advanced27 min studyLesson 462 in Physics

The hydrogen molecular ion and bonding

Use a nonorthogonal two-orbital variational model to explain one-electron bonding while comparing total molecular energy with a declared dissociation threshold.

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What you’ll understand

  • Normalize symmetric and antisymmetric combinations with orbital overlap.
  • Calculate their variational energies at a fixed geometry.
  • Distinguish electronic stabilization from a stable total molecular bond.

The hydrogen molecular ion contains two protons and one electron, making it a clean setting for studying a chemical bond without electron-electron repulsion. Its electron can occupy a coherent state spread over both centers. A simple linear combination of atomic orbitals captures the interference behind bonding and antibonding states, but it is still a restricted variational approximation. To decide whether a geometry is bound, we must also include proton repulsion and compare the total energy with the same separated-fragment energy reference.

One electron feels two attractive centers

At fixed proton separation R, use the electronic Hamiltonian excluding proton-proton repulsion, consistent with the preceding lesson. It contains electron kinetic energy and attraction to each proton. The electronic eigenvalue is then supplemented by positive nuclear repulsion to obtain the nuclear potential curve. At very small R the nuclear repulsion grows strongly; at large R the appropriate dissociation channel is a hydrogen atom plus a proton. Overall center-of-mass and finite-nuclear-mass corrections are omitted at this level.

Choose normalized 1s-like functions φA\varphi_A and φB\varphi_B centered on the two nuclei. They overlap, so their inner product S is generally nonzero. Their sum has enhanced amplitude between the nuclei, while their difference has a node on the median plane for equivalent centers. The sign of an orbital amplitude is not an electric charge: it specifies the relative quantum phase needed when amplitudes are added. Density is obtained only after squaring the complete normalized wavefunction.

He=−ℏ22me∇2−e24πϵ0(1rA+1rB),U±(R)=ε±(R)+e24πϵ0RH_e=-\frac{\hbar^2}{2m_e}\nabla^2-\frac{e^2}{4\pi\epsilon_0}\left(\frac1{r_A}+\frac1{r_B}\right),\qquad U_\pm(R)=\varepsilon_\pm(R)+\frac{e^2}{4\pi\epsilon_0R}
r A and r B are electron distances to the fixed nuclei, and R is the internuclear distance. The surface U contains the repulsive nuclear energy once; electronic energies and the dissociation threshold must share a reference.↔ Scroll wide equations horizontally.

Overlap changes both normalization and energy

For equivalent centers, symmetric and antisymmetric trial states have normalization factors involving 1+S1+S and 1−S1-S, respectively. Evaluating the electronic expectation value gives energies (HAA±HAB)/(1±S)(H_{AA}\pm H_{AB})/(1\pm S) when the matrix elements are real and equal-center symmetry applies. Dividing only by two would be incorrect for overlapping functions. The same calculation is a generalized eigenproblem with an overlap matrix, not an ordinary eigenproblem in an orthonormal atomic basis.

The symmetric state is lower than the diagonal electronic expectation when HAB−SHAA<0H_{AB}-S H_{AA}<0. The overlap contribution means that the sign of an off-diagonal element by itself is insufficient without its convention and normalization. A variational calculation provides an upper bound to the exact ground electronic energy for the chosen Hamiltonian and geometry. Enlarging the trial space can improve that bound, but a two-function result does not give an exact equilibrium bond length, dissociation energy, or complete spectrum.

ψ±=φA±φB2(1±S),S=⟨φA∣φB⟩,ε±=HAA±HAB1±S\psi_\pm=\frac{\varphi_A\pm\varphi_B}{\sqrt{2(1\pm S)}},\qquad S=\langle\varphi_A|\varphi_B\rangle,\qquad \varepsilon_\pm=\frac{H_{AA}\pm H_{AB}}{1\pm S}
The functions are individually normalized, S is real with magnitude below one for distinct linearly independent centers, and H AA and H AB are electronic energy matrix elements at the same R.↔ Scroll wide equations horizontally.

