Bonding and Antibonding States in H2+

The bonding/antibonding split derived here for one electron and two nuclei is the direct ancestor of every molecular orbital diagram drawn in modern chemistry — from the $\sigma/\sigma^*$ and $\pi/\pi^*$ manifolds used to rationalize UV-Vis and photoelectron spectra of organic chromophores, to the frontier orbital arguments (HOMO/LUMO) that underlie the Woodward–Hoffmann rules for pericyclic reactions, to the metal–ligand $\sigma$-donor/$\pi$-acceptor bonding schemes used throughout organometallic catalyst design. Density functional theory and Hartree–Fock calculations on real molecules still report canonical molecular orbitals that are, in the simplest diatomic fragments, recognizably bonding or antibonding combinations exactly like $\phi_\pm$ here, and computational chemists routinely inspect nodal structure between atoms — precisely the diagnostic emphasized in this treatment — to assign bond order and predict reactivity. In materials science, the same bonding/antibonding logic, generalized to extended solids, becomes the tight-binding picture of band formation: valence and conduction bands in semiconductors are, at heart, bonding and antibonding combinations of atomic orbitals spread over a periodic lattice, with the band gap playing the role of the $E_--E_+$ splitting computed above. Spectroscopically, the resonance integral's distance-dependence, $K(R)=-e^{-R}(1+R)$, is the qualitative template for why bond dissociation energies and vibrational force constants correlate with orbital overlap, a relationship exploited in interpreting infrared and Raman spectra of diatomics and in benchmarking new exchange-correlation functionals against the essentially exact H₂⁺ potential energy curve.

Ask the tutor: How would adding a variational orbital exponent (letting the 1s orbitals contract) change the predicted equilibrium bond length and binding energy of H₂⁺?

Picture the electron in H₂⁺ as being "shared" between two protons, A and B, separated by a distance $R$. If the electron sat purely on atom A, its wavefunction would be a hydrogen 1s orbital $1s_A = \left(\dfrac{1}{\pi a_0^3}\right)^{1/2} e^{-r_A/a_0}$, and likewise $1s_B$ if it sat on B. Since we do not know a priori which proton "owns" the electron, and since the true ground state must be symmetric under exchange of the two identical nuclei (up to a sign), the natural trial functions are the symmetric and antisymmetric combinations,

These are not arbitrary guesses. They are exactly the two combinations required by the symmetry of the nuclear framework: reflecting the molecule through the plane perpendicular to the bond at its midpoint swaps A and B, and $\phi_+$ is even (symmetric) under this operation while $\phi_-$ is odd (antisymmetric). Because the true molecular Hamiltonian commutes with this reflection, the true eigenfunctions must each have definite parity, and $\phi_\pm$ are the only parity-adapted combinations you can build from two 1s functions. The physical content emerges once you plot the probability densities $|\phi_+|^2$ and $|\phi_-|^2$ along the internuclear axis. For $\phi_+$, the two atomic orbitals add constructively in the region between the nuclei — the two waves are in phase there, so the resulting density exceeds what you would get from simply averaging $|1s_A|^2$ and $|1s_B|^2$ with no cross term. For $\phi_-$, the orbitals are exactly out of phase at the midpoint (where $1s_A = 1s_B$ by symmetry), forcing $\phi_- = 0$ there identically. There is a rigorous nodal plane perpendicular to the bond axis, at the midpoint, for every internuclear separation. Extra electron density between two positively charged nuclei is doubly stabilizing: it sits in a region where it is attracted simultaneously to both nuclear charges (lowering potential energy), and — this is the more subtle, quantum-mechanical part — building a smooth, node-free density between the nuclei is compatible with lower kinetic energy than piling density into a region bounded by a node. The node in $\phi_-$ forces the wavefunction to curve sharply near the midplane, which raises the kinetic energy contribution substantially. This is why the antibonding penalty in H₂⁺ is larger in magnitude than the bonding stabilization: nodes are kinetically expensive.

Now translate amplitude into energy. Because $\phi_+$ concentrates electron density in the attractive region between two nuclei of charge $+1$ each, its potential energy is lowered relative to a hydrogen atom plus a bare proton at infinity. Because $\phi_-$ is forced to zero exactly where that stabilization would occur, it loses the corresponding stabilization and, further, pays a kinetic penalty for curving around the node. The resulting energy ordering, $E_+ < E(H\ 1s) < E_-$, is the single most important qualitative fact about two-center one-electron bonding, and it is exactly what the secular equation below reproduces quantitatively.

