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What does the question mean by how many electron pairs are in 2p orbitals? How does one tell? Given that both the oxygen and the nitrogen are $\ce{sp^2}$ hybridized, wouldn't there be two 2p orbitals?

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In the resonance structure drawn at the top of your image, the $\ce{p}$ orbital on carbon and the $\ce{p}$ orbital on oxygen share a bonding pair of electrons, a lone pair of electrons reside in a nitrogen $\ce{p}$ orbital, two lone pairs reside in two oxygen $\ce{sp^2}$ orbitals. So for this case, the most commonly drawn representation, there is one lone pair in a $\ce{p}$ orbital. In your middle resonance structure, there is a lone pair in an oxygen $\ce{p}$ orbital, and another lone pair in the nitrogen $\ce{p}$ orbital; hence, 2 lone pairs in two different $\ce{p}$ orbitals. In your bottom resonance structure, there is a lone pair in an oxygen $\ce{p}$ orbital and a shared bonding electron pair in the $\ce{C-N}$ $\ce{pi}$ bond; therefor, one lone pair in a $\ce{p}$ orbital for this case.

This seems to be a poorly worded question. Since the real molecule is represented by a combination of the 3 resonance structures, the correct answer would be "somewhere between one and two lone pairs in $\ce{p}$ orbitals." Adding the detail contained in the above paragraph would probably help the grader understand your explanation.

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  • $\begingroup$ Thank you. How do pi electrons affect molecular geometry? $\endgroup$ – Dissenter Jun 9 '14 at 22:47
  • $\begingroup$ I'm not quite following, could you be more specific? $\endgroup$ – ron Jun 9 '14 at 23:06
  • $\begingroup$ Usually when assigning geometries we only consider sigma bonds; boron trifluoride is sp2 just as a carboxylic acid group's carbon is sp2. Do pi electrons affect geometry; if so, how? $\endgroup$ – Dissenter Jun 9 '14 at 23:19
  • $\begingroup$ I still might be missing your point. I'd say pi electrons affect geometry very much. The bond length in ethylene is shorter than in ethane. The single bond in 1,3-butadiene is shorter than in ethane. All of the C-C bonds in benzene have the same length, no alternation. That said, this could probably be a separate question. $\endgroup$ – ron Jun 9 '14 at 23:26

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