 3 O + ]2 0.1

=

(6.76 ×10 −5 ) 2 0 .1

= 4.57 ×10 −8

(b) Addition of ammonia results (overall) in: [Cu(OH2)6]2+(aq) + 4NH3(aq)

[Cu(NH3)4(OH2)2]2+(aq) + 4H2O(l)

Before the addition of NH3, [OH–] 99.9% label. Treat D2O with AlCl3 to prepare DCl: AlCl3 + 3D2O J Al(OD)3 + 3DCl Then treat Li[AlH4] with DCl. This reaction will liberate HD (but not H2 or D2). The gaseous product can be collected and a mass spectrum recorded: accurate masses of 1H = 1.008 and of 2H = 2.014. In the accurate-mass spectrum, a parent ion at m/z 3.022 is expected for HD. Any contamination with H2 or D2 will be seen from peaks at m/z 2.016 and 4.028 respectively. Fragmentation of HD, H2 and D2 leads to peaks at m/z 1.008 and 2.014. Alternatively, combustion of HD will give HDO. An accurate density measurement distinguishes this from H2O or D2O.

10.6

In dilute solutions (e.g. 0.01 mol dm–3), tert-butanol (Me3COH) is essentially monomeric, and the absorption at 3610 cm–1 is assigned to the stretch ν(OH) in an isolated molecule. The 1.0 mol dm–3 solution is concentrated enough that there will be intermolecular hydrogen bonding. Structure 10.3 illustrates this between two molecules, but hydrogen bonding can be more extensive, leading to

Hydrogen small aggregates in solution. The result of hydrogen bonding is that the covalent O–H bond is weakened, and hence there is a shift in ν (OH) to lower wavenumber (3610 to 3330 cm–1). The broadening of the band is characteristic of hydrogen bond formation since ν (OH) is no longer at one diagnostic value. 10.7

Me Me

155

Me

C O

H

O

H

C

Me Me

Me

(10.3)

As HCl is absorbed, the equilibrium: MCl(s) + HCl(g)

M[HCl2](s)

is established. The position of equilibrium is determined by the relative lattice energies of MCl and M[HCl2]. The relevant thermochemical cycle is:

MCl(s) + HCl(g)

ΔrHo(1)

ΔlatticeHo(M[HCl2], s)

ΔlatticeHo(MCl, s)

M+(g) + Cl–(g) + HCl(g) H H

ΔrHo(2)

M+(g) + [HCl2]–(g)

.

From this cycle:

O H

H O

H

O

H H

M[HCl2](s)

H

O H

(10.4)

X

δ+ H

(10.5)

Y

ΔrHo(1) = ΔrHo(2) + ΔlatticeHo(M[HCl2], s) – ΔlatticeHo(MCl, s) ΔrHo(2) is the enthalpy change for the reaction: HCl(g) + Cl–(g) J [HCl2]–(g) and is independent of the metal. ∴ ΔrHo(1) depends on the difference between the two lattice energies. Both lattice energies will be negative, but ΔlatticeHo(MCl, s) will be more negative than ΔlatticeHo(M[HCl2], s). Thus, the difference {ΔlatticeHo(M[HCl2],s) – ΔlatticeHo(MCl,s)} is always positive, meaning that the absorption of HCl is endothermic. However, the difference will be smallest (and so ΔrHo(1) least positive) for the larger M+ ion, i.e. for Cs+.

10.8

[H9O4]+ is a hydrogen-bonded species. Break the formula down into H2O and [H3O]+ units: (H2O)3([H3O]+). Suggested structure is then 10.4, with each O centre trigonal pyramidal.

10.9

Points to include: • Hydrogen bond (X–H.... Y) is formed between an H atom attached to an electronegative atom X, and an electronegative atom Y which possesses a lone pair of electrons. • Evidence for hydrogen bonding comes from physical data and structural features in the solid state. • Typical strength of a ‘normal’ hydrogen bond ≈ 20-25 kJ mol–1, but some are much stronger, e.g. 165 kJ mol–1 in [HF2]–. • The ‘normal’ hydrogen bond is considered as a weak electrostatic interaction (10.5) between H (δ+) and Y (lone pair). Some hydrogen bonds

156

Hydrogen (e.g. that in [HF2]–) are more covalent in nature; a 3c-2e bond description is appropriate. • Hydrogen-bonds may be asymmetrical (e.g. in ice or [H9O4]+, see 10.4) or symmetrical (e.g. in [HF2]–). • Formation of an X–H....Y hydrogen bond results in lengthening and weakening of the X–H covalent bond. This can be observed by a shift in IR spectroscopic band assigned to ν(XH) (see answer 10.6). • Hydrogen bonding may tend to cause association, e.g. formation of dimers (e.g. RCO2H, 10.6), or may lead to a polymeric (e.g. HF, 10.7) or a 3Dstructure (e.g. ice).

R C O

O

H H O

O C

F

R

F

H

(10.6)

H

H

H F

F

(10.7)

• Boiling points of hydrogen-bonded liquids (e.g. HF, H2O, NH3) are anomalously high compared to those of later group congeners (e.g. HF as compared to HCl, HBr, HI). Similarly, values of ΔvapH are high; melting points also affected. • Biological examples are crucial to life, e.g. formation of double-helical DNA through Watson-Crick or Hoogsteen hydrogen-bonded pairs of complementary bases. 10.10

(a)

KH + NH3 J KNH2 + H2 KH + EtOH J KOEt + H2

(b) KH is acting as a base; its conjugate acid is H2; the other conjugate acid-base pairs are NH3 with [NH2]–, and EtOH with [EtO]–. 10.11

10.12

(a)

2H2O J 2H2 + O2

(b)

2LiH J 2Li + H2

(c)

CaH2 + H2O J Ca(OH)2 + H2 (occurs at 298 K)

(d)

Mg + 2HNO3 J Mg(NO3)2 + H2 (occurs at 298 K)

(e)

2H2 + O2 J 2H2O (radical reaction that needs initiating with e.g. a spark)

(f)

CuO + H2 J Cu + H2O

(O2 formed at anode, H2 at cathode; condition is that an electrolyte is present) (H2 formed at anode, Li at cathode)

(requires heat)

Although H2O2 is thermodynamically unstable with respect to the reaction: H2O2 J H2O + 1/2O2

ΔGo = –116.7 kJ mol–1

it is kinetically stable. Ea is high, and a catalyst (e.g. MnO2) is needed to speed up the decomposition.

Hydrogen 10.13

(a) See Fig. 10.3. (b) Magnesium hydride has the formula MgH2; compare this to TiO2 (rutile). Mg2+ ions will take the sites of Ti4+ (see Fig. 10.3), therefore Mg2+ is 6-coordinate, octahedral. H– ions take the sites of O2– ions (see Fig. 10.3), and each H– ion is 3-coordinate, trigonal planar.

10.14 H H

Al

H H

H H

Al

Al H

H H

157

Fig. 10.3 A unit cell of rutile, TiO2; Ti are in the very central and the corner sites.

The text description (Section 10.7 in H&S): ‘AlH3 consists of a 3D-structure in which each Al(III) centre is in an AlH6-octahedral site; H atoms bridge pairs of Al centres.’ From this, sketch part of the structure, as in 10.8. The ratio of the coordination numbers is: Al : H = 6 : 2. It follows that the stoichiometry is: Al : H = 2 : 6 = 1 : 3 , i.e. AlH3

(10.8)

10.15

Fig. 10.4 Part of a chain in polymeric BeH2.

Figure 10.4 shows the structure of BeH2. Each Be is tetrahedrally coordinated, and each H atom bridges 2 Be. An sp3 hybridization scheme is appropriate for Be; Be–H–Be bridges formed by overlap of sp3 hybrid and H 1s orbitals. Each Be has 2 valence electrons, each H, one. This gives enough electrons to form 3c-2e bridges:

Be

H Be

Be

Be sp3 hybrid

H H

H H 1s atomic orbital

H H H

Ga

Ga H

H H

(10.9)

Ga2H6 (10.9) is isostructural with B2H6. The bridging H atoms are involved in 3c-2e interactions related to those in BeH2. 10.16

(a) The trend in bond angles can be rationalized in terms of the VSEPR model. CH4: C (group 14) uses all 4 valence electrons to form 4 C–H bonds, so no lone pairs. Tetrahedral molecule with ideal H–C–H bond angles of 109o.

N H

H H

(10.10)

NH3: N (group 15) uses 3 valence electrons for 3 N–H bonds, with one lone pair remaining. ‘Parent’ shape is tetrahedron; molecular shape is trigonal pyramidal (10.10). Lone pair–bonding pair > bonding pair–bonding pair repulsions, leading to H–N–H bond angles less than the ideal value. Consistent with observed 106.7o.

Hydrogen

158

O

H2O: O (group 16) uses 2 valence electrons for 2 O–H bonds, with 2 lone pairs remaining. ‘Parent’ shape is tetrahedron; molecular shape is bent (10.11). Lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair repulsions, leading to H–O–H bond angles less than the ideal value, and less than in NH3. Consistent with observed 104.5o.

H H

(10.11)

(b) For NH3, the dipole will act as shown in 10.12. This is determined by the greater electronegativity of N versus H, and the lone pair of electrons on N. In NH2OH (10.13), it is difficult to predict the direction of the dipole, but take into account the following: • electronegativities, χP, of N and O are 3.0 and 3.4 respectively; • N carries 1 lone pair, O has two; • the shape of NH2OH leads to the lone pairs of O facing away from the lone pair of N. The direction of the molecular dipole moment is the resultant of the bond dipole moments and those due to the lone pairs. This will not mimic that in NH3, nor will it be of the same magnitude.

δ– N H

H H

δ+

(10.12)

N

H O

H H

(10.13)

(c) Trouton’s rule: For liquid vapour:

ΔS vap =

ΔH vap bp

≈ 88 J K–1 mol–1

PH3, P2H4, SiH4, Si2H6 show ratios close to this value and obey Trouton’s rule. NH3 and N2H4 have higher values of ΔvapS due to effects of hydrogen bonding; hydrogen-bonded interactions are destroyed as the liquid vaporizes. The low value of ΔvapS for HCO2H implies vaporization is accompanied by more ordering of the system. Value lower than ≈ 88 J K–1 mol–1 implies more (or at least stronger) hydrogen bonding in vapour than in liquid; the vapour contains dimers. 10.17

The anions in [NMe4][HF2] and [NMe4][H2F3] are [HF2]– and [H2F3]–, respectively. From the data given in the question, the two H–F bonds ion [HF2]– are equal in length, and therefore this near-linear ion contains a symmetrical hydrogen-bonded interaction: F

X For details of the crystal structures of [NMe4][HF2] and [NMe4][H2F3], see: S.I. Troyanov et al. (2005) Z. Anorg. Allg. Chem., vol. 631, p. 1651.

H

–

F

The [H2F3]– ion contains short and long H...F separations. From Appendix 6 in H&S, rcov(H) = 37 pm and rcov(F) = 71pm. The distance of 89 pm corresponds to a covalent single bond; a significant ionic contribution accounts for the observed bond length being shorter than the sum of the covalent radii. Compare 89 pm with 92 pm in solid HF (Fig. 10.9 in H&S). The longer internuclear separations in [H2F3]– correspond to H...F hydrogen bonds. In [H2F3]–, the F...F...F bond angle is 125.9o. Therefore the ion can be described in terms of two HF molecules linked through an F– ion, and the structure is: 143 pm

F H F

125.9

o

H

89 pm F

Hydrogen 10.18 3' G T

A T A T A

Figure 10.13 in H&S shows the complementary base-pairs in DNA interacting through hydrogen bonds. The base-pairs are adenine and thymine (A–T) and guanine and cytosine (G–C): H

C

5'

O

N N

A

T A 3'

T G

H

N

H

H

N N

N

O

H

H

Me

N N

N O

O

H

N

C

N

N

5'

N

N

N

T

A

159

H Guanine–cytosine (G–C) base pair

Fig. 10.5 Oligonucleotides 5'CAAAGAAAAG-3' and 5'CTTTTCTTTG-3' assemble into a double helix.

10.19

The oligonucleotides 5'-CAAAGAAAAG-3' and 5'-CTTTTCTTTG-3' interact with one another through A–T and G–C base pairings (Fig. 10.5) and this results in the formation of a double helical assembly. (a) KMgH3 adopts a CaTiO3 (perovskite) structure. The unit cell is shown in Fig. 10.6. Each K+ ion is 12-coordinate with respect to the H– ions. Each Mg2+ ion is 6-coordinate with respect to the H– ions. Each H– ion is 6-coordinate (2 Mg2+ and 4 K+). (b) Construct a thermochemical cycle (Born-Haber cycle) for the formation of KMgH3. All the data needed are in the Appendices of H&S.

K(s)

ΔaHo(K, s) ΔaH (Mg, s)

IE1

K(g)

IE1 + IE2

o

Fig. 10.6 Unit cell of KMgH3. Code: K+, centre; Mg2+, dark grey; H–, pale grey.

Adenine–thymine (A–T) base pair

Mg(s) 3

/2H2(g)

3ΔaHo(H, g)

Mg(g) 3H(g)

3ΔEAHo(H, g)

K+(g) Mg2+(g)

ΔlatticeHo(KMgH3, s)

KMgH3(s)

.

–

3H (g) ΔfHo(KMgH3, s)

From the thermochemical cycle: ΔlatticeHo(KMgH3, s) = ΔfHo(KMgH3, s) – ΔaHo(K, s) – ΔaHo(Mg, s) – 3ΔaHo(H, g) – IE1(K, g) – IE1(Mg, g) – IE2(Mg, g) – 3ΔEAHo(H, g) = –278 – 90 – 146 – 3(218) – 419 – 738 – 1451 – 3(–73) = –3557 kJ mol–1 10.20

(a) From Appendix 11, the half-equations that are needed for this question are: Zn2+(aq) + 2e– 2H+(aq) + 2e– Cu2+(aq) + 2e–

Zn(s) H2(g) Cu(s)

Eo = –0.76 V Eo = 0.00 V Eo = +0.34 V

Now consider the reaction between Zn and a mineral acid, represented by H+: Zn(s) + 2H+(aq) J Zn2+(aq) + H2(g)

160

Hydrogen The value of Eocell for this reaction is given by: Eocell = 0.00 – (–0.76) = 0.76 V The reaction is spontaneous at 298 K because ΔGo(298 K) is large and negative: ΔGo = zFEocell = 2 × 96 485 × 0.76 × 10–3 = –147 kJ mol–1 If a similar reaction is considered between Cu and acid, it is found to be thermodynamically unfavourable since combination of the relevant half-equations shows Eocell for the following reaction is negative, and hence ΔGo is positive: Cu(s) + 2H+(aq) J Cu2+(aq) + H2(g) (b) The diagram on the right shows the structure of [H13O6]+. There is a symmetrical O.....H.....O in the central [H 5 O 2] + unit. This is connected through hydrogen bonds to 4 H2O units (asymmetrical interactions are likely).

Reaction does not occur H H

H

O

H

O

H

H

H O

H

H H

O

O H

O H

H

(c) The two structures for SbH3 that should be considered are: H

Sb

H

Sb

H

H

D3h Trigonal planar

X See Fig. 3.14 in H&S

10.21

H H

C3v Trigonal pyramidal

For the D3h structure, the symmetric stretch (all Sb–H bonds stretching in phase with each other) is IR inactive because this stretching mode is not accompanied by a change in dipole moment. D3h SbH3 is non-polar, and the symmetric stretch does not generate a dipole moment. For the C3v structure, the symmetric stretch is IR active. C3v SbH3 is polar, and the symmetric stretch leads to a change in dipole moment. The other three vibrational modes are IR active for both structures. (a) Set up the appropriate thermochemical cycle:

H2O(l) + H+(g)

ΔrHo = –690 kJ mol–1

[H3O]+(g) ΔsolvHo

ΔhydHo = –1091 kJ mol–1

[H3O]+(aq) = H+(aq) ΔsolvHo = –1091 – (–690) = –401 kJ mol–1 (b) The NiMH battery uses a metal alloy, LaNi5 or M'Ni5 (M' is misch metal, alloy of La, Ce, Nd and Pr). The alloy absorbs hydrogen and stores it as a hydride, e.g.

Hydrogen

161

LaNi5H6. The metal alloy is the cathode in an NiMH battery. The anode is Ni(OH)2, and the electrolyte is 30% aqueous KOH. Cell reactions (in which the alloy is M): Anode:

Ni(OH)2 + [OH]–

Cathode:

M + H2O + e–

Overall:

Ni(OH)2 + M

charge

discharge charge

discharge charge discharge

NiO(OH) + H2O + e–

MH + [OH]–

NiO(OH) + MH

During charging, hydrogen moves from anode to cathode and is stored in the metal alloy. During discharge, hydrogen is liberated from the alloy, and moves from cathode to anode. 10.22

Fig. 10.7 Unit cell of CaF2. The Ca2+ ions are shown in dark grey. The Ca2+ ion emphasized by the arrow is at the centre of one face of the unit cell (see text).

10.23

10.24

(a) Figure 10.7 shows a unit cell of CaF2. The Ca2+ ions lie at the corners of a cube and in the centre of each face. The arrow in Fig. 10.7 points at one ion in the centre of a face. This ion has four F– as nearest neighbours within the unit cell that is drawn, and four more in the next unit cell. The Ca2+ ion is therefore 8-coordinate and lies at the centre of a cubic arrangement of F– ions. Now consider the structure of Sr2RuH6. The structure is related to that of CaF2 but with Sr2+ ions replacing the F– ions, and [RuH6]4– ions replacing the Ca2+ ions. By analogy with the description above, each [RuH6]4– ion is at the centre of a cubic array of Sr2+ ions. (b)

SiCl4 + LiAlH4 J SiH4 + LiAlCl4 PPh2H + KH J H2 + K[PPh2] 4LiH + AlCl3

Et2O

LiAlH4 + 3LiCl

The correct pairings are shown in the table below: List 1 BeH2 [PtH4]2– NaH [NiH4]4– [PtH6]2– [TcH9]2–

List 2 Polymeric chain Square planar complex Saline hydride M(0) complex M(IV) complex Tricapped trigonal prism

HfH2.1 AlH3

Non-stoichiometric hydride 3D structure, octahedral Al

Comments See Fig. 10.15 in H&S Contains Pt(II) Group 1 metal hydride

See Fig. 10.14 in H&S for [ReH9]2– See answer 10.14, p. 157

(a) The formation of a solid solution indicates that ice and NH4F have related solidstate structures. Both crystallize with a wurtzite-like structure in which hydrogen bonding between H2O molecules or between [NH4]+ and F– ions is crucial to stabilizing the structures. (b) The structures of H3PO4, H2SO4 and HClO4 are shown in 10.14-10.16. The number of OH groups decreases on going from H3PO4 to H2SO4 to HClO4, and therefore, the number of possible hydrogen bonds formed per molecule decreases. The decrease in viscosity follows directly from this trend.

162

Hydrogen

O P HO

O

O OH OH

(10.14)

S O

Cl

OH OH

O

(10.15)

O OH

(10.16)

(c) For a discussion of Trouton’s constant, see answer 10.16c. For formic acid, there is stronger hydrogen bonding in the dimer in vapour phase than in the liquid and this lowers ΔvapS. (d) For a dibasic acid, it is usual that Ka(1) > Ka(2) and, therefore, pKa(1) < pKa(2). This trend O C C O is observed for both fumaric and maleic acids. – However, pKa(2) for maleic acid is larger (i.e. OH O the acid is weaker) than might be expected (10.17) because hydrogen-bonded interaction (structure + 10.17) hinders H dissociation. 10.25

(a) Appendix 12 gives bond enthalpy terms. The combustion of H2 gives water: 2H2 + O2 J 2H2O Break two H–H bonds: Break one O=O bond: Make four O–H bonds:

2 × 436 kJ mol–1 498 kJ mol–1 4 × 464 kJ mol–1

endothermic endothermic exothermic

Change in enthalpy = (2 × 436) + 498 – (4 × 464) = –486 kJ per 2 moles of H2 = –243 kJ mol–1 Mr H2 = 2.02 g

mol–1

Change in enthalpy = –243 kJ per 2.02 g = –120 kJ g–1 (b) Unit mass of compressed H2 gas and of liquid H2 refer to the same number of moles and therefore the energy storage capacities per unit mass are the same. A given number of moles of compressed H2 (at 35 MPa) occupies a greater volume than the same number of molecules of liquid H2 and therefore liquid H2 stores more energy per unit volume than compressed H2. (It is relevant to remember that both the compression and liquefaction of H2 require significant amounts of energy.) Although liquid H2 appears advantageous with respect to compressed gas, sustaining the low temperatures of cryogenic systems is probematical, resulting in necessary venting of gas as pressures increase upon warming. For the motor vehicle industry, compressed gas is superior to liquid H2. (c) The relative densities of gasoline and H2 mean that, for the same stored energy, a far greater volume of H2 than gasoline is required. Even after compression or liquefaction, there are significant volume disadvantages for H2. For a vehicle relying only on hydrogen fuel cells, the required frequency of refuelling stops is a disadvantage to consumers.

Hydrogen 10.26

163

(a) Mr NaAlH4 = 54.00 g mol–1 The %H content by weight =

4 × 1.008 × 100 = 7.47% 54.00

(b) If only steps (1) and (2) operate, then 3 moles of NaAlH4 release 4.5 moles of H2 (rather than 6 moles). The revised %H storage =

9 × 1.008 × 100 = 5.60% 3 × 54.00

(c) Thermodynamically, the decomposition is favoured but there is a large kinetic barrier. Adding the Ti dopant lowers the kinetic barrier, increasing the rate of reaction. Reversible processes are required to recycle the hydrogen storage material. 10.27

O

H H

(10.18)

(a) For the structure of ice, see Fig. 7.1 in H&S. Each H2O molecule forms four hydrogen bonds (10.18), and so each O atom is in a tetrahedral environment in a 3D-structure. The network is the same as diamond or wurtzite, and is very open. (b) Cow’s milk is >85% water. As a consequence of the open structure of ice described in (a), frozen milk has a relatively low density. On melting, the network partly collapses and some of the cavities in the structure become occupied by H2O molecules. Thus, the density increases. Conversely, on freezing, the density decreases. Thus for given mass, the volume increases upon freezing, and increases on melting and warming to room temperature. Quantitatively: Density of ice = 0.92 g cm–3 (see Section 7.2 in H&S) At 283 K, water has a density of 1.00 g cm–3 Volume =

Mass Density

At 283 K, 1 g of liquid water has a volume of 1 cm3. 1 = 1.09 cm 3. 0.92

On freezing, 1 g of ice has a volume of

10.28

(a) Nickel–metal hydride (Ni/MH) battery: at the anode:

Charge

Ni(OH)2 + [OH]–

at the cathode: M + H2O + e–

Discharge Charge Discharge

NiO(OH) + H2O + e–

MH + [OH]–

(b) The overall cell reaction is: Ni(OH)2 + [OH]– + M + H2O J NiO(OH) + H2O + MH + [OH]– i.e.

Ni(OH)2 + M J NiO(OH) + MH

Ni is oxidized from Ni(II) to Ni(III), and M is oxidized to M(I). Overall oxidation number change = +2

164

Hydrogen One H is reduced from +1 to –1. Overall oxidation number change = –2. Therefore the oxidation state changes balance. (c) The f-block metals comprise the anode ‘misch metal’, M, which can absorb hydrogen and store it in the form of a metal hydride, represented as MH in the equations above. (d) Fe is present as the steel case of the battery. It is not part of the electrochemical cell. (e) Both Co (in the misch metal alloy) and Ni (in the cell) are present in oxidized forms and can be recovered by electrodeposition, i.e. electrolysis and collection of the metal at the cathode: Co2+ + 2e–

Co

Ni2+ + 2e–

Ni

For more detail, see: T. Müller (2006) J. Power Sources, vol. 158, p. 1498.

165

11 Group 1: the alkali metals 11.1

(a)

Li Na K Rb Cs Fr

lithium sodium potassium rubidium caesium francium

(‘cesium’ in the US)

(b) ns1 11.2

X See Section 1.10 in H&S for further discussion of trends in ionization energies

Removal of the first electron is from a half-filled ns atomic orbital: M(g) J M+(g) + e–

involves: [X]ns1 J [X] where X = noble gas

Removal of the second electron is from a filled atomic orbital. For Li, this is from the 1s2 orbital; for the heavier group 1 metals, it is from an (n–1)p6 level: M+(g) J M2+(g) + e–

11.3

Fig. 11.1 A unit cell of a bodycentred cubic (bcc) structure.

11.4 X Values for the question are listed in Table 11.1 in H&S

(a) Points to include: • all group 1 metals have body-centred cubic structures; • sketch a unit cell (Fig. 11.1); • each atom is 8-coordinate (Fig. 11.1); • the lattice is not close-packed. (b) Points to include: • at 298 K, LiCl, NaCl, KCl, RbCl adopt an NaCl-type structure (see p. 88, Fig. 6.5a); • in NaCl structure, each Na+ is 6-coordinate (octahedral) with respect to Cl–, and each Cl– is 6-coordinate (octahedral) with respect to Na+ ; • NaCl structure can be considered an interpenetrating fcc Na+ and fcc Cl– ; • for CsCl, coordination number of each ion is 8 (see p. 88, Fig. 6.5b). See also Box 6.4 in H&S for a discussion of radius ratio rules which are relevant to this question, but use the rules with caution! (a) Melting points (Fig. 11.2). Points to include: • melting the solid refers to: M(s) Δ M(l) • all structures are the same type (see answer 11.3a) so a direct comparison of melting points is valid down the group; • all melting points are relatively low; • low values can be related to relatively weak M–M bonding; only one valence electron per M centre; • bcc lattice is non-close-packed, and this contributes to low melting points; • down the group, M–M bonding becomes weaker as ns–ns overlap becomes less effective (more diffuse orbitals as n increases). (b) Cation radii (Fig. 11.3). Points to include: • general trend is an increase in cation size down the group; • charge is constant (M+), so a direct comparison down the group is valid; • going from Li to Na, extra quantum shell is added; similarly from Na to K, K to Rb, and Rb to Cs.

Group 1: the alkali metals

166

175

450

150

400 125

350

100 75

300 Li Na K Rb Cs

Li Na

Fig. 11.2 Trend in the melting points of the alkali metals.

11.5 Li

Li

Li

Li

(11.1)

K Rb Cs

Fig. 11.3 Trend in the ionic radii of the alkali metals.

(a) Take Li as representative; bonding in all M2 species is similar as each M atom has one valence electron. Lewis structure for Li2 (11.1) indicates an Li–Li single bond. Valence bond theory gives a set of resonance structures 11.2, but the ionic structures will contribute very little. Li

Li+

Li

Li–

Li–

Li+

(11.2)

MO theory: construct an MO diagram using LCAO approach – consider only the valence atomic orbitals and electrons (Fig. 11.4). σu*

Energy

Fig. 11.4 MO diagram for the formation of Li2 from two Li atoms. Similar diagrams can be constructed for Na2 (using 3s valence orbitals), K2 (using 4s orbitals), Rb2 (using 5s orbitals), and Cs2 (using 6s orbitals).

2s

2s

σg Li

Li2

Li

The conclusion from the MO treatment is that there is one, fully occupied bonding MO, a bond order of 1, and the molecule is diamagnetic. (b) Bond dissociation energies decrease down the group: 110 kJ mol–1 for Li2, to 44 kJ mol–1 for Cs2. Each molecule has one filled σ MO, but ns–ns overlap depends on n (more diffuse orbitals as n increases): 2s–2s > 3s–3s > 4s–4s > 5s–5s > 6s–6s. The Li–Li bond is the strongest, and the Cs–Cs bond is the weakest. 11.6

(a)

electron capture 40 40 ⎯→18 Ar 19 K ⎯⎯⎯⎯⎯⎯

(b) Make the assumption that the decay follows only the path shown in part (a). This is not strictly correct (see eq. 11.2 in H&S). 1.0 g 40K = 0.025 moles

∴ Amount of Ar formed = 0.025 moles

Assume ideal gas at standard pressure (1 bar = 105 Pa) and temperature (273 K). Since the volume of 1 mole of gas under these conditions is 22.7 dm3: Volume of 0.025 moles = 0.025 × 22.7 = 0.57 dm3 (at 273 K, 1 bar)

Group 1: the alkali metals

167

or, use: PV = nRT to obtain a volume of gas at 298 K: V=

nRT 0.0025 × 8.314 × 298 = P 10 5

= 6.2 × 10–4 m3 = 0.62 dm3 (at 298 K, 1 bar) (c) K occurs in, e.g. some silicates. 40K decays: half-life 1.26 × 109 yr. Measure the radioactivity of mineral containing 1g of K; compare with that of any current material containing 1 g of K. Apply integrated rate equation as in worked example 3.5 in H&S. 11.7 X See also problem 11.24a

(a) Consider the Born-Haber cycle: 3ΔaHo(M, s)

3M(s)

ΔaH (N, g) o

1

/2N2(g)

3IE1

3M(g)

3M+(g)

Σ(ΔEAH ) o

N(g)

N3–(g)

ΔlatticeH(M3N, s)

M3N(s)

.

ΔfHo(M3N, s)

From this cycle: ΔfHo(M3N, s) = 3ΔaHo(M, s) + 3IE1 + ΔaHo(N, g) + Σ(ΔEAHo) + ΔlatticeH(M3N, s)

Fig. 11.5 Trends in the values of the contributions in the Born-Haber cycle for the alkali metal nitrides (see text).

Enthalpy change / kJ mol–1

Formation of N3– is independent of M, but requires a lot of energy (Σ(ΔEAHo) = 2120 kJ mol–1). This is the major factor for which the lattice energy must compensate. This is seen from Fig. 11.5; values relating to N are constant as the metal varies. Sum {3ΔaHo(M, s) + 3IE1 + ΔaHo(N, g) + Σ(ΔEAHo)} is always large and positive. Lattice energy is inversely proportional to internuclear separation; cation radii increase substantially down the group (see Fig. 11.3) and so the most negative lattice energy is for Li3N. This compound is predicted to be the most thermodynamically viable of alkali metal nitrides. (b) Assume all the M+ ions have a primary solvation number of 4 (Li+) or 6 (later M+, i.e. [M(OH2)6]+) in aqueous solution. The smallest hydrated ion with the highest charge density has the greatest attraction for further H2O solvent molecules, increasing the effective hindrance to ion movement through the solution. Cation radii (see Fig. 11.3) decrease down the group; charge density is highest for Li+. This rationalizes the order of the ionic mobilities: Li+ < Na+ < K+ < Rb+ < Cs+. (c) See answer 8.21, p. 130. 5000 4000

3ΔaHo(M, s) + 3IE1 + ΔaHo(N, g) + ΣΔEAHo

3000

2000

ΣΔEAHo

1000

3IE1 ΔaHo(N, g) 3ΔaHo(M, s)

0 Li Na K Rb Cs

168

Group 1: the alkali metals 11.8

From Table 11.2 in H&S: ΔfHo(LiI, s) = –270 kJ mol–1 ΔfHo(LiF, s) = –616 kJ mol–1

ΔfHo(NaI, s) = –288 kJ mol–1 ΔfHo(NaF, s) = –577 kJ mol–1

The possible reaction that could occur is: LiI(s) + NaF(s)

Δ

LiF(s) + NaI(s)

for which: ΔrHo(298 K) = ΔfHo(LiF, s) + ΔfHo(NaI, s) – ΔfHo(LiI, s) – ΔfHo(NaF, s) = –57 kJ mol–1 One should really look at ΔrGo, but since ΔrSo is negligible (a solid state reaction), a negative value of Δ r H o can be used to indicate that the reaction is thermodynamically favourable. 11.9

Grinding a metal halide or halo-anion with an alkali metal halide may lead to halide exchange (see problem 11.8, for example). Grinding [PtCl4]2– with KBr or KI may lead to anions such as [PtCl3Br]2– or [PtCl3I]2–. IR spectra that are recorded may therefore be of these salts (or of a mixture of anions), and may not represent IR spectroscopic data for [PtCl4]2–.

11.10

Consider the replacement of the organic Cl by F in terms of thermochemical cycle: C

Cl

+ MF

C

C+

F + MCl

+ Cl– + M+ + F–

The only variable when M changes is the difference between the lattice energies of MCl and MF. Use the lattice energy data in Table 11.2 in H&S: For KF: Difference in lattice energies = ΔU(KCl) – ΔU(KF) = –701 – (–808) = +107 kJ mol–1 For NaF: Difference in lattice energies = ΔU(NaCl) – ΔU(NaF) = –769 – (–910) = +141 kJ mol–1 Both oppose the reaction (i.e. positive enthalpy change), but for the larger metal ion (K+), the difference in lattice energies is smaller and so is less unfavourable. 11.11

The phase of the solid in equilibrium with the dissolved salt must alter at 305 K, for example: hydrate anhydrous salt The solubility of the lower temperature phase increases as the temperature is increased, and so (by Le Chatelier’s Principle) the dissolution must be an endothermic process. The solubility of the higher temperature phase decreases as the temperature increases, and so must be an exothermic process.

Group 1: the alkali metals 11.12

(1)

Li+ N3–

(2) Li+ (1)

Li+

Fig. 11.6 Part of the layer structure of lithium nitride. Layers are distinguished as being of type (1) or (2).

169

Consider ion sharing between the units of the lithium nitride structure shown in Fig. 11.6 (or Fig. 11.4a in H&S). (a)

N3– :

this lies wholly in the unit

N3–

Li+, layer (2): each is shared between 3 units as shown for the central Li+ atom in the diagram on the right: Thus, in layer (2): Number of N3– = 1 Number of Li+ = 6 × 1/3 = 2 ∴ Ratio Li+ : N3– = 2 : 1

Li+

(b) Li+ in layer (1) is shared between two layers, therefore: Number of Li+ = 2 × 1/2 = 1 For layers (1) and (2), the ratio Li+ : N3– = 3 : 1, so stoichiometry = Li3N. 11.13

Figure 11.7 shows the general MO diagram that you should construct (using the LCAO approach, see Chapter 2). Each O atom provides 6 valence electrons and so in O2 (which is paramagnetic), MOs are filled to give a configuration: σg(2s)2σu*(2s)2σg(2p)2πu(2p)4πg*(2p)2 – In [O2] , there is an extra electron to be accommodated giving: σg(2s)2σu*(2s)2σg(2p)2πu(2p)4πg*(2p)3 and MO theory rationalizes why [O2]– is paramagnetic. On going from [O2]– to [O2]2–, one more electron is added giving: σg(2s)2σu*(2s)2σg(2p)2πu(2p)4πg*(2p)4 and hence, MO theory is consistent with [O2]2– being diamagnetic.

11.14

Equation 11.21 in H&S is: 2Na Na+ + Na– This is disproportionation because it is the combination of: Na Na+ + e– – Na– Na + e

and

oxidation reduction

(a) [O2]– ; (b) [O2]2– ; (c) [O3]– ; (d) [N3]– ; (e) N3– ; (f) Na– .

11.16

Sources of information for this answer are Section 11.2 and Boxes 11.1-11.4 in H&S.

Fig. 11.7 MO diagram for the formation of [O2]n– with n = 0, 1 or 2 (see text for occupation of MOs).

Energy

11.15

σu*(2p) πg*(2p)

2p

2p

πu(2p) σg(2p) σu*(2s)

2s

2s

σg(2s) O

[O2]n–

O

170

Group 1: the alkali metals 11.17 C

N

(11.3) C

N

C

N

(11.4)

(a) The structure of [CN]– can be represented as in 11.3; 11.4 shows resonance structures. Using an MO approach, the bonding in [CN]– is described in the same way as that in CO because [CN]– and CO are isoelectronic. See Section 2.7 and Fig. 2.15 in H&S, and replace O by N–. (b) Consider a unit cell of NaCl (Fig. 6.5a, p. 88). If NaCN possesses the same structure, then Cl– ions in NaCl are replaced by [CN]–. This ion can only be treated as a sphere if it is rotating or occupies the lattice sites with random orientations; see Fig. 6.18 in H&S which shows how [NH4]+ is treated as a spherical ion. The structure of KOH was described similarly at the end of Section 11.6 in H&S.

11.18

Points to include (using information from Section 9.6 in H&S): • Dilute solutions of Na in NH3 are bright blue; the process taking place is: M J M+(NH3) + e–(NH3) where ‘(NH3)’ represents a solvated species in liquid NH3. • Very dilute solutions are paramagnetic; the magnetic susceptibility corresponds to one free electron per Na. • Saturated solutions are bronze and diamagnetic. • Increasing the concentration of Na in liquid NH3 leads to an initial decrease in molar conductivity, followed by an increase. The molar conductivity of a saturated solution is similar to that of Na metal. • Explanation of conductivity data: – at low concentrations: M J M+(NH3) + e–(NH3) – at concentrations ≈ 0.05 mol dm–3, association of M+(NH3) and e–(NH3) – metal-like behaviour at higher concentrations. • Magnetic susceptibility data at higher concentrations explained by the equilibria: 2M+(NH3) + 2e–(NH3) M2(NH3) – – M (NH3) M(NH3) + e (NH3) • Blue solutions of Na in liquid NH3 decompose slowly: 2NH3 + 2e– J 2[NH2]– + H2

11.19

(a)

NaH + H2O J NaOH + H2

(b)

(NaH is source of H–, so H2 formed by H+ + H– J H2) KOH + CH3CO2H J [CH3CO2]K + H2O (Brønsted base-acid)

(c)

2NaN3 J 2Na + 3N2

(d)

K2O2 + 2H2O J 2KOH + H2O2

Δ

but base catalyses the decomposition of H2O2: (e)

H2O2 J H2O + 1/2O2

NaF + BF3 J Na[BF4]

(NaF is a source of F–; BF3 is a Lewis acid and accepts F– to give [BF4]–) + – (f) Molten KBr, and so only K and Br ions available: At the cathode: At the anode:

K+ + e– J K 2Br– J Br2 + 2e–

(g) Aqueous solution, and so preferential release of H2 at the cathode: At the cathode: At the anode:

2H2O + 2e– J 2[OH]– + H2 2Cl– J Cl2 + 2e–

Group 1: the alkali metals

171

Fig. 11.8 Structures of M9O2 and M11O3 clusters in alkali metal suboxides. Colour code: M, black, pale grey, O.

11.20

(a) Na2O2 is colourless and the very pale yellow colour that is typically observed arises from the presence of small amounts of NaO2. (b) The paramagnetism of NaO2 arises from the presence of an unpaired electron in the [O2]– ion (see Fig. 11.7). The ground state electronic configuration of [O2]– is σg(2s)2σu*(2s)2σg(2p)2πu(2px)2πu(2py)2πg*(2p)2πg*(2p)1.

11.21

(a) Figure 11.8 shows the structures of the Rb9O2 and Cs11O3 clusters. Rb9O2 contains two octahedral units (each containing a central O atom) which are fused through one common face. Thus, 3 Rb atoms are shared, and the formula is Rb9O2. If 3 octahedra are shared through 3 common faces, a Cs11-cluster results. If each octahedral cavity contains an O atom, the result is the suboxide Cs11O3. (b) The suboxide Cs7O contains Cs atoms in addition to Cs11O3 clusters. The formula is more usefully written as Cs11O3.Cs10, i.e. Cs21O3 = Cs7O.

11.22

(a)

Li2CO3 sparingly soluble in water LiI very soluble in water Na2CO3 very soluble in water NaOH very soluble in water very soluble in water Cs2CO3 KNO3 very soluble in water Therefore, Li2CO3 is the least soluble in water. (b) RbOH very soluble in water, no decomposition NaNO3 very soluble in water, no decomposition very soluble in water, no decomposition Li2SO4 very soluble in water, no decomposition K2CO3 LiF sparingly soluble in water, no decomposition But: Na2O reacts with water: Na2O + H2O J 2NaOH (c) Li2CO3 is a sparingly soluble salt: Li2CO3(s)

2Li+(aq) + [CO3]2–(aq)

From the stoichiometry of the equation:

[Li+] = 2[CO32–]

172

Group 1: the alkali metals Ksp = 8.15 × 10–4 = [Li+]2[CO32–] = 4[CO32–]3 [CO32–] =

3

8.15 × 10 −4 4

= 0.0588 mol dm–3 Solubility of Li2CO3 = [CO32–] = 0.0588 mol dm–3 11.23

(a) 2KOH + H2SO4 J K2SO4 + H2O (b) NaOH + SO2 J NaHSO3

2NaOH + SO2 J Na2SO3 + H2O

or

(c) KOH + C2H5OH J K[C2H5O] + H2O (d) Na + (CH3)2CHOH J Na[(CH3)2HCO] + H2 (e) NaOH + CO2 J NaHCO3 450 K

(f) NaOH + CO

2NaOH + CO2 J Na2CO3 + H2O

or

HCO2Na

(g) H2C2O4 + CsOH J Cs2[C2O4] + 2H2O (h) 4NaH + BCl3 J NaBH4 + 3NaCl 11.24

(a) Construct an appropriate Born-Haber cycle:

3Na(s) 1

/2N2(g)

3ΔaHo(Na, s) ΔaH (N, g)

3Na(g)

o

3IE1 Σ(ΔEAH ) o

N(g)

3Na+(g)

ΔlatticeH(Na3N, s)

Na3N(s)

N3–(g)

.

ΔfHo(Na3N, s)

From this cycle: ΔfHo(Na3N, s) = 3ΔaHo(Na, s) + 3IE1 + ΔaHo(N, g) + Σ(ΔEAHo) + ΔlatticeH(Na3N, s) = 3(108) + 3(495.8) + 473 + 2120 – 4422 ≈ –18 kJ mol–1

This value shows that the formation of Na3N from its elements is exothermic and suggests that the reaction is favourable. For a true indication of the thermodynamic stability of Na3N, one needs to know a value of ΔfGo(Na3N, s). (b) A ccp arrangement of [NH2]– ions with Rb+ ions in octahedral holes corresponds to an NaCl-type structure. In fact, both ions are octahedrally sited and the positions

Group 1: the alkali metals

173

of the cations and anions can be interchanged in a sketch of the unit cell. Figure 11.9 shows a unit cell of RbNH2 with the ccp (fcc) arrangement of [NH2]– ions. To confirm the stoichiometry:

Fig. 11.9 A unit cell of RbNH2 with [NH2]– ions represented by dark grey spheres and Rb+ ions as pale grey spheres.

11.25

Site

Number of Rb+

Number of [NH2]–

Within unit cell

1

0

Edge site

12 × 1/4 = 3

0

Face site

0

6 × 1/2 = 3

Corner site

0

8 × 1/8 = 1

Total

1

1

The stoichiometry is therefore 1 : 1, and this corresponds to the formula RbNH2. (a) Li3N + 3H2O J 3LiOH + NH3 (b) M + O2 J M2O if M = Li, or M2O2 if M = Na, or MO2 if M = K, Rb or Cs. Whereas M2O reacts with water to give only MOH, M2O2 reacts with water to give MOH and H2O2, and MO2 reacts with water to give H2O2 and O2. Therefore, M = Li and A = Li2O. The reaction of Li with water is: 2Li + 2H2O J 2LiOH + H2 Therefore, B = H2. (a) An MO diagram for the formation of [Na2]2– is shown σu* opposite. The four valence electrons completely occupy the σg and 3s 3s σu* MOs. For the gas phase species (for which the MO σg calculation is valid), bond order = 0. In the solid state, [Na2]2– Na– Na– the [Na2]2– ion is actually a loosely associated pair of Na– ions (Na–Na = 417 pm) in the compound [Ba2+L–]Na– where HL is the cryptand 11.5. The [Na2]2– unit is stabilized by N–H.....Na– hydrogen-bonded interactions. Energy

11.26

H

H N

N

N

N

N

N H

H

N

N

H

H

(11.5)

(b) Values of enthalpies of hydration refer to the process: M+(g) J M+(aq) The charge is constant along the series Na+, K+, Rb+, but the size of the ion increases. The charge density (charge/surface area of the ion) therefore follows the order Na+ > K+ > Rb+. The hydration enthalpy is therefore the most negative for the smallest ion, Na+.

Group 1: the alkali metals

174

Fig. 11.10 The macrocyclic ether 18-crown-6. Partial rotation about the C–C and C–O single bonds leads to changes in ring conformation. Colour code: O, black, dark grey, C; pale grey, H.

11.27

O O

O

O

O O

(11.6)

X Remember! All metal nitrate salts are soluble in water

(a) The ligand 18-crown-6 is the macrocycle 11.6 (Fig. 11.10). The ligand is hexadentate and forms stable complexes with each of the alkali metal ions. The thermodynamic stabilities of the complexes [M(11.6)]+ in acetone are measured by K for the equilibrium: M+ + 11.6

[M(11.6)]+

ΔGo = –RTlnK

Values of log K increase from Li+ to K +, and then decrease from K + to Cs +, with the values for [Na(11.6)] + and [Cs(11.6)] + being the same. The radius of the cavity of ligand 11.6 is well matched to the radius of the K+ ion and this is generally cited as the reason for the optimum stability of [M(11.6)]+ complexes (M = group 1 metal) being for M+ = K+. However, one has to consider that as the conformation of the ligand changes, so too does its cavity size. For further discussion and alternative explanations, see: A.E. Martell et al. (1994) Coord. Chem. Rev., vol. 133, p. 39. (b) NaNO3, RbNO3, Cs2CO3, Na2SO4 and LiCl are soluble in water. Li2CO3 and LiF are insoluble or sparingly soluble. See also answer 11.22. The following thermochemical cycle is relevant for this question, although strictly, we should deal with changes in Gibbs energy rather than changes in enthalpy:

MX(s)

–ΔlatticeH(MX, s)

ΔsolH(MX, s)

M+(g) + X–(g) ΔhydH(M+, g) + ΔhydH(X–, g)

M+(aq) + X–(aq) The solubility of the salt is determined by the sign and magnitude of ΔsolG(MX, s), but values of ΔsolH(MX, s) also provide relevant information. From the cycle above: ΔsolH(MX, s) = ΔhydH(M+, g) + ΔhydH(X–, g) – ΔlatticeH(MX, s) For a salt M2X: ΔsolH(M2X, s) = 2ΔhydH(M+, g) + ΔhydH(X2–, g) – ΔlatticeH(M2X, s)

Group 1: the alkali metals

175

As examples, data for LiF and LiCl can be considered. Enthalpies of solvation of the ions are given in Table 7.6 in H&S. Lattice energies for alkali metal halides are given in Table 11.2 in H&S. For LiCl: ΔsolH(MX, s) = ΔhydH(M+, g) + ΔhydH(X–, g) – ΔlatticeH(MX, s) = –519 – 361 + 834 = –46 kJ mol–1 The dissolution of LiCl is exothermic and this suggests that the salt is soluble (but see below). For LiF: ΔsolH(MX, s) = ΔhydH(M+, g) + ΔhydH(X–, g) – ΔlatticeH(MX, s) = –519 – 504 + 1030 = +7 kJ mol–1 Dissolution of LiF is slightly endothermic, suggesting that the process is unfavourable. These results give the ‘right’ answers, i.e. the solubilities of LiCl and LiF are 1.5 mol per 100 g of water (at 273 K) and 0.01 mol per 100 g of water (at 291 K).

However, you must remember that some soluble salts dissolve in water in endothermic processes, e.g. at 298 K, the solubility of NaCl is 0.62 mol per 100g of water and ΔsolH(NaCl, s) = +3.9 kJ mol–1. Such difficulties are avoided if values of ΔG rather than ΔH are used. 11.28

List 1 Li3N

List 2 Formed by direct combination of the elements; has a layer structure

Comments See Fig. 11.4a in H&S

NaOH

Neutralizes aqueous HNO3; no evolution of gas

NaOH + HNO3 J NaNO3 + H2O

Cs

Reacts explosively with H2O

2Cs + 2H2O J 2CsOH + H2

Cs7O

A suboxide

See Section 11.6 in H&S, and answer 11.21

Li2CO3

Sparingly soluble in water

See answer 11.27b

NaBH4 Rb2O

Used as a reducing agent Basic compound with anti-fluorite structure

See Section 10.7 in H&S Anti-fluorite: see discussion accomanying Fig. 6.19a in H&S

Li

Has the highest IE1 of the alkali metals

See Section 11.3 in H&S

176

Group 1: the alkali metals 11.29

(a) Details for this answer come from Box 11.3 in H&S. A rechargeable lithiumion battery consists of positive LiCoO2 or LiFePO4 electrode separated from a graphite electrode by a solid electrolyte. Li+ ions migrate through the electrolyte and are intercalated into the graphite when the battery is charging. For LiCoO2: LiCoO2 + 6C(graphite)

Charge Discharge

CoO2 + LiC6

Lithium does not change oxidation state. During charging, Co is oxidized from Co(III) to Co(IV), and graphite is reduced. For the LiFePO4 electrode, Fe is oxidized from Fe(II) to Fe(III) during charging. (b) Hybrid electric vehicle (HEV) combines a lithium-ion battery to power an electric motor which works in conjunction with an internal combustion engine. The operating mode is computer controlled. A plug-in electric vehicle uses a lithium-ion battery to power an electric motor and the battery is recharged from an external power supply when the car is parked. 11.30

(a) Soda-lime glass is manufactured by fusing sand (SiO2), limestone (CaCO3) and Na2CO3 and the final glass contains 70-75% SiO2, 12-15% Na2O with added CaO and MgO. The presence of Na2O modifies the structure, converting some Si–O–Si bridges in the 3D-network of silica into terminal Si–O– units. The negative charges are balanced by Na+ ions. The properties of the glass are modified with respect to, e.g. borosilicate glass. (b) ‘Natural’ Na2CO3 is obtained from the mineral trona (Na2CO3.NaHCO3.2H2O) whereas ‘synthetic’ Na2CO3 is manufactured from limestone (CaCO3), NH3 and NaCl using the Solvay process. In this process, NH3 and CO2 are recycled as shown in Fig. 11.6 in H&S.

177

12 The group 2 metals 12.1

(a)

Be Mg Ca Sr Ba Ra

beryllium magnesium calcium strontium barium radium

(b) ns2 12.2

1 mole of dissolved Ca(OH)2 gives 1 mole of Ca2+ and 2 moles of [OH]– ions. ∴ Solubility = [Ca2+]

and

[OH–] = 2[Ca2+]

Ksp = [Ca2+][OH–]2 = 4[Ca2+]3 ∴ Solubility = [Ca2+] =

3

K sp 4

Similarly: Solubility of Mg(OH)2 =

3

=3 K sp 4

4.68 ×10 −6 = 1.05 × 10 − 2 mol dm −3 4 =3

5.61×10−12 = 1.12 ×10 − 4 mol dm −3 4

Relative solubilities of Ca(OH)2 : Mg(OH)2 ≈ 94 : 1. In extraction of Mg from seawater, Mg2+ is first precipitated as Mg(OH)2 by addition of Ca(OH)2. From the calculated solubilities, Mg(OH)2 is about 90 times less soluble than Ca(OH)2, and addition of Ca(OH)2 will provide [OH]– ions which then combine with Mg2+ and precipitate Mg(OH)2. 12.3

(a) Formation of magnesium nitride containing Mg2+ and N3– (but see Section 15.1 in H&S): 3Mg + N2

heat

Mg3N2

(b) Nitrides of group 2 (and group 1) metals liberate ammonia when they react with H2O (see Sections 15.5 and 15.6 in H&S): Mg3N2 + 6H2O J 2NH3 + 3Mg(OH)2 12.4

2–

(12.1)

(a) Consider the unit cell of NaCl (Fig. 6.5a, p. 88). In MgC2, Mg2+ ions replace Na+ ions in the NaCl lattice, and [C2]2– ions replace Cl–. A Cl– ion is spherical, but a [C2]2– ion (12.1) is not. Start with an NaCl structure – the unit cell dimensions along the three axes are the same. Now, replace the Cl– ions by [C2]2– ions so that all the [C2]2– ions are aligned in the same direction, coincident with one of the axes. The effect of packing the ions together in this way will be to elongate the unit cell in one direction. This is the direction along which the [C2]2– ions are aligned. (b) Free rotation of [CN]– or random orientations of the anions in NaCN means [CN]– is pseudo-spherical (see answer 11.17b).

178

The group 2 metals 12.5

(a) Reaction could be driven by increase in entropy: Δ [NH4]2[BeF4] ⎯⎯→ BeF2 + 2NH4F

(b) NaCl acts as a source of Cl–, and BeCl2 is Cl– acceptor, giving [BeCl4]2– : 2NaCl + BeCl2 J Na2[BeCl4] (c) Dissolving BeF2 in water leads to hydrated ions: water BeF2(s) ⎯ ⎯⎯ → [Be(OH2)4]2+(aq) + 2F–(aq)

12.6

(a) BeCl2 monomer is linear and the Be atom has two vacant 2p atomic orbitals. The left-hand diagram below shows one of these vacant orbitals. With the axis set shown, the other vacant orbital is 2py. Dimerization occurs by Cl lone pair donation into vacant orbital on Be: x

Cl Cl

Be

Cl

Cl Cl

Be

Cl

Be

Be

Cl

Cl

The dimer contains trigonal planar Be; an sp2 hybridization scheme for Be is appropriate. Each Be atom still has one vacant 2p atomic orbital. (b) Et2O is a Lewis base and BeCl2 is a Lewis acid; adduct formation occurs. BeCl2 has 2 vacant 2p orbitals (see above) and can accept 2 pairs of electrons:

OEt2 Be

Cl

Cl

z

Vacant 2px orbital

Et2O

Be

Cl Cl

BeCl2 + 2Et2O J BeCl2.2Et2O In BeCl2.2Et2O (12.2), Be is tetrahedrally coordinated and sp3 hybridization is an appropriate description. There are 4 localized σ-interactions.

(12.2)

12.7

(a) Figure 12.1 shows the unit cell of MgF2. (b) Per unit cell of MgF2 : Site

Number of Mg2+

Number of F–

Within unit cell

1

2

Corner site

8 × 1/8 = 1

0

In face of unit cell

0

4 × 1/2 = 2

Total

2

4

Stoichiometry is Mg2+ : F– = 1:2, or MgF2. 12.8

Fig. 12.1 Unit cell of MgF2 i.e. rutile-type structure. Mg2+ ions are shown as the small spheres.

The data in Table 12.4 in H&S give values of ΔfHo(298 K) for group 2 metal halides. The trends are represented here in Fig. 12.2. Points to note from data: • With a constant metal, the trend is always that ΔfHo(MF2, s) is more negative than ΔfHo(MCl2, s), than ΔfHo(MBr2, s), than ΔfHo(MI2, s). • With a constant halide (X = Cl, Br and I), the trend is for ΔfHo(MX2, s) to become more negative as group 2 is descended. • For MF2, ΔfHo(MF2, s) is least negative for M = Mg, and is approximately constant for the heavier metals.

{ΔaHo(M, s) + IE1 + IE2} / kJ mol–1

ΔfHo(298 K) / kJ mol–1

The group 2 metals

-400

MI2 -800

MBr2 MCl2

-1200

MF2 Mg

Ca

Sr

2400

2000

1600 Mg Ca

Ba

179

Sr

Ba

Fig. 12.3 Trend in values of {ΔaHo(M, s) + IE1 + IE2} for the group 2 metals.

Fig. 12.2 Trends in values of ΔfHo for group 2 metal halides.

Consider the Born-Haber cycle:

M(s) X For a diatomic X2 : 2ΔaHo(X, s) = D(X2,g)

X2(g)

ΔaHo(M, s) 2ΔaHo(X, s)

M(g)

IE1 + IE2 2ΔEAH (X, g) o

2X(g)

M2+(g)

ΔlatticeH(MX2, s)

2X–(g)

MX2(s)

.

ΔfHo(MX2, s)

ΔfHo(MX2, s) = ΔaHo(M, s) + IE1 + IE2 + 2ΔaHo(X, s) + 2ΔEAHo(X, g) + ΔlatticeH(MX2, s) With constant metal: Factors that influence ΔfHo(MX2, s) with varying X are 2ΔaHo(X, s), 2ΔEAHo(X, g) and ΔlatticeH(MX2, s). From Appendices 9 and 10 in H&S, the sum {2ΔaHo(X, s) + 2ΔEAHo(X, g)} becomes less negative from F to I. Values of ΔlatticeH(MX2, s) become less negative on going from F to I for a given M. Therefore, on going from F to I, values of ΔfHo(MX2, s) become less negative for a given M. With constant halide: Factors that influence ΔfHo(MX2, s) with varying M are ΔaHo(M, s), IE1, IE2 and ΔlatticeH(MX2, s). From Appendices 8 and 10 in H&S, the sum {ΔaHo(M, s) + IE1 + IE2} becomes less positive. The decrease from Mg to Ca is significantly larger than from Ca to Sr, or Sr to Ba (Fig. 12.3). This general trend corresponds to the observed trends in Fig. 12.2 for X = Cl, Br and I. The values of ΔlatticeH(MX2, s) become less negative from MgCl2 to BaCl2, from MgBr2 to BaBr2, and from MgI2 to BaI2. This offsets the sum {ΔaHo(M, s) + IE1 + IE2}, but the trend set by {ΔaHo(M, s) + IE1 + IE2} remains reflected in the trend in values of ΔfHo(MX2, s). For X = F, there is substantial variation in ΔlatticeH(MF2, s) on going from M = Mg to Ba. This offsets {ΔaHo(M, s) + IE1 + IE2} leaving ΔfHo(MX2, s) varying only by a small amount. 12.9

(a) Anhydrous CaCl2 is hygroscopic, and forms a hydrate, probably CaCl2.2H2O. In the presence of a lot of water, the hydrate becomes liquid (it is deliquescent). Action as a drying agent depends on the hygroscopic nature of anhydrous CaCl2. Anhydrous CaH2 reacts with water liberating H2 : CaH2 + H2O J Ca(OH)2 + H2

180

The group 2 metals

Cl Cl Cl

Be

Be Cl

n

(12.3)

12.10

CaH2 is used to remove H2O from, e.g., a solvent (suitable only if solvent does not react with CaH2). Caution! do not expose CaH2 to large amounts of water – it reacts vigorously. (b) Points to include: • BeCl2, colourless, deliquescent solid. CaCl2, colourless solid, hygroscopic, forming a hydrate which is deliquescent. • BeCl2 is a covalent solid with infinite chains (12.3); each Be is considered to be sp3 hybridized; localized bonding. • Anhydrous CaCl2 is an ionic solid; it adopts a distorted rutile structure; (MCl2, MBr2 and MI2 for M = Ca, Sr and Ba typically have layer structures or distorted lattice-types). • BeCl2 and CaCl2 are soluble in water forming [Be(OH2)4]2+ and [Ca(OH2)6]2+. • BeCl2 acts as Lewis acid; used as Friedel-Crafts catalyst; adducts illustrating Lewis acidity include [BeCl4]2– and BeCl2.2Et2O (see 12.2). • CaCl2 readily forms a hydrate – application as drying agent (see answer 12.9a). • Solubility of chlorides in polar solvents (e.g. Et2O) due to adduct formation. (a) MgCl2 + Mg J 2MgCl Mg2+ being reduced to Mg+, and Mg being oxidized to Mg+, so this is the reverse of a disproportionation reaction. Set up a suitable thermochemical cycle (see Section 6.16 in H&S): ΔrHo

Mg(s) + MgCl2(s) ΔlatticeHo(MgCl2, s) – ΔaHo(Mg, s)

Mg(g) + Mg2+(g) + 2Cl–(g)

2MgCl(s) 2ΔlatticeHo(MCl, s)

IE1 – IE2 (Mg, g)

2Mg+(g) + 2Cl–(g)

.

From the cycle: ΔrHo = IE1 – IE2 + 2ΔlatticeHo(MgCl, s) – ΔlatticeHo(MgCl2, s) + ΔaHo(Mg, s) From Appendices 8 and 10 in H&S: IE1 – IE2 (for Mg) = 737.7 – 1451 = –713.3 kJ mol–1 ΔaHo(Mg, s) = 146 kJ mol–1 ∴ IE1 – IE2 + ΔaHo = –567.3 kJ mol–1 Now consider the term {2ΔlatticeHo(MCl, s) – ΔlatticeHo(MgCl2, s)}. The lattice energy for MgCl2 will greatly exceed that of MgCl because: X Look at the Born-Landé equation in Section 6.13 in H&S

+ • z for Mg2+ is twice that of Mg+

• r0 for Mg2+ is smaller than that of Mg+ • Madelung constants for MX2 structures are ≈ 1.5 times those of MX lattices (see Table 5.4 in H&S). Thus, ΔlatticeHo(MgCl2, s) will be more negative than 2ΔlatticeHo(MCl, s), and the term {2ΔlatticeHo(MCl, s) – ΔlatticeHo(MgCl2, s)} will be significantly positive (this

The group 2 metals

181

can be quantified using the Born-Landé equation), sufficiently so to offset the –567.3 kJ mol–1 of the {IE1 – IE2 + ΔaHo} term. Thus: MgCl2 + Mg J 2MgCl is predicted to be endothermic, and an estimate of ΔrHo can be obtained using Born-Landé equation and cycle above. (b) To estimate ΔrHo for the reaction: CaCO3(calcite) J CaCO3(aragonite) dissolve each in dilute HCl, measure ΔsolHo for each, and apply a Hess cycle. In solution and after reaction, there will no longer be a distinction between the two phases of CaCO3, a common set of products being obtained.

CaCO3(calcite) + 2HCl(aq)

ΔrHo

CaCO3(aragonite) + 2HCl(aq)

ΔsolHo(1)

.

ΔsolHo(2)

Ca2+(aq) + 2Cl–(aq) + H2O(l) + CO2(g) From this cycle: 12.11

ΔrHo = ΔsolHo(1) – ΔsolHo(2)

(a) The reaction to be considered is: SrO2 + 2HCl J SrCl2 + H2O2 On the left-hand side, SrO2 acts as base, and HCl as acid. In the reverse direction. H2O2 acts as acid (forming [O2]2– and H+). Thus the conjugate acid-base pairs are: SrO2 + 2HCl J SrCl2 + H2O2 conjugate base(2)

12.12 Data: ΔfHo(SrO, s) = –592.0 kJ mol–1 ΔfHo(BaO, s) = –553.5 kJ mol–1 ΔfHo(Sr(OH)2, s) = –959.0 kJ mol–1 o ΔfH (Ba(OH)2, s) = –944.7 kJ mol–1 o ΔfH (H2O, l) = –285.5 kJ mol–1

conjugate acid (1)

conjugate base (1)

conjugate acid (2)

(b)

BaO2 + 2H2O J Ba(OH)2 + H2O2

(a)

MO(s) + H2O(l) J M(OH)2(s)

(reaction of base with a weak acid)

Only enough water is available for the stated reaction, so there is no dissolution of hydroxide. Sr(OH)2 and Ba(OH)2 dissolve in water and are strong bases. ΔrHo = ΔfHo(M(OH)2,s) – ΔfHo(MO,s) – ΔfHo(H2O,l) For M = Sr: ΔrHo = –959.0 – (–592.0) – (–285.5) = –81.5 kJ mol–1 For M = Ba: ΔrHo = –944.7 – (–553.5) – (–285.5) = –105.7 kJ mol–1 (b) For:

CaO(s) + H2O(l) J Ca(OH)2(s)

On going down group 2 from Ca to Ba, ΔrHo for : MO(s) + H2O(l) J M(OH)2(s) becomes more negative.

ΔrHo = –65 kJ mol–1

The group 2 metals

182

Set up an appropriate thermochemical cycle:

MO(s) + H2O(l)

ΔrHo

M(OH)2(s) ΔlatticeHo(M(OH)2, s)

ΔlatticeHo(MO, s)

M2+(g) + O2–(g) + H2O(l)

ΔrHo(1)

M2+(g) + 2[OH]–(g)

.

The value of ΔrHo(1) is independent of metal and so there is no need to consider this part of the cycle in more detail. As M varies, trend in ΔrHo depends on the difference between the lattice energies (or, associated enthalpy changes) of M(OH)2 and MO : ΔrHo = ΔrHo(1) + ΔlatticeHo(M(OH)2, s) – ΔlatticeHo(MO, s) 12.13

(a) Bubble CO2 through limewater; slaked lime is Ca(OH)2. (b) Reaction: Ca(OH)2(aq) + CO2(g) J CaCO3(s) + H2O(l) (c) Positive result for CO2 is the appearance of a white precipitate (CaCO3 formation), commonly referred to as a ‘milky’ appearance.

12.14 X See Section 7.12 in H&S for discussion of stability constants

O

N

O O

Mn+ log K K

Na+ 4.2 1.6 × 104

K+ 5.9 7.9 × 105

Rb+ 4.9 7.9 × 104

Mn+ log K K

Mg2+ 2.0 102

Ca2+ 4.1 1.3 × 104

Sr2+ 13.0 1013

Ba2+ >15 >1015

The ligand is 12.4, a cryptand that encapsulates Mn+. Values of K refer to the equilibrium:

O

O

Data to be discussed for the formation of the complexes [M(crypt-222)]n+ :

N

O

(12.4)

X For further details about this type of coordination, see the suggested reading at the end of Chapter 11 in H&S, under ‘Macrocyclic ligands’.

Mn+(aq) + crypt-222(aq)

[M(crypt-222)]n+(aq)

Trends to consider: • K for group 2 metal ions > for group 1 metal ions; • for group 1 M+, K varies within one order of magnitude – pattern irregular; • for group 2 M2+, K varies significantly; increases down group; Point (1): rationalized in terms of the charge on the ion; a higher charge gives a more stable complex. For points (2) and (3), consider the metal ion sizes: Mn+ Na+ rion / pm

K+ 102

Rb+ 138

Mg2+ Ca2+ 149 72

Sr2+ 100

Ba2+ 126

142

Cryptand has a characteristic cavity-size. For group 1 metals, most stable complex formed for K+, for which a good ion-size/cavity-size match is obtained; Na+ is rather small and Rb+, rather large, for crypt-222 cavity. Down group 2, rcation increases

The group 2 metals

183

with Ba2+ giving a match to the crypt-222 cavity; compare rion for K+ and Ba2+. All the group 2 metal ion complexes are relatively stable however – see point (1).

1

2

Li

Be

Na

Mg

K

Ca

12.15

Points to include: • Define ‘diagonal relationship’ in relation to periodic table (diagram 12.5). • Li shows ‘anomalous’ behaviour in group 1; it shows similarities to Mg. • ΔaHo values are similar: ΔaHo for Li and Mg = 161and 146 kJ mol–1, respectively; value for Li significantly higher than for Na and other group 1 metals. • Metallic radii are similar: rmetal(Li) = 157 pm; rion(Mg) = 160 pm. • Cation radii are similar: rion(Li+) = 76 pm; rion(Mg2+) = 72 pm. • Pauling electronegativities are similar, χP(Li) = 1.0, χP(Mg) = 1.3. • Chemical examples which relate Li to Mg rather than other group 1 metals: Li and Mg combine directly with N2 to form nitrides; Li and Mg do not form stable peroxides under conditions of combustion in O2; formation of many (synthetically important) organometallics (see Chapter 23 in H&S).

12.16

MgO is sparingly soluble in water; in terms of the common-ion effect, the presence of Mg2+ ions in the MgCl2 solution might be expected to suppress dissolution of MgO. The observation of increased solubility suggests complex formation is aiding dissolution, e.g. formation of a species such as [MgOMg]2+ or a hydrate thereof.

12.17

The ionic radius of Be2+ = 27 pm and of Mg2+ = 72 pm (see Appendix 6 in H&S). The small Be2+ ion can only accommodate 4 H2O molecules in its first coordination sphere, while 6 H2O molecules fit round the larger Mg2+ ion. The difference in coordination numbers is therefore attributed to steric effects.

12.18

The ligand H2L shown in the question is doubly deprotonated in the complex. The formula given in the question is [Ca(OH2)2(L)], but this crystallizes as a dimer that contains a centre-ofsymmetry (inversion centre, i). The Ca2+ ion is 8-coordinate, and there is a Ca2(μ-O)2 core. A possible structure is shown in 12.6, but no stereochemistry is shown. For an 8-coordinate structure, a dodecahedral or square antiprismatic geometry is most likely.

(12.5)

12.19

O O

O O

O

H2O

O Ca

O

i

H2O O

O

OH2

O

Ca

O O

OH2

O

O

(12.6)

(a) The energy released when one mole of crystalline MO (M = Mg or Ba) is formed from its constituent ions corresponds to the lattice energy, ΔU. MgO and BaO both adopt the same solid-state structure (NaCl type). Therefore the Madelung constant for the two compounds is the same. The lattice energy depends on the internuclear separation according to: ΔU ∝ −

1 r

r = rcation + ranion

184

The group 2 metals

ΔU ∝ −

1 rcation + ranion

Both compounds are oxides, therefore ranion is a constant. Therefore, since r(Mg2+) < r(Ba2+), ΔU(MgO) is more negative than ΔU(BaO). (b) When BeF2 dissolves in water, [Be(OH2)4]2+ and aquated F– ions are formed. Be2+ is small and the charge density of the ion is relatively high. Thus, the solvation energy is sufficiently large and negative to make the formation of [Be(OH2)4]2+ thermodynamically favourable. (c) At 298 K, Be has a coordination number of 12 in the hcp structure. If the coordination number is 8 when T > 1523 K, this indicates that a phase change occurs at 1523 K from hcp to bcc structure. 12.20

(a) A unit cell of CaF2 is shown in Fig. 6.6 on p. 89. Each Ca2+ ion is 8-coordinate and each F– ion is 4-coordinate. The stoichiometry of Na2S is 2 : 1, compared to 1: 2 for CaF2. Na2S adopts an antifluorite structure in which the Na+ ions occupy the 4-coordinate sites in a CaF2 structure, and the S2– ions occupy the 8-coordinate sites. (b) Isoelectronic means same number of electrons. The term is often more loosely used to refer to same number of valence electrons. C, N and O are in groups 14, 15 and 16 respectively, therefore C2–, N– and O are isoelectronic. Thus, [C3]4–, CO2 and [CN2]2– are isolectronic: 2–

C

C

C

2–

O

C

O

–

N

C

N

–

(c) In aqueous alkali, the following reaction occurs: Be(OH)2(s) + [OH]–(aq) J [Be(OH)4]2–(aq) Complex ion formation is therefore responsible for the dissolution of Be(OH)2. (d) A refractory material has a high melting point, low electrical conductivity, high thermal conductivty and is chemically inert. MgO has these properties. Two uses are in furnace linings (where MgO withstands the high temperatures without degrading) and in electric storage heaters. 12.21

(a) CaH2 + H2O J Ca(OH)2 + H2 (b) 2BeCl2 + LiAlH4 J 2BeH2 + LiCl + AlCl3 (c) CaC2 + 2H2O J C2H2 + Ca(OH)2 (d) BaO2 + H2SO4J BaSO4 + H2O2 (e) CaF2 + 2H2SO4(conc) J 2HF + Ca(HSO4)2

The group 2 metals

185

(f) MgO + H2O2 J MgO2 + H2O (g) MgCO3 (h) Mg in air 12.22

Δ

MgO + CO2 Δ

MgO + Mg3N2

(a) Since the question states that M is a group 2 metal, the flame test distinguishes M as Sr. Sr dissolves in liquid NH3 to give [Sr(NH3)6]. This decomposes according to: [Sr(NH3)6] J Sr(NH2)2 + 4NH3 + H2 A B C (b) Both MgCO3 and CaCO3 occur naturally and are widespread minerals. Mg metal is passivated and reacts with hot water, whereas Ca reacts with cold water. Metal X is therefore Ca, and the reaction with water is: Ca + 2H2O J Ca(OH)2+ H2 X D The saline hydride is CaH2: CaH2 + H2O J Ca(OH)2 + H2 The test for CO2 is: Ca(OH)2 + CO2 J CaCO3(s) + H2O

12.23 I THF

THF Ca

THF

THF I O

(12.7)

THF

I

THF

THF THF

Ba THF

THF

(a) The 6-coordinate structure is CaI2(THF)4 and the trans-isomer 12.7 is most likely on steric grounds. Since r(Ba2+) > r(Ca2+), it is possible to accommodate another THF ligand in the coordination sphere of the metal ion. Assuming a pentagonal bipyramidal structure for the 7-coordinate complex, the iodide ligands could occupy the axial or equatorial sites. On steric grounds, the trans-isomer 12.8 is the most likely. (b) The solubility of a salt is determined by the sign and magnitude of ΔsolG(MX, s) (see answer 11.27b). Sparingly soluble salts: BaSO4, MgCO3, Mg(OH)2, CaF2. Salts which are soluble in water without reaction: BeCl2, Mg(ClO4)2, BaCl2, Ca(NO3)2. Salts that dissolve but react with water: CaO, SrH2 CaO + H2O J Ca(OH)2

I

SrH2 + H2O J Sr(OH)2 + H2

(12.8)

12.24

List 1

List 2

CaCl2

Hygroscopic solid used for deicing

BeO

Crystallizes with a wurtzite-type structure

186

The group 2 metals

12.25

List 1

List 2

Be(OH)2

Amphoteric

CaO

Quicklime

CaF2

A prototype crystal structure

BaCl2

Used in qualitative analysis for sulfates

BeCl2

Polymeric in the solid state

MgO2

Strong oxidizing agent

Ca(OH)2/NaOH

Soda lime

(a)

MgCO3 + H2O2 J MgO2 + H2O + CO2 MgO + H2O2 J MgO2 + H2O

(b) pH dependence of the decomposition of MgO2: In water:

2MgO2+ 2H2O J 2Mg(OH)2 + O2

In acid:

MgO2 + 2H+ J 2Mg2+ + H2O2

In alkali, H2O2 forms [HO2]–: 12.26

then: H2O2J H2O + 1/2O2

H2O2 + [O2]– J [HO2]– + H2O

(a) Sulfur-containing emissions from coal-fired power stations arise from the sulfur content of the coal which is typically 90% Mg because Mg has a low density, making the camera body very lightweight. Mg alloys have high mechanical strength, and good corrosion resistance. (b) ‘Cathodic protection’ means providing a preferential oxidation pathway that protects the steel. The combination of water and O2 with sea-water acting as an electrolyte results in the corrosion of submerged steel or steel that is in the splash zone of waves. At pH 7, the half-equations are: Fe2+ + 2e–

Fe

E = –0.44 V

o

O2 + 2H2O + 4e–

4OH–

E[OH–] = 10–7 = +0.80 V

The net corrosion reaction is: 2Fe + O2 + 2H2O J 2Fe(OH)2 and the Fe(OH)2 precipitates (Ksp = 4.87 × 10–17). Further oxidation gives Fe2O3.H2O. If Mg blocks are attached to the steel, Mg is oxidized in preference to Fe: Mg2+ + 2e–

Mg

E = –2.37 V

More electropositive metals (e.g. Ca) are unsuitable because they react with cold water. (c) Mg burns with a brilliant white flame, the oxidation process being: 2Mg + O2 J 2MgO Mg is added to fireworks to give white sparks and to increase the brilliance of the overall display. (d) A change from steel to aluminium car-bodies has been made to reduce the weight of the vehicle and, therefore, reduces fuel consumption. Mg is added to the aluminium alloy to increase its mechanical strength and corrosion resistance, and to improve fabrication properties of the material.The Mg content in the alloy used in car manufacturing is in contrast to the >90% used in professional camera bodies (see part (a)). For mass markets, the relative prices of Al and Mg must be considered: Mg is a lot more expensive than Al. 12.28

(a) CaO and CaO.MgO are produced by calcining limestone (CaCO3) and dolomite (CaCO3.MgCO3): CaCO3

1100-1400 K

CaCO3.MgCO3

CaO + CO2

1100-1400 K

CaO.MgO + CO2

(b) Use Box 14.8 in H&S for this answer. CaO is one of the primary components of cement. In the manufacture of cement, CaO is made by calcining CaCO3 (see above) and is then heated ( ≈ 1600 K) with SiO2, Al2O3 and Fe2O3. The material

188

The group 2 metals becomes partially molten and forms a clinker composed of the mixed oxides 3CaO.SiO2, 2CaO.SiO2, 3CaO.Al2O3 and 4CaO.Al2O3.Fe2O3. The cement is hydrated and sets after an exothermic reaction, e.g. 2(3CaO.SiO2) + 6H2O J 3CaO.2SiO2.3H2O + 3Ca(OH)2 (c) Conversion of CaO to calcium carbide: CaO + 3C

2300 K

CaC2 + CO

The reaction of CaC2 with H2O gives ethyne and is used to manufacture C2H2 where coal is more important as a feedstock for the chemical industry than oil, e.g. South Africa and China. In Europe, the US and Japan, the production of CaC2 has declined as the dependence on oil in the chemical industries has increased. Production of ethyne from calcium carbide in the latter countries is for conversion to calcium cyanamide which is used as a fertilizer. (d) The wood chip to pulp conversion and recovery of NaOH and Ca(OH)2 is summarized as follows: NaOH, Wood chips

Spent liquor

aqueous Na2S

Pulp

heat

Na2CO3 Ca(OH)2 CaCO3 + 2NaOH heat CaO + CO2 H2O Ca(OH)2

189

13 The group 13 elements 13.1

(a)

B boron Al aluminium (aluminum in the US) Ga gallium In indium Tl thallium (b) Non-metallic: B; metallic: Al, Ga, In, Tl (c) ns2np1

13.2

Data from Table 13.1 in H&S are: Eo(Tl3+/Tl) = +0.72 V Eo(Tl+/Tl) = –0.34 V The potential diagram is therefore:

Potential diagrams: see Section 8.5 in H&S

Eo

Tl3+

–0.34

Tl+

Tl

+0.72

To find Eo(Tl3+/Tl+) (Eo on the potential diagram), find ΔGo for each step: For Tl3+ to Tl:

ΔGo1 = – zFEo = –(3)(0.72)F = –2.16F J mol–1

For Tl+ to Tl:

ΔGo2 = – zFEo = –(1)(–0.34)F = +0.34F J mol–1

For Tl3+ to Tl+:

ΔGo3 = ΔGo1 – ΔGo2 = –2.16F – 0.34F = –2.50F J mol–1

Eo(Tl3+/Tl+) = −

The data for the question are plotted in Figs. 13.1 and 13.2.

Ionization energy / kJ mol–1

13.3

ΔG o 3 (−2.50 F ) =− = +1.25 V zF 2F

4000 3000

IE3 IE2

2000 1000

IE1 0 B Al Ga In Tl

Fig. 13.1 Trends in values of the first three ionization energies for the group 13 elements.

Ionization energy / kJ mol–1

See Chapter 8 for more examples of this type of the calculation

2000

1500

1000

IE2

500

IE1 Be Mg Ca Sr Ba Ra

Fig. 13.2 Trends in values of the first two ionization energies for the group 2 elements.

190

The group 13 elements Group 13 elements: ground state electronic configuration ns2np1. IE2 refers to process: M+(g)

M2+(g)

and removal of one electron from an ns2 pair. Group 2 elements: ground state electronic configuration is ns2. IE2 refers to process: M+(g)

Screening effects: see Section 1.7 in H&S 13.4

M2+(g)

and removal of one electron from an ns1 configuration. Both involve removal of an electron from a singly charged ion, so a direct comparison is valid on charge grounds. The lower values of IE2 for the group 2 metals correspond to the fact that an electron is being removed from a singly occupied level. The decrease in values of IE2 on descending group 2 corresponds to an increased distance between nucleus and outer electron – ns orbital becomes more diffuse as n increases from 2 for Be to 6 for Ba. The trend in values of IE2 for the group 13 elements is less simple. Decrease from B to Al as for decrease from Be to Mg. On going from Ca to Ga, the 3d metals are present, from Sr to In, the 4d metals are present, between Ba and Tl, there are the 5d and 4f metals. Discontinuities in values of IE2 are due to the failure of the d and f-electrons (which have a low screening power) to compensate for the increase in nuclear charge. (a) Redox reaction: B reduced, Mg oxidized: B2O3(s) + 3Mg(s)

Δ

2B(s) + 3MgO(s)

(b) Al2O3 is amphoteric, Fe2O3 is basic; only Al2O3 reacts, leaving solid Fe2O3: Al2O3(s) + 3H2O(l) + 2NaOH(aq) (c) 13.5

2Na[Al(OH)4](aq) + CO2(g)

Nuclear spin data:

1H 31P 11B

H B H

H H

(13.1)

2Na[Al(OH)4](aq)

Al2O3.3H2O(s) + Na2CO3(aq) + H2O(l) I = 1/2 I = 1/2 I = 3/2

100% 100% 80.4%

(ignore 10B)

(a) [BH4]–, structure 13.1. 11B couples to 4 equivalent 1H to give a binomial quintet (Fig. 13.3). Measure JBH between any pair of adjacent lines.

