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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. I couldn't find $\ce{NO+}$ except in here, 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 $\ce{NO}$?

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  • $\begingroup$ There is sp mixing in all of the three compounds, but they are sometimes not showed to have simplicity. $\endgroup$ – DHMO Jan 19 '17 at 10:16
  • $\begingroup$ Note that $\ce{NO+}$ is isoelectronic with $\ce{N2}$ and $\ce{NO-}$ with $\ce{O2}$. $\endgroup$ – DHMO Jan 19 '17 at 10:18
  • $\begingroup$ "If it is correct, then, there is $s$-$p$ mixing in $\ce{NO+}$ but not in $\ce{NO}$?" what is the "it" referring to? $\endgroup$ – DHMO Jan 19 '17 at 10:22
  • $\begingroup$ Sorry, I know that sp mixing is present in essentially all of them, I wanted to see if they have the same order of energy of 2p sigma bmo and 2pi bmo. The "it" was the second link. I think the following example can clear my mind : is HOMO of CO an antibonding or bonding MO when it takes part in reactions and not ionization? $\endgroup$ – Mrigank Jan 19 '17 at 11:13
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    $\begingroup$ s-p mixing isn't an ON/OFF switch. It's always present but in O2, F2 it just isn't a sufficiently large effect such that it changes the relative ordering of the energy levels. In B2, C2, N2 it is large enough. You can generalise to the heteronuclear diatomics. $\endgroup$ – orthocresol Jan 19 '17 at 20:51
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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?
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


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.

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    $\begingroup$ Just a random side-question: What’s the dipole moment’s negative and positive sides in $\ce{NO+}$? $\endgroup$ – Jan Jan 19 '17 at 19:08
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    $\begingroup$ @Jan That's not easy to be answered since the calculated dipole for ions is dependent on the coordinate system. When running the calculations I did not pay attention to where I put the centre of mass. I'd need to rerun the calculations with that in mind, or think of another way to analyse. $\endgroup$ – Martin - マーチン Jan 19 '17 at 19:28
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    $\begingroup$ @Jan, for your entertainment: What is the dipole moment direction in the nitrosonium ion? $\endgroup$ – Martin - マーチン Jan 20 '17 at 15:01

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