 Solution

The physical curve has its bonding minimum near $c_2=+1/\sqrt{2}$, with ionic endpoints at $c_2=0,\pm 1$. Use electronic energies excluding the fixed $V_{NN}$. Define $a=h_{11}=h_{22}$, $b=h_{12}<0$, and $p=c_1{c}_2=c_2\sqrt{1-c_2^2}$. The one-electron contribution isThe Coulomb energy of the doubly occupied orbital isso $E_{\mathrm{phys}}=E_1+J_{\psi \psi }$. For localized orthogonal orbitals, a transparent illustrative approximation neglects integrals involving the product ${\varphi }_1{\varphi }_2$ and retains the on-site repulsion $U$ and intersite direct repulsion $V$:Since normally $U>V$, spreading the occupied orbital over both atoms reduces the repulsion. At equilibrium the bonding one-electron term favours $p=+1/2$, so the minimum represents the Restricted Hartree-Fock method approximation to $E({\mathrm{H}}_2)$. At $c_2=0$ or $\pm 1$, both electrons occupy one atom, giving the same ionic energy $E({\mathrm{H}}^{+}\cdots {\mathrm{H}}^{-})=2a+U$. At $c_2=-1/\sqrt{2}$ the orbital is antibonding and the one-electron contribution is raised.

The diagram below uses an illustrative set of integrals to show these landmarks; it is not a numerical calculation for the molecule. Zero orbital overlap alone does not force every exchange or three-index Coulomb integral to vanish, so the simplified expression is an explicit sketching approximation, not a consequence of orthogonality.