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Exercise · Q16

Q.Describe the structure of the HNO3\text{HNO}_3 molecule and explain, using resonance, why the two N-O bonds (other than N-OH) are found to be of equal, intermediate length.

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In HNO3\text{HNO}_3, nitrogen is sp2sp^2 hybridised, giving the molecule a planar, roughly trigonal shape. Nitrogen forms one ordinary single bond to the hydroxyl oxygen (N-OH\text{N-OH}) and is bonded to the remaining two oxygens through a system that can be drawn as two different canonical (Lewis) structures: in one, nitrogen forms a double bond (N=O\text{N=O}) to one of these oxygens and a single bond (carrying a formal negative charge) to the other; in the second canonical structure, these two assignments are exactly reversed. Because both structures are equally valid contributors and neither alone correctly describes the real molecule, the true bonding is a resonance hybrid of the two -- the extra π\pi-electron density is effectively shared/delocalised equally over both of these N-O linkages rather than being fixed as one double and one single bond. This is confirmed experimentally: both non-hydroxyl N-O bonds are measured to be of identical length (about 122 pm), intermediate between a typical N-O single bond (~136 pm) and N=O double bond (~115 pm) -- direct physical evidence for delocalisation rather than one fixed structure. [!ANSWER] HNO3 is planar (sp2 nitrogen); its two non-hydroxyl N-O bonds are equal in length because the molecule is a resonance hybrid of two canonical structures with the formal double bond alternating between the two oxygens, not one fixed N=O/N-O(-) arrangement.

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