Fig. 13.3 Simulated 11B NMR spectrum of [BH4]–.

Fig. 13.4 Simulated 1H NMR spectrum of [BH4]–.

(b) One signal arising from the 4 equivalent 1H; each 1H couples to the 11B nucleus; I = 3/2, so four possible spin states (+3/2, +1/2, –1/2, –3/2) giving rise to a 1 : 1 : 1 : 1 multiplet (Fig. 13.4). NB: this is not a binomial quartet. Measure JBH between any pair of adjacent lines; value must be the same as that obtained from the 11B NMR spectrum in part (a).

The group 13 elements

Me

H

Me

P

B H

(c) H3B.PMe3, structure 13.2. 11B couples to one 31P nucleus and to 3 equivalent 1H nuclei to give a doublet of binomial quartets (Fig. 13.5). See answer 4.34 (p. 54-55) for a more detailed explanation for the related H3B.PMe2Ph. (d) The notation 11B{1H} means ‘proton decoupled 11B’. 11B–1H spin-spin coupling is instrumentally removed from the 11B NMR spectrum. 11B{1H} NMR spectrum of THF.BH3 (13.3) is a singlet. See answer 4.34 (p. 54-55) for the 11B NMR spectrum of THF.BH 3

Me

H

(13.2)

H B H

O

H

(13.3)

13.6 Data (at 298 K): ΔfHo(Al2O3, s) = –1675.7 kJ mol–1 o ΔfH (Fe2O3, s) = –824.2 kJ mol–1 ΔfHo(Al, s) = 0 (standard state) ΔfHo(Fe, s) = 0 (standard state)

Fig. 13.5 Simulated 11B NMR spectrum of H3B.PMe3.

2Al(s) + Fe2O3(s)

The thermite process is:

191

Al2O3(s) + 2Fe(s)

To find ΔrHo, apply a Hess cycle. ΔrHo = ΔfHo(Al2O3, s) + 2ΔfHo(Fe, s) – ΔfHo(Fe2O3, s) – 2ΔfHo(Al, s) = –1675.7 – (–824.2) = –851.5 kJ mol–1 (per mole of Fe2O3) ΔfusHo(Fe, s) = 13.8 kJ mol–1; this gives the amount of heat energy needed to melt 1 mole of Fe. Enough energy is released in the thermite process to melt the Fe formed.

13.7

Scheme 13.4 shows BH3 acting: • as a Lewis base by donating a pair of electrons (a B–H bonding pair); • as a Lewis acid by accepting a pair of electrons. The result is the formation of two 3c-2e B–H–B bridges in B2H6. H H

H B

H

B

B

H

H

H

13.8

H H

B

H

H H

(13.4)

Ga2H6 is isostructural with B2H6, and analogous bonding pictures are appropriate. In Ga2H6, 2c-2e localized terminal Ga–H bonds and 3c-2e Ga–H–Ga bridge bonds: Overlap of H 1s orbital with two B sp3 hybrid

H H

Cl Cl

Ga

Ga

Cl Cl

(13.5)

Cl

Ga

Ga

H H

H

orbitals

H

Cl

In Ga2Cl6 (13.5), there are enough valence electrons to invoke all 2c-2e Ga–Cl interactions. The bonding is analogous to that in Al2Cl6.

The group 13 elements

192

13.9

The order of relative stabilities of adducts L.BH3 increases for L: Me2O < THF < Me2S < Me3N < Me3P < H–

Ph2P PPh2

(13.6)

Ph2 P

BH3 P Ph2

H3B

(13.7)

13.10 a Me2N

b

(a) Adding Me3N to THF.BH3 in THF: Me3N displaces THF to give Me3N.BH3 (or mixture of THF.BH3 and Me3N.BH3). In 11B NMR spectrum, the chemical shift of the signal changes; (observed change is from δ 0 ppm to δ –12 ppm). Each of THF.BH3 and Me3N.BH3 exhibits a binomial (1 : 3 : 3 : 1) quartet. (b) No displacement since Me2O is lower in the series than PMe3. Monitor with either 11B or 31P NMR spectroscopy. No change observed. (c) [BH4]– is an adduct of BH3 and H–. Since H– is higher in the series than THF, [BH4]– is stable in THF solution with respect to the formation of THF.BH3. Monitor by 11B NMR spectroscopy. No change in the signal. (d) Ph2PCH2CH2PPh2 (13.6) has two donor sites, each of which can coordinate to BH3 giving 13.7, the likely product if THF.BH3 is in excess. In 11B NMR spectrum, see a change from binomial quartet for THF.BH3 to a doublet of quartets (see Figure 12.5). If only one P B coordinate bond forms, the final 11B NMR spectrum will contain both a 1 : 3 : 3 : 1 quartet and a doublet of quartets at different chemical shifts. 31P NMR spectroscopy could also be used to follow the reaction. (a) The reaction to be considered is: N.GaH2Cl +

c H H

Ga

N

Me2N

(13.8)

For original data, see: B. Luo et al. (2005) Chem. Commun., p. 3463.

N Li Me2N

NMe2

Elimination of LiCl is likely with the tridentate ligand displacing the monodentate amine. Compound 13.8 can be proposed (m/z = 230). It has 3 13C environments (a, b, c in 13.8), consistent with the NMR spectroscopic data in the question. The 1H NMR spectroscopic data are also consistent: δ 4.90 ppm (s, 2H) = H atoms bonded to Ga δ 3.10 (t, 4H) and 2.36 ppm (t, 4H) = two CH2 groups (b, c) with coupling between them δ 2.08 (s, 12H) ppm = Me protons (a) (b) The tridentate ligand is an anion and retains a lone pair of electrons on the central N atom. The two terminal NMe2 groups use lone pairs to coordinate to the Ga(III) centre. Therefore, compound 13.8 can act as a Lewis base as follows: Me2N H H

Ga

Me2N

13.11 For a figure of the spectrum, see: E. Bernhardt et al. (2001) Chem. Eur. J., vol. 7, p. 4696.

Me2N N

Me3N.GaH3

H

Me3N

H

Ga

N

GaH3

Me2N

K[B(CF3)4] contains the [B(CF3)4]– ion with a tetrahedral B atom and 4 equivalent CF3 groups. 19F is 100 % abundant, I = 1/2. Therefore, the 11B nucleus couples to 12 equivalent 19F to give a binomial 13-line signal in the 11B NMR spectrum. From a Pascal’s triangle, the relative intensities of the central to outside lines in the multiplet is 924 : 1.

The group 13 elements 13.12

193

The solvolysis of K[B(CF3)4] in concentrated H2SO4 generates (F3C)3BCO. (a) The reaction is: conc. H2SO4

[B(CF3)4]– + [H3O]+

(F3C)3BCO + 3HF

(b) C3v symmetry would require the CF3 groups to be positioned so that the molecule contains a C3 rotational axis and three mirror planes containing the C3 axis. One of the two possible structures that has C3v symmetry is shown in two views below:

Looking down the OCB bonds (i.e. down the C3 axis)

If the CF3 groups are rotated by the same amount and in the same direction, the molecule retains a C3 axis but loses the σv planes. The symmetry is lowered to C3:

For spectroscopic and structural details, see: M. Finze et al. (2002) J. Am. Chem. Soc., vol. 124, p. 15385.

13.13

Looking down the OCB bonds (i.e. down the C3 axis)

(a) Can be rationalized in terms of smaller B atom being sterically protected by four OH groups from attack; larger Al allows formation of 5-coordinate intermediate. (b) For hydrolysis of B2H6:

(

Rate ∝ PB 2 H 6

)2 (PH O ) 1

2

Reaction scheme must show that the rate is order 1/2 with respect to B2H6, and first order with respect to H2O. Reaction steps consistent with this are: B2H6 BH3 + H2O

2BH3 slow

products

RDS

(c) In water, the equilibrium for B(OH)3 is: B(OH)3 + 2H2O

[B(OH)4]– + [H3O]+

This is neutral to bromocresol green (pH range 3.8-4.5).

pKa = 9.1

The group 13 elements

194

In water, [HF2]– is in equilibrium: [HF2]– + H2O

[H3O]+ + 2F–

pKa = 3.45

and appears acidic to bromocresol green. When excess B(OH)3 is added to K[HF2]: B(OH)3 + 2[HF2]–

[BF4]– + 2H2O + [OH]–

and the solution appears alkaline to the indicator. 13.14

13.15

13.16 3–

Cl Cl

Cl

M

Cl

Cl Cl

(13.9) 2–

Cl Cl

Cl

M

Cl Cl

(13.10) –

Br M Br

Br Br

(13.11)

(a)

BCl3 + EtOH

(b)

BF3 + EtOH

(c)

BCl3 + 3PhNH2

(d)

BF3 + KF

B(OEt)3 + 3HCl EtOH.BF3 B(NHPh)3 + 3HCl

KBF4

(a) Cryolite: Na3[AlF6]. (b) Perovskite: CaTiO3. (c) Rewrite Na3[AlF6] as Na2[NaAlF6] giving a formula that is equivalent to NaXF3 in which X is Na or Al. The formula NaXF3 can be related to CaTiO3. Cryolite possesses a perovskite lattice (Fig. 13.6) with 2/3 of the Na in the Ca sites, and the Al and 1/3 of the Na in the Ti sites.

(Ethoxide for chloride displacement) (Adduct formation) (B(NHPh)3 stabilized by B–N π-interactions) (Contains [BF4]– formed by BF3 accepting F–)

Ca2+

Ti4+

Fig. 13.6 Unit cell of perovskite, CaTiO3.

(a) Group 13 element has 3 valence electrons; add extra electron(s) from negative charges. [MBr6]3– Electrons in valence shell of M = 3 + 3 = 6 Number of bonding pairs (6 M–Br bonds) = 6 Total number of electron pairs = 6 (no lone pairs) ‘Parent’ shape = molecular shape = octahedral; see structure 13.9. [MCl5]2– Electrons in valence shell of M = 3 + 2 = 5 Number of bonding pairs (5 M–Cl bonds) = 5 Total number of electron pairs = 5 (no lone pairs) ‘Parent’ shape = molecular shape = trigonal bipyramidal; see structure 13.10. [MBr4]– Electrons in valence shell of M = 3 + 1 = 4 Number of bonding pairs (4 M–Br bonds) = 4 Total number of electron pairs = 4 (no lone pairs) ‘Parent’ shape = molecular shape = tetrahedral; see structure 13.11. (b) VSEPR suggests that structures should be trigonal bipyramidal. For 5-coordination, energy difference between trigonal bipyramidal and square-based pyramidal structures is usually small. Balance could be tipped in the solid state by crystal packing effects.

The group 13 elements

195

(c) Compounds contain chloro anions, and can be prepared from TlCl3 (Lewis acid-base reactions). A source of Cl– is needed, as well as a cation: TlCl3 + H2N(CH2)5NH2 + 2HCl

Ga Cl

Cl

2TlCl3 + 3CsCl

(13.12)

[H3N(CH2)5NH3]2+ + [TlCl5]2–

3Cs+ + [Tl2Cl9]3–

(d) Monomeric GaCl2 would be paramagnetic (13.12); since ‘GaCl2’ is diamagnetic, formulation Ga[GaCl4] containing Ga+ and [GaCl4]– ions is correct. 13.17 Cl Ge Cl

AlF3 + 3F– Cl Cl

Cl Ga

[AlF6]3–

On adding BF3, [BF4]– is formed in preference to [AlF6]3–, i.e. BF3 accepts F– from [AlF6]3–. This regenerates AlF3 which precipitates. (b) The data (i.e. 4 absorptions at similar wavenumbers, close values for Ga species, and shifted a little for Ge species) indicate a common species for GaCl2 and GaCl3/ HCl, and one which is isostructural with GeCl4 (13.13, Ge is group 14). The gallium species must be [GaCl4]– (13.14, Ga is group 13). GaCl3 reacts with Cl– to give [GaCl4]–, and GaCl2 exists as Ga[GaCl4], there being only one Ga–Cl containing species according to the Raman spectroscopic data. (c) Group 1 metal triiodides are M+[I3]–, so solid TlI3 is Tl+[I3]–. Hydrated Tl2O3 is insoluble, and oxidation of Tl+(aq) to solid Tl2O3 is very much easier than to Tl3+(aq) (see Section 8.3 in H&S). The oxidizing agent is I2, from:

(13.13)

Cl

(a) Dissolution requires source of F– to form complex ion (i.e. requires KF to be present in HF, see Section 8.7 in H&S for self-ionization of anhydrous HF):

Cl Cl

(13.14)

[I3]– 13.18

Fig. 13.7 The structure of [Al(BH4)4]– in the solid state as determined for [Ph3MeP][Al(BH4)4].

13.19

I2 + I–

In the solid state (Fig. 13.7), [Al(BH4)4]– has two H environments, terminal and bridging. The data in the question refer to [Al(BH4)4]– in solution. (a) 298 K to 203 K: one broad signal (in addition to cation protons) means that the terminal and bridging H atoms are involved in a dynamic process that makes them equivalent on the NMR spectroscopic timescale. (b) All 11B nuclei are equivalent giving rise to one signal in the 11B NMR spectrum. The quintet (analogous to Fig. 13.3, p. 190) is due to coupling of the 11B nucleus to 4 equivalent 1H nuclei, indicating that the dynamic process involves exchange of terminal and bridging H about a single B atom. Notice that each 11B nucleus ‘sees’ only 4 1H nuclei, and so the H atoms are not fluxional over the whole molecule (compare with [B3H8]– in answer 13.27b). (c) IR and NMR spectroscopic timescales are not the same. On the NMR timescale, the anion appears fluxional, but on the IR timescale, the anion appears to be static. The IR spectroscopic data are consistent with a structure like that in the solid state. The structure of one molecule of B2(OH)4 is shown in 13.15. Use Fig. 3.10 in H&S to determine the point group: • B2(OH)4 does not belong to a special point group. • The principal axis is C2. • There are no axes perpendicular to the principal axis. • There is a σh plane. The point group is C2h.

H O

H

B

O

O

B

H

O

(13.15)

H

The group 13 elements

196

13.20

(a) In aqueous solution, mineral acids such as HCl or H2SO4 dissociate by proton transfer to H2O, either fully or partially dissociated: HCl(aq) + H2O(l)

[H3O]+(aq) + Cl–(aq)

H2SO4(aq) + H2O(l)

[H3O]+(aq) + [HSO4]–(aq)

[HSO4]–(aq) + H2O(l) HO

–

OH

O

B

B

O

O B HO

O B

O

[H3O]+(aq) + [SO4]2–(aq)

In these examples, each acid behaves as a Brønsted acid. B(OH)3, on the other hand, behaves as a Lewis acid: B(OH)3(aq) + H2O(l)

– OH

(13.16)

13.21

[B(OH)4]–(aq) + [H3O]+(aq)

(b) Although the formula of borax is often written as Na2B4O7.10H2O, this is not informative in terms of structure. It is better written as Na2[B4O5(OH)4].8H2O. The solid state structure contains anions 13.16 and interconnected hydrated Na+ ions.

α-Aluminia (corundum): • hcp array of O2– ions with cations occupying 2/3 of the octahedral interstitial sites; • extremely hard; • relatively unreactive; • density = 4.0 g cm–3 γ-Aluminia (activated alumina) • defect spinel structure; • density = 3.5 g cm–3; • important for its catalytic and adsorbing properties; • amphoteric as shown by reaction with alkali and acid: γ-Al2O3 + 3H2O + 2[OH]–

2[Al(OH)4]–

γ-Al2O3 + 3H2O + 6[H3O]+ 13.22

2[Al(OH2)6]3+

(a) The dehydration of EtB(OH)2 by intermolecular H2O elimination leads to a cyclic product: O

Et

3EtB(OH)2

– 3H2O

B

Et B

O

O B Et

The reaction of ClB(NMe2)2 with Na results in reduction, NaCl elimination and the formation of a B–B bond: Me2N

2ClB(NMe2)2

NMe2

2Na B Me2N

+ 2NaCl

B NMe2

The group 13 elements

197

SbF5 is a fluoride ion acceptor, removing F– from [(C2F5)3BF]– : K[(C2F5)3BF] + SbF5

(C2F5)3B + K[SbF6]

(b) Dimer formation by PhB(OH)2 involves hydrogen bonds as shown in structure 13.17. This interaction is analogous to that found for carboxylic acids. Dimers further associate into a 3-dimensional network through further hydrogen-bonded interactions. Part of the solid state structure is shown in 13.18 (data: S.J. Rettig and J. Trotter (1977) Can. J. Chem., vol. 55, p. 3071).

H H

O

O B

B O

O

H

H

(13.17)

(13.18)

13.23

144 pm

(13.19) H H

H

B

H

Points to include: • Borazine, (HBNH)3, has planar structure 13.19 with equal B–N bond lengths. • Bonding can be represented by a pair of resonance structures of which 13.20 is one; B– and N+ are isoelectronic with C; (HBNH)3 is isoelectronic (and isostructural) with C6H6. • In C6H6, all C atoms carry same partial charge; in (HBNH)3, difference in electronegativities of B and N makes B δ+ and N δ–; B is susceptible to nucleophilic attack, and N susceptible to electrophilic attack. NB: compare actual partial charges with the formal charges in VB theory. • Unlike benzene (which is kinetically inert towards attack by e.g. HCl and H2O), borazine undergoes addition reactions, e.g.

N

N

(HBNH)3 + 3HCl

B

B

(HBNH)3 + 3H2O

N

H

H

(13.20)

(ClHBNH2)3

[H(HO)BNH2]3 (13.21) • Fully hydrogenated derivative made by substitution: (ClHBNH2)3

Na[BH4]

(H2BNH2)3

• Saturated products have chair-conformation of B3N3-ring; analogues of cyclo-C6H12. 13.24

H HO

OH H

B

H

N

HO

B N

H HO

N

OH

B

H OH

H

(13.21)

Each compound can be described in terms of localized 2c-2e bonds, utilizing a lone pair of electrons from each N and vacant orbital on each Al centre. Each Al and N centre can be considered to be sp3 hybridized, although bond angles in each

The group 13 elements

198

compound do not exactly match idealized hybrid model. Diagram 13.22 shows a bonding representation using N Al coordinate bonds. An alternative is to draw resonance structure 13.23. R Al

N R'

R' N

Al R

R' N

Al R

R Al

R Al

N R'

N R'

(13.22)

13.25

R' N

Al R

R' N

Al R

R Al

N R'

(13.23)

First, use the elemental analysis to obtain the C : H : Cl ratio in the product: C : H : Cl =

38 .74 7.59 19 .06 : : 12 .01 1.01 35 .45

= 3.23 : 7 .51 : 0 .54 = 6 : 14 : 1 Cl

H P

Cl

R

Ga Ga

H

Cl

The information in the question allows you to write the following:

P

2GaCl3 + 2KP(H)SitBu3

R

Cl

where X is a 4-membered, cyclic compound. It is reasonable to propose that the reactants in a 2 : 2 ratio in order to give a 4-membered ring. A likely structure for X is shown in 13.24. This is consistent with the elemental analysis: C : H : Cl = 24 : 56 : 4 = 6 : 14 : 1. The two isomers arise from the possible orientations of the substituents, above and below the ring, as shown in structures 13.25 and 13.26.

(13.25) Cl

H

Ga

P R

Cl

R

Cl P

Ga Cl

H

(13.26)

13.26

= BH

(13.27)

= BH

(13.28)

2KCl + X Cl2 Ga (tBu3Si)HP

PH(SitBu3) Ga Cl2

(13.24)

B5H9

There are 5 {BH}-units and 4 additional H atoms. Each {BH}-unit provides 2 electrons, each H atom provides 1 electron. Total number cage-bonding electrons = (5 × 2) + 4 = 14 electrons = 7 pairs B5H9 has 7 pairs of electrons with which to bond 5 {BH}-units. This means that there are (n + 2) pairs of electrons for n vertices. ∴ B5H9 is a nido-cage, derived from a parent 6-vertex deltahedron, see 13.27. 4 extra H atoms adopt B–H–B bridging positions around the square face.

[B8H8]2– There are 8 {BH}-units and no additional H atoms. Each {BH}-unit provides 2 electrons. 2 electrons from the 2– charge. Total number cage-bonding electrons = (8 × 2) + 2 = 18 electrons = 9 pairs [B8H8]2– has 9 pairs of electrons with which to bond 8 {BH}-units. This means that there are (n + 1) pairs of electrons for n vertices. ∴ [B8H8]2– is a closo-cage, an 8-vertex deltahedron (dodecahedron), see 13.28.

The group 13 elements 1 2

7 = BH = CH

12

(13.29)

= BH

(13.30)

199

C2B10H12 There are 10 {BH}-units and 2 {CH}-units. Each {BH}-unit provides 2 electrons; each {CH}-unit provides 3 electrons. Total number cage-bonding electrons = (10 × 2) + (2 × 3) = 26 electrons = 13 pairs C2B10H12 has 13 pairs of electrons with which to bond 12 cluster-units. This means that there are (n + 1) pairs of electrons for n vertices. ∴ C2B10H12 is a closo-cage, a 12-vertex deltahedron (icosahedron), see 13.29. Structure 13.29 shows the 1,2-isomer with the C atoms adjacent to each other. Cage atom numbering follows the scheme shown in 13.29, with atoms 2-6 being around the top pentagonal ring, and atoms 7-11 around the lower pentagonal ring. Other isomers of C2B10H12 are 1,7- and 1,12-C2B10H12 in which the C atoms are remote from each other. [B6H9]– There are 6 {BH}-units and 3 additional H atoms. Each {BH}-unit provides 2 valence electrons; each H provides 1 electron. 1 electron from 1– charge. Total number cage-bonding electrons = (6 × 2) + 3 + 1 = 16 electrons = 8 pairs [B6H9]– has 8 pairs of electrons with which to bond 6 {BH}-units. This means that there are (n + 2) pairs of electrons for n vertices. ∴ [B6H9]– is a nido-cage, derived from a parent 7-vertex deltahedron, see 13.30. 3 extra H atoms adopt B–H–B bridging positions around the open face. In a static model, two arrangements of the bridging H atoms could be suggested: Thick line represents B–H–B bridge

In practice, the anion is fluxional in solution. 13.27

−

+

e (a) Reaction to consider: B5H9 ⎯2⎯→ [B5H9]2– ⎯2H ⎯ ⎯ ⎯→ B5H11 From Wade’s rules, adding 2 electrons means parent deltahedron changes from being n = 6 (for B5H9, see answer 13.26) to n = 7 (for B5H11). Therefore, a change from a nido- to arachno-cage is predicted (and is observed).

2e–

Break one B–B edge

H

H B H

H H

B

B

H

H

(13.31)

H

(b) Diagram 13.31 shows the static structure of [B3H8]–. There are 2 B environments, therefore this is inconsistent with the solution 11B NMR spectrum. The anion must be undergoing a dynamic process in solution, with every B centre ‘seeing’ all 8 H atoms. This gives a nonet (Fig. 13.8).

Fig. 13.8 Simulated 11B NMR spectrum of [B3H8]–. The outer lines of the binomial nonet are just visible; relative intensities of lines 1 : 8 : 28 : 56 : 70 : 56 : 28 : 8 : 1.

200

The group 13 elements

Apical B atom

2B5H9 B10H16 + H2 c) Intermolecular elimination of H2: suggests cage-coupling by formation of inter-cage B–B bond: 2 different B sites in B5H9 (Fig. 13.9). Coupling may be Bapical–Bapical, Bbasal–Bapical or Bbasal–Bbasal, so 3 possible isomers of B10H16:

Basal B atom

Fig. 13.9 Structure of B5H9.

13.28

(a) Electrophilic substitution; initial attack at apical B atom (see Fig. 13.9): –HBr

B5H9 + Br2

1-BrB5H8

isomerization

2-BrB5H8

(b) PF3 is Lewis base; attacks B (Lewis acid), displacing H2; electron count for the cluster remains unchanged — no structural change for the cage: B4H10 + PF3

B4H8(PF3) + H2

(c) KH supplies H–, a base which deprotonates the cluster by removal of bridging H+: 1-BrB5H8

KH, 195 K

K+[1-BrB5H7]– + H2

(d) Reaction with ROH degrades cluster; B–H and B–B interactions are destroyed, but not the B–C bond; H2 is liberated: 2-MeB5H8 + 14ROH 13.29

13.30

4B(OR)3 + MeB(OR)2 + 11H2

The structure of Ag2[B12Cl12] is based on an antifluorite-type arrangement. Whereas a fluorite (CaF2) structure has a cation : anion ratio 1 : 2, an anti-fluorite has a cation : anion ratio of 2 : 1. Figure 13.10 shows a unit cell of CaF2. Taking this as a starting point, the unit cell of Ag2[B12Cl12] is drawn by placing Ag+ ions on the sites occupied by F– in Fig. 13.10, and placing [B12Cl12]2– ions on the sites occupied by Ca2+ ions. In the ideal lattice, each Ag+ ion is 4-coordinate (tetrahedral). (a) In each reaction, Ga(I) is oxidized to Ga(III): The second reactant therefore undergoes reduction: Ga+ + [I3]–

Ga3+ + 3I–

Ga+ + Br2

Ga3+ + 2Br–

Fig. 13.10 Unit cell of CaF2 (fluorite); Ca2+ ions are shown in dark grey.

Ga+

Reduction is: or Reduction is:

Ga3+ + 2e–

I2 + 2e– 2I– [I3]– + 2e– 3I– Br2 + 2e– 2Br–

Ga+ + 2[Fe(CN)6]3–

Ga3+ + 2[Fe(CN)6]4– Reduction is: [FeIII(CN)6]3– + e–

[FeII(CN)6]4–

Ga+ + 2[Fe(bpy)3]3+

Ga3+ + 2[Fe(bpy)3]2+ Reduction is: [FeIII(bpy)3]3+ + e–

[FeII(bpy)3]2+

88

Abundance / ppb

Fig. 13.11 Relative abundances of the group 13 elements in the Earth’s crust plotted (a) on a logarithmic scale and (b) in parts per billion.

log (abundance)

The group 13 elements

66 44

201

90 000 000

9000000

8000000

7000000

60 000 000

6000000

5000000

4000000

30 000 000

3000000

22

2000000

1000000

00

BB

AAl l

Ga Ga

InIn

0

T l Tl

0

B

B

Al

Al

(a)

Ga Ga

In

In

Tl Tl

(b)

(b) In the 205Tl NMR spectrum of solutions containing Tl3+ and 13C-enriched [CN]–, coordinate bond formation between Tl3+ and [CN]– results in signals for the 205Tl nuclei that exhibit coupling to the 13C (I = 1/ ) nuclei. The cyano ligands are 2 C-bonded. Since I = 1/2 for 13C, the observation of a binomial quintet indicates the presence of [Tl(CN)4]–. The quartet shows that Tl(CN)3 is present. 13.31

(a) Figure 13.11 illustrates clearly why a logarithmic scale is useful for displaying the relative abundances of the group 13 elements. The abundance of Al is ≅ 82 000 000 ppb (82 000 ppm), while that of In is only 49 ppb (0.049 ppm). From Figs. 13.1 and 12.2 in H&S, values of the abundances of Al ≅ 82 000 ppm and of Mg ≅ 24 000 ppm can be estimated. (b) Equation 13.18 in H&S is: 2H2Ga(μ-Cl)2GaH2 Ox. states: H –1 Ga +3 Cl –1 Oxidation: H Reduction: Ga

2Ga + 0

Ga+[GaCl4]– + 4H2 0 +1 +3 –1

8 × (–1 to 0) = +8 {2 × (+3 to 0)} + (1 × (+3 to +1)} = –8

Therefore, oxidation state changes for reduction balance oxidation. (c) The chair-conformation of the [B3N6]9– ion is shown in structure 13.32. Each B atom is approximately trigonal planar. Resonance structures are: N 2–

N 2–

B–

B–

N

(13.32)

2– N

N

N

–B

B– N

2– N

2– N

N–

N

N

–B

B– N

B–

–

2– N

– N

–B

– N B

N –

– – N

From Table 13.2 in H&S, a typical B–N single bond length ≅ 157-160 pm, while a typical double bond length ≅ 134-138 pm. The observed B–N bond lengths in the [B3N6]9– unit in crystalline La5(BN3)(B3N6) are 143 and 148 pm, indicating that these bonds contain some π-contribution, but are not formally of bond order 2. This is consistent with the net bonding suggested by the set of resonance structures above.

The group 13 elements

202

13.32

(a) The [BH4]– ligands can coordinate H H in a mono-, bi- or tridentate manner to H H B B the Al(III) centres. The most likely is H H bidentate, allowing Al to be octahedral H H H as shown on the right. Each B Al Al environment is the same, consistent H H H with the observation of one signal in H H the 11B NMR spectrum. The signal is a B B H H binomial quintet, indicating that each H H 11B nucleus couples to four equivalent protons. This can be explained if each [BH4]– ligand is fluxional on the NMR timescale. Terminal B–H and bridging B–H–Al hydrogen atoms are involved in the exchange process, but the Al–H–Al protons are not. The NMR spectroscopic data indicate that there is no exchange between protons attached to different boron atoms. (b) From the elemental analysis: Ratio B:Cl:C:O =

15.2 75.0 4.2 5.6 : : : = 1.4 : 2.1 : 0.35 : 0.35 10.8 35.5 12.0 16.0 = 4 : 6 :1 :1

The 11B NMR spectroscopic data show 2 B environments, ratio 1:3. The unique B is tetrahedral, the other 3 B atoms are trigonal planar. The absorption at 2176 cm–1 in the IR spectrum is consistent with a C ≡ O bond. The suggested structure is that of the adduct (Cl2B)3BCO, 13.33. O C B Cl2B

Each B attached to Cl is trigonal planar

BCl2 BCl2

(13.33)

13.33

O

C

O

O

N O O

C

C

N In

N

(13.34)

N

H

O

(a) By doping Si (group 14) with B or Ga (group 13), an acceptor band (see diagram on right) is created because a B or Ga atom has one less valence electron than an Si atom. On going from pure Si to B or Ga-doped Si, the band gap significantly decreases and the material becomes a better semiconductor. Doped Si is a p-type, extrinsic semiconductor.

Unoccupied MOs Acceptor level Small band gap Occupied MOs

(b) Ga(III) is likely to favour 6-coordination, octahedral or close to octahedral. The aza-macrocycle with pendant S-donors can bind to give a neutral complex containing 6 chelate rings (Fig. 13.12). Such a complex is expected to be thermodynamically stable. The complex has a fac-arrangement of donor atoms because the ligand is conformationally restricted. The second ligand has 7 donor atoms. It could use six of the donor atoms and coordinate with a fac-arrangement. The larger In(III) centre may accommodate seven donor atoms and in fact this is what is observed (structure 13.34).

The group 13 elements

203

S N

N N

N

S Ga

N

N

S S

S

S

Fig. 13.12 The hexadentate, aza-macrocycle for problem 13.33, and the structure of an octahedral Ga(III) complex involving this ligand.

13.34 H B

H

HB

BH HB

BH

B H

(a) In the solid state, the [B6H7]– ion has structure 13.35 in which each B atom has one terminal H atom attached to it, and one B3-face is bridged by an H atom. This leads to 2 B environments in a ratio 3 : 3 = 1 : 1. The 11B NMR data at 223 K are consistent with this. At 297 K, the 11B NMR spectrum shows that all B atoms are equivalent. The signal is a doublet with J = 147 Hz – typical of coupling to terminal 1H. J(BH term) >> J(BHcap) and only the dominant coupling is observed. The capping H atom is fluxional over the B6-cage at 297 K but there is no exchange with Hterm. (b) The reaction of Ga with NH4F is: 2Ga + 8NH4F

(13.35)

2[NH4][GaF4] + 3H2 + 6NH3

[NH4][GaF4] is compound X. Vertex sharing of octahedral GaF6-units occurs in one plane. Part of one layer of the structure is shown below: F The layer of vertex-shared F F octahedra can be represented as F F Ga F Ga F {GaF2}F4/2 where the ‘F4/2’ refers F F F to the shared F atoms. The 1:4 F F stoichiometry of GaF 4 is F F Ga therefore retained. An alternative F F F diagram of part of a layer using a F F Ga F polyhedral representation is F Ga F F shown in diagram 13.36. F

(13.36)

F

13.35

F

(a) Commercial uses of Ga (mainly as the semiconductors GaAs or GaN) are in integrated circuits and optoelectronic devices (laser diodes, light-emitting diodes, photodetectors and solar cells), and the manufacture of integrated circuits in particular has increased dramatically over the period 1975 to 2008. (b) Corundum, α-Al2O3, is colourless. If trace amounts of Cr(III) replace Al(III) in the crystal, the colour changes to deep red: Cr3+ has a d3 configuration and is in an octahedral O6-environment in the α-Al2O3 lattice. Crystal field strength determines energies of electronic transitions (see Chapter 20 in H&S); yellowgreen light is absorbed and the complementary colour (i.e. the observed colour, see Table 19.2 in H&S) is deep red. (c) H3N.BH3 contains 19.6% hydrogen by weight and is non-inflammable under ambient conditions. Possible methods of releasing H2 from H3N.BH3 are by heating or using a metal catalyst. For practical applications, H2 release must be reversible, and production of the thermodynamically stable boron nitride (BN) must be avoided. Thus only two-thirds of the hydrogen in H3N.BH3 is available.

204

The group 13 elements 13.36

(a)

O B

O

O

O

–

– O

B

O O

O B

Si O

O Tetrahedral B

O –

Trigonal planar B

O

O

–

Trigonal planar B [BO3]3–

Si

– O

O O

Tetrahedral Si

– – O O–

Tetrahedral Si [SiO4]4–

For tetrahedral B and Si, the units contain localized single bonds. For trigonal planar B, the B atom has an empty 2p orbital which will overlap with filled 2p orbitals on O leading to delocalization of electronic charge. (b) Examples of other structural units: – – O

O

– O

O

B

B

O

O

–

Si O

O

O O O

Si O

–

O O O–

Si O

– – O O–

(c) Na2O is ionic and addition to silica (SiO2) or boron oxide (B2O3) produces borates and silicates, the negative charges on which are balanced by Na+ ions. In pure silica, all Si units are tetrahedral Si(μ-O)4, and in pure B2O3, only planar B(μ-O)3 are present. Addition of Na2O converts some bridging O to terminal O–; possible building blocks are shown in parts (a) and (b). Similarly, aluminate units are formed when Na2O converts some bridging O in Al2O3 to terminal O–. From the % composition, you can give an estimate of the proportion of bridging O converted to terminal O–. 13.37

Points to include in your answer: • There are two polymorphs of BN: hexagonal and cubic; describe the structures. • Hexagonal (or α-) BN is graphite-like in its properties; cubic BN is diamond-like. • Layered structure of hexagonal BN (only van der Waals interactions between the layers) leads to lubricant properties. • Band gap in hexagonal BN > graphite, leading to it being an insulator. • 3D-structure of cubic (or β-) BN is very rigid; strong single covalent, polar bonds lead to high melting, robust material used in crucibles, cutting tools and as an abrasive.

205

14 The group 14 elements 14.1

(a)

C carbon Si silicon Ge germanium Sn tin Pb lead (b) Non-metallic: C; semi-metallic: Ge, Si; metallic: Sn, Pb (c) ns2np2

14.2

Figures 14.1-14.3 show the data to be discussed in this answer. (a) Melting points: see Fig. 14.1. Data for C refer to diamond; diamond, Si and Ge are isostructural with 3D-covalent diamond-type structures (see Fig. 6.4, p. 87). Melting the solid breaks down the structure; the trend in melting points for C, Si and Ge follows the decrease in strength of the covalent bonds (see Table 14.2 in H&S). White Sn (stable allotrope at 298 K) has a metallic lattice with coordination number of 6; Pb has a ccp structure (see Section 6.3 in H&S). (b) Values of ΔaHO(298 K): see Fig. 14.2. The enthalpy change refers to the process: E(s) J E(g)

or

En(s) J nE(g)

i.e. the solid state structure for element E is completely disrupted to give gas phase atoms. The decrease down group 14 corresponds to a decrease in strength of E–E bonding (see Table 14.2 in H&S). For the isostructural C, Si and Ge, the orbital overlap becomes less effective as the principal quantum number, n, increases from 2 (for C) to 4 (for Ge). The trend continues for the metallic members of the group. (c) Values of ΔfusHO(mp): see Fig. 14.3. Enthalpy change refers to the process: E(s) J E(l)

4000

2000

ΔfusHo(mp) / kJ mol–1

800 ΔaHo(298 K) / kJ mol–1

Melting point / K

i.e. the solid state lattice for element E collapses but is not completely disrupted. The value of ΔfusHO(mp) < ΔaHO(298 K) for a given element. The trend in ΔfusHO(mp) approximately mimics that for the melting points.

600

400

200

0 C Si Ge Sn Pb

Fig. 14.1 Trend in melting points for the group 14 elements.

100

50

0

C Si Ge Sn Pb

Fig. 14.2 Trend in standard enthalpies of atomization for the group 14 elements.

C Si Ge Sn Pb

Fig. 14.3 Trend in standard enthalpies of fusion for the group 14 elements.

206

The group 14 elements 14.3 142 pm

335 pm

Fig. 14.4 Part of the layered structure of graphite. The vertical lines emphasize which atoms lie over which; adjacent layers are mutually staggered.

14.4

Structure of α-graphite: see Fig. 14.4. Comment on the bonding: 3-coordinate C atoms within layers with delocalized π-bonding within each layer. (a) Interactions between the layers in the solid state are weak (van der Waals) enabling the layers to slide past one another easily. This slippage is the basis for the use of graphite powder as a solid lubricant. (b) Each C atom is 3-coordinate in a layer; consider as sp2 hybridized with 3 electrons used within σ-bonding framework and the fourth electron occupying a 2p atomic orbital involved in electronic delocalization within a layer:

Electrical conductivity of graphite is dependent on direction: high conductivity (resistivity is low) in a direction parallel to the layers, but perpendicular to this direction, conductivity is low. Use in electrodes follows from the high electrical conductivity, and the relative chemical inertness of graphite under ambient conditions. (c) Density of diamond > graphite (3.51 versus 2.25 g cm–3). At high pressures, graphite converts to diamond and this is a method for preparing artificial diamonds from graphite. Look at Fig. 14.10 in H&S. The fulleride anions are in an fcc arrangement. Use the method of working as for unit cells (see Chapter 6). Site

Number of [C60]n–

Number of K+

Central

0

9

Corner

8 × 1/8 = 1

0

Face

6 × 1/2 = 3

0

Edge

0

12 × 1/4 = 3

Total

4

12

The ratio of [C60]n– : K+ = 1 : 3, and the charge on the fulleride ion is 3–. 14.5

Selected reaction types that illustrate the carbon-carbon double bond character include: • addition of halogens; • cycloaddition reactions; • formation of epoxide C60O; • organometallic π-complex formation. For details of individual reactions, refer to Section 14.4 (subsection ‘Fullerenes: reactivity’) and Section 24.10 (subsection ‘Alkene ligands’, eqs. 24.82 and 24.83) in H&S.

14.6

(a) Carbides belong to different families, and their reactivity with water is a distinguishing feature. Mg2C3 and CaC2 contain [C=C=C]4– and [C ≡ C]2– ions respectively; reactions with water are: Mg2C3 + 4H2O J CH3C ≡ CH + 2Mg(OH)2 CaC2 + 2H2O J HC ≡ CH + Ca(OH)2

The group 14 elements

207

ThC2 contains [C2]4–. Hydrolysis gives a mixture of C2H2, C2H6 and H2 and distinguishes this carbide from one containing an acetylide ion, [C2]2–. TiC is an interstitial carbide, i.e. cubic close-packed metal atoms with C atoms in octahedral interstitial holes. Interstitial carbides are robust materials that are inert towards the action of water. X Review acid-base properties of non-aqueous solvents in Section 9.4 in H&S

(b) In liquid ammonia, [NH4]Br acts as an acid because it is a source of [NH4]+ ions. Magnesium silicide reacts with dilute acids to give silanes. Therefore, in liquid NH3, Mg2Si reacts with [NH4]Br according to: Mg2Si + 4[NH4]Br J 2MgBr2 + 4NH3 + SiH4 Reactions of Mg2Si with acids are usually non-specific, giving mixtures of silanes.

Me2 Si Me2Si

Me

Si

H

Si Me2

(14.1) Me2 Si Me2Si

Me

Si

D

Si Me2

(c) If hydrolysis of compound 14.1 involves cleavage of the Si–H bond, then the rate of alkaline hydrolysis of the deuterated compound 14.2 should be slower than the rate of alkaline hydrolysis of 14.1. In this case, the reaction would show a kinetic isotope effect (see Section 3.9 in H&S) because the bond enthalpy of the Si–D bond is greater than that of the Si–H bond. Since the rates of hydrolysis of 14.1 and 14.2 are the same, it follows that the rate-determining step is not the cleavage of the Si–H (or Si–D) bond. An alternative mechanism that can be suggested is attack by [OH]– at the Si centre in the rate-determining step.

(14.2)

14.7

Me3Si

N

SiMe3 SiMe3

(14.3)

O

C

(14.4)

O

(a) An explanation of the planarity of the NSi3-skeleton in 14.3 is not straightforward. A traditional explanation involves (p-d)π-bonding between vacant Si 3d and occupied N 2p orbitals, but the difficulty with this scheme is that it assumes that Si 3d orbitals lie at low enough energy to be accessible. Current arguments favour ‘negative hyperconjugation’ in which the N lone pair electrons are donated into the vacant Si–Cmethyl σ* orbitals rather than into the Si 3d orbitals. Another contributing factor is that the polarity of the N–Si bonds (χP(Si) = 1.9, χP(N) = 3.0) results in significant long-range electrostatic repulsions between the SiH3 groups. These are minimized if the NSi3-skeleton in N(SiMe3)3 is planar, rather than pyramidal. (b) At 298 K, CO2 exists as linear molecules (14.4) and SiO2 forms an infinite 3DSi structure (14.5). Both have O O Si O covalent bonding. Si The bonding in CO2 can be Si O described in terms of sp O O Si hybridized C, and the Si O formation of two C–O σ O O Si Si O O bonds and two C–O π-bonds O Si involving overlap between C Si 2p and O 2p orbitals. (14.5) The bonding in SiO2 is best described in terms of sp3 hybridized Si with Si–O σ-bonding. In terms of bond enthalpy terms:

The group 14 elements

208

Bond Bond enthalpy term / kJ mol–1 C=O 810 C–O 359 Si=O 642 Si–O 466 you can see that the (p-p)π contribution to the C=O double bond is much greater than to the Si=O bond, and this reflects the fact that 2p-2p overlap is more effective than 3p-2p overlap (better energy match). To consider the balance between discrete molecular and macromolecular CO2 on the grounds of bond enthalpies at 298 K, you must look at the difference between two C=O (1620 kJ mol–1) and four C–O bonds (1436 kJ mol–1): the discrete molecule with two C=O bonds wins. A similar exercise for SiO2 shows that the macromolecular structure with four Si–O wins. A phase of CO2 with a silica-like structure has been made by laser-heating at 1800 K and 40 GPa. A dense, amorphous, glass-like phase forms when molecular solid CO2 is compressed at 40-64 GPa and 564 K. 14.8

Cl

C

N

(14.6)

O

C

S

(14.7)

Si

H

H

H

(14.8)

Cl Cl

Cl

Sn

Cl Cl

(14.9)

Cl

O

Cl

Si Cl

Si

Cl

Cl

(14.10)

Cl

Use the VSEPR model to suggest shapes for the molecular species in the question. For full details of the method, refer to answer 2.17 (p. 20). (a) ClCN C: number of valence electrons = 4 Number of bonding ‘pairs’ (Cl–C and C ≡ N) = 2 No lone pairs on C ‘Parent’ shape = molecular shape = linear (see 14.6) (b) OCS C: number of valence electrons = 4 Number of bonding ‘pairs’ (C=O and C=S) = 2 No lone pairs on C ‘Parent’ shape = molecular shape = linear (see 14.7) (c) [SiH3]– Si: number of valence electrons = 4 1 electron from the negative charge Number of bonding pairs (3 Si–H) = 3 This leaves 1 lone pair on Si ‘Parent’ shape = tetrahedral Molecular shape = trigonal pyramidal (see 14.8) (d) [SnCl5]– Sn: number of valence electrons = 4 1 electron from the negative charge Number of bonding pairs (5 Sn–Cl) = 5 No lone pairs on Sn ‘Parent’ shape = molecular shape = trigonal bipyramidal (see 14.9) (e) Si2OCl6 First, recognize that the O atom bridges between the Si atoms. Si: number of valence electrons = 4 Number of bonding pairs (3 Si–Cl + 1 Si–O) = 4 No lone pairs on Si ‘Parent’ shape = molecular shape = tetrahedral at each Si (see 14.10) (f) [Ge(C2O4)3]2– First, note that you are dealing with 3 didentate oxalate ligands (see Table 6.7 in H&S) Ge: number of valence electrons = 4

The group 14 elements

209

2 electrons from the negative charge Number of bonding pairs (6 Ge–O) = 6 No lone pairs on Ge ‘Parent’ shape = molecular shape = octahedral (see 14.11) 2–

O C O

C

O C

O

O

Ge C O

O

O O C

C

O

2–

Cl Cl

O

O Cl

Pb

(14.11)

Cl

Cl

(g) [PbCl6]2– Pb: number of valence electrons = 4 2 electrons from the negative charge Number of bonding pairs (6 Pb–Cl) = 6 No lone pairs on Pb ‘Parent’ shape = molecular shape = octahedral (see 14.12) (h) [SnS4]4– Sn: number of valence electrons = 4 4 electrons from the negative charge Number of bonding ‘pairs’ (4 Sn=S) = 4 No lone pairs on Sn ‘Parent’ shape = molecular shape = tetrahedral (see 14.13)

Cl

(14.12)

4–

S Sn S

S S

(14.13)

14.9 X Look at worked example 14.4 in H&S

Apical Equatorial

(a) This question is about the application of Wade’s rules; these are discussed in detail in answer 13.26 (p. 198). Apply Wade’s rules to [Sn9Tl]3– 9 Sn atoms, and each provides 2 electrons (assume 1 lone pair per Sn) 1 Tl atom provides 1 electron (assume 1 lone pair on the Tl centre) 3 electrons from the 3– charge Total number cage-bonding electrons = (9 × 2) + 1 + 3 = 22 electrons = 11 pairs [Sn9Tl]3– has 11 pairs of electrons with which to bond 10 cluster atoms This means that there are (n + 1) pairs of electrons for n vertices ∴ [Sn9Tl]3– is a closo-cage, a 10-vertex deltahedron (bicapped square antiprism), see 14.14 (b) Isomers arise because the bicapped square antiprism has two different sites (see 14.14). In principle, the Tl atom can occupy either of these sites, and 2 isomers of [Sn9Tl]3– are possible.

(14.14)

210

The group 14 elements 14.10

(14.15)

For information on the structures and chemistries of hydrides of the group 14 elements, see Section 14.6 in H&S. The hydrides of B and Al (group 12) are discussed in Section 13.5 of H&S. Points that should be included in your answer: • Alkanes CnH2n+2 known for wide range of n; all have molecular structures with tetrahedral C; isomers with unbranched and branched chains, e.g. 14.15-14.16. • Silanes SinH2n+2 for 1≤n≤10; all have molecular structures; isomers with straight and branched chains; structural analogues of alkanes. • Comparing straight chain silanes with alkanes with n C or Si atoms: for a given n, boiling point of silane > alkane; silanes (but not alkanes, the difference being kinetic) are spontaneously explosively inflammable in air: SiH4 + 2O2 J SiO2 + 2H2O

(14.16)

• Germanes GenH2n+2 (straight and branched chain isomers) known for 1 ≤ n ≤ 9; structural analogues of alkanes and silanes. • Stannane SnH4; isostructural with CH4, but decomposes at 298 K: SnH4 J Sn + 2H2 • PbH4 is poorly characterized and may not actually exist. • Reactivities of hydrides of Si, Ge and Sn follow trend SiH4 > GeH4 < SnH4; GeH4 (like SiH4) is highly inflammable in air. • CH4, SiH4 and GeH4 all insoluble in water; CH4 kinetically stable with respect to hydrolysis by water, but SiH4 hydrolysed, also by alkalis: SiH4 + 2NaOH + H2O J Na2SiO3 + 4H2 • Reactions of SiH4 and GeH4 with alkali metals give synthetically useful salts of [EH3]– : liquid NH 3 ⎯→ 2M[GeH3] + H2 2GeH4 + 2M ⎯⎯ ⎯ ⎯

• BH3 versus CH4: hydrides of first member of groups 12 and 13 respectively. BH3 is trigonal planar, and B has sextet of valence electrons – unstable with respect to dimerization (see 13.4, p. 191). B2H6 reacts with Lewis bases, L, forming adducts LJBH3; B2H6 rapidly hydrolysed to B(OH)3 and H2. This contrasts with CH4 (tetrahedral) in which C has octet of electrons; no dimerization. Lewis acidity of BH3 is not mirrored in chemistry of CH4. • Similar comparison between AlH3 and SiH4 as between BH3 and CH4. 14.11

(a) Hydrolysis of group 14 halide liberates hydrogen halide: GeCl4 + 2H2O J GeO2 + 4HCl GeO2 is dimorphic, possessing rutile and quartz polymorphs (see answer 14.14). (b) Hydrolysis with aqueous alkali gives a silicate: SiCl4 + 4NaOH J Na4SiO4 + 4HCl Discrete, tetrahedral [SiO4]4– ions are not present; instead polymeric species form containing tetrahedral SiO4 units linked by Si–O–Si bridges.

The group 14 elements

211

(c) CsF is a fluoride donor and GeF2 is a fluoride acceptor. Products of such reactions depend on stoichiometry (as well as cation); with a 1:1 ratio: CsF + GeF2 J Cs[GeF3] The salt is expected to contain trigonal pyramidal [GeF3]– ions. Solid-state structures of related species such as salts of [SnCl3]– are cation-dependent. Discrete anions may be present, but in some salts, halide bridges connect Sn centres to give a polymer. (d) Hydrolysis of SiH3Cl liberates HCl; Si–O–Si bridge formation: 2SiH3Cl + H2O J (H3Si)2O + 2HCl

O H3Si

SiH3

Molecular species with structure 14.17. (e) Hydrolysis of SiF4 gives silica and HF, but SiF4 acts as a fluoride acceptor:

(14.17)

2SiF4 + 4H2O J SiO2 + 2[H3O]+ + [SiF6]2– + 2HF SiO2 has a 3D-structure (see answer 14.7, p. 207). [SiF6]2– is octahedral. (f) The stoichiometry stated as 2:1, indicating that SnCl4 accepts two Cl– : 2[Bu4P]Cl + SnCl4 J [Bu4P]2[SiCl6] Product is an ionic salt with tetrahedral [Bu4P]+ and octahedral [SiCl6]2–. Each spectrum is 119Sn observed, so all splittings are due to 119Sn-19F spin-spin couplings; each species is octahedral (use the VSEPR model). (a) [SnCl5F]2–, 14.18, has only one F and gives a doublet in the 119Sn NMR spectrum. (b)[SnCl4F2]2– has cis (14.19) and trans (14.20) isomers; each isomer has 2 equivalent F sites, giving a triplet in the 119Sn NMR spectrum; not possible to distinguish isomers A and B from these data alone. (c) [SnCl3F3]2– has fac (14.21) and mer (14.22) isomers. In 14.21, F atoms are all equivalent, and so a binomial quartet is seen in the 119Sn NMR spectrum; isomer B is the fac isomer. In 14.22, there are two F environments labelled a and b; coupling of 119Sn to F(a) gives a triplet, and further coupling to F(b) gives an overall signal that is a doublet of triplets. Isomer A is the mer-isomer. (d) [SnCl2F4]2– has trans (14.23) and cis (14.24) isomers. In 14.23, F atoms are equivalent, so a binomial quintet is observed in the 119Sn NMR spectrum; isomer A is the trans-isomer. In 14.24, there are two F environments labelled a and b; coupling of 119Sn to F(a) gives a triplet, and further coupling to F(b) gives an overall signal that is a triplet of triplets. Isomer B is the cis-isomer. (e) [SnClF5]2– (14.25) has two F sites (a and b). Coupling of 119Sn to F(a) gives a doublet; further coupling to F(b) gives an overall signal that is a doublet of quintets. (f) [SnF6]2– has 6 equivalent F atoms giving a binomial septet in the 119Sn NMR spectrum.

14.12 2–

F Cl

Sn

Cl Cl

Cl Cl

(14.18) 2–

Cl Cl

Sn

Cl

F F

Cl

(14.19) 2–

F Cl

Sn

Cl

Cl Cl

F

(14.20) 2–

F Cl

Sn

Cl

F

F

Cl

F

Cl

a

Sn

b F Cl

Cl

F

(14.21)

(14.22)

a

2–

2–

Cl F

Sn

F Cl

(14.23)

F

b

F

F

F

F b

a

Sn F

a

(14.24)

2–

Cl

b

Cl

F

Cl

F b

Sn Fa

(14.25)

b F F b

2–

The group 14 elements

212

14.13

(a) Concentrated NaOH oxidizes Sn and also provides [OH]– for complex formation: 2H2O + 2e–

H2 + 2[OH]–

[Sn(OH)6]2– + 4e–

Sn + 6[OH]–

Overall: Sn + 2[OH]– + 4H2O J [Sn(OH)6]2– + 2H2 or Sn + 2NaOH + 4H2O J Na2[Sn(OH)6] + 2H2 (b) SO2 reduces Pb(IV) to Pb(II), and is itself oxidized:

S–

SO2 + PbO2 J PbSO4

C S–

S

(c) Section 14.11 in H&S describes the reaction of H2O with CS2 to give H2S and CO2. With NaOH, reaction will give Na2CS3 and some Na2CO3. Na2CS3 contains the [CS3]2– ion, 14.26. (d) Hydrolysis of SiCl4 gives HCl and SiO2. Follow on from this to suggest that SiH2Cl2 hydrolyses according to:

(14.26)

O

O Si H

nSiH2Cl2 + nH2O J -(SiH2O)n- + 2nHCl

Si H

H

H

n

/2

(14.27)

-(SiH2O)n- is polymeric; structure 14.27 shows two repeat units. (e) Li[AlH4] acts as a hydrogenating agent, converting Si–Cl to Si–H. The product is ClCH2SiH3. 4RCl3 + 3LiAlH4 J 4RH3 + 3LiAlCl4

14.14

(a) To estimate ΔrHo for the reaction:

X Compare with answer 12.10b

GeO2(quartz) J GeO2(rutile)

where R = ClCH2Si

dissolve each in conc. HF, measure ΔsolHo for each, and apply a Hess cycle. In solution and after reaction, there will no longer be a distinction between the two phases of GeO2, a common set of products being obtained. From the cycle:

GeO2(quartz) + 6HF(conc)

ΔrHo

GeO2(rutile) + 6HF(conc) .

ΔsolHo(1)

ΔsolHo(2)

[GeF6]2–(aq) + 2[H3O]+(aq) apply Hess’s Law: ΔrHo = ΔsolHo(1) – ΔsolHo(2)

The group 14 elements

213

(b) The Pauling electronegativity of Si is estimated using the equation:

χP(X) – χP(Si) = X 1 eV = 96.485 kJ mol–1

ΔD

where ΔD is in eV and is given by: ΔD = D(Si–X) – 1/2{D(Si–Si) + D(X–X)}

(ΔD in kJ mol–1)

Choose X so that compounds containing Si–Si and Si–X bonds are readily accessible, e.g. X = H. Values of D(Si–H) and D(Si–Si) are listed in Appendix 12 in H&S (326 and 226 kJ mol–1 respectively), but comment on how these could be estimated: D(Si–H) = 1/4ΔaHo(SiH4, 298 K) 6D(Si–H) + D(Si–Si) = ΔaHo(Si2H6, 298 K) For D(H–H): X ΔaHo(H, 298 K) from Appendix 10 in H&S

D(H–H) = 2ΔaHo(H, 298 K) Values of ΔaHo(SiH4, 298 K) and ΔaHo(Si2H6, 298 K) determined from Hess cycles, exemplified below for SiH4. Experimentally measurable quantity is ΔcHo. ΔcHo SiH4(g) + 2O2(g) ΔfHo(1)

SiO2(s) + 2H2O(l)

SiH4(g)

ΔfHo(2)

ΔfHo(1)

Si(s) + 2H2(g) + 2O2(g)

ΔaHo(1)

Si(g) + 4H(g) .

ΔaHo(2)

Si(s) + 2H2(g)

(c) The purity of the samples can be assessed by using analytical methods. Analysis of Pb(MeCO2)4: determine Pb(IV) by allowing it to oxidize I– and titrating the I2 formed against thiosulfate: I2 + 2[S2O3]2– J 2I– + [S4O6]2– Or heat Pb(MeCO2)4 with HCl; Pb(IV) is reduced and Cl– oxidized to Cl2. Pass Cl2 into aqueous KI: Cl2 + 2I– J I2 + 2Cl– and titrate I2 formed against thiosulfate (see above). 14.15

Figure 14.5 shows the relevant part of the Ellingham diagram for this problem. The values of ΔfGo refer to the reactions: C + 1/2O2 J CO 1/ Sn 2

+ 1/2O2 J 1/2SnO2

The group 14 elements

Fig. 14.5 Ellingham diagram for CO and SnO2 (also see Fig. 8.6 in H&S).

ΔfGo / kJ per half-mole of O2

214

The reaction to be considered is:

-100

-150

-200

CO

-250 500 1000 Temperature / K

14.16

X Pyroxenes and feldspars are silicate minerals

C + 1/2SnO2 J CO + 1/2Sn

SnO2

At 1000 K, CO is more thermodynamically stable than SnO2 (CO has a more negative ΔfGo) and so C reduces SnO2 to Sn. At 500 and 750 K, CO is not more thermodynamically stable than SnO2, and at these temperatures, C cannot be used to extract Sn from SnO2.

(a) In the isomorphous (i.e. possess the same structures) pyroxenes CaMgSi2O6 and CaFeSi2O6, Fe2+ and Mg2+ occupy the same lattice positions. The ionic radii of Fe2+ and Mg2+ are 78 and 72 pm respectively (see Appendix 6 in H&S); the value for Fe2+ is for the high-spin ion since the coordination environment consists of weak-field O2– centres (see Chapter 20 of H&S). The similarity in size between Fe2+ and Mg2+ means that ion replacement causes little structural perturbation. (b) Consider NaAlSi3O8 to be the ‘host’ lattice. Going from this to CaAl2Si2O8 requires that Ca2+ replaces Na+, and at the same time Al3+ replaces Si4+. These simultaneous replacements allow electrical neutrality to be retained. Consider relevant ionic radii (see Fig. 14.22 in H&S): rion: Ca2+ = 100 pm;

Na+ = 102 pm;

Al3+ = 54 pm;

Si4+ ≅ 40 pm

Similarity in ion sizes of Ca2+ and Na+, and of Al3+ and Si4+ means ion replacement occurs with little or no structural perturbation. (c) Quartz is a polymorph of SiO2; when LiAlSi2O6 transforms to a quartz form, the Al3+ ions must take lattice sites adopted by Si4+ in SiO2 – similarity in size (see above) allows this to occur. Quartz lattice is a relatively open network and Li+ ions (rion = 76 pm) can occupy interstitial sites. Comparing LiAlSi2O6 with SiO2: • rewrite LiAlSi2O6 as Li+[AlSi2O6]– • think of [AlSi2O6]– as 3[Al/SiO2]1/3– • [Al/SiO2]1/3– compares directly with SiO2 • Li+ ions provide electrical neutrality. 14.17

(a) Molecules in question are all linear (D∞h). 2200 cm–1 is typical of ν(C ≡ N), and I is N ≡ C–C ≡ N. For CO2 and CS2, the lower wavenumber corresponds to the bond with lower force constant and higher reduced mass, so II is S=C=S, and III is O=C=O. (b) All are symmetrical molecules, therefore the symmetric stretch is IR inactive, and asymmetric stretch is IR active in each.

14.18

KCN(aq) is basic (see answer 7.7, p. 101) giving [OH]– in solution. This competes with [CN]– for Al3+. Al(OH)3 forms preferentially and precipitates.

14.19

(a)

Cyanic acid = HOCN Isocyanic acid = HNCO Thiocyanic acid = HSCN

HOCN + 2H2O J NH3 + H2CO3

H2CO3 J CO2 + H2O

(b) HNCO reacts to give the same products as HOCN (c)

HSCN + 2H2O J NH3 + H2CO2S

H2CO2S J OCS + H2O

The group 14 elements Fig. 14.6 Vibrational modes of [CSe3]2– (D3h). The + and – signs mean movement of the atoms up and down with respect to the plane of the paper.

+

+ Symmetric stretch (A1')

F

F

(14.28)

O

Deformation (E')

(a) The vibrational wavenumbers and assignments for [CSe3]2– are 802 (E', stretch), 420 (A2''), 290 (A1') and 185 (E', deformation) cm–1. The two common 3-coordinate structures are trigonal planar (D3h) and trigonal pyramidal (C3v). By looking at the relevant character tables in Appendix 3 in H&S, you can see that the assignments of the vibrational modes correspond to the D3h character table. Therefore [CSe3]2– is trigonal planar, belonging to the D3h point group. (b) See Fig. 14.6. (c) From the D3h character table, you can deduce that the A2'' and E' modes are IR active, because there is an x, y or z entry in the right-hand column in the table. Alternatively, you can consider each vibration in Fig. 14.6 and deduce whether there is a change in molecular dipole moment. The symmetric stretch does not lead to a change in dipole moment and is IR inactive. The other 3 modes are IR active.

14.21

(a) SiF4 (14.28). Use Fig. 3.10 in H&S.

START C

Asymmetric stretch (E')

Is the molecule linear?

No

Does it belong to one of the special groups?

Yes

STOP

Conclusion: the point group is Td. (b) [CO3]2– (14.29)

F

O

+

Deformation (A2'')

14.20

START

Si F

215

2–

O

(14.29)

Is the molecule linear?

No

Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Are there 3 C2 axes perpendicular to the principal axis?

No Yes: C3 axis

Is there a σh plane?

Yes

Yes

STOP

Conclusion: the point group is D3h. (c) CO2 (14.30) O

C

O

START

(14.30)

Is the molecule linear?

Yes

Is there a centre of inversion?

Yes

STOP

Conclusion: the point group is D∞h. (d) SiH2Cl2 (14.31) Cl

Cl Si H

H

(14.31)

START

Is the molecule linear?

No

Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Are there 3 C2 axes perpendicular to the principal axis? Is there a σh plane?

No Yes: C2 axis

Are there n σv planes?

Yes

Conclusion: the point group is C2v.

No No

STOP

The group 14 elements

216

14.22

Structure 14.32 shows C60 as drawn in the question. Below are two views of an icosahedron; the right-hand one views the icosahedron down one of the 5-fold axes.

(14.32)

By comparing the right-hand structure above with 14.32, you can see that C60 contains the same number of 5-fold axes as an icosahedron. The symmetry elements of C60 and 14.32 are the same, and therefore C60 belongs to the Ih point group. 14.23

X For more information, see: K. Kobayashi and S. Nagase (1997) Organometallics, vol. 16, p. 2489

(a) Take each Si as sp hybridized. One sp hybrid orbital is used to form a localized Si–H σ-bond. This leaves one sp hybrid orbital and two 3p orbitals per Si to form the triple bond. Following from the bonding description in Fig. 23.19 in H&S, use a doublet ground state for each SiH unit, i.e. one unpaired electron. A bonding description for Si2H2 is therefore: H Si

Si

H

Donation of 2 electrons from filled orbital on each Si to vacant 3p orbital on the other Si atom

F

C

O

+

(14.33)

F Sn

F

F

Si

H

Pairing of Si electrons to give one π-bond

(b) Use the VSEPR model to show that [FCO]+ has a linear structure: C: number of valence electrons = 4 Subtract an electron for the positive charge Number of bonding pairs (C–F and C=O) = 2 (count double bond as one group of electrons) No lone pairs on C ‘Parent’ shape = molecular shape = linear (see 14.33)

Sn

F

F Sn

F

H SiH3

F

Sn F

(14.34)

14.24

(c) The structure of cyclic α-SnF2 is shown in 14.34. Within the Sn4F4-ring, there are localized Sn–F single bonds; their formation uses a lone pair from each F as well as F and Sn electrons. Each Sn forms a localized Sn–F bond outside the ring. After bond formation, each Sn atom has a lone pair of electrons left and this means that the environment at each Sn centre is trigonal pyramidal. The ring is therefore puckered. The correct pairings are shown in the table below: List 1 List 2 Comments SiF4 Gas (298 K), tetrahedral molecules Si

Semiconductor; diamond-type structure

See Fig. 5.19a in H&S

The group 14 elements List 1 Cs3C60

List 2 Superconducting at 40 K

SnO

Amphoteric oxide

[Ge9]4–

Zintl ion

GeF2

Carbene analogue

[SiO4]4–

Ca2+ salt is component of cement

PbO2

Acidic oxide

Pb(NO3)2

Water-soluble salt not decomposed All metal nitrates are soluble on dissolution Sheet structure, octahedral Sn See structure 14.14 in H&S

SnF4

F F

Sn

217

Comments See Fig. 14.10 in H&S

Ge(II) compound with lone pair on Ge

14.25

(a) A repeat unit of polymeric [SnF5]– is shown in 14.35. The Sn atom is octahedrally sited and bridging occurs through 2 F atoms that are mutually cis. The diagram illustrates how a stoichiometry 1 : 5 is retained. Alternatively, the formula can be written in the form [SnF4F2/2]nn– . (b) Most lead(II) salts are sparingly soluble in water. Pb(NO3)2, PbCl2 and Pb(O2CCH3)2 are soluble in water; (all metal nitrates are soluble; the acetate salt is used for qualitative H2S test, see answer 14.27b). (c) ClCN has C∞v symmetry (linear). The atomic masses of Cl and N are significantly different, therefore absorptions in the IR spectrum can be assigned to C–Cl stretch, C ≡ N stretch and ClCN bend; each of the symmetric and asymmetric stretches is dominated by the stretching of one of the two bonds. The highest energy band (1917 cm–1) is assigned to ν(CN), the band at 1060 cm–1 to ν(CCl), and the lowest energy band (230 cm–1) to the bending mode.

14.26

(a)

GeH3Cl + NaOCH3 J NaCl + H3GeOCH3 (MeO– is a nucleophile)

(b)

CaC2 + N2

(c)

Mg2Si + 2H2O/H+ J 2Mg(OH)2 + SiH4 + higher silanes

(d)

K2SiF6 + 4K

(e)

1,2-(OH)2C6H4 + GeO2

n–

F

F F n

(14.35)

Δ

CaNCN + C (commercial synthesis of CaNCN, a fertilizer)

Δ

6KF + Si (historically, Berzelius’ isolation of Si) NaOH/MeOH

[Ge(1,2-O2C6H4)3]2–

(deprotonation of 1,2-(OH)2C6H4 and complex formation) (f)

(H3Si)2O + I2 J 2SiH3I + O2

218

The group 14 elements (g)

O

O

O

O

C60

296 K O2

Sn + 2[OH]– + 4H2O J [Sn(OH)6]2– + 2H2 (oxidation of Sn to Sn(IV) and stabilization by formation of complex anion as Na+ salt)

(h)

14.27

O3, 1,2-xylene, hν in the dark, 257 K

(a) KC8 can be written as K+[C8]–. KC8 is an intercalation compound (see structure 14.2 in H&S) in which K+ ions are intercalated between layers in the graphite host. Layers in KC8 are eclipsed compared to being staggered in graphite. Electrical conductivity KC8 > graphite is due to extra electrons delocalized in graphite layers. K3C60 is a fulleride salt K+3[C60]3– ; for structure, see Fig.14.10 in H&S; it becomes superconducting on cooling to 18 K. (b) Few lead(II) salts are water-soluble; the water-solubility of Pb(O2CCH3)2 makes it suitable for the qualitative test for gaseous H2S which produces a black precipitate: Pb(O2CCH3)2(aq) + H2S(g) J PbS(s) + 2CH3CO2H(aq) (c) Start by using the information in the question to draw the structures of [C2O4]2– and [C2S4]2–. O

2–

O C

C

O

C

or –

O

O–

O

–

O

C

O

O C

O

O

C O–

2–

S C

C

S S

S

To determine the point group, apply the strategy shown in Fig. 3.10 in H&S: For [C2O4]2– :

START

Is the molecule linear?

No

Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Are there 2 C2 axes perpendicular to the principal axis? Is there a σh plane perpendicular

No Yes: C2

to the principal axis? Conclusion: the point group is D2h.

Yes Yes

STOP

The group 14 elements

219

2–

S S C

C S

S

4

1

1 3

C2' 3

4 2

C2 (principal axis, coincides with C–C bond)

2

midpoint of C–C bond C2'

Fig. 14.7 The principal axis (C2) in [C2S4]2– runs along the C–C bond. The other two C2 axes can be defined by considering the relationship between the atomic positions 1-4 and the corners of a cuboid.

For [C2S4]2– (Fig. 14.7):

START

Is the molecule linear?

No

Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Are there 2 C2 axes perpendicular to the principal axis? Is there a σh plane perpendicular to the principal axis? Are there n σv planes containing

No Yes: C2

the Cn axis?

Yes No

Yes

STOP

Conclusion: the point group is D2d. 14.28

N

O

O

O

O

O

O

N

(a) crypt-222 is the cryptand ligand 14.36. This encapsulates the K+ ion and provides a large counter ion for the Zintl ion. (b) Data: 195Pt, 33.8% abundant, I = 1/2. The Pb atoms in [Pt@Pb12]2– are all equivalent and therefore there is one signal in the 207Pb NMR spectrum. This is a pseudo-triplet: 66.2 % Pb does not couple to 195Pt

(14.36) J207Pb195Pt = 3440 Hz 33.8 % of Pb couple to 195Pt to give a doublet

X For the original spectra, see E.N. Esenturk et al. (2004) Angew. Chem. Int. Ed., vol. 43, p. 2132.

(c) Data: 207Pb, 22.1% abundant, I = 1/2. 195Pt can couple to spin-active 207Pb nuclei, and therefore the appearance of the signal in the 195Pt NMR spectrum reflects the statistical distribution of 207Pb nuclei in the Pb12 cluster in which each site is structurally equivalent. There will be a singlet with a satellite pattern that consists of a doublet superimposed on a triplet, superimposed on a quartet, etc. Each coupling constant is 3440 Hz, and the separation between any pair of adjacent lines is 1720 Hz.

220

The group 14 elements 14.29 X

Data: ΔfGo(NaCl, s) = –384 kJ mol–1 ΔfHo(CCl4, l) = –128.4 kJ mol–1 So(CCl4, l) = 214 J K–1 mol–1 So(C, gr) = +5.6 J K–1 mol–1 So(Cl2, g) = 223 J K–1 mol–1

(a) Reaction: CCl4(l) + 4Na(s) J 4NaCl(s) + C(s) First, you need to calculate ΔfGo(CCl4, l) using: For the formation of CCl4(l):

ΔGo = ΔHo – TΔSo

C(gr) + 2Cl2(g) J CCl4(l) ΔfSo(CCl4, l) = So(CCl4, l) – So(C, gr) – 2So(Cl2, g) = 214 – 5.6 – 2(223) = –237.6 J K–1 mol–1 At 298 K: ΔfGo(CCl4, l) = ΔfHo(CCl4, l) – TΔfSo(CCl4, l) = –128.4 – 298(– 237.6 × 10–3) = –57.6 kJ mol–1 For the reaction of Na with CCl4: ΔrGo = 4(ΔfGo(NaCl, s)) – ΔfGo(CCl4, l) = 4(–384) – (–57.6) = –1478 kJ mol–1 (b) β-Cristobalite is a form of SiO2 and crystallizes with a structure that is related to that of diamond (see Fig. 6.20 in H&S). Each Si atom is tetrahedrally sited and the SiO4 tetrahedra are linked into a 3D-structure by sharing of O atoms between two tetrahedra. The structure is highly ordered. When liquid silica cools, it forms a non-crystalline glass consisting of an infinite structure made up of SiO4 tetrahedra connected in a random manner.

14.30

(a) SnO2 is a wide band-gap n-type semiconductor, and intrinsic oxygen deficiency makes the solid non-stoichiometric. This leads to electrical conductance. Adsorption of O2 onto the surface of solid SnO2 removes electrons from the conduction band. The operating temperature of an SnO2 sensor is 450–750 K and O2– and O– are present on the surface. When CO is present, it reacts with surface-bound O–: CO(g) + O–(surface) J CO2(g) + e– The electrons released enter the SnO2 conduction band, causing an increase in electrical conductivity. This change is the basis for the detection of CO. (b) In the presence of H+, the half-reactions are: Cathode:

O2 + 4H+ + 4e– J 2H2O

Anode:

CO + H2O J CO2 + 2H+ + 2e–

The current produced in the cell is proportional to the concentration of CO. The sensor can be used to measure [CO], and can be equipped with an alarm which sounds when [CO] exceeds a safe level.

The group 14 elements 14.31

221

(a) Zeolites are aluminosilicates. Al– is isoelectronic with Si and replacement of Si in silica by Al leads to: SinO2n J [AlSin–1O2n]– J [Al2Sin–2O2n]2– ......... The charge is balanced by, e.g., Na+ or H+ ions. Zeolites comprise 3D-networks containing large channels and cavities (shape and size are variable among different zeolites) which accommodate the cations and small molecules (e.g. H2O, MeOH,

hydrocarbons, aromatic molecules). Different zeolites have different Al : Si ratios; Al-rich systems are hydrophilic. (b) Hard water contains Ca2+ and Mg2+ ions. These replace pairs of Na+ (keeping charge neutrality) in NaxA, i.e. ion exchange occurs. Regeneration of the exchange resin (zeolite) is carried out by using an excess of concentrated aqueous NaCl.

14.32

(a) Silica (SiO2) is the most important starting non-metal oxide in the glass industry.

When liquid silica cools, it forms a non-crystalline glass consisting of a 3D structure made up of corner-sharing SiO4 tetrahedra which are linked in a random manner. This is an example of a network-forming oxide. Only a few oxides form glasses (e.g. B2O3, SiO2, GeO2, P2O5 and As2O5) because the criteria for a random assembly are: • the coordination number of the non-oxygen element must be 3 or 4 (a coordination number of 2 gives a chain and >4 gives too rigid a structure); • only one O atom must be shared between any two non-oxygen atoms, because greater sharing produces too rigid an assembly. (b) Na2O and CaO are ionic. When they are added to SiO2 or B2O3, some bridging O atoms are converted to terminal O–, producing anionic borates and silicates. The negative charges are balanced by Na+ or Ca2+ ions. In pure silica, all Si units are tetrahedral Si(μ-O)4, and in pure B2O3, only planar B(μ-O)3 are present. (c) SiO2 is neutral and possesses corner-sharing SiO4 units, i.e. the structural units are Si(μ-O)4 or can be represented simply as SiO4. In the equation: 2[SiO3/2O]– + 2Na+

Al2O3

–2SiO2

2[AlO4/2]– + 2Na+

the formulae [SiO3/2O]– and [AlO4/2]– represent the building blocks: terminal O –

X See also answer 13.36, p. 204 B

B

B

R

bridging

Si O

O O O

Al O

– O O

I

14.33

(a) The cell reaction in a lead-acid battery: PbO2(s) + Pb(s) + H2SO4(aq) + 2H+(aq) J Pb2+(aq) + 2H2O(l) + PbSO4(s) Oxidation state changes: Pb(IV) to Pb(II): Pb(0) to Pb(II):

reduction by two units oxidation by two units

222

The group 14 elements (b) The two half-cells are: PbSO4(s) + 2e–

Pb(s) + [SO4]2– (aq)

PbO2(s) + 4H+(aq) + 2e–

Pb2+(aq) + 2H2O(l)

(c) The operating voltage of 12 V is approximately 6× the value of Eocell. This is the result of coupling 6 cells in series. (d) The spontaneous reaction of the cell converts lead(IV) oxide and lead metal into lead(II) sulfate. The reaction is reversed by applying an external voltage with the anode and cathode reactions reversed with respect to their discharge reactions.

223

15 The group 15 elements 15.1

In assigning oxidation states to N and P, first assign oxidation states to O and H (if present) of –2 and +1 respectively. These are the most common oxidation states for these elements. But watch H: an oxidation state of –1 is assigned when it is combined with an electropositive element. (a) N2 Elemental form ∴ ox. state N = 0 (b) [NO3]– Overall charge of 1– ∴ ox. state N = +5 Overall charge of 1– ∴ ox. state N = +3 (c) [NO2]– (d) NO2 Neutral compound ∴ ox. state N = +4 (e) NO Neutral compound ∴ ox. state N = +2 (f) NH3 Neutral compound ∴ ox. state N = –3 (g) NH2OH Neutral compound 15.1 ∴ ox. state N = –1 (h) P4 Elemental form ∴ ox. state P = 0 3– Overall charge of 3– ∴ ox. state P = +5 (i) [PO4] (j) P4O6 Neutral compound ∴ ox. state P = +3 (k) P4O10 Neutral compound ∴ ox. state P = +5

15.2

For each part of the problem, the reaction to consider is:

H N H

O H

(15.1)

X Bond enthalpy terms (see Appendix 12 in H&S) / kJ mol–1 N ≡ N 946 N–N 160 P ≡ P 490 P–P 209 C ≡ C 813 C–C 346

2 X

X (g)

2D(X ≡ X)

X

ΔrHo

X

X

X (g)

X = N, P or CH

6D(X–X)

4X(g)

(b) X = P

ΔrHo = 2(946) – 6(160) = +932 kJ mol–1 (i.e. per mole of reaction shown) o –1 ΔrH = 2(490) – 6(209) = –274 kJ mol

(c) X = CH

ΔrHo = 2(813) – 6(346) = –450 kJ mol–1

(a) X = N

Comments: At 298 K, N2 is thermodynamically stable with respect to N4, but P4 is stable with respect to P2; N2 and P4 are elemental forms at 298 K. C2H2 does not spontaneously convert to C4H4; C2H2 is kinetically stable with respect to C4H4. 15.3

Allotropy: information in Section 15.4 of H&S. Points to include are: • N2; triply bonded diatomic; molecular structure; gaseous at 298 K. • Phosphorus has wide range of allotropes; white phosphorus (P4 molecules; solid at 298 K) is defined as standard state of the element, but is metastable; heating white phosphorus under inert conditions gives red phosphorus (complicated covalent lattice), and under pressure, white phosphorus converts to black polymorph (most stable allotrope with double-layer structure of puckered 6-membered rings). • At 298 K and 1 bar, As, Sb and Bi are solids with structures comparable to black phosphorus; coordination number changes from 3 for P to 6 for Bi as layer structure changes to 3D-structure.

