The Molecular-Orbital Method and H₂⁺

The bonding survey asserted that two hydrogen atoms bind through a symmetric electron wave function that concentrates charge between the nuclei. The molecular-orbital method turns that assertion into a calculation. Its simplest target is the hydrogen molecule ion ${\text{H}}_2^{+}$: one electron in the field of two protons. The problem is exactly solvable in prolate spheroidal coordinates, but the approximate treatment by a linear combination of atomic orbitals (LCAO) is what generalizes to every molecule and solid, and it introduces the overlap and exchange integrals that recur throughout the subject.

The two-center Hamiltonian

Label the protons $A$ and $B$, fixed a distance $R$ apart, and let $r_A$ and $r_B$ be the electron's distances from each. In the Born–Oppenheimer approximation the nuclei are clamped, and the electronic Hamiltonian is

with $k=1/4\pi {ϵ}_0$. The nuclear repulsion $k{e}^2/R$ is a constant for fixed $R$ and is added to the electronic eigenvalue at the end to give the total energy $E(R)$ whose minimum is the bond.

When the electron sits close to proton $A$, the term $-k{e}^2/r_B$ is a small perturbation and the electron occupies the hydrogen $1s$ ground state ${\varphi }_A$ centered on $A$; symmetrically it occupies ${\varphi }_B$ near $B$. A trial state that respects both limits is their linear combination.

The LCAO ansatz and the variational principle

The variational principle guarantees that for any trial $\psi$ the Rayleigh quotient

is an upper bound on the true ground-state energy, with equality only for the exact eigenstate. Minimizing $E$ over $c_A,c_B$ gives the best orbital of LCAO form. Setting $\partial E/\partial c_A=\partial E/\partial c_B=0$ yields the homogeneous linear system

in terms of the matrix elements $H_{ij}=⟨{\varphi }_i|\hat{H}|{\varphi }_j⟩$ and the overlap $S=⟨{\varphi }_A|{\varphi }_B⟩$ (with $⟨{\varphi }_A|{\varphi }_A⟩=1$). A nontrivial solution requires the secular determinant to vanish.

Expanding the determinant, $(H_{AA}-E{)}^2=(H_{AB}-ES{)}^2$, so $H_{AA}-E=\pm (H_{AB}-ES)$. The two roots are

with coefficient ratios $c_B/c_A=+1$ for $E_{+}$ and $c_B/c_A=-1$ for $E_{-}$. The lower root $E_{+}$ is the bonding orbital ${\psi }_{+}\propto {\varphi }_A+{\varphi }_B$, symmetric under the reflection that swaps the nuclei (labeled ${\sigma }_g$, even parity); the upper root $E_{-}$ is the antibonding orbital ${\psi }_{-}\propto {\varphi }_A-{\varphi }_B$, antisymmetric (labeled ${\sigma }_u$, odd parity), with a node on the midplane.

The three two-center integrals

The matrix elements reduce to standard integrals over hydrogen $1s$ orbitals. Use $\hat{H}={\hat{h}}_B-k{e}^2/r_A$, where ${\hat{h}}_B=-\frac{{\hslash }^2}{2m}{\nabla }^2-k{e}^2/r_B$ is the isolated-atom Hamiltonian with ${\hat{h}}_B{\varphi }_B=E_{1s}{\varphi }_B$ and $E_{1s}=-13.6\text{eV}$. Then

where the two integrals are

• Overlap integral $S$ — the geometric overlap of the two orbitals; it measures how much the atomic clouds share the same region and vanishes as $R\to \infty$.
• Coulomb integral $\mathcal{J}$ — the classical electrostatic attraction between the electron's charge cloud around $A$ and the second proton $B$. It is positive and, being a classical interaction, would be present even without quantum sharing.
• Exchange (resonance) integral $\mathcal{K}$ — an interference term with no classical analog, built from the product ${\varphi }_A{\varphi }_B$ in the overlap region. It is the integral responsible for the bonding-antibonding splitting.

In closed form, writing $\rho =R/a_0$ with $a_0$ the Bohr radius and measuring energies in units of $k{e}^2/a_0=27.2\text{eV}$,

Substituting into the two roots and adding the nuclear repulsion $k{e}^2/R$ gives the total energies

Both $\mathcal{J}$ and $\mathcal{K}$ are positive and decay with $R$. Because the bonding root subtracts $(\mathcal{J}+\mathcal{K})/(1+S)$ while the antibonding root subtracts $(\mathcal{J}-\mathcal{K})/(1-S)$, the exchange integral $\mathcal{K}$ lowers the bonding level and raises the antibonding level relative to the atomic value. The Coulomb term $\mathcal{J}$ alone, without exchange, produces almost no binding; the resonance integral is what makes the covalent bond.

