Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 2 / a / i / Solution

Suppress the filled 1s core pair, which contributes zero net bond order. For N${}_2$, use the valence orderingand occupy it as $({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p}{)}^4({\sigma }_{2p_z}{)}^2$.

For O${}_2$, the ordering of the bonding 2p levels is reversed: ${\sigma }_{2p_z}<{\pi }_{2p}$. Occupancy is $({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p_z}{)}^2({\pi }_{2p}{)}^4({\pi }_{2p_x}^{\ast }{)}^1({\pi }_{2p_y}^{\ast }{)}^1$, with parallel spins in the two degenerate antibonding pi orbitals.

Li${}_2$ has $({\sigma }_{2s}{)}^2$; Be${}_2$ has $({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2$ in the elementary valence description. Empty 2p-derived levels must also be included to discuss mixing and Be${}_2$'s weak bonding; their precise energies are not set by this qualitative diagram.