224

The group 15 elements 15.4

(a) Group 2 metal phosphide, best considered as ionic; liberates PH3 on hydrolysis: Ca3P2 + 6H2O J 3Ca(OH)2 + 2PH3 (b) Acid-base reaction of [NH4]+ with [OH]– : NaOH + NH4Cl J NaCl + NH3 + H2O (c) Aqueous NH3 produces [NH4]+ and [OH]– ; Mg2+ with [OH]– forms sparingly soluble Mg(OH)2: Mg(NO3)2 + 2NH3 + 2H2O J Mg(OH)2(s) + 2NH4NO3 (d) I2 is an oxidizing agent, able to oxidize As(–III) to As(V); in aqueous solution, an oxoacid of As(V) forms: AsH3 + 4I2 + 4H2O J H3AsO4 + 8HI (e) In liquid NH3, [NH2]– acts as a base and deprotonates PH3 liquid NH 3 ⎯→ KPH2 + NH3 PH3 + KNH2 ⎯⎯ ⎯ ⎯

15.5

(a) HCl dissolved in H2O is fully ionized: HCl(aq) + H2O(l) J [H3O]+(aq) + Cl–(aq) but aqueous NH3 contains dissolved NH3 which gives rise to the ‘ammonia smell’: NH3(aq) + H2O(l) [NH4]+(aq) + H2O(l)

[NH4]+(aq) + [OH]–(aq) K = 1.8 × 10–5 [H3O]+(aq) + NH3(aq) K = 5.6 × 10–10

(b) Ammonium carbamate is [NH4][NH2CO2], the salt of a very weak acid. Add moisture from the air on opening the bottle of smelling salts: [NH4][NH2CO2] + H2O [NH4]+(aq) + H2O(l)

[NH4][OH] + NH2CO2H [H3O]+(aq) + NH3(aq)

The NH3 formed results in the reviving smell. 15.6 X See Section 7.4 in H&S

Kb for NH3 refers to the equilibrium: NH3(aq) + H2O(l)

[NH4]+(aq) + [OH]–(aq)

pKb = 4.75

Ka for [NH4]+ refers to the equilibrium: [NH4]+(aq) + H2O(l)

NH3(aq) + [H3O]+(aq) pKa = ?

Now write out an expression for each K and combine them.

The group 15 elements

Kb =

[NH 4 + ][OH − ] [NH 3 ]

Ka =

225

[NH 3 ][H 3O + ] [NH 4 + ]

Combining these equations gives:

Kb [H 3O + ] [NH 4 + ] = = [NH 3 ] [OH − ] Ka ∴

K b × K a = [H 3O + ] × [OH − ] [H3O+] × [OH–] = Kw = 1 × 10–14

In water (pH 7): ∴

Ka × Kb = 1 × 10–14 pKa + pKb = 14

∴ 15.7

pKa = 14 – pKb = 14 – 4.75 = 9.25

Half-equations for oxidation of NH2OH to HNO3, and reduction of [BrO3]– (use Appendix 11 in H&S to help you): HNO3(aq) + 6H+(aq) + 6e–

NH2OH(aq) + 2H2O(l)

[BrO3]–(aq) + 6H+(aq) + 6e–

Br–(aq) + 3H2O(l)

Overall reaction is: NH2OH(aq) + [BrO3]–(aq) J HNO3(aq) + Br–(aq) + H2O(l) 15.8

(a) Preparation of sodium azide – use eq. 15.50 in H&S to help you: 3NaNH2 + NaNO3 J NaN3 + 3NaOH + NH3

X Na in liquid NH3: see Section 9.6 in H&S

(b) Sodium metal dissolved in liquid NH3 slowly liberates H2 and provides a method of making sodium amide: 2Na + 2NH3 J 2NaNH2 + H2 (Compare this to Na reacting with H2O to give NaOH.) (c) This reaction is a method of preparing lead azide: 2NaN3 + Pb(NO3)2 J Pb(N3)2 + 2NaNO3

15.9

(a) For species to be strictly isoelectronic, they must possess the same total number of electrons, not just the same number of valence electrons. A useful starting point for the answer is to note that C and N+ are isoelectronic, as are O and N–. Isoelectronic species include: O

C

O

N

N

N

N

C

N

O

N

O

O

C

N

The group 15 elements

Energy

σ*

Energy

226

σ*

π* 2py

2pz

2py

2py

2py

2py

N

N

non-bonding

2s

σ π σ

N

N

N

Fig. 15.1 Partial MO diagram to show σ-bond formation in [N3]– using a ligand group orbital approach.

N

N

(15.2)

N

N

Fig. 15.2 Partial MO diagram to show π-bond formation in [N3]– using a ligand group orbital approach. An analogous diagram can be constructed using 2px atomic orbitals (see text).

(b) The bonding scheme for [N3]– is similar to that for the isoelectronic CO2 (see Section 5.7 in H&S). You could begin with an atomic orbital basis set for each atom, but you can simplify the problem by considering the σ- and π-bonding separately. Lewis structure 15.2 is a good starting point to allow you to keep track of electrons. It is convenient to consider each terminal N atom as sp hybridized with one hybrid orbital accommodating one lone pair per N atom. Define an axis set: e.g. let the N atoms lie on the z axis. σ-Bond formation: Consider the interactions between in-phase and out-of-phase sp hybrid orbitals of the terminal N atoms (i.e. two ligand group orbitals) with the 2s and 2pz atomic orbitals of the central N atom. This is represented in the partial MO diagram in Fig. 15.1. Two σ-bonding MOs (with corresponding σ* MOs) result. Leave consideration of the electrons until later. π-Bond formation: Consider the interactions between in-phase and out-of-phase 2py orbitals of the terminal N atoms (i.e. two ligand group orbitals) with the 2py atomic orbital of the central N atom. This is represented in the partial MO diagram in Fig. 15.2. One π-bonding MO (with corresponding π* MO) results, as well as a non-bonding MO. An analogous partial MO diagram can be drawn using the 2px orbitals and again gives rise to one π-bonding MO (with corresponding π* MO) and a non-bonding MO. The πx and πy bonding MOs are degenerate; the πx* and πy* bonding MOs are degenerate. Electrons: The total number of valence electrons available is 16 (5 per N atom and 1 from the negative charge), i.e. 8 pairs. • 2 pairs are accommodated in outward-pointing sp hybrid orbitals on the terminal N atoms; • 2 pairs are accommodated in the two σ-bonding MOs (Fig. 15.1); • 2 pairs occupy the π-bonding and non-bonding MOs arising from interactions of the N 2py orbitals (Fig. 15.2); • 2 pairs occupy the π-bonding and non-bonding MOs arising from interactions of the N 2px orbitals. Check: There are 4 bonding (σ + π) pairs of electrons in [N3]– and this agrees with Lewis structure 15.2. There are 2 lone pairs per terminal N atom. The non-bonding electrons represent lone pairs in addition to those occupying the N sp hybrids.

The group 15 elements 15.10

Ni(B) Ni(A)

Fig. 15.3 Unit cell of NiAs; Ni centres are shown in dark grey.

(a) Figure 15.3 shows a single unit cell of NiAs. Consider the atoms labelled Ni(A) and Ni(B). Within this unit cell, Ni(A) has only one As neighbour, while Ni(B) has two. Ni(A) and Ni(B) lie in edge sites and each is shared between 4 unit cells. Figure 15.4 shows two adjacent unit cells; each of Ni(A) and Ni(B) is 3-coordinate. In the complete lattice, Ni(A) and Ni(B) are equivalent (to confirm this, ‘grow’ the structure in Fig. 15.4 by adding two more unit cells in front of those shown) and are 6-coordinate. The problem can be tackled in a similar manner by starting with a Ni centre in a corner site.

227

Ni(B) Ni(A)

Fig. 15.4 Two adjacent unit cells of NiAs; Ni centres are shown in dark grey.

(b) Per unit cell of NiAs (Fig. 15.3): Site

Number of Ni3+ Number of As3–

Within unit cell

0

2

Corner site

8 × 1/8 = 1

0

Edge site

4 × 1/4 = 1

0

Total

2

2

The stoichiometry is therefore Ni3+ : As3– = 2:2 = 1:1, or NiAs. 15.11

(a) Gas phase structures may be investigated using electron diffraction, or vibrational spectroscopy. (b) In the liquid phase, Raman spectroscopy can be used. IR spectroscopy is not appropriate because the symmetric stretching mode is IR inactive (no change in dipole moment).

15.12

Equation 15.66 in H&S:

F5S

Equation 15.67 in H&S:

Cl

SF5

Cl

2SF5

2Cl

N

F

Equation 15.68 in H&S:

NO is a radical

N

O

N

O

F Lewis structure of F2NNO

15.13

Fax P X

X X

(15.3)

Feq

P

Feq Feq

Fax

(15.4)

(a) Information is found in Section 15.7 of H&S. Points to include: • PX3 (X = F, Cl, Br, I) all molecular (15.3); • PF5 molecular (15.4) in solid, liquid and gas; fluxional in solution with Fax and Feq exchanging (Berry pseudo-rotation) on the NMR spectroscopic timescale; • PCl5 molecular (like 15.4) in gas phase; solid consists of [PCl4]+ (tetrahedral) and [PCl6]– (octahedral) ions; • Solid PBr5 is [PBr4]+Br– ; decomposes to PBr3 and Br2 in gas phase; • PI5 does not exist.

228

The group 15 elements (b) CsCl has structure shown in Fig. 6.5b (p. 88); Cs+ and Cl– ions are spherical. In [PCl4]+[PCl6]– , ions are tetrahedral and octahedral, respectively, but can be considered to be spherical if they occupy lattice sites in random orientations, or if each is freely rotating about the lattice site. [PCl4]+ and [PCl6]– are relatively large ions, and random orientations are more likely than rotational motion in the solid. 15.14

X Nuclear spin and abundance data: 31 P I = 1/2 100% 121 Sb I = 5/2 57.3% 123 Sb I = 7/2 42.7%

15.15 X Equation 15.70 in H&S is not a redox reaction

(a) [PF6]– is octahedral with 6 equivalent F atoms. Coupling of 19F nuclei to one nucleus gives a doublet in the 19F NMR spectrum. (b) [SbF6]– is octahedral with 6 equivalent F atoms. Coupling occurs to both 121Sb (57.3%) and 123Sb (42.7%) and the resulting signals are superimposed on top of each other. Coupling to 121Sb gives a 1 : 1 : 1 : 1 : 1 : 1 signal; 6 spin states (+5/2, +3/2, +1/2, –1/2, –3/2, –5/2) with equal probability of coupling to 19F. Coupling to 123Sb gives a 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 signal; 8 spin states with equal probability of coupling to 19F. Values of J(121Sb-19F) and J(123Sb-19F) are likely to be different, and exact appearance of signal in the 19F NMR spectrum depends on their values. 31P

Equation 15.64 in H&S: Oxidation states:

2NF3 + Cu J N2F4 + CuF2 +3 0 +2 +2 Cu oxidized, N reduced

Equation 15.73 in H&S: Oxidation states:

N2O4 + 2CoF3 J 2FNO2 + 2CoF2 +4 +3 +5 +2 N oxidized, Co reduced 3HNO2 J 2NO + HNO3 + H2O +3 +2 +5 N in one HNO2 oxidized, N in two HNO2 reduced

Equatio 15.111 in H&S: Oxidation states:

Equation 15.123 in H&S: Oxidation states:

15.16

Au + HNO3 + 4HCl J HAuCl4 + NO + 2H2O 0 +5 +3 +2 Au oxidized, N in HNO3 reduced

Assume static structures for the complexes. (a) cis- and trans-[PF4(CN)2]– : 31P couples to 19F (100 %, I = 1/ ). 2 For the trans-isomer, there is one Fa signal, a quintet in the 31P NMR b F spectrum (coupling to 4 19F). The P cis-isomer has 2 F environments F b (a and b). The 31P NMR spectrum Fa has one signal, a triplet of triplets cis (coupling to 2a and 2b). – (b) mer- and fac-[PF3(CN)3] The mer-isomer has 2 F Fa b environments (a and b). The 31P F NMR spectrum shows 2 signals P with coupling to 19F: a doublet NC (coupling to b) of triplets Fa (coupling to a). F atoms are mer equivalent in the fac-isomer. The 31 P NMR spectrum shows a quartet (coupling to 3 19F).

– CN

F

CN

–

CN P

F F

F CN

trans

– CN CN

–

F F

P

F CN

NC CN

fac

229

The group 15 elements 15.17

The three isomers of [PCl2F3(CN)]– are 15.5-15.7. Each isomer has 2 F environments. – – – Fb Fb b F

F a

Fb

P

CN Cl

a F

P

Cl

CN

Cl

Cl

F a

Fb

Cl

fac arrangement of F

mer arrangement of F cis arrangement of Cl

(15.5)

(15.6)

CN

P

Cl Fb

mer arrangement of F trans arrangement of Cl

(15.7)

If the structures are static in solution, the 19F NMR spectrum of each should show 2 signals. The observation that at room temperature, the 19F NMR spectra of 2 isomers exhibit 2 signals, while the spectrum of the third isomer shows only 1 signal means that 2 isomers are stereochemically rigid in solution but the third isomer is stereochemically non-rigid (fluxional). 15.18

(a) PCl5 donates Cl– to SbCl5 (not the other way around): PCl5 + SbCl5 J [PCl4]+[SbCl6]– (b) KF is source of F–, and AsF5 is an F– acceptor: KF + AsF5 J K+[AsF6]– (c) NOF donates F–, and SbF5 is a F– acceptor: NOF + SbF5 J [NO]+[SbF6]–

X See Section 9.9 in H&S

(d) HF donates F– to SbF5 (a good F– acceptor) and H+ combines with HF: 2HF + SbF5 J [H2F]+[SbF6]– In (c) and (d), there is also a tendency for Sb–F–Sb bridge formation, for example: SbF5 + [SbF6]– J [Sb2F11]–

15.19 F F

F

Sb

F

F

F F

F

Sb

F

F

F

(15.8) F F

– F

Sb

F

Sb

F F

(15.9)

F

(a) See structures 15.8 and 15.9. [Sb2F11]– has 6 pairs of electrons in the valence shell of each Sb(V) centre: the formal scheme shown on the right can be used to confirm this, although remember that the Sb centres are equivalent. By VSEPR, 6 pairs of electrons are consistent with an octahedral environment as observed. [Sb2F7]– has 5 pairs of electrons in the valence shell of each Sb(III) centre: the formal scheme shown on the right confirms this, but keep in mind that the Sb centres are actually equivalent. By VSEPR, 5 pairs of electrons are consistent with trigonal bipyramidal environments.

F F

F

Sb

F

F

F F

F

F

F F

– Sb

F

Sb

F

F F

Sb

F F

F

230

The group 15 elements (b) [{BiX4}n]n– and [{BiX5}n]2n– are chains (see right) with octahedral Bi(III). Stoichiometry ‘BiX4’ follows by having two X atoms in bridging sites and two terminal; stoichiometry ‘BiX5’ requires one bridging X.

Energy

15.20 4σ

π* 3σ

π

1σ

Fig. 15.5 Approximate ordering of the MOs in [NO]+.

15.21

X

n– X

Bi

X

Bi

X X

2n–

X

n

X X

n

[NO]+ is isoelectronic with CO. An MO diagram similar to that for CO can be constructed. Starting with the ordering of the MOs in CO, it is expected that the MOs in [NO]+ will lie in approximately the same order (Fig. 15.5). NO has one more electron than [NO]+ and this occupies a π* MO (add an electron to Fig. 15.5). On going from NO to [NO]+, the bond order in [NO]+ increases because occupation of the π* level cancels out the bonding contribution of one π electron. It follows that the bond distance decreases, and the bond strengthens. The relationship between wavenumber and force constant, k, for the bond is:

ν =

2σ

X

1 2πc

k

μ

where μ = reduced mass

Stronger bond, larger force constant, and higher value of ν .

For the reaction between oxalate and permanganate: 2CO2 + 2e– [C2O4]2– – + [MnO4] + 8H + 5e– Mn2+ + 4H2O Overall: 2[MnO4]– + 5[C2O4]2– + 16H+ J 10CO2 + 2Mn2+ + 8H2O The amount of [C2O4]2– present = 25 × 0.0500 × 10–3 = 1.25 × 10–3 moles ∴ Amount of [MnO4]– in 24.8 cm3 = 2/5 × 1.25 × 10–3 = 0.500 × 10–3 moles ∴ Concentration of KMnO4 solution, A =

0.500 ×10 −3 × 10 3 = 0.0202 mol dm–3 24.8

Now consider the oxidation of NH2OH. Fe3+ is reduced to Fe2+, and the amount of Fe2+ produced is equivalent to 24.65 cm3 of KMnO4 solution: 5Fe2+ + [MnO4]– + 8H+ J 5Fe3+ + Mn2+ + 4H2O Amount of [MnO4]– in 24.65 cm3 = 24.65 × 0.0202 × 10–3 = 4.98 × 10–4 moles ∴ Amount of Fe2+ = 5 × 4.98 × 10–4 = 2.49 × 10–3 moles ∴ Amount of Fe3+ reacting with NH2OH = 2.49 × 10–3 moles Amount of NH2OH used = 25.0 × 0.0494 × 10–3 = 1.24 × 10–3 moles ∴ Ratio of moles of Fe3+ : NH2OH = 2.49 : 1.24 = 2 : 1 The reduction of Fe3+ to Fe2+ is a 1-electron reduction, and (from stoichiometry above), the oxidation of NH2OH to B is a 2-electron oxidation per N, taking N from oxidation state –1 to +1. B may be N2O. The oxidation half-equation is: 2NH2OH N2O + 4H+ + H2O + 4e– Overall: 2NH2OH + 4Fe3+ J N2O + 4H+ + H2O + 4Fe2+

The group 15 elements 15.22 O– P

O + –

O

+ O

O P P +

O

P

O

+

Information for this answer is in Sections 15.10 and 15.11 in H&S. Points to include: • Structure of phosphorus(V) oxide (15.10); other polymorphs also contain {OP(–O)3} units; see H&S for discussion of valence stuctures. • Action of water on P4O10 gives phosphoric acid with tetrahedral PO4 unit (15.11), and salts retain this unit in [PO4]3–, [HPO4]2– and [H2PO4]–. O– O

O– P

O

HO

–

O

or

OH OH

HO

(15.11)

• Formation of condensed phosphates, e.g. H4P2O7 and H5P3O10; give a general representation to show retention of tetrahedral PO4 units:

O–

O

P

OH OH

(15.10) –

231

P+ O –

+ O –

O

P

P O

O

P

+

HO

P

P

+ H 2O

P O

O– O–

• Cyclic phosphates include [P3O9]3– (15.12), [P4O12]4– and [P6O18]6–. • Biological role of phosphates, e.g. ATP and ADP (see Box 15.11 in H&S) and DNA (Fig. 10.13 in H&S).

(15.12)

15.23 O

Fe

Sb Sb Fe

OH

+

O

(a) The unit cell of FeSb2O6 is shown in Fig. 15.6, and that of rutile in Fig. 6.5c (p. 88). Compare the two. The unit cell of FeSb2O6 consists of three stacked rutile unit cells (a triple-rutile lattice). (b) The three rutile-type subunits that make up one unit cell of FeSb2O6 are not identical: two subunits have 4 Sb and 4 Fe in the corner sites, while the central subunit has 8 Sb in the corner sites. Therefore, a unit cell consisting of all three subunits is needed to unambiguously define the 3D-structure of FeSb2O6. (c) O2– is 3-coordinate (trigonal planar); Fe(II) is 6-coordinate (octahedral); Sb(V) is 6-coordinate (octahedral). (d) Per unit cell of FeSb2O6: Site

Fig. 15.6 Unit cell of FeSb2O6.

Number of Fe(II)

Number of Sb(V)

Number of O2–

Within unit cell

1

2

10

Corner site

8 × 1/8 = 1

0

0

Edge site

0

8 × 1/4 = 2

0

Face site

0

0

4 × 1/2 = 2

Total

2

4

12

The stoichiometry is therefore Fe(II):Sb(V):O2– = 2:4:12 = 1:2:6, or FeSb2O6. 15.24

(a) Salts contain [P3O10]5– and [P4O13]6–, condensed phosphates (15.13) and 15.14).

–

O–

O–

O–

P+

P+

P

O

O O–

+ O–

O O–

(15.13)

O–

–

O–

O–

O–

P+

P+

P

O

O O–

+

(15.14)

P

+ O–

O

O O–

O–

O–

O–

232

The group 15 elements [P3O10]5– contains two P environments and gives two signals in the 31P NMR spectrum (relative integrals 2 : 1); [P4O13]6– also contains two P environments but in a 2 : 2 ratio, therefore gives two signals of equal integrals. (b) AsF5 has structure 15.15. A static structure would give two signals in the 19F NMR spectrum (relative integrals 3 : 2 for eq : ax). Dynamic behaviour exchanges Fax and Feq and if the rate of exchange is faster than the NMR timescale, only one signal is seen in the 19F NMR spectrum. Recording the spectrum over a range of temperatures may allow both the high-temperature limiting spectrum (one signal) and the low-temperature limiting spectrum (two signals) to be observed (or at least, some broadening of signals) but this depends on the activation energy of the exchange process. (c) There are three isomers:

Fax Feq

As

Feq Feq

Fax

(15.15)

Cl

Cl P

N N

P

N

P

NMe2

Me2N

NMe2

Cl

Cl

NMe2

Cl P

N N

P

Cl

Me2N

(15.16)

N

P NMe2

NMe2

Cl

Cl P

N N

P

NMe2

Cl

(15.17)

N

P NMe2

(15.18)

31P

or 1H NMR spectroscopy could be used. 15.16 has one P environment, 15.17 has two (although the signals may be near to coincident because the environments are very similar), and 15.18 has three. Assuming (on NMR spectroscopic timescale) free rotation about the P–N and C–N bonds within the PNMe2 units, the 1H NMR spectrum of 15.16 shows one signal, that of 15.17 two, and that of 15.18 three.

15.25

(a) From the reaction stoichiometry, 2-electron oxidation step is: 2Ti(III) J 2Ti(IV) + 2e– and this must be balanced by a 2-electron reduction. In NH2OH, oxidation state of N is –1, so oxidation state in product is –3. Product is NH3: NH2OH + H2O + 2e– J NH3 + 2[OH]– (b) From the reaction stoichiometry, 2-electron reduction step is: 2Ag+ + 2e– J 2Ag and this must be balanced by a 2-electron oxidation. ∴ P(III) oxidized to P(V); product is [PO4]3– : [HPO3]2– + H2O J [PO4]3– + 2e– + 3H+ (c) The 2-electron reduction: I2 + 2e– J 2I– occurs twice, and so the change in oxidation state for P is +1 to +3 to +5. This corresponds to H3PO2 J H3PO3 J H3PO4.

The group 15 elements 15.26 F N

F

F

F

(15.19)

233

(a) [NF4]+ N+, number of valence electrons = 4 Number of bonding pairs (4 N–F bonds) = 4 No lone pairs ‘Parent’ shape = molecular shape = tetrahedral, see 15.19. (b) [N2F3]+, consider as [F2NNF]+, derived from N2F4 with loss of F– : F

F N

–F

N F

–

F

F

F N

N

F

F N

F

N

F

(15.20)

Loss of F– generates a sextet N that is stabilized by π-donation to give planar structure 15.20. (c) NH2OH, consider N and O centres separately: N, number of valence electrons = 5 Number of electron pairs = 1 lone + 3 bonding pairs ‘Parent’ shape = tetrahedral; environment around N = trigonal pyramidal O, number of valence electrons = 6 Number of electron pairs = 2 lone + 2 bonding pairs ‘Parent’ shape = tetrahedral; environment around O = bent See structure 15.21. (d) SPCl3 P, number of valence electrons = 5 Number of bonding ‘pairs’ (1 P=S + 3 P–Cl bonds) = 4 No lone pairs ‘Parent’ shape = molecular shape = tetrahedral, see 15.22 (e) PCl3F2 P, number of valence electrons = 5 Number of bonding pairs (5 P–X bonds) = 5 No lone pairs ‘Parent’ shape = molecular shape = trigonal bipyramidal Three isomers are possible depending on positions of F and Cl atoms; on steric grounds, it is most likely that Cl atoms occupy equatorial positions, see 15.23.

H N

O

H

H

(15.21)

S P

Cl Cl

Cl

(15.22)

F Cl

P

Cl Cl

F

(15.23)

15.27 X For reduction by Zn, see eq. 15.17 in H&S

The fate of the 15N label in each step is critical; in each case, start with [15NO3]–. (a) Reduce [NO3]– to NH3 using Al or Zn; redox half-equations are: [15NO3]– + 6H2O + 8e– Al + 4[OH]–

15NH

3

+ 9[OH]–

[Al(OH)4]– + 3e–

Overall: 3[15NO3]– + 8Al + 5[OH]– + 18H2O J 3 15NH3 + 8[Al(OH)4]– Now react 15NH3 with sodium: 2 15NH3 + 2Na J 2Na[15NH2] + H2

234

The group 15 elements (b) First, prepare 15NH3 as in part (a). Oxidize 15NH3 with CuO or NaOCl: or

2 15NH3 + 3CuO J 15N2 + 3Cu + 3H2O 2 15NH3 + 3[OCl]– J 15N2 + 3Cl– + 3H2O

X See eqs. 15.97 and 15.98 in H&S

(c) Reduction of [15NO3]– with Fe2+ or Hg, e.g. 2[15NO3]– + 6Hg + 8H+ J 2 15NO + 3[Hg2]2+ + 4H2O To convert NO to [NO]+ requires an oxidizing agent, but choice of oxidant and conditions affect salt formed. Use Cl2 (which produces Cl–) in the presence of AlCl3 (Lewis acid which accepts Cl–): 2 15NO + Cl2 + 2AlCl3 J 2[15NO]+ + 2[AlCl4]–

15.28

X See Section 15.2 and eq. 15.11 of H&S

The fate of the 32P label in each step is critical; each reaction scheme starts with Ca3[32PO4]2 (phosphate rock). (a) Reduce [32PO4]3– to 32P4 and treat with NaOH(aq): 2Ca3[32PO4]2 + 6SiO2 + 10C 32P 4

1700 K

32P 4

+ 6CaSiO3 + 10CO

+ 3NaOH + 3H2O J 3NaH2PO2 + 32PH3

(b) First prepare 32P4 as above, then convert to PCl3 and treat with ice-cold water: 32P 4

+ 6Cl2 (limited supply) J 4 32PCl3

32PCl

3

+ 3H2O J H332PO3 + 3HCl

(c) First prepare 32P4 as above, then convert to 32P4S10, followed by treatment with Na2S: excess S 8; >570 K 32P 32P S 4 4 10 32P S 4 10

15.29

+ 16Na2S J 4Na332PS4 + 10H2S

For the reaction between oxalate and permanganate: [C2O4]2– 2CO2 + 2e– [MnO4]– + 8H+ + 5e– Mn2+ + 4H2O Overall: 2[MnO4]– + 5[C2O4]2– + 16H+ J 10CO2 + 2Mn2+ + 8H2O The amount of [C2O4]2– present = 25.0 × 0.0500 × 10–3 = 1.25 × 10–3 moles ∴ Amount of [MnO4]– in 24.8 cm3 = 2/5 × 1.25 × 10–3 = 0.500 × 10–3 moles ∴ Concentration of KMnO4 solution, C = Now consider the oxidation of N2H4.

0.500 ×10 −3 × 10 3 = 0.0202 mol dm–3 24.7

The group 15 elements

235

[Fe(CN)6]3– is reduced to [Fe(CN)6]4–, and the latter reacts with 24.80 cm3 of solution C: 5[Fe(CN)6]4– + [MnO4]– + 8H+ J 5[Fe(CN)6]3– + Mn2+ + 4H2O Amount of [MnO4]– in 24.80 cm3 = 24.80 × 0.0202 × 10–3 = 4.96 × 10–4 moles ∴ Amount of [Fe(CN)6]4– = 5 × 4.96 × 10–4 = 2.48 × 10–3 moles ∴ Amount of [Fe(CN)6]3– reacting with N2H4 = 2.48 × 10–3 moles Amount of N2H4 used = 25 × 0.0250 × 10–3 = 6.25 × 10–4 moles ∴ Ratio of moles of [Fe(CN)6]3– : N2H4 = 4 : 1 13

14

15

B

C

N

Al

Si

P

Ga

Ge

As

The reduction of Fe(III) to Fe(II) is a 1-electron reduction, and (following from the stoichiometry above), the oxidation of N2H4 to compound D is a 4-electron oxidation, i.e. a 2-electron oxidation per N. This takes N from oxidation state –2 to 0. D is N2. The oxidation step is: N2H4

N2 + 4H+ + 4e–

Overall: N2H4 + 4[Fe(CN)6]3– J N2 + 4H+ + 4[Fe(CN)6]4–

(15.24)

15.30

Write AlPO4 as 2{(Al,P)O2} to see relationship with SiO2. Think of the periodic relationship between Al, Si and P (15.24): replacement of 2Si(IV) by {P(V) + Al(III)} retains electrical neutrality. The sizes of Si(IV), P(V) and Al(III) centres are comparable, and therefore, the SiO2 macromolecular structure is not perturbed significantly on going from SiO2 to (Al,P)O2.

15.31

(a) The P–N bond lengths in phosphazenes such as [Cl3P=N–(PCl2=N)x=PCl3]+ are all similar (153 to 158 pm) , indicating that the bonding is delocalized. This is in contrast to the double and single bonds that are usually drawn in structural diagrams of these compounds. The bonding is best described in terms of contributions from charge-separated resonance structures (i.e. ionic bonding), and negative hyperconjugation involving to n(N)Jσ*(P–Cl) electron donation where n(N) represents the N lone pair. This is analogous to the negative hyperconjugation described for N(SiH3)3 in answer 14.7 (p. 207). (b) In [Cl3P=N–(PCl2=N)x=PCl3]+, the P–N–P bond angles are around 140-160o. The following resonance structure illustrates charge separation: Cl P

Cl Cl

15.32

(a) Data:

11B 31P

Br

Pr P Pr

13C

Br

(15.25)

–

N

P

+

–

N Cl

P

+ Cl Cl

I = 3/2 Abundance = 80.1% I = 1/2 Abundance = 100% I = 1/2 Abundance = 1.1%

Br

B

Pr

+

Cl

Cl

See structure 15.25. Ignore any coupling to 13C since sample is not enriched; coupling to 10B (I = 3) can also essentially be ignored (see case study 4 in Section 4.7 in H&S). The only spinspin coupling to consider is between 31P and 11B.

236

The group 15 elements In the 11B NMR spectrum, coupling to one 31P nucleus (two spin states: +1/2 and –1/2) gives rise to a doublet . In the 31P NMR spectrum, coupling to one 11B nucleus (four spin states: +3/2 , + 1/2, –1/2, –3/2) gives rise to a 1 : 1 : 1 : 1 (non-binomial) quartet (see Fig. 13.4, p. 190). The coupling constants in each spectrum must be the same. (b) Refer to Section 7.9 in H&S. Draw an appropriate thermochemical cycle:

[NH4][PF6](s)

–ΔlatticeGo(NH4PF6, s)

[NH4]+(g) + [PF6]–(g) ΔhydGo([NH4]+, g) + ΔhydGo([PF6]–, g)

ΔsolGo(NH4PF6,s)

[NH4]+(aq) + [PF6]–(aq) The solubility of the salt depends on the balance between the Gibbs lattice energy and the hydration energies of the ions; hydrogen bonding plays a part in the latter. ΔsolGo(NH4PF6, s) = ΔhydGo([NH4]+, g) + ΔhydGo([PF6]–, g) – ΔlatticeGo(NH4PF6, s) (c)

15.33

X Superacids: see Section 9.9 in H&S

[AsBr4][AsF6] J [AsBr4]F + AsF5

Reverse of the more usual reaction of AsF5 acting as F– acceptor

[AsBr4]F J AsBr2F + Br2

As(V) reduced to As(III); 2Br(–1) oxidized to Br2

AsBr2F + AsF5 J 2AsF3 + Br2

As(V) reduced to As(III); halide redistribution; 2Br(–1) oxidized to Br2;

3AsBr2F J 2AsBr3 + AsF3

No redox changes; halide redistribution

(a) PI3 + IBr + GaBr3 J [PI4]+[GaBr4]–

IBr is source of I+ and Br–; GaBr3 is halide acceptor

(b) POBr3 + HF + AsF5 J [P(OH)Br3]+[AsF6]–

HF/AsF5 is a superacid

(c) 2Pb(NO3)2 (d) 2PH3 + 2K

Δ

2PbO + 4NO2 + O2

liquid NH3

2K[PH2] + H2

In liquid NH3, PH3 acts as an acid

(e) Li3N + 3H2O J NH3 + 3LiOH (f) H3AsO4 + SO2 + H2O J H3AsO3 + H2SO4

H3AsO4 acts as oxidant

(g) BiCl3 + H2O J BiOCl + 2HCl

Hydrolysis of BiX3 for X = F, Cl, Br, I gives BiOX

(h) PCl3 + H2O J H3PO3 + 3HCl

Hydrolysis of PX3 gives H3PO3

The group 15 elements 15.34

(a) P

S

O

O N

O

O

O O

O

N

O

O

(15.26)

15.35

2–

F F F

F Bi F

F F

(15.27)

3–

F F

Sb

F

P

P P

O

Fe

O

–

N

O

S

S

F F

F

(15.28)

237

P group 15 5 valence electrons S group 16 6 valence electrons Each P atom forms 3 single bonds (obeys octet rule); each S atom forms 2 single bonds (obeys octet rule). All bonds are σ-bonds. Each atom can be considered as sp3 hybridized; each P has 1 lone pair, each S has 2 lone pairs of electrons.

(b) Bi behaves as a typical metal and its electrical resistivity increases as the temperature increases (see Fig. 5.9 in H&S). (c) Hydrated Fe(NO3)3 in hot 100% HNO3 in the presence of P2O5 gives a solid, ionic salt. P2O5 is a dehydrating agent and the salt [NO2][X] is likely to contain a complex anion containing Fe3+ with [NO3]– ligands. The most reasonable proposal for this complex is [Fe(NO3)4]–. The ionic compound is [NO2]+[Fe(NO3)4]–. The 8-coordinate Fe(III) centre arises because each [NO3]– ligand is bidentate (structure 15.26). (a) Each nucleus in [HPF5]– has I = 1/2 and is 100% (or for 1H, close to 100%) abundant. There are two F environments (ratio 1:4). The 31P NMR spectrum is a doublet (939 Hz) of doublets (731 Hz) of quintets (817 Hz). The spectrum is a 20line signal (Fig. 15.7). (b) [BiF7]2– Central atom is Bi (group 15); it has 5 valence electrons Add 2 more electrons from the negative charge Number of bonding pairs (7 Bi–F bonds) = 7 No lone pairs Total number of electron pairs = 7, so a pentagonal bipyramid (15.27) The observed structure is consistent with the predictions of VSEPR. [SbF6]3– Central atom is Sb (group 15); it has 5 valence electrons Add 3 more electrons from the negative charge Number of bonding pairs (6 Sb–F bonds) = 6 Number of lone pairs = 1 Total number of electron pairs = 7, so a pentagonal bipyramid The molecular shape is predicted to be derived from the parent shape, but is observed to be octahedral (15.28). It is therefore not consistent with the predictions of VSEPR, and the lone pair must be stereochemically inactive. J(31P–1H)

Fig. 15.7 Structure of [HPF5]– and simulation of its 31P NMR spectrum. The coupling pattern is a doublet of doublets of quintets, and measurements of the coupling constants are shown. Exercise: what other pairs of lines can be used to measure the three values of J ?

Feq

J(31P–19Feq)

–

H

J(31P–19Fax)

P

Feq Feq

Feq Fax

238

The group 15 elements (c) The reaction scheme to consider is:

X For more information, see: K.O. Christe (1995) J. Am. Chem. Soc, vol. 117, p. 6136

NF3 + NO + 2SbF5

420 K Step 1

[F2NO]+[Sb2F11]– + N2 Step 2

>450 K

F3NO + 2SbF5 Step 3 >520 K

[NO]+[SbF6]–

SbF5 Step 4

FNO + F2

Step 5

NF3, SbF5

[NF4]+[SbF6]– Step 1: redox reaction involving oxidation and reduction of N; +3 oxidation states of N NF3 NO +2 [F2NO]+ +5 N2 0 SbF5 acts as Lewis acid accepting F–, formally to give [SbF6]– (Lewis base) which combines with SbF5 (Lewis acid) to give [Sb2F11]–. Step 2: N remains in oxidation state +5; F– transfer from [Sb2F11]– to [F2NO]+ to give F3NO. Step 3: redox reaction: F3NO J FNO + F2 Ox. state of N: +5 +3 Reduction: ox. state change = –2 Ox. state of F: –1 –1 0 Oxidation of 2F: ox.state change = +2 Step 4: N stays in oxidation state +3, Sb remains in ox. state +5, F remains in ox. state –1; therefore, no redox chemistry. SbF5 acts as Lewis acid – accepts F– to give [SbF6]–. Step 5: N is oxidized from ox. state +3 in both FNO and NF3, to +5 in [NF4]+, but reduction product is not specified. SbF5 acts as Lewis acid, accepting F– to give [SbF6]–. 15.36

(a) For information on spinels, see Box 13.7 in H&S. Points to include: • spinels have the general formula AB2X4; usually, X = O, A and B are metals in ox. states +2 and +3 respectively, e.g. MgAl2O4, Fe3O4 = FeIIFeIII2O4. • ccp arrangement of O2– ions; 1/8 tetrahedral holes occupied by A2+ ions; 1/2 octahedral holes occupied by B3+ ions; unit cell contains 8 formula units, i.e. [AB2X4]8. • Structure of an inverse spinel is related to that of a spinel by exchanging the sites of the A2+ ions with half of the B3+ ions. • Sn3N4 is first nitride spinel that is stable under ambient conditions.

The group 15 elements

• Nitride spinels have the N atoms in ccp structure with 2/3 of the Si, Ge or Sn

O

atoms in octahedral holes and 1/3 in tetrahedral holes. Whereas oxide spinels have mixed oxidation state metals (M2+ and M3+), the group 14 nitride spinels have Si, Ge or Sn atoms in one (+4) oxidation state (+4).

As Cl

Cl Cl

(15.29) Cleq Cleq

Clax

O

i As

As Clax

239

Cleq Cleq

O

C2 axis passes through i

(15.30)

15.37

(b) C3v symmetry is consistent with monomer 15.29 and this species must be present in solution. The solid state structure at 153 K has C2h symmetry, i.e. a C2 axis, a mirror plane perpendicular to this axis, and an inversion centre in addition to the E operator. The molecule is a dimer with each As centre in a trigonal bipyramidal environment. In structure 15.30, the symmetry elements of the dimer are shown. There is an inversion centre at the midpoint of As...As and the C2 axis runs through this point, perpendicular to the plane of the paper; the As, Clax and O atoms lie in the σh mirror plane. (a) Fuming nitric acid appears orange because of photochemical decomposition that produces NO2: 4HNO3 J 4NO2 + 2H2O + O2 (b) An azeotrope is a mixture of two liquids that distils unchanged, the composition of liquid and vapour being the same. The composition of the azeotropic mixture depends on pressure. Concentrated HNO3 is the azeotrope containing 68% by weight of HNO3 and boiling at 393 K.

15.38

Remove 2 vertices

(a) The ‘@’ symbol in [Pd@Bi10]4+ shows that the cluster consists of a Bi10 cage that encapsulates a Pd atom. Make the assumption that the central M(0) atom does not contribute any valence electrons (ve) to cluster bonding. The cluster electron count is then: 10 Bi atoms (group 15, allocate one lone pair) = 10 × 3 = 30 ve Subtract 4 for the 4+ charge Total ve count = 26 = 13 electron pairs. The structure is based on a 12 vertex deltahedron with 2 vertices missing, i.e. a pentagonal antiprism (15.31). The structure is related to that of [Pd@Pb12]2– which is a closo, icosahedral cluster: (12 × 2) + 2 = 30 ve. (b) Data for the question: ΔfGo(N2O, g) = +104 kJ mol–1, ΔfGo(H2O, l) = –237 kJ mol–1, ΔfGo(NH4ClO4, s) = –89 kJ mol–1 4NH4ClO4(s) J 2Cl2(g) + 2N2O(g) + 3O2(g) + 8H2O(l)

(15.31)

ΔrGo = 2ΔfGo(N2O, g) + 8ΔfGo(H2O, l) – 4ΔfGo(NH4ClO4, s) = 2(104) + 8(–237) – 4(–89) = –1332 kJ mol–1 Four moles of a solid are converted to 7 moles of gas plus 8 moles of liquid. The change in entropy is highly positive, making the reaction entropically very favourable. 15.39

(a) Reaction of equimolar amounts of NaN3 and NaNO2 in acidic solution: [N3]– + [NO2]– + 2H+ J N2 + N2O + H2O

The group 15 elements

240

+ N

N

Collect the two gases and record the IR spectrum of the mixture. The stretching mode of the symmetrical N2 molecule is not IR active. The IR spectrum of N2O (15.32) shows bands at 2202 cm–1 (NN stretch), 1270 cm–1 (NO stretch) and 585 cm–1 (bend). (b) POCl3 reacts with an excess of Me2NH to give A. Use the elemental analysis data to obtain the ratio of C : N : H:

O

(15.32)

C:N:H =

40.21 23.45 10.12 : : 12.01 14.01 1.01

= 3.35 : 1.67 : 10.02 = 2 : 1 : 6

The reaction is likely to eliminate HCl, and the 1H and 13C NMR spectroscopic data show that there is only one H and one C environment. Possible products are PO(NMe2)3, PO(NMe2)2Cl and PO(NMe2)Cl2. Of these, only PO(NMe2)3 (15.33) has the correct elemental analysis. In resonance form 15.33, each C, N, P and O atom obeys the octet rule. The formation of B adds support for the formation of 15.33 because:

O P Me2N

+ NMe2 NMe2

(15.33)

R N H O

NMe2 P

Me2N

NMe2

R N

O 4Me2NH

P

Me2N

H

Me2N

NMe2

H

P

P

Me2N

O

NMe2

O

N R

H

B

N R

The miscibility of A with water arises from hydrogen bonding P–O...H. The formation of A: Me2NH + POCl3 J PO(NMe2)3 + 3Me2NH2Cl 15.40

F is more electronegative than Cl. Therefore the structures are: F Cl

P

F Cl

F

P

Cl

F

F F

P

Cl

Cl

P

F

Cl F

(15.34)

F

F

(15.35)

F Cl

Cl

P

Cl

F F

F

F

F

D3h

C2v

C2v

The point groups can be confirmed by using Fig. 3.10 in H&S. For PF2Cl3, the principal (C3) axis runs through the F–P–F bonds. For PF3Cl2, the principal (C2) axis bisects the Cl–P–Cl angle. For PF4Cl, the principal (C2) axis runs along the Cl–P bond. From the structures above, you can see that PF2Cl3 is non-polar, but both PF3Cl2 and PF4Cl are polar. Structures 15.34 and 15.35 show the directions of the dipole moments. By SI convention, the arrow points from δ – to δ +.

The group 15 elements 15.41

241

(a) Urea is (H2N)2CO. Reaction with nitrous acid: (H2N)2CO + 2HNO2 J 2N2 + CO2 + 3H2O (b) Reaction with sulfamic acid: [NO2]– + H2NSO3H J N2 + [HSO4]– + H2O (c) Under standard conditions (pH = 0), Eo for the following half-cell reaction is +0.93 V: [NO3]–(aq) + 3H+(aq) + 2e–

HNO2(aq) + H2O(l)

Variation in the reduction potential is given by the Nernst equation:

E = Eo –

= Eo –

RT ⎛ [reduced form] ⎞ ⎟ × ⎜ ln zF ⎜⎝ [oxidized form] ⎟⎠ [HNO2 ] ⎞⎟ RT ⎛⎜ × ln ⎜ zF ⎝ [NO3− ][H+ ]3 ⎟⎠

A change in [H+] alters the reduction potential. pH is related to [H+] by:

pH = –log [H+]

[H+] = 10–pH

The pH also affects the oxidizing ability of H2O2, [OCl]– and HOCl as these halfcells involve H+ ions. For the products of the reduction of H2O2 and HOCl in acidic solutions, look for relevant half-equations in Appendix 11 in H&S: H2O2(aq) + 2H+(aq) + 2e–(aq) 2HOCl(aq) + 2H+(aq) + 2e–(aq) or HOCl(aq) + H+(aq) + 2e–(aq)

2H2O(l) Cl2(aq) + 2H2O(l) Cl–(aq) + H2O(l)

In acidic solution, H2O2 oxidizes [NO2]– (as HNO2) as follows: H2O2(aq) + HNO2(aq) + H2O(l) J HNO3(aq) + 2H2O(l) In alkaline solution, [OCl]– oxidizes nitrite to nitrate: [OCl]– + [NO2]– J Cl– + [NO3]– 15.42

(a)

12P + 10KClO3 J 3P4O10 + 10KCl

Oxidation state changes: P 0 to +5 Total change = +60 Cl +5 to –1 Total change = –60 The glass powder provides the friction for lighting the match.

242

The group 15 elements (b) Oxidation states in starting materials (see structure 15.36 for P4S3): P4S3 P +3 3P +1 S –2 KClO3 Cl +5 (only Cl in KClO3 is involved in redox changes) O2 O –2

P

S

S S

P

KClO3 and O2 must be reduced. P4S3 must be oxidized.

P P

P4S3 + 5KClO3 + 1/2O2 J P4O10 + 3SO2 + 5KCl

(15.36)

Check oxidation state changes: P +3 to +5 Total change for 1P = +2 P +1 to +5 Total change for 3P = +12 S –2 to +4 Total change for 3S = +18 Total oxidation change = +32 Cl +5 to –1 Total change for 5Cl = –30 O 0 to –2 Change for O = –2 Total reduction change = –32 15.43

(a) See Section 15.13 in H&S, starting with eq. 15.144 in H&S plus details of polymerization reactions given in H&S. (b) OCH2CF3 P

N

OCH2CF3

A n

NaOCH2CF3 (NPCl2)n

NHMe MeNH2

N

P NHMe

H2NCH2CO2Et

B n

NHCH2CO2Et N

P n NHCH2CO2Et

15.44

C

(a) Phosphate rock is crude calcium phosphate: 2Ca3(PO4)2 + 6SiO2 + 10C

1700 K

P4 + 6CaSiO3 + 10CO

(b) Superphosphates: Ca(H2PO4)2 mixed with CaSO4 and other sulfates Triple superphosphates: Ca(H2PO4)2 (c) Condensation of phosphates is the elimination of water between two phosphates leading to P–O–P bridge formation and polyphosphate anions. See answers 15.22 and 15.24 on p. 231 for diagrams. (d) [P3O10]5– is a chelating ligand and binds Ca2+ and Mg2+ ions in hard water, solubilizing them and softening the water. Environmental concerns about phosphates (see Box 15.10 in H&S) have resulted in a reduction in the use of Na5[P3O10] and the current widespread use of zeolites for water softening. See answer 14.31 (p. 221) for an explanation of how zeolites remove Ca2+ and Mg2+ ions from hard water.

243

16 The group 16 elements 16.1

(a)

O S Se Te Po (b) ns2np4

oxygen sulfur selenium tellurium polonium

16.2

Start with 209Bi (Z = 83). In the (n,γ) reaction, a neutron reacts with the substrate and γ-radiation is released. Addition of the neutron converts 209Bi to 210Bi. Loss of a β-particle leads to an increase in Z by 1, but mass number stays the same: β− 209 210 210 83 Bi(n, γ ) 83 Bi ⎯⎯→ 84 Po

16.3

Aqueous alkali = aqueous solution of [OH]–, e.g. aqueous NaOH: At the anode: 4[OH]–(aq) J O2(g) + 2H2O(l) + 4e– At the cathode: 2H+(aq) + 2e– J H2(g)

16.4

For E = O: 8O(g) J 4O2(g)

S

S

O2 contains O=O bond; enthalpy change (exothermic because bonds are formed) is: ΔrHo = –4D(O=O) = –4(498) = –1992 kJ mol–1

S

S

S S

S

(16.1)

S

8O(g) J O8(g) Assume O8 is a structural analogue of S8 (16.1) with 8 O–O bonds; for this reaction: ΔrHo = –8D(O–O) = –8(146) = –1168 kJ mol–1 ∴ Formation of diatomic O2 is thermodynamically favoured over O8 ring formation. For E = S: 8S(g) J 4S2(g) ΔrHo = –4D(S=S) = –4(427) = –1708 kJ mol–1 8S(g) J S8(g) ΔrHo = –8D(S–S) = –8(266) = –2128 kJ mol–1 ∴ Formation of cyclic S8 is thermodynamically favoured over S2 formation.

244

The group 16 elements 16.5

X See: M. Laing (1989) J. Chem. Educ., vol. 66, p. 453; J.-P. Puttemans and G. Jannes (2004) J. Chem. Educ., vol. 81, p. 639; T.S. Carlton (2006) J. Chem. Educ., vol. 83, p. 477

(a) For O2, occupancies of the πg* level:

Ground state (triplet)

Second excited state (singlet)

The formal bond order is 2 in O2 in both the ground and these two excited states. (b) The simultaneous excitation of 2 electrons from ground to first excited states requires 2 × 94.7 kJ mol–1. E = 189.4 kJ mol–1 Per pair of molecules, E =

ν = λ=

16.6

First excited state (singlet)

189.4 × 10 3 6.02 × 10 23

J = 3.15 × 10–19 J

E (3.15 × 10 −19 J) = = 4.75 × 1014 s −1 h (6.626 × 10 −34 J s) c

ν

=

2.998 × 1010 4.75 × 1014

= 6.31 × 10 − 5 cm = 631 nm

(a) Reaction to consider: 2H2O2 J 2H2O + O2 which is the combination of half-equations:

X See Chapter 8 for further details of the method used in this answer

O2 + 2H+ + 2e– H2O2 + 2H+ + 2e– For the overall reaction:

2H2O2 2H2O

Eo = +0.70 V Eo = +1.78 V

Eocell = +1.78 – (+0.70) = 1.08 V ΔGo = – (2)(96 485)(1.08) × 10–3 = – 208 kJ mol–1

The large, negative value of ΔGo indicates that H2O2 is thermodynamically unstable with respect to H2O and O2. (b) ‘20 volume’ H2O2 : 1 volume of solution liberates 20 volumes of O2. Assume an ideal gas: 273 K and 1 bar pressure are standard conditions. Let 1 vol. of solution = 1 dm3 ∴ 20 vols. O2 = 20 dm3 For an ideal gas, 1 mole occupies a volume of 22.7 dm3 at 273 K and 1 bar pressure. ∴

20 dm3 O2 =

20 = 0.88 moles 22.7

From the equation in part (a), 2 moles of H2O2 produce 1 mole of O2 ∴ 0.88 moles of O2 are produced by 1.76 moles H2O2 To find the number of grams per dm3 : Mr(H2O2) = 2(1.008) + 2(16.00) = 34.016 Mass of H2O2 present in 1 dm3 = 34.016 × 1.76 = 59.9 g

The group 16 elements 16.7

245

Half-equations from Appendix 11 in H&S are needed for this problem. For H2O2: H2O2 + 2H+ + 2e– O2 + 2H+ + 2e–

2H2O 2H2O2

Eo = +1.78 V Eo = +0.70 V

(a) For the reaction of H2O2 with Ce4+, the relevant half-equation is: Ce4+ + e–

Ce3+

Eo = +1.72 V

The overall spontaneous reaction is: 2Ce4+ + H2O2 J 2Ce3+ + O2 + 2H+ (b) For the reaction of H2O2 with I–, the relevant half-equation is: I2 + 2e–

2I–

Eo = +0.54 V

The overall spontaneous reaction is: 2I– + H2O2 + 2H+ J I2 + 2H2O 16.8

The half-equation for the oxidation is: MnO2 + 2H2O + 2e– (a)

Mn(OH)2 + 2[OH]–

Eo[OH–] = 1 = –0.04 V

Mn(OH)2 + H2O2 J MnO2 + 2H2O

(b) MnO2 produced catalyses the decomposition of H2O2, so secondary reaction is: 2H2O2 J 2H2O + O2 16.9

16.10

catena-Se∞ is chiral because it is a helical chain and a helix has a handedness. See Fig. 3.20a in H&S. Use Fig. 3.10 in H&S.

START C3

(16.2)

Is the molecule linear?

No

Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Are there 3 C2 axes perpendicular to the principal axis? Is there a σh plane? Are there n (i.e. 3) σv planes containing

No Yes: C3 axis (see 16.2)

the Cn axis?

Yes No Yes

STOP

Conclusion: the point group is D3d. 16.11

(a) H2Se Se, number of valence electrons = 6 Number of bonding pairs (2 Se–H bonds) = 2 2 lone pairs ‘Parent’ shape = tetrahedral; molecule is bent, see 16.3.

Se H

H

(16.3)

246

The group 16 elements

S

(b) [H3S]+ S+, number of valence electrons = 5 Number of bonding pairs (3 S–H bonds) = 3 1 lone pair ‘Parent’ shape = tetrahedral Molecular shape = trigonal pyramidal, see 16.4. (c) SO2 S, number of valence electrons = 6 Number of bonding ‘pairs’ (2 S=O bonds) = 2 1 lone pair ‘Parent’ shape = trigonal planar Molecular shape = bent (16.5). (d) SF4 S, number of valence electrons = 6 Number of bonding pairs (4 S–F bonds) = 4 1 lone pair ‘Parent’ shape = trigonal bipyramidal Molecular shape = see-saw (16.6). (e) SF6 S, number of valence electrons = 6 Number of bonding pairs (6 S–F bonds) = 6 No lone pairs ‘Parent’ shape = molecular shape = octahedral, see 16.7. (f) S2F2 – for each S centre, assuming each S is in a similar environment: S, number of valence electrons = 6 Number of bonding pairs (1 S–S + 1 S–F bond) = 2 S 2 lone pairs ‘Parent’ shape = tetrahedral at each S S Molecular shape = bent (at each S), see 16.8. F There is, however, a second isomer (see Section 16.7 F (16.9) in H&S) which possesses structure 16.9.

H

H

H

(16.4)

S O

O

(16.5) F F

S

F F

(16.6) F F

F

S

F

F F

(16.7) F S

S

F

(16.8)

16.12

(a) SF4 (16.6) can act both as an F– donor or acceptor; behaviour depends on its reaction-partner. BF3 is an F– acceptor: SF4 + BF3 J [SF3]+ + [BF4]– CsF is ionic and an F– donor: SF4 + CsF J Cs+ + [SF5]– (b) SF4 is a selective fluorinating agent; converts C=O into CF2 group, and will fluorinate C–OH: RCO2H

16.13

SF4

RCF3

[SeI3]+ (structurally analogous to [H3O]+) is trigonal pyramidal with C3v symmetry. Figure 16.1 shows the vibrational modes for [SeI3]+. The assignment of symmetry labels is as described in Section 3.7 (‘XY3 molecules with C3v symmetry’) in H&S. In the C3v character table, the right-hand column has a product term (e.g. x2, xy) in each of the A1 and E rows and therefore, each of the 4 modes is Raman active.

The group 16 elements

Symmetric stretch (A1)

Asymmetric stretch (E)

Symmetric deformation (A1)

247

Asymmetric deformation (E)

Fig. 16.1 Vibrational modes of a C3v YX3 molecule. Each mode is both IR and Raman active.

16.14

X O

O

X X = H or F

(a) Approach the problem by first considering an appropriate MO diagram for O2, [O2]+ and [O2]2–, see answer 2.10, p. 17. From the MO diagram, the bond orders are O2, 2; [O2]+, 21/2; [O2]2–, 1, giving a trend of [O2]+ > O2 > [O2]2– consistent with the trend in O–O bond distances [O2]+ < O2 < [O2]2–. Now consider H2O2 and O2F2. Both have structure 16.10, and might be expected to possess similar O–O bond lengths consistent with single bonds. In H2O2, distance of 147.5 pm is similar to that in [O2]2– (149 pm) – single bond. In O2F2, O–O bond length of 122 pm is almost the same as that in O2, i.e. indicates double bond character; rationalize this in terms of the resonance forms 16.11. F–

(16.10) O F

F

O

O

O

F–

(16.11) S

S

S

S

S

S

(b) rcov(S) = 103 pm, therefore a single S–S bond distance is ≈ 206 pm. This is observed in S6 and H2S2 (isostructural with H2O2, 16.10). Consider remaining species: S2: 189 pm is consistent with a diatomic with S=S double bond. [S4]2+ : the cation is a planar square;equal bond lengths of 198 pm suggests S–S bond order between 1 and 2; 16.12 is one resonance structure; delocalized bonding is represented in 16.13. S2F2: 189 pm, i.e. the same as in S2, so indicates double bond character despite the fact that structure 16.8 shows a single bond. Rationalized in terms of contributions by ionic resonance structures as for O2F2 in 16.11. S2Cl2: 193 pm, not very different from that in S2F2, but less than 2rcov. S2F10: 221 pm (significantly > 2rcov) indicates a weak S–S bond; compound readily disproportionates with S–S cleavage (see Section 16.7 in H&S).

S 2+

(16.12)

S

(16.13)

16.15 F F

Se

F F

Se

F

F

F F

F

F

(16.14)

(16.15)

F S

F

S

F F

(16.16)

(16.17)

Cl Cl

The structures of the compounds are given in 16.14-16.19 with the resultant dipole moment indicated where appropriate. The Pauling electronegativities that need to be considered are: χP(Se) = 2.6, χP(S) = 2.6, χP(F) = 4.0, χP(Cl) = 3.2, χP(O) = 3.4. From the values given for the molecular dipole moments, you can comment to some extent on correlations between electronegativity differences, molecular shape and effects of lone pairs. SeF6 (16.14) is nonpolar. SeF4 and SF4 are isostructural, although bond lengths and angles will be different; values of χP(S) and χP(Se) are the same. The directions of the resultant dipole moments are shown in 16.15 and 16.16. In SCl2 (16.17), each S–Cl bond is polar in the sense Sδ+–Clδ–, and the resultant dipole of the two bond moments is larger than that due to the lone pairs.

S O

Cl Cl

(16.18)

O S O

Cl Cl

(16.19)

248

The group 16 elements Structure 16.18 shows the approximate direction in which the resultant dipole moment in SOCl2 acts; the resultant of the S=O and S–Cl bond moments compete with the effects of the lone pair. In SO2Cl2 (16.19), the resultant of the two S=O bond moments is greater than that of the two S–Cl moments. 16.16

First, consider the nuclear spin data: 19F 100% I = 1/2 125Te 7% I = 1/2 123Te 0.9% I = 1/2 125 In the Te NMR spectrum, the binomial octet is due to coupling to 7 equivalent 19F nuclei. [TeF ]– is pentagonal bipyramidal (16.20). NMR spectroscopic data 7 indicate that in solution at 298 K, the anion is stereochemically non-rigid (fluxional). In the 19F NMR spectrum, the singlet is due to the 92.1% 19F nuclei that are not attached to spin-active Te. 19F bonded to 125Te (I = 1/2) gives rise to a doublet, and 19F bonded to 123Te (I = 1/ ) also gives rise to a doublet. Coupling constants of 2876 2 and 2385 Hz are assigned to J(125Te-19F) and J(125Te-19F), but it is not possible from the data available here to say which is which.

16.17

Use the periodic table (16.21) to help you to identify isoelectronic species. Note that ‘isoelectronic’ is being used in respect only of the valence electrons.

Fax Feq Feq

Feq

Te

Feq

Feq Fax

(16.20)

14

15

16

C

N

O

Si

P

S

Ge

As

Se

Sn

Sb

Te

(a) [SiO4]4–, [PO4]3– and [SO4]2– are all isoelectronic and isostructural (tetrahedral). (b) CO2, SiO2 and [NO2]+ are isoelectronic; CO2 and [NO2]+ are isostructural (linear) but SiO2 has a macromolecular structure with tetrahedral Si. SO2 and TeO2 are isoelectronic but not isostructural; SO2 is molecular (16.22) but TeO2 adopts a 3D-structure with Te–O–Te bridges connecting TeO4 units (16.23). O

(16.21)

S O

O

Te

O O

O

O

O (TeO2)n

Se

O

(16.22)

O

Se O

Se O

O

Se O

(16.23)

O

O

O

O

(c) SO3, [PO3]– and SeO3 are all isoelectronic; SO3 and [PO3]– are isostructural (trigonal planar) but SeO3 is tetrameric (16.24). (d) [P4O12]4–, Se4O12 and [Si4O12]8– are isoelectronic and isostructural (cyclic as in 16.24).

(16.24)

16.18

S O

O–

O

–

(16.25)

X SO2 in aqueous solution: see Section 7.4 in H&S

(a) SO3, trigonal planar; [SO3]2–, trigonal pyramidal. Rationalized in terms of VSEPR: for SO3, all 6 valence electrons of S used in bonding, but in the dianion, there is a lone pair (16.25 shows one resonance form). (b) Dissolution of SO2 in water gives sulfurous acid, H2SO3, although the equilibrium constant shows that the solution contains mainly dissolved SO2: SO2(aq) + H2O(l)

H2SO3(aq)

K < 10–9

H2SO3 (cannot be isolated) dissociates in solution according to the following equilibria:

The group 16 elements H2SO3(aq) + H2O(l)

[H3O]+(aq) + [HSO3]–(aq)

[HSO3]–(aq) + H2O(l)

249

pKa = 1.82

[H3O]+(aq) + [SO3]2–(aq)

pKa = 6.92

Sulfite, [SO3]2–, is a reducing agent in alkaline solution: [SO4]2–(aq) + H2O(l) + 2e– X A commercial application of aqueous solutions of SO2 is in the wine industry where it is used to kill microorganisms and to stabilize the wine against oxidation (Box 16.7 in H&S).

[SO3]2–(aq) + 2[OH]–

Eo[OH–] = 1= –0.93 V

and in acidic solution, aqueous SO2 acts as a weak reducing agent: [SO4]2–(aq) + 4H+(aq) + 2e–

H2SO3(aq) + H2O(l)

Eo[H+] = 1 = +0.17 V

but value of E depends on the concentration of H+, and [SO4]2– can be reduced to SO2 in the presence of high [H+]: Cu + 2H2SO4(conc) J CuSO4 + SO2 + 2H2O Examples of aqueous SO2 operating as a reducing agent: H2SO3(aq) + H2O(l) + 2Fe3+ J [SO4]2–(aq) + 4H+(aq) + 2Fe2+(aq) H2SO3(aq) + H2O(l) + I2 J [SO4]2–(aq) + 4H+(aq) + 2I–(aq)

16.19

(a) All the structures are based on S8 (16.1) with NH replacing S; no N–N bonds (Fig. 16.2). (b) Preparation of S4N4 (16.26):

260 pm

S

S

6S2Cl2 + 16NH3 J S4N4 + 12NH4Cl + S8

163 pm N

N

N

S

S

N

(16.26)

Figure 16.2 Structures, and isomers where relevant, of S7NH, S6N2H2, S5N3H3 and S4N4H4.The rings are non-planar.

Shock sensitivity of S4N4 means it must be handled with care; pure samples tend to explode when struck. For a summary of reactions, see Fig. 16.20 in H&S. Examples chosen should illustrate: • reduction to S4N4H4 • halogenation • ring degradation to NSF and NSF3 • oxidation to [S4N4]2+ (contin)

S

H N

S

S S

S S

S

S

S NH

S S

N H

S

S

S

H N

S

S

NH

S S

H N

N H S NH

HN S

S

S

NH

S

H N

S S

S S

S

S S

S HN

H N

S NH

S H N

S

S

S

S

H N

N H

S

S

250

The group 16 elements • (SN)x polymer formation and should show changes in ring conformations that accompany reactions, e.g. unfolding of the cage in S4N4 to the crown ring of S4N4H4. 16.20

(a) The primary reaction is reduction of [SO4]2– to SO2 (see answer 16.18). For the formation of CuS, the reduction half-equation is: [SO4]2– + 8H+ + 8e–

S2– + 2H2O

Two factors to consider: (i) S2– combines with Cu2+ to give CuS which is sparingly soluble. Very low solubility of CuS (Ksp ≈ 10–44) facilitates reduction of sulfate to S2– (see Section 8.3 in H&S). (ii) Reduction potential is affected by concentration of H+: ⎧ RT ⎛ [reduced form] ⎞⎫ E = Eo – ⎨ ⎟⎟⎬ = Eo – × ⎜⎜ ln zF [oxidized form] ⎝ ⎠⎭ ⎩

Te

F

F

⎛ [S 2− ] × ⎜ ln ⎜ [SO 2− ][ H + ]8 4 ⎝

⎞⎫⎪ ⎟⎬ ⎟⎪ ⎠⎭

Very high [H+] makes Eo more positive, making [SO4]2– a stronger oxidizing agent. (b) Apply VSEPR model to rationalize the structure of [TeF5]– : Te–, number of valence electrons = 7 Number of bonding pairs (5 Te–F bonds) = 5 1 lone pair ‘Parent’ shape = octahedral Molecular shape = square-based pyramidal, see 16.27. (c) Initial precipitate is Ag2S2O3, and this dissolves in excess [S2O3]2– :

F F

⎧⎪ RT ⎨ ⎪⎩ zF

F

(16.27)

Na2S2O3(aq) + 2AgNO3(aq) J Ag2S2O3(s) + 2NaNO3(aq) Ag2S2O3(s) + 5[S2O3]2–(aq) J 2[Ag(S2O3)3]5–(aq) Black colour is Ag2S, formed by the disproportionation of [S2O3]2– : [S2O3]2– + H2O J S2– + [SO4]2– + 2H+ The sulfate gives a white precipitate with acidified aqueous Ba(NO3)2 : [SO4]2–(aq) + Ba(NO3)2(aq) J BaSO4(s) + 2[NO3]–(aq) 16.21 Values of Ar : Na 22.99 S 32.06 O 16.00

(a) 0.0261 g Na2S2O4 =

0.0261 = 1.50 × 10–4 moles 174.1

Ag is oxidized in HNO3 to Ag+ (i.e. a 1-electron oxidation) and this is equivalent to 30.0 cm3 0.10 M [NCS]– : Amount of [NCS]– = 30.0 × 0.10 × 10–3 = 3.0 × 10–3 moles Amount of Ag+ = 3.0 × 10–3 moles ∴ Ratio of [S2O4]2– : Ag+ = 1 : 2 2– Therefore, the [S2O4] undergoes a 2-electron oxidation. Reaction is: [S2O4]2– + 2Ag+ + H2O J [S2O5]2– + 2Ag + 2H+ (b) 0.0725 g Na2S2O4 =

0.0725 = 4.16 × 10–4 moles 174.1

The group 16 elements

251

Residual I2 is reduced by 23.75 cm3 0.1050 M [S2O3]2– : 2[S2O3]2– + I2 J [S4O6]2– + 2I– Amount of [S2O3]2– = 23.75 × 0.1050 × 10–3 = 2.49 × 10–3 moles Amount of residual I2 = 1.25 × 10–3 moles Initial amount I2 = 50 × 0.0500 × 10–3 moles = 2.50 × 10–3 moles Amount I2 reacted with Na2S2O4 = (2.50 – 1.25) × 10–3 = 1.25 × 10–3 moles ∴ Ratio [S2O4]2– : I2 = (4.16 × 10–4) : (1.25 × 10–3) = 1 : 3 Therefore, the reaction is:

NH2 O

C NH2

[S2O4]2– + 3I2 + 4H2O J 2[SO4]2– + 6I– + 8H+

(16.28)

16.22

Urea has structure 16.28. After reaction of urea with H2SO4, the product X reacts with NaNO2 and H+ to give N2, and adding BaCl2 yields BaSO4. For the latter: BaCl2(aq) + [SO4]2–(aq) J BaSO4(s) + 2Cl–(aq)

NH2 S

The formula of X is H3NO3S. Given that this is formed from OC(NH2)2 and H2SO4 suggests that H3NO3S is HOSO2NH2 (16.29, sulfamic acid). The reaction liberating N2 is:

OH

O O

(16.29)

HOSO2NH2 + NaNO2 J N2 + NaHSO4 + H2O 16.23

Oxoacids of sulfur are described in Section 16.9 in H&S. Points to include: • Acids that are not isolable but for which salts can be isolated are H2S2O4 (dithionous acid, 16.30), H2SO3 (sulfurous acid, 16.31), H2S2O5 (disulfurous acid) and H2S2O6 (dithionic acid). • Acids that can be isolated are H2SO4 (sulfuric acid, 16.32), HSO3F and HSO3Cl (fluoro- and chlorosulfonic acids, 16.33), H2SO5 (peroxomonosulfuric acid, 16.34) and H2S2O8 (peroxodisulfuric acid, 16.35). • Thiosulfuric acid (H2S2O3, 16.36) can be isolated but is very unstable. • Most important acid commercially is H2SO4; scale of manufacture is huge. • Structures: trigonal pyramidal or tetrahedral S in each oxoacid or oxoanion: • Show how each acid ionizes in aqueous solution, indicating whether strong or weak acids; give examples of salts formed, noting series of salts for polybasic acids. • Dithionate and dithionite anions possess weak S–S bonds. O

O S OH

S O

OH

S O

O

S

S

OH OH

O

OH

O

OH

(16.30)

(16.31)

OH O

(16.33)

O

O S

(16.32)

O OH

(16.34)

S O

O

S O

OH X X = F, Cl

O

S OH

OH

(16.35)

O

S O

OH OH

(16.36)

252

The group 16 elements • Concentrated H2SO4 as an oxidizing agent. • Use of liquid H2SO4 as a non-aqueous solvent (see Section 8.8 in H&S); self-ionization is: 2H2SO4

[H3SO4]+ + [HSO4]–

• ‘Superacid’ behaviour of HSO3F (see Section 9.9 in H&S). 16.24 S S

O

(16.37)

S– S –O

O O

(16.38)

N

S

S2O, 16.37. Structure analogous to that of SO2 (16.5) with S replacing O; bent rather than linear because of presence of lone pair on central S atom. [S2O3]2– – 16.38 shows one resonance structure. Analogue of [SO4]2– with tetrahedral S (6 valence electrons) in centre. NSF, 16.39, possesses an N ≡ S triple bond (but see question 16.29) and is bent because S (6 valence electrons) has a lone pair. On going from NSF to NSF3, S is oxidized. In NSF3, 16.40, all 6 valence electrons of S are used for bonding. Tetrahedral S, with N ≡ S triple bond (but see question 16.29). [NS2]+, 16.41, is linear. This is consistent with the VSEPR model; note that [NS2]+ is isoelectronic (with respect to its valence electrons) with [NO2]+ and CO2 (both linear). S2N2 is cyclic with equal S–N bond lengths which can be explained in terms of resonance structures, e.g. 16.42, or in terms of π-delocalization using N 2p and S 3p atomic orbitals. S2N2 has 22 valence electrons; allocate 8 electrons for 4 σbonds, and 1 lone pair per atom, leaving 6 electrons for π-bonding; these occupy the π-MOs shown in Fig. 16.3.

F

F

(16.39)

N

S

S

F

N

16.25

N

N

S

N

S

S

N

etc

S

F

(16.40)

S

(16.41)

(16.42)

[X]+ is a cycloaddition product of [NS2]+ (16.41) and RCN: S

+ N

S

S R

R

C

N

N

N S+

Fig. 16.3 π-Molecular orbitals in S2N2, illustrating that there is delocalization of π-electrons around the ring.

Energy

Allocate one lone pair outside the ring per N and S atoms. This leaves 6 valence electrons for the π-system. The cation is a Hückel (4n + 2) system and is planar.

N

S N

S

The group 16 elements 16.26

The correct pairings are shown in the table below: List 1 List 2 Comments Chiral polymer See Fig. 3.20 in H&S for S∞ analogous Se∞ [S2O8]2– Strong oxidizing agent, reduced See Appendix 11 in H&S to [SO4]2– – [S2] Blue, paramagnetic species See Section 16.6 in H&S S2F2 Na2O [S2O6]2– PbS H2O2

Exists as two monomeric isomers Crystallizes with antifluorite structure Contains a weak S–S bond, readily cleaved in acidic solution Black, insoluble solid

See structures 16.27 and 16.28 in H&S See Fig. 6.19 in H&S and accompanying discussion See eq. 16.104 in H&S See Table 7.4 in H&S

Disproportionates in presence of Mn2+ Reacts explosively with H2O

See answer 16.6a, p. 244

See Appendix 11 in H&S

H2S

Strong reducing agent, oxidized to [S4O6]2– A toxic gas

SeO3

Exists as tetramer in solid

HSO3Cl [S2O3]2–

16.27

253

Easily detected by its smell of rotten eggs See 16.24, p. 248

(a) The reaction of H2S with aqueous Cu(II) is: Cu2+ + H2S J CuS(s) + 2H+ When Na2S is added, soluble salt of complex anion is formed: CuS(s) + Na2S(aq) J Na2[CuS2](aq) (b) Sequence of reactions is: H2O + SO2 J H2SO3 SO2 + H2SO3 + 2CsN3 J Cs2S2O5 + 2HN3

Δ

(16.43)

Λ

16.28

(c) [Cr(Te4)3]3– is an octahedral complex containing three didentate ligands; the complex is chiral. The prefixes Δ and Λ are used to distinguish between the two enantiomers of the complex (16.43) which arise from the tris(didentate ligand) complex. Each chelate ring is puckered and can adopt different conformations distinguished by the labels δ and λ. See diagrams in Box 19.3 in H&S. The Δλλλconformation describes the conformations of each chelate ring and the arrangement of the ligands at the Cr3+ centre. liquid HF

(a)

SF4 + SbF5

(b)

SO3 + HF J HSO3F

[SF3]+[SbF6]–

The group 16 elements

254

16.29

(c)

Na2S4 + 2HCl J 2NaCl + H2S4

(d)

[HSO3]– + I2 + H2O J [HSO4]– + 2I– + 2H+

(e)

[SN][AsF6] + CsF

(f)

HSO3Cl + H2O2 J H2SO5 + HCl

(g)

[S2O6]2–

N

+ S

N F

S

S has 10 electrons in valence shell

S has 12 electrons in valence shell

(16.44)

(16.45)

SO2 + [SO4]2–

+ S F–

F F

F

in acidic solution

N

F S

NSF + Cs[AsF6]

(a) Structures 16.44 and 16.45 show the representations of NSF and NSF3 used in the main text of H&S, and in answer 16.24, p. 252. While commonly used to show the structures of these molecules, these representations imply that S expands its valence octet of electrons. Resonance structures that retain an octet are: –

N

Δ

F–

–

N

2+ S

F

–

F

N

2+ S

F

F

F

–

N

F–

2+ S

N

2+ S

F–

F

–

F–

F

F

F– N

2+ S F

F– F–

N

2+ S F

F –

(b) See Fig. 10.8 in H&S. Points to include: • one expects values of ΔvapH(bp) for members in a series of related molecular compounds to increase with increasing molecular size (increase in intermolecular dispersion forces); this is observed for H2S to H2Se to H2Te, but value for H2O is anomolously high; • extent of hydrogen bonding in liquid water greater than for later H2E molecules; values of χP for O, S, Se and Te are 3.4, 2.6, 2.6 and 2.1 respectively; • H2O is strongly hydrogen-bonded in the liquid but not in the vapour. (c) Of the compounds listed in the question, Al2Se3, SF4 and SeO2 react when they dissolve in water: Al2Se3 + 6H2O J 2H2Se + 2Al(OH)3 SF4 + H2O J SOF2 + 2HF

then

SOF2 + H2O J SO2 + 2HF

SeO2 + H2O J H2SeO3 SF6 is kinetically stable with respect to hydrolysis. As2S3, FeS2 and α-HgS occur naturally as the minerals orpiment, iron pyrites and cinnabar, respectively, and are insoluble in water.

The group 16 elements 16.30 Se

Se

+ Se

+ Se

Se +

Se +

Se

Se

Se

Se

Se

Se

Se

Se

+

Se

+

Se

+

+

(16.46)

16.31

F

255

(a) The fact that [Se4]2+ (equal Se–Se bond lengths) has D4h symmetry shows that it has a C4 principal axis and a σh plane perpendicular to this axis; it also has an inversion centre (centre of symmetry). This is enough information to conclude that the ion is square planar. (The full set of symmetry elements are listed in Appendix 3 in H&S.) (b) From Appendix 6 in H&S, rcov = 117 pm; d(Se–Se) < 2rcov, which suggests some π-character. (c) Resonance structures are shown in 16.46. (d) An appropriate MO diagram to describe the π-bonding in [Se4]2+ is similar to that for S2N2 in Fig. 16.3, p. 252. Six π-electrons occupy 1 bonding and 2 nonbonding MOs; π-bond order = 1 spread over the whole ring (1/4 per Se–Se). (a) See worked example 3.7 in H&S for the assignment of the D4d point group to S8. The structure of S8O is shown below:

F F

Te

CN

Te

F

CN

F

F

(16.47)

(16.48)

The molecule has no principal axis, but does possess a mirror plane. This leads to Cs symmetry (see Fig. 3.10 in H&S). (b) The reaction gives 3 products, TeF4–n(CN)n with n = 0, 2 and 3. These are shown in structures 16.47-16.49, but from the data given it is not possible to confirm the positions of substitution. The NMR spectroscopic data indicate that the F atoms in 16.47 are equivalent on the NMR timescale, and this is consistent with a stereochemically non-rigid species account for the observed spectroscopic data. 125Te NMR spectrum:

CN CN

Te

CN F

(16.49)

δ 1236 ppm (quintet, J 2012 Hz) assigned to 16.47 δ 816 ppm (triplet, J 187 Hz) assigned to 16.48 δ 332 ppm (doublet, J 200 Hz) assigned to 16.49 The 19F NMR spectrum of the mixture at 173 K exhibits 3 signals, each with 125Te satellites. For 16.47, the one F environment arises from a fluxional molecule. For 16.48, the one F environment could arise from either the structure shown or from an isomer with both F atoms in equatorial sites, or from a fluxional species. 16.32 Cl Cl

Te

Cl

2–

Cl Cl

Te

Cl

Cl Cl

Cl

(a) [Te2Cl10]2– contains 2 octahedral Te centres and therefore must contain bridging Cl atoms (16.50). (b) The formula of [(Ph3PO)2H]+ can be broken down into 2Ph3PO + H+. This suggests a hydrogen-bonded species, with the proton bridging between two Ph3PO (16.51) molecules. A symmetrical bridge might be expected (compare with [H5O2]+, see Fig. 10.1 in H&S) as shown in structure 16.52.

