Molecular orbital (MO) diagram for N2 and N2^-

Question

I have been taught that the MO diagram is different for molecules with 14 or less electrons than the one used for molecules with 15 or more electrons.

For $\ce{N2}$ the orbitals in increasing energy are:

$$\sigma_\mathrm{1s} < \sigma^*_\mathrm{1s} < \sigma_\mathrm{2s} < \sigma^*_\mathrm{2s} < \pi_{\mathrm{2p}_x}, \pi_{\mathrm{2p}_y} < \sigma_{\mathrm{2p}_z} < \pi^*_{\mathrm{2p}_x}, \pi^*_{\mathrm{2p}_y} < \sigma^*_{\mathrm{2p}_z}$$

because it has 14 electrons.

For $\ce{N2-}$ there are 15 electrons. Will the MO diagram be the same as that of $\ce{N2}$ or not? Why?

And because this is a good chance to ask, why is there this difference in orbital energies anyway?

Answer 1

I have been taught that the MO diagram is different for molecules with 14 or less electrons than the one used for molecules with 15 or more electrons.

This is (partly) wrong because the change in the order of $\mathrm{\sigma_{2p_{z}}}$ and $\mathrm{\pi_{2p_{xy}}}$ MOs to the left of $\ce{N2}$ is not directly related to the number of electrons. Rather, it is rationalized by a successive decrease of the $\mathrm{s}$-$\mathrm{p}$ interaction moving from $\ce{Li2}$ to $\ce{F2}$. The $\mathrm{s}$-$\mathrm{p}$ interaction is the bonding interaction between the $\mathrm{2s}$ orbital of one atom and the $\mathrm{2p_{z}}$ orbital of another atom which (among other things) increases the energy of the $\mathrm{\sigma_{2p_{z}}}$ MO relative to the hypothetical case without $\mathrm{s}$-$\mathrm{p}$ interaction.

Now the difference in energy between the $\mathrm{2s}$ and $\mathrm{2p_{z}}$ AOs increases from $\ce{Li2}$ to $\ce{F2}$ due to increasing nuclear charge and poor screening of the $\mathrm{2s}$ electrons by electrons in the $\mathrm{2p}$ subshell. As a result, the rising effect of $\mathrm{s}$-$\mathrm{p}$ interaction on $\mathrm{\sigma_{2p_{z}}}$ MO is getting less and less prominent, so that eventually to the right of $\ce{N2}$ $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\pi_{2p_{x,y}}}$ MO.

Image courtesy of the UC Davis ChemWiki.

Now, back to the question.

For $\ce{N2-}$ there are 15 electrons. Will the MO diagram be the same as that of $\ce{N2}$ (because it is actually an ionized molecule of $\ce{N2}$) or not? Why?

The short answer is: we could not tell it using the primitive molecular orbital theory introduced in the general chemistry courses. In exact same way we could not tell why $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\sigma_{2p_{z}}}$ MO to the left of $\ce{N2}$ and not to the left of, say, $\ce{C2}$. All this is simply because the primitive molecular orbital theory does not explain the things, rather it rationalizes them (for differences between explanation and rationalization, please, read here).

But then how do we know in the first place that $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\sigma_{2p_{z}}}$ MO to the left of $\ce{N2}$ so that we can start to rationalize it? Well, we know that from a more advanced molecular orbital theory which is based on the laws of quantum mechanics. And the same advanced theory should be applied for the case of $\ce{N2-}$ to get the MO picture of this chemical species.

Now note that even in this advanced molecular orbital theory a bunch of approximations is introduced, and the answer in general depends on at which level of theory calculations are done. The simplest picture would be the so-called frozen-orbital approximation (in the Hartree-Fock picture) in which MOs of an anion (as well as that of a cation) are assumed to be the same as for the corresponding neutral molecule. But this approximate picture might be far away from the reality.

Update

The question in fact does not make a lot of sense since $\ce{N2-}$ is known to be electronically unstable, i.e. its energetically less favourable then the system of $\ce{N2}$ plus a free electron. There exist few metastable resonance states of $\ce{N2-}$ that correspond to attachment of the scattered electron to excited valence states¹, but that's a different story which is beyond the scope of the question.

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1) Krauss, M., Neumann, D., Valence Resonance States of N2, JResNatlStandSecA, 77A, 4, 1973. PDF.

Answer 2

The reason for the special order for molecules with 14 or less electrons is due to the s-p mixing that arises due to the s and p atomic orbitals being close in energy. It is not actually possible to definitely answer your question without looking at the actual orbital energies.

Despite $\ce{N2-}$ being isoelectronic to $\ce{O2+}$, the lower effective nuclear charge on nitrogen should make its s orbitals a little closer to the energies of the p orbitals than they would be in oxygen. Hence s-p mixing should occur. However there are other considerations that we need to take in account, such as the extra electron repulsion that occur due to the extra electron. Yet, as I have stated above we can't definitively say which effect is more important without looking at the actual orbital energies.