(a) Complete the following reactions and draw the structure of the major product(s) formed: (i) 2XeF₆ + SiO₂ → ____ (50 °C) (ii)…

(a) Reactions and products

(i) 2XeF₆ + SiO₂ → 2XeOF₄ + SiF₄. XeOF₄ has five bond pairs and one lone pair on Xe (AX₅E), so it is square pyramidal, with O at the apex and the lone pair trans to O.

(ii) NH₄NO₃ → N₂O + 2H₂O. This is an internal redox reaction. N₂O is linear and unsymmetrical (N–N–O).

(iii) N₂O + 2NaNH₂ → NaN₃ + NaOH + NH₃ (Wislicenus method). The product is sodium azide, with the linear, symmetrical N₃⁻ ion [N=N=N]⁻.

(iv) Diborane and ammonia first give the diammoniate [BH₂(NH₃)₂]⁺[BH₄]⁻, which on heating to about 180 °C gives borazine, B₃N₃H₆, and H₂: 3B₂H₆ + 6NH₃ → 2B₃N₃H₆ + 12H₂. The printed ratio 2B₂H₆ : 6NH₃ does not balance for these products. Borazine is a planar six-membered ring of alternating B–H and N–H units, with sp² B and N ("inorganic benzene").

(v) 6SCl₂ + 16NH₃ → S₄N₄ + 2S + 12NH₄Cl, which balances for every element. S₄N₄ is a cage with the four N atoms in a square plane, the four S atoms above and below the plane in a flattened tetrahedron, and two transannular S···S contacts.

(b) Stereoisomers of [Co(CN)₂(H₂O)₂(NH₃)₂]⁺

This is an octahedral [Ma₂b₂c₂] complex, with five geometrical isomers, one of which is chiral, giving six stereoisomers.

1. All three pairs trans (achiral).
2. CN trans; H₂O and NH₃ each cis (achiral).
3. NH₃ trans; CN and H₂O each cis (achiral).
4. H₂O trans; CN and NH₃ each cis (achiral).
5. and 6. All three pairs cis: a non-superimposable pair of enantiomers, as it has no mirror plane.

(c) IUPAC names

• (i) tetraaquadichloridochromium(III) chloride.
• (ii) tris(ethane-1,2-diamine)cobalt(III) hexacyanidoferrate(III).
• (iii) sodium diaquabis(oxalato)chromate(III).
• (iv) aquadicarbonylchloridopalladium(II) ion.
• (v) sodium octaaquadi-μ-hydroxidodivanadate(4−). Charge balance on the anion gives V in the −1 state, which is chemically unusual and suggests the printed formula may be a misprint; the name follows the formula as printed.

Ligands are cited alphabetically (prefixes ignored), anionic ligands end in -o, bis/tris are used for complex ligands, an anionic complex ends in -ate, and the cation is named first.

(d) Rubredoxin

Rubredoxin is the simplest iron–sulfur protein, a small protein (about 54 residues, 6 kDa) from anaerobic bacteria.

• Structure: one Fe atom bound tetrahedrally to four cysteine thiolate S atoms, from two Cys-X-X-Cys segments, with no inorganic (acid-labile) sulfide.
• Redox and spectra: a one-electron carrier, Fe(III)/Fe(II), both high-spin; E° is near 0 V. The oxidised protein is red from S→Fe(III) charge transfer (about 490 and 380 nm) and gives an EPR signal near g = 4.3; the reduced form is almost colourless.
• Function: electron transfer in anaerobic metabolism.

Difference from ferredoxins: ferredoxins have two to four Fe atoms ([2Fe–2S], [3Fe–4S], [4Fe–4S]) held by cysteine and bridging inorganic sulfide, with antiferromagnetically coupled Fe, and their E° is markedly negative (about −0.25 to −0.65 V). They act in photosynthesis, N₂ fixation and respiration.

(e) Framework electrons and structures (Wade's rules)

Each B–H contributes 2 skeletal electrons; each extra H adds 1 and each negative charge adds 1. n pairs = n + 1 closo, n + 2 nido, n + 3 arachno, where n is the number of boron atoms.

(i) B₇H₇²⁻: 7 × 2 + 2 = 16 electrons = 8 pairs = n + 1, so closo: pentagonal bipyramid.

(ii) B₁₁H₁₃²⁻: 11 × 2 + 2 + 2 = 26 electrons = 13 pairs = n + 2, so nido: an icosahedron with one vertex missing, the extra H atoms bridging the open face.

(iii) B₂H₇⁻: 2 × 2 + 5 + 1 = 10 electrons = 5 pairs = n + 3, so arachno: [H₃B–H–BH₃]⁻, two BH₃ units held by one B–H–B bridge.