Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 2 / a / i / Solution

Carbon dioxide is linear, with two equivalent C–O bonds of formal bond order two; its Homo is predominantly oxygen nonbonding and its LUMO is C–O ${\pi }^{\ast }$ antibonding. The carbon uses two opposite $sp$ hybrids for its sigma bonds and two perpendicular p atomic orbitals for pi bonding. A localized $\mathrm{O}=\mathrm{C}=\mathrm{O}$ picture assigns each terminal oxygen an $s{p}^2$ sigma/lone-pair framework and one p orbital for its local pi bond.

In a symmetry-adapted description, each of the two perpendicular sets of oxygen and carbon p functions forms bonding, oxygen-centred nonbonding and antibonding combinations. The occupied nonbonding pair is labelled $1{\pi }_g$ and the lowest empty pi pair $2{\pi }_u^{\ast }$. The latter has appreciable amplitude at carbon, explaining its electrophilic reactivity. Hybridization of a terminal atom with only one bonded neighbour is a choice of localized basis, not an independently measurable bond angle; an alternative oxygen $sp$ plus p basis can represent the same delocalized system.

There are 16 valence electrons: four in the two sigma bonds, four in the two net pi bonds, and eight in the remaining oxygen lone-pair combinations. These local electron counts and the delocalized orbital description are alternative bookkeeping schemes, not additional electrons.