(a) (i) Predict the Lewis structures and geometry of the following as per VSEPR theory : (1) SO₂, (2) I₃⁻, (3) SF₄, (4) BrF₅, …

(a)(i) Lewis structures and VSEPR geometry (10 marks)

Count the valence electrons on the central atom plus those of the bonded atoms (and the charge), divide by two to get the electron pairs, place bond pairs and lone pairs, and pick the geometry from the number of electron domains, with lone pairs occupying the most spacious positions.

• SO₂: 6 + 2 × 6 = 18 electrons. S is doubly bonded to one O and singly to the other, by resonance equivalent to two S=O bonds, with one lone pair on S. Three electron domains give trigonal planar electron geometry, so the molecule is bent (V-shaped), O-S-O about 119°.
• I₃⁻: 7 + 2 × 7 + 1 = 22 electrons. The central I has two bond pairs and three lone pairs, five domains in a trigonal bipyramid, with the three lone pairs in the equatorial plane. The molecule is linear, I-I-I 180°.
• SF₄: 6 + 4 × 7 = 34 electrons. S has four bond pairs and one lone pair, five domains; the lone pair lies in an equatorial position, so the shape is see-saw (disphenoidal). The axial F-S-F angle is about 173° and the equatorial one about 102°.
• BrF₅: 7 + 5 × 7 = 42 electrons. Br has five bond pairs and one lone pair, six domains in an octahedron; the molecule is square pyramidal, with the four basal F bent slightly away from the lone pair (F-Br-F about 85 to 90°).
• XeF₄: 8 + 4 × 7 = 36 electrons. Xe has four bond pairs and two lone pairs, six domains in an octahedron, the lone pairs opposite each other (trans, to reduce repulsion). The molecule is square planar, F-Xe-F 90° and non-polar.

(a)(ii) Molecular orbital diagram of F₂ (10 marks)

Each F atom has 7 valence electrons (2s² 2p⁵), so F₂ has 14. Because the 2s-2p energy gap in fluorine is large, there is little s-p mixing, and the order of the molecular orbitals is σ2s < σ*2s < σ2p < π2p (two, degenerate) < π*2p (two, degenerate) < σ*2p.

Filling the 14 electrons gives the configuration σ2s² σ*2s² σ2p² π2p⁴ π*2p⁴, with σ*2p empty.

Bond order = (bonding − antibonding)/2 = (8 − 6)/2 = 1, a single F-F bond. All the electrons are paired, so F₂ is diamagnetic. The low bond order and the repulsion between lone pairs explain the weak F-F bond (about 158 kJ mol⁻¹).

(b)(i) Colours of [Co(H₂O)₆]²⁺ and [CoCl₄]²⁻ (10 marks)

Both are Co(II), d⁷, and their colour arises from d-d transitions, but the intensities differ greatly because of the symmetry of the complex.

[Co(H₂O)₆]²⁺ is octahedral and has a centre of symmetry. The d-d transitions (⁴T₁g(F) → ⁴T₁g(P), at about 19 500 cm⁻¹, in the green region) are Laporte-forbidden because they are g → g (no change of parity); they are spin-allowed, but they gain only a little intensity from vibronic coupling, so the molar absorptivity is very small (about 5 to 10 L mol⁻¹ cm⁻¹). The green light is weakly absorbed, so the solution is a pale pink.

[CoCl₄]²⁻ is tetrahedral and has no centre of symmetry. The d-orbitals can mix with the p-orbitals of the same metal, which relaxes the Laporte restriction, and the molar absorptivity is greater by a factor of about 100 (about 600 L mol⁻¹ cm⁻¹). In addition, Δt is only about 4/9 of Δo and the chloride is a weak-field ligand, so the absorption (⁴A₂ → ⁴T₁(P)) moves to lower energy, about 15 000 cm⁻¹, in the red and orange region. The red and orange light is strongly absorbed and the blue light is transmitted, so the complex is a deep blue.

(b)(ii) EAN rule and calculations (10 marks)

Sidgwick's effective atomic number (EAN) rule says that a metal in a complex tends to gain electrons from the ligands until the total number of electrons around it equals that of the next noble gas. EAN = (atomic number Z of the metal) − (oxidation state) + 2 × (number of ligands, for two-electron donors).

• [Cr(CO)₆]: Cr is 0, so EAN = 24 − 0 + 2 × 6 = 36 (krypton). The rule is obeyed.
• [Fe(CN)₆]⁴⁻: Fe is +2, so EAN = 26 − 2 + 12 = 36. The rule is obeyed.
• [Co(NH₃)₆]³⁺: Co is +3, so EAN = 27 − 3 + 12 = 36. The rule is obeyed.
• [Ni(NH₃)₆]²⁺: Ni is +2, so EAN = 28 − 2 + 12 = 38. This exceeds 36, so the rule is not obeyed. The complex is stable anyway (high-spin d⁸, octahedral), which shows that the rule is not general.

(c) Bonding in [Re₂Cl₈]²⁻ (10 marks)

Each Re is Re(III), d⁴, so the ion has eight metal d electrons. The ion has a very short Re-Re distance of about 224 pm, and the two ReCl₄ units are eclipsed (D₄h). With the z axis along the Re-Re bond, each Re uses its d(x²−y²), s and p orbitals for the four Re-Cl bonds, leaving d(z²), d(xz), d(yz) and d(xy) for metal-metal bonding:

• d(z²) + d(z²) overlap head-on to give a σ bond;
• d(xz) + d(xz) and d(yz) + d(yz) overlap sideways to give two π bonds;
• d(xy) + d(xy) overlap face-to-face to give a δ bond, which needs the eclipsed arrangement.

The eight electrons fill σ² π⁴ δ², leaving the antibonding δ*, π* and σ* levels empty. The bond order is (8 − 0)/2 = 4, a quadruple bond consisting of one σ, two π and one δ bond. The δ bond is weak, and the δ → δ* transition gives the blue colour. If the two ends were rotated by 45° into the staggered form the δ overlap would be lost; the eclipsed geometry is retained even though the Cl···Cl repulsions favour staggering, evidence for the δ bond. The complex is diamagnetic.