Question

Which of the following compound central atom has ${sp}^3{d}^3$ hybridised?

• A. $XeO{F}_5^{-}$
• B. $Xe{F}_3^{-}$
• C. $Xe{F}_4$
• D. $I{Cl}_4^{-}$

Answer 1

The correct option is A $XeO{F}_5^{-}$

Correct option-A.

Reason:-

A) $XeO{F}_5^{-}=\frac{8+6+35+1}{8}=\frac{50}{8}=6$

$\frac{2}{2}=1$ lone pair

$6+1=7=s{p}^3{d}^3$ hybridization.

B) $Xe{F}_3^{-}=\frac{8+21+1}{8}=\frac{30}{8}=3$

$\frac{6}{2}=3$ lone pair

$3+3=6=s{p}^3{d}^2$ hybridization.

C) $Xe{F}_4=\frac{8+28}{8}=\frac{36}{8}=4$

$\frac{4}{2}=2$ lone pair

$4+2=6=s{p}^3{d}^2$ hybridization.

D) $IC{l}_4^{-}=\frac{7+28+1}{8}=\frac{36}{8}=4$

$\frac{4}{2}=2$ lone pair

$4+2=6=s{p}^3{d}^2$ hybridization.

Option A is correct.