Question

The fluoride for which the dipole moment is not equal to zero, is :

• A. $Xe{F}_4$
• B. $C{F}_4$
• C. $S{F}_4$
• D. $P{F}_5$

Answer 1

Answer: (C)

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

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

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

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

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

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

${\left(\text{SF}\right)}_4\Rightarrow \text{H}=\frac{1}{2}\left(6+4\right)={\left(\text{sp}\right)}^3\text{d}\left(5\right)$
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.