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I'm confused about $\mathrm{sp^2}$ hybridization in the formation of $\ce{NO3-}$ ion. Valence electrons of N are $\mathrm{(2s)^2}$ and $\mathrm{(2p)^3}$.

Is one of $\mathrm{2s}$ electrons kicked up to one of the $\mathrm{2p}$ orbitals, and then the remaining $\mathrm{2s}$ electron hybridized with the two remaining $\mathrm{2p}$ orbitals which have unpaired electrons? If so, that gives $\mathrm{sp^2}$ orbitals to bond with $\mathrm{p}$ orbitals of the oxygen atoms. However, there is now that remaining unhybridized $\mathrm{p}$ orbital of nitrogen, which has 2 electrons, so how can it be involved in $\pi$ bonding?

I would appreciate someone explaining this. I'm sure I'm missing something simple, but my book doesn't have an explanation of $\mathrm{sp^2}$ hybridization for Group V elements.

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  • $\begingroup$ So the lone pair of nitrogen provides both electrons for the delocalized pi bond, and the oxygen atoms do not contribute an electron to the delocalized pi bond? $\endgroup$ – Michael Jul 24 '15 at 14:01
  • $\begingroup$ I withdrew my earlier comment. Although I could simply share with you that number of electron regions around an atom is an indication to quickly figure out its hybridization, I am not able to explain this in detail, because I have a sub question when looking at yours: I am not sure how the sp3 hybridized oxygen that has formal charge -1 is making a bond with sp2 nitrogen which has +1 formal charge. $\endgroup$ – bonCodigo Jul 26 '15 at 8:24
  • $\begingroup$ @bonCodigo, oxygen is thought as being unhybridized in this case. Rather, you think of the whole ion as being composed of $\mathrm{sp2}$-hybridized $\ce{N+}$, two unhybridized $\ce{O-}$, and one unhybridized $\ce{O}$. $\endgroup$ – Wildcat Jul 26 '15 at 12:54
  • $\begingroup$ @Wildcat this has been eating me. Almost for all ions $\ce{CO3^{-2}}$ etc $\endgroup$ – bonCodigo Jul 26 '15 at 13:00
  • $\begingroup$ @bonCodigo, well in principle, you can invoke $\mathrm{sp^3}$ hybridization for $\ce{O-}$ ions as well as $\mathrm{sp^2}$ hybridization for $\ce{O}$ atom. But, it won't change anything, so you don't need it. $\endgroup$ – Wildcat Jul 26 '15 at 13:06
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However, there is now that remaining unhybridized p orbital of nitrogen, which has 2 electrons, so how can it be involved in π bonding?

Everything is more-or-less "right" up to these point. The thing you're missing is that there is only one electron in the unhybridized $\mathrm{p}$ orbital of $\ce{N}$ atom since to get the following usual configuration of $\ce{NO3-}$ ion (image courtesy of Wikipedia)

enter image description here

with two negatively charged $\ce{O-}$ ions one electron has to be moved from $\ce{N}$ atom to $\ce{O}$ atom first. Now note that since all $\ce{O}$ atom are equivalent to each other we need to invoke the concept of the resonance, but for simplicity think first about any single of the resonance structure above.

So, formally we start by having:

  • one $\ce{N}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^3}$ electron configuration;
  • two $\ce{O}$ atoms with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
  • and one $\ce{O-}$ ion with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

In the first step we transfer one of the $p$-electrons from $\ce{N}$ atom to an $\ce{O}$ atom, so we end up with:

  • one $\ce{N+}$ ion with $\mathrm{(1s)^2 (2s)^2 (2p)^2}$ electron configuration;
  • one $\ce{O}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
  • and two $\ce{O-}$ ions with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

At the next step we invoke the $\mathrm{sp^2}$ hybridization of $\ce{N+}$ ion, so that we get:

  • one $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion with $\mathrm{(1s)^2 (sp^2)^3 (2p)^1}$ electron configuration;
  • one $\ce{O}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
  • and two $\ce{O-}$ ions with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

Finally, we form bonds as follows:

  • One $\sigma$ bond between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion and an $\ce{O-}$ ion in forming which we use 1 of the total 3 $\mathrm{sp^2}$-electrons of $\ce{N+}$ ion and the only unpaired $\mathrm{2p}$-electron of $\ce{O-}$ ion;
  • Another identical $\sigma$ bond between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ and another $\ce{O-}$ ion; here we use 1 of the remaining 2 $\mathrm{sp^2}$-electrons of $\ce{N+}$ ion and the only unpaired $\mathrm{2p}$-electron of $\ce{O-}$ ion;
  • Yet another $\sigma$ bond, but this time between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion and the $\ce{O}$ atom in forming which we use the last remaining $\mathrm{sp^2}$-electron of $\ce{N+}$ ion and 1 out of total 2 unpaired $\mathrm{2p}$-electrons of $\ce{O-}$ ion;
  • What remains now is one unhybridized $\mathrm{2p}$-electrons of $\ce{N+}$ ion and one unpaired $\mathrm{2p}$-electrons of $\ce{O-}$ ion which are used to form the $\pi$ bond.

Pictorially (image courtesy of Professor Peter Bird):

enter image description here

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  • $\begingroup$ Thank you, thank, thank you! I was so frustrated trying to figure this out, and couldn't find an explanation on the web or in other textbooks. I wish your clear, logical explanation was in texts; they seem to have only the simplest hybridization examples. I know this took some time to write your response; I, $\endgroup$ – Michael Jul 27 '15 at 12:50
  • $\begingroup$ (to continue) I'm very appreciative of your kindness. Am sure others looking for an explanation of NO3- on the web will benefit as well. $\endgroup$ – Michael Jul 27 '15 at 12:52

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