Take an example, $\ce{HNO3}$

The total valence of $\ce{HNO3}$ must be 24. By using formula $\pi =6\mathrm{n}+2-\mathrm{V}$ where $\pi$ is number of $\pi$ electrons, $\mathrm{n}$ is number of atoms excluding all hydrogen atom, and $\mathrm{V}$ is total valence, the $\pi$ is $2$. It means that there would be one double bond in the structure of $\ce{HNO3}$ and the remaining bonds other than $\pi$ bond are $\sigma$ bond.

I understand the double bond has 1 $\pi$ electron and 1 $\sigma$ electron. It means 10 electrons are used for the bonds and 14 electrons are remaining.

I know it would be impossible to violate octet rule because the calculation of $\pi$ already stated.

I have trouble how to spread the 7 lone pairs. Why does the $\ce{O}$ in the single bond $\ce{N-O}$ have 3 lone pairs, while the $\ce{O}$ in the double bond has 2 lone pairs? I don't expect the answer like "Just spread the lone pairs to somewhere".
I'm worried about the odd number of unshared electrons.

I know the $-$ means rich of electron and $+$ means lack of electron. To check the formal charge in each atom is easy, but how to calculate the formal charge, if I don't know where to put the $-$ and $+$.

I think this question is not a duplicate of The Lewis structure of HNO3. I already saw that question , but I'd like to ask here again, because I need the answer for my misunderstanding, even though the answer would be general structure of $\ce{HNO3}$.

To be precise, I just don't know where I need to put unshared electrons. It's like I'm missing the bridge. I know resonance, as arrow-pushing, and mechanisms are required in organic chemistry.


1 Answer 1


You have 24 valence electrons. Counting the electrons: 6 of these will go into the $σ$ framework and there are 14 electrons in non-bonding, either $sp^3$ or $sp^2$ hybrids, on the oxygens. This leaves 4 electrons for the $π$ system.

We can think of the Nitrogen and Oxygen in the double bond being as both being $sp^2$ hybridised, with vacant $p$ orbitals orthogonal to the plane of the molecule. These 4 remaining electrons will then go into the $π$ system formed by the overlap of these 2 orbitals, which overlap to form 2 molecular orbitals, the $1π$ and the $2π$ MOs.

We expect the $1π$ to have greater contribution from the Nitrogen due to the lower orbital energy of the $p$ in N compared to O, and the $2π$ to have greater contribution from the O instead of the N.

A full MO calculation of the molecule would also take into account the non bonding $sp^3$ electrons on the oxygen, of which one lobe will be able to interact slightly with the $ p$ orbitals, giving a delocalised system over the Nitrogen double and single bond, exactly as we see drawing resonance structure for the molecule.

  • $\begingroup$ This explanation with MO makes sense. Thank you. $\endgroup$
    – user28510
    Apr 2, 2016 at 23:20

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