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

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

NTA AbhyasNTA Abhyas 2022

Solution:

As per VSEPR Theory
$\left(\text{XeF}\right)_{4} \Rightarrow \text{H} \Rightarrow \frac{1}{2} \left(8 + 4\right) = \left(\text{sp}\right)^{3} d^{2} \left(6\right)$
Solution
Geometry square planar with 2 L.P.
$\overset{ \rightarrow }{\mu } = 0$ (symmetry)

$\left(\text{CF}\right)_{4} \Rightarrow \text{H} \Rightarrow \frac{1}{2} \left(4 + 4\right) = \left(\text{sp}\right)^{3} \left(4\right)$
Solution
Tetrahedral geometry
$\overset{ \rightarrow }{\mu } = 0$ (symmetry)

$\left(\text{PF}\right)_{5} \Rightarrow \text{H} = \frac{1}{2} \left(5 + 5\right) = \left(\text{sp}\right)^{3} \text{d} \left(5\right)$
Solution
Trigonal bipyramidal
$\overset{ \rightarrow }{\mu } = 0$ (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
$\overset{ \rightarrow }{\mu } \neq 0$ has a value
Solution
$\mu > 0$ due to two S - F bonds in equatorial plane.