• # question_answer The hybridization of atomic orbitals of nitrogen in $NO_{2}^{+},\,NO_{2}^{-}$ and $NH_{4}^{+}$are A)  $s{{p}^{2}},s{{p}^{3}}$ and $s{{p}^{2}}$ respectivelyB)  $sp,s{{p}^{2}}$ and $s{{p}^{3}}$ respectivelyC)  $s{{p}^{2}},sp$ and $s{{p}^{3}}$ respectivelyD)  $s{{p}^{2}},s{{p}^{3}}$ and $sp$ respectively

$NO_{2}^{+}=\frac{1}{2}[5+0+0-1]=2sp;$ $NO_{2}^{-}=\frac{1}{2}[5+0+1-0]=3s{{p}^{2}};$ $NO_{4}^{+}=\frac{1}{2}[5+4+0-1]=4s{{p}^{3}};$