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Can the structure of an azide ion ($\ce{N3-}$) be $\ce{N#N\bond{->}N^-}$?

I know the actual structure but was asking whether the single bond can be shifted to a N to make it a triple bond and a coordinate bond with other Nitrogen.

This is not a duplicate of Hybridization in azide ion? because I am asking whether it can have a coordinate bond or not while the other question asks what the hybridization of centre nitrogen in $\ce{N3-}$ is.

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    $\begingroup$ I have improved the formatting of your question using $\LaTeX$. For more information on how to do this yourself please see here and here. $\endgroup$
    – bon
    Jun 19, 2015 at 13:56
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    $\begingroup$ possible duplicate of Hybridization in azide ion? $\endgroup$
    – Mithoron
    Jun 19, 2015 at 14:01
  • $\begingroup$ Jan is right - it's usually written as single bond with formal charges, but you can write it as an arrow - it's the same. $\endgroup$
    – Mithoron
    Jun 19, 2015 at 16:27
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    $\begingroup$ related chemistry.stackexchange.com/questions/32528/… $\endgroup$
    – Mithoron
    Jun 19, 2015 at 16:34
  • $\begingroup$ Of course it can be a duplicate, as ron answered your question - wrote all mesomeric structures for azide, only used straight line and formal charges instead of arrow. $\endgroup$
    – Mithoron
    Jun 19, 2015 at 16:52

2 Answers 2

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You can’t really tell a dative bond from a ‘normal’ covalent bond until you break the bond. A dative bond will then dissociate heterolyticly while a covalent bond will dissociate homolyticly.

Just using that simplified way of putting it, you could say that there are many reactions of azides wherein $\ce{N2}$ is liberated leaving a nitrene or related structure; just one of many examples would be the Curtius rearrangement. However, that is not what is meant in the definition above; the definition talks about supplying the bond dissociation energy.

The BDE is not typically supplied in chemical reactions except in coordination compound ligand exchange and radical generation. If you were to apply it to $\ce{N3-}$, I predict homolytic dissociation. Thus, the bond between the nitrogens is more of a typical single bond.

Of course, aside from the chemical implications of dative versus covalent bond you may just have decided to use an arrow notation rather than formal charges. Technically nobody can stop you but still you shouldn’t call it a true dative bond.

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No.

The issue with using your proposal is very simple: $\ce{N_2}$ sucks at being reactive. It doesn't want to give up or share its electrons with any other atom. In fact, the azide ion is usually explosive because it keeps trying to form nitrogen gas. $\ce{N_2}$ has no real polarity and no particular motive to coordinate to anything, a single N included.

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    $\begingroup$ Sorry. you're wrong - structures with additional formal charges are more important in azide than those without them. $\endgroup$
    – Mithoron
    Jun 19, 2015 at 16:35
  • $\begingroup$ Formal charge of the azide has nothing to do with the stability of gaseous nitrogen. Azide is not formed directly from diatomic nitrogen and azide ions are terribly explosive. Sodium azide is used in airbags because it explodes so well, mercury azides detonate if you do nothing with them, etc. Nitrogen gas is extremely stable and will not try to coordinate to anything. Azide can adopt formal charges, yes, but that's distinct from a nitrogen coordinating to a nitride. $\endgroup$ Jun 19, 2015 at 16:42
  • $\begingroup$ Thanks that was pretty much good answer but is ther any other way to think about its formation rather than just thinking it will be unstable. $\endgroup$
    – Satyajeet
    Jun 19, 2015 at 16:46
  • $\begingroup$ see annswers to questions I linked, also one pair of N2 electrons can certainly be donated. $\endgroup$
    – Mithoron
    Jun 19, 2015 at 16:47
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    $\begingroup$ That definitive no is just very wrong. If one accepts donor-acceptor-bonds to be part of the Lewis formalism, then the indicated structure is indeed a resonance contributor. There is no such thing as a most stable resonance structure, as they are only present when coexistent. It is possible to have a highest contributor, but it is not the same thing. I am sorry, but since this is wrong (and accepted) I had to down vote it. $\endgroup$ Jun 20, 2015 at 6:26

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