Question

Which of the following molecule has $s{p}^3$ d-hybridization?

• A. $SO{F}_4$
• B. $S{F}_4$
• C. $Xe{F}_3^{+}$
• D. All

Answer 1

The correct option is D All

$SO{F}_4$ contains $5$ bonding domains and zero lone pairs. hence, it has $s{p}^{3}d$ hybridization with triagonal bipyramidal geometry.
$S{F}_4$ is see saw shaped due to $4$ bond pairs, $1$ lone pair and $s{p}^{3}d$ hybridization.
$Xe{F}_3^{+}$ has $2$ lone pairs and $3$ bond pairs. Hence, it undergoes $s{p}^{3}d$ hybridization and has T shaped geometry.