The energy splitting

Relative to the separated-atom energy $E_{1s}$, the two levels sit at

The splitting is asymmetric. Neglecting $\mathcal{J}$ against $\mathcal{K}$ near the bond length, the bonding level drops by roughly $\mathcal{K}/(1+S)$ while the antibonding level rises by $\mathcal{K}/(1-S)$. Since $0<S<1$, the antibonding level is pushed up more than the bonding level is pushed down.

Minimizing $E_{+}(R)$ numerically with the hydrogen $1s$ orbitals ($Z=1$) places the bonding minimum at $R_e=2.49a_0=0.132\text{nm}$ with a dissociation energy $D_e=1.76\text{eV}$. The exact solution of ${\text{H}}_2^{+}$ gives $R_e=2.00a_0=0.106\text{nm}$ and $D_e=2.79\text{eV}$; letting the orbital exponent vary (an effective nuclear charge $Z_{\text{eff}}\approx 1.24$ that contracts the atomic orbitals) recovers most of the discrepancy, showing that the atoms polarize as they bond. The antibonding curve $E_{-}(R)$ has no minimum and rises monotonically as $R$ decreases; an electron placed in ${\sigma }_u$ pushes the nuclei apart.

Charge density and the source of binding

Squaring the normalized orbitals gives the electron densities

The cross term $\pm 2{\varphi }_A{\varphi }_B$ is the interference. In the bonding orbital it adds density in the internuclear region where ${\varphi }_A{\varphi }_B>0$; in the antibonding orbital it subtracts, leaving a nodal plane of zero density between the protons.

The naive reading of this picture is electrostatic: the accumulated charge sits between the protons and screens their mutual repulsion, so the potential energy drops. A careful accounting through the virial theorem revises the story. At the equilibrium separation the binding is a lowering of the electron's kinetic energy: spreading the orbital over both nuclei lengthens its effective wavelength, and the kinetic energy falls faster than the potential energy rises. The two descriptions agree on the total but assign the binding to different terms; the exchange integral is the LCAO fingerprint of the delocalization that both describe.¹

Filling the orbitals and heavier diatomics

For ${\text{H}}_2^{+}$ the single electron occupies ${\sigma }_g$, giving one-electron binding. Adding a second electron with opposite spin produces neutral ${\text{H}}_2$: both electrons occupy ${\sigma }_g$, the binding roughly doubles, and the treatment of their mutual repulsion is the subject of the next lesson. A third electron would have to enter ${\sigma }_u$, whose antibonding character cancels the bond, so ${\text{H}}_3$ and ${\text{He}}_2$ do not form.

For homonuclear diatomics of the second row, the same construction applied to the $2s$ and $2p$ atomic orbitals produces the ordered set of molecular levels ${\sigma }_g(2s),{\sigma }_u^{\ast }(2s),{\pi }_u(2p),{\sigma }_g(2p),{\pi }_g^{\ast }(2p),{\sigma }_u^{\ast }(2p)$. Filling these levels with the available valence electrons and counting bonding minus antibonding occupancy gives the bond order, which predicts the trend across the row.

The molecular-orbital method thus reduces bonding to a diagonalization: choose an atomic basis, compute overlaps and matrix elements, and fill the resulting levels. The same three integrals — overlap, Coulomb, and exchange — reappear when the basis is a periodic array of atoms rather than two, where the discrete bonding and antibonding levels broaden into the energy bands of the tight-binding method.

Summary

• In the Born–Oppenheimer approximation the electronic problem is solved at fixed nuclear separation $R$, and the electronic energy plus $k{e}^2/R$ gives the potential curve $E(R)$.
• The LCAO trial function $\psi =c_A{\varphi }_A+c_B{\varphi }_B$ and the variational principle produce the secular equation, whose roots are $E_{\pm }=(H_{AA}\pm H_{AB})/(1\pm S)$.
• The matrix elements reduce to three two-center integrals: the overlap $S$, the Coulomb integral $\mathcal{J}$, and the exchange integral $\mathcal{K}$. The exchange integral produces the bonding-antibonding splitting; Coulomb alone gives almost no binding.
• The bonding orbital ${\sigma }_g$ concentrates charge between the nuclei and has a minimum at $R_e=2.49a_0$ ($D_e=1.76\text{eV}$ in the simplest LCAO); the antibonding orbital ${\sigma }_u$ has a node between the nuclei and is purely repulsive.