 Solution

Work with $M=(H-\alpha I)/\beta$. Its nearest-neighbour ring entries are $1$, and its diagonal entries are $1$ on the nitrogen sites and $0$ on the carbon sites. Because $M$ commutes with each molecular symmetry operation, it only mixes symmetry-adapted linear combinations with the same irreducible representation.

In the bases $(N_{+},C_0)$ and $(N_{-},C_y)$, the two nontrivial matrix blocks areTheir characteristic polynomials are $(1-\lambda {)}^2-2$ and ${\lambda }^2-3$. The remaining single-function blocks are $M_{B_{2g}}=1$ and $M_{A_u}=-1$. Thus, ordered from lowest to highest physical energy because $\beta <0$, the molecular orbitals areAll six levels are nondegenerate in this parametrization. The right-hand ladder of the diagram in part (ii) shows these levels and occupations. The occupied $B_{2g}$ level is the highest occupied molecular orbital, and the unoccupied $B_{1u}$ level immediately above it is the lowest unoccupied molecular orbital.