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In C, the answer states that electrons must come in pairs. And so C does not show resonance forms.

But here I read that if there is an unpaired electron, it is a free radical. Why can't free radicals be part of the resonance form?


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    $\begingroup$ (Di)radicals are legitimate resonance forms. $\endgroup$ – orthocresol May 12 '17 at 22:25
  • $\begingroup$ Thank you orthocresol. Just checking, are single radicals also resonance forms? $\endgroup$ – K-Feldspar May 12 '17 at 22:27
  • $\begingroup$ Yes, but you can't make a resonance form with only one unpaired electron from a resonance form with none - so in this case the usual depiction of O2 has no unpaired electrons, and any resonance form you draw can only have 0, 2, 4, ... unpaired electrons. If your original structure has one unpaired electron, then yes, certainly it can have other resonance forms that also have one unpaired electron. $\endgroup$ – orthocresol May 12 '17 at 22:29
  • $\begingroup$ Thanks @orthocresol, that makes sense. Mithoron, You are right, I do no quite understand what you mean by "something usually is or isn't a radical". Do you mean like a single hydrogen atom is a free radical but usually isn't a radical as it exists as H2? If not, do you mind giving me some search terms I can use to read more about what you are talking about (I don't really know what you mean so not sure what to google). Thank you. $\endgroup$ – K-Feldspar May 12 '17 at 22:37
  • $\begingroup$ In the case of dioxygen, I would not consider a diradical representation to be a resonance structure. The diradical representation is only attempting to account for the short comings of Lewis structures. $\endgroup$ – Zhe May 13 '17 at 0:41

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