(a) Explain the common structural features of two major iron-containing proteins—haemoglobin and cytochrome c present in the…

(a) Haemoglobin and cytochrome c: structure and coordination chemistry

Common features: both are haem proteins in which Fe is held by the four pyrrole nitrogens of a porphyrin ring (protoporphyrin IX) and completed by axial ligands from the protein. Haem is held in a hydrophobic pocket that protects Fe from water and from irreversible oxidation.

Differences: in haemoglobin (haem b, non-covalently bound) one axial site is the proximal histidine (F8) and the other is free for O₂; the distal histidine only H-bonds O₂. In cytochrome c (haem c, bound by thioether links to two cysteines) both axial sites are occupied (His18 and Met80); it cannot bind O₂ and functions as an electron carrier, cycling Fe(II)/Fe(III).

Deoxyhaemoglobin: Fe(II) (d⁶) is five-coordinate and high-spin (t₂g⁴e_g², four unpaired electrons), magnetic moment √24 ≈ 4.9 BM; it is larger than the porphyrin hole and lies about 0.4–0.6 Å above the ring plane.

Oxyhaemoglobin: O₂ binds as the sixth ligand; the ion becomes low-spin (t₂g⁶), diamagnetic (μ = 0), smaller, and moves into the ring plane, pulling the proximal histidine and triggering the T → R change. The formal oxidation number of Fe is +2 in both forms, although the bonding is better described as Fe(III)–superoxide with antiferromagnetic coupling.

(b) Ground-state configurations

• ₅₉Pr: [Xe] 4f³ 6s²
• ₆₃Eu: [Xe] 4f⁷ 6s² (half-filled f subshell)
• ₆₄Gd: [Xe] 4f⁷ 5d¹ 6s² (half-filled f plus one d electron)

(c) Bimolecular surface reaction (Langmuir–Hinshelwood)

A and B adsorb competitively on the same sites and react on the surface: A(ads) + B(ads) → products. Rate = kθ_Aθ_B.

Langmuir isotherms for competitive adsorption: θ_A = K_AP_A/(1 + K_AP_A + K_BP_B) and θ_B = K_BP_B/(1 + K_AP_A + K_BP_B).

Rate = kK_AK_BP_AP_B / (1 + K_AP_A + K_BP_B)²

(i) Weak adsorption of both: K_AP_A ≪ 1 and K_BP_B ≪ 1, so the denominator ≈ 1 and rate ≈ kK_AK_BP_AP_B, first order in each reactant (second order overall).

(ii) A much more strongly adsorbed: K_AP_A ≫ 1 + K_BP_B, so rate ≈ kK_BP_B/(K_AP_A): first order in B and negative first order in A, since A covers the surface and blocks B. The rate passes through a maximum as P_A increases.

(d) Electronic spectrum of [Ni(OS(CH₃)₂)₆]²⁺

Ni²⁺ is d⁸, octahedral, with six weak-field O-donor DMSO ligands. The ground term is ³A₂g, and three spin-allowed d–d bands are expected:

• ν₁: ³A₂g → ³T₂g(F), the lowest energy, near-infrared (equal to Δ₀).
• ν₂: ³A₂g → ³T₁g(F), in the visible/near-infrared.
• ν₃: ³A₂g → ³T₁g(P), the highest energy, in the visible to near-UV.

Weak spin-forbidden bands may also appear (e.g. to ¹E_g), often as shoulders.

(e) Reversible isothermal expansion of an ideal gas

Given: Vᵢ = 0.05 L, Pᵢ = 8 atm, V_f = 0.40 L, P_f = 1 atm, T = 298 K.

nRT = PᵢVᵢ = 8 × 0.05 = 0.40 L atm = 0.40 × 101.3 = 40.5 J.

Work: w = −nRT ln(V_f/Vᵢ) = −40.5 × ln 8 = −40.5 × 2.079 = −84.3 J (work done by the gas = 84.3 J).

ΔU and ΔH: for an ideal gas at constant temperature, ΔU = 0 and ΔH = 0.

Heat: q = −w = +84.3 J.

ΔS = q_rev/T = 84.3/298 = 0.283 J K⁻¹ (nR ln 8).

Is q equal to ΔH? No. ΔH = q only for a process at constant pressure with only pV work. Here the pressure falls from 8 atm to 1 atm, so q (84.3 J) differs from ΔH (0), which is zero since enthalpy of an ideal gas depends only on T.