Question

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

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

Answer 1

The correct option is B $sp,s{p}^2$ and $s{p}^3$ respectively

$N{O}_2^{+}$
$sp$ hybridisation and linear shape
Bond angle ${180}^0.$
$N{O}_3^{-}$
$s{p}^2$ hybridisation and triangular planar shape
Bond angle ${120}^0$
$N{H}_4^{+}$
$s{p}^3$ hybridisation and tetrahedral shape
Bond angle ${109}^0{28}^{\prime }$