Cl

Ph

O–

(16.50) P Ph

Ph +P

Ph Ph

(16.51)

Ph

Ph

O–

+ H

(16.52)

O–

P+ Ph

Ph

The group 16 elements

256

16.33

S H

O

(16.53)

H

S +

O

(16.54)

X See: P. A. Denis (2009) Spectrochim. Acta A, vol. 72, p. 720 X See answer 10.1 on p. 153

(a) S has 6 valence electrons. Three are used for bonding, leaving one lone pair and one odd electron. Thus, by VSEPR, HSO is bent (parent structure based on a tetrahedron, 16.53). (b) The problem with the structure above is that the valence shell of S contains 9 electrons. Resonance structure 16.54 describes the bonding in this radical. (c) All three modes of vibration (S–H and S–O stretches and H–S–O deformation) are IR active (all give a change in dipole moment). (d) The band at 2335 cm–1 is due to the S–H stretch. To check this: For 32S–1H:

1

μ1

=

1 1 + = 1.0312 32 1

ν1 μ2 = ν2 μ1

ν 2 = ν1 ×

For 32S–2H:

1

μ2

=

1 1 + = 0.5312 32 2

μ1 0.5312 = 2335 × = 1676 cm −1 1.0312 μ2

This is close to the experimentally observed value of 1704 cm–1. (e) HSO (which is an intermediate in the oxidation of atmospheric H2S to H2SO4) is a radical. It is isoelectronic with ClO and is involved in analogous radical reactions (see Box 14.6 in H&S). 16.34

(a) Acid-base chemistry involving CaCO3 (limestone) and CaCO3.MgCO3 (dolomite) and acid rain. H2SO4 is neutralized by basic carbonates: MCO3 + H2SO4 J MSO4 + H2O + CO2

M = Ca or Mg

Nitric acid will also be neutralized. (b) (i) log Ksp values of –8.6 and –8.2 correspond to Ksp values of 2.5 × 10–9 and 6.3 × 10–9 and therefore CaC2O4.H2O and CaC2O4.2H2O (calcium oxalate hydrates) are sparingly soluble salts and can immobilize excess Ca2+ ions. (ii) Gypsum is the hydrated sulfate CaSO4.2H2O and is formed by neutralization with acid rain: CaC2O4 + 2H2O + H2SO4 J CaSO4.2H2O + H2OC2O4 or you can start the equation with CaC2O4.H2O or CaC2O4.2H2O. 16.35 + O O

+ O O

O

(16.55) O

O

O

(a) Lewis structures for ozone and dioxygen are given in 16.55 and 16.56. An MO treatment for O2 shows that this is a diradical (see answer 2.29, p. 28 for complete answer). (b) Iron pyrites is FeS2 with Fe(II) and [S2]2–. The [S2]2– ion (isoelectronic with Cl2) contains a single bond (16.57). (c) The blue colour of lapis lazuli arises from [S3]– which absorbs at 595 nm. The colour is variable; if [S2]– is present (λmax = 370 nm), this gives a yellow colour. The structures of these radicals are 16.58 and 16.59.

(16.56) S S

S

S

S

(16.58)

S

S

(16.59)

(16.57)

(d) Elemental Te exists as helical chains containing Te–Te single bonds.

257

17 The group 17 elements 17.1

17.2

(a) Halogens (b) F Cl Br I At (c) ns2np5

fluorine chlorine bromine iodine astatine

(a) Displacement reactions (look at appropriate Eo values in Appendix 11 in H&S): 2X– + Cl2 J X2 + 2Cl–

X = Br or I

(b) Downs process is electrolysis of molten NaCl; overall: 2Na+(l) + 2Cl–(l) J 2Na(l) + Cl2(g) Products must be kept apart to prevent recombination and formation of NaCl. (c) Recombination of H2 and F2 in an explosive radical chain reaction: F2 + H2 J 2HF

δ– F

δ

–

V F

Lone pair-lone pair repulsions between lone pairs on O and F weaken bond. For O2F2, see resonance forms 16.11 (p. 247) for ionic contributions made in this case.

17.4

(a) pKa values for CF3CO2H and CH3CO2H are 0.23 and 4.75, respectively. A smaller pKa corresponds to a larger Ka (Ka = 10–pKa). Thus, CF3CO2H is a stronger acid than CH3CO2H. This is rationalized in terms of the inductive effect (17.1). See also answer 7.3 on p. 99.

O

V

F

V

δ–

17.3

C

C O

δ+

Overall inductive effect

(17.1)

C–H σ−orbital can overlap with one lobe of the 2p orbital H H H

(17.3)

H

(b) Both NH3 and NF3 are trigonal pyramidal molecules (17.2). This difference may be explained by considering the bond dipole moments and the effects of the N lone pair. χP(N) = 3.0, χP(H) = 2.2. Thus, each N–H bond is polar: Nδ––Hδ+. The resultant dipole moment acts in a direction that is reinforced by the N lone pair. In NF3, each N–F bond is polar in the sense Nδ+–Fδ– because χP(N) = 3.0, χP(F) =4.0. The resultant dipole moment opposes the effects of the lone pair, and NF3 is less polar than NH3.

N X=H

X X μ = 1.47 D

X=F

μ = 0.24 D

X

(17.2)

(c) Alkyl groups are electron releasing and donate electrons into the aryl ring by the inductive effect. Electrons are released through hyperconjugation (17.3). A Me group activates the ring towards electrophilic subsitution and directs the electrophile to the ortho- and para-positions. A CF3 group withdraws electrons from the aryl ring by the inductive effect (similar to 17.1) and this deactivates the ring towards electrophilic substitution. More electronic charge is removed from the ortho- and para-positions than the meta-positions, and therefore the CF3 group is meta-directing.

258

The group 17 elements 17.5

Hydrogen bonding for HF gives anomalously high values (see answer 10.9, p. 155); HCl, HBr and HI follow trend expected: increase with increasing molecular weight.

17.6

Estimate bond length from sum of covalent radii (all single bonds): ClF, 170; BrF, 185; BrCl, 213; ICl, 232; IBr, 247 pm. The agreement with tabulated data for XY is good where the difference in electronegativities {χP(Y) – χP(X)} is small. In the homonuclear diatomic molecule X2, contribution made to the bonding by covalent resonance structure X–X dominates, and there is negligible contribution from X+X–; similarly for Y2. This is also true for XY if χP(Y) ≈ χP(X), but if χP(Y) > χP(X) (e.g. BrF), then the ionic form contributes significantly (e.g. 17.4) and the bond is shorter than predicted from covalent radii. Br

F

Br+

F–

(17.4)

17.7

Br– F+ Negligible contribution because χP(F) > χP(Br)

(a) ClF3 is a very strong oxidizing agent, and therefore Ag+ is oxidized, rather than AgCl simply being fluorinated to AgF: 2AgCl + 2ClF3 J 2AgF2 + Cl2 + 2ClF (b) BF3 is an F– acceptor, and ClF donates F– but will not form ‘naked’ Cl+ : 2ClF + BF3 J [Cl2F]+[BF4]– (c) CsF is ionic and an F– donor; IF5 acts as an F– acceptor: CsF + IF5 J Cs+[IF6]– (d) SbF5 is an F– acceptor (will not act as a donor); the [SbF6]– ion formed can form an adduct with SbF5 : or

SbF5 + ClF5 J [ClF4]+[SbF6]– 2SbF5 + ClF5 J [ClF4]+[Sb2F11]–

(e) [Me4N]F is ionic and an F– donor, and IF7 acts as an F– acceptor: [Me4N]F + IF7 J [Me4N]+[IF8]– (f) 17.8

Δ K[BrF4] ⎯⎯→ KF + BrF3

Section 17.7 in H&S contains information for this problem. Points to include: • Self-ionization of BrF3 gives [BrF2]+ and [BrF4]–. • Interhalogens such as ClF and ClF3 act as F– donors or acceptors depending on their reaction partner, e.g. with ionic fluoride, the interhalogen accepts F– but with a potent fluoride acceptor (e.g. BF3, AsF5), the interhalogen donates F–. • Higher interhalogens are mainly fluorides, and hence F– donor/acceptor chemistry (as opposed to Cl–, Br– or I–) dominates. • Large cation (e.g. Cs+) needed to stabilize [XYn]– anion.

The group 17 elements 17.9

Cl

I

Cl

Cl Cl

(17.5)

F Cl

F F

F

(17.7)

Cl

I

Cl

Cl

I

Cl Cl

Cl

(17.9)

F F

Br

F F

F

(17.10)

17.10

259

Apply the VSEPR model; localize the overall charge on the central atom for purposes of electron counting. (a) [ICl4]– I–, number of valence electrons = 8 Number of bonding pairs (4 I–Cl bonds) = 4 2 lone pairs Molecular shape = square planar, see 17.5. F (b) [BrF2]+ Br+, number of valence electrons = 6 Br Number of bonding pairs (2 Br–F bonds) = 2 F 2 lone pairs Molecular shape = bent, see 17.6. (17.6) (c) [ClF4]+ Cl+, number of valence electrons = 6 F Number of bonding pairs (4 Cl–F bonds) = 4 F F 1 lone pair F I Molecular shape = disphenoidal, see 17.7. F F (d) IF7 F I, number of valence electrons = 7 Number of bonding pairs (7 I–F bonds) = 7 (17.8) No lone pairs Molecular shape = pentagonal bipyramidal, see 17.8. (e) I2Cl6; assume both I atoms are in the same environment and are connected by 2 bridging Cl atoms (must be 2, not 1, for symmetry). I, number of valence electrons = 7 Number of bonding pairs (3 I–Cl bonds) = 3 2 lone pairs One extra pair of electrons from the coordinate bond of μ-Cl Molecular shape = square planar at each I, see 17.9. (f) [IF6]+ I+, number of valence electrons = 6 Number of bonding pairs (6 I–F bonds) = 6 No lone pairs Molecular shape = octahedral. (g) BrF5 Br, number of valence electrons = 7 Number of bonding pairs (5 Br–F bonds) = 5 1 lone pair Molecular shape = square-based pyramidal, see 17.10. (a) Static structure of BrF5 (17.10) has two F environments: Fapical : Fbasal = 1 : 4. The 19F NMR spectrum contains 2 signals. Signal assigned to F apical is a binomial quintet (coupling to 4 equivalent 19Fbasal), and signal due to Fbasal is a doublet (coupling to 1 19Fapical). Relative integrals of quintet : doublet = 1 : 4. [IF6]+ is octahedral (see answer 17.9f). In the 19F NMR spectrum of [IF6]+, a singlet is observed because all 19F nuclei are equivalent. (b) The 19F NMR spectrum of BrF5 is expected to depend on temperature. Exchange of Fapical and Fbasal sites may occur. The high temperature limiting spectrum is a singlet; the low temperature limiting spectrum is as described in part (a). The 19F NMR spectrum of [IF6]+ is not expected to be temperature dependent.

The group 17 elements

260

17.11 Cl Cl Cl

Al

Al Cl

Cl Cl

(17.11)

(a) I2Cl6 is planar, with I in square planar environments; the planarity arises because of stereochemically active lone pairs (see answer 17.9e). Valence shell of each Al in Al2Cl6 has 8 electrons, all involved in bonding; each Al is tetrahedral (17.11). (b) Thermal decomposition of [Me4N]+[ClHI]– could lead to [Me4N]+I– and HCl, [Me4N]+Cl– and HI, or [Me4N]+H– and ICl. [Me4N]+ is a large cation, and therefore all three salts are expected to have about the same lattice energy. The bond strength of the second product is therefore the decisive factor. The bond enthalpies of HCl, HI and ICl are 432, 298 and 211 kJ mol–1 (value for ICl can be found from standard tables of data or estimated from Pauling electronegativity values of Cl and I, bond enthalpies of Cl2 and I2, see Section 2.5 in H&S). (c) The purple solution of I2 in n-hexane contains uncomplexed I2. In benzene, ethanol and pyridine, charge transfer complexes 17.12 form. Et I

I

I

O

I

I

I

N

I

I

H

(17.12)

Most stable complex is that with pyridine; pyridine preferentially complexes I2 in each of the solutions to which it is added, making each solution the same colour. The band at 368 nm (i.e. in the UV region) is assigned to a charge transfer band. The HOMO of I2 is the π* level (see Fig. 2.10 in H&S, MO diagram for the analogous F2). Excitation from this level to the next highest MO gives rise to a σ* ← π*(I2) transition, corresponding to the transition in the visible at 515 nm.

17.13

(a) Compound D (Fig. 17.1) contains S donor atoms. A charge transfer complex is formed by donation of S lone pair electrons into a vacant MO in I2 (see part (c)). (b) By the Beer-Lambert law, absorbance is proportional to concentration of solution. Therefore, changes in absorbance measurements during a reaction or for a series of samples can be used to monitor changes in concentration. From the maximum absorbance in the Job’s plot shown in Fig. 17.1, you can deduce that the ratio of D : I2 in the complex is 1 : 1. (c) The LUMO of I2 is a σ* orbital. Therefore, population of this orbital weakens the I–I bond resulting in a shift in the Raman band from 215 to 162 cm–1.

17.14

In each oxohalide: X=O double bonds and X–Y single bonds (X and Y = halogens). The halogen in the higher oxidation state is the central atom; F is always terminal (ox. state –1).

Fig. 17.1 Job’s plot using data from problem 17.13.

Absorbance

17.12

S

0.20

S

0.10

0.00 Ratio I2:D 0:10

S Donor, D

5:5

10:0

The group 17 elements F

(a) [F2ClO2]– Cl–, number of valence electrons = 8 Number of bonding ‘pairs’ (2 Cl–F + 2 Cl=O) = 4 1 lone pair Molecular shape = disphenoidal, probably with equatorial O, see 17.13. (b) FBrO3 Br, number of valence electrons = 7 Number of bonding ‘pairs’ (1 Br–F + 3 Br=O) = 4 No lone pairs Molecular shape = tetrahedral, see 17.14. (c) [ClO2]+ (note: isoelectronic with SO2) Cl+, number of valence electrons = 6 Cl Number of bonding ‘pairs’ (2 Cl=O) = 2 O O 1 lone pair (17.15) Molecular shape = bent, see 17.15. – (d) [F4ClO] Cl–, number of valence electrons = 8 Number of bonding ‘pairs’ (4 Cl–F + 1 Cl=O) = 5 1 lone pair Molecular shape = square-based pyramidal with axial O, see 17.16.

O

Cl

O F

(17.13) F Br O

O O

(17.14) O F

Cl

F

261

F F

(17.16)

17.15

(a) Section 17.9 in H&S has relevant information. In cold alkali: Cl2 + 2NaOH J NaCl + NaOCl + H2O but on heating, [OCl]– disproportionates and overall reaction in hot alkali is: 3Cl2 + 6NaOH J NaClO3 + 5NaCl + 3H2O (b) In the presence of excess I–, only 1 mole of I2 is produced per mole of [IO4]–. The oxidation step is: 2I– J I2 + 2e– Thus, reduction must be a 2-electron step: [IO4]– + 2e– + H2O J [IO3]– + 2[OH]– Overall: [IO4]– + 2I– + H2O J [IO3]– + I2 + 2[OH]– In the presence of the excess I–, the final step is: [IO3]– + 5I– + 6H+ J 3I2 + 3H2O (c) BaMnO4 contains [MnO4]2–, so [MnO4]– undergoes a 1-electron reduction. Amount of [MnO4]– used = 10 × 0.1 × 10–3 = 1 × 10–3 moles Amount of [I]– used = ∴

0.01587 ≈ 1.25 × 10–4 moles 126.90

Ratio [MnO4]– : I– = 1 × 10–3 : 1.25 × 10–4 = 8 : 1

I– must undergo an 8-electron oxidation and the product is [IO4]– (ox. state +7).

262

The group 17 elements 17.16

(a) ClO2 and [ClO2]– are bent, each with equivalent Cl–O bond lengths: 147 pm in ClO2 and 157 pm in [ClO2]–. Resonance structures can be drawn to rationalize the equivalence of the bonds: Cl

Cl O

O

O

Cl O

Fig. 17.2 π-molecular orbitals in ClO2.

17.17

Cl O

O

Cl

Cl O

O

O

O

O

O

Or, use MO theory to show delocalized bonding schemes analogous to π-allyl-type bonding. There are 12 atomic orbitals in the basis set of ClO2 (4 valence orbitals per atom) and therefore 12 MOs; these are occupied by 19 valence electrons. In [ClO2]–, 10 MOs are fully occupied. The π-MOs are shown in Fig. 17.2. In ClO2, the HOMO is the π* MO and is singly occupied. In [ClO2]–, this MO is fully occupied, giving more antibonding character to the Cl–O interactions. The MO scheme illustrates delocalization of electrons over the O–Cl–O framework, and rationalizes the lengthening of bonds on going from ClO2 to [ClO2]–. (b) Compare the ionic salt formulae: K+[ClO4]– and Ba2+[SO4]2–. [ClO 4]– is isostructural with [SO4]2–, similar sized ions which can be regarded as being spherical. Ionic radii of K+ (138 pm) and Ba2+ (142 pm) are almost the same. The two salts can crystallize with the same 3D-structure, i.e. they are isomorphous. Use reduction half-equations and Eo values from H&S Appendix 11 to help you. (a) [ClO3]– is a strong oxidizing agent: [ClO3]– + 6Fe2+ + 6H+ J Cl– + 6Fe3+ + 3H2O (b) Reduction of [IO3]– occurs in acidic solution, therefore the half-equation for [SO3]2– oxidation must also be in acidic solution (see Appendix 11 in H&S). Possible reaction is: [IO3]– + 3[SO3]2– J I– + 3[SO4]2– but partial reduction also possible: 2[IO3]– + 2H+ + 5[SO3]2– J I2 + 5[SO4]2– + H2O (c) A good starting point is reaction 17.75 in H&S in which [IO3]– and I– react to give I2. The reaction with Br– gives: [IO3]– + 5Br– + 6H+ J 2Br2 + IBr + 3H2O

17.18

(a) Note that it is an equilibrium being considered. Determine the total chlorine by addition of excess of I– and titration with thiosulfate: Cl2 + 2I– J I2 + 2Cl– I2 + 2[S2O3]2– J 2I– + [S4O6]2–

The group 17 elements

263

Of the products, only HCl is a strong acid, and so its concentration can be determined by pH measurements (not by titration which would upset the equilibrium). From these data, the equilibrium constant K at a given temperature can be found. Determine K at different temperatures, and from these data, ΔrHo can be found from: d ln K ΔH o = dT RT 2

(b) Take a weighed amount of I4O9 and dissolve in water. Neutralize the solution with NaHCO3 and titrate the I2 against thiosulfate (see part (a)). This gives the amount of I2 from the initial hydrolysis of I4O9. The reaction with thiosulfate produces I– which, under acidic conditions (created by adding, for example, excess dilute HCl) converts [IO3]– to I2: [IO3]– + 5I– + 6H+ J 3I2 + 3H2O The I2 formed in this step is then titrated against thiosulfate. (c) In the solidified noble gas matrix, species can be studied by vibrational spectroscopy. Raman (not IR) spectroscopy must be used to study a homonuclear diatomic because stretching mode is IR inactive. Going from Cl2 to [Cl2]– weakens the Cl–Cl bond because the extra electron occupies an antibonding MO (see Fig. 2.8 in H&S ). The formation of [Cl2]– from Cl2 should be accompanied by a shift to lower wavenumber for the absorption corresponding to the Cl–Cl stretch. 17.19

(a) HF remains strongly hydrogen-bonded (17.17) in the vapour state and therefore hydrogen bonds are not broken upon vaporization. Liquid H2O is hydrogen-bonded but on vaporization, these intermolecular interactions are broken. A greater amount of energy is therefore needed to vaporize H2O than HF. (b) The fact that AgCl and AgI are both soluble in saturated aqueous KI (in which I– is present), but insoluble in saturated aqueous KCl (in which Cl– is present) suggests that the complex formation between Ag+ and I– gives a more stable species than is formed between Ag+ and Cl–. Species that may be formed are [AgX2]– and [AgX3]2–.

17.20

(a) The structural descriptions liken the ammonium salts to NaCl, CsCl or ZnS and this arises from assumption that [NH4]+ can be treated as a spherical ion. In [NH4]+X– (X = Cl, Br, I), the ions are arranged in either an NaCl or CsCl structure. The preference for the wurtzite structure for [NH4]F arises from N–H......F hydrogen bond formation which gives a structure similar to that of ice. (b) The decomposition reaction is:

F H H F

(17.17)

[PH4]+X– J PH3 + HX For the product HX where X = F, Cl, Br or I, the weakest bond is H–I; in the absence of numerical data, you can only infer that this factor is the most important. For a detailed answer in terms of a thermochemical cycle, see answer 7.24, p. 107. 17.21

(a) Refer to Section 8.6 in H&S. Figure 17.3 shows the Frost-Ebsworth diagram for chlorine. Each of the species in the plot is also shown in the potential diagram (Fig. 17.4). For Cl in its standard state (Cl2), ΔGo = 0. In the Frost-Ebsworth diagram, values of ΔGo/F for the formation of M(N) (where N is the oxidation state) from M(0) are plotted against N. Therefore, in Fig. 17.3, each point shows the stability

264

The group 17 elements

Fig. 17.3 Frost-Ebsworth diagram for chlorine (pH 0).

[ClO4]–

– ΔG/F = zEo / V

10 8

[ClO3]–

6 HClO2

4 2

HOCl

0

–2

Fig. 17.4 Potential diagram for chlorine at pH 0.

[ClO4]–

Cl2 Cl

–

–1

+1.19

0

1

[ClO3]–

2

3

4

+1.21

5 6 7 Oxidation state, N

+1.64 HClO2

+1.61 HOCl

+1.36 Cl2

Cl–

+1.47

of a given species with respect to ΔGo/F = 0 for Cl2. In the potential diagram (Fig. 17.4), each reduction step is considered sequentially. For example, for the 1-electron reduction of HOCl to Cl2, the value of ΔGo/F in Fig. 17.3 is the same as the Eo value in Fig. 17.4. However, for HClO2, Fig. 17.4 considers only the 2-electron reduction of HClO2 to HOCl whereas Fig. 17.3 plots a point that corresponds to the 3-electron reduction from HClO2 to Cl2. (b) In Fig. 17.3 above, the lowest point represents the most stable oxidation state of Cl, and a move down the plot represents a thermodynamically favoured process. Therefore, Cl– is the most thermodynamically favoured species. (c) Species towards the top-right of the diagram are oxidizing, i.e. a move down the plot is thermodynamically favourable and the species is therefore easily reduced. In Fig. 17.3, the best oxidizing agent is therefore [ClO4]–. (d) If [H+] is involved in the half-equation, then by the Nernst equation the value of E depends on [H+]. See for example, answer 8.7 on p. 122. 17.22

(a)

10CsF + I2O5 + 3IF5 J 5Cs2IOF5 5CsF + I2O5 + 3IF5 J 5CsIOF4

Neither reaction involves reduction-oxidation; the oxidation state of I remains +5. (b) From Fig. 17.14 (at pH 0), the values of Eo needed are: [ClO4]– J [ClO3]– [ClO3]– J Cl2 Cl2 J Cl–

Eo = +1.19 V Eo = +1.47 V Eo = +1.36 V

and these refer to the half-equations: [ClO4]– + 2H+ + 2e– [ClO3]– + H2O – + – 2[ClO3] + 12H + 10e Cl2 + 6H2O Cl2 + 2e– 2Cl–

Eo = +1.19 V Eo = +1.47 V Eo = +1.36 V

The group 17 elements

265

From the last two values, calculate Eo for [ClO3]– J Cl– using ΔGo = –zFEo. For [ClO3]– J Cl2 ΔGo = –(10)(1.47)F = –14.7F For Cl2 J Cl– ΔGo = –(2)(1.36)F = –2.72F – – For [ClO3] J Cl ΔGo = –14.7F – 2.72F = –17.42F per 2 moles of [ClO3]– = –8.71F per mole of [ClO3]– ⎛ − 8.71F ⎞ ⎟ = +1.45 V Eo = − ⎜ ⎝ 6F ⎠

For the reaction: 4[ClO3]–

3[ClO4]– + Cl–

the half-equations to combine are: [ClO4]– + 2H+ + 2e– [ClO3]– + 6H+ + 6e– i.e.

4[ClO3]– + 6H+ + 3H2O

[ClO3]– + H2O Cl– + 3H2O

Eo = +1.19 V Eo = +1.45 V

3[ClO4]– + 6H+ + Cl– + 3H2O

Eocell = +1.45 – (+1.19) = 0.26 V ΔGo = –zFEocell = –(6)(96 485)(0.26) × 10–3 = –150 kJ mol–1 Although ΔGo is negative, the reaction does not occur at 298 K; this must be the result of a kinetic factor, i.e. high activation energy. (c) ClO2 is an ‘elemental chlorine-free’ bleaching agent and, on environmental grounds, is used in preference to Cl2. ClO2 is prepared from NaClO3, therefore consumption of NaClO3 gives a measure of the consumption of ClO2.

Fig. 17.5 Qualitative MO diagram for the formation of BrO from Br and O. The Br–O internuclear axis coincides with the z axis.

(a) Figure17.5 shows a qualitative MO diagram for BrO. An LCAO approach has been used. At such a qualitative level, it is not possible to say to what extent interaction between the Br 4s and O 2s will lead to σ and σ* MOs or to two MOs that are essentially non-bonding because of the energy mismatch between the ns σ* Energy

17.23

π* 4px, 4py 4pz

2px, 2py 2pz

π

σ 4s

2s Br

BrO

O

266

The group 17 elements

Cl

191 pm

Cl

+

O O 119 pm

orbitals. The diagram in Fig. 17.5 assumes no interaction. Thus, the lowest occupied MO is localized on O and the second lowest occupied MO is localized on Br. There are 13 electrons (6 from O and 7 from Br) and therefore the molecule is paramagnetic. From the MO diagram, the bond order is 1.5. (b) The structure of [Cl2O2]+ is shown in 17.18 (see H&S for more details). The bonding in this cation is described in terms of charge transfer between [Cl2]+ and O2. The HOMOs of these species are: O2

Cl----O = 242 pm

(17.18)

πg*(2px)1πg*(2py)1

See answer 2.10, p. 17

[Cl2]+ πg*(3px)2π g*(3py)1

Cl has one more valence electron than O

The degenerate HOMO of O2 is half-occupied, while that of [Cl2]+ contains 3 electrons. Interactions involving πg*(2px)1πg*(2py)1 of O2 and πg*(3px)2π g*(3py)1 of [Cl2]+ give out-of-plane (π-type) and in-plane (σ-type) bonding interactions: X For more information, see: T. Drews et al. (1999) J. Am. Chem. Soc., vol. 121, p. 4379

Cl

+

Cl O

O

πg*(3py)

πg*(3px)

πg*(2py)

πg*(2px)

17.24

(a) The data required are in the potential diagrams shown in Fig. 17.14 in H&S. For aqueous HOI at pH 0, the half-equations to describe its disproportionation are: [IO3]–(aq) + 5H+(aq) + 4e–

HOI(aq) + 2H2O(l)

Eo = +1.14 V

2HOI(aq) + 2H+(aq) + 2e–

I2(aq) + 2H2O(l)

Eo = +1.44 V

For the disproportionation: 5HOI(aq) J [IO3]–(aq) + 2I2(aq) + H+(aq) + 2H2O(l) Eocell = (+1.44) – (+1.14) = 0.30 V A positive Eocell corresponds to a negative ΔGo and to a spontaneous reaction. For HOI in aqueous HCl (pH 0), the half-equations to describe its disproportionation are: [IO3]–(aq) + 6HCl(aq) + 4e– [ICl2]–(aq) + 3H2O(l) + 4Cl–(aq) Eo = +1.23 V 2[ICl2]–(aq) + 4H+(aq) + 2e–

I2(aq) + 4HCl(aq)

Eo = +1.06 V

It is not possible to combine these half-equations to give the disproportionation: 5[ICl2]–(aq) + 4H+(aq) + 3H2O(l) J [IO3]–(aq) + 2I2(aq) + 10HCl(aq) such that Eocell has a positive value. Therefore, [ICl2]– is stable with respect to disproportionation under these conditions.

The group 17 elements

F F

F – Sb

F

+ Cl

F F

F

F F

F

(17.19)

(17.20)

F F

F F

F

Sb

F

F

Cl F

F F

(17.21)

Fax

Cl

Fax

(17.22)

(b) The structures of [ClF4]+ and [SbF6]– are shown in 17.19 and 17.20. These are consistent with the VSEPR model: [ClF4]+ Central atom is Cl Cl (group 17) has 7 valence electrons Subtract one electron for the positive charge Number of bonding pairs (4 I–F bonds) = 4 Number of lone pairs = 1 Total number of electron pairs = 5 = 1 lone and 4 bonding pairs ‘Parent’ shape = trigonal bipyramidal Molecular shape = disphenoidal [SbF6]– Central atom is Sb Sb (group 15) has 5 valence electrons Add one electron for the negative charge Number of bonding pairs (6 Sb–F bonds) = 6 No lone pairs Total number of electron pairs = 6 ‘Parent’ shape = molecular shape = octahedral In the solid-state structure of [ClF4][SbF6], the zigzag chains arise from the formation of Sb–F–Cl bridges. Each Cl and Sb atom is in an octahedral environment (17.21).

17.25

The correct pairings are shown in the table below: List 1 List 2 Comments HClO4 Strong acid in aqueous solution Adopts a prototype structure See Fig. 6.19a in H&S CaF2 I2O5 Anhydride of HIO3 Acid anhydride forms when one or more molecules of acid lose one or more molecules of H2O ClO2 Radical See answer 17.16a [BrF6]+ Formation requires a powerful See eq.17.33 in H&S fluorinating agent [IF6]– Distorted octahedral structure Apply VSEPR model; has stereochemically active lone pair HOCl Weak acid in aqueous solution C6H6.Br2 Charge transfer complex See Fig. 17.6 in H&S ClF3 Used in nuclear fuel industry See eq. 17.25 in H&S to fluorinate uranium RbCl Solid contains octahedrally RbCl has NaCl structure; sited chloride ion see Fig. 6.16 in H&S Halogen in square planar See structure 17.9, p.259 I2Cl6 environment

17.26

(a) Both ClF3 and BF3 are non-linear, so: Number of degrees of vibrational freedom = 3n – 6 = (3 × 4) – 6 = 6 Consider shapes of ClF3 (T-shaped, C2v, with bond angles < 90o, 17.22) and BF3 (trigonal planar, D3h, 17.23). The IR spectrum of ClF3 contains 6 absorptions, consistent with there being 6 normal modes of vibration, all IR active; these are approximately described as equatorial stretch, symmetric axial stretch, asymmetric axial stretch and three deformation modes.

F Feq

267

B F

(17.23)

F

The group 17 elements

268

X Caution! All perchlorate salts are potentially explosive

17.27

The IR spectrum of BF3 contains only 3 bands because the symmetric stretch is IR inactive; active modes are symmetric deformation, asymmetric stretch (doubly degenerate), asymmetric deformation (doubly degenerate) – see Fig. 3.14 in H&S. (b) Potentially explosive compounds: ClO2 (explodes on contact with organic material), KClO4 (potentially explosive on contact with organic material), Cl2O6 (explodes on contact with organic material), Cl2O (potentially explosive with a spark, with heat or on contact with organic material), Br2O3 (explosive), ClF3 (reacts explosively with water, ignites organic materials spontaneously), and BrF3 (reacts explosively with water and sets fire to asbestos; having a higher boiling point than ClF3, it is not vaporized as the reaction starts and so is more dangerous than ClF3). Hazards of other compounds: HF (causes severe skin burns). (a) A set of resonance structures for part of a chain in [(AgI2)n]n+ which illustrate I2 donating charge to Ag+ : I I

Ag+

+ Ag

I

I

I

I

Ag

I+

I

+

I

+I

I

I

I

[I5]+ has structure 17.24. The bond lengths are 264 pm (terminal) and 289 pm (middle) (see H&S Section 17.7). Resonance structures are: I

I

2+

I

I

I

I

(17.25)

F

SbF5 + 2e– F Sb F F

F

(17.26)

Sb

F F

F

(17.27)

F

I I

F

[SbF6]– + SbF5 J [Sb2F11]–

I2 is oxidized to [I2]+. This is paramagnetic (1 unpaired electron) and dimerizes to gives [I4]2+ : I2

F

F

I

For [Sb3F14]– : see text describing structure 17.15 in H&S

F Sb

F

F

+I

SbF3 + 2F–

SbF5 + F– J [SbF6]–

F F

+

(b) SbF5 acts as both an oxidizing agent (it is reduced to SbF3) and as a fluoride acceptor:

F Sb

I I

I

F

Ag

I

I

(17.24)

I

Ag+

I

I

F

+ Ag

[I2]+ + e–

2[I2]+ J [I4]2+

The structures of the product ions are shown in 17.25-17.27. Oxidation states: [I4]2+ : formally, each I is in oxidation state +1/2 [SbF6]– : Each F is in oxidation state –1; Sb in oxidation state +5. [Sb3F14]– : Each F is in oxidation state –1; Sb in oxidation state +3.

The group 17 elements 17.28

269

(a) BrO and HO2 are radicals and the first reaction involves homolytic cleavage of the HO bond in HO2 and formation of a new HO bond to give BrOH. Br

O +

O

O

BrOH + O

O

O

O

H

Reaction of HOBr with HBr is a redox reaction with Br being oxidized and reduced: oxidation states for Br: +1 in HOBr, –1 in HBr, 0 in Br2. Photolysis of Br2 is homolytic fission to give two Br radicals. Reaction of Br radical with ozone to give BrO radical and the diradical O2. (b) See Box 14.6 in H&S. Compare the equations given for reactions involving ClO and Cl with those in the question involving BrO and Br. 17.29

B

(17.28)

F B

Mes Mes

NMe3

(17.30)

17.30

(a) BMes3 has structure 17.28 with trigonal planar B and an empty 2p atomic orbital. It forms a Lewis acid-Lewis base complex with fluoride ion; F– donates a pair of electrons to B, and the hybridization of the B atom changes from sp2 to sp3 as the shape changes from trigonal planar to tetrahedral. The product is [Mes3BF]–. (b) Compound B in the question binds fluo– ride as a bridging unit between the two B Mes Mes atoms (structure 17.29). Chelation and B delocalization of negative charge stabilizes F the complex. B (c) A zwitterion carries both positive and Ph negative charges but is neutral overall, e.g. Ph – reaction of C with F gives compound 17.30. (17.29) (d) The structure shown in 17.30 for the complex formed between C and F– is not consistent with the 1H NMR spectroscopic data. Two signals are observed for the CH2 unit and one signal shows coupling to the BMes2 19F nucleus. This suggests formation of an F F...H interaction and the proposed structure H is 17.31. Me3N H (e) The use of the cationic trimethylammonium substituent results in (17.31) the formation of a system that binds fluoride ion in aqueous media. For more details for this answer, see: T. W. Hudnall et al. (2009) Acc. Chem. Res., vol. 42, p. 388. (a) I2 is insoluble in water, and iodine uptake by the human body requires a watersoluble form of I2. Reaction with iodide ion produces a water-soluble salt: I2 + KI J KI3 (b) HF is used commercially for etching patterns on glass, e.g. for security etching of car window glass. The reactions are: 4HF + SiO2 J SiF4 + 2H2O SiF4 + 2HF J H2SiF6

270

The group 17 elements (c) Perchlorate is the source of oxygen for oxidation of aluminium metal. The highly exothermic reaction is: 10Al + 6NH4ClO4 J 3N2 + 4Al2O3 + 2AlCl3 + 12H2O

271

18 The group 18 elements 18.1

(a) Noble gases (b) He helium Ne neon Ar argon Kr krypton Xe xenon Rn radon (c) a fully occupied quantum shell, 2s2 for He, and ns2np6 for the later elements.

18.2

See answer 2.9, p. 17.

18.3

XeF2

F Xe

XeF4

F

(18.1) F F

F

XeF6

Xe

F

F F

(18.3)

F F

F

F

F

Xe, number of valence electrons = 8 Number of bonding pairs (4 Xe–F) = 4 2 lone pairs Molecular shape = square planar, see 18.2.

F

F

Xe

F

F

(18.2)

Xe, number of valence electrons = 8 Number of bonding pairs (6 Xe–F) = 6 1 lone pair Molecular shape = distorted octahedral, see 18.3.

18.4

[XeF8]2– Xe2–, number of valence electrons = 10 Number of bonding pairs (8 Xe–F) = 8 1 lone pair The observation that the anion is square antiprismatic (18.4) indicates a stereochemically inactive lone pair.

18.5

(a) To determine a value of ΔfHo(XeF2, 298 K), measure the enthalpy change for the hydrolysis of XeF2:

F F

Xe

Xe, number of valence electrons = 8 Number of bonding pairs (2 Xe–F) = 2 3 lone pairs Molecular shape = linear, see 18.1.

F

(18.4)

2XeF2(s) + 2H2O(l)

ΔrHo(298 K)

2Xe(g) + 4HF(g) + O2(g)

.

4ΔfHo(HF, 298 K) (available from tables of standard data)

2ΔfHo(XeF2, 298 K) + 2ΔfHo(H2O, 298 K)

2Xe(g) + 2H2(g) + 2F2(g) + O2(g) 2ΔfHo(XeF2, 298 K) = 4ΔfHo(HF, 298 K) – ΔrHo(298 K) – 2ΔfHo(H2O, 298 K) available data

measured

available data

272

The group 18 elements (b) Set up the following thermochemical cycle (sub = sublimation), all data at 298 K:

XeF2(s)

ΔsubHo(XeF2)

XeF2(g) ΔaHo(XeF2) = 2D(Xe–F)

ΔfHo(XeF2, s)

Xe(g) + F2(g)

Xe(g) + 2F(g)

2ΔaHo(F2)

= D(F–F)

The value of ΔfHo(XeF2, s) comes from part (a) of this problem; ΔsubHo(XeF2) can be measured; D(F–F) = 158 kJ mol–1. The Xe–F bond energy can be found from: 2D(Xe–F) = D(F–F) – ΔfHo(XeF2, s) – ΔsubHo(XeF2) 18.6

Consider the reactions: Xe + Cl2 J XeCl2 Xe + F2 J XeF2 From the thermochemical cycle:

Xe(g) + X2(g)

ΔrHo

D(X–X)

XeX2(g) 2D(Xe–X)

Xe(g) + 2X(g) the enthalpy change for each reaction is: ΔrHo = D(X–X) – 2D(Xe–X)

X = Cl or F

The bond enthalpies (from Tables 17.1 and 18.2 in H&S) are: D(Cl–Cl) = 242 kJ mol–1 D(F–F) = 158 kJ mol–1 –1 D(Xe–F) = 133 kJ mol It is expected that D(Xe–Cl) < D(Xe–F) (e.g. see Tables 14.2 or 15.3 in H&S for trends in other halides). For X = F: ΔrHo = (158) – 2(133) = –108 kJ mol–1 For X = Cl, the reaction cannot be more exothermic than for X = F unless the Xe–Cl bond is stronger than the Xe–F bond which will not be the case. 18.7

Xe is next to Cs in the periodic table. Assume that the lattice energy of Xe+F– ≈ lattice energy of Cs+F–. Set up a Born-Haber cycle (e.g. see answer 6.15) and estimate ΔfHo(XeF, 298 K). Other data required including IE1 of Xe are obtained from tables of standard data.

The group 18 elements 18.8

[XeO6]4– Xe4–, number of valence electrons = 12 Number of bonding ‘pairs’ (6 Xe=O) = 6 No lone pairs Molecular shape = octahedral. XeOF2 Xe, number of valence electrons = 8 Number of bonding ‘pairs’ (1 Xe=O + 2 Xe–F) = 3 2 lone pairs Molecular shape = T-shaped, see 18.5; O atom is in equatorial site. XeOF4 Xe, number of valence electrons = 8 Number of bonding ‘pairs’ (1 Xe=O + 4 Xe–F) = 5 1 lone pair Molecular shape = square pyramidal, see 18.6; O atom is in apical site. XeO2F2 Xe, number of valence electrons = 8 Number of bonding ‘pairs’ (2 Xe=O + 2 Xe–F) = 4 1 lone pair Molecular shape = disphenoidal, see 18.7; O atoms are in equatorial sites. XeO2F4 Xe, number of valence electrons = 8 Number of bonding ‘pairs’ (2 Xe=O + 4 Xe–F) = 6 No lone pairs Molecular shape = octahedral; O atoms could be trans or cis. XeO3F2 Xe, number of valence electrons = 8 Number of bonding ‘pairs’ (3 Xe=O + 2 Xe–F) = 5 No lone pairs Molecular shape = trigonal bipyramidal, 18.8; F atoms are axial.

18.9

(a) CsF is ionic, and a source of F–. XeF4 acts as F– acceptor:

F O

Xe F

(18.5) O F

Xe

F

F F

(18.6) F O

Xe O F

(18.7) F O

Xe

273

O O

F

(18.8)

CsF + XeF4 J Cs[XeF5]

i.e. Cs+[XeF5]–

(b) F for O exchange; reaction driven by formation of SiF4 (formation enthalpy = –1615 kJ mol–1):

or

SiO2 + 2XeOF4 J SiF4 + 2XeO2F2 SiO2 + XeOF4 J SiF4 + 2XeO3

(c) SbF5 is very good F– acceptor; [SbF6]– forms adducts with SbF5. The [XeF]+ ion may associate with XeF2. Possible reactions are: or or

XeF2 + SbF5 J [XeF]+[SbF6]– 2XeF2 + SbF5 J [Xe2F3]+[SbF6]– XeF2 + 2SbF5 J [XeF]+[Sb2F11]–

274

The group 18 elements (d) Hydrolysis of XeF6 gives XeO3, but in alkaline solution this disproportionates to Xe and perxenate ion, [XeO6]4– . Overall: 2XeF6 + 16[OH]– J [XeO6]4– + Xe + O2 + 8H2O + 12F– (e) Hydrolysis of KrF2 is analogous to hydrolysis of XeF2: 2KrF2 + 2H2O J 2Kr + O2 + 4HF 18.10

Section 18.4 in H&S gives information for the answer. Points to include: • preparations, and difficulties of obtaining pure samples; • structures in gas phase and solid state; • hydrolysis by H2O; • reactions showing XeFn acting as oxidizing agents; • reactions with silica, and the unsuitability of silica glassware for studying reactions of Xe fluorides; • reactions illustrating F– donor and acceptor properties.

18.11

(a) CsRuF6 is ionic: Cs+[RuF6]–. The Raman spectrum shows bands arising from the Raman active vibrational modes of [RuF6]–. Thus, the product of the reaction also contains [RuF6]– : XeF2 + RuF5 J [XeF]+[RuF6]– The band in the Raman spectrum at 600 cm–1 arises from the Xe–F stretching mode. (b) F2 is an oxidizing agent, and is reduced to F–: F2 + 2e–

2F–

[XeF]+ is oxidized to Xe(VI) and accepts F–. The reaction is: [XeF]+[RuF6]– + excess F2 J RuXeF11 and RuXeF11 is [XeF5]+[RuF6]– with an inter-ion Xe....F–Ru interaction making the product monomeric. 18.12 X See also worked example 18.1 in H&S X 129Xe 19F,

26.4%, I = 1/2

100%, I = 1/2 B, 80 %, I = 3/2

11

No 19Fa-19Fb coupling is observed in this molecule

An addition reaction across the C=C bond is not likely. XeF2 can act as a source of F– and one plausible reaction to suggest is the removable of the BF2 unit and formation of [BF4]–. The 19F NMR spectrum shows that A has 3 F environments: 1 in the [BF4]– ion and 2 in the cation. Therefore, the cation does not contain a mirror plane containing the C=C bond. The 129Xe NMR spectrum shows that the 129Xe nucleus couples to 2 inequivalent 19F nuclei. The following product is consistent with the data: Fc couples to 11B to give a 1:1:1:1 (non-binomial) quartet. a F F c Cl 26.4% of Fa couples to 129Xe, and 26.4% Fb couples to 129Xe leading to B c each signal for Fa and Fb being a F c F F c b F Xe+ singlet with superimposed doublet. 129 Xe couples to Fa and Fb to give a doublet of doublets.

The group 18 elements 18.13 F F y

x F3As

F

Sb

F

F

Au

F

(18.9)

18.14

F Xe F F

F Cr

F

F

F F

F Cr

F

F n

Xe F

(18.11)

18.15

275

The reaction to consider is: [F3AsAu][SbF6] + Xe

in anhydrous HF/SbF5

[F3AsAuXe][Sb2F11]

[F3AsAu][SbF6] has structure 18.9. The hypothetical [AuF2]– ion is included in the question to show that the Au–F bond in [F3AsAu][SbF6] is relatively long. This suggests that [SbF6]– might be displaced fairly easily to give a [F3AsAu]+ unit to which Xe can add. If you started with AuF3, reduction in HF/SbF5 in the presence of Xe would give an Au(II) compound (see eq. 18.24 in H&S and discussion). (a) The structure of [Kr2F3]+ is shown in 18.10. There are two F environments, ratio 1 : 2. In the 19F NMR spectrum (19F, I = 1/ , 100%), coupling 2 between non-equivalent 19F nuclei gives rise to a doublet (for Fa) and a triplet (for Fb). The coupling constants for both signals must be the same.

b F

+ Kr

Kr

F a

F a

(18.10)

(b) Examples for this answer are in Section 18.4 of H&S. Points to include: • number of compounds involving Xe >> number involving Kr; • bridge formation appears to stabilize group 18 containing species in the solid state; • association between species by Xe–F–Xe or Kr–F–Kr, e.g. [XeF]+ and XeF2 leads to [Xe2F3]+ and similarly for [Kr2F3]+; • [XeCl]+ stabilized as in [XeCl][Sb2F11] by Cl–Xe–F bridge formation; • some bridge formations lead to more extended structures, e.g. in [XeF][CrF5], 18.11. Structure 18.11 gives one example where formula [XeF][CrF5] might indicate presence of [XeF]+ and [CrF5]– ions but structure is polymeric in the solid state. For other examples, see H&S. (a) 7KrF2 + 2Au J 2[KrF]+[AuF6]– + 5Kr

(See Section 18.5 in H&S)

(b) XeO3 + RbOH J Rb[HXeO4]

(See eq. 18.11 in H&S)

(c) [XeCl][Sb2F11]

298 K

Xe + Cl2 + [XeF][Sb2F11] + SbF5

(d) 3KrF2 + 2B(OTeF5)3 J 3Kr(OTeF5)2 + 2BF3 (See Section 18.5 in H&S) (e) C6F5XeF + Me3SiOSO2CF3 J [C6F5Xe]+[CF3SO3]– + Me3SiF (f) [C6F5XeF2]+ + C6F5I J [C6F5Xe]+ + C6F5IF2 18.16 X See: M. Gerken et al. (2002) Inorg. Chem., vol. 41, p. 198

XeO4 is thermodynamically unstable, and both solid and gas decompose explosively. Solutions of XeO4 in SO2ClF, HF or BrF5 at low temperatures are kinetically stable. Thus XeO4 can be prepared by: Na4XeO6 + 2H2SO4(100%) J XeO4 + 2H2O + 2Na2SO4 and is condensed at low temperatures in an appropriate solvent; explosions can still occur when cold solvent initially comes into contact with solid XeO4. Caution using BrF5 and HF (protective clothing and gloves). Always small scale reactions.

276

The group 18 elements Disposal of XeO4: only small quantities (approx. 50 mmol) to be disposed of at a given time by pouring (use long tongs to hold container) the cold solution of XeO4 in SO2ClF, HF or BrF5 into several litres of aqueous NaOH/ice in a fume cupboard which is shielded for protection against explosion. 18.17

KrF2 has D∞h symmetry (linear). Vibrational modes are the symmetric stretch, asymmetric stretch and bend (see Fig. 3.11 in H&S). The symmetric stretch is IR inactive (vibration does not give rise to a change in dipole moment). Of the vibrational modes at 590, 449 and 233 cm–1, that at 449 cm–1 must be due to the symmetric stretch and is not observed in the IR spectrum.

18.18

See Fig. 18.1 for a partial MO diagram for XeF2. There is one bonding MO, one non-bonding MO and one antibonding MO; bonding in XeF2 is described as a 3c2e interaction; formal Xe–F bond order = 1/2. Now consider [XeF]+; isoelectronic (in terms of only valence electrons) with F2. The MO diagram in Fig. 18.2 is constructed at a qualitative level, and it is difficult to assess the degree of interaction between the Xe 5s and F 2s orbitals because they are poorly matched in energy. The diagram in Fig. 18.2 assumes no interaction and the lowest lying MOs are non-bonding, being localized on F and Xe respectively. The net bonding in [XeF]+ arises from one σ-bonding MO, and the Xe–F bond order = 1.

X See Fig. 5.30 in H&S

18.19

(a) Boiling point of He = 4.2 K. See part (c). (b) Enthalpy of vaporization refers to: He(l) J He(g) Only van der Waals forces (induced dipole) operate between atoms of He and these are extremely weak, leading to the very small value of ΔvapH(bp). (c) The temperature of the super-cooled magnet is maintained using liquid He but without the surrounding liquid nitrogen (bp 77 K), He would vaporize too quickly leading to quenching of the magnet.

18.20

The ground state electronic configuration of He is 1s2. Excitation of an electron (by electrical discharge) from the ground to higher energy levels is followed by relaxation to the ground state and emission of energy. Of the possible transitions, some have energies that correspond to wavelengths outside the visible region, and some are of low probability. The yellow glow arises from emission(s) of energy corresponding to a wavelength(s) close to 600 nm.

Energy

σu*

5pz

σg

σ*

Energy

X See Fig. 1.3 in H&S for emission spectrum of atomic hydrogen

σu

π* 5px, 5py 5pz

2px, 2py 2pz

π

σg

σ 2s

σu Xe

XeF2

Fig. 18.1 Partial MO diagram for XeF2.

5s F----F

Xe+

[XeF]+

Fig. 18.2 MO diagram for [XeF]+.

F

277

19 d-Block metal chemistry: general considerations

4

19.1

(a) Detailed information about the s- and p-block elements is found by referring to Chapters 11-18 of H&S. Ignore oxidation state of 0 for the element itself. Points to include: • an s-block (group 1 or 2) element exhibits one oxidation state (+1 for group 1, +2 for group 2); exceptions are in group 1 metal alkalides containing M– ; • elements in groups 13-16 exhibit either one or two states, e.g. +3 or +1 in group 13 with +1 being available to the later elements owing to the 6s thermodynamic inert pair effect (see Box 13.4 in H&S); • in group 17, F shows oxidation state of –1, but variable oxidation states are observed for the remaining halogens with iodine showing the largest range (–1 to +7); • in group 18, only Xe has an extensive chemistry and exhibits oxidation states from +2 to +8. (b) In group 3, only the +3 state is stable, e.g. Sc(III); in group 12, Zn and Cd show only +2 while Hg shows +1 and +2; for the rest of the d-block, variable oxidation states are characteristic with the widest range observed in the middle of the block, e.g. Mn and Fe.

19.2

(a)

Group number 8 11

Ti

Fe

Cu

Zr

Ru

Ag

Hf

Os

Au

(19.1)

Sc Ti V Cr Mn Fe Co Ni Cu Zn

[Ar]4s23d 1 [Ar]4s23d 2 [Ar]4s23d 3 [Ar]4s13d 5 [Ar]4s23d 5 [Ar]4s23d 6 [Ar]4s23d 7 [Ar]4s23d 8 [Ar]4s13d10 [Ar]4s23d10

Group number 8 9 10 Ru Rh Pd Os

Ir

Pt

(19.2)

(b) Groups 4, 8 and 11: see 19.1. (c) Platinum group metals: see 19.2. 19.3

The trend in Eo values is irregular across the period. Factors that contribute to Eo are summarized in a thermochemical cycle analogous to that in answer 8.21. Use this answer for the way in which to tackle problem 19.3, and also refer to the comparison of Eo values for Cu2+/Cu and Zn2+/Zn given in Section 8.7 in H&S.

19.4

Relevant sections in H&S from which to draw material are: • interstitial hydrides: Section 10.7; • molecular hydride anions: Section 10.7 and Fig. 10.14 in H&S; • metal borides: Section 13.10 and Table 13.3; • interstitial carbides: Section 14.7; • steel (iron carbides and alloys): Section 6.7; • interstitial nitrides: Section 15.6. Include examples of syntheses as well as some structural data.

278

d-Block metal chemistry: general considerations 19.5

Properties that should be included; information from Section 19.5 of H&S: • metal atom has one or more occupied nd orbitals in the valence shell; • variable oxidation states possible with largest range being for metals in the middle of the row, e.g. Mn, Fe; • compounds often coloured – exceptions are d 0 , e.g. Sc(III), and d10, e.g. Zn(II); • compounds may be diamagnetic or paramagnetic depending on the electronic configuration and the coordination environment; • complex formation – this is not unique to d-block metal ions but is a common property; changes in coordination environment can lead to a change in colour, e.g. pink [Co(OH2)6]2+ to blue [CoCl4]2–, both Co(II).

19.6

(a) High coordination numbers are usually not feasible on steric grounds because cations of first row metals are too small to accommodate large numbers of donor atoms. Coordination number of 6 is common, e.g. in aqua ions. (b) A high oxidation state places a high formal charge on the metal centre; by the electroneutrality principle, the distribution of charge in the metal-containing species is such that the actual charge on the metal atom is no greater than ≈ +1. In [Ti(NO3)4], for example, the Ti(IV) centre is 8-coordinate (four bidentate [NO3]–); the ligands remove some of the positive charge from metal centre. (c) Ligands which are formally F– or O2– are highly electronegative and can remove positive charge from metal centres in high ox. states, see part (b); also, F2 and O2 are strongly oxidizing and are often associated with the formation of high ox. states.

19.7

Refer to Table 19.1 for ground state configurations of metal atoms. Refer to Table 7.7 in H&S for ligand abbreviations. (a) [Mn(CN)6]4– contains 6 [CN]– ligands; ox. state of Mn = +2; valence electronic configuration is d5. (b) [FeCl4]2– contains 4 Cl– ligands; ox. state of Fe = +2; valence electronic configuration is d6. (c) [CoCl3(py)3] contains 3 Cl– and 3 pyridine (neutral N-donor) ligands; ox. state of Co = +3; valence electronic configuration is d6. (d) [ReO4]– contains 4 O2– ligands; ox. state of Re = +7; Re is in group 7; valence electronic configuration is d 0 . (e) [Ni(en)3]2+ contains 3 en (neutral N,N′-donor) ligands; ox. state of Ni = +2; valence electronic configuration is d8. (f) [Ti(OH2)6]3+ contains 6 H2O ligands; ox. state of Ti = +3; valence electronic configuration is d1. (g) [VCl6]3– contains 6 Cl– ligands; ox. state of V = +3; valence electronic configuration is d 2 . (h) [Cr(acac)3] contains 3 [acac]– (O,O′-donor) ligands; ox. state of Cr = +3; valence electronic configuration is d3.

Table 19.1 Ground state electronic configurations of M. M

Configuration

Sc Ti V Cr Mn Fe Co Ni Cu Zn

[Ar]4s23d1 [Ar]4s23d2 [Ar]4s23d3 [Ar]4s13d5 [Ar]4s23d5 [Ar]4s23d6 [Ar]4s 23d 7 [Ar]4s 23d 8 [Ar]4s 13d10 [Ar]4s 23d10

19.8

Kepert model (see Section 19.7 in H&S) considers repulsions between ligands, but lone pairs of electrons on the metal centre do not influence the shape. (a) 2-coordinate predicted to be linear. (b) 3-coordinate predicted to be trigonal planar. (c) 4-coordinate predicted to be tetrahedral. (d) 5-coordinate predicted to be trigonal bipyramidal or square-based pyramidal (small energy difference). (e) 6-coordinate predicted to be octahedral.

d-Block metal chemistry: general considerations 19.9

Follow the method outlined in Chapter 3 of H&S and use the flow-chart in Fig. 3.10 in H&S. Trigonal bipyramid, 19.3:

START

(19.3)

279

Is the molecule linear?

No

Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Are there 3 C2 axes perpendicular to the principal axis?

No Yes: C3 axis Yes

Is there a σh plane?

Yes

STOP

Conclusion: the point group is D3h. Square-based pyramid, 19.4:

START

(19.4)

Is the molecule linear?

No

Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Are there 4 C2 axes perpendicular to the principal axis? Is there a σh plane? Are there n (i.e. 4) σv planes containing

No Yes: C4 axis

the Cn axis?

No No Yes

STOP

Conclusion: the point group is C4v. Square antiprism (two staggered squares form opposite faces), 19.5:

C2

START

Is the molecule linear?

No

No Does it have Td, Oh or Ih symmetry? Is there a Cn axis? Yes: C4 axis (in 19.5, perpendicular to the plane of paper) Are there 4 C2 axes perpendicular to the principal axis? Yes (one C2 axis is shown in 19.5) Is there a σh plane? No Are there n (i.e. 4) σv planes containing

(19.5)

the Cn axis?

Yes

STOP

Conclusion: the point group is D4d.

*

Dodecahedron, 19.6. Right-hand diagrams in 19.6 are related by rotation about an axis through the * points (using a molecular model or computer modelling package is beneficial).

START

* *

(19.6)

*

Is the molecule linear? Does it have Td, Oh or Ih symmetry? Is there a Cn axis?

No No Yes: C2 axis (vertical through the * points in 19.6)

Are there 2 C2 axes perpendicular to the principal axis? Yes (lower diagram in 19.6: (i) horizontal, (ii) through plane of paper) Is there a σh plane? No Are there n (i.e. 2) σv planes containing the Cn axis? Conclusion: the point group is D2d.

Yes

STOP

280

d-Block metal chemistry: general considerations

(Me3Si)3Si

19.10

(a) The static structure of Fe(CO)5 possesses 2 axial and 3 equatorial CO ligands, and therefore 2 axial and 3 equatorial C environments. (b) The molecule undergoes a low energy fluxional process; look at Fig. 4.5, p. 56; this should be included in answer 19.10. The process interconverts COax and COeq at a rate that is faster than the 13C NMR timescale; only one signal is observed in the spectrum.

19.11

The structures for this question are redrawn here in 19.7-19.9. Remember that these are solidstate data. The bond angles allow you to determine whether the metal is in a planar or non-planar environment: test for planarity is that the sum of bond angles is 360o. Compounds 19.7 and 19.8 have planar metal centres (but not strictly trigonal planar). In 19.9, Y is nonplanar (Σ angles = 345o); angles are equal, therefore the coordination environment is trigonal pyramidal. The deviation from planarity in 19.9 is probably caused by crystal packing forces and so may be a feature of only the solidstate structure. In 19.7, the steric crowding of the two bulky silyl groups causes ∠ Si–Fe–Si > ∠ Si–Fe–Cl. In 19.8, bite angle of chelating ligand is a constraint on the N–Cu–N angle.

111o

137o Fe Cl 112o (Me3Si)3Si

(19.7)

X See also Fig. 27.6 in H&S and the accompanying text

19.12

N N N

Cu N Cl

Ligand 19.10 is a tripodal ligand, i.e. three ‘arms’ with N-donors radiating from the central N-donor. The complex [Cu(19.10)Cl]+ is likely to be trigonal bipyramidal with the tetradentate 19.10 occupying four sites, and the Cl– ligand occupying one axial site as in 19.11.

(19.11)

Cl Cr

H2O

OH2 OH2

Cl

Cl Cr

H2O

Cl OH2

OH2

(19.13)

83o Cu 139

NCMe o

(19.8)

N(SiMe3)2 115o

Y

115o

115o

(Me3Si)2N

N(SiMe3)2

(19.9)

N

N N

N

(a) [Co(NH3)5Br][SO4] contains free [SO4]2– and coordinated Br–, whereas [Co(NH3)5(SO4)]Br contains free Br– and coordinated [SO4]2–. Aqueous solutions of BaCl2 and [Co(NH3)5Br][SO4] give BaSO4 precipitate; aqueous solutions of AgNO3 and [Co(NH3)5(SO4)]Br give AgBr precipitate. Only free ions are precipitated: Ba2+(aq) + [SO4]2–(aq) J BaSO4(s) Ag+(aq) + Br–(aq) J AgBr(s)

(19.12)

H2O

138o

(19.10)

19.13

H2O

+

(b) Distinguishing between [CrCl2(OH2)4]Cl.2H2O and [CrCl(OH2)5]Cl2.H2O needs quantitative study of precipitation of free Cl– by Ag+. [CrCl2(OH2)4]Cl.2H2O contains one equivalent of Cl–, but [CrCl(OH2)5]Cl2.H2O contains two. (c) [Co(NH 3 ) 5 Br][SO 4] and [Co(NH 3 ) 5 (SO 4 )]Br are ionization isomers; [CrCl2(OH2)4]Cl.2H2O and [CrCl(OH2)5]Cl2.H2O are hydration isomers (see Section 19.8 in H&S). (d) [CrCl2(OH2)4]+ is octahedral, therefore trans (19.12) and cis (19.13) isomers are possible.

d-Block metal chemistry: general considerations

(a) Coordination isomers are salts in which ligands are exchanged between the cation and anion. [Co(bpy)3]3+ contains neutral bpy (19.14) in which the N-donors can be only mutually cis. Coordination isomers of [Co(bpy)3]3+[Fe(CN)6]3– are: [Co(bpy)2(CN)2]+[Fe(bpy)(CN)4]– [Fe(bpy)2(CN)2]+[Co(bpy)(CN)4]– [Fe(bpy)3]3+[Co(CN)6]3– (b) trans- and cis-[Co(bpy)2(CN)2]+ in respect of the arrangement of the CN– ligands; cis-[Co(bpy)2(CN)2]+ posssesses enantiomers (19.15 and 19.16). trans- and cis-[Fe(bpy)2(CN)2]+; cis-[Fe(bpy)2(CN)2]+ possesses enantiomers. [Fe(bpy)3]3+ has enantiomers (look at Fig. 19.12 in H&S for related complex).

19.14

N

N

(19.14)

N

N N

Co

N

CN

NC

CN

NC

(19.15)

First, refer to Box 19.3 in H&S for the notation needed for this answer and for additional explanations. [Co(en)3]3+ contains 3 bidentate ligands and therefore has enantiomers labelled Δ-[Co(en)3]3+ and Λ-[Co(en)3]3+. Now consider the chelate ring conformations: using the notation from Box 19.3 in H&S, the Δ-[Co(en)3]3+ enantiomer can exist as (δδδ), (δδλ), (δλλ) or (λλλ), and similarly, Λ-[Co(en)3]3+ can adopt (δδδ), (δδλ), (δλλ) or (λλλ) configurations. All the isomers are related as diastereoisomers except in the cases where every chiral centre has changed configuration, e.g. Δ-(δδλ) and Λ-(λλδ).

Me

Me H2 N

Me H

N H2

H2 N

H

H2 N

H

H

Me H2 N

Ph H2 N

Ph

Pt2+ Me

Me

H2 N Pt2+

Me

N H2

N H2

N H2

N H2

Ph

(d)

(e)

H

(c) Me

Pt2+ N H2

N H2

N H2

Me

(b) H2 N

Me

H2 N Pt2+

N H2

N H2

H H2 N

N H2

H2 N Pt2+

(a)

H

(19.16)

19.16

Pt2+

Me

N

The chelate rings are 5-membered and for this answer, ignore the conformations of the chelate rings (see answer 19.16). (a) There are four isomers of [Pt(H2NCH2CHMeNH2)2]2+ depending on the orientations of the Me groups (Fig. 19.1a-d). (b) There are two isomers of [Pt(H2NCH2CMe2NH2)(H2NCH2CPh2NH2)]2+ depending on the relative positions of the Me and Ph groups (Fig. 19.1e-f).

H2 N

N H2

N

19.15

Me

H

Co N

N

H2 N

281

(f)

Fig. 19.1 (a)-(d) Isomers of [Pt(H2NCH2CHMeNH2)2]2+; (e)-(f) isomers of [Pt(H2NCH2CMe2NH2)(H2NCH2CPh2NH2)]2+.

Ph

d-Block metal chemistry: general considerations

282

+

O N

O

Co

O

N

N

+

O Co

N

N

O

N

N

N

(a)

(b)

Co

N

O

O

O

O

Cl

Cr OH2

(c)

(d)

PPh3

Cl

Pt

Cl

Cl

(f)

(g)

(h)

NH3

N

NH3

N

Co

O

O

Cr

H2O

O O

OH2

(e)

Ph2P

PPh3

+

NH3

O

OH2

OH2

Pt

–

–

O

PPh3 Cl

+

Cl N

Pt

Cr

O

PPh3 Cl

–

OH2

+

Cl

Cl

N

Cl

N

Co

PPh2

NH3 Cl

+

Cl H3N

Co

Cl

N N

Cl

NH3

NH3

NH3

(i)

(j)

(k)

(l)

Fig. 19.2 (a)-(b) Isomers of [Co(en)2(ox)]+, (c)-(e) isomers of [Cr(ox)2(OH2)2]–, (f)-(g) isomers of [PtCl2(PPh3)2], (h) the structure of [PtCl2(Ph2PCH2CH2PPh2)], and (i)-(l) isomers of [Co(en)(NH3)2Cl2]+.

19.17

NH2

O

O–

NH2

O

O–

(19.17)

(19.18)

19.18

X Pt 33.8% I = 1/2 103 Rh 100% I = 1/2 See Box 19.2 in H&S 195

19.19

N N

N

(19.19)

Ligands en and [ox]2– are drawn in 19.17 and 19.18; both are bidentate. (a) Octahedral [Co(en)2(ox)]+; 3 bidentate ligands; enantiomers (Fig. 19.2a,b). (b) Octahedral [Cr(ox)2(OH2)2]–; 2 monodentate and 2 bidentate ligands; cis- and trans-isomers; cis-isomer possesses enantiomers (Fig. 19.2c-e). (c) Pt(II), square planar; [PtCl2(PPh3)2] has trans- and cis-isomers (Fig. 19.2f,g). (d) Square planar with one bidentate ligand with a short chain length that forces a cis-arrangement (Fig. 19.2h). (e) Octahedral [Co(en)(NH3)2Cl2]+ may have trans-Cl, trans-NH3, or both sets mutually cis and for the latter, there are also enantiomers (Fig. 19.2i-l). (a) and (b) Pd(II) and Pt(II) form square planar complexes. Isomers of [PtCl2(PPh3)2] are shown in Fig. 19.2f-g. Symmetric and asymmetric stretches of PtCl2 unit are IR active for the cis-isomer; only the asymmetric stretch is IR active for trans-isomer (see Fig. 19.12 in H&S). Also applies to Pd(II) complex. For Pt(II) complex, 31P NMR spectra show satellite peaks from 195Pt-31P coupling, J(195Pt31P) cis > trans. (c) Use 31P NMR spectroscopy. fac-[RhCl3(PMe3)3] has one P environment, and gives one doublet due to J(103Rh31P). mer-Isomer has two P environments (label them P and P′); gives a doublet of triplets (rel. integral 1) and a doublet of doublets (rel. integral 2) due to J(103Rh31P) and J(31P31P′). Conformation change: rotation about inter-ring C–C bonds on going from free to coordinated tpy. All the complexes are octahedral; see Table 7.7 in H&S for ligands. (a) 6 monodentate ligands, therefore mer- and fac-isomers. (b) 2 monodentate and 2 bidentate ligands; cis- and trans-isomers; cis-isomer has enantiomers. (c) tpy (19.19) is tridentate and not very flexible; it is constrained to occupying coordination sites in a mer-arrangement; no isomers are possible.

d-Block metal chemistry: general considerations 19.20 H3C

CF3

283

(a) Ligand L– (19.20) is asymmetrical. The arrangements to give fac- and merisomers are shown below, using a schematic representation for the CF3 and CH3 ends of the ligands:

– O

O L–

Co

Co

fac

mer

(19.20)

(b) The presence of 3 chelating ligands leads to there being enantiomers:

Co

Co

Co

Co

mer

fac

(c) In the diagrams above, let the CF3 groups be represented by z. In the facisomer, each CF3 is in an identical environment, e.g. trans to a CH3. A is therefore the fac isomer because the 19F NMR spectrum shows a singlet. In the mer-isomer, each CF3 is in a different environment (look at the neighbours for each group) and this is consistent with the observation of 3 signals in the 19F NMR spectrum. B is the mer-isomer. 19.21

N N Ru

Cl

PPh2

Cl Ph2P

(19.21)

Compound X is [RuCl2(dppb)(phen)] and the fact that there is only one 31 P environment (one signal in the 31P NMR spectrum) shows that it is the trans isomer (19.21). On standing, the singlet is replaced by two doublets, each with the same coupling constant, i.e. the coupling arises from JPP, and there are two different 31P environments. Slow conversion of trans to cis-[RuCl 2 (dppb)(phen)] (19.22) is consistent with these data.

N N

Cl Ru Cl

PPh2 Ph2P

(19.22)

19.22

Pd(II) complexes are expected to be square planar, therefore the possibilities are cis- and trans-[PdBr2(NH3)2]. cis-[PdBr2(NH3)2] has two IR active Pd–N stretching modes observed at 480 and 460 cm–1. During isomerization to trans-[PdBr2(NH3)2], the band at 460 cm–1 disappears and that at 480 cm–1 shifts to 490 cm–1. This is consistent with trans-[PdBr2(NH3)2] possessing only one IR active mode arising from stretching of Pd–N bonds (the asymmetric stretch).

19.23

(a) The structure of [P3O10]5– is shown in 19.23. Bidentate coordination to a metal centre through O– on different P atoms is favoured (19.24 and 19.25). Coordination through 2 O-donors from one PO4 unit is possible but is less likely.

–

O–

O–

O–

P+

P+

P

O

O O–

+ O–

O O–

(19.23)

O–

284

d-Block metal chemistry: general considerations (b) The ligand is the conjugate base of a weak acid and a series of equilibria can be established for stepwise protonation or deprotonation of the terminal O– sites, ultimately to give H5P3O10. Species in solution are therefore dependent upon the pH, e.g. protonation shown in structures 19.24 and 19.25. O– HO

X See R.N. Bose et al. (2001) J. Chem. Educ., vol. 78, p. 83

NH3 H3N

P+

O

Co

H3N

O

+ P

O NH3 –

O

+ P

OH –

O

H3N

O

Co

H3N

O

P +

O

+ P

NH3

–

O

OH

(19.24)

19.24

O–

HO NH3

O

O– P+

O

O–

OH

(19.25)

(a) The ligands for the question are: O

O H2N

N

N

–O

F– Cl–

NH2

O–

H2O

N

Taking into account the oxidation state of each metal and charge of each ligand, the complexes can be formulated as: [Fe(bpy)3]2+ [Cr(ox)3]3– [CrF6]3– [Ni(en)3]2+ 2– [Mn(ox)2(OH2)2] [Zn(py)4]2+ + [CoCl2(en)2] (b) In [MnO4]–, Mn is in oxidation state +7, therefore a 100% ionic model for the bonding is unrealistic because the charges on the Mn and O atoms would be 7+ and 2– respectively; applying Pauling’s electroneutrality principle so that Mn has a resultant charge of +1: Charge on Mn + (4 × Charge on O) = overall charge on ion +1 + (4× Charge on O) = –1 Charge on O = –2/4 = –1/2 This charge distribution suggests that the bonding is largely covalent. 19.25

(a) cis-[CoCl2(en)2]+: chiral because 2 bidentate ligands and 2 cis-monodentate ligands (19.26). [Cr(ox)3]3–: chiral because 3 bidentate ligands. trans-[PtCl2(en) 2]2+: achiral Pt(IV) complex; octahedral with 2 didentate and 2 trans-monodentate ligands.

+

N N

Co

N

Cl

Cl

Cl

+

N Co

N

Cl

N

N Enantiomers

(19.26)

[Ni(phen)3]2+: chiral because 3 bidentate ligands. [RuBr4(phen)]–: achiral; 1 bidentate and 4 monodentate ligands.

N

d-Block metal chemistry: general considerations

285

cis-[RuCl(py)(phen)2]+: chiral because 2 bidentate ligands and 2 cis-monodentate ligands. (b) The isomers arise because of linkage isomerism:

Ph2 P

SCN

Ph2 P Pt

Pt P Ph2

Ph2 P

NCS

SCN

Both S-bonded

P Ph2

NCS Pt

NCS

Both N-bonded

P Ph2

SCN

One S-bonded and one N-bonded

Where thiocyanate ligands are equivalent, 31P NMR spectrum shows a singlet because PPh2 groups are equivalent. Where thiocyanate ligands are non-equivalent, spectrum has two doublets; PPh2 groups are non-equivalent, with 31P–31P coupling.

19.26

(a) The complex in the question is shown in 19.27. Square planar Pt(II), therefore chirality does not arise from coordination about metal centre. Each coordinated N atom is a chiral centre. Criterion for a molecule being chiral is that it possesses no plane of symmetry nor a centre of symmetry.

Me

H N

H

N

Pt

Cl

Cl

Me

(19.27)

(b) Both reactions involve complex formation: AgCl + 2NH3 J [Ag(NH3)2]Cl

Linear [Ag(NH3)2]+

Zn(OH)2 + 2KOH J K2[Zn(OH)4]

Tetrahedral [Zn(OH)4]2–

(c) [Cr(en) 3][Cr(ox) 3] and [Cr(en)(ox)2][Cr(en) 2(ox)] exhibit coordination isomerism: interchange of ligands between two metal centres. 19.27

(a) Td is consistent with a tetrahedral structure for [ZnCl4]2–; all bond lengths must be the same, and all angles must be 109.5o. If [AgCl3]2– has D3h symmetry, it must have a C3 principal axis with a σh plane perpendicular to this axis, i.e. [AgCl3]2– is trigonal planar. For [ZrF7]3–, possible structures are pentagonal bipyramid, monocapped octahedron or monocapped trigonal prism (Fig. 19.3); C2v point group corresponds to monocapped trigonal prism (C2 axis, two σv planes containing this axis).

Fig. 19.3 7-coordinate structures: (a) pentagonal bipyramid, (b) monocapped octahedron and (c) monocapped trigonal prism (C2v). In (c), the C2 axis passes through the two points marked *.

*

(a)

(b)

*

(c)

d-Block metal chemistry: general considerations

286

D3h [ReH9]2– must have a tricapped trigonal prismatic structure; the C3 axis runs through the triangular faces of the prism and σh plane perpendicular to this axis contains the 3 capping atoms. For a 4-coordinate MX4 structure, the choice is between tetrahedral (Td) and square planar (D4h); therefore [PtCl4]2– is square planar. If [AuCl2]– has D∞h symmetry, it must be linear; both bond lengths are equal. (b) A unit cell of CsCl is shown in Fig. 6.5b, p. 88. Each Cs+ is in a cubic coordination environment. In discrete 8-coordinate complexes, it is more usual to find dodecahedral, square antiprismatic and, less often, hexagonal bipyramidal coordination environments. The difference between the cube and square antiprism is a 45o twist of opposite square faces; this reduces steric interactions between ligands. 19.28

(a) Diastereoisomers are pairs of stereoisomers that are not enantiomers. DNA has a helical chain; right- and left-handed helices are labelled P and M. Combination of a P-helix with Δ- and Λ-complexes gives P,Δ- and P,Λ-species. Whereas Δ- and Λ-complexes are enantiomers, P,Δ- and P,Λ-species are diastereoisomers. (b) cis-[PtCl2(NH3)2] is achiral. The ligand PhMeCHNH2 contains a stereogenic centre (C atom) and cis-[PtCl2(PhMeCHNH2)2] has possible isomers (R,R), (R,S) and (S,S) depending on the stereochemistry at the stereogenic centres. (c) The first ligand has two stereogenic centres, but once coordinated, the N atoms also become stereogenic centres. Three possible stereoisomers are:

(S)

N (S)

(S)

Pt

H OH Cl

(R)

H2 N

Cl

(S)

H2 N

Pt

Cl Pt

Cl

N H2

N H2

(19.28)

(19.29)

Cl

N

(R)

(S)

H OH Cl

H Cl HO

(S)

Cl

NH2 Pt

Cl

Cl

(19.30)

(R)

(R)

H Cl HO

(R)

N (S)

(R)

Pt

H OH Cl

(R)

(R)

N

H Cl HO

Cl

Cl

N

NH2 Pt

H Cl

(S)

N

NH2 Pt

H Cl

(S)

(S)

N

NH2 Pt

H

H2 N

Pt

N

(R)

N

(R)

(R)

The left-hand pair, (S,S,S,S) and (R,R,R,R), are enantiomers. (S,S,S,S) and (R,S,R,R) are diastereoisomers, as are (R,R,R,R) and (R,S,R,R). All the pairs of enantiomers are: (R,R,R,R) and (S,S,S,S); (R,R,R,S) and (S,S,S,R); (R,S,S,R) and (S,R,R,S); (R,S,R,R) and (S,R,S,S). Other combinations of stereoisomers are diastereoisomers. (R,R,S,S) is a meso-compound, as is (R,S,R,S). The second ligand in the question has one stereogenic centre, and the complex has two stereoisomers 19.28 and 19.29 which are enantiomers. The third ligand contains one stereogenic centre, and once coordinated, the N atom also becomes stereogenic. The possible stereoisomers are:

(R)

(R)

N (R)

H Cl

NH2 Pt

Cl

Cl

(R,S) and (S,R) are enantiomers, as are (R,R) and (S,S). Any other pairs are diastereoisomers. The final ligand has two stereogenic centres that may be (S,S), (R,R) and (S,R). Structure 19.30 shows the (R,R)-complex. The complex has three stereoisomers: one pair of enantiomers, (R,R)/(S,S) and one meso-compound (R,S).

287

20 d-Block metal chemistry: coordination complexes

20.1

Energy

dz2

dx2–y2

eg

Δoct

dxy

dxz

dyz

t2g

(20.1)

H2N

20.2

Look at Table 19.2 in H&S. λmax is the wavelength of the absorption maximum; a value of λmax = 510 nm corresponds to absorption of green light and transmittance of red and violet, so solutions of [Ti(OH2)6]3+ look purple.

20.3

(a) en = 1,2-ethanediamine; N,N′-donor, 20.2. Usually bidentate; forms 5-membered chelate ring (chelate effect, see Section 6.12 in H&S). Occasionally monodentate. (b) bpy = 2,2′-bipyridine; N,N′-donor, 20.3. Bidentate; 5-membered chelate ring. (c) Cyanide, [CN]– (20.4); usually C-donor, monodentate; sometimes bridges in an M–C ≡ N–M mode (see examples in Chapter 21 of H&S). (d) Azide, [N3]–, (20.5); usually monodentate N-donor; sometimes bridges. (e) CO, (20.6); monodentate, C-donor (see Section 24.2 in H&S). (f) phen = 1,10-phenanthroline (20.7); N,N′-donor; bidentate forming 5-membered chelate ring. (g) [ox]2– = oxalate (20.8); O,O′-donor; bidentate forming 5-membered chelate ring. (h) [NCS]– (thiocyanate, 20.9) can be an N- or S-donor; usually monodentate but sometimes bridges in an M–N=C=S–M mode. (i) PMe3 (trimethylphosphine, 20.10) is usually a monodentate, P-donor, but see structure 24.18 in H&S and the accompanying discussion.

NH2

(20.2)

N

N

(20.3)

C

Refer to Section 20.3 in H&S. Points to include: • Gas phase metal ion Mn+ has degenerate nd atomic orbitals, and a valence configuration of nd x . • Consider formation of octahedral complex [ML6]n+. Crystal field theory treats metal ion and ligands as point charges; repulsions between electrons in Mn+ d orbitals and L donor electrons. • Ligands create a ‘crystal field’ around Mn+. In a spherical field, the energy of d orbitals is raised with respect to energy in gas phase Mn+. See left-hand side of Fig. 20.2 in H&S. • An octahedral crystal field leads to splitting of d orbitals into 2 sets (20.1): (i) higher energy dz2 and dx2–y2, and (ii) lower energy dxy, dxz and dyz. The dz2 and dx2–y2 orbitals point directly at the ligands while dxy, dxz and dyz orbitals point between the ligands. Therefore, repulsion between ligand electrons and electrons in dz2 and dx2–y2 is greater than between ligand electrons and electrons in dxy, dxz and dyz orbitals. See right-hand side of Fig. 20.2 in H&S. • The raising and lowering of energies is measured with respect to the energy level in the spherical crystal field, this is the barycentre (see diagram 20.14). • Include redrawn Figs. 20.2 and 20.3 from H&S in your answer.

