Is the azide ion ($\ce{N3-}$) paramagnetic?

How do I use molecular orbital theory to show that it is or isn't?

I drew a sine wave diagram and filled up the $1\pi$ and $2\pi$ MOs, and since there are no unpaired electrons, can I conclude that the ion is not paramagnetic?

  • $\begingroup$ Well, it isn't paramagnetic and even drawing few mesomeric structures looks enough for me $\endgroup$
    – Mithoron
    Commented May 22, 2015 at 22:37
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    $\begingroup$ @Mithoron Drawing mesomeric structures would not help, since Lewis structures do not contain information about spin. That's why we conveniently write $\ce{O=O}$ even though we know it is a paramagnetic triplett. $\endgroup$ Commented May 23, 2015 at 3:02
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    $\begingroup$ Since $\ce{N3^-}$ is isoelectronic to $\ce{CO2}$ you might find some answers in this related question. I am not absolutely certain if we have the same idea about the sine wave diagram (could you post a picture), but I assume you are on the right track. Your conclusion is sound. $\endgroup$ Commented May 23, 2015 at 3:09

1 Answer 1


Deriving a proper MO digram for N$_3^-$ is not completely trivial a task (e.g. a random google hit, which seems to me well done: www.chem.uci.edu/~lawm/Azide%20MO%20Diagram.pdf). In the end you will have to show that the degeneracies of the orbitals together with the electron count will not require open shells. Beware of Lewis-structures and mesomery, think of O$_2$, it looks perfectly o.k. to draw O=O, only the MOs and their degeneracies will reveal that its paramagnetic.


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