The variational treatment starts from the electronic Hamiltonian for a single electron moving in the field of two fixed protons separated by $R$ (Born–Oppenheimer, atomic units, energies in hartree, lengths in bohr):

where:

• $-\tfrac12\nabla^2$ is the electron kinetic energy operator,
• $r_A, r_B$ are electron–proton distances to nuclei A and B,
• $1/R$ is the fixed nuclear repulsion, a constant at given geometry, added to give the total electronic energy referenced to separated nuclei.

Expanding the trial wavefunction as $\psi = c_A(1s_A) + c_B(1s_B)$ and applying the linear variational method gives the $2\times 2$ secular determinant

where:

• $H_{AA} = \langle 1s_A|\hat H|1s_A\rangle$, $H_{BB} = \langle 1s_B|\hat H|1s_B\rangle$ are the diagonal (Coulomb) integrals, equal by symmetry: $H_{AA}=H_{BB}\equiv J'$,
• $H_{AB} = \langle 1s_A|\hat H|1s_B\rangle$ is the off-diagonal resonance integral, $H_{AB}\equiv K'$,
• $S = \langle 1s_A|1s_B\rangle$ is the overlap integral between the two atomic orbitals (not the identity, since $1s_A\neq 1s_B$),
• $E$ is the variational energy to be determined.

Because $H_{AA}=H_{BB}$ by the A↔B symmetry of the problem, the determinant factors cleanly, giving the two roots

where:

• $E_+$ (using the $+$ sign throughout) is paired with the symmetric combination $\phi_+ \propto 1s_A+1s_B$,
• $E_-$ (using the $-$ sign throughout) is paired with the antisymmetric combination $\phi_- \propto 1s_A - 1s_B$,
• $J'$ is a negative number close to (but not equal to) the hydrogen-atom energy, and $K'$ is negative and typically comparable in magnitude to or larger than $(E_{1s}-J')$, so the sign of $K'$ controls which root is lower.

It is standard (McQuarrie & Simon; Levine) to strip off the point-charge nuclear repulsion and hydrogen-atom reference energy and write $J'$ and $K'$ in terms of the more transparent Coulomb integral $J$ and resonance/exchange integral $K$, so that

where:

• $E_{1s} = -\tfrac12$ hartree is the isolated hydrogen 1s energy,
• $1/R$ is nuclear repulsion,
• $J(R) = \left\langle 1s_A\left|-\dfrac{1}{r_B}\right|1s_A\right\rangle$ is the Coulomb integral: the average interaction of the electron (localized in the "own" orbital) with the other nucleus,
• $K(R) = \left\langle 1s_A\left|-\dfrac{1}{r_B}\right|1s_B\right\rangle$ is the resonance (exchange) integral: it has no classical electrostatic analogue and exists purely because the electron is delocalized over both centers,
• $S(R) = \langle 1s_A|1s_B\rangle$ is the overlap.

All three integrals have closed forms (Levine, Quantum Chemistry; McQuarrie & Simon) in terms of $R$ in bohr:

where: all lengths are in bohr ($a_0$) and all energies in hartree; $J(R)\to 0$ and $S(R)\to 0$ as $R\to\infty$, correctly recovering the isolated-atom limit, while $K(R)$ also vanishes exponentially but does so more slowly than $S$ for large $R$ in the ratio that matters (both go as $e^{-R}$ times a polynomial, but $K$'s prefactor makes the resonance term dominate the near-degeneracy splitting).

Reading the equations. The bonding energy is $E_+ = E_{1s} + 1/R + (J+K)/(1+S)$ and the antibonding energy is $E_- = E_{1s} + 1/R + (J-K)/(1-S)$. Since $K<0$ for all $R>0$, the numerator $J+K$ is more negative than $J-K$, so before dividing by the overlap denominators, the bonding combination is already lower. Dividing by $(1+S)$ (bonding, $S>0$ so this denominator $>1$, which further lowers $E_+$ relative to what $J+K$ alone would give) versus $(1-S)$ (antibonding, denominator $<1$, which amplifies the destabilization) makes the antibonding penalty exceed the bonding stabilization in magnitude — the origin of the textbook statement that "antibonding is more antibonding than bonding is bonding."

