Question

The hybridisation of atomic orbitals of $N$ in $N{O}_2^{+},N{O}_3^{-}$ and $N{H}_4^{+}$ are respectively:

• A. $sp,s{p}^2,s{p}^3$
• B. $sp,s{p}^3,s{p}^2$
• C. $s{p}^2,sp,s{p}^3$
• D. $s{p}^2,s{p}^3,sp$

Answer 1

The correct option is A $sp,s{p}^2,s{p}^3$

In $N{O}_2^{+}$ it has only 2 bond pair and no lone pair so it is $sp$ hybridized.
In $N{O}_3^{-}$ it has only 3 bond pair and no lone pair so it is $s{p}^2$ hybridized.
In $N{H}_4^{+}$ it has only 4 bond pair and no lone pair so it is $s{p}^3$ hybridized.