University Physics IV · Molecular and Solid-State Physics · 12.1

Covalent Bonding & Molecular Orbitals

Two protons and one electron: the smallest molecule there is, and where chemistry stops being a rule and becomes an eigenvalue problem. Freeze the nuclei, mix the two 1s orbitals, let the variational principle pick the mixture. A bond length and a well depth come out — both wrong in instructive ways.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Nuclei are at least 1836 times heavier than electrons, so on the timescale an electron needs to rearrange, the nuclei barely move. Born–Oppenheimer takes that literally: clamp the nuclei at separation R, solve the electronic problem, add the nuclear repulsion, and the eigenvalue becomes a curve U(R) rather than a number. For H₂⁺ that electronic problem has no useful closed form, so you guess — and the variational principle makes guessing legitimate, because any normalised trial function returns an energy above the true ground state, never below.

Take the smallest honest basis, one 1s orbital on each proton, and write ψ = c_A φ_A + c_B φ_B. Symmetry does half the work: the protons are identical, so |c_A| = |c_B| and only ψ± = (φ_A ± φ_B)/√(2 ± 2S) survive. The secular determinant then gives E± = (H_AA ± H_AB)/(1 ± S) — one level down, one up, split by the resonance integral H_AB and skewed by the overlap S.

The cost shows in the numbers: this model binds H₂⁺ by 1.76 eV at 1.32 Å where experiment says 2.79 eV at 1.06 Å, and for H₂ the same single-configuration wavefunction dissociates to the wrong atoms.

Simple definition: A molecular orbital is a one-electron state that spans every nucleus in the molecule; in the LCAO approximation it is a weighted sum of atomic orbitals, with the weights fixed by minimising the energy.
Example: For H₂⁺ symmetry leaves only ψ± = (φ_A ± φ_B)/√(2 ± 2S). At R = 2.49 a₀ the overlap is S = 0.461, so 2 + 2S = 2.92 and the bonding orbital is 0.585(φ_A + φ_B), sitting 1.76 eV below H + H⁺.

Born–Oppenheimer potential curveU(R) = Eₑₗ(R) + e²/4πε₀R

Turns a four-body quantum problem into one curve you can plot, minimise, and test against a measured bond length.

R is held fixed while Eₑₗ is solved, so it is a parameter, not a coordinate; U in eV, R in m or a₀ = 0.0529 nm.

LCAO trial function for H₂⁺ψ± = (φ_A ± φ_B)/√(2 ± 2S), S = ⟨φ_A|φ_B⟩

Two identical nuclei force |c_A| = |c_B|, so the whole variational problem collapses to a choice of sign.

φ_A and φ_B are 1s orbitals on each proton; S is dimensionless, running from 1 at R = 0 to 0 at R = ∞.

Secular energies, overlap keptE± = (H_AA ± H_AB)/(1 ± S)

At R = 2.49 a₀: H_AA = −24.27 eV, H_AB = −14.14 eV, S = 0.461, giving E₊ = −26.29 eV and E₋ = −18.79 eV.

H_AA = ⟨φ_A|Ĥₑₗ|φ_A⟩ and H_AB = ⟨φ_A|Ĥₑₗ|φ_B⟩ are energies in J or eV; H_AB is negative at all bonding separations.

Overlap and resonance integrals, 1s basisS = e⁻ʷ(1 + w + w²/3), H_AB = E₁s S − (e²/4πε₀a₀)e⁻ʷ(1 + w)

Both decay exponentially, so covalent bonding is short-ranged: by w = 6 the overlap is down to 0.047.

w = R/a₀ is dimensionless, E₁s = −13.61 eV and e²/4πε₀a₀ = 27.21 eV. At w = 2.49, S = 0.461.

The split is not symmetric(E₋ − H_AA)/(H_AA − E₊) = (1 + S)/(1 − S)

Antibonding rises 2.7 times as far as bonding falls — which is why He₂⁺ is a molecule and He₂ is not.

Both shifts carry the same numerator, H_AA⋅S − H_AB; only the denominator differs. At S = 0.461 the ratio is 2.71.

Two electrons: exchange and the singletΨ = ψ₊(1)ψ₊(2) · [α(1)β(2) − β(1)α(2)]/√2

The chemist's electron pair is a corollary of Fermi antisymmetry, not an extra rule. H₂⁺ has no exchange term at all.

The spatial part is symmetric under 1 ↔ 2, so the spin part must be antisymmetric: total spin S = 0.

Clamping the nuclei is what makes U(R) exist

A proton is 1836 times an electron's mass, so at comparable momenta 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 11 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.

Symmetry writes down the trial function for you

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.

The resonance integral, not exchange, is what binds

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 is the whole source of the splitting: set H_AB = 0 and both levels collapse back onto H_AA. 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.

Numbers from the minimal basis, and how far off they are

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.

H₂: a second electron, and why its spins are paired

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 S = 0. 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.

The failure you can see at infinite separation

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 minimal basis 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 most of the 1.1 eV of correlation energy that separates the Hartree–Fock limit, 3.64 eV, from the measured 4.75 eV.

Change one variable at a time

Make the relationship visible.

Interactive model

Nuclear separation R2.50 a₀

Orbital exponent ζ1.00

At ζ = 1 drag R to 2.50 a₀ and read the minimum, 1.76 eV deep. Now raise ζ: both curves lift off the dashed line at large R, because a contracted orbital is a worse free atom — yet the well deepens to 2.35 eV and slides in to 2.00 a₀.

OVERLAP S0.458

RESONANCE H_AB-14.05 eV

σg CURVE U(R)-1.764 eV

σg–σu* GAP7.46 eV

Live interpretationOVERLAP S: 0.458. RESONANCE H_AB: −14.05 eV. σg CURVE U(R): −1.764 eV. σg–σu* GAP: 7.46 eV

Catch the common trap

Explain before calculating.

