H2+ Molecular Orbitals: Bonding and Antibonding

The bonding/antibonding split constructed here for H₂⁺ is the seed of essentially all of modern electronic structure theory. Every quantum chemistry package — from teaching-level Hartree–Fock codes to industrial density functional theory software used in pharmaceutical and materials research — represents molecular electrons in orbitals built as linear combinations of atomic-centered basis functions, and every one of those calculations produces filled bonding orbitals and empty (or higher-lying) antibonding orbitals whose qualitative character is exactly the $\psi_+/\psi_-$ pattern derived above, just dressed up with many more basis functions and electrons. In spectroscopy, the antibonding $\sigma_u^{*}$ orbital of H₂⁺ (and its analogue $\sigma_u^{*}$ in H₂) is the destination orbital in photoionization and Rydberg-state experiments, and its nodal structure directly explains why populating it — as happens, for instance, in electron-impact or photon-driven excitation of H₂ — weakens and eventually breaks the bond. This bonding/antibonding logic underlies why π orbitals in carbonyl and alkene chromophores are the acceptor orbitals in UV-visible spectroscopy, why σ orbitals along C–halogen bonds are the targets of nucleophilic backside attack in Sₙ2 reactions (hyperconjugative donation into σ* is a mainstay explanation in physical organic chemistry), and why frontier molecular orbital theory (HOMO/LUMO reasoning in Woodward–Hoffmann rules and Fukui's frontier orbital theory) can predict reactivity trends across huge swaths of organic and organometallic chemistry. Materials scientists use the same bonding/antibonding band-formation logic, scaled up to solids, to explain why partially filled bonding bands give metals their cohesion and why band gaps open when bonding and antibonding combinations split apart in energy — the tight-binding model of solid-state band structure is, structurally, an infinite generalization of exactly the two-orbital problem solved here.

Ask the tutor: How do the Coulomb integral $H_{AA}$ and resonance integral $H_{AB}$ combine with $S$ to actually produce the bonding and antibonding energies from these two orbitals?

Picture the electron in H₂⁺ as a wave that must be simultaneously "listened to" by both protons, A and B, held a fixed distance $R$ apart along the internuclear axis. Far from either nucleus the electron behaves like a hydrogen 1s electron centered on whichever proton it is near, so a natural trial wavefunction is a linear combination of atomic orbitals,

where $1s_A$ and $1s_B$ are normalized hydrogen 1s orbitals centered on nucleus A and nucleus B respectively. This is the LCAO Approximation applied to the simplest possible molecule. The key physical constraint is nuclear symmetry. Because both nuclei are protons, swapping labels A ↔ B must leave all observable properties — above all, the electron probability density $|\psi|^2$ — unchanged. If $|c_A| \neq |c_B|$, the electron would be found preferentially near one specific proton, which is unphysical for a homonuclear system with no reason to prefer either center. This symmetry argument, not an appeal to energy minimization, is what fixes $|c_A| = |c_B|$ and leaves only two allowed combinations: the in-phase sum and the out-of-phase difference. The Coulomb and resonance integrals that eventually assign energies to these two combinations belong to the analysis in Coulomb, Resonance, and Overlap Integrals; here the goal is only to build the orbitals and see their shapes. Take the sum first:

At every point in space, the two atomic wavefunctions are added in phase. Directly between the nuclei, both $1s_A$ and $1s_B$ are still appreciable (each nucleus is close enough that its 1s orbital has not fully decayed), so they add constructively — the resulting amplitude, and hence the probability density $|\psi_+|^2$, is enhanced relative to what you would get from the two isolated atomic densities added independently. This pile-up of negative charge in the internuclear region is precisely what is needed to screen the mutual repulsion of the two positively charged protons and glue them together: this is the bonding molecular orbital. Now the difference:

Along the perpendicular bisecting plane of the bond — the plane equidistant from A and B — $1s_A$ and $1s_B$ are numerically equal, so their difference vanishes identically. This defines an exact nodal plane perpendicular to the bond axis at the midpoint. Approaching this plane from either side, the wavefunction changes sign, and the probability density is driven to zero exactly at the midpoint and suppressed throughout the internuclear region. With the negative charge evacuated from between the nuclei, the bare protons repel each other with nothing to shield them: this is the antibonding molecular orbital.

