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There's a problem from an old exam that I was trying to do. The original structure is given on the left and they want all the possible resonance forms that have closed-shell atoms (octet for S) and formal charges of -1, 0, or +1 on all atoms. There's then a bold note that says, "As always, you should remember that linear atoms (sp hybridization) are not expected in rings of this size." So, my professor said that the circled electron pair in red is localized to the nitrogen atom, which therefore means that the resonance contributor on the right is invalid.

So, just to double check my logic, I'm assuming that the circled pair on right is localized because by moving them as if they were a delocalized pair, it would change the hybridization of nitrogen from sp2 to sp, which would make it's electronic geometry linear. Therefore, the pair on the nitrogen is localized because of the bold note (As always, you should remember that linear atoms (sp hybridization) are not expected in rings of this size) correct?

Resonance structures

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marked as duplicate by Mithoron, M.A.R., Todd Minehardt, pentavalentcarbon, Tyberius Feb 19 '18 at 4:49

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    $\begingroup$ More to the point: the lone pair on N is perpendicular to the pi bond and cannot overlap with it - hence no delocalisation. $\endgroup$ – orthocresol Jan 15 '17 at 19:21
  • $\begingroup$ I’m sure a permutation of this has been asked here before … The gist is that nitrogen’s lone pair is orthogonal and cannot participate in resonance. It’s not above or below the ring, it’s pointing away from it. $\endgroup$ – Jan Jan 15 '17 at 22:16