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(This question was on my last gen. chemistry exam.)

I counted 12 sigma bonds, but the answer was 13. I'm taking this course online so there is no explanation given. Could anyone offer a quick explanation on how this structure could give 13 sigma bonds?

enter image description here

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closed as off-topic by airhuff, Todd Minehardt, M.A.R., NotEvans., ron Dec 3 '17 at 14:58

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Welcome to chemistry.SE! If you have any questions about the policies of our community, please ‎visit the help center. $\endgroup$ – airhuff Dec 2 '17 at 19:38
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    $\begingroup$ You might also want to give the site's homework policy a read. In this particular post, simply counting the number of bonds would suffice as "effort" IMHO (I don't see what more could be done). Do bear in mind though, that homework-type questions (posted without any effort taken towards solving it) tend to get closed. Welcome to Chemistry.SE, cheers! :-) $\endgroup$ – paracetamol Dec 2 '17 at 20:21
  • $\begingroup$ The simple answer is that all bonds (in this molecule) contain $\sigma$ orbitals but a few also contain $\pi$ orbitals! $\endgroup$ – Gert Dec 2 '17 at 20:35
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There're are 13 sigma bonds there (count the red lines),

enter image description here


You've clearly tried to attempt this yourself, so good on you (and by virtue of that, it's equally clear that you counted the bonds too hastily).

I counted (quickly, because it looked simple enough) and got 12 too at first. Counted a little slower the second time (I was actually going to attempt to prove to you that the number of bonds is 12), turns out the ring is not entirely conjugated, so one carbon atom in it actually has two hydrogen atoms bonded to it. ;-)

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