The following image depicts resonance in $\ce{NO2}$ molecule:

resonance in NO2

I don't know if it is obvious or not, but I'm puzzled by this structure. I've read about the structure of the $\ce{NO2}$ compound, but it wasn't related.

In the first structure, nitrogen is having a single electron with a positive charge, and oxygen is having a negative charge. So, that means nitrogen has donated a single electron to Oxygen, without forming a bond. I know it can be possible, but I couldn't understand how it happens.

Also, I thought that nitrogen donates a lone pair to oxygen, as it can have a maximum co-valency of 4. As in the following image:

enter image description here

But, this isn't the case here. Why?

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    $\begingroup$ Possible duplicate of Structure of NO2 compound? $\endgroup$ Dec 11, 2018 at 13:46
  • $\begingroup$ But, @NilayGhosh that post asks about whether the single electron will stay on Nitrogen or on Oxygen. Whereas I want some insight on how a single electron stays on Nitrogen. $\endgroup$ Dec 11, 2018 at 13:54
  • $\begingroup$ I don't understand the question. There is an odd number of electrons, they have to be distributed as best as they can, preferably so that all elements get as close to 8 as possible (but not more). $\endgroup$ Dec 11, 2018 at 14:59
  • $\begingroup$ @Martin-マーチン I was saying that why did N donated only one electron (I've read that electrons are always donated in pair) to O, instead N could also make a bond with O, gaining a co-valency of four. $\endgroup$ Dec 11, 2018 at 15:06
  • $\begingroup$ Where ever you read that electrons are donated in pairs, you either have missed some context, or the book is grossly oversimplifying, or even completely wrong. $\ce{NO2}$ is an odd electron species, or a radical. It is also a very complicated bonding situation. The way you are trying to rationalise is insufficient for it. If there were a double bond between oxygen and nitrogen, then one of the elements would exceed the 8 valence electrons, which is not possible for these elements. Please edit your question with all the conceptual issues you have to make it clearer. $\endgroup$ Dec 11, 2018 at 15:13


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