Misconception check

The usual mental picture is a symmetric splitting: the bonding level drops by |H_AB|, the antibonding level rises by the same amount, so one electron in each cancels out. Read the formula again. The two shifts measured from H_AA share the numerator H_AA⋅S − H_AB and differ only in their denominator, 1 + S against 1 − S, and 1 − S is the smaller. At the H₂⁺ equilibrium overlap S = 0.461 the antibonding level is pushed up 2.71 times as far as the bonding level is pulled down. Filling both is not neutral, it is a net penalty — which is why He₂⁺, two bonding electrons against one antibonding, is a real molecule bound by about 2.5 eV, while He₂ with two of each is held only by dispersion, about 1 meV, at 5.2 Å. Drop S and you have Hückel theory: useful for counting nodes, useless for asking whether a molecule exists.

In the LCAO treatment of H₂⁺ the two levels are E± = (H_AA ± H_AB)/(1 ± S). Near the equilibrium separation H_AA = −24.3 eV, H_AB = −14.1 eV and the overlap is S = 0.46. Measured from H_AA, how do the two shifts compare?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo hydrogen 1s orbitals sit on protons separated by R = 2.00 a₀. The overlap integral is S = e⁻ʷ(1 + w + w²/3) with w = R/a₀. Find S, then the normalisation constants of the bonding and antibonding LCAO orbitals ψ± = N±(φ_A ± φ_B).

w is a pure number: w = R/a₀ = 2.00.
Bracket first: 1 + w + w²/3 = 1 + 2.00 + 4.00/3 = 4.3333. And e⁻² = 0.13534.
S = 0.13534 × 4.3333 = 0.5865 — these two clouds overlap heavily at this separation.
Normalisation: ⟨ψ±|ψ±⟩ = N±²(1 + 1 ± 2S) = 1, so N± = 1/√(2 ± 2S). Here 2 + 2S = 3.1729 and 2 − 2S = 0.8271.
N₊ = 1/√3.1729 = 0.561 and N₋ = 1/√0.8271 = 1.100.

AnswerS = 0.586; ψ₊ = 0.561(φ_A + φ_B) and ψ₋ = 1.100(φ_A − φ_B). N₋ exceeds 1 because the node at the midpoint removes amplitude that has to reappear on the atoms.

MediumAt R = 2.49 a₀ the H₂⁺ integrals are H_AA = −24.27 eV, H_AB = −14.14 eV and S = 0.461, and the nuclear repulsion at that separation is +10.93 eV. Take the hydrogen ground state as −13.61 eV. Find both molecular energies, the depth of the bonding curve below H + H⁺, and the height of the antibonding curve above it.

E₊ = (H_AA + H_AB)/(1 + S) = (−24.27 − 14.14)/1.461 = −38.41/1.461 = −26.29 eV.
E₋ = (H_AA − H_AB)/(1 − S) = (−24.27 + 14.14)/0.539 = −10.13/0.539 = −18.79 eV. Note it lies above H_AA even though both integrals are negative.
These are electronic energies only. Add the nuclear repulsion, which is identical for both states: U₊ = −26.29 + 10.93 = −15.36 eV and U₋ = −18.79 + 10.93 = −7.86 eV.
The dissociation limit is one H atom plus a bare proton, −13.61 + 0 = −13.61 eV. So Dₑ = −13.61 − (−15.36) = 1.76 eV, while σᵤ* sits −7.86 + 13.61 = 5.75 eV above the limit — repulsive at every R.

AnswerDₑ = 1.76 eV at R = 2.49 a₀ = 1.32 Å, against the measured 2.79 eV at 1.06 Å: the minimal basis recovers 63% of the bond. The antibonding curve lies 5.75 eV above the separated atoms.

HardThe H₂ molecular-orbital ground state is Ψ = ψ₊(1)ψ₊(2) with ψ₊ = N(φ_A + φ_B). Expand it in atomic-orbital products, say what each term describes, and use IE(H) = 13.60 eV and EA(H) = 0.75 eV to estimate the error this wavefunction makes at R → ∞. Then name the repair and what else it buys.

Expand: Ψ/N² = (φ_A(1) + φ_B(1))(φ_A(2) + φ_B(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).
The first two terms put one electron on each proton — covalent, H + H. The last two put both on the same proton — ionic, H⁺ + H⁻. A single MO configuration locks their weights equal at every R, including R → ∞ where they should not be equal at all.
Cost of the ionic limit: making H⁺ + H⁻ from two neutral atoms takes IE − EA = 13.60 − 0.75 = 12.85 eV.
A 50/50 mixture therefore dissociates to ½(0) + ½(12.85) = 6.42 eV above 2H(1s). A real restricted Hartree–Fock curve is worse still, about 7.7 eV, because a minimal basis describes the diffuse H⁻ badly.
Repair: add the doubly excited configuration ψ₋(1)ψ₋(2), whose expansion is (ionic) − (covalent). Writing Ψ = c₁ψ₊ψ₊ + c₂ψ₋ψ₋, the choice c₂ = −c₁ cancels the ionic terms exactly and leaves 2c₁(covalent).
At R → ∞ the variational minimum drives c₂/c₁ to −1, giving the right limit; at Rₑ the same mixing recovers most of the 1.1 eV separating the Hartree–Fock limit, 3.64 eV, from the measured Dₑ = 4.75 eV.

AnswerThe restricted MO curve dissociates about 6.4 eV too high because it forces 50% ionic character at infinite separation. Two-configuration CI, mixing in σᵤ*², removes that and recovers most of the 1.1 eV correlation energy.