(a) The decomposition of AB₂ to AB and B is a first-order reaction with k = 2·8 × 10⁻⁷ s⁻¹ at T = 1000 K. The atomic weights of A…

(a) First-order decomposition of AB₂

Given k = 2.8 × 10⁻⁷ s⁻¹ (at 1000 K; "1000 °C" in the question appears to be a slip, as no k is given at 1000 °C, so the given k is used). The atomic weights are not needed, since the reaction is first order and the answer is in mass ratios.

(i) t½ = 0.693/k = 0.693 / (2.8 × 10⁻⁷ s⁻¹) = 2.48 × 10⁶ s = 2.48 × 10⁶ / 86400 = 28.7 days.

(ii) ln(m₀/m) = kt, so t = (1/k) ln(1/0.6) = 0.5108 / (2.8 × 10⁻⁷) = 1.82 × 10⁶ s = 21.1 days.

(iii) t = 35 × 86400 = 3.024 × 10⁶ s; kt = 2.8 × 10⁻⁷ × 3.024 × 10⁶ = 0.847.

m = m₀e^(−kt) = 1 × e^(−0.847) = 0.43 g.

(b) Radiative and non-radiative processes; Jablonski diagram

In a singlet state, all electron spins are paired (S = 0, multiplicity 1); in a triplet state two electrons are unpaired with parallel spins (S = 1, multiplicity 3), and the triplet lies below the singlet of the same configuration.

Radiative processes emit a photon: fluorescence is S₁ → S₀, spin-allowed and fast (about 10⁻⁹–10⁻⁷ s); phosphorescence is T₁ → S₀, spin-forbidden and slow (10⁻³ s to seconds).

Non-radiative processes emit no photon: vibrational relaxation (within a state), internal conversion (between states of the same multiplicity, e.g. S₂ → S₁) and intersystem crossing (between different multiplicities, e.g. S₁ → T₁).

In the Jablonski diagram, energy increases upwards and singlet levels (S₀, S₁, S₂) stand on the left and triplet (T₁) on the right, each with vibrational levels. Absorption raises the molecule S₀ → S₁ or S₂ (straight upward arrows). It relaxes by vibrational relaxation and internal conversion to S₁ (wavy arrows), then emits fluorescence S₁ → S₀, or crosses to T₁ and emits phosphorescence T₁ → S₀.

(c) Geometry and hybridisation of carbon atoms in cytosine and thymine

Cytosine: ring carbons C2 (C=O), C4 (bearing NH₂ and double-bonded to N3), C5 and C6 (C5=C6) are all sp² hybridised with trigonal planar geometry (bond angles about 120°).

Thymine: ring carbons C2 and C4 (both C=O) and C5 and C6 (C5=C6) are sp² with trigonal planar geometry. The methyl carbon on C5 is sp³ with tetrahedral geometry (about 109.5°).

(d) Electronic transitions in UV–visible spectra of actinide ions

1. 5f → 5f transitions: within the 5fⁿ configuration. They are Laporte-forbidden and weak, but stronger and broader than lanthanide 4f–4f bands since the 5f orbitals are more extended and more involved in bonding.
2. 5f → 6d transitions: 5fⁿ → 5fⁿ⁻¹6d¹; Laporte-allowed, so intense and broad, and mainly in the UV and in the visible for ions easily oxidised or reduced, e.g. U⁴⁺, Np³⁺ and Pu³⁺.
3. Charge-transfer transitions: ligand-to-metal (LMCT), intense and broad, found in high-valent actinides such as the uranyl ion (UO₂²⁺), whose bands are in the near-UV and visible region.

(e) Identification of A and B; trans-effect by polarisation theory

[PtCl₄]²⁻ + NO₂⁻ → A = [PtCl₃(NO₂)]²⁻.

A + NH₃ → B = trans-[PtCl₂(NO₂)(NH₃)]⁻, in which NH₃ has replaced the chloride trans to NO₂, so NO₂ and NH₃ are trans to each other.

Justification (polarisation theory, Grinberg): Pt²⁺ polarises the ligand L (NO₂⁻) and induces a dipole in it. This in turn induces a dipole in the metal, oriented so that its positive end is next to L and its negative end faces the trans ligand X (Cl⁻). The induced charge on Pt repels the electron cloud of X, weakening the Pt–X bond and making X labile. The more polarisable the ligand L, the larger the dipole and the stronger its trans-effect. NO₂⁻ is more polarisable than Cl⁻ or NH₃, so the Cl trans to NO₂ is labilised and replaced by NH₃, and not the Cl cis to NO₂.