Question

The hybridization of central atom in $N_3^{-}$ and $IC{l}_4^{-}$ are respectively,

• A. $s{p}^3{d}^2,sp$
• B. $s{p}_3,s{p}^3{d}^2$
• C. $sp,s{p}^3{d}^2$
• D. $s{p}^3,s{p}^3{d}^2$

Answer 1

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


$IC{l}_4^{-}$ has four bond pairs and two lone pairs. Hence the hybridization is $s{p}^3{d}^2$ with square planar shape.
$N_3^{-}$ has two bond pairs and hence it is $sp$ hybridized.