N

(20.4)

O

O

P N

N

N

(20.5)

N

C

O

(20.6)

N

N

(20.7)

O

C

O

(20.8)

(20.9)

S

Me

Me Me

(20.10)

288

d-Block metal chemistry: coordination complexes 20.4

The order is as ligands appear in the spectrochemical series (Section 20.3 in H&S): Br– < F– < [OH]– < H2O < NH3 < [CN]– weak field

20.5

Factors to look for are different oxidation states of metal, different field strengths of ligands, or metals with same oxidation state and in the same triad. (a) [Cr(OH2)6]3+ should have larger Δoct than [Cr(OH2)6]2+ (+3 is higher ox. state). (b) [Cr(NH3)6]3+ should have larger Δoct than [CrF6]3– (both Cr(III), but NH3 is a stronger field ligand than F–). (c) [Fe(CN)6]3– is Fe(III), [Fe(CN)6]4– is Fe(II); [Fe(CN)6]3– will have larger Δoct. (d) [Ni(en)3]2+ should have larger Δoct than [Ni(OH2)6]2+ (en is stronger field ligand). (e) [MnF6]2– and [ReF6]2– both contain M(IV) with M from group 7; [ReF6]2– should have larger Δoct because Re is 3rd row metal, Mn is 1st row. (f) [Co(en)3]3+ and [Rh(en)3]3+ both contain M(III) with M from group 9; [Rh(en)3]3+ should have larger Δoct because Rh is 2nd row metal, Co is 1st row.

20.6

(a) Diagram 20.11 shows the d 8 configuration in an octahedral field. There is no vacant eg orbital and so no possibility of promoting an electron from a fully occupied t2g orbital to generate a high-spin configuration. ∴ Only one configuration. (b) Consider an example, e.g. octahedral d 4 , that can be low-spin (20.12) or highspin (20.13). Preference for high-spin or low-spin configuration depends on which configuration has the lower energy. This, in turn, depends on whether it is energetically preferable to pair the fourth electron (20.12) or promote it to the eg level (20.13). Need to consider the energy required to transform two electrons with parallel spins in different degenerate orbitals into spin-paired electrons in the same orbital – the pairing energy, P, depends on (i) the loss in the exchange energy on pairing the electrons (see Box 1.7 in H&S), and (ii) the coulombic repulsion between the spin-paired electrons. For high-spin: Δoct < P weak field For low-spin: Δoct > P strong field (c) Different numbers of unpaired electrons give rise to different effective magnetic moments (μeff), e.g. low-spin d 4 has 2 unpaired electrons, high-spin d 4 has 4. Use the spin-only formula to estimate the magnetic moment for n unpaired electrons:

eg t2g (20.11) eg t2g (20.12) eg t2g

20.7

eg +0.6Δoct –0.4Δoct (20.14)

μspin-only =

n(n + 2 )

In the case of an octahedral d 6 ion, low-spin is diamagnetic, high-spin is paramagnetic.

(20.13)

barycentre

strong field

t2g

For a given d n configuration, CFSE is difference in energy between d electrons in octahedral crystal field and d electrons in spherical crystal field. From diagram 20.14: CFSE = (–0.4Δoct)(number of electrons in t2g level) + (0.6Δoct)(number of electrons in t2g level) Examples of use of this equation: CFSE = (–0.4Δoct)(1) + 0 = –0.4Δoct d1 d 4 high-spin CFSE = (–0.4Δoct)(3) + (0.6Δoct)(1) = –0.6Δoct d 5 high-spin CFSE = (–0.4Δoct)(3) + (0.6Δoct)(2) = 0 But, for example, for low-spin d 4 , CFSE consists of two terms: the four electrons in the t2g orbital give rise to a –1.6Δoct term and a pairing energy, P, must be included to account for the spin-pairing of two electrons: CFSE = –1.6Δoct + P

d-Block metal chemistry: coordination complexes

289

Fig. 20.1 Rationalizing numbers of unpaired electrons in problem 20.8. For the ground state electronic configurations of the metal atoms, see Table 19.1, p. 278.

z

(b) [Mn(CN)6]2–

(c) [Cr(en)3]2+

(d) [Fe(ox)3]3–

octahedral Mn(II),

octahedral Mn(IV),

octahedral Cr(II),

octahedral Fe(III),

low-spin d 5

d3

high-spin d 4

high-spin d 5

(e) [Pd(CN)4]2–

(f) [CoCl4]2–

(g) [NiBr4]2–

d 8 , square planar Pd(II)

tetrahedral Co(II), d 7

tetrahedral Ni(II), d 8

20.8

See Fig. 20.1.

20.9

(a) Define an axis set; by convention, take the axial ligands to lie on the z axis (20.15 and 20.16). Trigonal bipyramid: the dz2 orbital points directly at 2 ligands and is destabilized the most (Fig. 20.2). The equatorial ligands lie in the xy plane, and the dx2–y2 and dxy orbitals are degenerate and higher in energy than the dxz and dyz orbitals which point between the ligands. Square-based pyramid: 1 ligand lies on the z axis, 2 lie ≈ along the x axis, 2 lie ≈ on the y axis; the dx2–y2 orbital (points ≈ at the basal ligands) is destabilized the most, the dz2 orbital is destabilized to a lesser extent (Fig. 20.2); since the basal ligands lie in the xy plane, the dxy orbital lies at higher energy than the dxz and dyz orbitals.

z

x y

(20.15)

(a) [Mn(CN)6]4–

(20.16)

dx2–y2

dz2

dz2

dxy dx2–y2 dxz

dyz

dxy dxz

dyz

trigonal

square-based

bipyramid

pyramid

Fig. 20.2 Crystal field splitting diagrams for trigonal bipyramidal and square-based pyramidal fields.

(b) Ni(II) is d 8 and both trigonal bipyramidal and square-based pyramidal complexes will be diamagnetic (place 8 electrons in the levels shown in Fig. 20.2). 20.10

(a) On going from gaseous Mn+ to complexed Mn+, interelectronic repulsion between metal d electrons decreases, i.e. pairing energies reduced. This is caused by an increase in effective size of metal orbitals: the nephelauxetic effect (‘cloud expanding’). For a common Mn+, nephelauxetic effect of ligands follows series: F– < H2O < NH3 < en < [ox]2– < [NCS]– < Cl– < [CN]– < Br– < I– For metal ions (with a common ligand) nephelauxetic effect follows series: Mn(II) < Ni(II) ≈ Co(II) < Mo(II) < Re(IV) < Fe(III) < Ir(III) < Co(III) < Mn(IV)

290

d-Block metal chemistry: coordination complexes Parameters for ligands (h) and metal ions (k) (see Table 20.10 in H&S) are used to estimate the reduction in electron-electron repulsion upon complex formation:

(

)

B0 − B ≈ hligands (k metal ion ) B0

where B is the Racah parameter, B0 is interelectronic repulsion in free ion. (b) From nephelauxetic series for ligands: F– < H2O < NH3 < en < [CN]– < I– 20.11

eg

t2g (20.19)

(a) [CoCl4]2– is Co(II), d7. [CuCl4]2– is Cu(II), d9. Tetrahedral Co2+ (20.17), t2 orbitals are all singly occupied; in tetrahedral Cu2+ t2 t2 (20.18), t 2 orbitals are asymmetrically filled. Thus, e e the complex suffers a JahnTeller distortion leading to (20.17) (20.18) the observed flattened tetrahedron. (b) Octahedral [CoF6]3– is Co(III), d6. F– is weak field ligand; ground state is highspin (20.19). A d6 ion might be expected to give a single absorption in the electronic spectrum (see diagram in Fig. 20.19 in H&S). The excited state of [CoF6]3– (t2g3eg3) suffers a Jahn-Teller effect because the eg level is asymmetrically filled. Small splitting of eg level leads to 2 possible transitions: 11500 and 14500 cm–1.

20.12

The electron configuration of Si is 1s22s22p63s23p2, but only the 3p2 (l = 1) configuration contributes to the term symbol. The working is as for carbon in Section 20.6 in H&S. There are 15 microstates. The table of microstates for a p2 configuration is given in Table 20.6 in H&S. The microstates are grouped according to values of ML and MS. Values of L and S are derived by looking for sets of ML and MS values: allowed values of ML : L, (L – 1), ..., 0, ... –(L – 1), –L S, (S – 1), ... –(S – 1), –S allowed values of MS : Term symbols are assigned as follows: • L = 2, S = 0 gives the singlet term, 1D; J can take values (L + S), (L + S – 1) ... L − S , so only J = 2 is possible; the term symbol is 1D2. • L = 1, S = 1 corresponds to a doublet term; possible values of J are 2, 1, 0 giving the terms 3P2, 3P1 and 3P0. • L = 0, S = 0 corresponds to a singlet term, and only J = 0 is possible; the term symbol is 1S0. The predicted energy ordering (from the rules above) is 3P0 < 3P1 < 3P2 < 1D2 < 1S0.

20.13

3F

and 3P are triplet terms. 1D, 1G and 1S are singlet terms. Electronic transitions obey the spin selection rule: ΔS = 0 i.e.transitions may occur from singlet to singlet, or triplet to triplet states, but a change in spin multiplicity is forbidden. The relative energies of the 3F, 3P, 1D, 1G and 1S terms are determined using Hund’s rules. The terms with the highest spin multiplicity are the 3F and 3P, and of these, the term with higher value of L has the lower energy. Therefore, 3F is the ground term. The other terms are singlets and their relative energies depend on the values

d-Block metal chemistry: coordination complexes

291

of L. The energy ordering of the terms for a d2 configuration is therefore 3F < 3P < < 1D < 1S. The most probable transitions originate from the ground state. The allowed transition (triplet to triplet in an absorption spectrum) is 3F ← 3P. Transition from ground to excited states are much more likely than from excited to excited states, and therefore you can ignore the latter on statistical grounds. 1G

20.14

d10 is a closed shell configuration and gives only a 1S term. For this term, L = 0, S = 0, and therefore J = 0. Using the following formula: S, P, D, F, G ... term Multiplicity of the term

(2S + 1)

LJ

J value

the ground (and only) term is therefore 1S0. Examples of d10 ions: Zn2+, Cu+. 20.15

Russell-Saunders coupling is a spin-orbit coupling scheme and it is not valid for all elements, in particular those with high atomic numbers. For heavier elements where spin-orbit coupling is particularly large, a more appropriate approach is to use a jjcoupling scheme. For detailed discussions, see: T.P. Softley (1994) Atomic Spectra, Oxford University Press, Oxford, p. 68; M. Gerloch (1986) Orbitals, Terms and States, Wiley, Chichester, p. 73.

20.16

The quantum number J takes values (L + S), (L + S – 1) ... |L – S|. For the 3F term, L = 3 and S = 1. Therefore, J = 4, 3, 2. Different values of J denote different levels within the term (diagram 20.20), and the energy differences between successive pairs of energy levels are 3λ and 4λ where λ is the spin-orbit coupling constant. The degeneracy of a J level is (2J + 1). For J = 4, degeneracy = 9; for J = 3, degeneracy = 7; for J = 2, degeneracy = 5. In a magnetic field, each J state splits into the (2J + 1) levels separated by gJμBB0 (gJ is the Landé g-factor, μB is the Bohr magneton, and B0 is the magnetic field). The splitting of the levels in a magnetic field (shown in diagram 20.21) is called the Zeeman electronic effect, and the small energy separations between these levels are the basis for EPR spectroscopy (see Chapter 4 in H&S).

Energy

3

F4

3

F 3

F3

3

F2

(20.20)

20.17 Term

Components in octahedral field

S P D F G

A1g T1g Eg + T2g A2g + T2g + T1g A1g + Eg + T2g + T1g

3

3

F

F4

3

F3

3

F2

LS coupling

(20.21)

The splitting of terms in an octahedral field is tabulated in the margin. (a) The 2D term will split to 2T2g and 2Eg. (b) The 3P term does not split but becomes the 3T1g. (c) The 3F term splits to become the 3T1g, 3T2g and 3A2g. See Fig. 20.20 in H&S.

effect of magnetic field

292

d-Block metal chemistry: coordination complexes Table 20.1 Table of microstates for a d 1 ion. –2 ↑

+1 ML

ml

↑ ↑

+2 +1 0 –1 –2

↑

–1

↑

0 +2

–2

↑

–1 ml

Table 20.2 Table of microstates for a d 2 ion.

ML

↑

↑

0

↑

↑

+1

↑

+2

↑

↑

↑

↑

↑ ↑

↑ ↑ ↑

↑

↑

↑

↑ ↑

+3 +2 +1 0 –1 –2 –3 +1 0 –1

D

2

3

F

3

P

20.18

The construction of tables of microstates is described in detail in Section 20.6 in H&S. Assume for this answer that you need only consider the weak field limit, i.e. terms of maximum spin multiplicity (see discussion in H&S, Section 20.6). Set up the table of microstates as in Table 20.1. The left-hand column in the top part of Table 20.1 gives ml values for a d orbital; the row in the lower part of Table 20.1 gives ML values for a d 1 ion. In a tetrahedral field, the 2D term has E and T2 components (see Orgel diagram in Fig. 20.19 in H&S). The multiplicity of the term (the 2 in the 2D symbol) is determined from 2S + 1; for 1 electron, S = 1/2. (b) Set up the table of microstates for the d 2 ion remembering that the electrons singly occupy orbitals (Table 20.2). The ML values are obtained by summing the ml values in each column. Your table may not look identical to Table 20.2 because the columns may come in a different order; the columns here are arranged to give the values of ML in an order from which the term symbols are easily found. The 3P term does not split (see answer 20.17) and gives a T1 component in a tetrahedral field, and a T1g component in an octahedral field. In an octahedral field, the components of the 3F term are A2g, T2g and T1g; in a tetrahedral field, they are A2, T2 and T1.

20.19

(a) Equations needed:

X Remember: lower wavenumber corresponds to longer wavelength X Visible range of spectrum and absorption/transmittance of light: see Table 19.2 in H&S

20.20

Wavenumber (in cm −1 ) =

1 Wavelength (in cm)

1 cm = 107 nm

10 000 cm–1 = 1000 nm; 20 000 cm–1 = 500 nm; 30 000 cm–1 = 333 nm. (b) Visible range ≈ 400-750 nm (25 000-14 285 cm–1). (c) Look at λ corresponding to the visible part of spectrum. [Ni(OH2)6]2+ absorbs ≈ 660 nm, so appears green; [Ni(NH3)6]2+ absorbs ≈ 570 nm, so appears purple. (d) Consider positions of NH3 and H2O in spectrochemical series: H2O is a weaker field ligand than NH3. The relative energies of transitions are estimated from an Orgel diagram (Fig. 20.20 in H&S) and for the weaker field ligand, the transition energies are lower, i.e. [Ni(OH2)6]2+ < [Ni(NH3)6]2+. Since E is proportional to wavenumber, the absorptions for [Ni(OH2)6]2+ appear at lower wavenumbers than those for [Ni(NH3)6]2+. (a) Cr(III) is d 3 . Sketch an Orgel diagram for an octahedral d 3 ion. This corresponds to the left-hand side of Fig. 20.20 in H&S. Three absorption transitions are expected: T 2g I A 2g T 1g (F) I A 2g T 1g (P) I A 2g

d-Block metal chemistry: coordination complexes

centre of symmetry F N

Co

N

N N

F

(20.22)

293

(b) trans-[Co(en)2F2]+ (20.22) has a centre of symmetry, but the cis-isomer does not; loss of centre of symmetry permits greater p-d mixing and, therefore, a greater probability of transitions, leading to more intense colour for the cis-isomer. Comparing chlorido and fluorido trans-complexes: charge transfer from Cl– to Co3+ (LMCT) accounts for the more intense colour of chlorido complex; LMCT for F– is unlikely. LMCT occurs when a ligand that is easily oxidized is bound to a metal centre (usually one in a high oxidation state) that is readily reduced.

20.21

(a) LMCT bands are at 282 nm for [OsCl6]3– and at 348 nm for [RuCl6]3–. The LMCT band moves to longer wavelength (smaller energy) because it is easier to reduce Ru(III) than Os(III). (b) The bpy ligand easily accepts an electron and therefore charge transfer will be in the direction M(II) to L, not L to M(II), i.e. an MLCT band rather than an LMCT band is observed.

20.22

[Ti(OH2)6]3+ is a d1 ion. Its electronic spectrum consists of two bands close together (observed as one band with a shoulder) because of a Jahn-Teller effect in the excited state, t2g0eg1. Single occupancy of the eg level lowers the degeneracy, but the energy separation between the t2g and eg levels is small. Two ‘d-d’ transitions from ground to excited state are therefore possible, but they are close in energy and give rise to absorptions at similar wavelengths. [Ti(OH2)6]2+ is a d2 ion. Three ‘d-d’ transitions are predicted (Fig. 20.20 in H&S): T 2g I T 1g (F)

A 2g I T 1g (F)

T 1g (P) I T 1g (F)

but only two, well separated bands are observed in the absorption spectrum (see answer 20.38b). 20.23

Figure 20.23b in H&S shows the 4F and 4P terms arising from a d3 configuration. There are three transitions and their energies, measured from the electronic spectrum, are given by: I 4 A 2g 4 1g (F) I A 2g 4 T (P) I 4 A 1g 2g 4T 4T

X See: A.B.P. Lever (1968) J. Chem. Educ., vol. 45, p. 711 20.24

(20.23)

2g

E = Δ oct E = 1.8Δ oct – x E = 1.2Δ oct + 15B + x

Limitations: (i) The figure refers only to the extreme weak field limit and this is a severe limitation of the method outlined here for determining the Racah parameter B. (ii) The method is easiest to apply if all three transitions are observed. If only two bands are observed (e.g. one absorption is hidden under an intense charge transfer band), it becomes more difficult to abstract a value for B from the observed data. (a) Co(II) is d 7, and the Orgel diagram for octahedral d 7 is the same as for octahedral d 2 (Fig. 20.20 in H&S). Absorption data in the question are reported in cm–1, and the smallest wavenumber corresponds to the lowest energy transition. H2O is a mid-field ligand; an assumption has to be made about the crossing of the lines in the Orgel diagram. [Co(OH2)6]2+ is high-spin (20.23) with 3 unpaired electrons: Multiplicity = (2 × 3/2) + 1 = 4

d-Block metal chemistry: coordination complexes

294

The assignments are: 8100 cm–1 16000 cm–1 19400 cm–1

I 4T1g(F) 4 1g(P) I T1g(F) 4A I 4T (F) 2g 1g 4T

4T

2g

(b) From Fig. 20.23a in H&S: 0.8Δoct + x = 8100 cm–1

(i)

1.8Δoct + x = 16000 cm–1

(ii)

Equation (ii) – (i) gives: Δoct = 16000 – 8100 = 7900 cm–1 This value does not agree with the value listed in Table 20.2 because the method is applicable only to a limiting case where field strength of the ligands is extremely weak. 20.25

Racah parameters provide information about electron–electron repulsions. On going from a gaseous metal ion to a metal complex, there is an effective expansion of ‘metal’ orbitals associated with metal–ligand bond formation. This results in a reduction in the electron–electron repulsions and, therefore, a reduction in the Racah parameter B. Parameters for ligands and metal ions (see answer 20.10a) may be used to estimate this reduction.

20.26

Use the spin-only formula to find the number of unpaired electrons, or fit values calculated from the spin-only formula to the experimental values (assume you can ignore contribution of the orbital angular momentum to the magnetic moment ):

Table 20.3 Calculated values of μ(spin-only) for n unpaired electrons. n 1 2 3 4 5

μ(spin-only) / μB 1.73 2.83 3.87 4.90 5.92

μ(spin-only) =

n ( n + 2)

From values in Table 20.3: (a) [VClx(bpy)] has 1 unpaired electron; d 1 corresponds to V(IV), therefore x = 4. (b) Kx[V(ox)3] has 2 unpaired electrons; d 2 corresponds to V(III); the complex anion is therefore [V(ox)3]3– and so x = 3. (c) [Mn(CN)6]x– has 3 unpaired electrons; d 3 corresponds to Mn(IV); since the cyano ligand is [CN]–, the overall charge must be 2–, i.e. x = 2.

20.27

For an electron to have orbital angular momentum, it must be possible to transform the orbital containing the electron into an equivalent and degenerate orbital by rotation. For a fuller explanation, refer to Section 20.10 in H&S, the subsection entitled ‘Spin and orbital contributions to the magnetic moment’.

20.28

(a) K3[TiF6] contains [TiF6]3–, therefore Ti(III), d 1 .

μ(spin-only) = n( n + 2) = 3 = 1.73 μB

d-Block metal chemistry: coordination complexes

295

(b) In a d 1 ion (t2g1), there is one electron in one of the dxy, dxy or dyz orbitals. These orbitals can be interconverted by rotations through 90o, and therefore the electron has orbital angular momentum. This results in an orbital contribution to the magnetic moment. 20.29

Octahedral Ni2+ (d 8 ) should have no orbital contribution, and μeff is expected to be close to spin-only value. Tetrahedral Ni2+ has an orbital contribution because ground state configuration is e4t24, and so spin-orbit coupling occurs. This results in μeff > μ(spin-only). In a square planar Ni2+ complex, all electrons are paired leading to a diamagnetic complex. Look at Fig. 20.1e: the promotion energy to the dz2 orbital (highest level) is too large for a high-spin complex to form.

20.30

Consider for which ions you expect there to be an orbital contribution to the magnetic moment. Orbital contributions to the magnetic moment are important only for the t2g1 (see answers 20.28b and 20.35c), t2g2, t2g4eg2 and t2g5eg2 configurations. (a) [Cr(NH3)6]3+ (b) [V(OH2)6]3+ (c) [CoF6]3–

octahedral d 3 (t2g3) octahedral d 2 (t2g2) octahedral, high-spin d 6 (t2g4eg2)

no orbital contribution orbital contribution orbital contribution

Therefore, only for (a) will the spin-only formula give a reasonable estimate of the magnetic moment. An octahedral d 3 ion has a 4A 2g ground term, and a better estimate of the magnetic moment can be obtained by using eq. 20.23 in H&S. 20.31

All the examples in the question are octahedral complexes. (a) [Co(OH2)6]3+ Co(III), low-spin d 6 , diamagnetic. 3– (b) [CoF6] Co(III), rare example of high-spin d 6 , paramagnetic. 2– Ni(IV), low-spin d 6 , diamagnetic. (c) [NiF6] 3– (d) [Fe(CN)6] Fe(III), strong-field ligand, low-spin d5, paramagnetic. 4– Fe(II), strong-field ligand, low-spin d6, diamagnetic. (e) [Fe(CN)6] (f) [Mn(OH2)6]2+ Mn(II), high-spin d 5 , paramagnetic.

20.32

(a) rion is estimated from ionic lattice; rion values for high-spin, octahedral ions are: Ti2+ d2 t2g2eg0

V2+ d3 t2g3eg0

Cr2+ d4 t2g3eg1

Mn2+ Fe2+ d5 d6 3 2 t2g eg t2g4eg2

Co2+ d7 t2g5eg2

Ni2+ d8 t2g6eg2

Cu2+ Zn2+ d9 d 10 6 3 t2g eg t2g6eg4

86

79

80

83

75

69

73

78

74pm

Plot of rion against number of d electrons shows an ‘inverse’ double-humped curve. Radii increase at points in series when electrons enter dz2 or dx2–y2 orbitals – these point directly at the ligands, and interelectronic repulsion increases. (b) The trend is the inverse of that for lattice energies (Fig. 20.35 in H&S). Lattice energy is inversely proportional to the internuclear separation and is therefore inversely related to rcation. The double hump is rationalized in terms of variation in LFSE (include a graph or table of LFSE vs d n electrons, Fig. 20.34 in H&S). Hydration enthalpies (Fig. 20.36 in H&S) behave similarly to lattice energies. 20.33

Normal spinel has tetrahedral Ni2+ (d8) and 2 octahedral Mn3+ (d 4 ); inverse spinel has tetrahedral Mn3+, octahedral Mn3+ and octahedral Ni2+. One octahedral Mn3+ is

296

d-Block metal chemistry: coordination complexes common to both types of spinel and can be neglected. Compare LFSE values: X Δtet ≈ 4/9Δoct

LFSE tet. Ni2+ + oct. Mn3+ = –(0.8 × 4/9 × 8500) – (0.6 × 21 000) = –15 622 cm–1 LFSE oct. Ni2+ + tet. Mn3+ = –(1.2 × 8500) – (0.4 × 4/9 × 21 000) = –13 933 cm–1

X Spinels AB2O4: see Box 13.7 and Section 20.11 in H&S

Therefore, one predicts a normal spinel; the factor not taken into account is the Jahn-Teller effect for Mn3+ (d4); although one predicts a normal spinel by a small margin, the structure is, in practice, an inverse spinel.

20.34

(a) LFSEs can be estimated as in answer 20.7 – see also Table 20.3 in H&S. The difference in LFSE on going from octahedral [Co(OH2)6]2+ to tetrahedral [CoCl4]2– is much less for Co2+ (d 7 ) than for Ni2+ (d 8 ). Remember that Δtet ≈ 4/9Δoct. (b) Data are consistent with H4[Fe(CN)6] being a weak acid with respect to the 4th acid dissociation constant; H+ complexing of [Fe(CN)6]4– makes reduction easier. (c) LFSE plays only a minor part. There is a loss of LFSE on reduction of Mn3+ (d 4 ), a gain on reduction of Fe3+ (d 5 ), and a loss on reduction of Cr3+ (d 3 ); the decisive factor is the large value of the 3rd ionization energy for Mn.

X Refer to Section 8.3 in H&S X Refer to Section 20.13 in H&S 20.35

(a) Information for this answer comes from Section 20.3 in H&S. Points to include in your answer: • relationship between x, y, z axes and positions of 6 L in octahedral [ML6]n+; • crystal field theory assumes point charges; • metal orbitals that point directly at ligands are dz2 and dx2–y2; • repulsions between electrons in dz2 and dx2–y2 orbitals and ligand electrons greater than between electrons in dxz, dxy and dyz orbitals and ligand electrons; • include a sketch of Fig. 20.2 in H&S; • energies of dz2 and dx2–y2 orbitals raised with respect to barycentre, and energies of dxz, dxy and dyz orbitals lowered. (b) First, note that each complex in the question contains Fe. To determine the ordering, consider the oxidation state of the Fe centre, and the field strength of the ligands: CN– is a much stronger field ligand than H2O; Δoct is largest for combination of strong field ligand and higher oxidation state, and smallest for weaker field ligand and lower oxidation state. Order is: [Fe(CN)6]3– > [Fe(CN)6]4– > [Fe(OH2)6]2+

t2 e (20.24)

20.36

(c) Tetrahedral d8 has configuration e4t24 (20.24). For an electron to have an orbital angular momentum contribution, it must be possible to transform the orbital it occupies into an equivalent, degenerate orbital by rotation. i.e. the three t2 orbitals can be interconverted by rotations through 90o. If all the t2 orbitals in the tetrahedral complex are either singly or doubly occupied (i.e. t23 or t26), an electron cannot be transferred from one of the t2 orbitals to another. In the tetrahedral d8 ion, there is one fully and two singly-occupied t2 orbitals and so there is an orbital contribution to the magnetic moment. Refer to Section 20.10 in H&S for greater detail. (a) The complexes in the question are all octahedral. Jahn-Teller distortions are observed for d9 and high-spin d4 configurations where there is asymmetric filling of the eg level: [CrI6]4– Cr(II) high-spin d4 Jahn-Teller distortion expected

d-Block metal chemistry: coordination complexes

297

Cr(II) low-spin d4 no Jahn-Teller distortion expected [Cr(CN)6]4– [CoF6]3– Co(III) high-spin d6 no Jahn-Teller distortion expected 3– [Mn(ox)3] Mn(III) high-spin d4 Jahn-Teller distortion expected (b) In [Et4N]2[NiBr4] and K2[PdBr4], complex ions are [NiBr4]2– and [PdBr4]2–: Ni(II) d8 [NiBr4]2– 2– [PdBr4] Pd(II) d8 Two possible geometries: tetrahedral and square planar. The splitting of d orbitals and electronic configuration depends on the arrangement of the ligands. Splitting diagrams and orbital occupancies for tetrahedral and square planar d8 configurations are shown in 20.25 and 20.26, respectively. If [NiBr4]2– is paramagnetic, then it must have a tetrahedral structure. Since [PdBr4]2– is diamagnetic, it is square planar. (c) Information for this answer is in Section 20.4 in H&S. Points to include: • valence orbitals of Ni are 3d, 4s and 4p; • each ligand provides one orbital (consider as outward pointing hybrid of NH3 that contains the lone pair); • assume Oh symmetry for the complex; • in Oh point group, metal s orbital has a1g symmetry, p orbitals are degenerate with t1u symmetry, d orbitals split into two sets, eg (dz2 and dx2–y2) and t2g (dxy, dyz, dxz); • in Oh point group, 6 LGOs for the L6-fragment have a1g, t1u and eg symmetries (see Figs. 5.27 and 20.12 in H&S); • by matching symmetries, form 6 bonding MOs (a1g, t1u and eg), 6 antibonding MOs (a1g*, t1u* and eg*) and 3 non-bonding MOs (t2g); • effectively, the d orbitals of Ni split into two levels, lower energy t2g and higher energy eg*.

(20.25)

(20.26)

20.37

2+

(a)

2+

N

H2 N

N NH2

Fe

N

H 2N

N

H2 N

Fe

N

N

H 2N

NH2

Enantiomers of complex with pyridine N donors fac, and NH2 groups fac 2+

H2 N

N Fe

N

(20.27)

2+

N

N NH2

H 2N

NH2

H 2N

Fe

H2 N N

N

Enantiomers of complex with pyridine N donors mer, and NH2 groups mer

(20.28)

(b) Complex contains Fe(II), d6; for octahedral complex, low-spin d6 is diamagnetic (20.27) and high-spin is paramagnetic (20.28). At 120 K, the compound undergoes a change from low to high-spin; specifying a single temperature rather than a range implies that the change is abrupt rather than gradual (see Fig. 20.30 in H&S).

d-Block metal chemistry: coordination complexes 20.38

Cl L

L L

Ni

L

Ni

Cl trans

Cl cis

(20.29)

(20.30) Cl Ni

L

Cl L

(20.31)

Fig. 20.3 Orgel diagram that includes a d2 ion in an octahedral field.

Cl

(a) Both [CoCl4]2– and [Co(OH2)6]2+ contain Co(II), d7; d–d transitions are spinallowed, but Laporte forbidden: Laporte selection rule: Δl = ± 1 Spin-allowed d–d transitions that are Laporte forbidden can be observed because of vibronic coupling. Molecular vibrations cause temporary loss of centre of symmetry in octahedral complex and at this moment, mixing of p and d orbitals can occur; thus, the so-called d–d transition actually involves an orbital with pd character and the transition is no longer Laporte forbidden. The probability of such a transition occurring remains low and the absorption is of low intensity (low value of εmax). The difference in εmax values between tetrahedral and octahedral complexes indicates that the intensity of absorption, and therefore the probability of transition, is greater for [CoCl4]2– than [Co(OH2)6]2+. In non-centrosymmetric [CoCl4]2–, p–d mixing occurs to a greater extent than in centrosymmetric octahedral complex. (b) [V(OH2)6]3+ contains V3+, i.e. d2. From the appropriate Orgel diagram (Fig. 20.3), three transitions are predicted from the 3T1g(F) ground state: 3T I 3T (F) lowest energy transition 2g 1g 3T (P) I 3T (F) 1g 1g 3A I 3T (F) 2g 1g The question states that there is no absorption assigned to the 3A2gI 3T1g(F) transition and this therefore suggests that this is the highest energy transition and that it is hidden under a higher energy charge transfer band (CT bands are intense). The bands at 17200 cm–1 and 25600 cm–1 can be assigned to 3T2g I 3T1g(F) and 3T (P) I 3T (F) transitions respectively. 1g 1g (c) Possible structures for [NiCl2(PPh2CH2Ph)2] are shown in 20.29-20.31 (L = PPh2CH2Ph). Complexes contain Ni(II), d8. See answer 20.36b; square planar complex is diamagnetic, tetrahedral complex is paramagnetic. The observed value

Energy

298

A2g or A2 T1g or T1 T1g or T1

P T1g or T1

T2g or T2

F T2g or T2

T1g or T1

A2g or A2 Δ 2

Δ

0 7

d , d tetrahedral d3, d8 octahedral

2

7

d , d octahedral d3, d8 tetrahedral

d-Block metal chemistry: coordination complexes

299

of μeff = 3.18 μB is consistent with 2 unpaired electrons (spin-only value is 2.83 μB). [NiCl2(PPh2CH2Ph)2] therefore changes from a red, square planar form to a blue-green, tetrahedral form. The magnetic moment measurements cannot distinguish between the two square planar isomers 20.29 and 20.30. On steric grounds, the trans-isomer should be favoured. 20.39

20.40

(a) For a representative Kotani plot, see Fig. 20.29 in H&S. Although this plot is for a t2g4 configuration, a similar plot can be drawn for a t2g1 configuration. Ions with a d1 configuration include Ti3+ and V4+ from the first row of the d-block, and Re6+, W5+ and Tc6+ from the second and third rows. Note that these positive charges are formal charges and such highly charged free ions are unlikely to exist; species with metals in these oxidation states have high degree of covalent bonding. The Kotani plot is of μeff against kT/λ; λ is small for first row metal ions, but is large for second row metal ions, and larger still for third row ions. (i) Ti3+, V4+ should possess room temp. values of μeff on near-horizontal part of Kotani plot; ions show little variation in value of μeff with temperature; (ii) Re6+, W5+, Tc6+ lie on steep part of curve and temperature has a significant effect on values of μeff for these ions. (b) F–, σ and π-donor; CO, σ-donor, π-acceptor; NH3, σ-donor. Orbitals available on ligands: F– inward pointing (radial) hybrid with lone pair; 2p (tangential) orbital with lone pair; CO inward pointing HOMO of CO, mainly C orbital character; LUMO, C–O π* MO; NH3 inward pointing, sp3 hybrid with lone pair. Diagrams that describe the overlap between metal orbitals and LGOs should be taken from Fig. 20.12 in H&S for the σ-bonding MOs, and from Fig. 20.14 in H&S for π-interactions (see also the diagrams below). Also include diagrams from Box 20.2 in H&S to describe how 3 LGOs arise from 6 ligands for the π-donor or acceptors.

pz

dxz

pz

F–

Mn+

F–

dxz

π* CO

M/Mn+

π* CO

(a) In metal complexes, intense absorptions can arise from transfer of electronic charge between ligand and metal orbitals. These can be either the transfer of an electron from an orbital with mainly ligand character to one with mainly metal character (ligand-to-metal charge transfer, LMCT), or transfer of an electron from an orbital with mainly metal character to one based on the ligand(s) (metal-toligand charge transfer, MLCT). Because CT transitions are not restricted by the selection rules that apply to ‘d–d’ transitions, the probability of them occurring is high, and absorption bands are intense. M-to-L electron transfer corresponds to metal oxidation and ligand reduction. Therefore, an MLCT transition occurs when a readily reducible ligand is coordinated to an M centre that is easily oxidized. LMCT occurs when an easily oxidized ligand is coordinated to a metal that is easily reduced.

d-Block metal chemistry: coordination complexes

300

Examples of LMCT: [MnO4]– , [CrO4]2–, [FeCl4]2–, [OsCl6]2–. These all contain high oxidation state metals that are readily reduced. Examples of MLCT: [Fe(bpy)3]2+, [Ru(bpy)3]2+, [Fe(phen)3]2+. These are M(II) complexes, and the metal d-orbitals are relatively close to the ligand π* orbitals giving rise to an MLCT absorption energy corresponding to the visible part of the spectrum. (b) Tanabe-Sugano diagrams are a means of representing the energies of terms, but differ from Orgel diagrams because the energy of the ground state is set to zero for all field strengths. The energies of all other terms and their components are then plotted with respect to the ground term. A discontinuity appears on a Tanabe-Sugano diagram if there is a change in ground term as the field strength increases. A TanabeSugano diagram is a plot of E/B against Δ/B, and it is therefore possible to use these diagrams to obtain a value of the Racah parameter B for a complex from the electronic transition energies. (Worked example 20.5 in H&S shows the method.) Usually, the small size of diagrams reproduced in texts only allows approximate values of B to be estimated. 20.41

N N N

N Cu

N

N N

N

(20.32)

20.42

(a) Deprotonation of the two NH groups in H2Pc gives a dianion that binds Cu2+ in a square planar environment (20.32). Four chelate rings are formed and the stability constant of [Cu(Pc)] is high. (b) [Cu(Pc)] is blue (or blue-green) because the absorption maximum is at 620 nm, meaning that it absorbs red light (see Table 19.2 in H&S). (c) π* I π means a transition from a lower energy π-orbital to higher energy π∗orbital. The ligand has a conjugated π-system and electronic transitions originate from high-lying occupied, ligand-based MOs in the complex to low-lying unoccupied, ligand-based MOs. (d) In [M(Pc)], the lower wavelength absorption band tails into the visible and this absorption (violet light with complementary observed colour of yellow, see Table 19.2 in H&S) adds to the absorption at 600 nm. The net observed colour is green. (e) A red shift of the lowest energy band means that the band at 620 nm shifts to longer wavelength. Assuming that the absorption remains in the visible region, the colour of the dye is green (see Table 19.2 in H&S). (f) Sulfonate groups are introduced to enhance solubility in H2O. (a) The conformational change is rotation about the C–C bond between the two pyridine rings so that the ligand can form a chelate ring when it binds a metal ion. (b) The sensitizer is designed so that it harvests the maximum amount of light. Hence absorption over the whole visible range is highly desirable. The value of εmax is a measure of the intensity of absorption. High values of εmax are therefore required. (c) The low lying π* MO is ligand-based and originates from the antibonding combinations of 2p orbitals of atoms in the bpy ligands. This is a metal-to-ligand charge transfer (MLCT) transition; the metal must be one that is readily oxidized, in this case Ru(II) to Ru(III).

301

21 d-Block metal chemistry: the first row metals

21.1

21.2 X See Section 19.2 in H&S

O

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

M

4s23d1

4s23d2

4s23d3

4s13d5

4s23d5

4s23d6

4s23d7

4s23d8

4s13d10 4s23d10

Zn

M 2+

3d 1

3d 2

3d 3

3d 4

3d 5

3d 6

3d 7

3d 8

3d 9

3d 10

See Table 19.3 in H&S: the range of oxidation states is greatest for metals in the middle of the row with the highest ox. state exhibited by Mn (+7). The +2 state is exhibited by all metals except Sc, and +3 for all metals except Zn. Sc and Zn each shows only one oxidation state in their compounds: Sc3+ is d0 (empty d shell) and Zn2+ is d10 (full d shell).

21.3

MeOCH2CH2OMe (21.1) is an O,O′-donor, giving a 5-membered chelate ring. [BH4]– can be mono-, bi- or tridentate (structures 13.7-13.9 in H&S). If the Ti3+ centre is 8-coordinate, the complex probably contains bidentate 21.1 and 3 bidentate [BH4]–.

21.4

(a) MgO crystallizes with an NaCl structure. Therefore, to form a continuous range of solid solutions with MgO, Li2TiO3 must also have an NaCl structure, i.e. it is [Li+]2Ti4+[O2–]3. Substitution is possible because Li+, Ti4+ and Mg2+ are about the same size, and because electrical neutrality can be maintained. (b) From eq. 21.11 and Section 8.2 in H&S:

O

(21.1)

X Solid solution: see Section 6.7 in H&S

[TiO]2+ + 2H+ + e– 2H+ + 2e–

Ti3+ + H2O

H2

Eo = +0.1 V at pH 0 Eo = 0 V at pH 0

From these values, no (or little) reaction at pH 0 occurs provided everything stays in solution. But TiO2 is extremely insoluble, and its insolubilty is increased by the presence of [OH]–. Further, the insolubility of H2 in H2O encourages reaction. 21.5

Ammonium vanadate in acidic solution contains [VO2]+. Amount of [VO2]+ = 25.00 × 0.1000 × 10–3 = 2.500 × 10–3 moles After reduction by SO2, vanadium is oxidized back to [VO2]+ by [MnO4]–. Amount of vanadium to be oxidized = 2.500 × 10–3 moles Amount of [MnO4]– = 25.00 × 0.0200 × 10–3 = 0.500 × 10–3 moles ∴ Ratio [MnO4]– : vanadium = 1:5 [MnO4]– undergoes a 5-electron reduction, and therefore vanadium undergoes a 1-electron oxidation. [MnO4]– + 8H+ + 5e– [VO2]+ + 2H+ + e–

Mn2+ + 4H2O [VO]2+ + H2O

Hence, reduction of [VO2]+ by SO2 gives [VO]2+. Reaction of [VO2]+ with Zn reduces it to Vn+ – the need to add this immediately to [VO2]+ arises from its instability in air. Oxidation of Vn+ to Vm+ takes place, and at

d-Block metal chemistry: the first row metals

302

X Alternative method is to notice that in 2nd titration, volume of KMnO4 is 3× that in the first titration, and so, in the 2nd titration, V undergoes a 3-electron reduction

the same time, some of the excess [VO2]+ is reduced to either [VO]2+ or V3+ which is then oxidized by [MnO4]–. The uncertainty about these vanadium oxidation states does not matter since the net change in oxidation states equals the change when [VO2]+ is reduced to Vn+. Amount of [VO2]+ initially = 2.500 × 10–3 moles = amount of Vn+ to be oxidized. Amount of [MnO4]– = 75.4 × 0.0200 × 10–3 = 1.49 × 10–3 moles ∴ Ratio [MnO4]– : vanadium = 3:5 [MnO4]– undergoes a 5-electron reduction, and therefore vanadium undergoes a 3-electron oxidation, and Vn+ is V2+ (i.e. [V(OH2)6]2+). [MnO4]– + 8H+ + 5e– [VO2]+ + 4H+ + 3e–

21.6 X See V(III) and V(IV) subsections in Section 21.6 in H&S 21.7

V2+ + 2H2O

Heating VBr3 causes disproportionation:

2VBr3 J VBr4 + VBr2

VBr4 formed decomposes on heating:

2VBr4 J 2VBr3 + Br2

If Br2 is removed, the final product will be VBr2. [NH4]V(SO4)2.12H2O is an alum. Use μeff = 2.8 μB to find the number of unpaired electrons and, hence, the oxidation state of V:

μ(spin-only) =

X But see answer 20.38b

Mn2+ + 4H2O

n ( n + 2)

A value of n = 2 gives μ(spin-only) = 2.83 μB. Ground state electronic configuration of V = 4s 23d3, therefore 2 unpaired electrons corresponds to V3+ (d2). The alum contains octahedral [V(OH2)6]3+. Sketch an Orgel diagram for an octahedral d 2 ion. This corresponds to the right-hand side of Fig. 20.3, p. 298. Three absorption transitions are expected: T 2g I T 1g (F)

21.8 N

N

N

The ligand L is a hexadentate N,N′,N′′,O,O′,O′′-donor. It is likely to form a 6-coordinate, octahedral complex with Cr3+, and the complex is therefore neutral [Cr(L)]. The three N atoms are in a macrocycle and, therefore, only the fac-isomer 21.2 is possible.

T 1g (P) I T 1g (F)

A 2g I T 1g (F) CO2– N

–O C 2

N

N

CO2–

L

(21.2)

21.9

Use Appendix 11 in H&S for data: Cr2+ + 2e– Cr Cr3+ + 3e– Cr Cr3+ + e– Cr2+ 2H+ + 2e– H2 O2 + 4H+ + 4e– 2H2O 2– + – [Cr2O7] + 14H + 6e 2Cr3+ + 7H2O

Eo = –0.91 V Eo = –0.74 V Eo = –0.41 V Eo = 0 V Eo = +1.23 V Eo = +1.33 V

d-Block metal chemistry: the first row metals

303

Aqueous HClO4 is kinetically very inert to reduction (see the discussion accompanying eqs. 17.77 and 17.78 in H&S). Oxidation of Cr by [ClO4]– can therefore be ignored. The purpose of HClO4 is to make the solution acidic. Cr is oxidized by H+, and it will be oxidized to Cr3+ rather than Cr2+. Even though Eo for oxidation to Cr2+ is larger than for oxidation of Cr to Cr3+, ΔGo per mole of Cr is more negative for oxidation to Cr3+. Look at worked example 8.4 in H&S, and include a similar calculation in your answer. In air, Cr3+ will not be further oxidized. 21.10

(a) The reaction occurring is: 2[MnO4]– + 16H+ + 5[C2O4]2– J 2Mn2+ + 8H2O + 10CO2 Follow the reaction by colorimetry (Beer-Lambert law, loss of absorbance from [MnO4]–) or gas evolution (production of CO2). (b) The reaction rate increases after Mn2+ begins to be present in solution. This is autocatalysis in which Mn2+ catalyses the reaction between [MnO4]– and [C2O4]2–.

21.11 O– P

+

N

Ph Ph

Ph

N

(21.3)

N

(21.4) Se Se–

Se Se–

(21.5)

N N

N

(21.12) CO2–

CO2– N N –O

CO2–

2C

(21.13)

Consider the modes of bonding within the complexes listed. Other binding modes may be possible in other complexes. OPPh3 (21.3) is O-bound, monodentate. [N3]– (azide, 21.4), monodentate. [Se4]2– (21.5) bidentate; donor atoms are cis, giving 5-membered chelate ring. [pc] 2– (21.6) has a non-flexible framework (related to a porphyrin); Mn2+ is coordinated approximately within the plane of the central four N-donors. OC(NHMe) 2 (21.7) is monodentate through O-donor. N(CH2CH2NMe2)3 (21.8) is tripodal and tetradentate, i.e. 3 ‘arms’ radiating from central N-donor; in a trigonal bipyramidal complex, 21.8 favours occupying axial and 3 equatorial sites as in 19.11, p. 280. THF (21.9) is monodentate, O-bound. Hpz (21.10) is monodentate, N-bound. bpy (21.11), bidentate, N atoms must be mutually cis. [NCS]– may be N- or S- bonded, but the notation [NCS-N]– shows it coordinates through N. HOCH 2 CH 2 OH, bidentate and the O atoms must be mutually cis. tpy (21.12), tridentate; N-donors are mer. [EDTA] 4– (21.13) hexadentate, but relatively flexible allowing octahedral coordination. 12-crown-4 is a tetradentate macrocycle but is conformationally restricted.

N N–

N

N

N –N

N N

(21.6) NHMe O

C NHMe

(21.7) NMe2

N Me2N

NMe2

(21.8)

O

HN N

(21.9)

(21.10)

N

N

(21.11)

304

d-Block metal chemistry: the first row metals 21.12

(a) Fe(II) and Fe(III) can be distinguished by using Mössbauer spectroscopy (refer to Section 4.9 in H&S for discussion of this technique). (b) For hard and soft metal ions, refer to Table 7.9 in H&S. Look also at the stability constant data in Table 7.8 in H&S. It can be shown that Fe3+ is a hard cation by observing that Fe3+(aq) changes colour at high [Cl–] and changes colour if Cl– is displaced on addition of F–. Both Cl– and F– are hard anions; hardness F– > Cl–. (c) Mn(V) disproportionates unless it is in strongly alkaline solution. Treat a known amount of the precipitate with acid to give MnO2 (as a precipitate) and [MnO4]– (in solution). Separate MnO2 and [MnO4]– by filtration, and determine the amount of each by titrating against oxalic acid in strongly acidic (H2SO4) solution. Results will allow determination of oxidation state changes and hence confirm Mn(V).

21.13

(a) and (b) See beginning of Section 21.9 in H&S for reactions of halogens with Fe; Cl2 oxidizes Fe to Fe(III), but I2 oxidizes it to Fe(II): 2Fe + 3Cl2 J 2FeCl3 Fe + I2 J FeI2 (c) Concentrated H2SO4 acts as oxidizing agent, Fe(II) J Fe(III): 2FeSO4 + 2H2SO4 J Fe2(SO4)3 + SO2 + 2H2O (d) [SCN]– acts as a ligand; Fe3+ is a hard metal ion and therefore expect that [SCN]– will coordinate through hard N-donor rather than soft S-donor:

O–

O

[Fe(OH2)6]3+ + [SCN]– J [Fe(OH2)5(SCN-N)]2+ + H2O

C

(e) K2C2O4 provides the oxalate ligand (21.14) which is bidentate with 2 hard Odonors; on standing, Fe3+ oxidizes oxalate:

C O–

O

(21.14)

[Fe(OH2)6]3+ + 3[C2O4]2– J [Fe(C2O4)3]3– + 6H2O 2Fe3+ + [C2O4]2– J 2Fe2+ + 2CO2 (f) FeO is basic and reacts with acids to give a salt and water: FeO + H2SO4 J FeSO4 + H2O (g) Addition of alkali to Fe(II) salts precipitates white hydroxide which is oxidized in air to give mixed Fe(II)Fe(III) and Fe(III) oxides. The reaction with alkali is: FeSO4 + 2NaOH J Fe(OH)2(s) + Na2SO4 21.14

(a) Figure 21.1 shows the trend in lattice energies for first row metal(II) chlorides. A similar plot applies to fluorides. The ‘double hump’ is attributed to crystal (or ligand) field stabilization energy contributions. For 0, 5 and 10 d electrons in highspin configurations, CFSE = 0, and interpolation from these points on the graph allows an estimate of (lattice energy – CFSE). The CFSE for FeF2 is therefore estimated from the difference between the lattice energy value determined from a

–Lattice energy / kJ mol–1

d-Block metal chemistry: the first row metals

305

Born-Haber cycle, and a value interpolated from the lattice energies of MnF2 (highspin d 5 ) and ZnF2 (d 10). (b) Construct a thermochemical cycle:

2800

2600

ΔGo2

Co3+(aq) + e–

Co2+(aq)

ΔGo1

2400

ΔGo3 ΔGo4

[Co(NH3)6]3+ + e–

2200 0

2

4

6

8

10

Number of d electrons

Fig. 21.1 Trend in lattice energies for first row MCl2.

[Co(NH3)6]2+

.

Values of ΔGo2 and ΔGo4 can be found from the Eo values in the question: ΔGo2 = –zFEo = – (1)(96 485)(+1.92) × 10–3 = –185 kJ mol–1 ΔGo4 = –zFEo = – (1)(96 485)(+0.11) × 10–3 = –11 kJ mol–1 Use the overall formation constant for [Co(NH3)6]2+ to find ΔGo3 : ΔGo3 = – RT ln K = – (8.314)(298)(ln 105) × 10–3 = –28.5 kJ mol–1 From the cycle above: ΔGo1 = ΔGo2 + ΔGo3 – ΔGo4 = –202.5 kJ mol–1 For the overall formation constant of [Co(NH3)6]3+ :

K

⎛ ⎞ ⎜ ΔG o ⎟ −⎜ ⎟ ⎜ RT ⎟ ⎠ =e ⎝

=

⎛ ⎞ −202.5 ⎟ −⎜⎜ −3 × 298 ⎟⎠ 8.314 × 10 ⎝ e

K ≈ 1035 21.15

Preferences between spinel structures: see end of Section 20.11 in H&S. The distribution of metal ions between tetrahedral and octahedral sites in a spinel can be rationalized in terms of LFSEs. In a normal spinel AIIBIII2O4, tetrahedral sites are occupied by the A2+ and octahedral sites by B3+. An inverse spinel has metal ions arranged (BIII)tet(AIIBIII)octO4. Now consider Co3O4: rewrite as CoIICoIII2O4. Co3+ is d6 and low-spin, therefore LFSE is greater if the Co3+ ions occupy octahedral rather than tetrahedral sites. This means that a normal spinel is favoured.

21.16

(a) The ions present are [Co(en)2Cl2]+ and [CoCl4]2–. Octahedral Co(III) complex cation is low-spin d6 (21.15), so diamagnetic. Tetrahedral Co(II) complex anion is d7 with configuration e4t23, so no orbital contribution is expected. For μ(spin-only) with 3 unpaired electrons:

eg

μ(spin-only) = n( n + 2) = 3.87 μB t2g (21.15)

This is only slightly larger than the observed value of 3.71 μB. (b) The value of μeff = 5.01 μB for [CoI4]2– is significantly greater than μ(spin-only) due to spin-orbit coupling. It is also greater than μeff for [CoCl4]2–, e.g. value in part (a). Although the following equation is only applicable to A and E ground terms: ⎛

μ eff = μ (spin - only)⎜⎜1 − ⎝

αλ ⎞

⎟ Δ oct ⎟⎠

(where λ is negative for a d 7 ion)

it shows that μeff is inversely related to ligand field strength, and Δoct for Cl– > I–.

306

d-Block metal chemistry: the first row metals 21.17

(a) The reaction to give the green precipitate is Ni2+(aq) + 2[CN]– J Ni(CN)2(s)

(forms as hydrated compound)

Dissolution to give the yellow and then red solutions is due to complex formation; note that the hexacyanido ion does not form: Ni(CN)2 + 2[CN]– J [Ni(CN)4]2– [Ni(CN)4]2– + [CN]– J [Ni(CN)5]3–

(yellow) (red)

(b) The yellow salt from part (a) is K2[Ni(CN)4]. Sodium in liquid NH3 is a reducing agent. Ni(II) may be reduced to Ni(I) or Ni(0) forming K4[Ni2(CN)6] or K4[Ni(CN)4]. 21.18 Ph Ph (or H) H 2N H H 2N

H (or Ph)

(21.16)

21.19

Aqueous NiCl2 contains [Ni(OH2)6]2+ which reacts with 21.16 to give a paramagnetic complex which contains H2O. The value of 3.30 μB (larger than spin-only value) corresponds to 2 unpaired electrons (d 8 ). The complex is probably trans[Ni(21.16)2(OH 2)2]Cl2. This loses H2O to give square planar, diamagnetic [Ni(21.16)2]2+. Isomerization cannot be square planar/tetrahedral since a tetrahedral Ni(II) complex would be paramagnetic. Now consider [Ni(21.16)2]2+ in detail. You can readily see that isomers can arise from the relative positions of the substituents. However, 21.16 contains two stereogenic centres, and therefore should be labelled (R,S)-, (R,R)- or (S,S)-21.16. For the complex, the possibilities are: • [Ni{(R,R)-21.16}{(S,S)-21.16}]2+ • [Ni{(R,R)-21.16}2]2+ and [Ni{(S,S)-12.16}2]2+ (pair of enantiomers) • [Ni{(R,R)-21.16}{(R,S)-21.16}]2+ and [Ni{(S,S)-21.16}{(S,R)-21.16}]2+ (enantiomers) • [Ni{(S,R)-21.16}2]2+ (cis and trans isomers) Another possibility for the product is cis-[Ni(21.16)Cl2] which has 3 stereoisomers: cis-[Ni{(R,S)-21.16}Cl2], [Ni{(R,R)-21.16}Cl2] and [Ni{(S,S)-21.16}Cl2] (a) Addition of alkali to Cu(II) salts precipitates blue copper(II) hydroxide: CuSO4 + 2NaOH J Cu(OH)2(s) + Na2SO4 (b) Reduction of Cu(II) in presence of Cl– leads to copper(I) chloride: CuO + Cu + 2HCl J 2CuCl + H2O (c) Concentrated HNO3 is an oxidant: Cu + 4HNO3(conc) J Cu(NO3)2 + 2H2O + 2NO2 (d) Complex formation with NH3 as the ligand, and dissolution of Cu(II) complex: Cu(OH)2 + 4NH3 J [Cu(NH3)4]2+ + 2[OH]– (e) Addition of alkali gives a precipitate of Zn(OH)2, but this dissolves in the presence of excess [OH]– as [Zn(OH)4]2– forms: ZnSO4 + 2NaOH J Zn(OH)2(s) + Na2SO4 Zn(OH)2(s) + 2NaOH J Na2[Zn(OH)4]

d-Block metal chemistry: the first row metals

307

(f) Action of acid on a metal sulfide: ZnS + 2HCl J H2S + ZnCl2 21.20 Me

O

N

Me

N

H

O H

(21.17)

21.21 X See Section 8.2 in H&S for the effects of pH on Eo values, and Section 8.3 in H&S for the effects of complex formation on Eo values

21.22

(21.18)

(21.19)

(a) Structure 21.17 shows H2dmg (dimethylglyoxime). Refer to Figs. 21.30 and 21.33b in H&S which show the solid state structures of [Ni(Hdmg)2] and [Cu(Hdmg)2]. In [Ni(Hdmg)2], there is strong hydrogen bonding between the two [Hdmg]– ligands and this helps to stabilize both the square planar coordination environment and planarity of the ligand. Molecules stack in the solid state with relatively close Ni...Ni contacts. In [Cu(Hdmg)2], there is also association between the ligands through hydrogen bonding, but in the solid state, dimers exist with each Cu(II) being 5-coordinate. (b) Ni and Pd are in group 8. Pd(II) usually has a strong preference for square planar coordination (large CFSE for d 8 ). Therefore one expects [Pd(Hdmg)2] to be structurally similar to [Ni(Hdmg)2], exhibiting stacks of planar molecules in the solid state. This is actually observed with close Pd...Pd contacts of 325 pm. SO2 is oxidized to sulfate and the half-equation is: [SO4]2– + 4H+ + 2e–

SO2 + 2H2O

Eo for this reduction is pH-dependent. Conc. HCl also provides Cl– which acts as a ligand to Cu2+ forming, for example, [CuCl4]2–. The reduction potential for Cu2+/ Cu of +0.34 V refers to aqueous Cu2+. Complex formation alters the Eo value. To test this reasoning, try the effect of replacing HCl by (i) saturated LiCl or any other very soluble chloride which will alter the concentration of Cl– present, and (ii) HClO4 or another very strong acid, the anion of which is a very poor complexing agent and difficult to reduce (see answer 21.9). Steric factors always favour tetrahedral over square planar coordination. In the absence of any steric control, electronic factors dictate the geometry. (a) Pd(II) is d8 and a square planar arrangement maximizes LFSE: Pd(II) and also Pt(II) complexes are usually square planar. (b) and (c) Cu(I) and Zn(II) are d10 and so LFSE = 0 for all crystal fields. Therefore, there is no electronic driving force for a particular structure and tetrahedral coordination is favoured over square planar on steric grounds. Ni(II) is d8. Consider the crystal field splitting of the d orbitals: 21.18 and 21.19 show that a square planar complex is diamagnetic, but a tetrahedral complex is paramagnetic.

21.23

In giving your answer, check that the oxidation states match those in the question. (a) [MnO4]– (also called permanganate ion) (b) [MnO4]2– (c) [Cr2O7]2– (d) [VO]2+ (e) [VO4]3– (ortho); [VO3]– (meta) (f) [Fe(CN)6]3–

21.24

Detailed information is given in Chapter 21 in H&S. Points to include: • Variation in oxidation states across row; thus, oxides formed by each metal. • Basic (e.g. CuO), amphoteric (e.g. V2O5) or acidic (e.g. CrO3) behaviour; reactions of oxides with water, alkalis and acids. • Oxidizing properties of CrO3 and MnO2. • Highest ox. state metal oxide is Mn2O7 (it is highly explosive). • Mixed oxidation state oxides, e.g. Fe3O4, Co3O4 (spinels). • Hydrated oxides, e.g. Fe2O3.H2O. • Example of non-stoichiometric metal(II) oxides (see Section 6.17 in H&S).

308

d-Block metal chemistry: the first row metals 21.25

Detailed information is given in Chapter 21 in H&S. Points to include: • Sc chemistry limited; hard Sc3+ forms discrete [ScF6]3– and [ScF7]4–. • Halido anions for oxidation states > +4 are very rare indeed (e.g. [VF6]–). • Octahedral anions common, e.g. [TiCl6]2–, [TiCl6]3–, [VF6]3–, [CoF6]3–, [NiF6]2–; fluorido ligands often associated with high ox. states. • Tetrahedral anions for the later metal ions, e.g. [FeCl4]2–, [FeBr4]2–, [CoCl4]2–, [NiCl4]2–. • Halido bridges lead to multinuclear anions, e.g. [V2Cl9]3– with octahedral coordination. • Halido bridges may lead to chains or 3D-structures in examples such as CsNiCl3, Na2MnF5, [Me2NH2][MnCl3], Cr(II) halido anions; octahedral metal sites often observed.

21.26

Start with analysis: X contains 11.4% Fe and 53.7% ox2–

X

ox2– : [C2O4]2–

3–

O O

Fe

3–

O Fe

O

This suggests a complex [Fe(ox)3]n– where n = 4 or 3. Since the reaction conditions are Fe(ox)2 + H2O2 (oxidizing agent) + H2ox (source of ox2–) + K2ox (source of ox2– and K+ cation), it is reasonable to suggest that Fe(II) is oxidized to Fe(III) giving K3[Fe(ox)3]. The remaining 34.9% (look at the analysis results) is K+ plus (most likely) water of crystallization: K3[Fe(ox)3].3H2O

O

O

11.4 53.7 : = 0.20 : 0.61 = 1 : 3 55.85 88.02

O O

O

Ratio Fe : ox2– =

11.4% Fe, 53.7% ox2–, 23.9% K, 11.0% H2O = 100%

Therefore, X is K3[Fe(ox)3].3H2O. Its reaction with alkali (see the discussion following eq. 21.64 in H&S) and photochemical decomposition are:

O

[Fe(ox)3]3– + 3[OH]– J 1/2Fe2O3.H2O + 3ox2– + H2O

O O

2K3[Fe(ox)3] J 2Fe[ox] + 3K2[ox] + 2CO2

(21.20)

[Fe(ox)3]3– is chiral (3 bidentate ligands, octahedral, 21.20), but the reaction with [OH]– suggests that [Fe(ox)3]3– may be too labile to be resolved into enantiomers. 21.27 O S Me

Me

(21.21)

DMSO (21.21) acts as a ligand; the [ClO4]– anion acts as a counterion. With Co2+, DMSO forms [Co(DMSO) 6 ] 2+ , and compound A is the 1:2 electrolyte [Co(DMSO)6][ClO4]2. With CoCl2, DMSO will give the same complex cation but the Cl– ions also act as ligands. B is the 1:1 electrolyte [Co(DMSO)6][CoCl4]: 2CoCl2 + 6DMSO J [Co(DMSO)6]2+[CoCl4]2–

21.28

H2S with aqueous Cu2+ precipitates CuS which has a very low solubility (Ksp = 6.0 × 10–37); it precipitates even in acidic medium: Cu2+ + H2S J CuS(s) + 2H+

X See Sections 8.2 and 8.3 in H&S for factors that influence Eo values

With hot conc. H2SO4, Cu is oxidized and S reduced. High [H+] and low solubility of CuS affect the Eo values to an extent that makes both of the following reductions possible: [SO4]2– + 4H+ + 2e– J SO2 + 2H2O [SO4]2– + 8H+ + 8e– J S2– + 4H2O (which gives CuS)

d-Block metal chemistry: the first row metals 21.29

309

(a) The information for this answer is in Box 21.5 in H&S: 2BaFeO4 + 3Zn J Fe2O3 + ZnO + 2BaZnO2 (b) Ligand to metal charge transfer (LMCT) involving a filled orbital on the ligand (formally O2–) and an empty orbital on the Mn centre. Mn is in a high oxidation state in both [MnO4]– (Mn(VII)) and [MnO4]2– (Mn(VI)). The absorption moves to lower energy as the oxidation state of the metal increases. (c) FeS2 (iron pyrites) is an Fe(II) compound formulated as Fe2+(S2)2– (it is not an Fe(IV) compound, Fe4+(S2–)2). Thus, the ions are in a 1:1 ratio and the compound can adopt an NaCl structure.

21.30

eg

t2g (21.22)

21.31

eg

t2g (21.23) eg

t2g (21.24)

(a) F– is a weak-field ligand and octahedral [CoF6]3– contains high-spin Co3+; electronic configuration is t2g4eg2 (21.22). The spin-only value of the magnetic moment is: μ(spin-only) = n( n + 2) = 4.90 μB The t2g4 configuration means (see answer 20.35c) that there is an orbital contribution to μeff and for a more than half-filled shell, μeff > μ(spin-only). (b) [(NC)5CoOOCo(CN)5]6– is a Co(III) complex with an [O2]2– bridging ligand. Assume oxidation occurs at the [O2]2– ligand; 1-electron oxidation removes an electron from the πg*(2px)2πg*(2py)2 level giving πg*(2px)2πg*(2py)1. The bond order increases (see answer 2.10). (c) A racemate is a 1:1 mixture of enantiomers, therefore you are looking for complexes that are chiral: [Ni(acac)3]– octahedral; 3 bidentate, chelating ligands, therefore chiral; [CoCl3(NCMe)3]– tetrahedral; achiral; cis-[Co(en)2Cl2]+ cis-octahedral with 2 chelating ligands, therefore chiral; trans-[Co(en)2Cl2]+ octahedral; 2 bidentate, chelating ligands with trans monodentate ligands; achiral. (a) [Ni(DMSO)6]2+ contains Ni2+ (d8) and is octahedral. From the Orgel diagram in Fig. 20.3, p. 298, an octahedral d8 complex is expected to show three absorptions in its electronic spectum. The left-hand side of Fig. 20.3 is relevant to this question and from Fig. 20.3, the three absorptions at 7 728, 12 970 and 24 038 cm–1 can be assigned as follows: 3T I 3A lowest energy transition, 7 728 cm–1 2g 2g 3T (F) I 3A 12 970 cm–1 1g 2g 3T (P) I 3A highest energy transition, 24 038 cm–1 1g 2g (b) CuF2 is Cu(II), d9; in the solid state structure (rutile type) of CuF2, each Cu(II) centre is octahedrally sited (see Fig. 6.5c, p. 88). Splitting of d-orbitals in an octahedral field gives electronic configuration 21.23 with an asymmetrically filled eg level. Lengthening of two (opposite) Cu–F bonds reduces the symmetry at the Cu(II) centre and removes the degeneracy of the eg orbitals (Jahn-Teller effect). Octahedral [CuF6]2– and [NiF6]3– are both low-spin d7 with electronic configuration 21.24; this has an asymmetrically filled eg level. Jahn-Teller distortion of the structure by elongation (2 long and 4 shorter bonds) or compression (2 short and 4 longer bonds) removes the degeneracy of the eg orbitals and stabilizes the system. (c) 'VBr3.6H2O' contains V(III), d2. Octahedral cation expected. Possibilities are: • [V(OH2)6]3+(Br–)3 cation has a centre of symmetry 2+ – . • [VBr(OH2)5] (Br )2 H2O cation does not have a centre of symmetry

d-Block metal chemistry: the first row metals

310

• [trans-VBr2(OH2)4]+Br–.2H2O cation has a centre of symmetry • [cis-VBr2(OH2)4]+Br–.2H2O cation does not have a centre of symmetry • [fac-VBr3(OH2)3].3H2O cation does not have a centre of symmetry • [mer-VBr3(OH2)3].3H2O cation does not have a centre of symmetry Assuming coordination of Br– occurs under the reaction conditions, the complex cation consistent with the structural data is [trans-VBr2(OH2)2]+. 21.32 NPh

PhN PhN

Ph N

V

V

N Ph

PhN

NPh

NPh

[V2L4] + KC8 THF

(21.25)

X See F.A. Cotton et al. (2003) J. Am. Chem. Soc., vol. 125, p. 2026 21.33 X See also answer 21.31b

– N S

V

S

S N

Structure 21.25 shows the complex [V2L4] in the question. As drawn, the V–V bond order is not specified (see below). (a) Each ligand is L–, therefore the complex formally contains a [V2]4+ core. V is in group 5 and has 5 valence electrons; number of electrons available for V–V bonding = 5 + 5 – 4 = 6. These electrons occupy σ and π-orbitals to give a configuration σ2π 4; formal bond order = 3. (b) KC8 is an intercalation compound, K+[C8]– and acts as a reducing agent: K(THF)3[V2L4] + 8C(graphite)

(c) During reduction of [V2L4] to [V2L4]–, the addition of an electron to the lowest lying unoccupied MO leads to the electronic configuration σ2π 4δ1. This increases the V–V bond order to 3.5, and the bond is expected to shorten. (a) [NiL2]2[S2O6]3.7H2O contains [NiL2]3+, i.e. d7 ion. [NiL2][NO3]Cl.H2O contains [NiL2]2+, i.e. d8 ion. Low-spin d7 complex will be Jahn-Teller distorted; elongation of octahedral structure to give 2 Ni–N bond lengths of 211 pm and 4 bond lengths in range 196-199 pm removes degeneracy of eg level. In the d8 complex, the bond lengths are all approximately the same (likely error of measurement within ± 1pm). (b) [Fe(bpy)3]2+ is Fe(II), d6; low-spin complex is diamagnetic. Partial oxidation of the sample or the presence of other Fe(III) impurities leads to the presence of paramagnetic d5 ions. Bulk sample therefore possesses a very low magnetic moment. (c) The species in equilibrium in the question are tautomers, related by transfer of proton between sites and associated rearrangement of multiple bonds. Structure 21.26 shows mer-[VL3]– in which the S-donors (or the N-donors) are mutually mer. Isomers of [V(Me2NCH2CH2NMe2)L2] are:

N

S

(21.26)

N

S

V

N

N

N

N

N

N

V

N

S

S

S-donors of L– trans; enantiomers N N

V

N S S

N N

S

V

S

S N N

N

N-donors of L– trans; enantiomers

N

V

S N S

N N

N

V

S

N N

N

N-donors (and S-donors) of L– cis; enantiomers

d-Block metal chemistry: the first row metals 21.34

311

(a) A square-based pyramidal structure with the oxido ligand in the apical site. Structure 21.27 shows one isomer (Me groups may be on the same side of the basal plane as shown, or on opposite sides). It is also possible that an H2O molecule coordinates to give a 6-coordinate complex, see part (c). O O O

V

O O

O

O

(21.27)

X See: T. Kiss et al. (2000) J. Inorg. Biochem. vol. 80, p. 65

21.35

H 2O

(a) A low pH value corresponds to a high [H+] and this allows [Cr(OH2)6]3+ to exist in aqueous solution. The presence of H+ prevents the dissociation of H+ from [Cr(OH2)6]3+ (Le Chatelier).

2+

O H2O

(b) [VO]2+ is an abbreviation for [V(O)(OH2)5]2+. (c) The speciation curves in Fig. 21.40 in H&S refer to the species in Fig. 21.2. Note the following points: • The structure shown for [VOL2] assumes sequential replacement of pairs of H2O ligands by L–. • Some structures have other possible isomers, e.g. in [VOL]+ defined in Fig. 2.12, L– could coordinate with an O-donor trans to the oxido ligand. (d) Diabetes is a disease in which levels of insulin in the human body are low, and this leads to disruption of sugar metabolism. The fact that vanadium(IV) complexes mimic insulin explains their potential role as anti-diabetic drugs.

V OH2

OH2

H2O

OH2

H2O

OH2

O

O O

O

O

V OH2

O

O

O

O

O

O

O

[VO(OH)L2]–

OH2

V OH2

O H O O H

V OH2

O O

[{VOL}2(OH)2] O HO HO

O OH2 H2O

V OH2

O

O

O

–

O

V

O

[VOL]+ O

O

O

O

V

[VO]2+

O

+

O

O H

V OH2

OH OH

[(VO)2(OH)5]–

Fig. 21.2 Structural interpretation of the abbreviated formulae given in Fig. 21.40 in H&S.

O

d-Block metal chemistry: the first row metals

312

OH2 H2O

Cr

H2O OH2

OH2

H O

Cr

O H

OH2

H O O H

5+

OH2 Cr

OH2

OH2

H2O

OH2

H2O

OH2

H O

Cr

O H

OH2

OH2 Cr OH2

H O O H

OH2 Cr OH2

H O

6+

OH2 OH2

Cr

O H

OH2 OH2

Fig. 21.3 Linear tri- and tetranuclear condensed cations of chromium(III).

(b) The linear tri- and tetranuclear species are shown in Fig. 21.3. (c) Collagen is a fibrous protein, i.e. a polypeptide. The amino acid residues in the protein which contain carboxylic acid functionalities are derived from glutamic acid and aspartic acid: H H2N

H

R

O Cr (OH2)5

H H 2N

Glutamic acid

CO2H CO2H

Aspartic acid

If carboxylate donors compete with hydroxido ligands, replacement in the bridging positions of the Cr-containing cations occurs, for example:

(OH2)5 Cr O

CO2H

H2N

CO2H

General amino acid structure

O

CO2H

O

OH2 H2O

Cr

H2O

X

OH2

Cr

X OH2

(21.28)

4+

OH2

X = OH– or RCO2–

OH2 OH2

When both bridging positions are substituted by collagen carboxylate groups, crosslinking of protein strands occurs (21.28). (d) The three Cr(VI) species are likely to be [Cr2O7]2–, [CrO4]2– and [HCrO4]–, equilibria being: [Cr2O7]2– + H2O

2[HCrO4]–

[Cr2O7]2– + H2O

2[CrO4]2– + 2H+

Reduction of Cr(VI) to Cr(III), and precipitation as Cr(OH)3 is a method of removal. 21.36

X See: X. Lou et al. (2008) Chem. Commun., p. 5848

(a) Deprotonation sites in zincon are circled; state of deprotonation depends on pH and hence the charge on the complex formed with M2+ varies. M2+ is bound in the O,N,N',O'-cavity. (b) Solubility in water is enhanced. (c) 463 nm is in visible region and arises from the N=N chromophore. (d) Both 463 and 600 nm absorption maxima are in the visible and the change in λmax gives a change in colour (see Table 19.1 in H&S). (e) [CN] – is a strong field ligand; ligand displacement assuming zincon is in the form of L3–: [CuL]– + 4[CN]– J [Cu(CN)4]2– + L3–

HO3S

N

N

OH

N

NH

CO2H

313

22 d-Block metal chemistry: the heavier metals

22.1

(a) The aim of this question is to give you an easy way to learn the positions of the metals in the periodic table: learn by building up each triad from the first row. Group:

3 4 5 6 7 8 Sc Ti V Cr Mn Fe Y Zr Nb Mo Tc Ru La Hf Ta W Re Os (b) Lanthanoid series comes between La and Hf.

Metallic radius / pm

22.2 200

3rd row 2nd row

180

1st row

160 140 120 3

6 9 12 Group number

Fig. 22.1 Trends in metallic radii of the d-block metals on crossing the periods.

22.3

9 Co Rh Ir

10 Ni Pd Pt

11 Cu Ag Au

12 Zn Cd Hg

Data for this answer are found in Tables 6.2, 21.1 and 22.1 in H&S. (a) The trends in metallic radii for 12-coordination are plotted in Fig. 22.1. Although not all the metals adopt close-packed structures at 298 K (see below), radii for different coordination numbers can be adjusted as shown in Section 6.5 in H&S. Figure 22.1 shows that, in general, values of metallic radii: • show little variation across a given row of the d-block; • are greater for second and third row metals than for first row metals: • are similar for the second and third row metals in a given triad. The last fact is due to the lanthanoid contraction (see Section 27.3 in H&S). (b) Comparisons between values of ΔaHo are valid if the metals are structurally similar. Close-packed lattices (hcp or ccp) are usual for d-block metals except for groups 5 and 6 (bcc) and group 12 metals. There is a rough correlation between the number of unpaired electrons and the value of ΔaHo, with the maximum values being for middle d-block metals. The trend in values of ΔaHo for the third row metals was shown in Fig. 6.3 (p. 86); this general trend is followed for each row. Down each triad, the trend is ΔaHo(third row) > ΔaHo(second row) > ΔaHo(first row); the differences are greatest for the middle group metals. (a) Use data from Appendices in H&S. The question gives a value for ΔfHo(CrCl2) and so this gives you a clue where to start. Assume CrCl2 and WCl2 (both group 6 metal(II) chlorides) have the same structure type. Calculate ΔlatticeHo(CrCl2) from a Born-Haber cycle: ΔlatticeHo = ΔfHo(CrCl2, s) – ΔaHo(Cr, s) – IE1(Cr, g) – IE2(Cr, g) – D(Cl2, g) – 2ΔEAH(Cl, g) = –397 – 397 – 652.9 – 1591 – 242 – 2(–349) = –2582 kJ mol–1 A value of ΔlatticeHo(WCl2) can be estimated by using the fact that ΔU (ΔU ≈ ΔlatticeHo) is inversely proportional to internuclear distance: Δ lattice H o ( WCl 2 ) o

Δ lattice H (CrCl 2 )

=

r (Cr 2 + ) + r (Cl − ) r ( W 2 + ) + r (Cl − )

314

d-Block metal chemistry: the heavier metals Since r(Cl–) >> r(Cr2+) or r(W2+), the difference between the radii of Cr2+ and W2+ will not affect the ratio of internuclear distances very much. Therefore:

⎛ r (Cr 2 + ) + r (Cl − ) ⎞ ⎟Δ Δ lattice H o ( WCl 2 ) ≈ ⎜ H o (CrCl 2 ) ≈ Δ lattice H o (CrCl 2 ) ⎜ r (W 2 + ) + r (Cl − ) ⎟ lattice ⎝ ⎠ ≈ 1 (or slightly larger than 1) The radius of W2+ will be slightly larger than that of Cr2+, so ΔlatticeHo(WCl2) will be a little less negative than ΔlatticeHo(CrCl2). A reasonable estimate is ΔlatticeHo(WCl2) ≈ –2450 to –2500 kJ mol–1. Now use these values in a Born-Haber cycle to estimate ΔfHo(WCl2, s). For ΔlatticeHo(WCl2) = –2450 kJ mol–1 : ΔfHo = ΔlatticeHo(WCl2, s) + ΔaHo(W, s) + IE1(W, g) + IE2(W, g) + D(Cl2, g) – 2ΔEAH(Cl, g) = –2450 + 850 + 758.8 + 1700 + 242 + 2(–349) = +403 ≈ +400 kJ mol–1

X See Section 22.7 in H&S for the structure of WCl2 22.4

ρ=

M X V

X For an alternative way of representing the structure of NbCl4 or NbBr4, see structure 22.11 in H&S

Similarly, a value of ΔlatticeHo(WCl2) = –2500 kJ mol–1 gives ΔfHo(WCl2, s) ≈ +350 kJ mol–1. (b) The high endothermic value of ΔfHo(WCl2, s) shows that an ionic compound is unlikely to form. The factors that contribute to the endothermic formation reaction (as opposed to the exothermic formation of CrCl2) are the higher ionization energies of W than Cr, and the higher enthalpy of atomization of W versus Cr. (a) Hf(IV) and Zr(IV) oxides are isostructural and have structures in which Zr and Hf are 7-coordinate. The lanthanoid contraction results in the Hf and Zr centres being approximately the same size, and the unit cell sizes for MO2 are the same. The big difference in densities (ρ(HfO2) is almost double ρ(ZrO2)) is because the atomic weight of Hf >> Zr (178.5 versus 91.2). There are 14 lanthanoid elements between La and Hf in the periodic table. F (b) Nb(IV) is d 1 , and in NbF4 (22.1) F F F Nb each Nb(IV) centre has an unpaired F F F F electron and the compound is Nb Nb F F F F paramagnetic. In NbX4 (22.2, X = Cl Nb F F or Br), the Nb centres (and therefore F the odd electrons) are paired up with F Nb–Nb interactions; the compound is (22.1) therefore diamagnetic. X X

X X X

X X

X

X Nb

Nb

X

X

Nb

Nb X

X

X X

X

(22.2)

22.5

(a) CsBr is ionic and provides Br–; NbBr5 acts as a bromide acceptor: CsBr + NbBr5 J Cs[NbBr6] (b) Molten KF contains K+ and F–; TaF5 acts as a fluoride acceptor. Under these conditions (and given that high coordination numbers are possible for the heavier,

d-Block metal chemistry: the heavier metals

315

early d-block metals, and F is not sterically demanding) the most likely reactions are: 2KF + TaF5 J K2[TaF7] or 3KF + TaF5 J K3[TaF8] rather than: KF + TaF5 J K[TaF6] (c) Several products are possible; a sensible prediction is the addition of bpy and retention of all 5 F atoms (F is small and high coordination number is possible). NbF5 + bpy J [Nb(bpy)F5] (d) MF5 (M = Nb, Ta) is tetrameric (22.3) and NbBr5 is a dimer (22.4). F F

F F

F M

F

F F

(22.5)

F M F

F

F

M

F

Br Nb

F

F

Br Br

M

F

F

Br

F

Br

Br Nb

Br

F

Br

Br Br

F

(22.3)

(22.4)

[NbBr6]–

Products: is a monocapped octahedron (ligand polyhedron is square antiprism (ligand polyhedron, 22.6); [Nb(bpy)F5] is 7-coordinate; bpy is a bidentate ligand with a fairly restricted bite angle, so a pentagonal bipyramid with bpy in the equatorial plane is plausible. is octahedral; [TaF7]2– shown in 22.5); [TaF8]3– is a

(22.6)

22.6

TaS2 is a Ta(IV) sulfide (S2–) whereas FeS2 contains Fe(II) and [S2]2–. TaS2 adopts a structure in which the ratio of M : S sites is 1 : 2 as in CdI2, whereas FeS2 adopts a structure in which the ratio of cations : anions is 1 : 1 as in NaCl.

