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Lynn Holland

The molecular orbitals of lithium hydride

Daniel Iyamuremye  Follow

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...

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Gilbert Mendez  Follow
I think it qualifies as an answer. I dont think it can really be "answered" completely as it stands, the numbers seem off and this is quite possibly because they come from different sources, so its fair to ask for where they come from.More
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Dare Robertson  Follow

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.

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