# Why is there a difference between O2 and B2 sigma 2p molecular orbitals in diagrams?

The molecular orbital diagram for $\ce{O2}$ says that the sigma 2p bonding molecular orbital is lower in energy than the pi 2p bonding molecular orbital. Why is this not the case in the $\ce{B2}$ MO diagram?

• It is, though, isn't it? – mikhailcazi Oct 25 '13 at 14:48

This topic is really fun. You see in case of boron molecule $\sigma_{2s}$ and $\sigma_{2p_z}$ (being on the intermolecular axis) will have the same symmetry and hence the orbitals will overlap and result in the formation of 2 bonded $sp_z$ orbitals. So the $\sigma_{2p_z}$ orbital will obtain somewhat $s$ character and it will become unstable or in other words will be of higher energy. The $\sigma_{2s}$ MO will obtain somewhat $p$ character. So in a energy level MO diagram $\sigma_{2p}$ bonding orbital is placed above the $\pi_{ 2p}$ orbital.
However, in case of $\ce{O2}$, due to the higher nuclear charge of oxygen, the energy difference of $2s$ and $2p$ is large so there is no such kind of intermixing between the bonded MOs. Whereas in boron, the energy difference of $2s$ and $2p$ is less, facilitating intermixing.