The AntiBonding Orbital with Shrinking Interatomic Distance

Question

I guess this is more of a chemistry question, but whatever. I think it's interesting.

Suppose you had two bare atomic nuclei. For concreteness, lets assume the nuclei are the same with atomic number $Z$. Lets bring in a single electron and focus on the ground states of the nuclei.

When the nuclei are far apart, the ground states are degenerate. When we bring the nuclei together, the ground state splits into the bonding and anti-bonding orbitals. Let $\Delta E$ represent some measure of the energy difference between the bonding and anti-bonding orbitals.

From intuition, I would expect $\Delta E$ to increase with decreasing internuclear distance $R$. What happens as $R$ shrinks to zero?

I expect the bonding orbital to become the ground state of an "atom" with charge $2Z$. Is that correct? More importantly, what happens to the anti-bonding orbital?

This isn't an exercise in the Born-Oppenheimer Approximation. I magically hold the nuclei at a distance $R$, so their repulsion doesn't matter. Also, electron-electron repulsion doesn't matter because I only introduce one electron.

Answer 1

The split into the bonding and anti-bonding orbitals comes from the LCAO (linear combination of atomic orbitals) approximation, and this approximation breaks down long before the two nuclei merge. I'm not sure it makes sense to ask what becomes of the antibonding orbital when the nuclei get close.

Later: I looked up the LCAO approach in my venerable (1978!) copy of Atkins' book Molecular Quantum Mechanics. The energy of the two states is given by:

$$ E = \frac{\alpha \pm \beta}{1 \pm S} $$

where:

$$ S = \int d\tau_1 \phi_1 \phi_2 $$ $$ \alpha = \int d\tau_1 \phi_1 \hat{H} \phi_1 $$ $$ \beta = \int d\tau_1 \phi_1 \hat{H} \phi_2 $$

Assuming the approximation remains valid up to zero separation the energy will diverge to (I think) infinity. I say "I think" because it actually ends up as zero divided by zero at zero separation.

Answer 2

The binding orbital is parity invariant to reflections around the center between the two nuclei, while the anti-binding orbital gets a minus sign under this reflection, from the fact that it is mixing the two ground states with a minus sign phase. This property is preserved as you adiabatically bring the nuclei close, and you can trace the level you get using only this fact.

The spin of the electron is decoupled and irrelevant, the binding ground state is necessarily adiabatically linked to the 2Z atom bound state, and the anti-binding state must become the lowest parity odd state. The obvious candidate is the p-orbital (n=2 l=1) which is aligned along the axis of separation of the two nuclei (as the separation becomes infinitesimal). In order for this to not be the endstate of the adiabatic flow of eigenstates, there must be another eigenstate which becomes degenerate with the anti-binding configuration. This is not going to happen, because the states which come from the excited states of atoms (at infinite separation) are always higher in energy than the anti-binding state until the separation gets to zero (this is intuitively clear, although it would be hard to prove). At exact collision, you get a degeneracy of the antibinding state with 2 excited states, since there are 3 degenerate p orbitals at n=2. This is the first degeneracy that develops in the system, which is clear from going the other way.

If you split the nuclear charge of 2Z into two nearby Z nuclei, you add a dipole moment to the nucleus. This dipole moment on average doesn't affect the n=2 S-orbital at first order in perturbations, only at second order. But at first order, this perturbation splits the n=2 p levels so that the one where the p-lobes are aligned with the separation of the nuclei is lower energy. This is the state that's turning into the antibinding.