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I have tried to apply the rules and basics that I learnt so far. But I am confused about the hybridization of oxygen atoms which are making the single bond in nitrate ion.

Following are the steps I used. The average bond order of 4/3 makes the molecule unstable. enter image description here

Based on the steps, the single bonds are made of $\ce{sp^3}$ of oxygen and $\ce{sp^2}$ of nitrogen, the bonds are sigma bonds. However that doesn't seem like the case. If I look at $\ce{HNO3}$ dissociation equation in the water, lone electron pair of nitrogen is entirely shared with (co-ordinative covalent bond) one of the oxygens making a single bond.

$\ce{ HNO3(aq) -> H+(aq) + NO3^{-}(aq) }$

But considering $\ce{NO3}$ ion individually, without thinking about nitric acid, I can't find the long pair to be in the same type of co-ordinative bond. Isn't there an unbonding sp2 electron with nitrogen (probably one from the lone pair)?

The other issue, why isn't oxygen hybridizing to be $\ce{sp^3}$ here? What am I doing wrong here?

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  • $\begingroup$ You indeed can think that these two $\ce{O-}$ are sp3-hybdiridzed. But I would rather explicitly call them $\ce{O-}$ rather then just $\ce{O}$. $\endgroup$ – Wildcat Jul 26 '15 at 13:09
  • $\begingroup$ nintrate-ion is isoelectronic to carbonate ion. The bonding is very similar except that in nitrate due to positive charge on nitrogen, the bonds are stronger. $\endgroup$ – permeakra Jul 16 '16 at 10:58
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The actual structure of the nitrate ion is symmetrical. It forms a trigonal planar molecule.

This means that the bonds between oxygen and nitrogen are all the same. The resonance structures are a simple way to explain that structure using two-electron bonds. In reality the "extra" electron is delocalised across all three bonds (this will involve the nitrogen lone pair and nitrogen will be sp2 with no free lone pair).

The average bond order will be, as you calculate 4/3 but this doesn't, as you claim, make the molecule unstable.

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