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

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?





> 
> 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.


[![enter image description here](https://i.stack.imgur.com/4KdEO.jpg)](https://i.stack.imgur.com/4KdEO.jpg)


Image courtesy of the [UC Davis ChemWiki](https://web.archive.org/web/20150706110342/http://chemwiki.ucdavis.edu/Wikitexts/UC_Davis/UCD_Chem_2A/UCD_Chem_2A:_Hayashi/Unit_IV:_Electronic_Structure_and_Bonding/10:_Bonding_II:_Additional_Aspects/10.5:_Molecular_Orbital_Theory). 


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](https://chemistry.stackexchange.com/a/34327/186)).


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.




---


1) Krauss, M., Neumann, D., Valence Resonance States of N2, *JResNatlStandSecA*, 77A, 4, 1973. [PDF](http://nvlpubs.nist.gov/nistpubs/jres/77A/jresv77An4p411_A1b.pdf).


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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.


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