Covalent bonding and molecular orbitals

A proton is 1836 times an electron's mass, so at comparable kinetic energies the nuclei move some √1836 ≈ 43 times more slowly. Born–Oppenheimer promotes that to an approximation: write Ψ ≈ χ(R)ψₑₗ(r; R), solve the electronic Schrödinger equation with R frozen as a parameter rather than treated as a coordinate, and add the nuclear repulsion e²/4πε₀R. The result, U(R), is a curve rather than a number, and only then is R released — U(R) becomes the potential in which the nuclei vibrate and rotate, its minimum the bond length and its curvature the force constant. The approximation is excellent while electronic states stay well separated: in H₂ the first electronic gap is over 10 eV against a vibrational quantum of 0.55 eV. It breaks where two electronic surfaces approach. At a conical intersection the nuclear and electronic motions couple strongly, and most of photochemistry happens exactly there.

The variational principle says ⟨Ĥ⟩ ≥ E₀ for any normalised trial function, with equality only for the true ground state — so a guess can only be too high, and better means lower. Take the minimal basis, one real 1s orbital on each proton, ψ = c_A φ_A + c_B φ_B. Minimising ⟨Ĥ⟩ under normalisation gives (H_AA − E)c_A + (H_AB − ES)c_B = 0 and its mirror; setting the 2×2 determinant to zero returns E± = (H_AA ± H_AB)/(1 ± S). You can shortcut the algebra entirely. The protons are identical, so Ĥ commutes with the operator that swaps them, and the eigenfunctions must be even or odd under that swap — which forces c_B = ±c_A before a single integral is evaluated. The even combination is σ_g, nodeless between the nuclei; the odd one is σᵤ*, with a nodal plane exactly at the midpoint. At R = 2.00 a₀ the normalisation constants are 0.561 and 1.100: the node removes amplitude from between the protons, and it has to reappear on them.

H_AB = ⟨φ_A|Ĥₑₗ|φ_B⟩ has no classical counterpart. It is an energy built from a product of two orbitals centred on different nuclei, so it is non-zero only where both have amplitude — the region between the protons. It is negative, and it drives the splitting: both levels move away from H_AA in proportion to H_AA⋅S − H_AB, so they would collapse back onto H_AA only if H_AB fell to H_AA⋅S. The name matters here. This integral is often called the exchange integral, and for H₂⁺ that is simply wrong — one electron has nothing to exchange with. Fermi exchange first appears in H₂, where antisymmetrisation produces a genuine two-electron exchange integral K alongside the Coulomb integral J. Be careful with the energetics too. At the equilibrium separation the virial theorem forces E = −⟨T⟩, so the kinetic energy at Rₑ is higher than in the free atoms while the potential energy falls by twice as much. The picture of charge simply piling up between the nuclei gets the sign of the total right and the bookkeeping wrong.

Evaluate the integrals for 1s orbitals with the hydrogenic exponent ζ = 1 and minimise U(R). The minimum sits at R = 2.49 a₀ = 1.32 Å, 1.76 eV below H + H⁺. Experiment gives 1.06 Å and 2.79 eV, so the minimal basis recovers 63% of the binding and puts the bond about 25% too long. The variational principle also shows how to improve it without enlarging the basis: let the orbital exponent float, φ ∝ e^{(−ζr/a₀)}, and minimise over ζ as well as R. The best value is ζ = 1.24 at R = 2.00 a₀, where the well deepens to 2.35 eV — now at the measured bond length. Read that twice. Contracting the orbital raises the energy of a free hydrogen atom, costing about 0.78 eV at infinite separation, and still lowers the molecular energy at Rₑ. That contraction is the half of the bonding mechanism the naive charge-buildup story leaves out, and it is what the second slider in the figure does.

Put both electrons in σ_g and the spatial function ψ₊(1)ψ₊(2) is symmetric under exchange of the labels. The total wavefunction must be antisymmetric, so the spin function has to be the singlet [α(1)β(2) − β(1)α(2)]/√2, with total spin zero. That is the origin of the chemist's paired-spin covalent bond: not a rule about electrons wanting to pair, but the only spin state antisymmetry permits above a symmetric spatial function. The energy now needs two-electron integrals — a Coulomb integral J for the repulsion between the two charge clouds, and an exchange integral K, which is genuinely present here because there are two electrons to permute. Minimal-basis MO theory gives Dₑ ≈ 2.7 eV at 0.85 Å; Heitler–London valence bond, which keeps only the covalent product, gives 3.14 eV at 0.87 Å. Experiment gives 4.75 eV at 0.741 Å.

Expand the MO ground state and the flaw is visible without any integral. ψ₊(1)ψ₊(2) ∝ [φ_A(1)φ_B(2) + φ_B(1)φ_A(2)] + [φ_A(1)φ_A(2) + φ_B(1)φ_B(2)]: a covalent piece with one electron on each proton, and an ionic piece with both on the same one, weighted equally at every R. At R → ∞ the molecule is therefore described as half H + H and half H⁺ + H⁻, and making H⁺ + H⁻ from two hydrogen atoms costs IE − EA = 13.60 − 0.75 = 12.85 eV. The curve dissociates roughly 6.4 eV too high; a real restricted Hartree–Fock calculation gives about 7.7 eV, worse still because a single orbital, shared with the neutral atoms, describes the diffuse H⁻ badly. The cure is to stop insisting on one configuration. Mix in ψ₋(1)ψ₋(2), whose expansion is ionic minus covalent, and let the coefficient vary with R: it tends to −1 at large R, cancelling the ionic terms exactly, and at Rₑ it supplies just under half of the 1.1 eV of correlation energy that separates the Hartree–Fock limit, 3.64 eV, from the measured 4.75 eV.