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Here are the MO schemes of $\ce{N2}$ (left) and $\ce{O2}$ (right).

Why is the $\sigma$-MO formed by the $p$ AOs energetically above the $\pi$-MO for $\ce{N2}$ but not for $\ce{O2}$?

Can it be explained this way:
In second period elements with more than a half filled $\ce{p}$ orbitals, the energy pattern of the MOs is regular.
In second period elements with less than or half filled $\ce{p}$ orbitals, as the $\ce{s}$ and $\ce{p}$ atomic orbitals have similar energies, so the formed $\ce{\sigma_{g}}$ MO would have similar energy as $\ce{\sigma_{u}}$, but because of electron cloud repulsion it has a big increase in energy, hence it is above the $\pi$ orbitals? And this phenomenom is known as sp mixing.

Or am I wrong somewhere? Is my reasoning corret?

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  • $\begingroup$ The answer will take some time and I won't be able to get to it before I finished work. So, for the time being: In short, your explanation contains some good and valid thoughts but if you want to really understand the energetic orderings you will have to dive into pertubational MO theory. $\endgroup$ – Philipp Jul 18 '14 at 12:02
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This phenomenon is explained by s-p mixing. All the elements in the second period before oxygen have the difference in energy between the 2s and 2p orbital small enough, so that s-p mixing (combination) can occur lowering the energy of the σ(2s) and σ*(2s) and increasing the energy of the σ(2p) and σ*(2p) molecular orbitals. By moving towards right in a period, the s orbital gets more stabilized than the p orbital and the difference in their energies increases, making the s-p mixing for oxygen much smaller.

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    $\begingroup$ I think it is an exaggeration to say s-p mixing for oxgyen is negligible in comparision to nitrogen. Quantitatively, for N2 the sigma bond is about 27% 2s and 73% 2p and O2 is about 13% 2s and 87% 2p. So there is less mixing for O2, but there is still mixing. pubs.acs.org/doi/abs/10.1021/ja00544a007 $\endgroup$ – DavePhD Mar 19 '15 at 18:36
  • $\begingroup$ “All the elements in the second period before oxygen” does it means all molecules have 1 atom before oxygen or a molecule only have atoms before oxygen? Is it only works with diatomic molecules? $\endgroup$ – Chao Song May 11 '17 at 7:44
  • $\begingroup$ All homonuclear diatomic molecules before oxygen. $\endgroup$ – RBW May 11 '17 at 12:38
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S-p mixing is the primary cause of the difference in the molecular orbitals of nitrogen and oxygen, which is influenced by the initial atomic orbital energies.

The lighter second period elements (prior to oxygen) have a relatively small difference in energy between the 2s and 2p orbitals. This allows sufficient s-p mixing to lower the energy of the σ(2s) and σ*(2s) molecular orbitals, and is energetically offset by an increase in energy of the σ(2p) and σ*(2p) molecular orbitals. In the dinitrogen molecular orbital scheme, the dashed lines are there to represent s-p mixing influencing the energy of the four molecular orbitals involved.

The effective nuclear charge increases to the right of the period, stabilizing the 2s orbital more drastically than the 2p orbital. This can be seen qualitatively in the first figure here.

The more stabilized 2s orbital does not s-p mix as effectively, due to the greater energy difference between the 2s and 2p orbitals. As nuclear charge increases, s-p mixing becomes less significant. The change of the molecular orbital ordering between nitrogen and oxygen is the manifestation of this decreased s-p mixing. In the dioxygen molecular orbital scheme the s-p mixing effect is no longer significant enough to alter the relative orbital arrangement.

More information about the details of this difference can be found in most inorganic chemistry textbooks. There are good diagrams showing the gradual change in energy differences across the second row range of homonuclear diatomic molecules. This document highlights the progressive lowering of molecular orbital energies due to the the atomic orbital changes, and presents a correlation diagram linking internuclear separation with molecular orbital energies as well.

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The strength of interaction of sigma and pi-kind developing differently with distance between two atoms. The pi-interaction, which usually is weaker, is favoured at shorter atomic distance, because it leads to better side overlapping of orbitals. The maximum of energy for pi is by shorter atomic distance than for sigma. $\ce{N2}$ molecule has very short $\ce{N-N}$ distance, which leads to much better p-orbitals overlap and as consequence to lower energy of pi-orbital.

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  • $\begingroup$ Okay, thank you. Why would pi orbital have bigger energy than sigma in oxygen and not in boron molecule? Oxygen has a double bond. $\endgroup$ – RBW Aug 3 '14 at 14:54

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