-1
$\begingroup$

enter image description here

These are the two resonating strctures of NO given in my textbook. 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 NO where its not possible to avoid an unpaired electron, cannot oxygen donate only one electron to nitrogen. If it does that, we could have this structure. Is this structure possible?

enter image description here

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.

| improve this question | |
$\endgroup$
1
$\begingroup$

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).

| improve this answer | |
$\endgroup$
  • $\begingroup$ On the assumption of last conclusion (nitrogen dioxide as a molecular diradical), we can redraw our structure like that: $\endgroup$ – impulsgraw Dec 9 '16 at 19:07
  • $\begingroup$ link $\endgroup$ – impulsgraw Dec 9 '16 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 Dec 10 '16 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 Dec 10 '16 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 Dec 10 '16 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.