---
title: The Molecular-Orbital Method and H₂⁺
module: Molecules and Chemical Bonding
moduleNumber: 1
lessonNumber: 2
order: 102
summary: >
  The hydrogen molecule ion is the two-center problem that fixes the language of
  chemical bonding. This lesson builds the molecular orbital as a linear
  combination of atomic orbitals, minimizes the energy through the variational
  secular equation, and reduces the result to three two-center integrals: the
  overlap, the Coulomb term, and the exchange (resonance) integral. The bonding
  and antibonding levels, their potential-energy curves, and the charge piled
  between the nuclei follow from those integrals.
topics: [Molecules and Chemical Bonding]
draft: false
sources:
  - book: Ashcroft & Mermin
    ref: "Ch. 32 — Electron Interactions and Magnetic Structure; bonding appendix"
  - book: Hook & Hall
    ref: "Ch. 1 — The Physics of Bonding"
  - book: Simon
    ref: "Ch. 5–6 — The Chemical Bond, LCAO"
---

The [bonding survey](/condensed-matter/molecules-and-bonding/bonding-mechanisms)
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

$$
\hat H = -\frac{\hbar^2}{2m}\nabla^2 - \frac{ke^2}{r_A} - \frac{ke^2}{r_B},
$$

with $k = 1/4\pi\epsilon_0$. The nuclear repulsion $ke^2/R$ 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.

> **Definition (Born–Oppenheimer separation).** Because the nuclei are thousands
> of times heavier than the electron, they move slowly enough that the electron
> adiabatically follows them. The electronic Schrödinger equation is solved at
> each fixed nuclear configuration $R$, and the resulting electronic energy plus
> the nuclear repulsion serves as the potential energy $E(R)$ governing nuclear
> motion.

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

## The LCAO ansatz and the variational principle

> **Definition (LCAO molecular orbital).** A molecular orbital is approximated as
> a linear combination of atomic orbitals centered on the constituent nuclei. For
> $\text{H}_2^+$ the trial function is
> $$
> \psi = c_A\,\phi_A + c_B\,\phi_B,
> $$
> with $\phi_A,\phi_B$ the normalized hydrogen $1s$ orbitals and $c_A,c_B$
> variational coefficients.

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

$$
E[\psi] = \frac{\langle\psi|\hat H|\psi\rangle}{\langle\psi|\psi\rangle}
$$

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

$$
\begin{aligned}
(H_{AA}-E)\,c_A + (H_{AB}-E S)\,c_B &= 0,\\
(H_{BA}-E S)\,c_A + (H_{BB}-E)\,c_B &= 0,
\end{aligned}
$$

in terms of the matrix elements $H_{ij} = \langle\phi_i|\hat H|\phi_j\rangle$ and
the **overlap** $S = \langle\phi_A|\phi_B\rangle$ (with $\langle\phi_A|\phi_A\rangle
= 1$). A nontrivial solution requires the secular determinant to vanish.

> **Definition (Secular equation).** The condition for a nontrivial solution of
> the LCAO variational equations is
> $$
> \det\!\begin{pmatrix} H_{AA}-E & H_{AB}-E S\\ H_{AB}-E S & H_{AA}-E\end{pmatrix}=0,
> $$
> using $H_{BB}=H_{AA}$ and $H_{BA}=H_{AB}$ from the $A\leftrightarrow B$
> symmetry of the homonuclear ion.

