I have to find all of the possible canonical structures of this molecule
And here's what I did
Are they correct and if yes, did I miss any possibility?
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$\begingroup$ These are correct, and I see one more fully octet structure that is closely related to a non-octet structure. $\endgroup$ – jerepierre Mar 23 '16 at 19:48
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$\begingroup$ Would it be this one that I've added on the top row? $\endgroup$ – p-bromonitrobenzene Mar 23 '16 at 19:53
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$\begingroup$ I added it in the question. $\endgroup$ – p-bromonitrobenzene Mar 23 '16 at 19:54
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4$\begingroup$ No, in that drawing oxygen has 10 electrons. I would only draw two, to be honest. You're already delocalising a lone-pair on oxygen in one direction. You could also delocalise it in the other direction. $\endgroup$ – Brian Mar 23 '16 at 19:56
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1$\begingroup$ If oxygen is neutral in the last structure you drew, it would still have 10 electrons. We only observe elements capable of an extended octet in the 3rd period. However, if it has 8 electrons, its charge would be 2+. Oxygen is very electronegative, so that doesn't sound so likely, right? I will post a more detailed answer, or at least resources, a bit later. Meanwhile, please edit your post, so that it doesn't look confusing to someone new coming in. As you did in your original post :-)! $\endgroup$ – Brian Mar 23 '16 at 20:58
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May it look like this? In those two possibles ways?
answered Mar 23 '16 at 23:06
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1$\begingroup$ Your top row is not possible. Look at the last three rows: there are octet violations $\endgroup$ – Ben Norris Mar 24 '16 at 1:23
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$\begingroup$ And I have done the same things in the bottom row $\endgroup$ – p-bromonitrobenzene Mar 24 '16 at 1:30
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$\begingroup$ Oh, you are correct! You have an octet rule violation in the bottom row also. $\endgroup$ – Ben Norris Mar 24 '16 at 1:32
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$\begingroup$ But how? Is it because of the triple bond of the oxygen? $\endgroup$ – p-bromonitrobenzene Mar 24 '16 at 1:33
