Is the $\ce{NO-}$ a radical species? If so, can someone please explain why because $\ce{NO}$ has an odd number of electrons (11), so adding an additional electron would give 12 valence electrons and a formal charge of -1 on the nitrogen. Is my general chemistry failing me or?

  • $\begingroup$ Your title doesn't match your question. Hyponitrite is $\ce{N2O2}^{2-}$. As for the short-lived intermediate $\ce{NO-}$, I don't know whether or not it is a radical. Note, however, that it is formally isoelectronic to $\ce{O2}$, which surely is. $\endgroup$ – Ivan Neretin Dec 9 '16 at 6:33
  • $\begingroup$ "nitric oxide anion in its triplet ground state" scitation.aip.org/content/aip/journal/jcp/114/18/10.1063/… $\endgroup$ – DavePhD Dec 9 '16 at 14:32

The nitroxyl anion is isoelectronic to oxygen. It has pretty much the same orbitals as oxygen does and it also features the same number of electrons. The principal difference is the orbital shape: since oxygen and nitrogen are two different atoms with a different number of protons, there is no plane of symmetry bisecting the $\ce{N=O}$ bond axis (in group theory terms: the point group is $C_{\infty \mathrm v}$ rather than $D_{\infty \mathrm{h}}$). This difference means that bonding orbitals will be more oxygen-centred while antibonding ones will be more nitrogen-centred.

However, if you take a look at the molecular orbital scheme, the key orbitals are identical to those in oxygen. The highest occupied molecular orbitals are two symmetry-equivalent π* orbitals and the unoccupied σ* orbital is higher in energy. Since the two π* orbitals are symmetry equivalent, there is no way the molecule can pair the spins; the highest energy electrons will be parallel and occupy one of the π* orbitals each.

Considering that we have exactly the same situation as oxygen in the key orbitals, this also means that $\ce{NO-}$ features exactly the same states. Most notably, the ground state is a triplet $\ce{^3NO-}$ state, while a higher-energy singlet $\ce{^1NO-}$ state also exists — entirely analogously to the singlet $\ce{^1O2}$ and triplet $\ce{^3O2}$ states of dioxygen.

When calling the nitroxyl anion a radical, one should remember that it is, like the oxygen molecule, more accurately described as a diradical. Its properties are distinct from those of monoradicals such as $\ce{NO}$, which feature a doublet state.


I agree with the answer above (the foundamental state of nytroxyl anion is a triplet diradical). Chemically, though, as shown by its strong tendency to dimerize in the absence of coordination centers, it behaves more as a nitrenoid synton (-)O-N: As such it tends to couple with itself to form hyponitrite (-)O-N=N-O(-)

In this aspect it recalls somehow the double nature of CO : an hybrid of a oxonium carbanion and a carbene carbonyl.


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