The reason both combinations require careful normalization rather than a naive $1/\sqrt2$ prefactor is that $1s_A$ and $1s_B$ are not orthogonal — they overlap in space, and this overlap is quantified by the overlap integral

$S$ depends on the internuclear distance $R$: at $R \to 0$ the two orbitals coincide and $S \to 1$; at $R \to \infty$ they no longer overlap and $S \to 0$. At typical bonding distances in H₂⁺ ($R \approx 2.00\ a_0 \approx 106$ pm), $S$ is a substantial positive fraction, of order $0.46$–$0.6$ depending on how it's evaluated — this nonzero overlap is exactly what forces the normalization constants for $\psi_+$ and $\psi_-$ to differ from each other and from the naive guess $1/\sqrt2$.

The upshot: the two LCAO combinations of a single pair of 1s orbitals already reproduce the entire qualitative content of the covalent bond — an attractive, density-enhancing combination and a repulsive, density-depleting combination — with no need to invoke electron-electron interactions at all, since H₂⁺ has only one electron. The energetics that turn "more density between the nuclei" into "lower total energy" are worked out quantitatively once the Coulomb integral $H_{AA}$ and resonance integral $H_{AB}$ are introduced; the symmetry labels $\sigma_g$ and $\sigma_u^{*}$ used loosely above are made precise in the treatment of Sigma and Pi Molecular Orbitals and Symmetry Labels.

The trial wavefunction, forced by A↔B symmetry to have $|c_A| = |c_B|$, reduces to exactly two possibilities:

where:

• $\psi_+$ is the symmetric (bonding) combination, $\psi_-$ the antisymmetric (antibonding) combination
• $1s_A, 1s_B$ are normalized 1s hydrogenic orbitals centered on nuclei A and B
• $N_\pm$ are real normalization constants (assuming real, non-negative $1s_A, 1s_B$, so $S>0$)

Reading: there are only two ways to combine one orbital from each of two equivalent centers while keeping $|\psi|^2$ symmetric under nuclear exchange — add them, or subtract them. Normalization requires $\int |\psi_\pm|^2 d\tau = 1$, which brings in the overlap integral explicitly:

where:

• $S$ is the (dimensionless) overlap integral, a function of $R/a_0$
• the $+$ sign in $N_+$ goes with $\psi_+$, the $-$ sign in $N_-$ goes with $\psi_-$

Reading: because $1s_A$ and $1s_B$ overlap ($S>0$ for $R$ finite), the bonding combination needs a smaller normalization constant than $1/\sqrt2$ (the two lobes reinforce, so less amplitude per orbital is needed to reach total probability 1), while the antibonding combination needs a larger one (the lobes partially cancel, so more amplitude per orbital is needed). The electron density built from each MO is

where:

• $1s_A^2$ and $1s_B^2$ are the densities the electron would have if localized purely on A or purely on B
• the cross term $2\,1s_A1s_B$ is the interference term, present only because the electron is genuinely delocalized over both centers

Reading: without the cross term, $|\psi_\pm|^2$ would just be an average of two atomic densities — no chemistry would happen. The entire physical content of "bonding" versus "antibonding" lives in the sign and magnitude of this cross term.

Example 1 — Foundational: normalization constants at the equilibrium-like separation of H₂⁺. Take $R = 2.00\,a_0$ (close to the true equilibrium bond length of H₂⁺, $R_e \approx 2.00\,a_0 \approx 106$ pm) and use the standard exact overlap integral for two 1s Slater orbitals with effective charge $\zeta=1$,

with $R$ in units of $a_0$. Substituting $R=2.00$:

Since $e^{-2.00} = 0.1353$,

Now the normalization constants:

Sanity check on the asymmetry: $N_- / N_+ = 1.0996/0.5614 = 1.959$ — the antibonding orbital's prefactor is nearly double the bonding one's at this fairly short distance, exactly as expected since $S$ is large enough here that $(1-S)$ is small. As $R \to \infty$, $S\to 0$ and both constants converge to the textbook value $1/\sqrt2 = 0.7071$, recovering the naive normalization for non-overlapping, effectively orthogonal orbitals — a useful limiting-case check.