22.7

Chlorido-bridges support the dinuclear framework in [Cr2Cl9]3– (22.7) and there is no Cr–Cr bonding. Each Cr(III) is octahedral d 3 , a n d h a s a magnetic moment arising from 3 unpaired electrons. The tendency for metal-metal bonding increases down the triad. In [W2Cl9]3– (which also has 3 chlorido bridges), the 6 electrons from the two W(III) centres pair up to give a W ≡ W bond and salts of [W2Cl9]3– are diamagnetic.

22.8

Cl Cl

(22.8)

Cl Cl

Cr Cl

Cl Cl

3–

Cl

Cl

Cl

Cr Cl

(22.7)

(a) The formula [Mo6Cl8]Cl2Cl4/2 means that the compound can be formally ‘broken down’ into the units: • [Mo6Cl8]4+ core (the 4+ charge balances the (2 + 4/2) Cl–); • 2 terminal Cl; • 4 Cl bridges (2 Cl ‘belonging to each Mo6 unit), each bridging between 2 Mo6 units giving sheet structure 22.8. The [Mo6Cl8]4+ core has 8 triangular faces, and the 8 Cl atoms are in μ3-modes. ∴[Mo6Cl8]Cl2Cl4/2 = [Mo6Cl8]Cl2+2 = Mo6Cl12 = MoCl2 (b) How many valence electrons (ve) are available? W = s 2 d 4 = 6ve, Cl = 1 ve Total ve = 36 (from 6W) + 8 (from 8Cl) – 4 (for 4+ charge) = 40 16 ve are used for 8 M–Cl (remember that this assumes nothing about the bonding modes of the Cl atoms), leaving 24 ve for 12 W–W bonds, i.e. bond order = 1.

316

d-Block metal chemistry: the heavier metals 22.9

Refer to Section 22.8 in H&S for more details. Points to include: • Oxohalides: tetrahedral TcOCl3 and ReOX3 (X = F, Cl, Br). • Large numbers of Tc(V) and Re(V) complexes with one oxido ligand; often square-based pyramidal with oxido group in apical site, e.g. [TcOCl4]–, [TcO(ox)2]–. • Examples of octahedral complexes: [ReOCl5]2–, trans-[TcO2(py)4]+; the trans-dioxido group appears in a range of complexes; protonation equilibria such as: trans-[ReO2(en)2]+

H+ –H+

trans-[ReO(OH)(en)2]2+

H+ –H+

trans-[Re(OH)2(en)2]3+

• ReOCl3(PPh3)2 is an important starting material in Re(V) chemistry; made from [ReO4]– and PPh3 in alcohol in presence of HCl. • Tc(V) oxido complexes are important in development of medical imaging agents. Refer to Sections 21.8 and 22.8 in H&S for greater detail. Points to include: • Mn occurs naturally (β-MnO2 is major ore). Tc is a man-made element and is a β-emitter. Chemistry of Mn is more widely studied than that of Tc. • Range of Mn oxidation states is 0 to +7, with +2 and +7 being of major importance; range for Tc also 0 to +7, but higher oxidation states (e.g. +5) more important for Tc than Mn; there is a wide range of Tc(V) complexes. • [TcO4]– more stable with respect to reduction than [MnO4]–; [TcO4]– is an important starting material in Tc chemistry. • Chemistry of Mn-containing cations exceeds that of Tc-containing cations. • Tc–Tc bond formation more important than Mn–Mn bond formation.

22.11

Diagram 22.9 shows the structure of [Re2Cl8]2–, with eclipsed ligands and unsupported (i.e. no μ-ligands) metal–metal bond as the important features. Take the Re atoms to lie on the z axis. Each Re atom uses four atomic orbitals (s, px, py, dx2–y2) to form Re–Cl bonds. Mixing of pz and dz2 orbitals gives two hybrids pointing along z axis; each Re atom uses four orbitals for Re–Re bonding: dxz, dyz, dxy and one pzdz2 hybrid (the other pzdz2 hybrid is non-bonding). The bonding can be considered in terms of the interactions shown in Fig. 22.2. Each Re(III) provides 4 electrons for Re–Re bond formation, giving a σ 2 π 4 δ 2 configuration, i.e. quadruple bond. The presence of the δ -component forces the two ReCl4-units to be eclipsed.

22.12

The shortest Re–Re distances (224 pm) are in [Re2Cl8]2– (quadruple bond, see above) and [Re2Cl4(μ-Ph2PCH2CH2PPh2)2]. This contains two Re(II) and has 10 electrons for Re–Re bonding and a σ 2 π 4 δ 2 δ * 2 configuration, therefore Re ≡ Re bond. The fact that the bond length is the same in [Re2Cl8]2– ({Re2}6+ core with quadruple bond) and [Re2Cl4(μ-Ph2PCH2CH2PPh2)2] ({Re2}4+ core with triple bond)

(22.9)

Fig. 22.2 Formation of an Re ≡ Re quadruple bond in [Re2Cl8]2–.

dxy

δ interaction

2 π interactions

dxz (also dyz)

Energy

22.10

σ* π* δ* δ π

σ interaction

pzdz2 Re

Re

z

σ

d-Block metal chemistry: the heavier metals

Cl

Cl

Cl

Cl

Cl

Re

Re Cl

Cl

Cl

n

(22.10)

22.13

L Cl

L Os

L

L N

3+

L Os

Cl

L L

L

(22.11)

Rh

22.15

(a) Section 22.10 in H&S contains details needed. Points to include: • Neutral Rh(IV) and Ir(IV) halides include only fluorides; RhF4 is made from RhBr3 or RhCl3 with BrF3; IrF4 made from IrF6 or IrF5 and Ir. • [MX6]2– ions are known for M = Rh, X = F, Cl and M = Ir, X = F, Cl, Br. • [RhCl6]3– is fluorinated and oxidized to [RhF6]2– by BrF3. • [RhCl6]2– is prepared from [RhCl6]3– and Cl2. • [IrF6]2– is made by treating [IrF6]– with water. • Na2[IrCl6] (useful synthon) is made from NaCl, Ir and Cl2; reaction of [IrCl6]2– with Br– converts it to [IrBr6]2–. (b) Octahedral [IrH3(PPh3)3] has fac- and mer-isomers (22.12 and 22.13). You can use 31P{1H} NMR spectroscopy to distinguish between them. The fac-isomer shows a singlet (one P environment). The mer-isomer has two P environments and shows a triplet and doublet (JPP). 1H NMR spectroscopy could also be used: hydride signals are diagnostic – the fac-isomer shows one resonance with JPH to one trans and two cis 31P, while the mer-isomer exhibits two resonances with both JHH and JPH. To work out these latter couplings, consider H(a) in 22.13: it couples to two H(b), to one trans-31P and to two cis-31P. (To what does H(b) couple?)

22.16

(a) [Ir(CN)6]3– has a regular octahedral structure. The point group is Oh. (b) Of the T1u, A1g and Eg modes of vibration, only the T1u stretching mode is associated with x, y or z in the right-hand column of the Oh character table (Appendix

PPh3 PPh3

fac

(22.12) Ha

bH

Rh

Ph3P

Inspection of the formula of the Os-containing species suggests (i) ‘N9H24’ replaces 3Cl–, and (ii) two Os units come together when A is formed. The reaction with HI indicates that there are two types of chloride, probably 3 non-coordinated Cl– and 2 coordinated Cl–. Now consider the reaction of A with KOH: since 1 equivalent of A contains 9 equivalents of N, the results show that all the N is liberated as NH3. The group ‘N9H24’ could contain 8NH3 + N3–. The nitrido ligand is likely to be the linking group between the two Os centres and 22.11 is a proposed structure of the cation: A is [22.11]Cl3. The vibrational spectra show that the cation has a centre of symmetry, and so the bridge is linear. This is consistent with bridging N3–; bonding is analogous to that described in Fig. 22.19 in H&S for a bridging O2– ligand; A contains Os(III), π-bond order = 0.5. Refer to Section 22.9 in H&S for greater detail. Points to include: • For oxidation states of ≥ +6, only RuF6 and OsF6 (octahedral monomer). • For M(V): RuF5 and OsF5 (both tetrameric), and OsCl5 (dimeric). • For M(IV): RuF4, OsF4, OsCl4 and OsBr4 (all polymeric). • For M(III), Ru halides outnumber Os halides: RuF3, RuCl3, RuBr3, RuI3, OsCl3, OsI3; +3 is lowest ox. state for which binary halides are well established. • Give syntheses and some properties of each compound mentioned.

H

Ph3P

may be an effect of the bridging ligands ‘clamping’ the metals together and opposing the lowering in bond order. Next in the bond length sequence is Re3Cl9 (249 pm); triangular Re3 unit with Re=Re double bonds. This has an {Re3}9+ core with 12 valence electrons for Re–Re bonding. The longest bonds (273 pm in ReCl4 and 270 pm in [Re2Cl9]–) are single bonds. ReCl4 is polymeric (22.10) and [Re2Cl9]– has an {Re2}8+ core and a triple bond might be suggested. But, the bond is long (given that there are 3 μ-Cl) and suggests a single bond, with 2 electrons per Re not used for Re–Re bonding.

22.14

H H

317

PPh3 Hb

PPh3

mer

(22.13)

d-Block metal chemistry: the heavier metals

318

3 in H&S). Therefore, only the T1u mode is IR active, and only one absorption is observed between 2200 and 2000 cm–1 in the IR spectrum. 22.17

X is a non-electrolyte, and so a neutral molecule. The IR spectroscopic data indicate that an Rh–H bond is present in X, the H originating from H3PO2. Use of D3PO2 produces an Rh–D bond, the vibrational wavenumber of which is lower than that of the Rh–H bond. MePh2As acts as a ligand. Therefore X is RhBr2H(AsMePh2)3. Reaction with Br2: RhBr2H(AsMePh2)3 + Br2 J RhBr3(AsMePh2)3 + HBr Treatment with mineral acid regenerates RhBr3 and free MePh2As.

22.18 Cl

Cl

Cl

Pd Cl

Pd Cl

Cl

Pd

Cl

Pd Pd Cl

Pd

Cl Cl

Cl

Cl

(22.14) Nb Nb

Nb Nb

Nb

Nb

(22.15)

(a) β-PdCl2 is a hexamer (22.14) with Pd atoms arranged in an octahedron but at non-bonded separations. The [Nb6Cl12]2+ unit is present in Nb6Cl14 (Fig. 22.6 in H&S) and contains an octahedron of Nb atoms in which there is Nb–Nb bonding (22.15); along each Nb–Nb edge there is a μ-Cl. The arrangement of Cl atoms is the same in both 22.14 and 22.15. (b) For details, see Section 22.11 in H&S. Points to include: • Compounds which appear to contain Pd(III) tend to be mixed valence species, e.g. ‘PdF3’ is Pd[PdF6] with Pd(II) and Pd(IV). • Examples of dinuclear complexes formally containing {Pd2}6+ core with Pd–Pd bond could be classed as Pd(III), but Pd–Pd bonding makes ox. state assignment ambiguous. (c) Ni2+, Pd2+ and Pt2+ are d8 and CFSE effects mean that a square planar complex is favoured on electronic grounds, especially for Pd2+ (2nd row metal) and Pt2+ (3rd row) for which field strengths are large. Ni2+ shows examples of both tetrahedral and square planar complexes depending on field strengths and steric effects of ligands (e.g. [Ni(CN)4]2– is square planar, [NiCl4]2– is tetrahedral). Steric effects always favour tetrahedral over square planar coordination sphere, but competing electronic effects (e.g. strong field [CN]–) can tip the balance in favour of square planar.

22.19

(a) The results of a single crystal X-ray diffraction study are definitive, but it is not always possible to grow suitable crystals. cis- and trans-[PtCl2(NH3)2] can be distinguished by their dipole moments (cis is polar, trans is non-polar) and by IR spectra. Because the trans-isomer is non-polar, only the asymmetric PtCl2 stretch is IR active, but for the cis-isomer, both symmetric and asymmetric stretches are IR active. [Pt(NH3)4][PtCl4] contains the ions [Pt(NH3)4]2+ and [PtCl4]2– and so is a 1:1 electrolyte. [PtCl2(NH3)2] is a non-electrolyte. (b) [(H 3 N) 2 Pt(μ-Cl) 2 Pt(NH 3 ) 2 ]Cl 2 would contain the ions [(H 3 N) 2 Pt(μCl)2Pt(NH3)2]2+ and 2Cl– and would be a 1:2 electrolyte. It does not contain terminal Pt–Cl bonds and, therefore, no associated IR absorptions are observed.

22.20

(a) Halide exchange reaction, with excess I– : K2[PtCl4] + 4KI J K2[PtI4] + 4KCl

[PtI4]2– is square planar (d8)

(b) NH3 substitutes for Cl–, but only for two Cl which must be in cis positions because of the trans effect (see Chapter 26); the product is square planar: [PtCl4]2– + 2NH3(aq) J cis-[PtCl2(NH3)2] + 2Cl–

d-Block metal chemistry: the heavier metals

319

(c) phen (22.16) is bidentate and binds cis; only two Cl– are displaced (see above) and the product is square planar: N

[PtCl4]2– + phen J cis-[PtCl2(phen)] + 2Cl–

N

(22.16)

(d) tpy (22.17) is tridentate. With the chelate effect in its favour, it displaces three Cl– from [PtCl4]2–; product cation is square planar: [PtCl4]2– + tpy J [PtCl(tpy)]+Cl– + 2Cl– Note that [PtCl(tpy)]Cl can also be made by reacting PtCl2 with tpy. (e) [CN]– (strong field ligand) displaces all four Cl–; product [Pt(CN)4]2– is square planar and, in the solid state, forms stacks with Pt–Pt interactions:

N N

N

(22.17)

K2[PtCl4] + 4[CN]– J K2[Pt(CN)4] + 4Cl– 22.21

Complexes of type [PtCl2(R2P(CH2)nPR2)] and [PdCl2(R2P(CH2)nPR2)] contain square planar M(II) centres. The coordination mode of the bidentate bisphosphine ligand depends on the length of the (CH2)n backbone. For a small chain, a chelating mode, as shown in 22.18 for n = 2, will force a cis arrangement of Cl– ligands. For trans-[PtCl2(R2P(CH2)nPR2)] to form, the (CH2)n (CH2)n backbone must be long, e.g. n = 12 would PR2 R2P give a long enough chain to ‘reach’ over between Cl Cl the two trans coordination sites. For intermediate Pt Pt chains, e.g. n = 7 or 9, it is likely that steric factors Cl Cl will hinder the formation of the cis-monomer, and PR2 R2P (CH2)n a dimer will form instead. Of the possible cis or (22.19) trans arrangements, the trans-form is favoured as shown in 22.19.

22.22

(a) [Pt(NH3)4][PtCl4] (Magnus’s green salt) forms chains of alternating [Pt(NH3)4]2+ and [PtCl4]2– ions (22.20) with a degree of Pt–Pt interaction that affects the electronic spectrum of [PtCl4]2–. In [Pt(EtNH2)4][PtCl4], the absorption spectrum shows the presence of [Pt(EtNH2)4]2+ and [PtCl4]2– ions, indicating that the discrete ions are present and the salt is structurally dissimilar from [Pt(NH3)4][PtCl4]. The EtNH2 ligand is significantly more sterically demanding than NH3, and the more bulky amine ligand forces the Pt atoms so far apart that interaction between them is impossible. (b) Both AgCl and AgI are sparingly soluble in water. The increased solubility of AgI in saturated AgNO3 must be due to complex formation. Ag+ is a soft metal centre, as is I–, while Cl– is hard (see Table 7.9 in H&S). In the presence of an excess of Ag+ in solution, AgI forms [Ag2I]+, thereby enhancing the solubility of AgI. An analogous complex for AgCl is not favoured and AgCl remains sparingly soluble in saturated AgNO3 solution. (c) If the equilibrium is: Hg2+ + Hg 2Hg+

R2 P

Cl Pt Cl

P R2

(22.18)

H3N

Pt

H3N Cl

NH3 Pt

Cl H3N

Cl Cl

Pt

H3N Cl

NH3

NH3 NH3

Pt

Cl

Cl Cl

(22.20)

the equilibrium constant is:

K=

[Hg + ]2

[Hg 2 + ] and the ratio of [Hg(I)]/[Hg(II)] depends on [Hg(II)]. Mercury(I) forms the dinuclear species [Hg2]2+. If the equilibrium is: Hg2+ + Hg [Hg2]2+

320

d-Block metal chemistry: the heavier metals

the equilibrium constant is:

K=

[Hg2 2+ ] [Hg2+ ]

and so the ratio [Hg(I)]/[Hg(II)] is always constant.

S

S

22.23

Refer to Sections 21.12 and 22.12 in H&S for detailed discussion. Points to include: • Cu(I) and Cu(II) are the important oxidation states for Cu; disproportionation of Cu(I) can be prevented by choice of suitable ligands. • A few complexes of Cu(III) and Cu(IV) are known, e.g. [CuF6]3–, [CuF6]2–. • Ag(I) dominates chemistry of Ag. • Au(I) and Au(III) important for Au, Au(III) being the more stable; this is rationalized in terms of relativistic effects (see Box 13.3 in H&S). • Compare Eo values for Ag+/Ag, Cu+/Cu, Cu2+/Cu+, Au+/Au couples. • High oxidation states: Au(V) and Ag(III).

22.24

Refer to Section 22.13 in H&S for detailed discussion. Points to include: • Group 12 metals are Zn, Cd, Hg; each metal has an (n+1)s2nd10 configuration. • Hg atypical of d-block because liquid at 298 K; amalgams (e.g. Na/Hg) important. • For Zn and Cd, only the +2 oxidation states (d10) are stable, but for Hg, Hg(I) is stable in the form of [Hg2]2+. • Zn and Cd are unlike metals in groups 4 to 11 because they do not exhibit variable oxidation states. • In triads of groups 4-11, 2nd and 3rd row metals similar as a consequence of lanthanoid contraction, but Cd and Hg are somewhat unlike each other, the effects of lanthanoid contraction being relatively unimportant. • Stability of halide complexes: F– complex most stable for hard Zn2+, but I– complexes most stable for soft Cd2+ and Hg2+

22.25

(a) Hg(II) is a class b, of soft, metal ion (a soft acid) forms the most stable complexes with soft ligands (soft bases) such as sulfur. (b) The 1H NMR spectrum shows that the four CH2CH2CH2 units in ligand 22.21 are equivalent and therefore a 4-coordinate (tetrahedral) Hg(II) centre is proposed. This assumes that the complex is non-fluxional and that there are no solvent molecules coordinated. Hg(II) is d10 and does not show a strong preference for a particular geomerty; it could be 4-, 5- or 6-coordinate. (c) The sketch of the 1H NMR spectrum should show a triplet of relative integral 2 (at δ 3.40 ppm, J = 6.0 Hz) and a quintet of relative integral 1 (at δ 2.46 ppm, J = 6.0 Hz). 16.6% of the α-protons couple to 199Hg giving rise to satellite peaks around the triplet.

α β

S

S

(22.21)

22.26

Refer to Chapters 19 and 22 in H&S for relevant examples/discussions. Outline answer: • M–M bonding: High values of ΔaHo (see answer 22.2) indicate tendency to form strong M–M bonds. M–M bond strength increases down a triad. 2nd and 3rd row metals form more compounds with M–M bonds than do 1st row metals; multiple bonds are also important (see answers 22.11 and 22.12). • High coordination numbers: 2nd and 3rd row metal ions larger than 1st row congeners; examples of high coordination number include complexes with macrocyclic ligands which occupy equatorial plane in, e.g. pentagonal bipyramid. • Metal halido clusters: focus on those with M–M bonding; show formation of 2D-layered structures, 3D-networks and discrete clusters; illustrate electron counting schemes to work out M–M bond orders (see answer 22.8).

d-Block metal chemistry: the heavier metals

321

• Polyoxometallates: within 2nd and 3rd rows, you should discuss Nb, Ta, Mo and W-containing species; of these, Mo and W species have the wider chemistries; show solution equilibria, structural families including Lindqvist, α-Keggin and α-Dawson anions. 22.27

F

Os

O

Ox. states:

–

F

(a) At 570 K:

F

2ReCl4 + 2PCl5 J [PCl4]+2[Re2Cl10]2– +4 +5 +5 +4

In this reaction, no redox changes occur. At lower temperatures and starting with ReCl5 and excess PCl5, [PCl4]3[ReCl6]2 and Cl2 form. Formation of Cl2 indicates a redox reaction must occur since oxidation state of Cl in starting materials is –1 and in Cl2 is 0. [PCl4]3[ReCl6]2 contains [PCl4]+ ions, and therefore the overall charge for the two [ReCl6]n– ions is –3. The product is [PCl4]+3[ReCl6]2–[ReCl6]– :

O O

(22.22)

2ReCl5 + 3PCl5 J [PCl4]+3[ReCl6]2–[ReCl6]– + 1/2Cl2 Ox. states: Re/P +5 +5 +5 +4 +5 Cl –1 –1 –1 –1 –1 0

J(19F–187Os)

Fig. 22.3 Simulated 19F NMR spectrum of fac-[OsO3F3]– (see text).

22.28

X See Fig. 19.8 in H&S

Reduction of one Re from +5 to +4 is balanced by oxidation of one Cl from –1 to 0. (b) The F environments in fac-[OsO3F3]– (22.22) are equivalent and give rise to one signal in the 19F NMR spectrum. 187Os is spin-active (I = 1/2) but occurs only to an extent of 1.64%. 98.36% of F atoms appear as a singlet = central line in Fig. 22.3. 1.64% of F couple to 187Os giving rise to a doublet that is superimposed on the dominant singlet; the outer lines in Fig. 22.3 are the doublet. Since the abundance of 187Os is so low, the relative intensity of the doublet is actually smaller than is shown in Fig. 22.2; intensity has been enhanced so the doublet is visible in the figure. The value of J(19F– 187Os) is measured from the doublet as shown in Fig. 22.3. (a) Oxidation states of Zr and Hf are the same, both +4. [NH4]3[ZrF7] contains 7coordinate [ZrF7]3– ion. In [NH4]3[HfF7], the Hf(IV) is octahedral, so this suggests that the “compound” is a mixture: [NH4]2[HfF6] + NH4F. Possible 7-coordinate structures are:

Monocapped octahedron Br Br

Br Br

Nb Cl

Br Nb

Br Br

Br Br

(22.23)

22.29

Monocapped trigonal prism

Pentagonal bipyramid

Pentagonal bipyramid is not usually adopted unless imposed by ligand, e.g. macrocyclic, pentadentate ligand may occupy equatorial sites. Examples of both the monocapped octahedron and monocapped trigonal prism are known, and distorted structures lying in between these extremes blur the distinction between them. (b) NbCl5 is Cl-bridged dimer with one 93Nb environment; similarly NbBr5 is dimeric (see structure 22.4). Halide exchange forms species such as 22.23. In 22.23, there are two 93Nb environments and the 93Nb NMR spectrum will contain two signals. (a) Figure 22.4 reproduces the structure of ReO3 given in Fig. 22.33 in H&S. This is a polyhedral representation of the structure showing that each Re atom is in an octahedral environment, surrounded by 6 O atoms and that O atoms are shared

322

d-Block metal chemistry: the heavier metals

Fig. 22.4 (a) Polyhedral representation of the solid-structure of ReO3; (b) a unit cell of ReO3 using a ‘ball-and-stick’ representation; Re, dark grey; O, pale grey.

(a)

(b)

between octahedra. The unit cell derived from this diagram is shown in Fig. 22.4b. Each Re atom is at a corner of a cube and each O atom lies on an edge; only 12 of the O atoms in Fig. 22.4a are shown in Fig. 22.4b. The octahedral environment of each Re atom can be seen by extending the structure. To confirm the stoichiometry from Fig. 22.4b: Site

Number of Re(VI)

Number of O2–

Central

0

0

Corner

8 × 1/8 = 1

0

Edge

0

12 × 1/4 = 3

Total

1

3

Therefore, stoichiometry Re:O = 1:3. (b) See eq. 22.36 in H&S for corresponding tungstate. The yellow precipitate is [NH4]3[PMo12O40]. 12[MoO4]2– + [HPO4]2– + 23H+ J [PMo12O40]3– + 12H2O 22.30

(a) Explanations for diamagnetism: [Os(CN)6]4– Os(II), d6, strong field ligand, low-spin octahedral (22.24). 2– Pt(II), d8, square planar (22.25). [PtCl4] OsO4 Os(VIII), d0. 2– trans-[OsO2F4] Os(VI), d2, distorted octahedral with short Os=O bonds; place these on z axis and energy level diagram is as in 22.26.

dx2–y2 eg

dxy t2g

3–

N C NCSe

Rh

NCSe

SeCN SeCN

C N

(22.27)

dx2–y2 dxz dyz

dz2 dxz, dyz

(22.24)

dz2

(22.25)

dxy (22.26)

(b) Nuclear spin data: 103Rh, I = 1/2, 100%; 13C, I = 1/2, 1.1%; 77Se, I = 1/2, 7.6%. The observed coupling patterns in 13C and 77Se NMR spectra arise from 13C–103Rh and 77Se–103Rh spin-spin coupling, respectively. Octahedral [Rh(SeCN-Se)6]3–, all Se equivalent; one signal in 77Se NMR spectrum, J(77Se–103Rh) = 44 Hz. 13C NMR spectrum shows one singlet because all 13C equivalent and two-bond Rh–C separation does not lead to observable J. After assigning the spectrum for this complex, it is possible to assign the 13C NMR

d-Block metal chemistry: the heavier metals

323

spectrum of [trans-Rh(CN)2(SeCN)4]3– (22.27), making the assumption that the singlet in the 13C NMR spectrum arises from SeCN and doublet from cyanido ligands i.e. J(13CCN–103Rh). All Se atoms are equivalent and there is therefore one signal which is a doublet, J(77Se–103Rh) = 36 Hz. 22.31

(a) The structure for the question is shown in 22.28.

N

Me

X See: J. Cámpora et al. (2003) Organometallics, vol. 22, p. 3345

H2 C

Me

N

BH

N

N Pd

N

Me

N

H2 C

Me

N

N

N

N

BH

N N

NO

(22.28)

N

N N

The complex is neutral. The oxidation state of +4 for Pd is partially balanced by the formal charges on the non-nitrosyl ligands (above) leading to the conclusion 2+ that the nitrosyl ligand is formally [NO]–. At the end of Section 20.4 in H&S, the coordination modes of neutral NO (a radical, left diagram below) were considered:

N

Mo

NCMe

Mo

NCMe NCMe NCMe

(22.29)

O C

Mo

O

Mo

O

Mo

C

O

O

M

N

O

M

Linear M–NO

N N represents the formamidine ligand

Mo

O N

(22.30)

X See: F.A. Cotton et al. (2001) Inorg. Chem., vol. 40, p. 478

N

O

M

N

Bent M–NO

When NO is treated as a neutral ligand, it acts as 3-electron donor when in a linear mode, and 1-electron donor when in a bent mode. If we formally consider the ligand in complex 22.28 to be [NO]–, then the bent mode confirmed by the structural data ( ∠ Pd–N–O = 118o) is consistent with [NO]– acting as 2-electron donor. The value of ν (NO) = 1650 cm–1 is consistent with a formal N=O double bond. (b) Starting material is shown in 22.29. MeCN ligands displaced by oxalate, but [C2O4]2– has the possibility of acting as a bridging ligand 22.30. Now consider the geometrical constraints of each cis-L2Mo2 unit where L– is the formamidine ligand; bridging units by oxalate ligands can lead to a “molecular square”. This is proposed for A; another way of describing the N N structure is as a [4+4] assembly, i.e. 4 N [Mo2] units and 4 [C2O4]2– bridges. N N Mo O O Mo N Assembly of such species sometimes C C N O O Mo N Mo gives mixed products, e.g. mixture of O O [3+3] and [4+4] assemblies; in this O O case, the [4+4] compound is more C likely because the geometry at the C [Mo2] units is a controlling factor; C C [3+3] and [4+4] assemblies require 60 O and 90o angles, respectively, subtended O O between the bridging units. Elemental O Mo N O O Mo N analysis cannot distinguish between C C [3+3] and [4+4] products. If parent ions N Mo O O N Mo N N are observed in the mass spectrum, ESIN N MS is the best way to distinguish between possible products.

d-Block metal chemistry: the heavier metals

324

P

22.32

(a) The melting points of W and Cu are 3695 and 1358 K, respectively; W can be used at higher temperatures than Cu. (b) Heating W in O2 causes oxidation. Use of noble gas prevents W reacting. (c) Evaporation of W slowly occurs in incandescent light bulbs, W is lost from the filament, and a thin film of W is deposited on the inside surface of the bulb. In the presence of Br2 or I2, vaporized W atoms react to give tungsten halides. In the high temperature zone around the filament, WXn decomposes to W and X2, and W is redeposited on the filament; this increases the lifetime of the bulb. F2 and Cl2 are not used because the decomposition temperatures of the halides are higher than bromide and iodide: bond strength W–F > W–Cl > W–Br > W–I.

22.33

(a) A helical chain is chiral. A right-handed helical chain recognizes another righthanded chain giving a right-handed (P) double helix; left-handed chains give a lefthanded (M) double helix (22.31). So, the double helix in Fig. 22.34 in H&S is chiral. (b) The ligand contains a stereogenic centre (22.32). (c) The (R)- and (S)-thiomalates each form an enantiomerically pure helix (the Pand M-helices, respectively, see R. Bau (1998) J. Am. Chem. Soc., vol. 120, p. 9380). This produces two distinct double helices (P and M) in the unit cell. (d) Au...Au contacts (ca. 300 pm) are important in solid state structures of Au(I) compounds. Au(I) has a closed shell d10 configuration and interactions between the centres are relatively strong (similar to a hydrogen bond) due to relativistic effects. (e) Gold(I) complexes contain labile ligands and are prodrugs; they are not themselves pharmacologically active, but undergo ligand exchange to produce an active drug in vivo. Soft cysteine residues in proteins displace soft S-donor ligands in the prodrug Myochrysine allowing uptake of Au(I). See Box 22.11 in H&S.

M

(22.31) CO2– Na+

*

H

HS

CO2– Na+

(22.32)

22.34

(a) Electro-, thermo- and photochromic materials exhibit colour changes triggered by application of a potential difference, heat or light, respectively. (b) Pure WO3 is transparent and becomes blue when a lithium tungsten bronze forms. The change is reversible. For full details, see Box 22.4 in H&S. (c) WO3 is a cathodic electrochromic material; darkening of the electrochromic film occurs at the cathode: WVIO3 + xLi+ + xe–

LixWVI1–xWVxO3

IrO2 differs because (i) it is an anodic electrochromic material, and (ii) reversible colour change depends on H+ migration. 22.35 Cl N

Cl

Ru

O

N O

(22.34) Cl N

Ru

N

Cl Cl

N

S

N

Ru

Cl O

O

S

(22.35)

(22.36)

(a) Ru oxidation states: +3 in [RuCl4(im)(DMSO)]–, +3 in [RuCl4(Ind)2], +2 in [RuCl2(DMSO)2(Biim)], and +3 in [RuCl3(DMSO)(Biim)]. S (b) Isomers arise from: O • cis/trans, or mer/fac arrangements • DMSO (22.33) can coordinate through O or S atom (22.33) • enantiomers. Stereoisomers: 2 for [RuCl4(Ind)2] (cis and trans N); 4 for [RuCl4(im)(DMSO)]– (cis and trans N and S donors, or cis and trans N and O donors); 6 for [RuCl3(DMSO)(Biim)] (mer Cl with O- or S-DMSO; fac-Cl with O- or S-DMSO plus enantiomers); 14 for [RuCl2(DMSO)2(Biim)] (trans-Cl with O/O or S/S or S/ O-DMSO ligands, cis-Cl with 2 O-DMSO in arrangement 22.34 + enantiomers, or similarly with 2 S-DMSO, cis-Cl with S/O-DMSO in arrangement 22.35 or 22.36 + enantiomers; trans O/O or S/S or S/O-DMSO ligands.

325

23 Organometallic compounds of sand p-block elements

23.1

23.2

(23.1)

Et2O

(a)

MeBr + 2Li

(b)

Na + (C6H5)2

(c)

nBuLi

(d)

Na + C5H6 J Na+[C5H5]– + 1/2H2

THF

(lithiation; ΔfHo of LiBr drives reaction) (Na reduction of biphenyl)

MeLi + LiBr Na+[(C6H5)2]–

+ H2O J nBuH + LiOH

(hydrolysis of a lithium alkyl gives corresponding alkane) (route to prepare synthetically important Na[Cp])

In the tetramer (MeLi)4 (23.1), each C atom appears to be 6-coordinate. A localized bonding scheme within the C4Li4 unit is not appropriate. Consider each C atom as sp3 hybridized; allocate localized 2c-2e C–H bonds. Each Me group provides one sp3 hybrid and one electron to the bonding in the C4Li4 unit. Each Li atom has one valence 2s electron, but can make use of empty 2p atomic orbitals to form sp3 hybrids, three of which are used to overlap with C sp3 hybrids. Total number of valence electrons available = 4 (from Li) + 4 (from C) = 8. In the diagrams below, first note the positions of the C and Li atoms at the corners of an approximate cube; the Li atoms define a tetrahedron, and so do the C atoms. Each C atom caps one Li3 face. Each C atom provides one sp3 hybrid pointing into an Li3 face. Each Li atom provides three sp3 hybrids, each pointing into a different Li3 face. The right-hand diagram below shows the result for one Li3 face. One of three Li

H 3C

CH3

C

Li Li

C H3 C H3

Li

Li

Li Li C

C

sp3 hybrids on Li

Li

C

sp3 hybridized C

Overlap of the hybrid orbitals gives a 4c-2e interaction over each Li3 face. This model allows for a degree of Li–Li bonding as well as Li–C interactions. 23.3 H3C

Be

CH3

(23.2)

H3 C Be

Be C H3

(23.3)

H3 C C H3 n

Gas phase: Me2Be consists of monomeric, linear molecules (23.2). Bonding described in terms of sp hybridized Be with 2 valence electrons, and sp3 hybridized C. Three localized 2c–2e C–H bonds in each Me group, leaving one sp3 hybrid and one valence electron for localized σ-bond formation with Be. This is analogous to descriptions of the bonding in monomeric BeH2 and BeCl2. Solid state: polymeric chain structure (23.3). The bonding can be described in the same way as in polymeric BeH2 (see answer 10.15) using a C sp3 hybrid instead of an H 1s atomic orbital. In polymeric BeCl2, lone pairs of electrons from Cl supply enough electrons so that all Be–Cl interactions are localized 2c-2e (see answer 12.9b).

Organometallic compounds of s- and p-block elements

326

23.4

(a)

(b) or

Mg + 2C5H6 J (η5-C5H5)2Mg + H2 (i.e. Cp2Mg)

(reduction of cyclopentadiene and oxidation of Mg)

MgCl2 + LiR J RMgCl + LiCl

(preparation of Grignard reagent, and magnesium dialkyl)

MgCl2 + 2LiR J R2Mg + 2LiCl (c)

23.5

RBeCl

LiAlH4

RBeH

The coordination number of Mg is likely to be 4; shape: tetrahedral. THF is a monodentate ligand, and O-donor. The part of the formula ‘(Me3Si)2C(MgBr)2’ shows that C is bonded to 2 MgBr units. This leaves Mg with 2 vacant coordination sites. ∴ n = 4, see structure 23.4.

(reduction using [AlH4]– )

O Mg Me3Si Me3Si

O Br

C

Br Mg

O

O

(23.4)

23.6 H3 C H 3C H 3C

Al

Al

CH3 CH3

C H3

(23.5) Cl Cl Cl

Al

Al Cl

Cl Cl

(a) For the equilibrium:

Al2R6

2AlR3

the smaller value of K corresponds to the smaller steric demands of R. For small Me group, a dimer is strongly favoured. For bulky Me2CHCH2 group, the dimer is less favoured. (b) The terminal Me groups in Al2Me6 (23.5) involve localized 2c-2e Al–C bonds. Each Al is sp3 hybridized with 3 valence electrons; after terminal Al–C bond formation, one electron is left for bridge bonding. Each Al–C–Al bridge is a 3c-2e interaction. In Al2Cl6, each bridging Cl provides three valence electrons to give localized 2c-2e interactions (23.6). In Al2Me4(μ-Cl)2, all bonds are localized 2c2e.

(23.6)

23.7

(a)

Al2Me6 + 6H2O J 2Al(OH)3 + 6CH4

(hydrolysis of aluminium alkyl releases alkane)

(b)

nAlR3 + nR′NH2 J (RAlNR′)n + 2nRH

(RH elimination to give AlN ring or cage)

(c)

Me3SiCl + Na[C5H5] J Me3Si(η1-C5H5) + NaCl (elimination of NaCl with formation of Si–C σ-bond)

(d)

2Me2SiCl2 + Li[AlH4] J 2Me2SiH2 + LiCl + AlCl3 (reduction by [AlH4]– )

Organometallic compounds of s- and p-block elements 23.8 Ph C

Me Me

π

C

σ Ga

Me

Ga

Me

C C Ph

(23.7)

Pb

Cl Ph

Cl

(23.10) Ph

Ph

Cl Pb

Pb Ph Ph

Al

Ph Ph

(23.8)

This question is tackled by (i) considering the availability of halido-substituents to act as bridging groups between group 14 elements, (ii) considering the likely coordination number of the group 14 element, and (iii) checking that the proposed structure retains the correct stoichiometry. (a) General formula R2EX2 suggests a chain with double halido-bridges, i.e. 23.10 with octahedral Pb. (b) General formula R3EX with an R group which is not too bulky suggests a chain with single halido-bridges, i.e. 23.11 with trigonal bipyramidal Pb. (c) In (2,4,6-Me3C6H2)3PbCl, the aryl substituents are very sterically demanding. Therefore, (2,4,6-Me3C6H2)3PbCl is likely to be a monomer with tetrahedral Pb. (d) [PhPbCl5]2– is 6-coordinate and unlikely to increase its coordination number; therefore monomeric, with octahedral Pb.

Ph

Ph

Ph

Al

23.10 Cl

Pb

Ph

Going from (η1-C5Me5)2SiBr2 to (η5-C5Me5)2Si reduces Si(IV) to Si(II). The reducing agent is prepared by treating anthracene (23.9) with K. Anthracene undergoes a 1-electron reduction to give the salt K+[(23.9)]–. The radical anion [(23.9)]– acts as a reducing agent, regenerating anthracene as it is oxidized. The second product of the reaction is KBr.

Ph Cl

ipso-C

23.9

(23.9)

Ph

(a) Section 23.4 in H&S contains relevant information. Points to include: • monomeric R3B with trigonal planar B, and 2c-2e B–C bonds; • dimeric Al2R6 with tetrahedral Al, and 3c2e bridging interactions; • monomeric GaR3, InR3 and TlR3 with trigonal planar group 13 element, but tendency for interactions between molecules in solid state. (b) [Me2(PhC2)Ga]2 (23.7) exhibits σ,π-mode of bonding for the bridging PhC2-units. In [Ph3Al]2 (23.8), two Ph groups bridge between Al atoms with the ipso-C atoms ≈ tetrahedrally sited.

327

Cl Ph

(23.11)

23.11

(a) Hydrolysis of Sn–Cl bond: Et3SnCl + H2O J Et3SnOH + HCl or 2Et3SnCl + H2O J (Et3Sn)2O + 2HCl (b) Elimination of NaCl; formation of Sn–C σ-bond: Et3SnCl + Na[Cp] J (η1-Cp)Et3Sn + NaCl (c) Elimination of NaCl; formation of Sn–S bonds: 2Et3SnCl + Na2S J (Et3Sn)2S + 2NaCl (d) Elimination of LiCl; formation of Sn–C σ-bond: Et3SnCl + PhLi J Et3PhSn + LiCl (e) Oxidation of Na; Sn–Sn bond formation: 2Et3SnCl + 2Na J Et3SnSnEt3 + 2NaCl

Organometallic compounds of s- and p-block elements

328

23.12

R R

R

R

R

α

Sn

R

R

R

R R

(a) Structure 23.12 shows the general structure of (η5-C5R5)2Sn. The tilt angle α increases as the steric demands of R increase: α = 125o for R = H, 144o for R = Me, 180o for R = Ph. (b) Structure of (η5-C5Me5)2Mg is as expected and the C5 rings are parallel. It is difficult to rationalize why the heavier group 2 metals show tilted structures; no simple explanation can be put forward but it may be a solid-state (crystal packing) effect. Furthermore, while Cp2Mg is monomeric and soluble in hydrocarbon solvents, Cp2Ca, Cp2Sr and Cp2Ba are polymeric and insoluble in ethers and hydrocarbons.

(23.12)

23.13

X See: J.A. Nobrega et al. (1998) Chem. Commun., p. 381 for full information

InBr contains In(I) and oxidation will give In(III), so A is an In(III) compound. Oxidative addition of C–Br bond occurs. In the 1H NMR spectrum, the multiplet at δ 3.6 ppm (rel. integral 8) is assigned to 1,4-dioxane, singlet at δ 5.36 ppm (rel. integral 1) is a CH unit. Suggests A is the adduct Br2InCHBr2.C4H8O2. A contains two more C–Br bonds which react with 2 equivalents of InBr in the same manner as the first reaction. Each In(III) centre can behave as a Lewis acid, reacting with the Br– that is added: H C Br2In

3–

H InBr2 InBr2

+ 3Br

–

C Br3In

InBr3 InBr3

Product is stabilized by a large counterion, [Ph4P]+. Check the analysis of [Ph4P]3[HC(InBr3)3] (Mr = 2093.56): In =

Br =

344.46 ×100 = 16.45% 2093.56 719.1 × 100 = 34.35% 2093.56

Signals in 1H NMR spectrum of B at δ 8.01-7.71 ppm (rel. integral 60) are assigned to Ph protons; singlet at δ 0.20 ppm (rel. integral 1) assigned to CH proton. B is therefore [Ph4P]3[HC(InBr3)3]. 23.14

5p orbital (vacant) R R Sn

Sn

R R

sp2 orbital with lone pair

(23.14)

2– (a) B is in group 13; [R2BBR2]2– is isoelectronic with an R R alkene R2C=CR2, and contains a B=B double bond (23.13). B B The B2C4-framework is planar; bulky R groups needed to stabilize compound. Bonding: sp2 hybridized B, overlap R R giving B–B σ-bond; π-bond from overlap of remaining 2p (23.13) atomic orbitals; 4 valence electrons per B–. (b) Ga is in group 13; [R2GaGaR2]– is related to [R2BBR2]2–, but has one electron less for bonding. Ga–Ga σ-bonding MO is fully occupied, but π-bonding MO is half-occupied giving a Ga–Ga bond order of 1.5. (c) Sn is in group 14; by analogy with an alkene, R2SnSnR2 might be thought to have an Sn=Sn double bond, but distannenes (and digermenes) have non-planar frameworks. Bonding described in terms of sp2 hybridized Sn(II), localized Sn–CR bonds and a lone pair occupying the remaining sp2 hybrid orbital. Donation of lone pair into vacant 5p atomic orbital on adjacent Sn atom (23.14). (d) Ge is in group 14; R3GeGeR3 is an analogue of an alkane R3CCR3. Ge is sp3 hybridized; Ge–Ge single bond; staggered conformation.

Organometallic compounds of s- and p-block elements

(e) As is in group 15; RAsAsR (23.15) is an analogue of RN=NR; As is sp2 hybridized, giving localized As–CR bonds and a lone pair occupying one hybrid orbital; overlap of sp2 hybrid orbitals gives the As–As σ-bond, and overlap of singly occupied 4p atomic orbitals gives the As–As π-bond. A trans-arrangement of R groups is sterically favoured.

R As

329

As

R

(23.15)

23.15

(a)

2Me3Sb + B2H6 J 2Me3Sb.BH3

(Me3Sb acting as a Lewis base)

(b)

Me3Sb + H2O2 J Me3SbO + H2O

(oxidation of Sb(III) to Sb(V) with accompanying O transfer)

(c)

Me3Sb + Br2 J Me3SbBr2

(oxidative addition of Br2)

(d)

Me3Sb + Cl2 J Me3SbCl2

(oxidative addition of Cl2)

(e)

Me3SbCl2 + 3MeLi J Li+[Me6Sb]– + 2LiCl (methylation; Li+ salt converted to a salt with a large cation for isolation) Me3Sb + MeI J Me4SbI (oxidative addition of MeI)

(f)

Me3Sb + Br2 J Me3SbBr2

(oxidative addition of Br2)

Me3SbBr2 + 2Na[OEt] J Me3Sb(OEt)2 + 2NaBr (elimination of NaBr and formation of an ethoxy derivative) 23.16

Sections 23.4-23.6 in H&S contain relevant information. Points to include: • group 13, oxidation state +3, with +1 becoming stable for the heaviest elements, exemplified by cyclopentadienyl derivatives: R3B, R3Al but forms dimers R2Al(μ-R)2AlR2, R3Ga, R3In, R3Tl, CpGa, CpIn, CpTl; • group 14 (ignoring C), oxidation state +4 with +2 becoming stable for the later elements: R4Si, R4Ge, R4Sn, R4Pb, R2Si (highly reactive), R2Ge and R2Sn, R2Pb (stabilized or stability increased by bulky R groups); • group 15 (ignoring N and P), oxidation states +3 and +5, R3E and R5E with R3E being sensitive to oxidation; • comment on relative stabilities among groups of similar compounds.

23.17

(a)

GeCl4 + 4RMgCl GeCl4 + 4RLi

Et2O

Et2O

R4Ge + 4MgCl2

R4Ge + 4LiCl

(Grignard reagent) (organolithium reagent)

A range of derivatives can be made by the above methods. 1/ B H + 3RCH =CH J B(CH CH R) (b) (hydroboration; 2 2 6 2 2 2 2 3 R = alkyl, aryl) Et2O. BF3 + 3RMgCl

Et2O

R3B + 3MgClF + Et2O

(R = alkyl, aryl)

(c)

GaCl3 + 3Li[C5R5] J (C5R5)3Ga + 3LiCl

(d)

nR2SiCl2 + 2nNa J cyclo-(R2Si)n + 2nNaCl (small R favours large ring; bulky R favours small ring)

Organometallic compounds of s- and p-block elements

330

(e) Requires oxidation of R3As; cannot be formed from AsX5, X = halogen: R3As

R

2RLi

R5As

(f)

2R2AlCl + 2K J R2AlAlR2 + 2KCl

(g)

SbCl3 + 3RMgCl

Et2O

R3Ge + 3MgCl2

(only works for very bulky R group, e.g. (Me3Si)2CH) (Grignard reagent)

Relevant information is found in Sections 23.4-23.6 in H&S. Points on which to base an answer: • η1, η3 and η5-modes of bonding; • differences between gas phase and solid state structures, e.g. CpTl is a monomer in gas phase, but polymeric in solid, and similarly Cp2Pb. • monomer versus polymer formation for related species in solid, e.g. Cp3Ga is monomeric, Cp3In forms chains; • in group 14, tilted versus parallel C5R5 rings in (η5-C5R5)2E; • fluxional properties in solution observed by using (typically) 1H NMR spectroscopy.

23.19

Relevant information is found in Sections 23.4-23.6 in H&S. Points to include: • in group 13, compound types are R2E–ER2 and reduced derivatives; • in group 14, compound types are R3E–ER3 (ethane analogue), R2E=ER2 (ethene analogue) and R2E–ER2 (non-planar E2C4 framework, see 23.14). • in group 15, compound types are R2E–ER2 and RE=ER (see 23.15); • sterically demanding substituents vital to stabilization of these compounds except for R2E–ER2 where E = group 15 element; • common bulky substituents include 23.16 and 23.17; • examples: B2R4 and [B2R4]2– (ethene analogue); Al2R4 and [Al2R4]– (radical anion); Ga2R4, [Ga2R4]– and [Ga2R4]2– ; R2Si=SiR2, R2Ge=GeR2, but R2Sn–SnR2 (see 23.14); R2E–ER2 for E = As, Sb, Bi and R = Ph; RE=ER for E = As, Sb and R is bulky; • selected examples of preparations of these compounds.

23.20

Main points to include (give examples of specific reactions in each case): • general use of organolithium reagents; • general use of Grignard reagents; • transmetallation using organomercury compounds, e.g. for group 1 and 2 metal M–C bonds; • hydroboration for B–C bonds; • for cyclopentadienyl compounds, use of Na[C5R5] or Li[C5R5].

23.21

Points to include: • Distinguish between directly observing a nucleus and observing coupling to heteronuclear NMR-active nuclei in, e.g. 1H or 13C NMR spectra. • Give examples of heteronuclei, e.g. 7Li, 92.5%, I = 1/2; 29Si, 4.7%, I = 1/2; 117Sn and 119Sn, 7.7 and 8.6%, I = 1/ . 2 • Summarize types of information that can be obtained, e.g. structural, stereochemically rigid or non-rigid; speciation in solution (to compare with solid state structural data from X-ray diffraction). • Specific examples: see detailed discussions in Chapter 23 in H&S which include (RLi)n, [Ph3SiH2]–, organotin halides.

R R = Me, iPr

(23.16) C

R3AsCl2

23.18

R

Me3Si

Cl2

SiMe3

H

(23.17)

Organometallic compounds of s- and p-block elements 23.22

R

R

R E

E

R

E R

R

A

R

R

E B

(23.19)

23.23 –

Ar B Ar

Ar Ar

Ar = C6F5

(23.20)

23.24

Pb

(23.21)

Pb

(23.22)

23.25

An R2E unit (e.g. a stannylene, analogue of a carbene) can be represented as in 23.18. This shows a lone pair of electrons on atom E. Assuming sp2 hybridized E, then a valence state could involve both non-bonding electrons in an sp2 hybrid orbital (a singlet state, 23.18), or one electron in the sp2 hybrid orbital and one promoted to an np atomic orbital (a triplet state). Combining two triplet R2E units to form A in 23.19 gives a σ -bond and a π-bond as in ethene. Combining two singlet R2E units to form B in 23.19 is the description detailed in answer 23.14c.

331

R E R

(23.18)

[B(C6F5)4]– (23.20) and [CHB11Me5Br6]– (carbaborane anion related to those in Figure 9.6 in H&S) are very weakly coordinating anions. The conjugate acid of [CHB11Me5Br6]– is one of a family of superacids. Used for the stabilization of unusual cations in, for example, the following salts: • [Mes3Si][CHB11Me5Br6] which contains well separated ions in the solid state; [Mes3Si]+ is the first example of a free silylium ion; • [(2,4,6-iPr3C6H2)3Sn][B(C6F5)4] • [(η5-C5Me5)Si][B(C6F5)4] (see Fig. 23.15 in H&S). Each of the cations above would interact with an anion that had the ability to coordinate. Hence, the choice of extremely weakly coordinating anions to permit the isolation of truly free cations of this type. (a) Coparallel rings (23.21) result in non-polar molecule. If the rings are tilted, there is a resultant dipole moment (23.22). The observation of a polar molecule can be used as evidence that the molecule does not contain coparallel rings. (b) In (η5-C5Me5)2Be, all Me Me Me groups are equivalent and Me Me 1H only one NMR H spectroscopic signal is Be observed. Diagram 23.23 Me represents the solid-state Me structure of (η5-C5HMe4)(η1H 1 Me C5HMe4)Be. The solution H Me Me b NMR spectrum exhibits 3 b Me (23.23) signals in a ratio 6 : 6 : 1; these Me Me a a correspond to 2 Me H environments and equivalent Be CH protons. The molecule is H therefore fluxional with Me a a Me equivalent rings, i.e. average Me b b Me structure (shown with (23.24) staggered rings) is as in 23.24. CF3SO3Me (methyl triflate) is a source of Me+ and the starting material, RP=PR (R is 2,4,6-tBu3C6H2) is methylated at one P atom: RP=PR + CF3SO3Me J [RP=PRMe]+[CF3SO3]–

(compound A)

RP=PR has one P environment, therefore one 31P NMR signal; [RP=PRMe]+ has 2 P environments, so two 31P NMR signals with 31P–31P coupling. Methyl lithium

Organometallic compounds of s- and p-block elements

332 R

R R

Me P

P Me Me Me

R

(23.25)

(23.26)

23.26

provides Me– and reaction of [RP=PRMe]+ with MeLi leads to RMeP–PMeR (compound B). The low temperature 31P NMR spectrum can be explained by the presence of 2 stereoisomers, shown in 23.25 and 23.26 as Newman projections. B contains a P–P single bond; each P atom is in a trigonal pyramidal environment with a lone pair. (a) The precursor is structurally set up to form Sb–Sb bonds as follows: SbCl2

SiMe3 C

Me

C O Ba O

O O 141o

O O

C

SbCl2 SbCl2

Sb Na Me

C

Sb

+ 6NaCl

Sb

In this reaction, Na is oxidized to Na+, and Sb (because element–element bonds are formed) formally undergoes reduction from +3 to +1. (b) Compound 23.27 possesses a macrocyclic ligand (crown ether) with hard donor atoms that coordinate to Ba2+. The large size of the metal ion permits a high coordination number; the crown ether occupies equatorial sites in a hexagonal bipyramidal coordination sphere; axial sites occupied by [Me3SiC ≡ C]– ligands. Non-linear C ≡ C–Ba units suggest bonding is not a simple σ-bond; possibility that there is a weak π-contribution (see answer 23.8b). Ba2+ is large enough to accommodate this interaction. (c) The steric demands of C5H5 (Cp) versus C5Me5 affect the intermolecular interactions in each compound. Polymeric Cp2Ba is insoluble in common organic solvents, consistent with a polymeric structure; steric demands of C5Me5 result in weaker intermolecular interactions although in the solid state, the large size of Ba2+ allows some intermolecular interactions to occur and a polymeric structure results; as a consequence of intermolecular interactions in (C5Me5)2Ba being weak, polymeric structure is not retained in solution and the compound is soluble in aromatic solvents. In contrast to (C5Me5)2Ba, (C5Me5)2Be is monomeric in the solid state, presumably because a combination of a small Be(II) centre and steric demands of C 5Me 5 prevents the occurrence of Be---C5Me 5 intermolecular interactions.

C

C Me3Si

(23.27)

23.27

(a) (η5-C5Me5)GeCl acts as Cl– donor and MCl2 (M = Ge, Sn) as Cl– acceptor: (η5-C5Me5)GeCl + MCl2 J [(η5-C5Me5)Ge]+[MCl3]–

– M Cl

Cl Cl

M = Ge, Sn

(23.28)

[A]+ = [(η5-C5Me5)Ge]+ [B]– = [GeCl3]– [C]– = [SnCl3]– (b) The 13C NMR spectrum shows two signals, one for Me and one for C5-ring carbons: δ 121.2 (Cring), 9.6 (CMe) ppm. (c) The common ion seen in the mass spectra of the compounds is assigned to [(η5C5Me5)Ge]+; Ge has 5 isotopes (see Appendix 5 in H&S) and the molecular ion will reflect this and also the presence of 1.1% 13C. (d) The structure of [MCl3]– (M = Ge or Sn) is trigonal pyramidal; use the VSEPR model: M is in group 14, therefore 4 valence electrons. Add one electron for negative charge. Three bonding pairs and one lone pair of electrons. Molecular shape (23.28) derived from tetrahedral arrangement of electron pairs. (e) See Box 23.1 in H&S.

Organometallic compounds of s- and p-block elements 23.28 Et I I

Bi

(a) The reactions are: BiCl3 + 3EtMgCl J Et3Bi + 3MgCl2 Et3Bi + 2BI3 J 3EtBiI2

Et I I

Bi n

(23.29)

333

Compound X = Et3Bi Compound Y = EtBiI2

Two repeat units in the polymeric structure of EtBiI2 are shown in 23.29; each Bi is square-based pyramidal and the stoichiometry EtBiI2 is retained by having both I atoms involved in bridges in a chain. Possible for Et groups to be on the same or opposite sides of the ---Bi---Bi--- chain. (b) Reaction sequence to account for observed products and 1 : 4 ratio of reagents: TeCl4 + 3ArLi J Ar3TeCl + 3LiCl Ar3TeCl + ArLi J Ar4Te + LiCl TeCl4 + 2ArLi J Ar2TeCl2 + 2LiCl These are Te(IV) products, and disproportionation is the next step:

then:

Ar4Te + Ar2TeCl2 J Ar4TeCl2 + Ar2Te Ar4TeCl2 + 2LiAr J Ar6Te + 2LiCl

(c) RLi reacts with R'SbCl2 eliminating LiCl and forming RR'SbCl. In addition to the R–Sb bond, a coordinate bond can form using the pendant NMe2 group; the two enantiomers of the product are: NMe2 C

Me2N

Sb

Sb

Cl

Cl

CH(SiMe3)2

(Me3Si)2HC

23.29

C

The equilibrium to consider is: R Sn R

Sn Ph

2RPhSn Ph

R = C6H3-2,6-(C6H2-2,4,6-iPr3)2

The red solution exhibits an 119Sn NMR signal at a similar chemical shift to that of Sn(C6H-2-tBu-4,5,6-Me3)2, and this indicates that the equilibrium is shifted fully to the right-hand side. Thus, a red colour is associated with the presence of RPhSn. The two 119Sn NMR signals at δ 246 and 2857 ppm arise from the two different Sn environments in RSnSnRPh2. The changes in relative integrals of these and the signal at δ 1517 ppm can be explained in terms of red RPhSn being in equilibrium with green RSnSnRPh2. Lowering the temperature shifts the equilbrium to the left, and warming the solution shifts the equilibrium to the right. 23.30

(a) Elemental analytical data allow you to deduce the empirical formula of a compound, or at least, the ratio of atoms of elements for which analyses have been determined (routinely C, H and N). These data cannot distinguish between monomer, dimer, trimer ... polymer because the elemental analyses of, for example, PhSeCl3, (PhSeCl3)2, (PhSeCl3)3 ... , (PhSeCl3)n are all identical.

334

Organometallic compounds of s- and p-block elements (b) The polymeric structure must have each Se in a square-based pyramidal environment, and overall stoichiometry (PhSeCl3)n. There must be two Cl per Se that are not involved in bridging interactions. A simple representation is:

X The actual structure is not as simple as depicted here: see N.A. Barnes et al. (2005) Dalton Trans., p. 1759.

Ph

Ph

Cl

Cl Se

Se

Cl

Cl

Cl

Cl n

(c) The observed Se–Cl bond distances are 220, 223, 263 and 273 pm. The two shortest distances are assigned to terminal Se–Cl bonds, and longer distances to bridge bonds. 23.31

k=

ln 2 t1/ 2

ln 2 k

t1/ 2 =

For TBT:

For DBT:

t1/ 2 =

ln 2 = 2 .1 y 0.33

t1/ 2 =

For MBT: ln 2 = 1.9 y 0.36

t1/ 2 =

ln 2 = 1 .1 y 0.63

23.32

(a) Volatile organometallics are used as sources of group 13 and 15 elements for deposition of III-V semiconductors. Refer to eq. 23.30 in H&S and discussion, and to Section 28.6 in H&S, subsection “III-V Semiconductors”. (b) Et3Al is a component of Ziegler–Natta catalysts which are important in the polymer industry. Your answer should include details from Section 25.8 in H&S.

23.33

(a)

2RSH J RSSR + 2H+ + 2e–

(b)

Me2AsO(OH) + 2H+ + 2e– J Me2AsOH + H2O

(c)

H3AsO4 red

MeAsO(OH)2

ox

H3AsO3

red

ox

MeAs(OH)2

Me2AsO(OH)

Me3AsO

ox

red Me2As(OH)

red Me3As methylation

ox = oxidation of As(III) to As(V)

[Me4As]+

red = reduction of As(V) to As(III)

The final step is addition of Me+ to Me3As suggesting the methoxy group is transferred as Me+. Going from H3AsO3 to MeAsO(OH)2 involves the following reaction which is consistent with transfer of Me+ from S-adenosylmethionine: OH HO

As

O OH

+ Me+

HO

As

OH

+ H+

Me

Each of the remaining methylations is analogous to the reaction above.

335

24 Organometallic compounds of d-block elements

24.1

(a) The μ notation refers to ‘bridging’ ligands; η notation gives the ‘hapticity’ of the ligand = number of atoms in the ligand that interact with M (see Box 19.1 in H&S). μ-CO bridges two M atoms (24.1); μ4-PR bridges four M (24.2); η5-C5Me5 has all five C atoms interacting with M (24.3, Me groups omitted); η4-C6H6 is coordinated so only four of the C atoms interact with M (24.4); μ3-H bridges three M (24.5). P R

O C M O M

M

24.2 σ-donation π-back donation O

(24.7)

24.3

2 3 4

(24.8)

M

M

M M

(24.3)

(24.4)

(24.5)

A synergic effect is one in which there is cooperation between two (or more) effects. With respect to metal carbonyl bonding, the synergic effect (or Dewar-ChattDuncanson model) refers to the two complementary components of the M–CO interaction (scheme 24.7): • donation of a lone pair of electrons from CO to vacant M orbital (e.g. dz2); this provides the metal with an excess of electron density; electroneutrality principle indicates that this charge distribution is unsatisfactory; • donation of electrons from filled M orbital (e.g. dxz) to vacant CO π* orbital; this reduces the negative charge on the metal, and at the same time strengthens the M–C interaction. Apply the bonding model in answer 24.2, remembering that back donation weakens the C–O bond because electrons occupy the CO π* orbital. (a) [V(CO)6]– and Cr(CO)6 are isoelectronic, but the formal negative charge in [V(CO)6]– leads to greater back donation, thereby weakening the C–O bond. The vibrational wavenumber is related to the force constant, k, of the bond:

ν = P

H

(b) The ligand can coordinate through any number of C atoms and therefore can adopt an η1, η2, η3, η4 or η5-mode. CO can be terminal or bridging, and the bridging mode may be μ- or μ3. Other modes are also possible for CO, e.g. 24.6. (c) PPh3 almost always coordinates in a terminal mode, but see structure 24.18 in H&S and accompanying discussion.

M

C

(24.2)

M

Cp–

(24.6)

M

M

(24.1)

C M

M

M

M

M

1 2πc

k

μ

and the weaker the bond, the lower the wavenumber. The increased back bonding strengthens the M–C bond, and the vibrational wavenumber shifts to a higher value on going from Cr(CO)6 to [V(CO)6]–. (b) PPh3 is shown in 24.8. The Tolman cone angle gives a measure of the steric demands of the ligand when coordinated to the metal. Taking PPh3 as a starting point, the introduction of Me groups in the 4-positions does not affect the steric demands of the coordinated ligand, but if Me groups are in the 2-positions, the ligand becomes more bulky, as is shown by the increased value of the cone angle.

Organometallic compounds of d-block elements

336

(c) MeCN is a more labile ligand than CO and replacement of one or two CO in Ru3(CO)12 by MeCN gives preferential sites of substitution for PPh3. Direct reaction of Ru3(CO)12 with MeCN does not occur; Me3NO acts as an oxidizing agent: H

C

C

H

H

Me3NO + CO J Me3N + CO2

H C

C

creating a vacant coordination site that is occupied by MeCN. This is then readily replaced by PPh3. (d) Free HC ≡ CH is linear with sp hybridized C. The η2-coordination results in back donation from Os to the C ≡ C π* orbital reducing the C-C bond order (24.9) and making the C atoms more sp2-like. Therefore, the Os–C–H bond angle increases.

Os

(24.9)

24.4

a

Me

(24.10)

The wavelength of X-rays is ca. 10–10 m and this is the same order of magnitude as internuclear distances in molecules. X-rays are scattered by electrons surrounding nuclei, but the scattering power of an atom depends on the number of electrons and therefore the most difficult atoms to locate are H atoms, especially in the presence of heavy atoms. In a neutron diffraction experiment, neutrons are diffracted by atomic nuclei. Thus, neutron diffraction data give more precise information about the location of H atoms than do X-ray diffraction data.

X See R. Bau et al. (2003) Inorg. Chem., vol. 43, p. 555

24.7

The hydride in 24.12 couples to one 31P trans to H and one 31P cis to H to give a doublet (200 Hz) of doublets (17 Hz). Approximately one-third of this signal is further split as a result of coupling to with 195Pt nucleus (33.8 % I = 1/2). The final signal is a doublet (200 Hz) of doublets (17 Hz), upon which is superimposed a doublet (1080 Hz) of doublets (200 Hz) of doublets (17 Hz).

24.8

(a) Lengthening from 135 to 151 pm is significant and indicates substantial back donation into the C=C π* orbital. The CN substituents are electron withdrawing and encourage back donation, i.e. back donation for C2(CN)4 > C2H4. (b) THF is a labile ligand and is readily replaced by PPh3:

C 6F 5 Pt

(24.12)

C H2

24.6

SiMe3

(24.11)

P Me2

H

PMe2 CO Ru CO PMe2

Each C5Me4SiMe3 contains 3 Me environments and these should appear as 3 singlets of relative integrals 36 : 12 : 12 for SiMe3, Me(a)ring and Me(b)ring, respectively (24.11). You can therefore assign the signals at δ 0.53, 2.25 and 2.36 ppm to these protons; you cannot unambiguously distinguish between Me(a)ring and Me(b)ring. The multiplets at δ 1.41 and 3.59 ppm arise from the two different CH2 environments in THF. One signal remains (δ 4.29 ppm) and this must be due to the metal hydrides. The formula in the question shows two different types of H atoms, but the NMR spectrum shows only one environment. The signal is a quintet, indicating that the hydrides are fluxional on the NMR timescale and ‘see’ all 4 89Y nuclei. (See: K.C. Hultzsch et al. (2003) Z. Anorg. Allg. Chem., vol. 629, p. 1272.)

Me b

Me2 P

Me2P OC Ru OC Me2P

H2 C O C

24.5 Me a

Me b

(a) See structure 24.10; the prefix μ tells you the ligand is bridging; other ligands are terminal. (b) Negative δ is consistent with a metal hydride and the δ –10.2 ppm signal can be assigned to the Ru–H–Ru hydride. 1H nucleus couples to 4 equivalent 31P nuclei (100%, I = 1/2) to give a binomial quintet.

H

Mo(CO)5(THF) + PPh3

hν

Mo(CO)5(PPh3) + THF

Organometallic compounds of d-block elements

337

Fig. 24.1 Orbital interactions in a metal π-allyl complex; the symmetry labels apply to a C2v allyl ligand (see H&S, Fig. 24.8). pzdz2

dxz

ψ1(b1)

dyz

ψ2(a2)

ψ3(b1)

In the 31P NMR spectrum, the signal at δ –6 ppm is assigned to free PPh3. On coordination, the signal shifts to higher frequency (more +ve δ) and the signal at δ +37 ppm is assigned to Mo(CO)5(PPh3). (c) Absorptions around 2000 cm–1 are assigned to CO stretching modes. PPh3 is a poorer π-acceptor than CO and, compared to Fe(CO)5, in Fe(CO)3(PPh3)2 more charge is available on Fe for back donation to the remaining 3 COs. This lowers the C–O bond strength and lowers ν (CO). The fact that there is one band in Fe(CO)3(PPh3)2 and two in Fe(CO)5 reflects a change in molecular symmetry. See the discussion of metal carbonyls in Section 3.7. 24.9

The bonding scheme is shown in Fig. 24.1. The allyl ligand has 3 π orbitals, ψ1 (Fig. 24.1) is the lowest in energy and is occupied; electrons are donated to the metal. In C3H5, MO ψ2 (non-bonding) is singly occupied; ψ3 is vacant and back donation into this MO weakens the C–C bonds.

24.10

(a) (η5-Cp)Rh(η2-C2H4)(PMe3) Electron count = 5 + 9 + 2 + 2 = 18 (b) (η3-C3H5)2Rh(μ-Cl)2Rh(η3-C3H5)2 (bridging Cl is a 3-electron donor) Electron count per Rh = (2 × 3) + 9 + 3 = 18 (c) Cr(CO)4(PPh3)2 Electron count = 6 + (4 × 2) + (2 × 2) = 18 (d) Fe(CO)3(η4-CH2=CHCH=CH2) Electron count = 8 + (3 × 2) + 4 = 18 Electron count per Fe = 8 + (3 × 2) + 3 + 1 = 18 (e) Fe2(CO)9 (structure 24.13) (1 electron per metal from an M–M bond) (f) [HFe(CO)4]– Electron count = 1 + 8 + (4 × 2) + 1 = 18 (1 electron from the –ve charge) (g) [(η5-Cp)CoMe(PMe3)2]+ Electron count = 5 + 9 + 1 + (2 × 2) – 1 = 18 (subtract 1 electron for +ve charge) (h) RhCl(H)2(η2-C2H4)(PPh3)2 Electron count = 9 + 1 + (2 × 1) + 2 + (2 × 2) = 18 (terminal Cl is a 1-electron donor)

X Each metal is taken as being in oxidation state zero; all ligands are treated as being neutral O C

OC OC

Fe OC

C O

CO CO

Fe CO

C O

(24.13)

24.11 OC CO OC

Fe OC

2–

CO Fe

Raman data show an unbridged Fe–Fe bond is present. Each Fe centre will obey the 18 electron rule and so the anion will be symmetrical and each Fe will have the same number of CO ligands: [(CO)xFe–Fe(CO)x]2–. To find x, apply the 18 electron rule, allocating one electron from the –ve charge per Fe:

CO

Electron count per Fe = 18 = 8 + 1 + 2x + 1

OC CO

x=4

[A]2– is [Fe2(CO)8]2–. The likely structure is 24.14 with staggered ligands.

(24.14)

24.12 X See also answer 22.31a, p. 323

NO can be a 1- or 3-electron donor, but to achieve an 18 electron count per Fe, each NO must act as a 3-electron donor. 24.15 and 24.16 are possible structures in which

ON ON

NO Fe

Fe

ON

O N

ON

NO NO

(24.15)

2+

Fe ON

NO Fe

N O

(24.16)

NO

2+

Organometallic compounds of d-block elements

338

each Fe obeys the 18 electron rule. They could be distinguished by using IR spectroscopy; terminal and μ-NO can be distinguished in the same way as one tells terminal and μ-CO apart, i.e. μ-NO at lower wavenumber. 24.13 π-electron donation π-back donation CR2 M CR2

(24.17)

48 60 62 64 72 74 86 90

Fe

CO

H

(24.18)

24.15

Total valence electron (ve) counts shapes are listed in Table 24.1. (a) [Ru6(CO)18]2– (b) H4Ru4(CO)12 (c) Os5(CO)16 (d) Os4(CO)16 (e) Co3(CO)9(μ3-CCl) (f) H2Os3(CO)9(μ3-PPh) (g) HRu6(CO)17B

Table 24.1 Total valence electron (ve) counts for Mx polyhedra. Triangle Tetrahedron Butterfly Square Trigonal bipyramid Square-based pyramid Octahedron Trigonal prism

CO

Using Wade’s rules, break the cluster formula down into Os(CO)3 units, extra (or fewer) CO ligands, and the charge (if applicable). Each Os(CO)3 unit provides 2 electrons for cluster bonding, and each extra CO, 2 electrons. Os7(CO)21 7 Os(CO)3 = (7× 2) = 14 electrons = 7 pairs 7 pairs of electrons are consistent with a ‘parent’ deltahedron with 6 vertices = octahedron. There are 7 Os atoms to be accommodated, and so this is a capped closo structure. (No extra electrons are associated with the addition of the capping unit.) A capped octahedron (24.19) is suggested for Os7(CO)21. [Os8(CO)22]2– 8 Os(CO)3 = (8× 2) = 16 electrons Subtract 2 CO = – 4 electrons 2– charge = 2 electrons Total = 16 – 4 + 2 = 14 electrons = 7 pairs 7 pairs of electrons are consistent with a ‘parent’ octahedron. There are 8 Os atoms to be accommodated, and so this is a bicapped closo structure. Structure 24.20 is suggested for [Os8(CO)22]2–; this is one of three possible isomers depending on the positions of the two capping units.

(24.20)

ve

(b) In (η5-Cp)(η1-Cp)Fe(CO)2 (24.18), the 5 protons of the η5-Cp ring are equivalent and give one signal in the 1H NMR spectrum. The η1-Cp ring in the static structure contains 3 H environments, but in solution at 303 K, the observation of only one proton signal for this Cp ring indicates that the {(η5-Cp)Fe(CO)2}-unit is ‘hopping’ from one CH unit to the next in the η1-Cp ring at a rate that is faster than the NMR timescale.

24.14

(24.19)

Cluster shape

(a) The vibrational wavenumber of 1652 cm–1 reflects the strength of the C=C bond in the free ligand. Coordination to the metal in [PtCl3(η2-MeCH=CH2)]– involves donation of π-electrons and back donation of metal electrons into the C=C π* orbital (24.17). This weakens the CC interaction, shifting the vibrational wavenumber to 1504 cm–1.

(h) Co3(η5-Cp)3(CO)3 (i) Co3(CO)9Ni(η5-Cp) 24.16

for selected low oxidation state metal cluster Total ve = (6 × 8) + (18 × 2) + 2 = 86 Total ve = (4 × 1) + (4 × 8) + (12 × 2) = 60 Total ve = (5 × 8) + (16 × 2) = 72 Total ve = (4 × 8) + (16 × 2) = 64 Total ve = (3 × 9) + (9 × 2) + 3 = 48 Total ve = (2 × 1) + (3 × 8) + (9 × 2) + 4 = 48 Total ve = 1 + (6 × 8) + (17 × 2) + 3 = 86 (interstitial B) Total ve = (3 × 9) + (3 × 5) + (3 × 2) = 48 Total ve = (3 × 9) + (9 × 2) + 10 + 5 = 60

(a) Number of valence electrons available in Os5(CO)18 = (5 × 8) + (18 × 2) = 76 Now consider the Os5 framework shown at the right-hand side:

Organometallic compounds of d-block elements

339

Number of valence electrons required for a framework of 3 edge-sharing triangles = (3 × 48) – (2 × 34) = 76 for 3 triangles

for 2 shared edges

The number of valence electrons required for the raft cluster = number available. (b) [Ir8(CO)22]2– (24.21) consists of two tetrahedra connected by an Ir–Ir bond which is a localized 2c-2e interaction. Write the formula as [{Ir4(CO)11}2]2– and work out an electron count for each sub-cluster (the two sub-clusters are identical). Total ve per sub-cluster = (4 × 9) + (11× 2) + 1 + 1 = 60 from the charge per sub-cluster

from Ir–Ir bond between sub-clusters

(24.21)

24.17

(a) Ligand substitution reaction, ethene displacing CO:

CO CH2 Fe

Fe(CO)5 + C2H4

Re2(CO)10

(24.22)

2Na[Re(CO)5]

NO CO

CO OC

Na[Mn(CO)5] + ONCl J NaCl + Mn(CO)4(NO) + CO

NO Mn

CO CO

CO

If NO is a 3-electron donor, CO must be lost so that Mn obeys 18-electron rule; product is trigonal bipyramidal, one of two isomers 24.23. (d) The carbonyl anion is protonated by phosphoric acid:

(24.23)

Na[Mn(CO)5] + H+ J HMn(CO)5 + Na+

H OC

Na/Hg

[Re(CO)5]– is isoelectronic and isostructural with Fe(CO)5, i.e. trigonal bipyramidal. (c) ONCl provides [NO]+; it reacts with [Mn(CO)5]– to give a nitrosyl complex:

CO Mn

Product structure 24.22.

(b) Na/Hg is a reducing agent; Re–Re bond in Re2(CO)10 is cleaved:

CO

OC

Fe(CO)4(η2-C2H4)

CO CO

CH2

hν

Mn

OC

see structure 24.24

(e) Ligand substitution reaction. More than one CO could be displaced, but on steric grounds, more than two may be unfavourable. Ni(CO)4 is tetrahedral; the phosphine substituted products are also tetrahedral.

CO CO

CO

Ni(CO)4 + PPh3 J Ni(CO)3(PPh3) or

(24.24)

24.18

Ni(CO)2(PPh3)2

In Fig. 24.15 in H&S, a Me group migrates to a position cis to the initial acyl group. The distribution of products is shown in Fig. 24.15. To examine the CO insertion mechanism, start with the labelled compound shown in Fig. 24.15 in H&S. The migration of the ‘inserted’ CO leaves the Me group in the position originally occupied by the acyl group. The CO migrates to position cis to the initial acyl group, with concomitant loss of a terminal CO. One of the cis sites contains 13CO while 3 sites contain unlabelled CO: O

Me OC

Mn

OC

CO CO

CO A

Me Me

C – 13CO

* OC

Mn

* OC

CO * 13

CO

CO

– CO*

OC

Mn

OC

CO 13

CO

CO B

The product distribution is therefore 75% B and 25% A: the Me group can never end up trans to the 13CO ligand. Compare this to Fig. 24.15 in H&S for methyl migration.

