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

(a) In benzene, each C is sp² hybridised; the unhybridised 2p orbitals overlap sideways to give a delocalised π cloud. Its six π electrons satisfy Hückel's 4n+2 rule with n = 1, so benzene is a classical aromatic with equal C–C bond lengths and high resonance stabilisation. Borazine, B₃N₃H₆, is isoelectronic with benzene: B and N are approximately sp², and the ring has six π electrons. However, B–N bonds are polar, so π electron density is uneven, largely on N; borazine is aromatic but weaker and not a true benzene analogue. Cyclic phosphazenes, e.g. N₃P₃Cl₆, have a P–N σ ring. Each N has a lone pair, which can donate into phosphorus orbitals. In the older view this is dπ–pπ bonding using P 3d orbitals; in the modern view it is negative hyperconjugation into P–N σ* or P-substituent orbitals. Thus the π delocalisation is not a simple carbon-like pπ–pπ system, even though six π electrons may be counted. It is substituent-dependent, electron density remains high on N, and the ring is best called inorganic aromatic rather than benzene-like or borazine-like.

(b) Collision theory treats reacting molecules as hard spheres. It assumes reaction occurs when two molecules collide with energy greater than or equal to activation energy Ea. Its basic rate constant is k = Z e^(−Ea/RT), so Rate = Z e^(−Ea/RT)[A][B], where Z is the collision frequency. Its limitations are:

• It ignores molecular geometry, orientation and the steric factor P. Observed rate = P Z e^(−Ea/RT), but P is not predicted.
• It considers only translational collision energy and ignores internal vibrational, rotational and electronic energy.
• It assumes every collision with E ≥ Ea reacts, whereas reaction must pass through a saddle point on a potential energy surface.
• It cannot explain unimolecular reactions, solvent effects, quantum tunnelling, or pressure-dependent reactions without extra assumptions.
• It gives no structural information about entropy of activation or mechanism.

Transition state theory (TST) explains these failures. It postulates a quasi-equilibrium between reactants and an activated complex at the highest point of the reaction coordinate: A + B ⇌ AB‡. The equilibrium constant is K‡ = [AB‡]/([A][B]). The activated complex decomposes to products with universal frequency ν = k_B T/h, where k_B is Boltzmann constant and h is Planck constant. Therefore, Rate = ν[AB‡] = (k_B T/h)K‡[A][B]. For Rate = k[A][B], this gives k = (k_B T/h)K‡. Using ΔG‡ = −RT ln K‡, so K‡ = e^(−ΔG‡/RT) (for a bimolecular reaction, a standard-state concentration factor is absorbed into K‡; the exponential form is unchanged), the Eyring equation is k = (k_B T/h)e^(−ΔG‡/RT) = (k_B T/h)e^(ΔS‡/R)e^(−ΔH‡/RT). Thus TST explains collision-theory limitations: the energy barrier is replaced by ΔG‡ or ΔH‡; orientation, structure and steric restrictions appear in ΔS‡ and partition functions; the steric factor P is approximately e^(ΔS‡/R); internal states and the potential energy surface are included.

(c) Langmuir's adsorption isotherm is derived under these assumptions: adsorption is monolayer; surface sites are identical; adsorbed molecules do not interact laterally; adsorption and desorption are dynamic; equilibrium exists.

Let θ be the fraction of surface sites covered, so vacant fraction = 1 − θ. The rate of adsorption is proportional to pressure P and vacant sites: R(ads) = k₁P(1 − θ). The rate of desorption is proportional to covered sites: R(des) = k₂θ. At equilibrium, R(ads) = R(des): k₁P(1 − θ) = k₂θ. Expanding: k₁P − k₁Pθ = k₂θ. Collecting θ terms: k₁P = θ(k₂ + k₁P). Hence θ = k₁P/(k₂ + k₁P). Divide numerator and denominator by k₂: θ = (k₁/k₂)P / [1 + (k₁/k₂)P]. Let b = k₁/k₂, with units Pa⁻¹. Then θ = bP/(1 + bP). If x/m is mass adsorbed per unit mass of adsorbent and is proportional to θ, then x/m = a bP/(1 + bP). Here θ is dimensionless, P is in Pa, b is in Pa⁻¹, k₁ in Pa⁻¹ s⁻¹ and k₂ in s⁻¹.

Limiting cases:

• Low pressure, bP ≪ 1: θ = bP/(1 + bP) ≈ bP. Also R(ads) ≈ k₁P. Therefore adsorption is first-order in pressure.
• High pressure, bP ≫ 1: θ = bP/(1 + bP) ≈ 1. Here 1 − θ ≈ 1/(bP), so R(ads) = k₁P(1 − θ) ≈ k₂, independent of P. Therefore adsorption is zero-order in pressure.

Thus Langmuir adsorption follows first-order kinetics at low pressure and zero-order kinetics at high pressure.