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I am a bit confused about the lewis structure of $\ce{NO}$ The answer from my worksheet is $\ce{N=O}$ with nitrogen having a free radical. But I drew the structure as N triple bonded to $\ce{O (:N#O:)}$ and oxygen has the free radical. I am confused as to why my answer is not the case

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marked as duplicate by Tyberius, Mithoron, A.K., a-cyclohexane-molecule, Todd Minehardt Jul 31 '18 at 23:34

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    $\begingroup$ You'd get more then 8 electrons on N. $\endgroup$ – Mithoron May 2 '18 at 0:09
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    $\begingroup$ It's always a good idea to check formal charges. In N=O, the formal charge is zero on each atom. In N triple bonded to O with a single electron on N, the formal charge is 1 and -1 on the N and O respectively. Thus, N double bonded to O is correct. $\endgroup$ – Dr. J. Jun 1 '18 at 10:34
  • $\begingroup$ related: Where does the 9th electron go in a N=O bond?, Is Bond Formation “Strictly” Exothermic? $\endgroup$ – mykhal Jul 31 '18 at 14:37
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As an extension of Mithoron's comment, if you drew the structure as N triple bonded to O, you would fulfil both the octets without placing the free radical electron.

Now, you'd have to extend the octet of either nitrogen or oxygen, but you have no free orbitals to place the extra electron.

Hence, the first one is the correct Lewis structure.

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