= Solution
\b[The physical curve has its bonding minimum near $c_2=+1/\sqrt2$, with ionic endpoints at $c_2=0,\pm1$.] Use electronic energies excluding the fixed $V_{NN}$. Define $a=h_{11}=h_{22}$, $b=h_{12}<0$, and $p=c_1c_2=c_2\sqrt{1-c_2^2}$. The one-electron contribution is
$$
E_1=2a+4bp.
$$
The Coulomb energy of the doubly occupied orbital is
$$
J_{\psi\psi}=\frac{e^2}{4\pi\epsilon_0}\iint
\frac{|\psi(\mathbf r)|^2|\psi(\mathbf r')|^2}{|\mathbf r-\mathbf r'|}\,d^3r\,d^3r'>0,
$$
so $E_{\rm phys}=E_1+J_{\psi\psi}$. For localized orthogonal orbitals, a transparent illustrative approximation neglects integrals involving the product $\phi_1\phi_2$ and retains the on-site repulsion $U$ and intersite direct repulsion $V$:
$$
J_{\psi\psi}\simeq U(c_1^4+c_2^4)+2Vc_1^2c_2^2
=U+2(V-U)p^2.
$$
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 $\pm1$, both electrons occupy one atom, giving the same ionic energy $E(\mathrm H^+\cdots\mathrm H^-)=2a+U$. At $c_2=-1/\sqrt2$ 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.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2a-orbital-mixing.png]
{title=Physical and noninteracting electronic energies as the occupied orbital changes}
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