Organometallic compounds of d-block elements

340

24.19 β-H atoms

LnM

H

C

C H

H

R' H

α-H atoms

(24.25) H LnM Ph2P

(24.26)

24.20 X OC

M

CO CO

OC CO C4v

For specific examples to include in your answer, see Section 24.7 in H&S. (a) Oxidative addition involves, for example, addition of XY with cleavage of the X–Y single bond, addition of a multiply bonded species with reduction in the bond order and formation of a metallacycle, addition of a C–H bond in an orthometallation step. The metal increases its oxidation state by 2 units and its coordination number by 2. Initially, a cis product results from oxidative addition; rearrangement may occur. (b) Reductive elimination is the reverse of oxidative addition. (c) The position of α-H atoms with respect to M are illustrated in 24.25. Abstraction of one α-H atom from LnMCHR2 gives a carbene complex (M=CR2) and abstraction of two α-H atoms from LnMCH2R gives a carbyne complex (M ≡ CR). (d) The position of β-H atoms with respect to M are shown in 24.25. β-Hydrogen elimination involves transfer of a β-H atom from (for example) an alkyl group to M, and the conversion of the σ-alkyl group to a π-bonded alkene. (e) Alkyl, R, migration involves the migration of an R group to a cis CO ligand and formation of an acyl group; the reaction is facilitated by an incoming ligand (e.g. CO) which fills the coordination site vacated by R. The process is concerted. (f) Orthometallation is a type of oxidative addition reaction involving the ortho C–H bond of a phenyl (or similar) ring. A cyclized product such as 24.26 results. Cr(CO)6 undergoes substitution reactions with Ph2P(CH2)4PPh2. This ligand can be mono- or bidentate. The IR spectra in the CO stretching region are almost identical, suggesting that the Cr(CO)x units are essentially the same in A and B. There are 3 absorptions in the IR spectrum of A and of B, and this is consistent with local C4v symmetry (24.27) at each Cr centre (see Table 3.5 in H&S). Therefore suggest the structures below rather than one monodentate and one chelating Ph2P(CH2)4PPh2 which would give complexes with different IR spectra. CO

(24.27)

OC

Cr

OC CO

CO

PPh2

P Ph2

CO OC

Cr

OC CO

24.21

Ph2 P

CO P Ph2

31P

2 signals A

31P

1 signal B

CO Cr

OC

CO CO

CO

(a) Oxidative addition of Pd0(PPh3)2 to give compound A: Br Ph3P Pd(PPh3)2 +

Pd Ph3P

Pd(0)

Ph3P Pd

and/or Br

Pd(II) d8 square planar cis-isomer

Br

PPh3

trans-isomer likely to be favoured on steric grounds

Assume that A is the trans-isomer (but no information in question to distinguish between trans and cis-isomers). (b) A is a 16-electron complex and undergoes 2-electron addition (alkene addition). Pd(II) favours square planar, 16-electron complex, therefore after addition, phenyl migration occurs to give compound B:

Organometallic compounds of d-block elements

341

O O PPh3 Br

O +

Pd

addition

Ph3P

O

Br

PPh3

Pd

PPh3 18-electron, not favoured for Pd(II)

O O O Ph3P Br

Ph3P

phenyl migration

Pd

Br PPh3

O Pd PPh3 16-electron

B

(c) B undergoes β-elimination to give a coordinated alkene that dissociates (alkene D) to leave a square planar, 16-electron complex C. O

O O

Ph

O Ph3P Br

Hβ Ph

Ph3P H β

24.22

X

More examples in Section 24.10 in H&S)

O O

Pd

D

Ph3P Br

Ph

Pd H Ph3P

Ph3P

H Pd

Br

PPh3

C

18-electron, not favoured for Pd(II)

(a) Iron tricarbonyl complexes of 1,3-dienes are stable under various reaction conditions, and the Fe(CO)5 precursor is cheap. The Fe(CO)3 group acts as a protecting group for CO2Et the diene group and this permits Fe reactions to be carried out on other (CO)3 parts of the organic molecule (e.g. (24.28) 24.28). The diene group can be deprotected in a later step in the reaction sequence. The presence of the Fe(CO)3 group facilitates reactions of the 1,3-diene with nucleophiles with stereochemical control; the nucleophile is only able to attack the coordinated diene at the side away from the metal centre. (b) Fullerenes (of which C60 and C70 have been the most studied) are carbon cages with C–C and C=C bonds. The C=C bonds function like alkenes, forming η2complexes, e.g. Pt(η2-C2H4)(PPh3)2 + C60 J C2H4 + Pt(η2-C60)(PPh3)2 (c) Two fragments are isolobal if they exhibit frontier MOs with the same symmetries, approximately the same energies, and containing the same number of electrons.

Organometallic compounds of d-block elements

342

OC

H C

OC

CO Mn

H H

OC

CO

(24.29)

24.23

CH3 and Mn(CO)5 are isolobal: each has one frontier MO of σ -symmetry, containing 1 electron (24.29). Combination of two CH3 fragments gives C2H6 with a C–C single bond (overlap of σ-orbitals and pairing of the 2 electrons). Similarly, two Mn(CO)5 fragments combine to give Mn2(CO)10 with an Mn– Mn single bond. Diagram 24.30 illustrates the bonding mode of an η3-allyl ligand to a metal centre. The plane containing the C and H atoms is approximately perpendicular to an axis connecting M and the centroid of the C3-unit. Rotation of the allyl ligand about this axis (as in 24.30) can never exchange atoms H(1) and H(2). Since 1H NMR spectra show these protons to be equivalent on the NMR timescale, the rotation shown in 24.30 cannot be the mechanism for fluxionality.

H(2)

H(1)

(24.30)

24.24

See Section 24.12 in H&S for more detail. Points to include: • Fischer-type carbene possesses a low oxidation state metal, a heteroatom (e.g. O) and an electrophilic carbene centre; • Schrock-type carbene contains an early d-block metal in a high oxidation state, and shows nucleophilic character; • draw representative structures; • give representative syntheses; • give representative reactions.

24.25

(a) KOtBu deprotonates the heterocyclic cation in 1,3-dimethylimidazolium iodide:

Me

N

+

tBuO–

N

Me

– tBuOH

H

(b) The carbene substitutes for a CO ligand in Ru3(CO)12. The singlets in the 1H NMR spectrum arise from Me (δ 3.80 ppm) and alkene CH ( δ 7.02 ppm). Assume monosubstitution to give A = Ru3(CO)11L where L = carbene. (c) See structure 24.31. Axial and equatorial isomers are possible; equatorial is more likely (24.31) on steric grounds. 24.26

Me

N

N

Me

N-heterocyclic carbene

OC N

CO CO

OC Ru

N OC

Ru OC

CO CO

Ru CO CO CO

(24.31)

The two C5 rings in Cp2Fe could be eclipsed or staggered; in this answer, they are taken as being eclipsed (as in the gas phase). The five π-MOs of C5H5 are drawn in Box 23.1 in H&S. For a bonding scheme in ferrocene, consider the orbital interactions between the Fe atom and a pair of Cp ligands – draw out the 10 ligand group orbitals that arise from taking combinations of the MOs of Cp. The left-hand column in Fig. 24.2 shows the in-phase and out-of-phase combinations of the lowest energy π-MOs of the Cp ligands, and matches each (by symmetry) to one of the atomic orbitals of Fe. This operation is repeated using pairs of Cp π-MOs (see Box

Organometallic compounds of d-block elements Fig. 24.2 Orbital interactions between atomic orbitals of an Fe atom and the ligand group orbitals (LGOs) of a set of two eclipsed Cp ligands. Only six of the ten LGOs are shown in the diagram; each of LGOs ψ3, ψ4, ψ7 and ψ8 is one of a degenerate pair of orbitals; ψ5, ψ6 and ψ9 interact with the Fe py, dyz and dx2–y2 atomic orbitals, while ψ10 (like ψ8) has no match.

ψ4

ψ2

px

pz

s

ψ1

ψ3

Fe

No match

ψ7

dxy

dxz

dz2 Cp

ψ8

Cp

Fe

343

Cp

Fe

Ph

Fe

23.1 in H&S). In Fig. 24.2, the middle column shows in-phase and out-of-phase combinations of the second π-MO of Cp and matches each (by symmetry) to an Fe atomic orbital; see also Fig. caption. Of the last two LGOs shown in Fig. 24.2, only one can be matched by symmetry to an Fe atomic orbital; LGO ψ8 becomes nonbonding in ferrocene; see also Fig. caption.

O

(24.32)

24.27

(a) FeCl3 oxidizes Fe(0) to Fe(I) forming [(η5-Cp)2Fe]+[FeCl4]– and FeCl2. (b) Friedel-Crafts acylation of the Cp ring to give 24.32, or of both rings to give (η5-C5H4C(O)Ph)2Fe. (c) In the presence of Al and AlCl3, toluene displaces a Cp ligand. Since toluene provides 6 π-electrons, the product is cationic with Fe retaining an 18-electron centre: product is [(η5-Cp)Fe(η6-C6H5Me)]+[AlCl4]– containing cation 24.33. (d) NaCl is eliminated, and a Co–Fe bond forms, giving (η5-Cp)(CO)2Fe–Co(CO)4. This actually has two μ-CO ligands but this cannot be deduced from the data given.

24.28

The singly substituted product is structurally related to 24.32 with Ph replaced by Me; in the doubly substituted product, both ligands are identical. Use 1H NMR spectroscopy to distinguish between the products. The η5-Cp ring gives a singlet ( ≈ δ 5 ppm); η5-C5H4C(O)Me gives a singlet for the Me group and two multiplets for the ring protons (coupling between H(a) or H(a)′ and H(b) and H(b)′, see 24.34). The compounds could also be distinguished by using 13C NMR spectroscopy, or mass spectrometry; elemental analysis would distinguish compositions.

Me Fe

(24.33)

H(a) H(b)

Me

H(b)'

O H(a)'

(24.34)

24.29

Cl Cl Ru

Ru Cl

Cl C6Me6 represented as C6H6

(24.35)

C6Me6 is a π-ligand, donating 6 (or 4 or 2) electrons to Ru. In [(C6Me6)RuCl2]2, use the 18-electron rule to work out the bonding modes of the Cl atoms and whether or not there is an Ru–Ru bond. Per Ru: Ru (8 ve), η6-C6Me6 (6 ve), one terminal Cl (1 ve), 2 μ-Cl (3 ve) gives 18 ve, suggesting structure 24.35 for A. Product B contains [(C6Me6)2Ru]2+. Use 18electron rule to work out the hapticity of the 2+ ligands: Ru (8 ve) needs 10 more electrons so, given there is a 2+ overall charge, each Ru Ru C6Me6 must provide 6 ve and be η6-bonded (structure 24.36). Conversion of B to C is a reduction step (Na in liquid NH3), suggesting C is (C6Me6)2Ru. Apply 18-electron rule: Ru C6Me6 represented as C6H6 (8 ve) needs 10 more electrons so one ligand (24.36) (24.37) is η6- and one is η4-bonded (23.37).

Organometallic compounds of d-block elements

344

EMO

EMO

pzdz2 (a1)

ψ4 (b2u)

ψ2, ψ3 (eg)

dxz, dyz (e)

Fe OC

ψ1 (a2u) CO CO

Fig. 24.3 Frontier orbitals of a C3v Fe(CO)3 unit and of a D4h cyclobutadiene ligand; symmetry labels refer to the isolated fragments.

Fe

24.30

(a) 2,5-Norbornadiene could act as a 2- or 4-electron donor although it is conformationally restricted and usually binds in an η4-mode. Use the 18-electron rule to confirm that this is satisfactory in Fe(CO)3L: Fe (8 ve), 3CO (6 ve) leaving 4 ve to be contributed by L. Structure 24.38 is suggested. Similarly, for Fe(CO)3L where L = heptatriene, an η4-mode is expected on the basis of the 18-electron rule. Structure 24.39 is H H proposed. (b) Going from Fe to Mo, number of electrons contributed by metal decreases by 2. To maintain 18electron count at Mo, heptatriene must provide an extra 2 electrons and changes from η4 (Fe) to η6 (Mo). (c) The reagent [Ph3C]+ abstracts H– from the organic Fe CO OC ligand, converting heptatriene to [C7H7]+. Neutral η7CO C7H7 is a 7 π-electron ligand and the product is [(η7+ (24.39) C7H7)Mo(CO)3] – check the electron count at Mo: Mo (6 ve), η7-C7H7 (7 ve), 3CO (6 ve), +ve charge (–1 ve) = 18 electrons.

24.31

Draw the frontier MOs of C3v Fe(CO)3 and cyclobutadiene (C4H4) ligand (Fig. 24.3). The Fe(CO)3 unit provides 2 electrons and the C4H4 ligand has 4 π-electrons. Match fragment orbitals by looking at their symmetries: the Fe pzdz2 hybrid overlaps with ψ1 of the organic ligand, and the Fe dxz and dyz orbitals overlap with ψ2 and ψ3 of the ligand. Orbital ψ4 becomes non-bonding in (η4-C4H4)Fe(CO)3. In the complex, there are 3 bonding MOs involving Fe–C4H4 character and the 6 electrons fully occupy these MOs – electrons are paired and the complex is diamagnetic.

24.32

(a) Re2(CO)10 contains an Re–Re single bond; each Re obeys 18-electron rule with {7 + (5 × 2) + 1} electrons. There are 5 terminal CO ligands per Re and they adopt a staggered arrangement to minimize steric interactions (24.40). [Re2Cl8]2– contains a rhenium–rhenium quadruple bond and the δ-component of the bond forces the ReCl4-units to be eclipsed. (b) Apply the 18-electron rule to [M(CO)4]n– for M = Co and Fe to find n: [Co(CO)4]n– Electron count = 18 = 9 + (4 × 2) + n n=1 [Fe(CO)4]n– Electron count = 18 = 8 + (4 × 2) + n n=2 (c) The information for this answer is in Fig. 24.16 in H&S; the production of PhCDO is explained as shown in the following scheme.

CO

OC

CO

(24.38)

CO

OC

CO

CO OC

Re OC CO

Re OC

(24.40)

CO CO

Organometallic compounds of d-block elements –

H OC

MeOD

CO

Cr

OC

CO

Cr

OC

CO

–

D OC

345

CO CO

PhCOCl

CO

–

Cl OC

Cr

OC

CO

+ PhCDO

CO CO

(a) Valence electron count for H2Os3(CO)11 is 2 + (3 × 8) + (11 × 2) = 48. This electron count can be determined without knowing anything about the mode of bonding of the ligands, e.g. a CO ligand provides 2 electrons irrespective of whether it is terminal or bridging. Valence electron count for H2Os3(CO)10 is 2 less than for H2Os3(CO)11, i.e. 46. The compound is an unsaturated 46-electron species and formally contains an Os=Os bond (structure 24.41). (b) The question states that in each isomer of H2Os3(CO)11, the two H atoms are attached to the same Os. 1H NMR spectroscopic data show that in each isomer the H atoms are inequivalent. Isomer (a) in Fig. 24.4 corresponds to the structure given in the question; for this isomer, 1H NMR signals are δ –10.46 ppm (doublet, J = 2.3 Hz, terminal H) and δ –20.25 ppm (doublet, J = 2.3 Hz, bridging H); the H atoms in isomer (a) are mutually cis. For the second isomer, the larger coupling constant of 17.1 Hz suggests that H atoms are mutually trans; the signals at δ –12.53 and –18.40 ppm can be assigned to terminal and bridging H atoms respectively. Structure (b) in Fig. 24.4 is consistent with these data. Keeping the bridging H in the same position, a third isomer (c) can be drawn with a cis relationship between the H atoms; this isomer differs from (a) in that the terminal H is axial in (a) and equatorial in (c); signals at δ – 8.64 and –19.42 in (c) assigned to terminal and bridging H atoms respectively.

24.33 (CO)4 Os (OC)3Os

Os(CO)3 H

H

(24.41)

X See S. Aime et al. (2000) Inorg. Chem., vol. 39, p. 2422

(a) The structure of H3Os6(CO)16B can be considered in terms of Wade’s rules as follows: 3H provide 3 electrons 6 Os(CO)2 units provide 0 electrons (see Table 24.4 in H&S) 4 extra CO ligands provide 8 electrons Interstitial B provides 3 electrons Total electron count = 14 electrons = 7 pairs

24.34

H

Os

Os

Os H

(a)

Os

Os

H Os

Os H

(b)

H

Os

Os H

(c)

= CO

Fig. 24.4 Structures of the solution isomers of H2Os3(CO)11. See text in answer 24.33 for an explanation of 1H NMR spectra.

346

Organometallic compounds of d-block elements By Wade’s rules, this is consistent with an octahedral, closo-cage containing an interstitial B atom, whereas the experimental structure is a nido-cage based on a closo, 7-vertex cluster. For the total valence electron count: Number of valence electrons = (6 × 8) + (16 × 2) + (3 × 1) + 3 = 86 From this electron count, an octahedral structure is expected (see Table 24.5 in H&S). For reasons not readily understood, H3Os6(CO)16B is an exception to both sets of electron counting rules. (b) Using the Dewar-Chatt-Duncanson model, M–CO bonding can be described in terms of σ-donation and π-back donation (see answer 24.3). The extent of back donation is greater in [W(CO)6] than in [Ir(CO)6]3+ because formal tripositive charge on Ir prevents electrons being back-donated to the CO ligands; σ-donation dominates in [Ir(CO)6]3+; νCO for cation > neutral complex because back donation in [W(CO)6] populates C–O π* MOs, weakening the C–O bonds. 24.35

Reduction of Ir4(CO)12 with Na in THF gives Na[Ir(CO)x] (A) which has a strong absorption in its IR spectrum (THF solution) at 1892 cm–1. This value is relatively low and is consistent with the negative charge on [Ir(CO)x]– resulting in significant π-back donation. Use the 18-electron rule to find x: Electron count = 18 = 9 + 2x + 1 x = 4 [Ir(CO)x]– Based on steric arguments, [Ir(CO)4]– is expected to have a tetrahedral structure. [Ir(CO)4]–

Na, liquid NH3; Ph3SnCl, Et4NI

[Et4N][B] + nCO

From the elemental analysis of [Et4N][B]: Ratio C:H:N = (51.1/12.01) : (4.55/1.008) : (1.27/14.01) = 47:50:1 The [Et4N]+ cation accounts for C8H20N, leaving C39H30 for the anion. The IR spectrum indicates that some CO ligands are retained in [B]–; the fact that CO is lost means [B]– must contain ≤3 CO. The incorporation of two Ph3Sn groups accounts for C36H30. 1H NMR spectroscopic data are consistent with two Ph3Sn groups (δ 7.1 and 7.3 ppm, 30 H, aromatic protons) and 4 Et groups (δ 3.1 ppm, quartet, 8H; δ 1.2 ppm, triplet, 12 H). Propose that [B]– is [Ir(CO)3(SnPh3)2]–. A likely structure is a trigonal bipyramidal arrangement of ligands, and isomerism can arise because of axial and equatorial sites. Possible isomers have the two SnPh3 groups (i) both axial, (ii) both equatorial, or (iii) one axial and one equatorial. The isomer with trans-SnPh3 (SnPh3 in axial sites) is most likely on steric grounds. 24.36

(a) [Ph3C]+ abstracts H–, then β-hydrogen elimination and conversion of σ-bonded alkyl to π-bonded alkene. The charge on the complex changes from 0 to +1, and Fe retains an 18-electron count: + [Ph3C][BF4] OC OC

Fe C Me

Me H

OC OC

Fe

HC H2C

Me

[BF4]– + Ph3CH

Organometallic compounds of d-block elements

347

(b) Substitution of CO by an alkyne. The π-bonded alkyne acts as 2-electron donor. The Ti centre retains an 18-electron count:

PhC

Ti

OC

CPh

Ti

OC

OC

+ CO

Ph Ph

(c) Ag+ abstracts I–, and the vacant coordination site taken by MeCN. The complex changes from being neutral to cationic; counterion is [BF4]–. The W centre retains an 18-electron count with the organic ligand being a 5-electron donor: +

[BF4]– + AgI

AgBF4 in MeCN W OC OC

W CO

OC

I

CO

OC

NCMe

(d) Cyclooctatetraene can act as a 4-electron donor and replaces 2 CO ligands. The Co centre retains an 18-electron count:

OC

hν

+

Co

Co

CO

(e) Protonation of coordinated buta-1,3-diene leads to a coordinated allyl ligand; change from organic 4- to 3-electron donor is compensated by the addition of the 1-electron Cl (electron counting with formally neutral ligands): Me HCl Fe

Fe

OC OC

CO

OC CO

Cl CO

(f) Fischer-type carbene complex converted to carbyne (alkylidyne) complex:

Ph (OC)5W

C OMe

BBr3

Br(OC)4W

C

Ph + BBr2OMe + CO

348

Organometallic compounds of d-block elements 24.37

Ferroquine has planar chirality because one Cp ring is unsymmetrically substituted. The structures of the enantiomers: Cl

Cl

NMe2

Me2N

(R)

(S)

Fe

HN N

N

(R)-Ferroquine

24.38

CO OC

Ir

OC

Fe

NH

(S)-Ferroquine

(a) cis-[IrI2(CO)2]– contains Ir(I) which has a d8 configuration. The complex is square planar (see Section 20.3 in H&S). (b) Oxidative addition of MeI involves cleavage of C–I bond, oxidation of Ir(I) to Ir(III), and change in coordination number of Ir from 4 to 6. The product is fac-A:

I

I OC

I Me

Ir

I

OC

MeI

I

OC

I

Ir

I

OC

(24.42)

Me

(c) fac-B has structure 24.42. Me group migrates from Ir to an adjacent CO with formation of C–C bond. A vacant coordination site is created, and reaction with I– follows. (d) cis-[IrI2(CO)2]– Ir has16 valence electrons Ir(I) [IrI3(CO)2Me]– (fac-A) Ir has18 valence electrons Ir(III) Ir has18 valence electrons Ir(III) [IrI2(CO)3Me] (fac-B) IrI2(CO)2(COMe) Ir has16 valence electrons Ir(III) Ir has16 valence electrons Ir(I) cis-[IrI2(CO)2]– 24.39

(x–1)+

x+

(a) Ru N

+ H+

Ru

OH2

OH

N

N

N

(b) Electron-withdrawing substituents in the arene ring make the Ru centre more δ+ and coordinated H2O more readily loses H+. (c) The equilibrium in (a) is driven to the right-hand side when the pH is high (Le Chatelier) and to the left-hand side at high [H+] (i.e. low pH). 24.40

(a) Li

2BuLi, TMEDA

X TMEDA = Me2NCH2CH2NMe2

Fe

R2SiCl2

Fe

Fe

SiR2

Li

(b) n

Fe

SiR2

ROP Fe

Reaction is driven by the ring strain in the ferrocenophane

Si R2

n

349

25 Catalysis and some industrial processes

25.1

(a)

Oxidative addition

PPh3

of R′X R'

Pd

CH2=CHR

X

R'X

R′ migration

PPh3

Formation of active

(‘alkene insertion’)

catalytic species PPh3

X See Section 24.8 in H&S for discussion of reaction types

2PPh3

Pd(OAc)2

NEt3

RHC

Pd(PPh3)2

Pd

R'H2C

X

PPh3

[Et3NH]X PPh3 Reductive elimination of HX

NEt3

H

Pd

β-Elimination and loss of alkene

X

E-CHR=CHR'

PPh3

(b) When R′ is vinyl, benzyl or aryl, there is no β-hydrogen present in the R′ group. See Section 24.8 in H&S for further discussion. 25.2

(a) Cross-metathesis between alkenes: –C2H4 R

+

R'

R'

R

CM

(b) Alkyne metathesis using a metal alkyidyne complex: L nM

R

R

R

L nM

L nM

R Ln M +

+ R'

P(C6H11)3 Cl Cl

R'

R'

R'

R'

R'

R'

R'

(c) ROMP (ring opening metathesis polymerization):

Ru Ph n

P(C6H11)3

X

(25.1)

X

n

25.3

The reaction in the question is an example of ROMP (see above) and is catalysed by a Grubbs’ catalyst, e.g. 25.1. For the initial steps in the mechanism of the reaction, see Fig. 25.5 in H&S.

25.4

(a) A catalyst precursor is not the catalytically active species but is one added to the system; it undergoes e.g. loss of CO or another ligand to form the active catalyst. HCo(CO)4 contains an 18-electron Co and is not catalytically active; loss of CO generates HCo(CO)3 with a 16-electron centre – coordinatively unsaturated.

350

Catalysis and some industrial processes (b)

Formation of active

HCo(CO)4

catalytic species

X See Section 24.8 in H&S for discussion of reaction types

– CO CH2=CHCH2OH

HCo(CO)3

CH3CH=CHOH β-H atom elimination (H

Alkene adds to 16-

migrates to Co); conver-

electron Co giving 18

sion of σ-complex to π-

electron π-complex

bound ligand which is

CH3

eliminated; catalyst reforms (CO)3Co

CH2 H(CO)3Co

CH

CHCH2OH CH2OH

H atom migration and conversion of πcomplex to σ-complex

25.5

For complete details, see Section 25.5 in H&S. Points to include: • Monsanto process converts MeOH to MeCO2H; catalyst is cis-[Rh(CO)2I2]–. • MeOH first converted to MeI; MeI oxidatively adds to cis-[Rh(CO)2I2]–. • Me group migrates to the Rh-bound CO giving a σ-bound C(O)Me; CO fills vacant coordination site. • Reductive elimination of MeC(O)I, then reaction with H2O gives MeCO2H. • Draw the cycle in Fig. 25.9 in H&S to illustrate overall process. • Tennessee-Eastman process is closely related to Monsanto process; converts MeCO2Me to (MeCO)2O using cis-[Rh(CO)2I2]– catalyst; role of Rh(I) catalyst is the same as in Monsanto process with MeI (formed along with MeCO2Li from MeCO2Me) being converted to MeC(O)I; MeC(O)I reacts with MeCO2Li to yield (MeCO)2O. • Illustrate with a diagram how the secondary cycle in the Tennessee-Eastman process differs from that in the Monsanto process.

25.6

(a) If hydrogenation of the alkene can, in theory at least, lead to enantiomeric products, the alkene is prochiral. The alkenes to consider are 25.2-25.5. Ph

Ph

Ph

H

H

Me

Ph

H

Ph

H

H

H

H

H

NHC(O)Me

*

* + (E)-isomer

+ (E)-isomer

(25.2)

(25.3)

(25.4)

CO2H

(25.5)

The C atoms labelled * will become chiral after hydrogenation, e.g. 25.3 will become PhMeCHCH2Ph. Therefore, 25.3 and 25.5 are prochiral. (b) ⎛ R −S ⎞ R −S 85 ⎟ ×100 %ee = ⎜ ∴ = = 0.85 ⎜ R +S ⎟ R + S 100 ⎝ ⎠ The question states that %R > %S-enantiomer, so you can now proceed knowing that R > S. For amounts given in %: R + S = 100 R – S = 85 Combining these expressions gives: amount of R-enantiomer ≈ 92% amount of S-enantiomer ≈ 8%

Catalysis and some industrial processes 25.7 X Compare this proposed cycle with the inner catalytic cycle in Fig. 25.11 in H&S

(a) Proposed cycle:

HRh(CO)(PPh3)3 PPh3

Oxidative addition

Formation of active catalytic species

– PPh3

R

PPh3

RCH2CH2CHO OC

of H2; elimination

Rh

Alkene adds to 16electron Rh giving 18

H2

electron π-complex

PPh3 OC

Rh

CH2

C H

H

PPh3

of aldehyde

351

O

R H2 C

C C H2

PPh3

R OC

Alkyl migration (or

Ph3P

‘CO insertion’) OC

giving acyl complex

HC

Ph3P

Rh

H 2C

H

R

Ph3P

CO

CO adds; at the same

CH2

time, H migration and conversion of π-complex

CO

and 16-electron Rh

CH2

Rh

Ph3P

to σ-complex

(b) Regioselectivity of hydroformylation of RCH=CH2 is the selectivity towards n or i-isomers; given by the n : i ratio. RCH2CH2CHO (n-isomer) RCH=CH2 R(CH3)CHCHO (i-isomer) As the temperature increases, n : i ratio decreases, showing that there is a greater selectivity for the linear isomer at lower temperatures. 25.8 X Alkene isomerization: see answer 25.4

25.9

Hydroformylation of CH3CH2CH=CHCH3 can lead to I, II and/or III. The active catalyst is HCo(CO)3 which catalyses alkene isomerization and hydroformylation. Steric effects influence the product distribution. I is expected to be formed in the highest yield, and III in the lowest yield. Suggest that the 35 : 12 : 5 ratio corresponds to I : II : III. See eq. 25.21 and accompanying discussion in H&S.

CHO

I CHO

II

III CHO

(a) See Figs. 25.7 and 25.8 in H&S which show the hydrogenation of an alkene RCH=CH2 (R = Me for propene) catalysed by RhCl(PPh3)3 and HRh(CO)(PPh3)3, respectively. (b) Orthometallation involves one PPh3 ligand:

H Ph3P Ph3P

Ru

R Cl PPh3

H Ph2P

+ H

H

Ru

Ph3P

This reaction removes the active catalyst from the system.

Cl PPh3

+ RCH2CH3

352

Catalysis and some industrial processes 25.10 PPh2 P

Ph

Ph

NMe3

Ph

(25.6)

(25.7)

25.11

X Also see answers 25.5- 25.8

25.12

(a) Ligands 25.6 and 25.7 both coordinate through a P-donor. In the complexes Fe(CO)4(PPh3) and [Fe(CO)4(25.7)]+, the fact that the IR spectral absorptions occur at very similar wavenumbers indicates that the amount of back donation into the CO π* orbitals is similar in each complex; there must be a similar charge distribution with respect to the Fe centre in each complex. (b) The biphasic catalyst must be water-soluble. The choice of cation is critical. Na+ will favour solubility of aqueous solution, whereas [nBu4N]+ and [Ph4P]+ will increase the likelihood of the salt being soluble in an organic solvent. Therefore, the salt to choose is Na[RuL3]. Choose processes from those in Section 25.5 in H&S. Suggested plan for answer: • Need for an organometallic complex containing a metal which can easily go from 16 to 18 to 16-electron centre; active catalyst must be coordinatively unsaturated, e.g. HCo(CO)3, [Rh(CO)2I2]–. • An example of alkene hydrogenation, including a brief discussion of asymmetric hydrogenation and importance to drug industry. • Monsanto process for making MeCO2H; related manufacture of (MeCO2)2O. • Hydroformylation and problems of regioselectivity and chemoselectivity. (a)

Ni{P(OR)3}4

Equil. const. K

16-electron Ni

18-electron Ni

For

Ni{P(OR)3}3 + P(OR)3

R = 4-MeC6H4 R = iPr R = 2-MeC6H4

Greater amount of Ni{P(OR)3}3 formed, i.e. ligand loss and formation of 16-electron Ni favoured.

K = 6 × 10–10 K = 3 × 10–5 K = 4 × 10–2

The active catalyst is the 16-electron species, formed by loss of P(OR)3. Data show that the dissociation step depends on steric demands of R and is favoured when R is bulky. (b) Catalytic cycle up to last 2 steps: L

HC≡N Ni L

L

Oxidative addition of C–H bond of HCN

A

NiL2 CN H

(25.8) NC

NiL2

Ni NC

CN

(25.9)

Addition of alkene πdonor and migration

L

L L

L

L

Loss of L leaves a H

16-electron Ni centre Ni

of H gives η3-MeCHCHCH2

Ni

L L

L

NC

(c) The formation of A in the cycle above involves transfer of CN with A being either 25.8 or 25.9. In the last step, addition of L releases the alkene and regenerates the active catalyst. For commercial purposes, the linear alkene 25.8 is required.

Catalysis and some industrial processes 25.13 X Valence electron (ve) counts: see Table 24.1, p. 338

353

(a) Cluster electron count in H2Os3(CO)10 = (2 × 1) + (3 × 8) + (10 × 2) = 46 For an Os3 triangle, a 48 ve count is expected, and so H2Os3(CO)10 is unsaturated. This can be represented by altering structure 25.35 in H&S to show an Os=Os double bond. (b) The first step in the process is addition of the alkene (2 electron π-donor) to give a 48-electron Os3-complex. The cluster-bound H atoms facilitate the isomerization as shown in the following cycle: R

R

R

H

or

H

R

H

R

H

R

or

H

H

H H

R H = Os(CO)4 =Os(CO)3

25.14 adatom

kink

step

(25.10) step

Ni has an fcc structure and so one could draw an atom arrangement as shown in Fig. 6.1 (p. 85). In reality, the surface has imperfections such as those in 25.10; also atoms missing in regions of flat surface creating ‘holes’. Diagram 25.11 illustrates part of two layers of close-packed atoms, and shows a step. The 4 dark grey atoms define an M4 ‘butterfly’ site; a triangular M3 site is present on the top surface. Any ‘flat’ surface on clean Ni (assuming no adatoms or holes) is made up of triangular sites. In discrete cluster molecules, CO usually adopts terminal or bridging modes, and such modes can also be adopted on a surface (25.12-25.14). At a step site, a CO ligand can bond in mode 25.15 and the C–O bond is weaker than in modes 25.12-25.14 – CO is ‘activated’. O C M

(25.11)

(25.12)

25.15

O C

O C M

M

(25.13)

M

M

(25.14)

M

M

C M

M

O

M

(25.15)

(a) Rh (a platinum-group metal) is rare and expensive; use of alumina support reduces cost. In the bulk metal, only the surface atoms are available for catalysis; small metal particles dispersed on γ-alumina (has a large surface area) gives a catalyst in which a high proportion of Rh atoms in the system are available. The γalumina support may play active role in the catalysis, e.g. in hydrocarbon reforming: • Rh (or other platinum-group metal) catalyses alkane dehydrogenation; • alkene isomerization is facilitated by the acidic γ-alumina surface; • Rh catalyses conversion of isomerized alkene to more highly branched alkane.

354

Catalysis and some industrial processes (b) Oxidations in catalytic converter are not all catalysed by the same metal: 2CO + O2 J 2CO2 CnHm + (n + m/4)O2 J nCO2 + m/2H2O 2NO + 2CO J 2CO2 + N2 2NO + 2H2 J N2 + 2H2O 2SO2(g) + O2(g)

25.16

X For further details, see ‘Production of SO3 in the Contact process’ in Section 25.8 in H&S

1/ O 2 2

N2(g)

N(ad)

N(ad)

(25.16)

25.18 X For detailed structural data of the MgCl2-supported catalyst, see B.L. Goodall (1986) J. Chem. Ed., vol. 63, p. 191.

ΔrHo = –96 kJ per mole of SO2

(a) This is a gaseous equilibrium. In the forward reaction, 3 moles give 2 moles of gases, i.e. a decrease in pressure. By Le Chatelier’s principle, increasing the external pressure encourages the forward reaction. (b) By Le Chatelier, an increase in external temperature will be opposed; since the forward reaction is exothermic, the back reaction is endothermic and so is favoured. (c) Lower temperatures favour the forward reaction, but rate of SO3 formation is slow (not useful commercially). Rate depends on temperature, but in the above equilibrium, the back reaction is favoured by an increase in temperature (not useful commercially). The rate is increased by using a V2O5 catalyst on SiO2 carrier with K2SO4 promoter, passing reactants through several catalyst beds; action of catalyst: SO2 + V2O5

25.17

2SO3(g)

catalysed by Pd and Pt catalysed by Pd and Pt catalysed by Rh catalysed by Rh

(a)

2VO2 + SO3

+ 2VO2 J V2O5

N2(g) + 3H2(g)

2NH3(g)

ΔrHo = –92 kJ per mole of N2

The forward reaction is exothermic and proceeds with reduction in number of moles of gas; rate of formation of NH3 is slow. As in answer 25.16, a higher temperature decreases yield of NH3 and so is not a viable means of increasing the rate of formation of NH3. Increasing the external pressure increases yield of NH3 (application of Le Chatelier). Optimum conditions are 723 K and a high pressure in presence of Fe3O4/ K2O/SiO2/Al2O3 catalyst; the active catalyst is α-Fe. The catalyst facilitates dissociation of N2 and H2; diatomics are adsorbed on the catalyst surface and dissociate giving adsorbed N and H atoms which associate forming adsorbed NH, NH2 and finally NH3. NH3 then desorbs. The rate of formation of NH3 is sensitive to the metal catalyst. (b) The rate determining step is the adsorption of N2 (25.16). V (early d-block metal) tends to retain adsorbed N atoms, blocking surface sites and hindering NH3 formation. For Pt (late d-block metal), adsorption of N2 has a high ΔG‡ and the reaction is slow. Fe is relatively cheap, but Os (one of Pt group metals) is rare and expensive. (a) MgCl2 has a layer structure which matches that of the active titanium chloride catalyst; Ti4+ and Mg2+ are similar sizes. Coordinatively unsaturated Ti must be present in surface sites – a vacant lattice site must be cis to the surface Cl (replaced by alkyl group during catalysis) as shown at the left-hand side of Fig. 25.1. (b) Et3Al (used with the MgCl2 supported catalysts) and Et2AlCl are co-catalysts which alkylate the surface Ti (Fig. 25.1). (c) Part of the chain-growth (Cossee-Arlman mechanism) is shown in Fig. 25.1; surface alkylation, alkene addition cis to the alkyl group, followed by alkyl migration – then addition of the alkene again, alkyl migration and so on. Stereoregular products are formed; this is an essential feature of the commercial catalyst system.

Catalysis and some industrial processes

Et3Al

Cl Cl

Cl

Ti

Cl

Ti

Cl

Cl

Cl

C2H4

Cl

Ti

Cl

Cl

Cl

Alkyl migration Cl

Cl

355

Ti

Cl

Cl

Cl

Cl

Surface site with vacant coordination site

Fig. 25.1 Representation of alkene polymerization using an MgCl2-supported TiCl4 Ziegler-Natta catalyst.

25.19

(a) Metallocene catalysts are homogeneous. These are easier to study than the heterogeneous Ziegler–Natta catalysts. (b) RCH=CH2 + Zr

+ Zr

Me

Me R

Coordinatively unsaturated Methyl migration

RCH=CH2 + Zr

Alkyl migration

etc.

Me R Coordinatively unsaturated

25.20

Choose examples from Section 25.8 in H&S. Suggested plan for answer: • Heterogeneous catalyst is one which is in a different phase from the reaction it is catalysing, e.g. solid catalyst, gaseous reactants; examples: metal surfaces, alumina, silica, zeolites. • Fischer-Tropsch chain growth: CO/H2 converted to hydrocarbons using Fe or Co catalyst; only commercially viable when oil stocks are limited/expensive; adsorbed CO and H2 dissociate to give adsorbed atoms; adsorbed C and H combine to give CH and CH2 (H2O desorbed) and chains build up from association on surface between adsorbed (ad) units: H(ad) + CH2(ad) J CH3(ad) CH3(ad) + CH2(ad) J CH3CH2(ad) CH3CH2(ad) + CH2(ad) J CH3CH2CH2(ad) ... etc Combination of alkyl group and H or two alkyl groups release an alkane; βhydrogen elimination from adsorbed alkyl group releases an alkene. Vinyl groups may also be involved as surface species. • Ziegler-Natta catalysts: for alkene polymerization with stereoregular products, e.g. isotactic polypropene. • Haber process for NH3 production: see answer 25.17 for details. • Contact process for SO3 production (first step in H2SO4 manufacture): see answer 25.16 for details.

356

Catalysis and some industrial processes 25.21

Give general comments about structural properties of zeolites: aluminosilicates with macromolecular structures containing channels and pores of specific shapes and sizes depending on the zeolite. (a) Separation of n- and iso-alkanes follows from the ability of a zeolite to be ‘shape selective’, e.g. n-butane is a linear chain, but iso-butane is branched and more bulky. The pore size of 430 pm is selective in that zeolite 5A adsorbs nalkanes but iso-alkanes are too sterically demanding to enter the pores. (b) The Al sites in an aluminosilicate act as Brønsted acids: O

O Si O

Al O O

O Si O O H

Al

Si

O

O

O

O O O

O

Si O O H

O

and therefore zeolites are acid catalysts. The level of catalytic activity depends on the Al:Si ratio which varies with the zeolite. Zeolite ZSM-5 (Nan[AlnSi96–nO192]. ≈ 16H2O where n < 27) is used commercially as an acid catalyst for isomerization of 1,3-xylene to 1,4-xylene – isomer selectivity presumably controlled by shape selectivity of ZSM-5 (see part (a)). Conversion of benzene to ethylbenzene requires an acid catalyst (compare use of EtCl/AlCl3 in a Friedel-Crafts alkylation of C6H6). Adsorption of organic substrate onto surfaces within zeolite channels exposes substrate to highly active acid catalyst. 25.22

For full details, see Sections 25.7 and 25.8 in H&S. Suggested plan for answer: • a ‘3-way’ catalytic converter includes Rh/Pd/Pt as catalyst to facilitate oxidation of hydrocarbons and CO and reduction of NO (see answer 25.15b). • Metal particles supported on Al2O3 washcoat give high surface area; metal particle size important for optimal catalytic activity. • High operating temperatures degrade metal particles, causing Al2O3 to undergo phase change reducing its surface area. • Light-off temperature is the minimum at which the catalyst reaches 50% efficiency; during cold start, catalytic converter needs secondary heating. • Air-fuel ratio has to be optimized: < 14.7 : 1 means too little O2 for hydrocarbon and CO oxidation, while > 14.7 : 1 means too much O2 and competition for the H2 which is needed to reduce NO. • Storage of O2 achieved using CeO2 as the ‘storage vessel’: when O2 is needed, reduce CeO2 to Ce2O3; conversely, oxidize Ce2O3 to CeO2 to store excess O2.

25.23

(a) The ligand in the question is shown in 25.17. The ligand has been designed for use in Ru-based catalysts for hydrogenation reactions; 25.17 can coordinate to Ru through the P-donors. The ligand is chiral by virtue of the twist at bond C(a)–C(b) labelled in the diagram; the twist is caused by steric interactions between the bulky PPh2 groups. A Ru-based catalyst incorporating 25.17 is therefore suitable for asymmetric hydrogenations.

RO

RO

OR

OR

OR

RO

O

O NH

HN

a Ph2P

b PPh2

R = CH2C6H2-3,4,5-(OC10H21)3

(25.17)

Catalysis and some industrial processes

357

(b) BINAP is shown in 25.18. The question states that the hydrogenations are carried out in an EtOH/hexane solvent system which separates into two phases when a small amount of water is added. The functionalization on going from BINAP to 25.17 involves the incorporation of long aliphatic tails, designed to make the catalyst soluble in hexane. Partitioning the solvent into two phases allows the catalyst to be recovered after phase separation. Ph2P

PPh2

(25.18)

25.24

(a)

4NH3 + 6NO J 5N2 + 6H2O

Reduction: 6 × (+2 to 0) for NO to N2 Oxidation: 4 × (–3 to 0) for NH3 to N2

Therefore the redox changes balance. (b) A ‘first-generation’ Grubbs’ catalyst is shown in structure 25.19; it is a carbene complex. The general equation for ring closure metathesis is:

P(C6H11)3 Cl

Total = –12 Total = +12

Ru

Cl

Ph

RCM (ring-closing metathesis)

X

P(C6H11)3 C6H11 = cyclohexyl

(25.19)

+ C2H4 X

Starting from the tetraene given in the question, ring closure has to occur between a pair of alkenes at the left and right sides of the molecule, because ring closure between the two left-hand side alkenes would give an unfavourable 3-membered ring. The proposed reaction is therefore: Grubbs' catalyst

O

O + 2C2H4

O

A

25.25

O

(a) [Rh(nbd)(Ph2PCH2CH2PPh2)]+ is a Rh(I) complex, d8 and square planar – structure 25.20. Reaction with 2 equivalents of H2 reduces the coordinated diene to a bicyclic alkane, causing dissociation from the metal:

+ HPh2 P

2H2

Rh P HPh2

coordinated

(25.20)

(b) [Rh(Ph2PCH2CH2PPh2)(solv)2]+ contains Rh(I), d8 and is square planar, structure 25.21.

+ PPh2 Ph2P

Rh solv

(25.21)

solv

358

Catalysis and some industrial processes (c) The catalytic cycle, starting with substitution of solvent molecule for alkene, is: + PPh2 Ph2P

RCH2CH3

Rh

solv RCH=CH2

solv solv

16e Rh(I) solv

Ph2 H P solv Rh CH2 P Ph2 solv CH2R

+

+

PPh2 Ph2P

CH2

Rh

CHR

solv

16e Rh(I) Ph2 H P H Rh CH2 P Ph2 solv CHR

H2

+

18e Rh(III)

25.26

(a) Each Rh atom in complex 25.39 in H&S is an 18-electron centre (9 + 2 + 2 + 2 + 2 + 1) and, therefore, the complex is unlikely to be catalytically active. (b) The advantage of the complex over a mononuclear species is the presence of four potentially catalytically active sites in one species; cooperativity effects might be observed which might enhance the activity of the catalyst. (c) The hydroformylation of pent-1-ene could give aldehydes A, B and C below: CHO CHO CO, H2 catalyst

A

B

CHO C

Aldehydes A and B arise directly from pent-1-ene; A is a linear product, while B is branched; “selectivity for branched over linear aldehyde” indicates that B is formed in preference to A. C and further amounts of B follow after isomerization of the alkene to pent-2-ene. Formation of C adds to amount of branched isomers formed in the reaction. One could represent the selectivity as follows: y% z%

x%

Ratio branched : linear = (y + z) : x

Catalysis and some industrial processes 25.27

359

(a) Acetic acid is manufactured using the Cativa process. Acetic acid is precursor to acetyls, e.g. vinyl acetate which is used to manufacture polyvinylacetate (PVA). (b) HI required to convert MeOH (raw material) to MeI. (c) Water-gas shift reaction:

CO + H2O

iron oxide catalyst, 700 K

CO2 + H2

(d) From Fig. 25.22 in H&S:

A = [IrI3(CO)2H]– B = H2 C + D = CO2 + 2HI Conversion of [IrI2(CO)2]– to [IrI3(CO)2H]– is oxidative addition. Oxidation states and electron counts: [IrI2(CO)2]– Ir(I) 16 electron Ir Ir(III) 18 electron Ir [IrI3(CO)2H]– [IrI4(CO)2]– Ir(III) 18 electron Ir

25.28

PPh2 PPh2

(25.22)

25.29

(a) Production of aldehydes from alkenes (hydroformylation) uses homogeneous Co or Rh-based catalysts: HCo(CO)4, HCo(CO)3(PR3), HRh(CO)(PR3)3. Details in Fig. 25.11 in H&S, Table 25.5 in H&S and discussion. (b) Production of polypropene is Ziegler-Natta catalysis (heterogeneous) or group 4 (Ti, Hf, Zr) metallocene catalysts (homogeneous). Details in Section 25.8 in H&S. (c) Tennessee-Eastman process is used to manufacture acetic anhydride. Homogeneous catalysis using [Rh(CO)2I2]]–. See Fig. 25.10 in H&S and discussion. (d) Manufacture of Naproxen requires asymmetric hydrogenation of the alkene precursor (25.40 in H&S). Homogeneous catalysis using Ru{(S)-BINAP}Cl2 ((S)BINAP = 25.22); the chiral ligand makes the synthesis enantiomerically selective. (a) Scrubbing industrial waste gases to remove SO2 is acid-base chemistry and is not catalysis: SO2 + H2O J H+ + [HSO3]– H+ + [HSO3]– + 1/2O2 J 2H+ + [SO4]2– or

2H+ + [SO4]2– + Ca(OH)2 J CaSO4.2H2O 2H+ + [SO4]2– + CaCO3 J CaSO4.2H2O + CO2

See Box 12.2 in H&S for details. (b) Reduction of NO to N2 is catalysed (heterogeneous) by Rh as the emission gases pass through the vehicle’s catalytic converter: 2NO + 2H2 J N2 + 2H2O 25.30

(a) Examples of uses of sulfuric acid: • manufacture of superphosphate fertilizers (reaction of H2SO4 with phosphate rock, Ca3(PO4)2) is the major use; • see Fig. 16.3 in H&S for other uses. (b) Oxidation of sulfur to SO2 is the first step, followed by oxidation to SO3. Although thermodynamically favourable, the latter step is very slow under ambient conditions

360

Catalysis and some industrial processes and is catalysed by V2O5 on a SiO2 carrier with K2SO4 promoter. See Section 25.8 in H&S for complete details. The final step is the conversion of SO2 to H2SO4. Direct reaction: SO3 + H2O J H2SO4 is violently exothermic, and so SO3 is passed through oleum (concentrated sulfuric acid) and the mixture diluted with water to generate commercial H2SO4. (c) By-product sulfuric acid is the acid produced using SO2 produced as a byproduct during the roasting of metal sulfides to extract the metal, e.g.: CuS + O2 J Cu + SO2 The raw material is therefore not elemental sulfur as in the process described in part (b).

361

26 d-Block metal complexes: reaction mechanisms

(a) A complete reaction pathway from reactants to products usually consists of a series of elementary steps, e.g. in a free radical reaction, the pathway involves initiation, propagation and termination steps. (b) The elementary step with the highest activation energy is the rate determining (or slow) step. The molecularity of this step determines the observed kinetics of the overall reaction. (c) The activation energy, Ea, is Transition state the energy barrier that must be Transition state overcome for a reaction, or step Ea(1) Ea(2) in a reaction pathway, to proceed (see Fig. 26.1). Intermediate (d) An intermediate lies in a Reactants local energy minimum along a reaction pathway (Fig. 26.1); it can be detected (e.g. by Products spectroscopic methods) and, possibly, isolated. Reaction coordinate (e) A transition state lies at an Fig. 26.1 Reaction profile for a two-step energy maximum along a reactpathway; Ea = activation energy. ion pathway (Fig. 26.1) and cannot be isolated. (f) A rate equation shows the dependence of the rate of reaction on the concentrations of species in the reaction; usually expressed in terms of rate of loss of a starting material or rate of formation of a product. (g) For a reaction: A J products a zero order dependence on A means that the rate of reaction is not affected by variation in [A]: Energy

26.1

−

d[A] =k dt

where k = rate constant

The integrated form of the rate law is: [A] = [A]0 – kt

where [A]0 = [A] at time t = 0

If the rate is first order with respect to A: −

d[A] = k[A] dt

or

ln [A] = ln [A]0 – kt

If the rate is second order with respect to A: 1 1 d[A] or = + kt = k[A]2 [A] [A]0 dt (h) A nucleophile is a species which donates electrons (e.g. Cl–), being attracted to a centre such as a metal which is δ +. −

d-Block metal complexes: reaction mechanisms

kobs / s–1

362

26.2

The profiles should be of the same form as shown in Fig. 26.1 on the previous page. For the dissociative process, the reactants are MLxX and Y, and the intermediate is MLx (with Y still as a reactant and X having departed); the products are MLxY and X. For the associative process, the reactants are MLxX and Y, and the intermediate is MLxXY; the products are MLxY and X.

26.3

For further details, see Section 26.3 in H&S. Points to include and discuss are: • Negative values of ΔV‡ and ΔS‡. • Rate constants for substitution of H2O for Cl– in [PtCl4-x(NH3)x](x–2) are about the same. • Solvent effects (see eq. 26.16 in H&S). • Effects of steric demands of entering and leaving groups. • Retention of stereochemistry.

26.4

Reaction is substitution by py in [Rh(cod)(PPh3)2]+, and from the general rate law for a square planar complex: d[Rh(cod)(PPh 3 ) 2 + ] = k[Rh(cod)(PPh 3 ) 2 + ] + k 2 [Rh(cod)(PPh 3 ) 2 + ][py] dt where: kobs = k1 + k2[py]

50

−

40

Data suggest that the pathways are: k2

[Rh(cod)(PPh3)2]+ + py competes with:

30

0

0.02 0.04 0.06 [py] / mol dm–3

Fig. 26.2 Plot of the kinetic data for answer 26.4.

26.5

k1

[Rh(cod)(PPh3)2]+ + S

[Rh(cod)(PPh3)(py)]+ + PPh3

.

[Rh(cod)(PPh3)S]+ + PPh3

[Rh(cod)(PPh3)S]+ + py fast [Rh(cod)(PPh3)(py)]+ + S

Plot values of kobs against [py] (Fig. 26.2) and confirm a linear relationship between kobs and [py]. This corresponds to the equation for kobs given above. The gradient of the line in Fig. 26.2 = k2 = 322 dm3 mol–1 s–1; intercept = k1 = 25 s–1. (a) The chosen routes to cis and trans-[PtCl2(NH3)(NO2)]– depend on the relative trans-effects of the Cl–, NH3 and [NO2]– ligands. The relative abilities of the ligands to direct trans-substitution are [NO2]– > Cl– > NH3. To prepare the cis-isomer, NH3 must be introduced first so that Cl– directs substitution trans to itself and cis to NH3: Cl Cl

Pt

2– NH3

Cl

–

NH3

– Cl–

Cl

Cl

Pt

Cl

–

NH3 [NO2]– – Cl–

Cl

Pt

NO2

Cl

Cl

To prepare the trans-isomer, [NO2]– must be introduced first so that this ligand directs substitution trans to itself: 2–

Cl Cl

Pt

Cl

[NO2]– – Cl–

Cl

2–

NO2 Cl

Pt Cl

Cl

–

NO2 NH3 – Cl–

Cl

Pt NH3

Cl

d-Block metal complexes: reaction mechanisms

363

(b) The first step is formation of [PtCl3(PEt3)]–. Of Cl– and PEt3, PEt3 has the greater trans-effect, and directs substitution of the next PEt3 ligand trans to the first PEt3: Cl Cl

Pt

2–

–

PEt3 PEt3

Cl

–

Cl

Pt

PEt3

Cl

– Cl–

– Cl Cl

26.6

PEt3 Cl

Pt

Cl

PEt3

Cl

(a) Proposed associative mechanism for trans-[PtL2Cl2] going to trans-[PtL2ClY]+: L Cl

Pt

L

L Cl

Y Cl

Cl

Pt

– Cl

Cl

Pt

Y

Y L

L intermediate

L

(b) If the 5-coordinate intermediate is sufficiently long-lived, it may be stereochemically non-rigid in solution (Berry pseudo-rotation); this would prevent the stereoselectivity shown above and both cis and trans-isomers would form.

26.8

The two term rate law arises because there are two competing pathways (see eqs. 26.12 to 26.14 in H&S and accompanying discussion). A plot of kobs against [tu] is linear and k1 is the intercept (see Fig. 26.3b in H&S). The fact that the plot passes close to the origin means that the value of k1 is very small and therefore the k1 pathway (i.e. the solvent pathway) is not very important in this reaction. Use the data in the question to plot the Eyring plot shown on the right. Gradient = –5200 = −

0.00325 –3.0 ln (k2/T)

26.7 X See: R. Romero et al. (1974) Inorg. Chim. Acta, vol. 11, p. 231

ΔH ‡ R

–3.5

ΔH‡ = 5200 × 8.314 × 10–3 = 43 kJ mol–1 (to 2 sig. fig.)

For the activation entropy:

–4.0

–4.5

Intercept on the vertical axis when 1/T = 0 is +13.6 ‡ ⎛ k ' ⎞ ΔS 13 .6 = ln ⎜ ⎟ + R ⎝h⎠

⎛ 1.381 × 10 − 23 = ln ⎜ ⎜ 6.626 × 10 − 34 ⎝

⎞ ΔS ‡ ⎟+ ⎟ 8.314 ⎠

Δ S ‡ = (13 .6 − 23 .77 )× 8.314 = −84 .1 J K −1 mol −1

0.00345 (1/T) / K–1

364

d-Block metal complexes: reaction mechanisms [Co(NH3)5(OH2)]3+ + X– J [Co(NH3)5X]2+ + H2O

26.9

Rate law is:

d[Co(NH 3 ) 5 X 2+ ] = k obs [Co(NH 3 ) 5 (OH 2 ) 3+ ][X − ] dt

The fact that ΔV‡ is positive indicates a dissociative (D or Id) mechanism, but the the rate law suggests an associative mechanism. Application of Eigen-Wilkins mechanism rationalizes this apparent contradiction. An ‘encounter complex’ is formed between [Co(NH3)5(OH2)]3+ and X– in a pre-equilibrium step (equilibrium constant KE); H2O then leaves in the rate-determining step. Apply eqs. 26.26 to 26.33 in H&S to establish that second order kinetics (i.e. the rate equation above) hold at low concentrations of X– when KE[X–] H2O. Substitution of H2O by Cl– in 26.1 is directed by the coordinated Cl– and the product is trans-[RhCl2(OH2)4]+. For the next step, all the H2O ligands are equivalent and the product has to be the mer-isomer:

OH2

Cl

OH2

H2O

(26.1)

Rh

H2O

OH2 OH2

Cl

Cl– – H2O

H2O*

Cl Rh

H2O

OH2 Cl

Cl

In the last step, the site of substitution is controlled by the trans-effect of the Cl– ligand trans to a H2O ligand (marked by * above) giving trans-[RhCl4(OH2)2]–. (b) [RhCl5(OH2)]2– can be prepared from: or:

trans-[RhCl4(OH2)2]– + Cl– J [RhCl5(OH2)]2– + H2O [RhCl6]3– + H2O J [RhCl5(OH2)]2– + Cl–

+

Cl H2O

Rh

H 2O

Cl

Preparation of cis-[RhCl4(OH2)2]– cannot start from mer-[RhCl3(OH2)3] because of the stronger trans-effect of Cl– with respect to H2O. A suitable reaction is: [RhCl5(OH2)]2– + H2O J cis-[RhCl4(OH2)2]– + Cl–

OH2 OH2

Similarly, fac-[RhCl3(OH2)3] cannot be prepared by treating cis-[RhCl2(OH2)4]+ (26.2) with Cl– because the trans-effect of coordinated Cl– will direct substitution to give the mer-isomer. A suitable synthesis is:

(26.2)

cis-[RhCl4(OH2)2]– + H2O J fac-[RhCl3(OH2)3] + Cl– 26.11

Anation is the substitution of an uncharged ligand (here H2O) by an anionic ligand such as Cl–. Co, Rh and Ir are all group 9 metals, and M3+ is octahedral d 6 . CFSE increases down a triad consistent with the rate of substitution following the trend [Co(OH2)6]3+ > [Rh(OH2)6]3+ > [Ir(OH2)6]3+.

d-Block metal complexes: reaction mechanisms K

[Co(NH3)5X]2+ + [OH]–

26.12

k2

[Co(NH3)4(NH2)X]+

[Co(NH3)4(NH2)X]+ + H2O [Co(NH3)4(NH2)]2+ + X– fast

[Co(NH3)4(NH2)]2+ + H2O K=

365

[Co(NH3)5(OH)]2+

[Co(NH 3 ) 4 (NH 2 )X + ][H 2 O] [Co(NH ) X 2+ ][OH − ]

([H2O] ≈ 1 in dilute solution)

3 5

∴ [Co(NH 3 ) 4 (NH 2 )X + ] = K [Co(NH 3 ) 5 X 2 + ][OH − ]

As [Co(NH3)4(NH2)X]+ is formed, it is used in the rate-determining step, but is also reformed in the pre-equilibrium. The rate can be written in terms of kobs or k2:

(

d[Co(NH 3 ) 5 (OH) 2+ ] = k obs [Co(NH 3 ) 5 X 2 + ] + [Co(NH 3 ) 4 (NH 2 )X + ] dt = k 2 [Co(NH 3 ) 4 (NH 2 )X + ]

)

Substituting for [Co(NH3)4(NH2)X]+ from above and rearranging gives:

k obs =

=

∴

26.13

k 2 K [Co(NH3 ) 5 X 2+ ][OH − ]

[Co(NH3 ) 5 X 2+ ] + K [Co(NH3 ) 5 X 2+ ][OH − ] k 2 K [Co(NH3 )5 X 2+ ][OH − ]

[Co(NH3 ) 5 X 2+ ](1 + K [OH − ])

d[Co(NH3 ) 5 (OH) 2+ ] k 2 K [OH − ][Co(NH3 ) 5 X 2+ ]total = dt 1 + K [OH − ]

NR1R2R3 is trigonal pyramidal and, therefore, optically active; the left-hand diagram below shows enantiomers A and B. Inversion at N is generally facile and this process interconverts A and B (right-hand diagram) preventing resolution of enantiomers.

N R1

R3 R2

R3 R2

A

N

N R1

B

R1

R3 R2

A

R1 N

R2 R3

B

mirror plane

26.14

L– derived from HL (defined in question 26.14 in H&S) are acac-type ligands. Look at scheme 26.48 in H&S which shows racemization occurring through a 5coordinate intermediate. The chelating ligands in the question have the possibility of coordination through 3 different pairs of O,O′-donors. A mechanism involving dissociation of one end of chelate and reformation of Co–O bond will exchange C(O)CH3 and C(O)CD3 groups, allowing isomerization as well as racemization.

366

d-Block metal complexes: reaction mechanisms 26.15

From the reaction scheme:

k1

[Fe(OH2)6]3+ + [SCN]–

[Fe(OH2)5(SCN)]2+ + H2O

k –1

K1

K2 k2

[Fe(OH2)5(OH)]2+ + H+ + [SCN]–

[Fe(OH2)4(OH)(SCN)]+ + H+ + H2O

k –2

Let [Fe] represent [Fe(OH2)63+], and [Fe(SCN)] represent [Fe(OH2)5(SCN)2+]. −

d[SCN− ] = k1[Fe][SCN− ] − k −1[Fe(SCN)][H 2 O] dt + k 2 [Fe(OH2 )5 (OH)2+ ][SCN− ] − k −2 [Fe(OH2 ) 4 (OH)(SCN)+ ][H 2 O]

K1 =

[Fe(OH 2 ) 5 (OH) 2+ ][H + ] [Fe]

K2 =

[Fe(OH2 ) 4 (OH)(SCN)+ ][H+ ] [Fe(SCN)]

∴[Fe(OH 2 ) 5 (OH) 2+ ] =

K1[Fe] [H + ]

K 2 [Fe(SCN)]

∴ Fe(OH2 ) 4 (OH)(SCN)+ ] =

[H + ]

Make substitutions into the rate equation:

−

X Take [H2O] ≈ 1, which is approximately true for dilute solutions

d[SCN − ] = k1[Fe][SCN − ] − k −1[Fe(SCN)][H 2 O] dt ⎛ K [Fe] ⎞ ⎛ ⎞ ⎟[SCN − ] − k − 2 ⎜ K 2 [Fe(SCN)] ⎟[H 2 O] + k2 ⎜ 1 ⎜ [H + ] ⎟ ⎜ ⎟ [H + ] ⎝ ⎠ ⎝ ⎠ ⎛ ⎛ k K ⎞ k K = ⎜ k1 + 2 1 ⎟[Fe][SCN − ] − ⎜ k −1 + − 2 2 + ⎟ ⎜ ⎜ [H ] ⎠ [H + ] ⎝ ⎝

26.16

+

NH3 H3N H3N

Co

O O

C

O

H218O

+

NH3 H3N H3N

NH3

CO2

O

Co

⎞ ⎟[Fe(SCN)] ⎟ ⎠

H+

18

NH3

OH2

H

NH3 H3N H3N

3+

NH3 H3N H3N

Co NH3

OH2 18OH

2

H3N H3N

Co NH3

2+ CO2

18

NH3

OH2

2+

NH3 H+

Co

O

OH 18OH

– CO2 2

The unlabelled H2O is essentially H216O, i.e. natural abundance O. The scheme is consistent with H216O and H218O being present in the product in a 1:1 ratio.

.

d-Block metal complexes: reaction mechanisms

Look at the product enantiomers in Fig. 26.9 in H&S; each red line represents an edge of the octahedron involved in a chelate ring in M(L–L)3. Start with the product in Fig. 26.9b (shown in A below with a numbering scheme to match that in H&S). By rotating this structure about a horizontal axis in the plane of the paper (i.e. apex 5 moves towards you), the enantiomer can be drawn in a new orientation (diagram B). Now rotate this structure about a vertical axis in the plane of the paper to generate the structure in orientation C, i.e. a structure identical to that of the product enantiomer in Fig. 26.9a in H&S. Since rotations merely take the molecule from one orientation to another, the scheme below confirms that the enantiomers formed in Figs. 26.9a and b in H&S are the same. 5 5 5

6

4

4

4

6

6

A 26.18 SO3–

N

N

B

L2– (26.3) is an N,N′-donor, and the Fe(II) complex [FeL3]4– is a tris-chelate. The two sets of experimental data are the same within experimental error, meaning that the rate constants for dissociation and racemization are the same. To find ΔH‡ and ΔS‡, construct an Eyring plot. Figure 25.3 shows this for values of k r; an almost identical plot is obtained using kd data. ‡ (a) Gradient = –15 300 = − ΔH R

SO3–

(26.3)

∴

C -14

ln(k/T)

26.17

367

-15 -16 -17 -18 0.0032

0.0034

/T / K

1

ΔH‡ = 15 300 × 8.314 × 10–3 = 128 kJ mol–1

–1

Fig. 26.3 Plot of the racemization data for answer 26.18.

Extrapolation of the line in Fig. 26.3 allows you to find the intercept when 1/T = 0.

⎛ k ' ⎞ ΔS Intercept = 35.2 = ln⎜ ⎟ + R ⎝h⎠

‡

where k′ = Boltzmann constant h = Planck constant

⎧ 1.381×10 −23 ⎫⎪ ⎛ k ' ⎞⎫ ⎪⎧ ∴ ΔS‡ = R ⎨35.2 − ln⎜ ⎟⎬ = 8.314⎨35.2 − ln ⎬ = 95 J K–1 mol–1 ⎪⎩ 6.626 × 10 −34 ⎪⎭ ⎝ h ⎠⎭ ⎩ (b) kr = kd, so data are consistent with dissociative mechanism for racemization. 26.19 X Liquid NH3: see Sections 9.4 and 9.6 in H&S 26.20

In liquid ammonia, self-ionization gives: 2NH3 [NH4]+ + [NH2]– – KNH2 in liquid NH3 provides [NH2] and is analogous to the presence of [OH]– in aqueous solution. The fact that [NH2]– catalyses the displacement of Cl– by NH3 in [Cr(NH3)5Cl]2+ is therefore consistent with a conjugate base (Dcb) mechanism in which [NH2]– is the base, removing H+ from coordinated NH3. Reaction 26.51 in H&S provides an example of an inner-sphere mechanism: [Co(NH3)5Cl]2+ + [Cr(OH2)6]2+ + 5[H3O]+ J [Co(OH2)6]2+ + [Cr(NH3)5Cl]2+ + 5[NH4]+

d-Block metal complexes: reaction mechanisms

A

(b) Electron transfer in B has the highest Ea (most common). Energy

X R = reactants P = products A, B and C = transition states

Energy

(a) Bridge formation to form A has highest Ea. B C

P Reaction coordinate

X For more detail, see eqs. 26.59 and 26.61 in H&S

X See Fig. 26.10 in H&S

26.23

C

C

A B

R P

Reaction coordinate

P Reaction coordinate

Refer to Section 26.5 in H&S for details. Points to include: • Outer-sphere mechansm does not involve covalent bond formation between reactants; occurs when metal centres are kinetically inert; mechanism is important for long-range electron transfer in biological systems. • In an inner-sphere mechanism, electron transfer occurs between reactants across a covalently-bound bridging ligand. • Agreement with Marcus-Hush theory is test for outer-sphere mechanism – this relates data for self-exchange reactions and the corresponding cross-reaction. • Self-exchange reaction is electron transfer between like metal centres in different oxidation states, e.g. [Ru(bpy)3]3+ with [Ru(bpy)3]2+. For such a reaction, ΔGo = 0. Gibbs energy of activation, ΔG‡, for self-exchange is related to the rate constant by: k

26.22

A

R

R

26.21

B

(c) Bridge cleavage in complex C has highest Ea. Energy

368

⎛ − ΔG ‡ ⎞ ⎜ RT ⎟⎠ = κZe ⎝

where κ ≈ 1; Z ≈ 1011 dm3 mol–1 s–1

I and III are self-exchanges, and II is the corresponding cross-reaction. In I, k is relatively large and indicates fast electron transfer. A second row metal has a relatively large Δoct, and both Ru3+ (d 5 ) and Ru2+ (d 6 ) are low-spin, differing only in an extra non-bonding electron in a t2g orbital. For ground state Ru3+ and Ru2+ complexes, the Ru–N bond distances are similar. In III, [Co(NH3)6]3+ is low-spin d6, but [Co(NH3)6]2+ is high-spin d7 and has longer Co–N bonds. Electron-transfer between [Co(NH3)6]3+ and [Co(NH3)6]2+ occurs between vibrationally excited states having similar Co–N bond lengths, i.e. an ‘encounter complex’. The greater the changes in bond lengths needed to establish the encounter complex, the slower the rate of electron transfer. In III, the value of k indicates a relatively slow reaction. In cross-reaction II, Co3+ to Co2+ involves low to high-spin change. Unlike selfexchanges I and III where ΔGo = 0, the cross-reaction is assisted by the change in Gibbs energy of reaction. (a) When an electron transfer reaction is accompanied by ligand transfer, the proposed pathway is an inner-sphere mechanism. The steps in such a mechanism are bridge formation between the two metal centres by a ligand, electron transfer, and finally cleavage of the bridge. Examples of ligands that can form bridges in such reactions are halides, [CN]– and [NCS]–. See also answer 26.20. (b) A self-exchange reaction between low-spin Os(II) and Os(III) involves kinetically inert d6 metal centre. By the Franck–Condon approximation, the nuclei are essentially stationary during electron transfer between them; electron transfer can only occur between two vibrationally excited states with identical structures. This pair of structures is the ‘encounter complex’. Electron transfer between low-spin Os(II) and Os(III) involves two metal centres that differ only by one electron in the

d-Block metal complexes: reaction mechanisms

369

t2g set of orbitals (low-spin d6 versus low-spin d5); the Os–L bond lengths are quite similar. Since only small changes in bond lengths are needed to achieve the encounter complex, electron transfer is rapid. 26.24 X See also answer 26.5

(a) The first substitution step can only have one outcome. For the second substitution, take into account that the trans-effect of Cl– > NH3: Cl Cl

2–

Pt

NH3

Cl

–

NH3 Cl

– Cl–

Cl

Pt

Cl

NH3 NH3

Cl

– Cl–

NH3

Pt Cl

Cl

cis-Isomer

(b) cis-[Co(en)2Cl(OH2)]2+ is formed: +

N N

N

Co

N

2+

N – Cl–

Cl

N

N

Co

H2O

N N

Cl

N

2+

N

Cl cis-Isomer

Co

N

Cl OH2 cis-Isomer

(c) NO displaces H2O: [Fe(OH2)6]2+ + NO J [Fe(NO)(OH2)5]2+ + H2O 26.25 X For further discussion and literature citations, see B.J. Coe et al. (2000) Coord. Chem. Rev., vol. 203, p. 5, (Section 5.1 in this review)

+ N

N

(a) The reaction in the question occurs by a dissociative mechanism and therefore there is a 5-coordinate transition state: X OC

Cr

X CO

–L

CO

OC

OC OC

Cr

X CO CO

L

CO

OC

Cr

CO CO

OC CO

The reaction follows first order kinetics and the rate determining step is the loss of L. The rate constant k1 depends on X, varying with X in the order CO < P(OMe)3 ≈ P(OPh)3 < PnBu3. The enhanced rate for the phosphine ligands arises from stabilization of the transition state by these σ-donor ligands. (b) Ligand must undergo a change from trans,trans (with respect to the relative orientations of the pyridine rings) to a cis,cis-conformation to form complexes such as 26.4. For the reaction of 26.4 with pyridine (py), the expression for the rate constant is: kobs = k1 + k2[py] For substitution in a square planar complex, there are usually 2 competing pathways and in this case (where tpy = 2,2':6',2''-terpyridine, S = solvent) this will give: k2

[Pt(tpy)Cl]+ + py J [Pt(tpy)(py)]2+ + Cl– Pt

N

(26.4)

Cl

k1' [Pt(tpy)Cl]+ + S J [Pt(tpy)(S)]2+ + Cl– fast

[Pt(tpy)(S)]2+ + py J [Pt(tpy)(py)]2+ + S

Rate = k 2 [py][Pt(tpy)Cl + ] + k1 '[S][Pt(tpy)Cl + ] leading to a rate equation with two terms in the expression for kobs: = k 2 [py][Pt(tpy)Cl + ] + k1[Pt(tpy)Cl+ ]

370

d-Block metal complexes: reaction mechanisms 26.26

The reactant and product contain a tridentate P,P',P''-donor ligand and the reaction can be monitored by 31P NMR spectroscopy. The change from RS– to halide ligand will influence δ 31P and J(31P-31P). Electronic spectroscopy (UV-vis) could also be used to follow the reaction; each complex has a characteristic absorption spectrum. Although an experimental method might be chosen based on a spectroscopic property of a complex, its suitability for following a reaction to gain kinetics data will depend on the relative timescale of the experimental method compared to the rate of the reaction. If the reaction occurs too rapidly, it may not be possible to monitor the reaction by, e.g., NMR spectroscopy.

26.27

(a) The reaction to consider involves complex 26.5 with L = Me2SO going to 26.5 with L = py. Both complexes are square planar. The lack of any dependence on the incoming ligand suggests that the rate determining step is loss of Me2SO, i.e. a dissociative process. The positive value of entropy of activation supports a dissociative mechanism. (b) The following is an electron transfer reaction:

Me Ph3P

Pt

Me

L

[Co(NH3)X]2+ + [Cr(OH2)6]2+ + 5[H3O]+ J [Co(OH2)6]2+ + [Cr(OH2)5X]2+ + 5[NH4]+

L = Me2SO, py

(26.5)

Co(III) reduced to Co(II). Cr(II) oxidized to Cr(III). NH3 ligands are substituted by H2O. The reaction is accompanied by migration of X– from Co to Cr, suggesting that the reaction pathway is an inner-sphere electron transfer mechanism, with a Co–X–Cr bridge formed during the reaction. The rate determining step is electron transfer; electron transfer is faster across the iodide than the chloride bridge.

X RDS = rate determining step

[Co(NH3)5X]2+ + [Cr(OH2)6]2+ RDS [(NH3)5CoIII(μ-X)CrII(OH2)5]4+ [(NH3)5CoII(μ-X)CrIII(OH2)5]4+

[(NH3)5CoIII(μ-X)CrII(OH2)5]4+ + H2O [(NH3)5CoII(μ-X)CrIII(OH2)5]4+ [Co(NH3)5]2+ + [Cr(OH2)5X]2+ fast hydrolysis

[Co(OH2)6]2+ 26.28

Reaction is: CoIIIL5X + Y J CoIIIL5Y + X where L, X and Y are general ligands and the reaction is in aqueous solution. Two competing steps (ignoring overall charges as these are general ligands) with respective rate constants are: k1

CoIIIL5X

k2

CoIIIL5X

H2O –X Y –X

CoIIIL5(OH2) CoIIIL5Y

Y (fast)

CoIIIL5Y

dissociative associative

Substitution in octahedral complexes is more usually dissociative.Two other possible pathways involve ligand deprotonation (resulting from reaction with [OH]–) or protonation of X prior to it dissociating; let these two steps have associated rate constants of k3 and k4. The overall rate equation (assuming all steps are operative) is: Rate = k1[CoL5X] + k2[CoL5X][Y] + k3[CoL5X][OH–] + k4[CoL5X][H+] If the reaction is carried out under slightly acidic conditions, protonation of the leaving group will be a dominant pathway. If the reaction is carried out in alkaline solution, then the k3 step (conjugate-base mechanism) obviously becomes important.

d-Block metal complexes: reaction mechanisms 26.29 Cl H2N

Pt

Cl

371

(a) en is a chelating ligand, and so [Pt(en)Cl2] must have a cis-arrangement of Cl atoms (29.6). The ligand dach contains two stereogenic centres and so three stereoisomers (R,R), (S,S) and (R,S). Unless specified, the ligand dach can be assumed to be the racemate shown in 26.7.

NH2

H2 N

(26.6) (R)

H2 N

Cl (S)

Pt

(R)

N H2

Cl Pt

(S)

N H2

Cl

Cl

(26.7)

Methylcysteine is chiral, but no enantiomer is specified so assume a racemate is used. The ligand is in the deprotonated form, and the anionic complex is 26.8. O2C

O2C (S)

NH2

Cl

(R)

NH2

Pt S

Cl Pt

S

Cl

Cl

(26.8)

(b) L-Histidine and 1,2,4-triazole are biologically relevant ligands. (c) 310 K and pH 7.2 represent physiological conditions. (d) First, notice that in the table of data in the question, the multiplying factors with values of k are not all the same; if 103 k2 = 8.0 dm3 mol–1 s–1, then k2 = 8.0 × 10–3 dm3 mol–1 s–1. Values of the second order rate constants k2 and k4 characterize the direct nucleophilic attack of L-histidine and 1,2,4-triazole on the substrate. For the first step (k2), conclude that the order of reactivities for L-histidine is: [Pt(MeCys)Cl2]– >> cis-[Pt(NH3)2Cl2] ≈ [Pt(en)Cl2] > [Pt(dach)Cl2] and for 1,2,4-triazole is: [Pt(MeCys)Cl2]– >> cis-[Pt(NH3)2Cl2] > [Pt(en)Cl2] > [Pt(dach)Cl2] The fact that [Pt(MeCys)Cl2]– shows the greatest reactivity arises from the translabilization of the S-donor atom; this is consistent with the first nucleophile entering trans to S. For the three complexes with only N-donors, steric factors are important; steric effects are greatest for dach and least for NH3. The second step (k4) is slower than the first step (k2) in all eight reactions. The orders of reactivity for the second step are the same as for the first step for both entering ligands. Compared to the first step, the second step does not show such a markedly high reactivity for [Pt(MeCys)Cl2]–, and this is consistent with the second entering ligand going trans to N (rather than S), i.e. for the second step in each reaction, the incoming nucleophile enters trans to N.

372

d-Block metal complexes: reaction mechanisms Comparison of values of k2 for L-histidine and 1,2,4-triazole shows that 1,2,4triazole is a better nucleophile than L-histidine. Values of k1 and k3 tell you the extent to which the back reactions compete with the forward reactions.

373

27 The f-block metals: lanthanoids and actinoids

27.1

(a) Lanthanoids (Ln) are the 14 elements which follow lanthanum in the periodic table. They lie between La and Hf, and their valence electrons occupy the 4f level. Lanthanum, [Xe]6s 2 5d 1 4f 0 , is strictly not a lanthanoid but is often classed with these metals. Physical and chemical properties of the lanthanoids tend to be similar; chemistry is generally that of the +3 oxidation state. (b) See Section 27.5 in H&S for details. Points to include: • Ln3+ ion sizes are similar, as are their properties, so separation is difficult. • Separation is by solvent extraction using (nBuO)3PO or cation-exchange. A solution containing Ln3+ ions is poured on to cation-exchange resin column, where Ln3+ exchanges with H+ or Na+ ions; resin-bound Ln3+ removed using, for example, [EDTA]4– because formation constants of [Ln(EDTA)]– complexes (unlike those of resin-bound Ln3+ which are nearly constant) increase regularly from La3+ to Lu3+, and the complexes can be eluted preferentially.

27.2

Ce3+ is 4f 1 . For the f orbital, quantum number l = 3 and the single electron may occupy one of 7 atomic orbitals. For the metal ion, L = 3 (see answer 20.18 and apply the same method of working). For a single electron, spin quantum number S = 1/2 and the spin multiplicity is (2S+1) = 2. Quantum number J takes values (L+S) ... | L–S|; for L = 3 and S = 1/2, J = 5/2 or 7/2; the ground state has the lower value of J. The full term symbol for the ground state is 2F5/2. For calculating the magnetic moment, the spin-only formula is not applicable to lanthanoid ions. The appropriate formula is:

X See also Section 20.6 in H&S

μ eff = g J J ( J + 1)

gJ = 1+

S ( S + 1) − L ( L + 1) + J ( J + 1) 2 J ( J + 1)

⎛1 3⎞ ⎛5 7⎞ ⎜ × ⎟ − (3 × 4) + ⎜ × ⎟ 2 2⎠ ⎝2 2⎠ = = 1+ ⎝ ⎛5 7⎞ 2⎜ × ⎟ ⎝2 2⎠

μ eff = g J J ( J + 1) =

6

5

7

2

×

7 2

6 7

= 2.54 μ B

27.3 X See answers 6.17b and 11.10 for related lattice energy discussions

Consider the reaction: 3LnX2 J 2LnX3 + Ln For a given metal, the relative stabilities of LnX2 and LnX3 depend on the difference between the lattice energies of 3LnX2 and 2LnX3. For a given lanthanoid, the variable that is of importance is the anion radius (lattice energy is inversely proportional to internuclear separation) and the difference in lattice energies is smallest for the largest anion, i.e. I–.

27.4

A ‘saline’ iodide LnI2 would be of the form Ln2+(I–)2 whereas a ‘metallic’ iodide is formulated Ln3+(e–)(I–)2. The electrical conductivity of the metallic iodide is higher than expected for a saline formulation. Saline iodides form for Ln = Sm, Eu, Yb, and metallic diiodides for Ln = La, Ce, Pr, Gd.

