Molecular orbital diagram for nitrogen monoxide, the nitrosyl cation and the nitrosyl anion

As I understand it, all of them should have same order of energy of molecular orbitals as a model for $\ce{NO}$ should withstand ionization and adding an electron.


JD Lee Pg 109 shows the same diagram for $\ce{NO}$ as the one quoted in [here](https://chemistry.stackexchange.com/a/15317/24577). I couldn't find $\ce{NO+}$ except in [here](http://www.uky.edu/~holler/orbitals/no/no.html), where $\sigma\_{2p\_z}$ has a higher energy than $\pi\_{2p\_x}$ and $\pi\_{2p\_y}$.


If it is correct, then, there is $s$-$p$ mixing in $\ce{NO+}$ but not in $\ce{NO}$?




As usual with these simple molecules: It's not as simple as it seems. Here it actually doubles down quite a bit more.  

The short and very dis-satisfactory answer is: **It is very, very complicated.** For this reason, I am reluctant to present a molecular orbital diagram. However, I will provide calculated MOs. While it may be tempting to just look at the pictures, the rationalisation behind it is more important. What will follow now are a few attempts at that rationalisation.


In the nitric oxide series, $\ce{NO+}$ is surprisingly the simplest. Why is that surprising? Because it is isoelectronic with $\ce{CO}$; and this molecule is one of the most complicated ones there are. You can read a bit more about it here: [How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?](https://chemistry.stackexchange.com/q/30797/4945)  

I do compare $\ce{NO+}$ more with $\ce{CO}$ rather than $\ce{N2}$ to which it is also isoelectronic, because of the asymmetry introduced by the different nuclei. The molecule is the simplest to explain, because it is a singlet ground state. All electrons are neatly paired, and spacial symmetry can easily be matched with the degeneracy of the orbitals.  

The π-orbitals are not the HOMO, instead it is the σ lone pair at nitrogen. 


The next more complicated is $\ce{NO-}$, which is isoelectronic to $\ce{O2}$. In this case it is hard to find a less symmetric isoelectronic version. It is well known, that the ground state is a triplet because of the degeneracy of the π-orbitals. While we would expect the orbital ordering to be the same as in $\ce{NO+}$, it is not. This is mainly due to the fact, that occupied orbitals behave differently than unoccupied orbitals.  

In some way, because of the paramagnetic character, we have a situation wich cannot be described by a single molecular orbital description. You need at least two: one for α-electrons (or spin up) and one for β-electrons (or spin down). The good thing however is, that both of these descriptions may still be matched with spatial symmetry.  

The bonding π orbitals become higher in energy than the σ lone pair because the anti-bonding counterparts are partially filled.


The most complicated molecule in the series is the radical $\ce{NO}$. There is absolutely no simple description. Because of the unpaired electron, you need at least two descriptions again. However, this is not enough because now one of the spin-state description cannot match the spatial symmetry. This is because one electron cannot occupy two orbitals at the same time in an MO description. Because of that (from a technical point of view) the orbital degeneracy has to be broken. Physically this is not possible.  

The molecule has to be treated with a multi-reference approach, i.e. one MO description for every possible configuration. It is probably safe to assume that the orbitals of π symmetry don't split that much and are higher in energy than the σ lone pair.


Below you can find the calculated MO. The level of theory used for $\ce{NO+}$ was RMP2/def2-QZVPP, because this is equivalent to the level the other molecules were treated, which is ROMP2//UMP2/def2-QZVPP. One can clearly see how the orbital symmetry for $\ce{NO}$ breaks the degeneracy of the pi orbitals.  

For $\ce{NO+}$ MOs 5,6 and 8,9 are degenerated. For $\ce{NO-}$ MOs 6,7 and 8,9 are degenerated.  

Colour code: blue/orange, α, β-occupied; purple/yellow-orange, α/β-occupied; red/yellow, unoccupied; contour value, 0.1.


[![molecular orbitals for NO+, NO, NO- at ROMP2//UMP2/def2-QZVPP](https://i.stack.imgur.com/ehxdm.jpg)](https://i.stack.imgur.com/ehxdm.jpg)




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With a little more literature research (prior to answering the question) one would have found the following publication:
[Renato P. Orenha and Sérgio E. Galembeck. Molecular Orbitals of $\ce{NO}$, $\ce{NO+}$, and $\ce{NO–}$: A Computational Quantum Chemistry Experiment. *J. Chem. Educ.* **2014,** *91* (7), 1064–1069.](https://dx.doi.org/10.1021/ed400618j)


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