Why does the mixing of sigma 2s and 2p orbitals lower the energy of the sigma 2s orbital and raise the energy of the sigma 2p orbital?

My general chemistry textbook (General Chemistry Principles and Modern Applications, Tenth Edition) says that for bonding in diatomic molecules with Z ≤ 7, orbital mixing occurs between the $\sigma_{2s}$ and $\sigma_{2p}$ molecular orbitals because they are close enough in energy. As a result, the energy of the $\sigma_{2s}$ molecular orbital is lowered and the energy of the $\sigma_{2p}$ molecular orbital is raised.

My question is why: if the $\sigma_{2s}$ orbital mixes with the $\sigma_{2p}$ orbital, wouldn't the modified $\sigma_{2s}$ orbital be higher in energy, as a result of mixing with an orbital slightly higher in energy?

Likewise, wouldn't the modified $\sigma_{2p}$ orbital be lower in energy as a result of mixing with an orbital slightly lower in energy? Why is it the other way around?

• Nov 21, 2015 at 19:10

This is a frequent source of confusion in my experience - you'll see there are at least two other very similar questions here already.

Let's take the first point: why do the $\sigma_{2s}$ orbitals get lower in energy when "mixing" with the $\sigma_{2p}$?

I'm going to ignore "mixing" and use the word "interact" instead. If I consider the 1s orbitals in $\ce{H2}$, I take two 1s atomic orbitals and they (of course) interact. So one orbital goes down in energy, and one orbital goes up in energy.

This is a common theme - if there's any degree of overlap between two orbitals, one will become more stable and one will become less stable. In the diagram, we claim "bonding" and "anti bonding" but in more complex molecules, these terms become hard to identify (e.g. benzene).

The theme remains: if orbitals interact (i.e., have the same symmetry), some will go down in energy, some will go up in energy.

That's the answer to your first question. Since the $\sigma_{2s}$ orbitals do interact with $\sigma_{2p}$ the 2s will go down in energy.

Your second question is about the $\sigma_2p$ going up in energy.

I could answer with a formula from quantum mechanics, but let's argue qualitatively. Here's a diagram for two interacting orbitals:

This is much like our case with the 2s and 2p orbitals. We see that the result is a system that's most likely more stable - the orbital lower in energy is even lower because of the interaction.

If we consider the reverse case - that the 2s goes up in energy, and the 2p goes down in energy, we would have a system that's less stable. In theory, this could happen - if the interaction is negative. In other words, the system would be more stable without the interaction than with.

In short, the "mixing" or interaction between 2s and 2p orbitals occurs because the species becomes more stable.

• So, the reason the σ2s orbital goes down in energy and the σ2p orbital goes up in energy has to do with the fact that 2s orbitals are more inclined to be lower in energy and 2p orbitals are inclined to be higher in energy? Nov 21, 2015 at 20:29
• I'm not sure I'd use "inclined," but yes - the 2s is lower in energy to begin with, so it gets even more stable because of the favorable interaction with the 2p. Nov 21, 2015 at 21:18
• But doesn't the fact that the 2p orbital increases in energy make it somewhat less favorable? Nov 21, 2015 at 21:41
• The question is whether the total energy is favorable. Nov 21, 2015 at 23:23