QJEEM21Ch08e7: JEE Main 2021 Chemistry - Class 11 JEEM21Ch08e7 Solution: The hybridisations of the atomic orbitals of nitrogen in NO 2 − …

Hybridisation = Number of sigma bonds + number of lone pair of electrons + number of coordinate bond.

For $$N{O}_2^{-}$$ hybridisation = 2 sigma NO bonds + 1 lone pair on N = 3; so $$N{O}_2^{-}$$ is sp²-hybridised.


For $$N{O}_2^{+}$$ hybridisation = 2 sigma NO bonds = 2; so $$N{O}_2^{+}$$ sp-hybridised.


For $$N{O}_4^{+}$$ Hybridisation = 4 sigma NH bonds = 4; so $$N{O}_4^{+}$$ is sp³-hybridised.


Therefore, hybridisation of N in $$N{O}_2^{-}$$, $$N{O}_2^{+}$$, $$N{O}_4^{+}$$ are sp², sp, sp³ respectively.