Example 2 — Applied: electron density at the bond midpoint versus far outside the molecule. Continue with $R = 2.00\,a_0$ and locate the midpoint, a distance $r = 1.00\,a_0$ from each nucleus. For a hydrogenic 1s orbital ($\zeta=1$, in atomic units),

At the midpoint, $1s_A = 1s_B = \phi_0 = \dfrac{1}{\sqrt\pi}e^{-1.00} = \dfrac{0.3679}{1.7725} = 0.2076$. Bonding density at the midpoint (using Step 3 of the second derivation):

Antibonding density at the same point: exactly $0$, as shown generally above. Now compare to a point outside the bond, say $2.00\,a_0$ beyond nucleus B (so $3.00\,a_0$ from B, $5.00\,a_0$ from A) — far into the antibonding-favored region. There, $1s_A(5.00) = \frac{1}{\sqrt\pi}e^{-5.00} = 0.003800$ and $1s_B(3.00) = \frac{1}{\sqrt\pi}e^{-3.00} = 0.02807$. The cross term $1s_A 1s_B = 1.067\times10^{-4}$ is tiny compared to $1s_B^2 = 7.879\times10^{-4}$, so both densities are dominated by the single-atom tail:

Outside the bond, the antibonding density is now larger than the bonding density (unlike at the midpoint) — the antibonding orbital does not destroy density everywhere, it specifically redistributes it away from the internuclear region and into the outer lobes, consistent with the qualitative picture and confirming the redistribution is a shift, not a uniform suppression.

Example 3 — Exam/competition level: how much extra charge sits between the nuclei for $\psi_+$, and the total-probability bookkeeping. A common exam question asks: integrate $|\psi_+|^2$ over all space and show it equals 1, then quantify the "overlap population" — the fraction of total probability that is attributable purely to the cross term. From the second derivation,

confirming normalization holds globally, as it must by construction — the local enhancement at the midpoint found in Example 2 is compensated by a slight depletion elsewhere so the grand total is unchanged. Now define the "overlap population" as the contribution of the cross term alone to the total probability:

At $R=2.00\,a_0$ with $S=0.5865$:

So at this separation, roughly 37% of the total electron probability in the bonding MO is carried by the cross (interference) term rather than by the "atomic" $1s_A^2$ and $1s_B^2$ terms individually — a striking quantitative statement that the covalent bond in H₂⁺ is not a minor perturbation on two separate atoms but is dominated by genuine quantum delocalization at this distance. For the antibonding orbital the analogous quantity is negative,

whose magnitude exceeding 1 in size (though the total must still integrate to $+1$ once combined with the positive atomic terms) reflects how strongly antibonding character concentrates the deficit right where $1s_A$ and $1s_B$ overlap most, requiring a correspondingly large compensating normalization $N_-^2$. As a limiting-case check, letting $S\to 0$ (large $R$) gives $P_\text{overlap}\to 0$ for both orbitals — at large separations there is no interference left to speak of, and each MO reduces to an equal, incoherent mixture of the two atomic densities.

The LCAO treatment of H₂⁺ traces to the earliest days of applying the new quantum mechanics to chemical bonding, in the years immediately following Schrödinger's 1926 papers. The molecular-orbital viewpoint — building wavefunctions delocalized over the whole molecule from atomic building blocks — is inseparable from the names Friedrich Hund and Robert Mulliken, who independently developed the molecular orbital framework as an alternative to the valence-bond picture that Heitler and London had pioneered for H₂. Mulliken in particular pushed hard for the view that electrons in molecules occupy orbitals extending over all the nuclei, exactly the picture realized concretely by $\psi_+$ and $\psi_-$ for H₂⁺, and he later systematized the symmetry notation ($\sigma_g$, $\sigma_u^{*}$, etc.) that chemists still use today.