Chemistry - CHEMICAL BONDING Question with Solution | TestHub

As per VSEPR Theory

${\text{XeF}}_4\Rightarrow \text{H}\Rightarrow \frac{1}{2}\left(8+4\right)={\text{sp}}^3{d}^2\left(6\right)$

Geometry square planar with 2 L.P.
$\overrightarrow{\mu }=0$(symmetry)


${\text{CF}}_4\Rightarrow \text{H}\Rightarrow \frac{1}{2}\left(4+4\right)={\text{sp}}^3\left(4\right)$

Tetrahedral geometry
$\overrightarrow{\mu }=0$(symmetry)


${\text{PF}}_5\Rightarrow \text{H}=\frac{1}{2}\left(5+5\right)={\text{sp}}^3\text{d}\left(5\right)$

Trigonal bipyramidal
$\overrightarrow{\mu }=0$(Symmetry)


${\text{SF}}_4\Rightarrow \text{H}=\frac{1}{2}\left(6+4\right)={\text{sp}}^3\text{d}\left(5\right)$
With one L.P see saw geometry
$\overrightarrow{\mu }\ne 0$has a value



$\mu >0$due to two S - F bonds in equatorial plane.