A question in my exercise book demonstrates that nitrogen cannot form non-octet structures since the most common examples — nitrogen dioxide and nitrogen monoxide — are unstable/reactive. For example, $\ce{NO2}$ quickly dimerizes to $\ce{N2O4}.$ $\ce{NO}$ is quickly oxidized to $\ce{NO2}.$ Other elements, like phosphorus, on the other hand, can form stable phosphorus pentachloride.

I do not think the question is precisely set. Despite the instability of nitrogen dioxide and nitrogen monoxide, it does not mean nitrogen “cannot” form non-octet structures.

Could anyone give me examples of stable nitrogen compounds where the nitrogen atom does not achieve an octet? I just want to prove that the answer is wrong.

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    $\begingroup$ What do you mean when you say NO2/NO are unstable? Do you mean reactive? $\endgroup$
    – porphyrin
    Commented Nov 7, 2023 at 9:57
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    $\begingroup$ 1. "Unstable" and "Reactive" are two different things. 2. What's the basis of comparing nitrogen oxides with phosphorus halides? $\endgroup$ Commented Nov 7, 2023 at 11:29

1 Answer 1


Your comparison of the nitrogen oxides with phosphorus pentachloride is not apt. The former have unpaired electrons whereas the latter has multicener bonding with all electrons paired. These represent different departures from the octet rule.

Nitrogen forms trifluoramine oxide, $\ce{NOF3}$, whose structure is described here. The nitrogen is bonded tetrahedrally to the other atoms, and the normal octet rules "should" allow only single bonds between all the connected atom pairs. But empirically the nitrogen-oxygen linkage shows double-bond character as if the nitrogen were forming five bonds. "Outer $d$" valence orbitals can't explain this when the atoms are all in Period 2. Molecular-orbital theory can.


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