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

$$
E_+ = \frac{H_{AA}+H_{AB}}{1+S},\qquad
E_- = \frac{H_{AA}-H_{AB}}{1-S},
$$

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
$\psi_+ \propto \phi_A+\phi_B$, symmetric under the reflection that swaps the
nuclei (labeled $\sigma_g$, even parity); the upper root $E_-$ is the
**antibonding** orbital $\psi_- \propto \phi_A-\phi_B$, antisymmetric (labeled
$\sigma_u$, odd parity), with a node on the midplane.

```algorithm
LCAO variational solution of a two-center molecular orbital.

1. Choose atomic basis orbitals phi_A, phi_B on the two nuclei.
2. Compute the overlap S = <phi_A | phi_B>.
3. Compute the Hamiltonian matrix elements H_AA and H_AB.
4. Form the secular determinant and set it to zero.
5. Solve for the roots E_+ and E_- of the secular equation.
6. For each root, back-substitute to get the coefficient ratio c_B / c_A.
7. Add the nuclear repulsion k e^2 / R to obtain the total energy E(R).
8. Repeat over R and locate the minimum of the bonding curve.
```

$$
% caption: Two 1s orbitals on separate nuclei combine, in phase, into the
% bonding sigma_g orbital that piles amplitude between them, and out of phase
% into the antibonding sigma_u orbital with a node on the midplane.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % bonding row
  \begin{scope}
    \draw[black] (-2.6,0) -- (2.6,0);
    \fill[black] (-1.1,0) circle (2pt);
    \fill[black] (1.1,0) circle (2pt);
    \draw[acc, very thick, domain=-2.5:2.5, samples=100, variable=\x]
      plot ({\x},{1.3*exp(-(\x-1.1)*(\x-1.1)/0.5) + 1.3*exp(-(\x+1.1)*(\x+1.1)/0.5)});
    \node[acc, anchor=south] at (0,1.55) {bonding};
    \node[black, anchor=north] at (0,-0.1) {amplitude added};
  \end{scope}
  % antibonding row
  \begin{scope}[xshift=7.0cm]
    \draw[black] (-2.6,0) -- (2.6,0);
    \fill[black] (-1.1,0) circle (2pt);
    \fill[black] (1.1,0) circle (2pt);
    \draw[black, very thick, dashed, domain=-2.5:2.5, samples=140, variable=\x]
      plot ({\x},{1.3*exp(-(\x-1.1)*(\x-1.1)/0.5) - 1.3*exp(-(\x+1.1)*(\x+1.1)/0.5)});
    \node[black, anchor=south] at (0,1.55) {antibonding};
    \draw[black, dashed] (0,-1.4) -- (0,1.4);
    \node[black, anchor=north] at (0.05,-0.1) {node};
  \end{scope}
\end{tikzpicture}
$$

## The three two-center integrals

The matrix elements reduce to standard integrals over hydrogen $1s$ orbitals. Use
$\hat H = \hat h_B - ke^2/r_A$, where $\hat h_B = -\tfrac{\hbar^2}{2m}\nabla^2 -
ke^2/r_B$ is the isolated-atom Hamiltonian with $\hat h_B\phi_B = E_{1s}\phi_B$
and $E_{1s} = -13.6\ \text{eV}$. Then

$$
H_{AA} = E_{1s} - \mathcal J,\qquad H_{AB} = E_{1s}\,S - \mathcal K,
$$

where the two integrals are

$$
\mathcal J = \left\langle \phi_A\left|\frac{ke^2}{r_B}\right|\phi_A\right\rangle,
\qquad
\mathcal K = \left\langle \phi_A\left|\frac{ke^2}{r_B}\right|\phi_B\right\rangle.
$$

- **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 $\phi_A\phi_B$ 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 $ke^2/a_0 = 27.2\ \text{eV}$,

$$
S(\rho) = e^{-\rho}\!\left(1+\rho+\tfrac13\rho^2\right),\quad
\mathcal J(\rho) = \frac{ke^2}{a_0}\!\left[\frac1\rho - e^{-2\rho}\!\left(1+\frac1\rho\right)\right],\quad
\mathcal K(\rho) = \frac{ke^2}{a_0}\,e^{-\rho}\!\left(1+\rho\right).
$$

Substituting into the two roots and adding the nuclear repulsion $ke^2/R$ gives
the total energies

$$
E_\pm(R) = E_{1s} + \frac{ke^2}{R} - \frac{\mathcal J \pm \mathcal K}{1\pm S}.
$$

