Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2021 / Paper 2A / Section A / 19 / Solution

Each site has a pi-electron population $4/3$, and each adjacent pi bond order is $2/3$. The two carbon sites supply one pi electron each and the nitrogen lone pair supplies two, giving four electrons in two occupied molecular orbitals. Orthogonality to the given Homo and LUMO fixes the remaining occupied normalized orbital, up to overall sign, asFor doubly occupied orbitals, the Hückel orbital charge density and Hückel pi bond order follow fromThe diagonal sums to four electrons. The negative $P_{13}$ is an off-diagonal density-matrix element between nonadjacent sites, not an additional directly connected chemical bond.

Relative to neutral pi contributions $(1,2,1)$, the pi-system charges are $(-e/3,+2e/3,-e/3)$: electron density has moved from nitrogen toward the two carbon sites. Thus the physical electric dipole moment vector points from the midpoint of the carbon pair toward nitrogen along the molecular symmetry bisector. The conventional chemical dipole arrow, with a crossed positive tail, points oppositely: from nitrogen toward the carbon pair. This is only the pi contribution, not a prediction of the full sigma-plus-pi molecular dipole.