Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section H / 33 / a / Solution

Each $\mathrm{C}{\mathrm{p}}^{\ast }\mathrm{Al}$ fragment leaves two electrons after allocating one aluminium valence electron to its monovalent cyclopentadienyl ligand. Four fragments therefore give eight cluster electrons. In the tetrahedral arrangement, fragment lone-pair/donor and vacant acceptor orbitals combine into delocalized cluster orbitals. The occupied low-energy set has $a_1^2{t}_2^6$ symmetry, analogous to a filled 1S and 1P shell; higher combinations remain empty. This is a multicentre description, not six independent two-electron Al–Al bonds, which would require twelve electrons.

A schematic molecular-orbital diagram should show the occupied nondegenerate $a_1$ level below the occupied triply degenerate $t_2$ set, separated from the empty higher levels. Detailed energies and ordering of the empty levels require an orbital calculation. On descending Group 13, stabilization of the $n{s}^2$ pair and reduced favourable $s$–$p$ mixing can weaken the tendency towards the same compact Al-like bonding pattern; metal–metal distances and preferred aggregation change. Ligand size, donor/acceptor properties and hapticity also tune fragment orbital energies and steric access, so neither substitution of the metal nor of Cp* is electronically innocent.