= Solution
\b[Two <electrons> can occupy a fully bonding orbital delocalized equally over all three <hydrogen> atoms.] Three $1s$ <atomic orbitals> on an equilateral triangle form one totally symmetric $a_1'$ combination and a degenerate pair $e'$ of higher orbitals. In an orthonormal equal-interaction model with diagonal energy $\alpha$ and off-diagonal $\beta<0$,
$$
E(a_1')=\alpha+2\beta,\qquad E(e')=\alpha-\beta.
$$
The lower orbital is proportional to $1s_A+1s_B+1s_C$, with constructive overlap along every edge. Its two <electrons> give a <three-centre two-electron bond> and equivalent interactions between every pair of nuclei. At the equilibrium distance, this bonding competes with nuclear <Coulomb force> to favour the symmetric triangular arrangement. The occupied shell is nondegenerate and closed, so there is no first-order Jahn-Teller incentive to distinguish the three edges. A straight-line or localized single-bond description would miss that equally shared bonding pattern.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1-triangular-orbitals.png]
{title=Original three-site orbital filling for the two-electron cation and the effect of a third electron on a degenerate triangular shell}
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