I have a confusion related to the structure of $\ce{N2O3}.$ I was taught 1 is the correct one, but I believe 2 is better since it has no formal charge on it. Why is 1 correct and 2 wrong?

Structures of N2O3


2 Answers 2


What really happens in nature has higher priority than our believe what happens or should happen. See Dinitrogen_trioxide with the structure, bond lengths and angles.

Be aware both $\ce{NO}$ and $\ce{NO2}$ are radicals with an unpaired electron.

$\ce{N2O3}$ being( $\ce{ON-NO2}$ ) as 2 paired radicals is the direct analogy
to the dimer $\ce{N2O4}$ being $\ce{O2N-NO2}$, both being formed at low temperatures.

  • $\begingroup$ So it’s just what it is? $\endgroup$
    – Rasputin
    Aug 20, 2020 at 10:44
  • 3
    $\begingroup$ @Rasputin Yes, it is what it is. Also keep in mind you're using a model, which is not the same thing as reality. Some times, models do not give the prediction they ought to give. $\endgroup$
    – Zhe
    Aug 20, 2020 at 13:42

I think you assume that Nitrogen Sesquioxide is formed by a dehydration reaction of an acid of Nitrogen, hence necessitating the presence of a bridged oxygen. This isn't the case. It is created due to the addition of the radical on the Nitrogen of $\ce{NO}$ to the radical on the Nitrogen on $\ce{NO2}$. Hence why the structure is the way it is.


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