(a) How does the bonding in cyclic phosphazene differ from that of benzene and borazine? (10 marks) (b) What are the limitations…

(a) Bonding in cyclic phosphazene versus benzene and borazine Benzene (C₆H₆) is a planar ring with six equal C–C bonds (1.39 Å). Each carbon has one p-electron, and six π electrons are delocalised continuously in a pπ–pπ cloud above and below the ring (Hückel 4n+2), giving true aromaticity. Borazine (B₃N₃H₆, “inorganic benzene”) is isoelectronic and planar with equal B–N bonds (1.44 Å), also pπ–pπ bonded, but N donates its lone pair into empty B p-orbitals. The B–N bond is polar, so delocalisation is weaker, and borazine adds HCl or water across B=N. Cyclic phosphazenes (NPX₂)ₙ, e.g., (NPCl₂)₃, have alternating P and N with equal P–N bonds of about 1.58 Å, shorter than a P–N single bond (1.77 Å). The σ-framework uses sp³-type P and sp²-type N. The extra π-bonding uses N p-orbitals with P d-orbitals (pπ–dπ) (or, in the modern view, negative hyperconjugation), which overlap in separated three-centre “islands” (Dewar) rather than a continuous ring current. Hence the ring need not be planar (the tetramer is puckered) and phosphazene is not aromatic in the benzene sense; the π bonding is highly polar and involves d-orbitals.

(b) Limitations of collision theory and the transition state theory Collision theory, k = Z·exp(−E_a/RT), assumes that molecules are hard spheres, and that every collision with energy ≥ E_a leads to reaction. Limitations: (i) it ignores molecular orientation and structure, so calculated rates exceed experimental ones, particularly for complex molecules, and a steric factor P (<1) must be added empirically; (ii) it cannot predict E_a or the pre-exponential factor; (iii) it fails for reactions in solution and for unimolecular reactions; (iv) it ignores internal (vibrational/rotational) energy and entropy changes. Transition state (activated complex) theory proposes that reactants A + B pass through a high-energy activated complex at the saddle point of the potential energy surface, A + B ⇌ [AB]‡ → products, in a quasi-equilibrium with constant K‡, and the complex breaks down at the frequency k_BT/h. In the energy profile, the difference between the energy of the reactants and the top of the barrier is the activation energy (not a bond energy).

k = κ(k_BT/h)K‡ = (k_BT/h) exp(ΔS‡/R) exp(−ΔH‡/RT) (Eyring equation), with E_a ≈ ΔH‡ + RT. The steric factor appears as the entropy of activation: a more ordered complex has negative ΔS‡, so P < 1. It thus explains orientation requirements and gives rate in terms of thermodynamic quantities.

(c) Langmuir adsorption isotherm Assumptions: adsorption is confined to a monolayer, the surface is uniform with identical sites, molecules do not interact, and adsorption is dynamic. Let θ be the fraction of sites covered, so (1 − θ) is free, and P the gas pressure. Rate of adsorption = k_a P (1 − θ); rate of desorption = k_d θ. At equilibrium k_aP(1 − θ) = k_dθ. With K = k_a/k_d: KP(1 − θ) = θ, so θ = KP/(1 + KP). Equivalently, x/m = (x/m)_max KP/(1 + KP). Limiting cases: at low pressure KP ≪ 1, θ ≈ KP, so coverage and rate of adsorption are directly proportional to P (first order). At high pressure KP ≫ 1, θ → 1; the surface is saturated, and the rate is independent of P (zero order). In between, the order is fractional (0 to 1). Surface catalysed reactions use this to explain order changes.