= Solution
\b[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, as
$$
\psi_1=\frac1{\sqrt6}(\phi_1+2\phi_2+\phi_3).
$$
For doubly occupied orbitals, the <Hückel orbital charge density> and <Hückel pi bond order> follow from
$$
P_{ij}=2\sum_{a\in\mathrm{occupied}}c_{ia}c_{ja},\qquad
P=\begin{pmatrix}
4/3&2/3&-2/3\\
2/3&4/3&2/3\\
-2/3&2/3&4/3
\end{pmatrix}.
$$
The 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 \b[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.
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