= 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 are
$$
M_{B_{1u}}=\begin{pmatrix}1&\sqrt2\\\sqrt2&1\end{pmatrix},
\qquad
M_{B_{3g}}=\begin{pmatrix}1&\sqrt2\\\sqrt2&-1\end{pmatrix}.
$$
Their <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> are
$$
\boxed{\begin{array}{c|c|c}
\varepsilon&\text{irrep}&\text{electrons}\\\hline
\alpha+(1+\sqrt2)\beta&B_{1u}&2\\
\alpha+\sqrt3\beta&B_{3g}&2\\
\alpha+\beta&B_{2g}&2\\
\alpha+(1-\sqrt2)\beta&B_{1u}&0\\
\alpha-\beta&A_u&0\\
\alpha-\sqrt3\beta&B_{3g}&0
\end{array}}
$$
All 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>.
Back to article page