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

$$
\Delta E_\pm = E_\pm - E_{1s} = \frac{ke^2}{R} - \frac{\mathcal J \pm \mathcal K}{1\pm S}.
$$

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.

$$
% caption: The overlap denominators split the two atomic 1s levels
% asymmetrically: the antibonding level (divided by 1 minus S) rises more than
% the bonding level (divided by 1 plus S) falls.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % atomic levels left and right
  \draw[black, thick] (-0.2,0) -- (1.4,0);
  \node[black, anchor=east] at (-0.25,0) {atomic 1s};
  \draw[black, thick] (7.2,0) -- (8.8,0);
  \node[black, anchor=west] at (8.85,0) {atomic 1s};
  % molecular levels middle
  \draw[black, thick, dashed] (3.6,1.35) -- (5.0,1.35);
  \node[black, anchor=south] at (4.3,1.4) {antibonding};
  \draw[acc, thick] (3.6,-0.95) -- (5.0,-0.95);
  \node[acc, anchor=north] at (4.3,-1.0) {bonding};
  % connectors
  \draw[black] (1.4,0) -- (3.6,1.35);
  \draw[black] (1.4,0) -- (3.6,-0.95);
  \draw[black] (7.2,0) -- (5.0,1.35);
  \draw[black] (7.2,0) -- (5.0,-0.95);
  % reference dashed
  \draw[black, dashed] (1.4,0) -- (7.2,0);
\end{tikzpicture}
$$

Minimizing $E_+(R)$ numerically with the hydrogen $1s$ orbitals ($Z=1$) places
the bonding minimum at $R_e = 2.49\,a_0 = 0.132\ \text{nm}$ with a dissociation
energy $D_e = 1.76\ \text{eV}$. The exact solution of $\text{H}_2^+$ gives
$R_e = 2.00\,a_0 = 0.106\ \text{nm}$ 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.

$$
% caption: Total energy of H2-plus versus internuclear separation: the bonding
% sigma_g branch develops a minimum at the equilibrium bond length, the
% antibonding sigma_u branch is purely repulsive.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,-1.7) -- (0,2.6) node[left] {$E(R)$};
  \draw[->, black] (0,0) -- (7.0,0) node[below] {$R$};
  \draw[black, dashed] (0,0) -- (7.0,0);
  \node[black, anchor=south east] at (7.0,0.05) {separated atom};
  % antibonding: repulsive
  \draw[black, very thick, dashed, domain=0.7:6.6, samples=120, variable=\x]
    plot ({\x},{4.5*exp(-\x/1.2)});
  \node[black, anchor=west] at (2.2,1.15) {antibonding};
  % bonding: well
  \draw[acc, very thick, domain=0.85:6.6, samples=140, variable=\x]
    plot ({\x},{7.2/(\x*\x) - 5.0/\x});
  \node[acc, anchor=north west] at (3.0,-0.55) {bonding};
  \fill[acc] (1.44,-0.868) circle (2pt);
  \draw[black, dashed] (1.44,-0.868) -- (1.44,0) node[above right] {$R_e$};
\end{tikzpicture}
$$

## Charge density and the source of binding

Squaring the normalized orbitals gives the electron densities

$$
|\psi_\pm|^2 = \frac{1}{2(1\pm S)}\left(|\phi_A|^2 + |\phi_B|^2 \pm 2\phi_A\phi_B\right).
$$