A bond is an energy comparison across geometry

Enhanced internuclear electron density can favor bonding, but the complete energy balance also includes electronic kinetic energy, electron-nucleus attraction, and nuclear repulsion. A bound minimum requires the total surface to fall below an appropriate separated-fragment threshold over a range of geometries. Comparing one electronic eigenvalue with a bare atomic number while omitting nuclear repulsion can falsely suggest excessive binding. Even a total surface minimum lies below the lowest vibrational state by its zero-point energy.

The elementary molecular-orbital bond-order convention assigns one half to a single electron in a bonding orbital, but that count is a qualitative model label, not a directly measured bond energy. Spectroscopy tests actual energy differences and nuclear motion. For the example below, all matrix elements and the separated threshold are deliberately hypothetical to make the arithmetic transparent. They illustrate the calculation’s logic rather than quote computed or measured constants of the physical hydrogen molecular ion.

Electronic lowering is not the full bond energy

At one hypothetical geometry use HAA=−10 eVH_{AA}=-10\,\mathrm{eV}, HAB=−3 eVH_{AB}=-3\,\mathrm{eV}, and S = 0.20. Nuclear repulsion is +1 eV, and the stated separated-fragment threshold is −10 eV. Find the two electronic energies and decide whether the lower total energy at this geometry is below threshold.

  1. The symmetric electronic energy is (−10−3)/(1+0.20)=−65/6 eV≃−10.833 eV(-10-3)/(1+0.20)=-65/6\,\mathrm{eV}\simeq-10.833\,\mathrm{eV}. Its normalization includes constructive overlap.

  2. The antisymmetric electronic energy is (−10+3)/(1−0.20)=−8.75 eV(-10+3)/(1-0.20)=-8.75\,\mathrm{eV}. It is higher in this restricted model.

  3. Adding nuclear repulsion gives the lower total energy −65/6+1=−59/6 eV≃−9.833 eV-65/6+1=-59/6\,\mathrm{eV}\simeq-9.833\,\mathrm{eV}. This lies 1/6 eV above the specified −10 eV threshold, despite electronic lowering relative to H AA.

The result

The symmetric electronic state is lower, but this geometry does not bind relative to the stated threshold. Determining a bond requires scanning total energy versus separation, not inspecting electronic interference alone.

Explore the model

Interactive lab

Interactive lab · waves

Relative phase changes an amplitude sum

Use the two fixed-frequency waves as an interference analogy. Reset, set Phase difference = 0° and Amplitude ratio = 1, and predict doubled wave amplitude. For a separate run Reset, set Phase difference = 180° and Amplitude ratio = 1, and predict cancellation.

Relative phase changes an amplitude sumTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe lab sums traveling sine waves, not overlapping atomic orbitals. It does not normalize a molecular wavefunction, integrate electronic energy, or predict a bond; the relative-phase idea is the only intended connection.

Check your understanding

Why does lowering a fixed-geometry electronic energy not by itself establish a stable molecular bond?

Try it yourself

For a different hypothetical geometry take S = 0, HAA=−5 eVH_{AA}=-5\,\mathrm{eV}, and HAB=−1 eVH_{AB}=-1\,\mathrm{eV}. Nuclear repulsion is +0.5 eV and the separated threshold is −5 eV. Find total symmetric-state energy and its stabilization below threshold.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Orthogonal trial functions remove the overlap denominator, but nuclear repulsion still belongs in total energy.

Compare with the solution

The symmetric electronic energy is −5−1=−6 eV-5-1=-6\,\mathrm{eV}. Including repulsion gives U+=−5.5 eVU_+=-5.5\,\mathrm{eV}, which is 0.5 eV below the stated threshold. This geometry is energetically favorable in the hypothetical model. Establishing an equilibrium distance or a nuclear bound level still requires a potential curve and nuclear kinetic energy, not this one point alone.

Take it with you

  • Atomic-orbital overlap enters both normalization and the variational eigenproblem.
  • Relative orbital phase controls interference, while probability density is the squared full state.
  • Bonding requires a total-energy curve and a consistent dissociation reference, beyond an electronic energy lowering.

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