The hybridization states of the atomic orbitals of the central Nitrogen atom in the , , and ions are — Chemical Bonding Chemistry Question

💡 Solution & Explanation

For $NO_3^-$: $S.N. = 3$ $\sigma$ bonds + $0$ lone pairs = $3$ ($sp^2$). For $NO_2^+$: $S.N. = 2$ $\sigma$ bonds + $0$ lone pairs = $2$ ($sp$). For $NH_4^+$: $S.N. = 4$ $\sigma$ bonds + $0$ lone pairs = $4$ ($sp^3$).