# In molecular orbital theory, why does s-p mixing in the dinitrogen molecule not effect the 1σᵤ orbital?

s-p mixing in $\ce{N2}$ causes the $1\sigma _g$ orbital to decrease in energy, the $2 \sigma _g$ orbital to increase in energy (to higher than the $1 \pi _u$) yet the $1 \sigma _u$ is left unchanged by the interaction. Why is this?

My lack of understanding may be as a result of the fact that I am unclear as to what exactly it is that is mixing. Clearly they are the s and p orbitals but is this within the atom or is the s orbital on one atom interacting with the p on the other?

Some clarification as to why the $1\sigma _g$ orbital to decreases in energy, the $2 \sigma _g$ orbital to increases in energy would also be much appreciated.

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Rob, if you poke around on the internet you can find many sites that show diagrams for how the $\ce{N2}$ molecular orbitals are formed from two sets of nitrogen ($\ce{N1}$) atomic orbitals. Many of the diagrams are different one from another - that's a tip off that s-p mixing is not something that can always be predicted in advance. With that understanding, let's step through your diagram and questions.

It might be useful to take a look at the $\ce{N2}$ MO diagram here. It's just like your diagram. but they break down the steps and it might be a bit less confusing.

In any case we have a nitrogen atom with 5 valence electrons, 2 in the 2s atomic orbital and 1 each (total of 3) in the 3 degenerate p orbitals. If we move a second nitrogen atom into the vicinity, these atomic orbitals will hybridize in order to start the bond formation process.

$\ce{N2}$ is analogous to acetylene, they both have a triple bond that is made up of 1 sigma bond and 2 pi bonds. In acetylene - and nitrogen - the sigma bond is $\ce{sp}$ hybridized. We form this bond by

• Mixing the s and a p atomic orbitals (the p orbital whose lobe is linear with the s orbital [see the diagrams in the link I supplied above, they call it the $\ce{p_{z}}$ orbital]) on one of the nitrogens creates a bonding and antibonding $\ce{sp}$ orbital on that nitrogen atom.
• Same thing happens on the other nitrogen atom
• As the 2 nitrogen atoms begin to interact and form bonds, the 2 $\ce{sp}$ orbitals on each nitrogen interact (split further).
• The bonding $\ce{sp}$ orbital on each nitrogen interacts to form a lower energy and higher energy pair of bonding molecular orbitals (the $\ce{1\sigma _{g}}$ and $\ce{1\sigma _{u}}$ orbitals in your diagram - BTW, there should be a dashed line from the p orbitals to the $\ce{1\sigma _{g}}$ orbital - it is missing in your diagram, maybe that has led to some of your confusion).
• The antibonding $\ce{sp}$ orbital on each nitrogen also interact in the same manner to create 2 antibonding sigma molecular orbitals ($\ce{2\sigma _{g}}$ and $\ce{2\sigma _{u}}$)

Now for pi bonds,

• The remaining two p atomic orbitals on each nitrogen are orthogonal to the s and $\ce{p_{z}}$ orbital - they can't mix with those orbitals.
• These atomic p orbitals $\ce{p_{x}}$ and $\ce{p_{y}}$ can only mix with their like counterpart on the other nitrogen atom. This gives rise to a bonding and antibonding $\ce{p_{x}}$ pi molecular orbital and a bonding and antibonding $\ce{p_{y}}$ pi molecular orbital

With that background, let's now look at your questions

s-p mixing in N 2 causes the 1σ g orbital to decrease in energy, the 2σ g orbital to increase in energy (to higher than the 1π u ) yet the 1σ u is left unchanged by the interaction. Why is this?

I'd say it a bit differently as I did above. Mixing of the 2 bonding $\ce{sp}$ orbitals, one on each nitrogen atom, causes them to split (one is raised and one is lowered). The same thing happens to the antibonding $\ce{sp}$ orbital on each nitrogen - one is raised, one is lowered.

Clearly they are the s and p orbitals but is this within the atom or is the s orbital on one atom interacting with the p on the other?

Hopefully I answered this above. I tried to distinguish between atomic and molecular orbitals. If I haven't been clear, post back.

Some clarification as to why the 1σ g orbital to decreases in energy, the 2σ g orbital to increases in energy would also be much appreciated.

Hopefully what I wrote above clarified this as well, let me know if it didn't.

I can see how Ron's answer fully addresses the effect of sp-mixing, I would just like to add a probable cause for this effect: I'm not sure about this theory, but could it have anything to do with the relative penetrations of the n(sigma bonding and antibonding) orbitals being much more than that for the higher energy n+1(sigma bonding and antibonding) orbitals? The ones in the lower energy state (n'th quantum number) being more stabilized due to its greater capacity at penetration, which leads to it being more stable (it is closer to the positive core of each atom). The decrease in the stability of the n+1 orbitals can thusly be accounted for, since the effective positive charge from the nucleus is decreased.