The Hydrogen Molecule Ion H₂⁺

Introduction

The hydrogen molecule ion, H₂⁺, is two protons sharing one electron. It is the simplest molecule of all, and with the nuclei held fixed its electronic Schrödinger equation can be solved exactly in elliptical coordinates. That makes it the perfect testing ground for the linear variation method: we can build approximate molecular orbitals from two hydrogen 1s orbitals, solve a 2 × 2 secular determinant, and then check how close the answers come to the exact bond length and bond energy.

Core explanation

The Hamiltonian. Label the protons A and B, separated by R, with the electron at distances r A and r B from them. In the Born–Oppenheimer approximation the nuclei are fixed and, in atomic units,

Ĥ = −½∇² − 1/r A − 1/r B + 1/R

The last term is the proton–proton repulsion, a constant for each R.

The trial function. Near nucleus A the electron should look like a hydrogen 1s electron on A, and similarly for B. The linear variation function is

ψ = c A 1s A + c B 1s B

Solving the secular problem. Because the two nuclei are identical, H AA = H BB = α and the secular determinant is exactly the homonuclear case. The roots are

E₊ = (α + β)/(1 + S) and E₋ = (α − β)/(1 − S)

with the normalised wavefunctions

ψ₊ = (1s A + 1s B)/√[2(1 + S)] (bonding, σg) ψ₋ = (1s A − 1s B)/√[2(1 − S)] (antibonding, σu )

Here α, β and S all depend on R. The overlap, for example, is S(R) = e^(−R)(1 + R + R²/3) in atomic units: close to 1 at small R and falling towards 0 as the atoms separate.

Electron density. For ψ₊ the density is proportional to 1s A² + 1s B² + 2(1s A)(1s B). The cross term adds density in the region between the nuclei, where the electron is attracted by both protons at once. For ψ₋ the cross term is subtracted, removing density between the nuclei and creating a nodal plane perpendicular to the bond axis, half-way between them.

Potential energy curves. Plotting E₊ and E₋ (including 1/R) against R gives two curves that both approach −0.5 Eₕ, the energy of H + H⁺, at large R. The σg curve dips to a minimum: a bound molecule. The σu curve rises steadily as R decreases: purely repulsive, with no bound state.

How accurate is it? The simple LCAO calculation with unmodified hydrogen 1s orbitals gives:

- equilibrium bond length R e ≈ 2.49 a₀ (132 pm) - dissociation energy D e ≈ 1.76 eV

The exact values are R e = 2.00 a₀ (106 pm) and D e = 2.79 eV. The LCAO model predicts a bond of the right order of magnitude but too long and too weak. Letting the orbital exponent vary with R (optimum about 1.24 at equilibrium) improves D e to about 2.35 eV and R e to 2.00 a₀: the orbitals contract towards the nuclei, as expected for an electron attracted by two protons.

Step-by-step reasoning

1. Fix the nuclei at separation R. 2. Build the trial function from 1s orbitals on each nucleus. 3. Use symmetry: α A = α B, so the homonuclear secular determinant applies. 4. Obtain E₊ and E₋ and their coefficient ratios c A = ±c B. 5. Repeat for many R values to trace the potential energy curves. 6. Locate the minimum of the bonding curve to find R e and D e.

Visual explanation

Picture contour plots of the two orbitals. The σg orbital looks like a single elongated cloud enclosing both nuclei, fattest at the centre. The σu orbital has two lobes of opposite phase, one around each nucleus, separated by a flat nodal plane where the density drops to zero.

Real-world analogy

Two ripples from stones dropped in a pond meet in the middle. If they are in step, the crests add and the water rises between the stones — like the bonding orbital. If they are out of step, crest meets trough and the water in the middle stays flat — like the antibonding node.

Real-world example

H₂⁺ is abundant in interstellar clouds, formed when cosmic rays ionise H₂. It reacts rapidly with H₂ to give H₃⁺, a key ion that drives much of the chemistry of space. The exact H₂⁺ energies also calibrate high-precision measurements of the proton-to-electron mass ratio.

Why?

Why does one electron create a bond at all? In the bonding orbital, density concentrated between the protons is attracted by both of them. Detailed analysis shows that the orbital also contracts near each nucleus, lowering the potential energy by more than enough to offset the proton–proton repulsion at R e.

Common misconception

"A bond needs a pair of electrons." H₂⁺ has a genuine bond with a single electron, bond order ½ and D e = 2.79 eV. Electron pairs strengthen bonds, but the essential ingredient is constructive overlap of orbitals.

Worked example

Question: Calculate the overlap integral S for two hydrogen 1s orbitals at R = 2.00 a₀ and hence the normalisation constant of ψ₊.

Reasoning: S = e^(−2.00)(1 + 2.00 + 4.00/3) = 0.1353 × 4.333 ≈ 0.586. Then N₊ = 1/√[2(1 + 0.586)] = 1/√3.172 ≈ 0.561.

Answer: S ≈ 0.586 and ψ₊ ≈ 0.561(1s A + 1s B).

Quick check

1. Why does the antibonding orbital of H₂⁺ have zero electron density at the mid-point of the bond? Answer: At the mid-point 1s A and 1s B have equal values, so their difference, and therefore the wavefunction, is exactly zero.

Exam focus

Be ready to write the H₂⁺ Hamiltonian, build normalised ψ₊ and ψ₋, sketch both potential energy curves and compare predicted and exact R e and D e. State clearly that the LCAO energy is an upper bound, so the predicted bond is weaker than the true one.

Advanced insight

The labels g and u (gerade and ungerade) describe behaviour under inversion through the bond centre: σg is unchanged, σu changes sign. Electric-dipole transitions connect g and u states only, so the σg → σu absorption is allowed. Because σu is repulsive, absorbing a photon of this energy breaks the ion apart: photodissociation.

Summary

H₂⁺ consists of two protons and one electron. Combining the two 1s orbitals gives a bonding σg orbital, (1s A + 1s B)/√[2(1 + S)], with density built up between the nuclei, and an antibonding σu orbital with a nodal plane. The bonding curve has a minimum at about 2.5 a₀ with D e ≈ 1.8 eV, compared with the exact 2.00 a₀ and 2.79 eV; optimising the orbital exponent improves both.

Practice questions

1. Write the electronic Hamiltonian for H₂⁺ in atomic units and state the approximation used. Answer: Ĥ = −½∇² − 1/r A − 1/r B + 1/R, within the Born–Oppenheimer approximation, which treats the nuclei as fixed. 2. Why does the energy of both LCAO states approach −0.5 Eₕ at large R? Answer: At large separation the system becomes a hydrogen atom plus a bare proton, and S, β and 1/R all tend to zero, leaving the hydrogen 1s energy. 3. Explain why the simple LCAO dissociation energy of 1.76 eV is smaller than the exact 2.79 eV. Answer: The LCAO function is only an approximation, so its energy lies above the exact energy at every R; the minimum is therefore shallower than the true well. 4. What happens to the orbital exponent when it is optimised at the equilibrium geometry, and why? Answer: It rises to about 1.24, above 1, because the electron is attracted by two protons and its orbital contracts towards the nuclei.