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Hi so I was studying chemical bonding where i encountered a problem which is stated below.

When we talk about ${NO_3}^-$ we draw its structure as following enter link description here But the thing which I dont understand is that in two of the O atoms there are three lone pairs(ie: $6 e^-$) but as we know from O's $e^-$ config. (which is $1s^2 2s^2 2p^4$) there are two unpaired $e^-$ which will take part in bonding as there is no vacant orbitals for $e^- $ to get excited to . Now if in this O atom only one sigma bond is present then technically we are left with $5e^-$ and not $6e^-$ which is shown in the figure.

Similarly in the N atom (whose $e^-$ is $1s^2 2s^2 2p^3$ ) does not have any other vacant orbital to excite its $e^-$ thus showing that it can make only 3 bonds then why is it making 4 bonds( 1 pi and 3 sigma) ?

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marked as duplicate by Mithoron, Mathew Mahindaratne, Buttonwood, M.A.R., M. Farooq Jul 28 at 1:32

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The problem lies in the electron count. Each bond contains 2 electrons and the pair is counted twice, once for being around each atom in the bond. And don't forget the extra electron to make the ion.

The + and - are from the point of view of the atom nucleus: e.g., consider the nitrogen to share 2 electrons with the oxygen on the right, to give 2 to the oxygen above, and to give 1 to the oxygen on the left. The other electron in that bond is one that makes the anion negative overall.

Then nitrogen gains its octet of 8 electrons, but formally only shares 50% of those electrons, so it started out with 5 but only controls 4 - it now has a positive charge. The right oxygen shares, so no charge. The top oxygen got 2 from nitrogen, so is negative, and the left oxygen got 1 from nitrogen and one from the metal of which this is the anionic part. And then, draw up the other two diagrams that have the "neutral oxygen" in the other positions, add them all and average them. Then each oxygen winds up with a 2/3 negative charge (but the nitrogen is still +1), while the anion is -1 and planar.

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I think you should not consider the orbital theory in every detail, it just makes things overcomplicated. You have nitrate which is very stable and there is a reason for it.

In a non oxidizated state you have 6 valence electrons (outer s and p orbitals) for oxygen. Oxygen has a higher electronegativity than nitrogen and therefore it takes one electron and gets reduced to -1 to have a full filled p orbital (8 valence electrons). The other oxygen takes 2 electrons and the oxidation state does not need to be changed. The oxidized nitrogen (+1) is more stable because of the half filled p orbital (Hund rule). Pi bonds are only made between p-p orbitals and therefore can only exist in the double bond because the s-p bond is prefered (1 overlap) and more stable.

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