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

$$
% caption: Electron-density contours: the bonding orbital concentrates charge in
% the internuclear region, screening the proton repulsion, while the antibonding
% orbital is depleted there and split by a nodal plane.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % bonding: merged contours
  \begin{scope}
    \fill[black] (-0.9,0) circle (2pt);
    \fill[black] (0.9,0) circle (2pt);
    \draw[acc] (0,0) ellipse (1.7 and 0.9);
    \draw[acc] (0,0) ellipse (1.25 and 0.62);
    \draw[acc, thick] (0,0) ellipse (0.8 and 0.4);
    \node[acc, anchor=south] at (0,1.15) {bonding};
    \node[black, anchor=north] at (0,-1.15) {charge between nuclei};
  \end{scope}
  % antibonding: two separate lobes with node
  \begin{scope}[xshift=6.2cm]
    \fill[black] (-0.9,0) circle (2pt);
    \fill[black] (0.9,0) circle (2pt);
    \draw[black] (-0.9,0) circle (0.75);
    \draw[black, thick] (-0.9,0) circle (0.45);
    \draw[black] (0.9,0) circle (0.75);
    \draw[black, thick] (0.9,0) circle (0.45);
    \draw[black, dashed] (0,-1.05) -- (0,1.05);
    \node[black, anchor=south] at (0,1.15) {antibonding};
    \node[black, anchor=north] at (0,-1.15) {nodal plane};
  \end{scope}
\end{tikzpicture}
$$

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.[^ruedenberg]

[^ruedenberg]: Ashcroft & Mermin, Ch. 32, and Hook & Hall, §1.3, present the LCAO
integrals; the kinetic-energy interpretation of covalent binding through the
virial theorem is developed in Simon, Ch. 5.

## 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](/condensed-matter/molecules-and-bonding/hydrogen-molecule-and-exchange).
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.

$$
% caption: Correlation diagram for a second-row homonuclear diatomic: atomic 2s
% and 2p levels on each side split into the ordered molecular levels in the
% center, filled in order to set the bond order.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % atomic 2s
  \draw[black, thick] (0,0) -- (1.0,0);
  \node[black, anchor=east] at (-0.05,0) {2s};
  \draw[black, thick] (7.6,0) -- (8.6,0);
  % atomic 2p
  \draw[black, thick] (0,2.4) -- (1.0,2.4);
  \node[black, anchor=east] at (-0.05,2.4) {2p};
  \draw[black, thick] (7.6,2.4) -- (8.6,2.4);
  % molecular from 2s
  \draw[acc, thick] (3.8,-0.55) -- (4.8,-0.55);
  \node[acc, anchor=north] at (4.3,-0.62) {bonding 2s};
  \draw[black, thick, dashed] (3.8,0.6) -- (4.8,0.6);
  \node[black, anchor=south] at (4.3,0.66) {antibonding 2s};
  % molecular from 2p
  \draw[acc, thick] (3.8,1.5) -- (4.8,1.5);
  \node[acc, anchor=south] at (4.3,1.56) {bonding 2p};
  \draw[black, thick, dashed] (3.8,3.1) -- (4.8,3.1);
  \node[black, anchor=south] at (4.3,3.16) {antibonding 2p};
  % connectors
  \draw[black] (1.0,0) -- (3.8,-0.55);
  \draw[black] (1.0,0) -- (3.8,0.6);
  \draw[black] (7.6,0) -- (4.8,-0.55);
  \draw[black] (7.6,0) -- (4.8,0.6);
  \draw[black] (1.0,2.4) -- (3.8,1.5);
  \draw[black] (1.0,2.4) -- (3.8,3.1);
  \draw[black] (7.6,2.4) -- (4.8,1.5);
  \draw[black] (7.6,2.4) -- (4.8,3.1);
\end{tikzpicture}
$$

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](/condensed-matter/band-theory/tight-binding-method).

## Summary

- In the Born–Oppenheimer approximation the electronic problem is solved at fixed
  nuclear separation $R$, and the electronic energy plus $ke^2/R$ gives the
  potential curve $E(R)$.
- The LCAO trial function $\psi = c_A\phi_A + c_B\phi_B$ 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 $R_e = 2.49\,a_0$ ($D_e = 1.76\ \text{eV}$ in the simplest LCAO);
  the antibonding orbital $\sigma_u$ has a node between the nuclei and is purely
  repulsive.