If we try to calculate the oxidation state of nitrogen in $\ce{N2O}$ using the familiar algebraic method, we get oxidation state $+1$ for both nitrogen atoms and that's what I found when I looked it up on the internet.

I tried to do it with the structure and that's where I got confused:


The $\ce{N=N}$ bond is a coordinate bond, so that should give a $-1$ oxidation state for the left $\ce{N}$ and $+1$ for the middle one. Since the middle one has a double bond with oxygen, it gets additional $+2$ for a total of $+3.$

But if I consider this resonance structure:


The left $\ce{N}$ is getting an oxidation state of $0$ and the middle one gets $+2.$ So I am confused as to what is actually correct.

Is it possible for atoms to exist in superpositions of oxidation states? Or maybe I am overlooking some basic concept?

  • 2
    $\begingroup$ It is possible for molecule to be in a superposition of different quantum states involving different electronic structure. Why should not be possible to be in superposition of oxidation states, what are, in contrary to the former, just a thought conceptual construct ? See also wikipedia: Nitrous oxide $\endgroup$
    – Poutnik
    Commented Jan 29, 2021 at 7:05
  • 1
    $\begingroup$ Some jurisdictions set the age of consent to 16 years, some to 17, and some to 18. Which is actually correct? Er, well... Same thing here. Oxidation states are human concept; atoms don't care about them. $\endgroup$ Commented Jan 29, 2021 at 8:10
  • $\begingroup$ related: chemistry.stackexchange.com/a/31272/102629 $\endgroup$
    – cngzz1
    Commented Jan 30, 2021 at 10:19

1 Answer 1


If we try to calculate the oxidation state of nitrogen in $\ce{N2O}$ using the familiar algebraic method, we get oxidation state $+1$ for both nitrogen atoms and that's what I found when I looked it up on the internet.

Well … you get an average oxidation state. This calculation arguably implicitly assumes that all nitrogen atoms be equivalent. In some cases (e.g. hydrazine) they are and the result you get algebraicly is as you would expect from a Lewis depiction. In other cases (e.g. here) this is not the case as the nitrogens are not equivalent (only one is bound to oxygen). Thus, it is obvious that a simple algebraic approach should fail.

But what about the ‘real’ result? What about the resonance structures? Well, this is where things get really difficult. You essentially have two π systems orthogonal to each other and each occupied by four electrons which can manifest as a lone pair on either end and a multiple bond to the other atom. If you really wanted to play the game, you could include another resonance structure as shown below, where all π lone pairs are centred on the terminal nitrogen:

$$\ce{N#\overset{+}{N}-\overset{-}{O} <-> \overset{-}{N}=\overset{+}{N}=O <-> \overset{2-}{N}-\overset{+}{N}#\overset{+}{O}}$$

(It is clear that this third resonance structure contributes least to the overall picture as it has a greater charge separation and charges are separated opposite to what electronegativity would predict.)

The experimental structure shows that the $\ce{N-N}$ distance is slightly shorter than the $\ce{N-O}$ distance which one could use to assume that the $\ce{N-N}$ bond have a slightly larger bond order than the $\ce{N-O}$ bond. But ultimately, they are still very similar (the difference is merely $\pu{4pm}$) so equal bond orders might also be an option. Long story short: short of calculating the electronic distribution (i.e. solving the Schrödinger equation) you won’t be able to arrive at a definite answer for the ‘real’ oxidation states.

So what can you do on paper? Comparing the resonance structures, the leftmost as I have ordered them is slightly better than the central one as the formal charges are distributed according to the different electronegativities. Thus, I would be inclined to give it a slightly higher weight and – in a classroom setting – use it to determine oxidation states. That said, it is too much of an ambiguous example to be seriously used in any examination unless the goal of said examination is to develop the chain of arguments that formed this answer.

  • $\begingroup$ Actually this was a question from an exam and the given answer was +1 but what you said makes sense. I could challenge the answer key, I suppose. $\endgroup$
    – Physicsa
    Commented Jan 29, 2021 at 14:35
  • $\begingroup$ @Physicsa Well … did they ask or imply using the algebraic method? Or did they strongly suggest using the structure? Because if the exam question intended to test the algebraic method, that’s okay imho. $\endgroup$
    – Jan
    Commented Jan 29, 2021 at 15:46
  • $\begingroup$ They didn't imply using any one method over another. $\endgroup$
    – Physicsa
    Commented Feb 5, 2021 at 8:17

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