These are the two resonating structures of $\ce{NO}$ given in my textbook:

$$\ce{:\!N=\overset{\Large .\!\!.}{\underset{\Large .}{O}}\!: <-> :\!\overset{\Large .}{N}=\overset{\Large .\!\!.}{O}\!:}$$

In the first resonating structure, oxygen is shown to have 9 electrons. How can that be possible since oxygen does not have the third shell so it has no orbital available to accommodate any more electron?

Also, why don't elements donate only one electron in such cases? I know unpaired electrons cause repulsions but in cases like $\ce{NO}$ where it's not possible to avoid an unpaired electron. Cannot oxygen donate only one electron to nitrogen? If it does that, we could have another structure. Is this structure possible?

$$\ce{:\!\overset{\Large .\!\!.}{N}=\overset{\Large .}{O}\!:}$$

Also, just out of curiosity, between oxygen and nitrogen, on whom will be one electron pair more stable? I am not able to decide this on the basis of electronegativity.


1 Answer 1


There is actually an erroneous explanation in your textbook. There is how the nitrogen monoxide molecular structure is explained in school: as there are no extended methods known by pupils or moreover the teacher, it's ok children to think that third unpaired electron on nitrogen atom is resonating between two atomic kernels as it's presented on picture upstaris. But it's wrong. Of course, that third electron cannot entirely belong to an oxygen atom (look to the right structure of first picure), because it (ox. atom) simply doesn't have enough free "quantum cells" to take it up to.

Things became much easier ever since people could describe the situation using molecular orbitals. Consider the molecular orbital diagram below: MO diagram for nitrogen monoxide

It's to say, that

  1. As the modern theory of molecular orbitals considers, there is some inter-atomic space between two atoms, where is the bonding electron allocation is the most likely situated in by the moment. There is exacly where our "third" elecron situated towards to that molecule. It doesn't ever belong to oxygen atom, but, technically, we can consider that structer as a resonant. It would me more correct, if to represent it's structure like that:

    enter image description here

  2. As that molecule has one antibonding electron π*2px it is a molecular diradical and it has an unstable structure. That also follows because of it's real physical-chemical properties (semi-stable only under low temperatures, and is very chemically active molecule).

  • $\begingroup$ On the assumption of last conclusion (nitrogen dioxide as a molecular diradical), we can redraw our structure like that: $\endgroup$
    – impulsgraw
    Commented Dec 9, 2016 at 19:07
  • $\begingroup$ link $\endgroup$
    – impulsgraw
    Commented Dec 9, 2016 at 19:09
  • $\begingroup$ So will this unpaired electron will be drawn more toward oxygen atom because of the higher electronegativity of oxygen!!?? $\endgroup$
    – Arishta
    Commented Dec 10, 2016 at 3:52
  • $\begingroup$ As I already said, no because it (oxygen atom) simply doesn't have enough free "quantum cells" to take it up to. $\endgroup$
    – impulsgraw
    Commented Dec 10, 2016 at 13:51
  • $\begingroup$ Since we have not been taught molecular orbital theory in detail, it might take some time for me to fully understand your answer. There is a question in my textbook : On which atom is the unpaired electron in more delocalized? The ànswer is oxygen, so how else will we justify that? $\endgroup$
    – Arishta
    Commented Dec 10, 2016 at 14:02

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