374

The f-block metals: lanthanoids and actinoids 27.5

(a) The reaction to consider is: [Ln(OH2)y]3+ + [EDTA]4– J [Ln(EDTA)(OH2)x]– + (y – x)H2O

X See also Section 8.3 in H&S

The negligible variation in values of ΔHo for this reaction as Ln3+ varies arises because of the similarity in size of the metal ions across the first row of the fblock; given a constant set of ligands as here, and a fairly constant cation size, then the metal ion–ligand interactions will be similar. The fact that ΔHo ≈ 0 indicates that solvent–ligand and metal ion–ligand interactions are similar for both reactants and products. (b) In aqueous solution, the equilibrium to be considered is: Ce4+(aq) + e–

Ce3+(aq)

but in acidic solution (at constant pH), anions are present which may act as ligands to the metal centres. Complex formation by anions follows the series Cl– > [SO4]2– > [NO3]– > [ClO4]–, and the differences in coordinating ability affect the ease of reduction of Ce4+. (c) Perovskite is CaTiO3 – a mixed metal oxide (see Figs. 6.7 and 6.8, p. 89) containing Ca2+, Ti4+ and O2–. BaCeO3 must also be a mixed metal oxide and contain Ba2+, Ce4+ and O2– ; it is not a salt of type Ba2+[CeO3]2–. 27.6

For details, see Section 27.4 in H&S. Points to include: • Spin-orbit coupling is of greater importance than crystal field splitting; terms differing only in J value are sufficiently different in energy to be separated in the electronic spectrum; high values of L for some f n ions; a large number of electronic transitions possible, so a large number of lines in the spectrum. • 4f electrons are well shielded and little influenced by ligand environment (in contrast to d electrons); this leads to sharp lines associated with f-f transitions, with wavenumbers similar to those of the gas phase Ln3+ ions. • Weak absorptions reflect fact that probabilities of f-f transitions are low (there is little d-f mixing; mixing is needed for selection rule Δl = ± 1). • Absorptions due to 4f-5d transitions are broad and influenced by ligand environment.

27.7

For further details, see Sections 27.3 and 27.7 in H&S. Points to include: • Ln3+ ions are large and high coordination numbers (>6) are possible. • Coordination number is controlled by steric effects, not f n configuration. • Ln3+ ions are hard and favour hard donors such as O. • Coordination numbers typically range from 6 to 12; aqua ions typically 9coordinate; highest numbers tend to involve bidentate ligands, e.g. [NO3]–. • Low coordination numbers can be stabilized using e.g. [R2N]– ligands, but R must be sterically demanding.

27.8

First, note that [NCS]– could coordinate through N or S, but since Ln3+ ions are hard, N- rather than S-coordination is expected. [Ln(NCS)6]3– is likely to be octahedral. [Ln(NCS)7(OH2)]4– is 8-coordinate and could be dodecahedral, square antiprismatic, cubic or a distorted version of one of these; a hexagonal bipyramid is less likely for monodentate ligands; this structure tends to be imposed by a macrocyclic ligand (but see answer 27.13a). 7-Coordinate structures are hard to predict and [Ln(NCS)7]4– could be pentagonal bipyramidal, capped octahedral or a distorted version of one of these.

X See Figs. 20.8 and 20.9 in H&S

The f-block metals: lanthanoids and actinoids 27.9

375

(a) For further details, see Section 27.8 in H&S. Points to include: • Lithium alkyls generally used to introduce σ-bonded alkyls: 3LiR + LnCl3 J 3LiCl + LnR3

O

N

O O

THF

N

DME

In coordinating solvents (e.g. THF, DME or TMEDA), LnR3 can react with excess LiR to give [LiLn]x[LnR3+x] where L is the solvent. • Cp derivatives prepared by general reaction: 3Na[Cp] + LnCl3 J 3NaCl + Cp3Ln

TMEDA

or Cp2LnCl or CpLnCl2 formed if stoichiometry of reactants is altered. • η5-Cp mode usual, but complexes may be monomeric, e.g. Cp3Tm, or polymeric, e.g. Cp3Pr, but in presence of a donor solvent, monomeric product may be isolated, e.g. Cp3Pr(NCMe)2. (b) K2[C8H8] provides the [C8H8]2– ligand which can coordinate in an η8-mode, forming sandwich complexes. Reactions proposed are: SmCl3 + 2K2[C8H8] J 3KCl + K[(η8-C8H8)2Sm] SmI2 + 2K2[C8H8] J 2KI + K2[(η8-C8H8)2Sm]

27.10 Ce4+ + e–

Ce3+ Eo = +1.72 V (27.1)

O2 + 4H+ + 4e–

2H2O Eo = +1.23 V

(27.2)

27.11

(a) Fig. 27.9 in H&S shows potential diagrams for U, Np, Pu and Am. The Eo values show that Am3+ is much more difficult to oxidize than U, Np or Pu in oxidation states +3 to +5. The first step in separation could therefore be to oxidize U, Np, Pu to [MO2]2+ using Ce4+ (see 27.1) and precipitate Am(III) as AmF3 along with CeF3. To separate AmF3 from CeF3, Am3+ could be oxidized by [S2O8]2– and the product extracted. Alternatively, Am3+ could be separated from Ce3+ using a cation-exchange column, eluting with H4EDTA (see answer 27.1b). (b) NpO2(ClO4)2 contains [NpO2]2+. Zn amalgam is a good reducing agent and should reduce [NpO2]2+ to Np3+ (see Fig. 27.9 in H&S). At pH 0 (i.e. 1 M HClO4), O2 (see 27.2) should oxidize Np3+ to [NpO2]+ (with some being oxidized to [NpO2]2+) although oxidation might be slow. Solution X contains 21.4 g of U(VI) per dm3 ∴ Amount of U =

21.4 = 9.00 × 10–2 mol dm–3 238.03

After Zn amalgam reduction followed by O2 oxidation to Un+, the 25.00 cm3 aliquot is oxidized to [UO2]2+ by 37.5 cm3 0.1200 mol dm–3 Ce4+ (see 27.1). Amount of Un+ in 25.00 cm3 = 25.00 × 10–3 × 9.00 × 10–2 = 2.25 × 10–3 moles Amount of Ce4+ needed to oxidize Un+ to U(VI) = 37.5 × 0.1200 × 10–3 = 4.50 × 10–3 moles n+ 4+ Ratio of moles U : Ce = 1 : 2 The Ce4+ reduction to Ce3+ is a 1-electron process, and therefore the oxidation of Un+ to U(VI) must be a 2-electron process. Hence, Un+ is U4+. In the next set of reactions, 100 cm3 of X is converted to U4+ : Amount of U4+ = 0.1 × 9.00 × 10–2 = 9.00 × 10–3 moles Treatment with aq. KF precipitates UF4, deduced from the amount of substance: 2.826 Amount of UF4 = = 9.00 × 10–3 moles 238.03 + (19.00 × 4)

The f-block metals: lanthanoids and actinoids

376

UF4 is now heated at 1070 K with O2 and yields 1.386 g of product, an aqueous solution of which contains F–, precipitated as 2.355 g PbClF. Amount of F– = amount of PbClF =

2.355 207.19 + 35.45 + 19.00

= 9.00 × 10–3 moles ∴ Not all the fluorine present in the products from 9.00 × 10–3 moles of UF4 is present as F– in solution. Possible products from the reaction of UF4 with O2 are UF6 (a volatile solid which would vaporize at 1070 K) and UO2F2 (ionizes to [UO2]2+ and 2F– in solution). 9.00 × 10–3 moles F– could originate from 4.50 × 10–3 moles UO2F2. This would suggest that 4.50 × 10–3 moles UF6 were also formed, but were not collected in the solid product (see above). Check the amount of substance: Mass of 4.50 × 10–3 moles UO2F2 = 4.50 × 10–3 × {238.03 + (2 × 19.00) + (2 × 15.99)} = 1.386 g This matches that observed, and so the reaction of UF4 with O2 is: 2UF4 + O2 J UF6 + UO2F2 27.12

(a) F2 oxidizes U(IV) to U(VI) on heating: UF4 + F2 J UF6 (b) SOCl2 acts as a chlorinating agent, then H2 as a reducing agent: Pa2O5 + 5SOCl2 J 2PaCl5 + 5SO2 2PaCl5 + H2 J 2PaCl4 + 2HCl (c) H2 reduces UO3 on heating: UO3 + H2 J UO2 + H2O (d) UCl5 disproportionates on heating:

O– Me

Me

2UCl5 J UCl4 + UCl6 (e) NaOC6H2-2,4,6-Me3 is a source of an aryloxide ligand (27.3) which is sterically demanding, and reaction with UCl3 gives a 3-coordinate complex:

Me

UCl3 + 3NaOC6H2-2,4,6-Me3 J 3NaCl + U(OC6H2-2,4,6-Me3)3

(27.3)

27.13 H H

H

B

Np(BH4)3 H

(27.4)

(a) Cs2[NpO2(acac)3] contains Cs+ and [NpO2(acac)3]2– ions; [acac]– is bidentate, and [NpO2(acac)3]2– is 8-coordinate: hexagonal bipyramidal with axial oxido-ligands. (b) Np(BH4)4 contains 4 [BH4]– ligands which can coordinate through one, two or three H atoms (i.e. η1, η2 or η3-modes). Given that Np4+ can tolerate a high coordination number, suggest that Np(BH4)4 contains four η3-[BH4]– ligands (27.4). (c) Discrete guanidinium cations and [ThF3(CO3)3]5– anions are present. Planar [CO3]2– acts as a bidentate ligand, giving 9-coordinate [ThF3(CO3-O,O′)3]5–. (d) [Li(DME)]3[LuMe6] contains [Li(DME)]+ and octahedral [LuMe6]3– ions. Likely to be cation-anion interactions as shown in Fig. 27.10a in H&S.

The f-block metals: lanthanoids and actinoids

(e) Sm{CH(SiMe3)2}3 is a 3-coordinate organometallic complex with sterically demanding R groups. Trigonal planar might be expected, but by analogy with similar complexes, trigonal pyramidal is possible in the solid state. (f) By analysis, the compound is [UO2][CF3SO3]2.2(18-crown-6).5H2O. It is reasonable to assume [CF3SO3]– ions are present. This leaves [UO2]2+ which will not be present as a discrete entity. It will have trans-oxido ligands and could be coordinated by the 18-crown-6 (27.5) giving a hexagonal bipyramidal complex with 5H2O of crystallization, or by the H2O ligands with the crown ether as solvate of crystallization. The complex cation is actually [UO2(OH2)5]2+.

O O

O

O

O O

(27.5)

27.14

(a) Step 1: gain a neutron: A = 239U; step 2: lose a β-particle: B = 239Np; step 3: lose a β-particle: C = 239Pu. (b) Step 3: lose a β-particle: F = 242Cm; step 2: gain a neutron: E = 241Am; step 1: lose a β-particle: D = 241Pu.

27.15

Balance each equation using the values of Z and mass numbers.

27.16 O

377

2+

U O

(27.6)

(a)

253 4 99 Es + 2 He

→

256 1 101Md + 0 n

(b)

244 94 Pu

+

16 8O

→

255 1 102 No + 5 0 n

(c)

249 98 Cf

+

11 5B

→

256 103 Lr +

(d)

248 18 96 Cm + 8 O

261 → 104 Rf + 501 n

(e)

249 98 Cf

→

+

18 8O

263 106 Sg +

4 01 n

4 01 n

(a) The formulae suggest Th(II), Th(III) and Th(IV) compounds, but ThI2, ThI3 and ThI4 are all Th(IV) iodides. ThI4 is a saline halide; ThI2 and ThI3 are metallic halides formulated as Th4+(I–)2(e–)2 and Th4+(I–)3(e–). See also answer 27.4. (b) In the solid state, discrete [UO2]2+ (27.6) ions are not present but are complexed by further ligands around the equatorial plane, e.g. [UO2(OH2)5]2+. (c) UCl4 + 4NaOR J 4NaCl + U(OR)4 Product is usually of the form U(OR)4 only if R is very bulky, e.g. R = 2,6-tBu2C6H3; with less bulky R groups, donor solvents, L, may coordinate giving e.g. U(OR)4Lx.

27.17

(a) In each example, NaCl or LiCl is eliminated as the driving force for reaction: (i)

(η5-Cp)3ThCl + Na[Ru(CO)2(η5-Cp)] J NaCl + (η5-Cp)3Th–Ru(CO)2(η5-Cp)

(ii)

(η5-Cp)3ThCl + LiCHMeEt J LiCl + (η5-Cp)3ThCHMeEt

(iii)

(η5-Cp)3ThCl + LiCH2Ph J LiCl + (η5-Cp)3ThCH2Ph

(b) (η5-Cp)2ThCl2 can undergo a redistribution reaction: 2(η5-Cp)2ThCl2

THF

(η5-Cp)3ThCl + (η5-Cp)ThCl3(THF)2

Replacing H atoms by Me groups gives the more sterically hindered compound (η5-C5Me5)2ThCl2 which does not undergo an analogous redistribution reaction.

378

The f-block metals: lanthanoids and actinoids (c) Elimination of KI occurs; [C8H8]2– coordinates in an η8-mode. It is likely that the coordination sphere will be too crowded if 3 THF ligands are retained. Propose that the product is (η5-C5Me5)(η5-C8H8)U(THF)x with x = 1 or 2. In practice, x = 1. 27.18

(a) This is a U(IV) complex and so start with UCl4. Treatment with the Grignard reagent C3H5MgCl in an ether solvent: UCl4 + 4C3H5MgCl

Et2O

4MgCl2 + U(η3-C3H5)4

(b) Treatment with H+ removes propene, leaving a vacant coordination site to be filled by Cl–: U(η3-C3H5)4 + HCl J U(η3-C3H5)3Cl + CH3CH=CH2 (c) In dimer 27.7, the presence of the bridging chlorido ligands increases the Th coordination number with respect to a monomer (η5-C5Me5)(η8-C8H8)ThCl. In the presence of a coordinating solvent such as THF, the monomer is stabilized because the solvent can fill a vacant coordination site as in 27.8. This is a general strategy for stabilizing monomeric organothorium and uranium complexes.

Cl Th O

(27.8)

27.19

Cl Th

Th Cl

(27.7)

(a) In a non-stoichiometric oxide, vacant O2– sites need to be countered by an appropriate decrease in oxidation state of some metal ions so that the solid retains electrical neutrality. An excess of O2– ions can similarly be offset by the presence of some higher oxidation state metal ions. For actinoids, a range of oxidation states with similar stabilities is available, making the formation of non-stoichiometric oxides common. For lanthanoids, the +3 oxidation state dominates, with the result that variation in oxidation state is not favourable. (b) [NpO6]5– contains Np(VII). This needs to be stabilized against reduction which involves loss of oxido ligands in the presence of H+, e.g. [NpO6]5– + 8H+ + e–

[NpO2]2+ + 4H2O

In strongly alkaline solution, no H+ is available, and [NpO6]5– is stabilized with respect to reduction. (c) Fig. 27.9 in H&S shows that aqueous Pu 4+ should contain significant concentrations of disproportionation products. The fact that this does not occur in the presence of an excess of molar H2SO4 indicates that [SO4]2– must stabilize Pu4+ more than it does [PuO2]2+, [PuO2]+ or Pu3+ – a consequence of charge effects. 27.20

Refer to Sections 27.8 and 27.11 in H&S for specific details. Major differences between d- and f-block organometallics to highlight are: • CO complexes are extremely common within the d-block, but rare within the f-block. • Large size of the f-block metals allows higher coordination numbers, e.g. (η5-Cp)4Th but (η5-Cp)2Fe, and involvment of larger organic ligands in metallocenes, e.g. (η8-C8H8)2U.

The f-block metals: lanthanoids and actinoids 27.21

R S

O

R R = C6H11

(27.9)

X Hard and soft metal ions and ligands: see Section 7.13 in H&S 27.22 NMe NMe N Me

Sc

Cl

379

(a) Most of the chemistry of the lanthanoid metals involves the +3 oxidation state. Ln2+ is known for Eu, Sm and Lu; oxidation to M3+ occurs readily, e.g. Eo for Sm3+/Sm2+ = –1.5 V. Thus, the M2+ ions are strong reducing agents. (b)The difference in solid-state structures between members of the Cp2YbX(THF) family (X = F, Cl, Br) can be attributed to steric effects. Dimer formation through bridging X atoms increases the coordination number of the metal centre when compared to the corresponding monomer. For small F, dimer formation is possible; for larger Cl and Br, the coordination sphere of the Yb(III) centre is not sufficiently uncrowded to allow the expansion of the coordination number. Hence monomers are observed for Cp2YbCl(THF) and Cp2YbBr(THF). (c) Sulfoxide ligand 27.9 can coordinate through O- or S-donor and linkage isomerism can be observed. [Th(NO 3 -O,O') 3 {(C 6 H 11 ) 2 SO} 4 ] +[Th(NO 3 O,O')5{(C6H11)2SO}2]– contains Th(IV), i.e. hard metal centre. This prefers to bind hard rather than soft donor atoms; O-donor is hard, S-donor is soft, therefore ligand 27.9 preferentially adopts an O-bonded mode. (a) The ligand is a small macrocyclic ligand that, in an octahedral complex, is restricted to binding fac. Complex A is therefore likely to be fac-[LScCl3] (27.10). The reaction with MeLi eliminates LiCl: fac-[LScCl3] + 3MeLi J fac-[LScMe3] + 3LiCl

Cl Cl

Compound B is therefore fac-[LScMe3]. The oxidation state of Sc in each complex is +3; this is the common oxidation state for scandium, yttrium and the lanthanoid metals in their compounds. (b) For [(η 5 -Cp) 2 La{C 6 H 3 -2,6NMe2 (CH2NMe 2) 2}], assume each η 5-Cp – NMe2 ligand occupies one site in the 5Cp La coordinate complex (see Box 19.1 in Cp H&S for notation). The structure of the NMe – 2 [C 6 H 3 -2,6-(CH 2 NMe 2 ) 2 ] ligand is NMe2 shown in 27.11; it has the potential to be (27.11) (27.12) a tridentate ligand. Propose structure 27.12 with the Cp– ligands occupying equatorial sites. The bite size of the tridentate ligand restricts it to an axial-equatorial-axial coordination mode.

(27.10)

27.23 X See worked examples 27.1 and 27.2 in H&S

X For further reactions, see M.P. Wilkerson et al. (1999) Inorg. Chem., vol. 38, p. 4156

(a) A value of μeff can be calculated from:

μ eff = g J J ( J + 1)

gJ = 1+

S ( S + 1) − L ( L + 1) + J ( J + 1) 2 J ( J + 1)

S = 6/2 = 3 L = 3 Eu3+ , f 6 : Sub-shell is less than half-filled, therefore J = (L – S) = 0. Since J = 0, μeff = 0. Experimental values of μeff > 0; Eu3+ has low-lying excited states (7F1 and 7F2) which are thermally populated to some extent and give rise to observed μeff. (b) Loss of labile THF from O [UO2Cl2(THF)3] gives a coordinatively O O O unsaturated U(VI). Thus, dimer 27.13 can Cl U Cl Cl U form. This is A. Re-formation of the Cl O O monomer is a pathway to B and C in which O O U(VI) is octahedrally sited and contains trans-oxido ligands: (27.13)

The f-block metals: lanthanoids and actinoids

380

tBu

O O U

O Cl

2K[O-2,6-tBu2C6H3]

Cl

O

O U

– 2KCl

O

O

O

2Ph3PO – 2THF

O Ph3PO

Et2N

N

NEt2

(27.14)

Bu

t

Bu

B

tBu

Cl U Cl O

NH

t

O

Ph3PO

27.24

O

C

(a) Ligand 27.14 is HL. Reaction with MeLi deprotonates HL, forming LiL (A) and CH4. Reaction of TbBr3 with LiL gives B (highest mass peak in its mass spectrum = m/z 614). B = TbBr2C17N4H35 = [TbBr2L]. From Appendix 5 in H&S, you can see that Tb is monotopic (only one isotope); Br has 2 isotopes, 79Br and 81Br in ≈ 1:1 ratio. The appearance of the envelope of peaks at m/z 614 reflects the presence of these isotopes: [Tb(79Br)2L], [Tb(81Br)2L] [Tb(79Br81Br)L]. (b) The ligand is 27.15 and is potentially Ph Ph Ph Ph tridentate: hard O,O',O''-donor set matches P P hard Pu(IV). In [Pu(27.15)2(NO3)2]2+, a O N+ O coordination number of 10 is achieved if – O 27.15 is tridentate and [NO3]– is bidentate. (27.15)

27.25

(a) n = 6 (five CO2H, one OPO3H). (b) The diphenylcyclohexyl group is lipophilic, i.e. enables the molecule to dissolve in fats, hydrocarbon solvents etc. (c) Possible 9-coordinate geometry is a tricapped trigonal prism (or distorted analogue). The donor atoms are six carboxylate O, three N, plus O from an H2O molecule (27.16). (d) HnL contains a stereogenic centre (27.17). On OR" coordination to Gd3+, the central N atom becomes stereogenic. The Δ and Λ enantiomers refer to the R'RN * N(CH2CO2H)2 manner in which Ln– wraps around Gd3+. Thus (R)L n– can form ΔR,R, ΛR,R, ΔR,S and ΛR,S (27.17) diastereoisomers. (f) For M = Na3GdL(H2O): m/z 981 = [M – H2O + Na]+; m/z 959 = [M – H2O + H]+; m/z 937 = [M – H2O – Na + 2H]+; m/z 915 = [M – H2O – 2Na + 3H]+.

27.26 X For more details for the answer, see C. A. Barta et al. (2007) Dalton Trans., p. 5019

(a) Hard ligand donor atoms such as O are compatible with hard Ln3+ ions. (b) Base deprotonates phenolic OH to give anionic ligand, [ma]–. (c) For La3+ and Eu3+ complexes, peaks at m/z 537 and 551, respectively, arise from [M + Na]+ where M is Ln(ma)3. (d) Mr for Eu(ma)3 is 527.3, and %C = 41.00, %H = 2.87. The experimental data are low in C and high in H. Try a hydrate: Eu(ma)3.H2O: %C = 39.65, %H = 3.14 which fit the experimental results. (e) La3+ has an electronic configuration of f0 and is diamagnetic. The remaining ions are paramagnetic. (f) Bone is made up of collagen and hydroxyapatite, Ca5(PO4)3(OH). Binding studies of the Ln3+ complexes with hydroxyapatite allow a study of Ca2+/Ln3+ exchange.

N N

N O O

O

O

O O H2

(27.16)

X See Box 15.11 in H&S

381

28 Inorganic materials and nanotechnology 28.1

At the anode: At the cathode:

Solid AgI at 450 K Ag anode

Diagram 28.1 shows the electrical circuit described in the question; AgI remains solid at 450 K. When current flows, the observation that one electrode gains in mass and one loses mass is explained by:

Ag cathode

(28.1)

Ag J Ag+ + e– Ag+ + e– J Ag

electrode mass loss electrode mass gain

The Ag+ ions must be transported through solid AgI, illustrating the property of an ion conductor, i.e. AgI conducts electricity by carrying Ag+ ions rather than electrons.

28.2

At 300 K, NaCl is solid. The very low conductivity typifies an ionic solid which is an insulator. In the NaCl lattice, the ions lie in planes; planes perpendicular to one another have the same structure and so the conductivity is not direction dependent. For the other solids, conductivities are orders of magnitude higher than for NaCl indicating that they are electrical conductors. Na β-alumina is Na+ ion conductor; spinel-type layer structure with Na+ ions in interlayer channels; conducts only in the direction of these channels, so conductivity is direction dependent. Li3N has a layer structure with Li+ ions occupying sites within and between layers. A deficiency of Li+ ions within layers leads to Li+ ion conduction in a direction parallel to layers; Li+ ions between the layers are not involved in conduction; conductivity is direction dependent.

28.3

Battery construction: Li and V6O13 electrodes separated by solid polymer electrolyte. Consider first the oxidation of Li to Li+. Then Li+ ions transported across solid electrolyte to V6O13 where they are reversibly intercalated: xLi+ + V6O13 + xe–

LixV6O13

Charging and discharging of the battery corresponds to Li+ ions moving in opposite directions across electrolyte. 28.4

TiO, VO, MnO, FeO, CoO and NiO crystallize with NaCl structures but are nonstoichiometric, e.g. Fe0.96O-FeO0.89. In TiO and VO, there is overlap of metal t2g orbitals giving a partially occupied band E and allowing electrical conduction. Unoccupied Crossing the d-block leads to increase band in effective nuclear charge, with electrons tending to be localized on the Metal conduction metal centres. This rationalizes why band MnO is an insulator at 298 K. FeO, CoO and NiO also show low conductivities Oxygen 2p valence band at 298 K. Raising the temperature increases the conductivities because of ‘electron-hopping’ from M2+ to M3+, the latter being present because of nonstoichiometry. In the presence of O2, heating the oxide converts more M2+ to M3+, and more positive holes lead to higher electrical conductivity.

382

Inorganic materials and nanotechnology 28.5

(a) In a degenerate semiconductor, the Fermi level has moved into the conduction band of the host resulting in metallic-like electrical conductivity. In a TCO (transparent conducting oxide), the band gap must be ≥3 eV. (b) A TCO is a semiconducting material with a high electrical conductivity and a high optical transparency. Examples: tin-doped In2O3 (ITO) in which charge carriers are created by replacing In(III) by Sn(IV), and fluorine-doped SnO2 (FTO) in which charge carriers arise when O 2– replaced by F –. The charge carrier concentration is high (e.g. in ITO, the concentration of Sn(IV) centres is ca.1020 cm–3).

28.6

(a) Details of dye-sensitized solar cells (DSCs) are given in Section 28.3 in H&S. Points to include: • nanocrystalline anatase (TiO2) is an optically transparent semiconductor and therefore a dye that absorbs in the visible range is needed to absorb photons; • DSC comprises two glass plates, each coated with a thin film of a TCO; • one plate is the working electrode and is coated with a mesoporous TiO2 nanoparticles + dye; • the second plate is the counter electrode and is coated with thin Pt film; • between the two plates is the electrolyte (e.g. I–/I3– redox couple); • upon excitation, the dye injects an electron into the conduction band of the semiconductor; the reduced state of the dye is regenerated using the electrolyte redox couple; • draw a diagram of the DSC, and comment on the redox cycles involved and any competitive processes; • DSC converts solar energy to electrical energy without net chemical change in the cell. (b) Wide range of absorption maxima with high εmax is needed to ensure a highly efficient dye, i.e. maximum number of photons absorbed.

28.7

An exciton formed on electron/hole recombination (see answer 28.8) is in either a singlet (S = 0) or triplet (S = 1) state; statistically, the singlet : triplet ratio is 1 : 3. In an OLED comprising only π-conjugated organic materials, only the singlet state excitons decay radiatively (fluorescence) and the triplet states are wasted because a change in spin multiplicity (to give phosphorescence) is very inefficient. The problem can be overcome by incorporating neutral, heavy-metal complexes into the host organic matrix. The complex is designed so that the triplet state of the inorganic complex lies at lower energy than that of the organic material. This provides a pathway for the triplet excitons from the matrix to migrate to the metal complex molecules. The metal must be a heavy d-block metal so that spin–orbit coupling allows mixing of singlet and triplet states, and the triplet excitons can then undergo radiative decay.

X Fluorescence and phosphorescence: see Section 20.8 in H&S

28.8

(a) Electroluminescence is the emission of light which results when a direct current is passed through the material and electrons (injected from the cathode) and holes (injected from the anode) recombine. (b) The diagram should be similar to Fig. 28.9a in H&S, with the label given as “electroluminescent material” altered to illustrate that this material is an organic host coupled with the iridium(III) complex shown in the question. This complex contains a heavy metal as required to access the triplet states of the emitter (see answer 28.7). Functions of the layers:

Inorganic materials and nanotechnology

383

• Al cathode (with low work function): electrons are injected from the cathode into the LUMO of the organic matrix in the electroluminescent material; • electroluminescent material consists of organic matrix and inorganic complex, the combination of which allows one to harvest emission from both singlet and triplet excited states (see answer 28.7); • at the ITO anode, electrons are removed from the HOMO of the electroluminescent material. See Section 28.3 in H&S for complete details. 28.9 X Look at Figs. 28.12 and 28.13 in H&S

(a) YBa2Cu3O7 is high-temperature superconductor; consists of stacked perovskitelike units cells. Prototype perovskite is CaTiO3 – compare YBa2Cu3O7 to CaTiO3 by making replacements: Ba2+ and Y3+ for Ca2+, and Cun+ for Ti4+. Lattice is oxygendeficient. Structure described in terms of CuO2-BaO-CuO2-Y-CuO2-BaO-CuO2 layers, with non-CuO2 oxide layers isostructural with layers in NaCl. Combination of these relationships leads to the description given in the question. (b) MRI scanners use NbTi (Tc = 9.5 K) multicore conductors. Although replacement by high-Tc superconductors would be financially beneficial, problems must be overcome: (i) costs of cooling material to Tc; situation improves as higher values of Tc are obtained; NbTi requires He coolant, but if Tc > 77 K, liquid N2 could be used; (ii) bulk cuprate superconductors lose their superconductivity after material has carried only limited amount of current; specialized fabrication techniques required.

28.10

See Fig. 28.16 in H&S and accompanying discussion of Chevrel phases.

28.11

(a) MgO doping of ZrO2 parallels doping using CaO (refer to answer 6.27b). (b) CaF2 is doped with LaF3 to form Ca1–xLaxF2+x containing excess F– which occupy interstitial sites in the host lattice. Ca1–xLaxF2+x is an F– ion conductor. (c) Doping Si with B introduces electron deficient sites giving unoccupied level in band structure; band gap between this and occupied band beneath it is small, making thermal population of acceptor level possible. Positive holes left in valence band when electrons move to acceptor level act as charge carriers; gives p-type (extrinsic) semiconductor. (d) Doping Si with As introduces electron rich sites; extra electrons occupy discrete level below conduction band with small band gap which allows thermal population of the conduction band; gives n-type (extrinsic) semiconductor.

28.12

(a) FeSe adopts a PbO layer structure (Fig. 28.1). Each atom is 4-coordinate; each Se is tetrahedral and each Fe is in the apical site of a square-based pyramid (4 Se in the basal sites). The Se atoms reside in the middle of each layer, and the Fe atoms on the upper and lower surfaces. (b) Se is isoelectronic with As–. NaFeAs has a layer structure in which [FeAs]– layers are structurally like FeSe, but lie further apart with Na+ ions in between the [FeAs]– layers. (c) LaOFeAs consists of alternating [La2O2]2+ and [Fe2As2]2– layers. This compound is not superconducting, but F-doping in the La2O2 layers to give LaO1–xFxFeAs introduces charge carriers into the Fe2As2 layers. 5–11% of the O2– sites need to be doped with F– for the onset of superconductivity.

Fig. 28.1 Part of one layer of FeSe (PbO structure-type). Fe, small atom; Se, larger atom.

28.13

(a) Graphite has a layered structure, in which adjacent planar layers of edge-sharing C6-rings are staggered with respect to each other (Fig. 14.4 in H&S). Each C atom is sp2 hybridized and bonding is delocalized. The structure of MgB2 consists of

384

Inorganic materials and nanotechnology planar layers of edge-sharing B6-rings; layers lie directly over one another and are separated by sheets of close-packed Mg atoms. Electron transfer from Mg to B leads to the boron framework being formally negatively charged. Describe the bonding in terms of band theory: MgB2 has two π-bands, one occupied (electron type) and one unoccupied (hole type). In graphite, the σ-bands are lower than the Fermi level, whereas in MgB2, the σ-bands cross the Fermi level producing holes in the σ-bands. 28.14

(a) I– oxidized to I2; some V(V) reduced to V(IV); Li+ ions intercalated in host vanadate: xLiI + V2O5

Δ

LixV2O5 + x/2I2

(b) Formation of [WO4]2– (no redox involved); CaWO4 is the mineral scheelite: CaO + WO3

Δ

CaWO4

(c) Products are mixed metal oxides – do not contain discrete anions: 4SrO + Fe2O3 28.15

Δ, O2

2Sr2FeO4 (or SrFeO3)

(a) BiCaVO5 contains Bi(III), Ca(II), V(V); reagents should have same ox. states: Bi2O3 + 2CaO + V2O5 J 2BiCaVO5 (b) If CuMo2YO8 contains Mo(VI), then Cu is Cu(I) since Y has to be Y(III): Cu2O + 4MoO3 + Y2O3 J 2CuMo2YO8 (c) Li3InO3 contains Li+ and In(III): 3Li2O + In2O3 J 2Li3InO3 (d) The only available ox. state for Y is Y(III), therefore Ru2Y2O7 contains Ru(IV): 2RuO2 + Y2O3 J Ru2Y2O7

28.16

Choose a CVD process from Section 28.6 in H&S, and give a sketch of the apparatus used. For uses in the semiconductor industry, include: • High-purity Si from HSiCl3 and H2, or thermal decomposition of SiH4. • B-doped Si produced by depositing film of α-BN on Si wafer and heating to facilitate diffusion of B atoms into Si. • Formation of high purity SiC from e.g. SiCl4 and NH3. • III-V semiconductors: e.g. GaAs from Me3Ga and AsH3

28.17

(a) Wear-resistant coatings (e.g. for high-speed cutting tools) include Al2O3 (deposited from AlCl3, CO2 and H2), nitrides of Ti, Hf and Zr (from volatile metal chloride, H2 and N2), and TiC (deposited using TiCl4, CH4 and H2). (b) GaAs (intrinsic semiconductor) deposited from volatile Me3Ga and AsH3. (c) Only light source for glass reflectors is car headlights; visibility depends on weather conditions. Energy source for LEDs includes sunlight; advantage of LEDs is that energy can be stored, and enhanced intensity of emission gives better visibility range.

X For more details for parts (a), (b) and (d), see Section 28.6 in H&S

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385

(d) BaTiO3 deposited using Ti(OiPr)4 and Ba2+ β-diketonates; volatile complexes are difficult to find; F-substituents introduced to increase volatility, but films are contaminated with F. For YBa2Cu3O7 formation studies have included β-diketonate complexes of Ba2+, Cu2+ and Y3, but no results of commercial significance yet available. 28.18

Structure 28.2 shows part of a graphene sheet; the atoms shown as filled circles make up one triangular sublattice and the open circles comprise an identical sublattice. Interpenetration of the two sublattices generates the graphene sheet. The dotted lines define the unit cell; there are two axes only (not a 3D lattice) since the graphene sheet is planar. This repeat unit contains two C atoms, one from each sublattice.

(28.2)

28.19

Ways of solubilizing graphene: • acid oxidation (e.g. KMnO4 and H2SO4) gives oxidized graphene sheets that contain carboxylate, epoxy and hydroxyl substituents and this leads to hydrophilic sheets, soluble in a variety of solvents; • using pyrene or perylene solvents functionalized with hydrophilic substituents; pyrene and perylene are extended aromatic systems which solubilize graphene via face-to-face π-stacking interactions.

28.20

Three types of edge structure. Cutting the sheet in one of two directions (28.3) gives zigzag or armchair edge structures, but reconstruction of the edge can produce an edge structure made up of pentagons and heptagons (28.4).

armchair zigzag

28.21

(28.3)

(a) The classes of singlewalled carbon nanotube are defined according to the vectors shown opposite. Starting with a graphene sheet, vectors are drawn that define the axes of the zigzag and armchair carbon nanotubes. Vectors perpendicular to the first vectors then define the open ends of the tubes.

(28.4)

Tube axis for zigzag

(0, 0)

Tube axis for armchair

(n, 0) Zigzag (θ = 0o)

(n, m) Armchair (θ = 30o)

386

Inorganic materials and nanotechnology (b) Use of the arc discharge method favours the formation of multi-walled carbon nanotubes (MWNT). SWNTs formed by the laser vaporization technique usually form in bundles (van der Waals forces operate between the surfaces of adjacent tubes). A significant number of tubes are capped. (c) Purification of carbon nanotubes involves oxidation using HNO3 or heating in air; this removes amorphous carbon and metal catalysts. Tubes are then sonicated and purified by chromatography. Treatment with oxidizing agents such as HNO3/ H2SO4 is a means of ‘cutting’ long carbon nanotubes into shorter lengths. It also introduces carbonyl and carboxylate functionalities that can be used to prepare derivatives of SWNTs. 28.22

X For evidence for this ordering, see G.A. Nazri et al. (1994) Solid State Ionics, vol. 70/71, p. 137.

28.23

(a) For the structure of lithium nitride, see answer 11.12, or Fig. 10.4a in H&S. It is a layer structure with Li+ ions within and between the layers; a deficiency of Li+ ions within the layers results in Li+ ion conduction in a direction parallel to the layers; only the Li+ ions within the layers are involved in ion conduction. The ionic character in lithium nitride, phosphide and arsenide is expected to follow the order Li3As < Li3P < Li3N. (b) Epitaxal growth of a crystal on a substrate crystal is such that the growth follows the crystal axis of the substrate. Thin films of MgB2 fabricated by passing B2H6 over Mg (supported on Al2O3 or SiC) heated at 970 K under an H2 atmosphere have superconducting properties as do single crystals of MgB2; Tc = 39 K and this is the highest Tc observed for a metal boride. (a) Information for this answer can be obtained from Section 28.6 in H&S; MOCVD stands for metal–organic chemical vapour deposition and therefore the precursors are specifically organometallic compounds or complexes containing organic ligands. Points to include: • precursor(s) must be volatile; • delivery by uniform mass transport of a volatile precursor to a hot surface on which thermal decomposition of a precursor (or reaction between two precursors) occurs to give solid product; • product is deposited as a thin film; • control of temperature is critical; • relate technique to the example given in the question: Al(OiPr)3 is decomposed to α-Al2O3 with gasous propene and H2 being released. (b) The reaction in the question is: 2H2NNMe2 + IntBu2(N3) J InN + 2Me2NH + 2tBuH + 2N2 The likely role of H2NNMe2 is as an H atom donor to facilitate tBuH elimination; if H2NNMe2 were the main source of N for InN, indium nitride should also be formed in the reaction in which tBu3In is used in place of the azide, IntBu2(N3). The experimental data indicate that the [N3]– ligand is the source of N for the formation of InN. For information on LEDs, see Table 28.5 in H&S and associated discussion; the colour of emitted light depends on the band gap of the semiconductor.

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(a) Doping CaF2 with 1% NaF results in the solid containing F– vacancies (‘holes’) because of the differences in the stoichiometries of the two compounds. The electrical conductivity of 0.1 Ω–1 m–1 is orders of magnitude greater than that of CaF2 itself, and arises from migration of F– 0 93 300 through the ‘holes’ in the solid. Temperature / K (b) The change in electrical resistivity for a superconductor shows a dramatic Fig. 28.2 Sketch of the change in electrical resistivity against temperature for fall in resistivity (i.e. an increase in YBa Cu O ; T = 93 K. 2 3 7 c conductivity) at the critical temperature as the material is cooled (Fig. 28.2 shows this for YBa2Cu3O7). The electrical resistivity of a typical metal increases with increasing temperature (see Fig. 6.10 in H&S), while that of a semiconductor decreases with increasing temperature (see Fig. 6.11 in H&S).

28.25

Solid-state lighting includes light-emitting diodes (LEDs, Section 28.6 in H&S), organic light-emitting diodes (OLEDs, Section 28.3 in H&S), and light-emitting electrochemical cells (LECs, Section 28.3 in H&S). • LED: semiconductor junction device constructed from inorganic materials such as GaAs1–xPx and InxGa1–xAs, the composition of which determines the band gap and the colour of the emitted light. • OLED: your answer should include a diagram similar to Fig. 28.9a in H&S and a brief description of the operation of the device (see answers 28.7 and 28.8); OLEDs use neutral electroluminescent materials. • LEC: LECs use ionic electroluminescent materials and your answer should include a diagram based on Fig. 28.10a in H&S. Give a brief description of the operation of the device using information from the LECs subsection in Section 28.6 in H&S.

28.26

Points to include: • Superconducting magnets in highfield NMR spectrometers including MRI scanners (NbTi multicore conductors, Tc = 9.5 K). MgB2-based MRI magnet has recently been developed. • Josephson junctions. • SQUID systems for magnetic susceptibility measurements, in biomedical imaging and in undersea mine detection. • MAGLEV trains depend upon superconducting electromagnets providing near frictionless motion. • Particle accelerators: the Large Hadron Collider particle accelerator at CERN uses 10000 NbTi superconducting magnets. • Power transmission - bulk application is not yet realized.

28.27

Rechargeable lithium-ion battery: two host materials and migration of Li+ ions between them. Graphite intercalates Li+ ions which migrate out of LiCoO2 during battery charging; this results in oxidation of Co(III) to Co(IV) and reduction of graphite (formally represented as C6– in the reduced state). During discharging, the Li+ ions migrate back to the metal oxide, as Co(IV) is reduced to Co(III):

Resistivity

28.24

388

Inorganic materials and nanotechnology

LiCoO2 + 6C(graphite)

charge discharge

LiC6 + CoO2

Other hosts such as LiFePO4 may replace LiCoO2. The electrolyte is usually LiPF6 in an alkyl carbonate material. The cell potential is 3.6 V. Major applications are in small electronic devices (e.g. laptop computers, mobile phones, MP3 players); there is a growing demand in the automobile industry. First mass-produced hybrid electric vehicle (Mercedes-Benz) uses a 120 V lithium-ion battery pack to power an electric motor working in tandem with an internal combustion engine. Regenerative braking converts kinetic energy to electrical energy which is stored in the battery, and the electrical motor also recovers energy during deceleration. 28.28

Carbon fibres are manufactured by thermal degradation of pitch, rayon or polyacrylonitrile (PAN): see Fig. 28.27 in H&S. Your answer should include details of impurities and how to expel them, graphene sheet and graphite-like material production, and melt spinning. Carbon–carbon composites are carbon-fibre reinforced materials. Lightweight and strong material, corrosion resistant; examples of applications: military aircraft body parts, racing cars and other top of the range cars, sports equipment, camera (e.g. tripod) equipment.

389

29 The trace metals of life 29.1

Table 29.1 Examples of naturally occurring α-amino acids. R (in 29.1)

α-Amino acid

H CH2CO2H CH2SH CH2CH2CO2H CH2CHMe2 CH2CH2SMe CH2OH

glycine L-aspartic acid L-cysteine L-glutamic acid L-leucine L-methionine L-serine

(a) The general formula of an amino acid is shown in 29.1. A peptide (or polypeptide) is produced by condensation of the NH2 + CO2H groups of different amino acids (29.2). A peptide chain has an N-terminus (NH2 group) and C-terminus (CO2H group). peptide link R'

R C

N

N n+

Fe

N

N

(29.3)

H

CO2H

H

H

O

N

C N

C

H

O

C C

NH2

(29.1)

R

H

(29.2)

(b) 20 naturally occurring (common) α-amino acids containing particular R groups (examples in Table 29.1). Except for glycine, all naturally occurring amino acids are chiral, but Nature is enantiomerically specific. (c) A protein is a polypeptide with very high Mr; structures are complex; proteins classified as fibrous or globular. Prosthetic group in a protein is non-amino acid unit. In a metalloprotein, the prosthetic group binds a metal centre and this group is essential for biological activity of the protein, e.g. acts as redox-active centre or Lewis acid. Examples of metalloproteins: ferredoxins, haemoglobin, cytochrome c. (d) If the metal ion is removed from a metalloprotein, the remaining organic molecule is the apoprotein. (e) Haem is the prosthetic group in e.g. haemoglobin and myoglobin. Part of coordination environment about Fe is porphyrin group (29.3) with substituents which depend upon metalloprotein; axial ligand(s) have role in connecting haem to protein.

29.2

Refer to Section 29.2 in H&S for full details. Structures are complex, but answer should include representative examples of metal-binding sites. Points to discuss: • Uptake of Fe into the blood as Fe3+-containing transferrins; in this form, Fe is transported to protein ‘storage vessel’ called ferritin (e.g. in liver and spleen). • Ferritin has very high Mr ( ≈ 445 000 for horse spleen apoferritin) and a complex structure – hollow shell accommodating ≤4500 high-spin Fe3+ as microcrystalline oxohydroxophosphate (e.g. of biomineralization). Iron probably enters cavity as Fe2+, being oxidized once inside. • Serum transferrin transports Fe to bone marrow; transferrin consists of single, high Mr polypeptide chain containing 2 pockets in which Fe3+ is coordinated by amino acid O- and N-donors; [CO3]2– or [HCO3]– must also be present. • Aerobic microorganisms store and transport Fe as Fe3+ using siderophores – O-donor ligands; complexes have very high stability constants. • In mammals, thioneins contain soft S-donors and transport soft metal ions e.g. Hg2+ and Cd2+; act against toxic metals.

29.3

Model ligand L2– is catecholate 29.4. The ligand [L′]6– (29.5, conjugate base of enterobactin) contains three catecholate domains and binds one Fe3+. The coordination mode of 29.5 in [FeL′]3– could be structurally modelled by using 3 catecholate ligands

The trace metals of life

390

–O –O

O–

O –

O NH

–O

–

(29.4)

O

NH O

HN O

in the complex [FeL] 3–. Fe 3+ is d 5 whereas Cr3+ is d 3 and kinetically inert. Use of Cr3+ in place of Fe3+ permits solution studies to be carried out – rates of substitution at Cr3+ are slow. 29.4 X Soft metal ions and ligands: see Table 7.9 in H&S

N

N H2C

N H

N H

(29.6)

(29.7)

29.5

N(His) (Porph)N

N(Porph)

Fe3+

(Porph)N

N(Porph) O O

(29.9)

O– O–

(29.5)

(a) Thioneins are sulfur-rich proteins with the protein chains folded into pockets, exhibiting soft S-donor binding sites. Cd2+ is a soft metal ion. Stability constant of complex formed between soft metal ion and soft atom donor set is high. Other metal ions favoured by thioneins are Cu+, Zn2+ and Hg2+. (b) [Cu4(SPh)6]2– contains a cluster of Cu+ ions bridged by [PhS]– ligands. Thionein pockets are rich in Cys residues and tend to bind multinuclear metal units. Thus, [Cu4(SPh)6]2– (a discrete complex, more readily studied by e.g. spectroscopic methods than a metalloprotein) is a suitable model for the Cu-containing metalloprotein in yeast. H (c) The R group in the amino acid histidine is 29.6. The heterocyclic ring is similar to that in B imidazole (29.7) and trispyrazolylborate (29.8). N N N Coordination complexes featuring 29.7 or 29.8 N N are suitable models for metalloproteins with His N residues. (29.8) Fuller details are given in Section 29.3 in H&S; suggested answer plan is as follows: (a) Single haem unit contains a functionalized porphyrin (protoporphyrin IX) which is an N4-donor (29.3). In oxyhaemoglobin, iron is bound as Fe3+ with axial His residue from protein chain and axial O2 unit (29.9); deoxyhaemoglobin contains Fe2+ and binding of O2 is accompanied by electron transfer oxidizing Fe2+ to Fe3+, and reducing O2 to [O2]–. Each of low-spin Fe3+ (d 5 ) and [O2]– has an unpaired electron; antiferromagnetic coupling makes oxyhaemoglobin diamagnetic. (b) ‘Picket fence’ porphyrins are used in model complexes; they have bulky substituents that prevent formation of [O2]2– bridged species. The latter form in less sterically hindered complexes, e. g. [Fe(TPP)] (H2TPP = tetraphenylporphyrin): 2[FeII(TPP)] + O2 J [(TPP)FeIII–O–O–FeIII(TPP)] (c) Haemoglobin is tetrameric and the 4 haem units communicate with each other via changes in the conformation of the protein chains; once one O2 is bound, the affinity of the remaining Fe sites for O2 increases dramatically, and similarly, release of the first O2 triggers release of the other three.

The trace metals of life

391

(d) Change from high-spin Fe2+ (octahedral d 6 ) to low-spin Fe3+ (octahedral d 5 ) which is antiferromagnetically coupled to [O2]–: deoxy- to oxy-form is a change from paramagnetic to diamagnetic species. More details for this answer are found in Section 29.3 in H&S; include the following points plus comments on model compounds: (a) In mammals, myoglobin binds O2 in the same way as haemoglobin (see answer 29.5 for details) except for the fact that myoglobin is monomeric and shows no cooperativity. Supporting evidence comes from model/structural studies using picket fence porphyrins and from magnetic data. (b) Haemerythrin is a non-haem Fe-containing O2 carrier in marine invertebrates. Active site is dinuclear (crystallographic data for both deoxy- and oxy-forms); magnetic data show antiferromagnetically coupled Fe(II) centres in the deoxyform. O2 binding involves coordinatively unsaturated Fe(II) and μ-OH:

29.6

(His)N

Fe (His)N

(His)N

N(His)

CuII

CuII

N(His)

.

O (His)N

N(His)

(29.10)

29.7 N(His) Cu (His)N Type 1

S(Cys) S(Met)

(29.11)

OH2 (or OH–) Cu Type 2 (His)N ((His)N)3Cu Type 3

II

II

N(His) Cu(N(His))3 O (or OH–)

(29.12)

29.8

.

N(His)

Fe O H

O (His)N

Asp O O

Glu OO

N(His) N(His)

O2 O2

Asp O O

Glu OO

(His)N

FeIII

(His)N

FeIII O

O O

H

N(His)

N(His) N(His)

(c) Haemocyanins are Cu-containing O2 carriers in e.g. arthropods; no haem unit. Colourless Cu(I) deoxy-form turns blue on O2 uptake as Cu(I) oxidized to Cu(II). Active site contains 2 Cu(I)(His)3 units in close proximity (Cu---Cu = 354 pm, non-bonded) and oxy-form contains an Cu(II)–[O2]2––Cu(II) unit (29.10) in which Cu(II) centres are antiferromagnetically coupled (superexchange). Evidence for structures and bonding modes: magnetic data, Raman spectroscopic data for ν(O–O), and crystallographic data for both deoxy- and oxy-forms. For additional details, see Section 29.4 in H&S. Points to include: • Cu centres in blue copper proteins fall into 3 classes with spectroscopic (as well as structural) distinctions; proteins contain at least one Type I centre. • Function: redox centres utilizing Cu2+/Cu+ couple. • Type 1 centre exhibits intense absorption in electronic spectrum with λmax ≈ 600 nm (εmax ≈ 100× that of Cu2+(aq)) assigned to charge transfer from Cys ligand to Cu2+; EPR spectrum shows narrow hyperfine splitting (Cu2+ is d9). • Electronic and EPR spectra of Type 2 centre typical of simple Cu2+ coordination complexes. • Type 3 centre exhibits absorption with λmax ≈ 330 nm; consists of 2 Cu2+ centres, antiferromagnetically coupled giving a diamagnetic system. The Cu2-unit acts as 2-electron transfer centre and is involved in O2 reduction. • Protein crystallographic data provide structural information: Type 1 centre in plastocyanin is 4-coordinate (29.11) (see answer 29.8 for more detail); ascorbate oxidase (catalyses O2 reduction to H2O) contains one Type 1 centre (as 29.11), one Type 2 and one Type 3 centre – Type 2 and Type 3 Cu atoms form Cu3 unit (29.12); combination of Type 1, 2 and 3 centres in ascorbate oxidase facilitates electron transfer from organic substrate (at Type 1 Cu) and reduction of O2 (at Type 2/3 Cu site). Plastocyanin is a blue copper protein containing Type 1 Cu (see answer 29.7) . His, Cys and Met residues on protein backbone provide coordination site 29.11;

The trace metals of life

392

Cu(II) and Cu(I) prefer 4- and 3-coordination respectively: although 29.11 is 4coordinate, Cu–S(Met) > Cu–S(Cys), so provides a compromise environment for Cu(I) and Cu(II); rapid electron transfer is possible. Reduction of Cu(II) to Cu(I) accompanied by lengthening of Cu–N and Cu–S bonds. The fact that in 29.11, one Cu–S bond is long implies tending towards 3-coordination, i.e. favouring Cu(I); supported by high reduction potential for plastocyanin (+370 mV at pH 7). 29.9

Ascorbate oxidase is a blue copper protein (crystallographic data available) and contains Type 1, 2 and 3 centres: • Type 1 Cu (see 29.11) is remote from other 3 Cu sites but able to communicate with them via protein chain; coordination is by a Cys, Met and two His residues (see also answers 29.7 and 29.8). • Type 2 centre (single Cu) and Type 3 centre (Cu2 unit with metal atoms antiferromagnetically coupled through μ-O or μ-OH) form a (Cu)(Cu2) unit (see 29.12); all protein residues are His. • Function: reduction of O2 to H2O coupled with oxidation of organic substrate (a phenol): O2 + 4H+ + 4e–

2H2O

4RH + O2 J 4R• + 2H2O

• All Cu centres involved in electron transfer via Cu(II)/Cu(I) couple; Type 1 Cu involved in electron transfer from organic substrate; O2 reduction occurs at Type 2/3 site. 29.10 (Cys)S Fe S

S Fe

S Fe

S(Cys)

(a) Blue colour arises from Cu2+ and represents the oxidized form of the blue copper protein; reduced form contains Cu+ and is colourless. (b) [4Fe-4S] ferredoxin (29.13) contains Fe3+/Fe2+ centres; 1-electron redox process involves Fe4 unit – not localized at one Fe centre but can be represented as: 2Fe(III).2Fe(II) + e–

Fe(III).3Fe(II)

Fe S

S(Cys)

Cluster is held in a pocket of the protein chain; changes in conformation of metalbinding pocket alter the Fe coordination environment and affect reduction potential. (c) O2 acts as a π-acceptor when it binds to Fe(II) in deoxyhaemoglobin to form the Fe(III) oxy-complex. CO is also a π-acceptor ligand and therefore competes for Fe(II) in the same binding site as O2. Once bound, CO blocks the haem site preventing O2 coordination. [CN]– is also a π-acceptor ligand, but favours higher oxidation state metal centres than CO; [CN]– binds to Fe(III) in cytochromes (involved in electron transfer processes).

29.11

For full details, see ‘The mitochondrial electron-transfer chain’ in Section 29.4 in H&S. Points to include: • Mitochondrial electron-transfer chain is the means of transferring electrons in living cells. • Draw out the chain shown in Fig. 29.15 in H&S. • Emphasize range of reduction potentials that must be covered by biological systems at pH 7: –414 mV for H+ reduction to H2, to +815 mV for O2 reduction to H2O; each member of the chain operates within a very small potential range. Hence the need for a chain of electron transfer mediators. • Electron transfer involving a metal centre in metalloprotein is a 1-electron process; transfer involving organic substrate is usually a 2-electron step.

(Cys)S

(29.13)

The trace metals of life

393

• Quinones play vital role in overcoming the 1-electron/2-electron mismatch since they can undergo both 1- and 2-electron processes:

O

H. 1e-

O

O

Quinone

29.12

S Ph Ph

Fe S

S S

Fe

Hydroquinone

μ(spin-only) high-spin Fe3+ = n ( n + 2) = 5(5 + 2) = 5.92 μ B

S

2–

(b) Spinach ferredoxin is a [2Fe-2S] system with an Fe2(μ-S)2{S(Cys)}4 core; the model complex 29.15 is structurally related to this, with Ph substituents replacing Cys residues. While 29.15 models the active site, it cannot provide information about the effects of the protein chain (e.g. its conformation). (c) Nitrogenase contains two types of clusters which are active sites: (i) Cys bridged double [4Fe-4S] site (‘P-cluster’); (ii) FeMo-cofactor. The discrete complex in the question models half of FeMo-cofactor (Fig. 29.1). The identity of atom X in Fig. 29.1 is ambiguous, but of the possibilities (C, N, O) the most likely atom is N. For the model, Mössbauer data consistent with delocalization of charge, i.e. 2.67 is an average oxidation state of Fe(II).2Fe(III), a model that is appropriate for the active site.

29.13

Although charge is not localized, redox reactions in [4Fe-4S] protein summarized as:

SPh Fe

PhS

Semiquinone

OH

μ(spin-only) high-spin Fe2+ = n( n + 2) = 4( 4 + 2) = 4.90 μ B

Ph

(29.14)

PhS

HO

(a) [Fe(SPh)4]2– (29.14) contains tetrahedral Fe(II) with [PhS]– ligands. Rubredoxin contains a tetrahedral Fe{S(Cys)}4-site with the S(Cys) residues attached to the protein chain. Thus, 29.14 is a good structural model. For the model compound, observed values of μeff are 5.05 and 5.85 μB for reduced and oxidized forms. These compare well with μ(spin-only) for tetrahedral high-spin Fe2+ (d 6 ) and Fe3+ (d 5 ):

2–

Ph

OH

H. 1e-

SPh

S

(29.15)

4Fe(III)

3Fe(III).Fe(II)

2Fe(III).2Fe(II)

A

B

Fe(III).3Fe(II)

C

4Fe(II)

D

Each of steps A to D is a 1-electron reduction or oxidation. (a) Under physiological conditions, couples B and C are accessed; E′ values must fall within appropriate limits at pH 7 (see answer 29.11); typically, E′ for C ≈ –350 to –450 mV, and for B, ≈ +350 mV. H N Et

S

S

S Fe

Fe

Et

S

3–

S (Cys)S

S Et

Fe S

Fe

S S

Fe

Fe

S X

Fe Fe

S S

S

Fe

Fe S

X is most probably N

O

Mo

S Mo (CO)3

N

S

C

O C

CH2CH2CO2– CH2CO2–

O

Fig. 29.1 Model compound (left-hand diagram) for part of the FeMo cofactor in nitrogenase; the part modelled is indicated with the hashed circle in the right-hand diagram. See also Fig. 29.23b in H&S.

The trace metals of life

394

X See Fig. 29.15 in H&S

29.14

29.15 N(His) (Porph)N

Fe2+

(Porph)N

N(Porph) N(Porph)

(29.18) N(His) (Porph)N

Fen+

(Porph)N

N(Porph) N(Porph)

(b) HIPIP (high potential protein) corresponds to couple B; HIPIP is the highest oxidized state of the Fe4 unit under physiological conditions. (c) See part (a) for typical E′ values. Mitochondrial chain ranges from –414 mV for H+ reduction to H2, to +815 mV for O2 reduction to H2O. The [4Fe-4S] ferredoxins are involved in electron transfer at the most negative potential end of the chain, while HIPIP operates further along the chain (more +ve potential). Combining B and C to give a single 2-electron reduction or oxidation does not occur. Active site of [2Fe-2S] ferredoxin is represented in 29.16 and that in Rieske protein in 29.17; structural difference is coordination by His (29.17) rather than Cys residues. At physiological pH, E′ values of [2Fe-2S] ferredoxins are negative (depends on protein: ≈ –300 to –400 mV), whereas Rieske protein has a positive potential (+290 mV for protein isolated from spinach). Their roles in the mitrochondrial chain (see answer 29.11) are therefore different: electron transfer in Rieske protein is coupled to oxidation of plastoquinol to plastosemiquinol which releases H+; electron transfer in ferredoxins is utilized at the most negative potential end of the chain.

(Cys) S

S

(Cys)S Fe S (Cys)

Fe S(Cys)

S

(29.16)

(His) N

S

(Cys)S Fe S (Cys)

Fe S

N(His)

(29.17)

(a) Haem unit contains Fe2+ or Fe3+ coordinated within N4-donor set of porphyrin group (see structure 29.3). Deoxymyoglobin contains protoporphyrin IX and binds Fe(II); does not become Fe(III) until O2 bound in oxymyoglobin. Cytochrome c is an electron transfer centre, and Fe(II) or Fe(III) is present depending on redox state. Deoxymyoglobin contains 5-coordinate Fe(II) with axial site occupied by His residue (29.18) from protein chain. In cytochrome c, Fe is 6-coordinate and haem unit is bound to the protein chain through axial His and Met residues, (29.19), and through 2 Met residues covalently linked to the porphyrin ring (refer to Fig. 29.24 in H&S). (b) Cytochrome c is a 1-electron transfer centre, utilizing Fe3+/Fe2+ redox couple. In the mitochondrial electron-transfer chain (see answer 29.11) cytochrome c accepts an electron from cytochrome c1 and transfers it to cytochrome c oxidase: Cyt c1+

Cyt c ox+

Cyt c –

e

–e

–

S(Met)

–e–

n = 2 or 3

Cyt c

Cyt c1

(29.19)

+

e– Cyt c ox

Electron transfer between these metalloproteins is proposed to occur by a mechanism that involves tunnelling through the edge of the haem unit. Electrons so transferred are finally used in the 4-electron reduction of O2 to H2O. 29.16

(a) Cytochrome c oxidase is the terminal member of mitochondrial electron transfer chain and catalyses reduction of O2 to H2O (electron transfer is coupled to H+ pumping): O2 + 4H+ + 4e–

2H2O

The trace metals of life

395

(b) Four active sites (structurally confirmed) are CuA, CuB, haem a and haem a3: • CuA contains a Cu2{μ-S(Cys)}2 site; • haem a contains 6-coordinate Fe with axial His residues; • CuB is 3-coordinate (3 His residues) with Cu atom facing (but not bonded to) Fe in haem a3; • haem a3 contains 5-coordinate Fe with the vacant coordination site facing CuB (Fe- - -Cu = 450 pm); haem a3 and CuB are antiferromagnetically coupled. Proposed operation: • Electron transfer involves CuA and haem a; electrons are transferred from cytochrome c (see scheme in answer 29.15) to CuA and are passed to haem a. • Haem a3 and CuB together are the O2 binding site; at this site, O2 is reduced to H2O and the site is also involved in pumping H+ across the mitochondrial inner membrane. • The 4 sites work together even though CuA is ≈ 2000 pm away from haem a and haem a3, and haem a is 900 pm distant from haem a3; a hydrogenbonded network involving protein chain residues, haem propanoate substituents, and a CuA-bound His residue is thought to provide an electrontransfer route between CuA and haem a. 29.17

(a) Haemoglobin utilizes haem-Fe for O2 binding; haem unit provides a versatile 5- to 6-coordinate binding site with the N4-donors of the porphyrin ring allowing the Fe2+ or Fe3+ centre to lie out of or in their coordination plane. Fe in cytochromes undergoes less marked structural changes than Fe in haemoglobin; the coordination sphere (e.g. 6-coordinate in cytochrome c) tolerates reversible Fe3+ to Fe2+ change as an electron is transferred. (b) See answer 29.16b for details; emphasize the fact that metal centres present must be able to transfer electrons, bind O2 and pump H+. (c) 5-Coordinate Fe2+ is present in deoxy-form of haemoglobin; vacant site is filled by O2; binding is concomitant with 1-electron transfer to give Fe3+ and coordinated [O2]–; this is reversible, with the 5-coordinate ‘rest state’ being reformed when O2 is released. Cytochrome c does not need to bind another ligand (although it does when [CN]– poisons the system) since its role is one of electron transfer with the 6-coordinate Fe centre reversibly going between Fe2+ and Fe3+. (d) NP1 is a haem protein in salivary glands of blood-sucking insect Rhodnius prolixus. Fe(III) in NP1 binds NO 10× more tightly at the pH of the insect’s saliva than at the pH of the insect’s victim. Once insect saliva is released into the victim, NO is released causing expansion of the blood vessels and inhibiting blood clotting.

29.18

Full details are given in Section 29.5 in H&S. Suggested plan for answer: • Zn2+ is a hard metal centre; coordinated by hard N- and O-donors; highly polarizing and able to act as a Lewis acid. • Metalloenzymes containing Zn2+ include carbonic anhydrase II (CAII) and carboxypeptidase A (CPA). • CAII catalyses reversible hydration of CO2:

O C O

H O Zn

H2O + CO2

2+

(His)N

N(His) N(His)

(29.20)

H+ + [HCO3]–

In the active site, Zn2+ ion is bound by three protein His residues and [OH]– (29.20); [OH]– is hydrogen bonded to Glu and Thr residues. CO2 enters a hydrophobic pocket and is converted to [HCO3]– which is released; H2O enters and is deprotonated, regenerating the active site.

396

The trace metals of life • CPA catalyses cleavage of peptide link in polypeptide: O H N

R' N H

O CO2H

CPA

+ H2O

R'

H N

OH

R

+ H2N

CO2H

R

Specific cleavage at C-terminus and selective for C-terminal amino acid with large aliphatic or Ph substituent. In active site, Zn2+ bound by O,O′Glu, 2 His residues and H2O. Use Fig. 29.30 in H&S to describe the mechanism of catalysed peptide link cleavage, emphasizing not only the Lewis acidity of Zn2+, but also the cooperation between protein residues and the role of hydrogen bonding. • The answer could also be extended to include carboxypeptidase G2 in which the active site contains 2 Zn2+ centres. 29.19 X N Zn2+

Rate of hydrolysis depends on concentrations of acid anhydride, Zn2+ and [OH]–, showing that all are involved in rate determining step. Tetrahedral environment for Zn2+ is likely; propose initial complex 29.21 where X could be H2O or an O-donor from the acid anhydride. A plausible mechanism involves attack by coordinated [OH]– at carbonyl C atom followed by ring opening: H

OH NH

O

X

O

O

O

O

–

O

X

N

NH

N

O

O

+

Zn2+

Zn2+

O

O H NH

O

– H+

(29.21)

O X

Zn2+ N

O

CO2–

NH

The degree of protonation depends on pH. 29.20

For full details, see the end of Section 29.5 in H&S. Points to include: • Carbonic anhydrase contains Zn2+, d10 ion; electronic spectroscopic and magnetic studies (common techniques for investigating metalloproteins or model compounds) are not appropriate for a d10 ion – filled d level means diamagnetic complexes in all geometries and no 'd-d ' electronic transitions. Therefore need to substitute Zn2+ by a metal ion that can be a spectroscopic and magnetic probe, but substitution must not perturb the coordination environment. • Co2+ substitution is suitable: Co2+ is d7 and so gives electronic spectroscopic and magnetic data; ionic radii of Co2+ and Zn2+ are similar; Co2+ and Zn2+ can be accommodated within similar coordination geometries; replacement of Zn2+ in a protein by Co2+ often has only small effects on protein conformation.

The trace metals of life 29.21

The macrocycle in the question is Me shown in 29.22 = H4L. The structure N of desferrioxamine is shown in Fig. NH N OH 29.5 in H&S. The features in O O common are the -C(O)N(OH)groups which, once deprotonated, HO N N N act as chelates. The macrocyclic unit acts as a scaffold that provides Me Me O N a degree of preorientation of the 3 chelating groups towards the OH (29.22) formation of an octahedral complex with Fe3+. In [Fe(HL)], [HL]3– acts as a trischelate, binding through three O,O'-donor sets to an octahedral Fe3+ centre. The conformation of the macrocycle can alter to accommodate what is needed to bring the three hydroxamate arms into the correct orientation for chelation.

29.22

(a) The Zn4(Cys)9(His)2 cluster in the metalloprotein indicates that this is a metallothionein. Typically, metallothioneins contain only Cys residues which bind metals centres such as Zn(II), Cu(II), Cd(II) and Hg(II) through soft S-donor atoms. Zn(II) is intermediate in hard-soft classification (see Table 7.9 in H&S); in the metalloprotein in the question, the tetranuclear unit is supported by soft S and hard N-donors. (b) For the general part of the answer, see Section 29.3 in H&S, subsection “Cytochromes P-450”. Important points to include: • cytochromes P-450 catalyse insertion of O into a C–H bond of an aromatic or aliphatic hydrocarbon; • converts RH to ROH; • active site represented by 29.23 with 5-coordinate high-spin Fe(III). Using the information in H&S, construct the following catalytic cycle:

S(Cys) (Porph)N

Fe(III)

(Porph)N

397

N(Porph) N(Porph)

(29.23)

X See M.C. Feiters et al. (2000) Chem. Soc. Rev., vol. 29, p. 375 for further information

S(Cys) N

ROH

Fe(III)

N

H2O

N

RH

H2O

N OH2

S(Cys) N

S(Cys) N

Fe(IV)

N O

RH

Fe(III)

N

N

RH

N

RH is shown to be in close association with the active centre

N N e–

H2O

S(Cys)

H+

N

S(Cys) N

Fe(III)

N O

RH

N

N N

S(Cys) N

HO e–,

H+

N

Fe(III)

N

N RH O2 peroxido complex

Fe(II) RH O2

N N

The trace metals of life

398

29.23 t

Bu

N P N i

Pr

3

(29.24)

29.24

H2 C

2– CH2

H 2C S

S

NC

CO Fe

OC

Fe

CO

OC

CN

(29.26)

29.25

The tripodal ligand in the question is drawn here as structure 29.24. The proposed structure for [Zn(29.24)(OH)]+ contains approximately tetrahedral Zn(II) = structure 29.25. (a) The active site in carbonic anhydrase involves 3 His residues (see structure 29.20); the imidazole rings in ligand 29.24 mimic His residues. (b) The tripodal ligand has only limited flexibility and is compatible with the formation of tetrahedral [Zn(29.24)(OH)]+.

+

i

Pr

i

Pr P

N

N

N

N t

N Bu

Pr

t

Bu

N

Zn OH

i

t

Bu

(29.25)

(a) Compounds such as 29.26 are used as models for the active sites of [NiFe] and [FeFe]-hydrogenases. Crystal structure data for [NiFe]-hydrogenase from the bacterium D. gigas show that the active site contains Ni and Fe atoms bridged by peroxido ligand and 2 Cys residues; the Fe centre is ligated by CO or [CN]–. Crystallographic data for the [FeFe]-hydrogenases from D. desulficans and C. Pasteurianum, confirm the presence of an Fe4S4 cluster coupled to an Fe2S2 unit in which each Fe carries CO and [CN]– ligands (see Figs. 29.19 and 29.21 in H&S and accompanying discussion for more details). (b) See Fig. 29.1 in this book, and Fig. 29.23 in H&S for structural details of the FeMo cofactor in nitrogenase. Before 2002, the central atom (X in Fig. 29.1) had not been observed. This meant that the inner Fe atoms in the FeMo cofactor appeared to be 3-coordinate; such coordinatively unsaturated Fe centres were thought to be possible sites for addition of N2. However, the extra interactions between the Fe atoms and the 6-coordinate X atom means that the Fe centres are 4-coordinate and can no longer be considered to be coordinatively unsaturated. (a) Haemoglobin contains 4 haem units (see Fig. 29.7a in H&S). The equilibrium: Haemoglobin + O2

(Haemoglobin)(O2)

K1 ≈ 10

refers to binding of the first molecule of O2 to tetrameric haemoglobin; equilibrium: (Haemoglobin)(O2)3 + O2

(Haemoglobin)(O2)4

K4 ≈ 3000

refers to binding the last molecule of O2. There is cooperativity between the haem units, with each O2 binding step triggering the next; this leads to K4 > > K1; communication between the haem units that leads to this effect is made possible by conformational changes in the protein chains. (b) Photosystem II (PSII) catalyses oxidation of H2O to O2 in green plants and algae. The cofactor shown on the right contains Mg 2+ bound within a functionalized porphyrin ligand; the cofactor is chlorophyll (see Fig. 12.10 in H&S).

The trace metals of life

399

29.26

(a) Active site of [FeFe]-hydrogenases contains an Fe4S4 cluster linked to an Fe2S2 unit in which each Fe carries CO and [CN]– ligands (see Fig. 29.21 in H&S). The Fe atoms are bridged by an S....S chain; the chain could be SCH2CH2CH2S, SCH2NHCH2S or SCH2OCH2S. In X-ray diffraction, the scattering powers of the C, N and O atoms are similar and a crystallographer has difficultly distinguishing between CH2, NH or O using only diffraction data. The models shown in the question mimic the Fe2S 2 unit in [FeFe]-hydrogenases. The neutral model compounds are isoelectronic because CH2, NH and O are isoelectronic. CO and [CN]– are isoelectronic, and replacing two CO ligands in the neutral Fe2(CO)6S2cluster by [CN]– ligands generates an isoelectronic dianion. (b) In electrospray mass spectrometry, proton addition (or Na+ addition) occurs. If M corresponds to myoglobin, molecular mass 16950, then peaks are: m/z 1413 [M + 12H]12+, 1304 [M + 13H]13+, 1212 [M + 14H]14+, 1131 [M + 15H]15+.....808 [M + 21H]21+. The mass difference is not a constant value because the charge on each ion is different, and the observed peaks correspond to mass/charge values.

29.27

(a) Haemoglobin is tetrameric.The 4 haem units communicate with each other via changes in the conformation of the protein chains; the conformation changes as the heam units bind O2. Once one O2 is bound, the affinity of the remaining Fe sites for O2 increases dramatically, and similarly, release of the first O2 triggers further release of O2 from the tetramer. (b) Figure 29.8 in H&S shows that haemoglobin has a low affinity for O2 at low pressures of O2 (e.g. in mammalian tissues), but the affinity rises rapidly as the pressure increases to 0.02 bar. The highest pressure of O2 in a mammal is in the lungs where O2 binds to haemoglobin. The shape of the O2-binding curve for myoglobin shows that myoglobin is fully saturated with O2 in the lungs (highest pressures in the graph) but that the protein will not release enough O2 in the tissues (lower pressures, around 0.03 bar) to sustain metabolism.

29.28

Haemoglobin: mammals. [NiFe]-hydrogenases: microorganisms (anaerobic bacteria). Rubredoxins: bacteria. Plastocyanins: higher plants and blue-green algae. Active sites: Haemoglobin: tetrameric protein, each subunit contains one haem unit as the O2binding site. Cooperative, reversible binding of O2 (see answer 29.27b). Fe(II) in the rest state, oxidized to Fe3+ when O2 binds and the latter is reduced to [O2]–. See Section 29.3 in H&S for more details. [NiFe]-hydrogenases: see answer 29.24 for structural description. Microorganisms use H2 as a reducing agent, and reversible reduction of H+ to H2 is catalysed at the active site in hydrogenases. Rubredoxin: the active site is a mononuclear FeS(Cys)4 unit. Acts as a 1-electron transfer site, using the reversible Fe3+/Fe2+ redox couple, the reduction potential of which depends on the conformation of the protein chain to which the Cys residues are attached. See Section 29.4 in H&S for more details Plastocyanins: blue copper proteins containing one Type I copper centre. In spinach plastocyanin, Cu2+ is bound by 2 His, a Cys and a Met residue, i.e. CuN2S2 coordination sphere.The active centre catalyses 1-electron transfer, using the reversible Cu2+/Cu+ redox couple. See Section 29.4 in H&S for more details.

400

The trace metals of life 29.29

The protonation state of phytic acid depends on the pH. Acid dissociation of the OP(O)(OH)2 units produces a series of anions which act as chelating ligands. The hard O-donors target hard metal ions such as Fe3+. Free Fe(III) is readily complexed by phytic acid (complex has a high stability constant) and the uptake of Fe3+ by the body is thereby inhibited.

29.30

(a) Figure 29.32 in H&S shows disulfide units in the protein albumin. Disulfide bridges are formed by oxidative coupling of thiol units: 2RSH

RSSR + 2H+ + 2e–

Cys residues are thiols, so oxidative coupling leads to cross-linking or folding of chains:

HS SH

S S

(b) Cysteine, glutathione and albumin all act as soft S-donor ligands, compatible with soft metal ions such as Cd2+ (hard-soft acid-base principle). (c) Other soft metal ions targeted by metallothioneins are Hg2+ and Zn2+.

116

118 [289]

[288]

[291]

[294]

115 [284]

114

Uuo

113

Uut Uuq Uup Uuh

*Achary* HOUS_A01, 4/2/12, 16, 16:10, page: 1

‡

6.022 136 7  1023 mol1 1.380 658  1023 J K1 9.648 530 9  104 C mol1 8.314 510 J K1 mol1 6.626 075 5  1034 J s 1.097 373 15  107 m1 0.022 711 08 m3 mol1 ¼ 22:711 08 dm3 mol1 2.997 924 58  108 m s1 9.109 389 7  1031 kg 1.602 177 33  1019 C 1.672 623 1  1027 kg 1.674 928 6  1027 kg 1.660 540 2  1027 kg 5.291 772 49  1011 m 8.854 187 816  1012 F m1 9.274 015 4  1024 J T1 3.141 592 653 59

Value and SI units

An electron volt is a non-SI unit with a value of 1:602 18  1019 J; to compare eV and kJ mol1 units, it is necessary to multiply by the Avogadro number.

Standard state pressure 1 atmosphere pressure (non-SI) Energy‡

105 Pa ¼ 102 kPa ¼ 1 bar 1 atm ¼ 101 325 Pa 1 eV ¼ 96:4853 kJ mol1

L k F R h R Vm c me e mp mn mu ¼ 1 u a0 "0 B p

Avogadro constant Boltzmann constant Faraday constant Molar gas constant Planck constant Rydberg constant Molar volume of an ideal gas at 105 Pa (1 bar) and 273 K Speed of light in a vacuum Electron rest mass Charge on an electron (elementary charge) Proton rest mass Neutron rest mass Atomic mass unit Bohr radius Permittivity of a vacuum Bohr magneton Ratio of circumference to diameter of a circle

Conversions

Symbol

Physical constant

Physical constants

*Achary* House_Ibc, 2/4/12, 10, 10:17, page: 1

Symbol

Ac Al Am Sb Ar As At Ba Bk Be Bi B Br Cd Cs Ca Cf C Ce Cl Cr Co Cu Cm Dy Es Er Eu Fm F Fr Gd Ga Ge Au

Element

Actinium Aluminium Americium Antimony Argon Arsenic Astatine Barium Berkelium Beryllium Bismuth Boron Bromine Cadmium Caesium Calcium Californium Carbon Cerium Chlorine Chromium Cobalt Copper Curium Dysprosium Einsteinium Erbium Europium Fermium Fluorine Francium Gadolinium Gallium Germanium Gold

Elements

89 13 95 51 18 33 85 56 97 4 83 5 35 48 55 20 98 6 58 17 24 27 29 96 66 99 68 63 100 9 87 64 31 32 79

Atomic number

227.03 26.98 241.06 121.75 39.95 74.92 210 137.34 249.08 9.01 208.98 10.81 79.91 112.40 132.91 40.08 252.08 12.01 140.12 35.45 52.01 58.93 63.54 244.07 162.50 252.09 167.26 151.96 257.10 19.00 223 157.25 69.72 72.59 196.97

Relative atomic mass / g mol 1

Hafnium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium Lead Lithium Lutetium Magnesium Manganese Mendelevium Mercury Molybdenum Neodymium Neon Neptunium Nickel Niobium Nitrogen Nobelium Osmium Oxygen Palladium Phosphorus Platinum Plutonium Polonium Potassium Praseodymium

Element

Hf He Ho H In I Ir Fe Kr La Lr Pb Li Lu Mg Mn Md Hg Mo Nd Ne Np Ni Nb N No Os O Pd P Pt Pu Po K Pr

Symbol

72 2 67 1 49 53 77 26 36 57 103 82 3 71 12 25 101 80 42 60 10 93 28 41 7 102 76 8 46 15 78 94 84 19 59

Atomic number

178.49 4.00 164.93 1.008 114.82 126.90 192.22 55.85 83.80 138.91 262 207.19 6.94 174.97 24.31 54.94 258.10 200.59 95.94 144.24 20.18 237.05 58.69 92.91 14.01 259 190.23 16.00 106.42 30.97 195.08 239.05 210 39.10 140.91

Relative atomic mass / g mol 1

Promethium Protactinium Radium Radon Rhenium Rhodium Rubidium Ruthenium Samarium Scandium Selenium Silicon Silver Sodium Strontium Sulfur Tantalum Technetium Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium

Element

Pm Pa Ra Rn Re Rh Rb Ru Sm Sc Se Si Ag Na Sr S Ta Tc Te Tb Tl Th Tm Sn Ti W U V Xe Yb Y Zn Zr

Symbol

61 91 88 86 75 45 37 44 62 21 34 14 47 11 38 16 73 43 52 65 81 90 69 50 22 74 92 23 54 70 39 30 40

Atomic number

146.92 231.04 226.03 222 186.21 102.91 85.47 101.07 150.35 44.96 78.96 28.09 107.87 22.99 87.62 32.06 180.95 98.91 127.60 158.92 204.37 232.04 168.93 118.71 47.90 183.85 238.03 50.94 131.30 173.04 88.91 65.41 91.22

Relative atomic mass / g mol

1

*Achary* House_Ifc, 2/4/12, 10, 10:17, page: 1