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Question:

The hybridization of orbitals of N atom in NO₃⁻, NO₂⁺ and NH₄⁺ are respectively:

sp2, sp, sp3

sp, sp2, sp3

sp, sp3, sp2

sp2, sp, sp

Solution:

Maximum covalency of nitrogen is 5.
In NO₃⁻, there are 3 bond pairs and hence, sp² hybridization.
In NO₂⁺, there are 2 bond pairs and hence, sp hybridization.
In NH₄⁺, there are 4 bond pairs and hence, sp³ hybridization.
Hence, option B is correct.