The molecular orbitals of lithium hydride

Question

So my understanding of molecular orbitals is as follows: the molecular orbitals of $\ce{LiH}$ can be thought of as being formed from the interaction/overlap of two atomic orbitals, one from lithium and one from hydrogen in this case. I also understand that the atomic orbitals have to be quite close in energy for any significant interaction to take place.

The occupied atomic orbitals of lithium are 1s and 2s, with energies of around -66 eV and $\pu{-5.3eV}$ respectively. The occupied atomic orbital of hydrogen is 1s, with an energy of -13.6 eV. I believe that it's safe to assume that the 1s orbital of lithium is far too low in energy to interact with the 1s orbital from hydrogen, meaning that only the interactions are between the 2s orbital of lithium and the 1s orbital of hydrogen, producing a bonding MO and an anti-bonding MO.

Wikipedia states that "Bonding MOs are lower in energy than the atomic orbitals that combine to produce them." In other words, the 2$\sigma$ bonding MO should have an energy less than -13.6 eV, which is the energy of the 1s hydrogen orbital. However, computer calculations of the MOs of LiH show that this bonding MO has an energy of -8eV, which would seem to contradict the statement from Wikipedia.

So assuming there is no fault in my reasoning so far, we can safely say that this Wikipedia statement is only generally true, and false in this particular case. So why is this?

I was trying to rationalise this as follows: generally, the reason why bonding MOs are lower in energy than the atomic orbitals used to create them is that bonding MOs concentrate electron density between the two nuclei, resulting in the electrons being attracted by both nuclei (as opposed to just one), leading to the MO being lower in energy than the original AOs. However, in LiH, due to the energy of the 1s AO on hydrogen being significantly lower than that of the 2s AO on lithium, the 1s AO will be the major contributor to the bonding MO, meaning that the electron density in the bonding MO will be significantly closer to hydrogen's nucleus than lithium's nucleus.

Now, originally, there is one electron each in the 1s hydrogen AO and 2s lithium AO, and when they come together, two things change: Firstly, there is electron-electron repulsion between these two electrons. Secondly, the electrons are now bound by two nuclei rather than 1. However, due to the high polarity of the bond, this second change is minimal, such that the first change (introduction of repulsion) actually outweighs the second change, meaning that the resulting bonding MO is actually higher in energy than the hydrogen 1s AO.

Is my rationalization of this accurate/correct? If not, please suggest alternative explanations/ improvements to my thought process because I'm frankly not fully convinced by myself.

Answer 1

We may want to consider also the energy change in the lithium 1s orbital. Formation of the molecular orbitals also changes the energy level of the core orbitals even though these do not participate appreciably in the bonding.

This is because what I call "reverse shielding" occurs. In a lithium atom in the ground state, there are of course two 1s electrons and one 2s electron. We commonly suppose that the 1s electrons, packed close to the nucleus, partially shield the 2s electron from the nuclear charge and raise the energy level of the latter. But, because the probability distributions of the 1s and 2s clouds overlap, there is a small probability that the 2s electron is instead shielding one or both of the 1s electrons. The effect on the 1s wavefunctions is small, due to the small probability, but the effect on lithium 1s energy can still register in electron volts because the 1s wavefunction is crowded into a region of strong Coulonb potential.

Then, when the hydrogen atom is added and the bonding orbital forms, what had been the 2s electrons are drawn away from the lithium atom and thus the reverse shielding of the lithium 1s electrons is diminished. Therefore, the lithium 1s orbital is lowered in energy, possibly enabling a stable molecule even if the bonding molecular orbital is higher than the isolated hydrogen 1s orbital as the OP suggests.

Answer 2

generally, the reason why bonding MOs are lower in energy than the atomic orbitals used to create them is because bonding MOs concentrate electron density between the two nuclei, resulting in the electrons being attracted by both nuclei (as opposed to just one), leading to the MO being lower in energy than the original AOs

That's not how bonding works. Actually having the two electrons "crunched together" in the middle is not that great at first because although they are now being attracted by both nuclei, they also repel one another and because the average distance between the electrons is smaller than their distance to the nuclei that repulsion is very noticeable. Therefore the overall potential energy of the electrons actually increases.
The gain in energy is due to the fact that the kinetic energy is being lowered due to the electrons now having "more space to move in" (if you refer to the particle in a box you'll see where that is coming from). A more concrete explanation is that the overall wave-function describing the two electrons has less curvature and as the curvature of a wave-function is directly connected to its kinetic energy (see Schrödinger-equation) this results in an overall lower kinetic energy.

Now that this is out of the way let me comment on your original question:

Bonding MOs are lower in energy than the atomic orbitals that combine to produce them.

That is true as long as you are viewing orbitals with roughly the same energy level (covalent edge case). If you are viewing two orbitals with a high energy difference (ionic edge case) you'll get a binding MO that is roughly at the level of the lower AOs. Therefore you could say that the electrons are getting transferred from the higher AO into the lower AO creating two ions that now bind together in a crystal-structure.
However that does not fit to the values you have provided either, so I would ask you to reveal where you got them from and unless someone has a better explanation I'd guess that one of those values is wrong.

I'm sorry for writing an answer that doesn't really answer your question but it was too much for a comment but (in my opinion) still something worth pointing out...

Answer 3

You have to be careful about the context of the statements like the one you quote from Wikipedia, and since you don't give the full details (and Wikipedia is transient anyway) I have to guess.

It seems likely that the statement that the energy of a stable MO must be less than the energy of the orbitals of the constituents refers to the Linear Combination of Atomic Orbitals (LCAO) theory. This simple statement probably only refers to molecules formed from a pair of atoms with a single valence electron (like H and Li), and it is only true in the Born-Oppenheimer approximation where the ions are at a fixed separation, and before you take account of the Coulomb repulsion between the ions (which is responsible for the fact that there is an optimum bond length). With this context, the result of LCAO states that the MO with a single electron has lower energy than either of the constituent atomic orbitals with a single electron.

In other words, AB$^+$ is deeper than A or B. In the example of LiH, LCAO predicts that the binding energy of LiH$^+$ is more than the binding energies of either Li or H.

The value of 8eV you found is the binding energy of LiH, not the binding energy of LiH$^+$, which is much larger. I found a value of 243kJ/mole = 2.5eV for the bond formation energy of LiH, i.e. removing both electrons from LiH costs 5.4+13.6+2.5 = 21.5 eV. If the binding energy of LiH is 7.8eV (which is the first ionisation energy) then the binding energy of LiH$^+$ must be 21.5-7.8 = 13.7 eV (which is the second ionisation energy). Clearly, this is only 0.1eV more than the binding energy of H, but nevertheless, the LCAO statement is correct.

Of course, these values for the molecular ionization energies are experimental, and therefore the value of 13.7eV takes into account the ion-ion Coulomb repulsion and the change in bond length from LiH to LiH$^+$. These corrections are explicitly neglected in the LCAO statement. If these corrections are removed then the LCAO prediction would be even deeper than 13.7eV. While LCAO can be extended to take account of these Coulomb corrections, the statement that the energy of AB$^+$ is below A or B is only necessarily true without them.