Example 1 — Foundational: evaluate all integrals and both energies at the true equilibrium bond length. The experimental/variational equilibrium distance for H₂⁺ in this LCAO treatment is close to $R = 2.00\,a_0$ (the exact non-LCAO equilibrium is $R_e = 2.00\,a_0$ as well, to three figures; the LCAO treatment slightly overestimates it, but $R=2.00\,a_0$ is the standard textbook check point). Evaluate $S$, $J$, $K$, and both energies.

With $e^{-2.00}=0.13534$,

Now the bonding and antibonding energy shifts (each relative to $E_{1s}=-0.5000$ hartree):

So at $R=2.00\,a_0$: $E_+ = -0.5537$ hartree, $E_- = -0.1608$ hartree (both measured as the full molecular electronic energy including nuclear repulsion, relative to $E_{1s}=-0.5$ giving the shifts above). The bonding state is stabilized by about $1.46\,\mathrm{eV}$ relative to $H+H^+$ at infinite separation, while the antibonding state is destabilized by $9.23\,\mathrm{eV}$ — roughly six times larger in magnitude, exactly the asymmetry argued for in Step 4 of the second derivation. (The crude LCAO treatment underestimates the true binding energy of $2.79\,\mathrm{eV}$ found from the exact H₂⁺ solution — a known, well-documented shortcoming of the minimal 1s-only basis, fixed by adding a variational orbital exponent or polarization functions — but the qualitative bonding/antibonding asymmetry is already fully captured.)

Example 2 — Applied: how does the splitting evolve as the bond stretches, and where does resonance stop mattering? Consider $R=3.50\,a_0$, a stretched, non-equilibrium, deliberately asymmetric distance (chosen off any "nice" round number relative to $R_e$) to see how quickly the resonance-dominated picture decays.

With $e^{-3.50}=0.030197$,

Compare the two distances: bonding stabilization only dropped from $1.46\,\mathrm{eV}$ ($R=2.00$) to $1.31\,\mathrm{eV}$ ($R=3.50$) — a modest change — while the antibonding destabilization collapsed from $9.23\,\mathrm{eV}$ to $2.31\,\mathrm{eV}$, a factor of four. This shows explicitly that the asymmetry between bonding and antibonding effects is largest at short range (where overlap $S$ is large, amplifying the $1/(1-S)$ denominator for the antibonding state) and shrinks — though never fully vanishes at finite $R$ — as the bond stretches, consistent with both curves converging to $\Delta E_\pm \to 0$ as $R\to\infty$.

Example 3 — Exam/competition level: extracting the resonance integral from an observed splitting, and testing the S→0 idealization. A common exam-style question gives you a splitting between bonding and antibonding levels and asks you to back out an effective resonance integral under the simplifying (and only approximately valid) assumption $S\approx 0$, which is the same zero-overlap idealization used in simple Hückel theory. Under $S=0$, the formulas collapse to

so the full splitting is $E_- - E_+ \approx 2K$ — this is the origin of the ubiquitous Hückel-theory statement "the bonding/antibonding gap is $2\beta$." Suppose a photoelectron-type experiment (hypothetically, for a diatomic one-electron model system analogous to H₂⁺ but with a different effective nuclear charge) reports $\alpha = -13.0\,\mathrm{eV}$ and a bonding-to-antibonding gap of $E_- - E_+ = 7.40\,\mathrm{eV}$. Under the zero-overlap approximation,

Now test how good the $S=0$ idealization really is by re-deriving what $S\neq 0$ would predict for the same underlying $\alpha,K$: using the true formulas $E_\pm = (\alpha\pm K)/(1\pm S)$ with, say, $S=0.30$ (a realistic value at moderate bond length),

Notice the qualitative danger revealed by this check: once $S$ is included honestly, $E_+ = -12.85\,\mathrm{eV}$ is now higher than $\alpha=-13.0\,\mathrm{eV}$ used naively, i.e. the zero-overlap shortcut can even get the sign or ordering details wrong if $\alpha, K, S$ are not mutually consistent (in a real molecule $\alpha$ itself already contains $J(R)$, which shifts with $R$ together with $S$ and $K$ — they are not independent free parameters, as the closed forms above make explicit). The lesson for an exam: the $S\to0$ shortcut is useful for a fast estimate of the splitting $E_--E_+\approx 2K$, but never trust it for the absolute positions of $E_+$ and $E_-$ relative to $\alpha$ without also checking the size of $S$ at the relevant $R$.