Question

How many bonds in $XeO{F}_4$ possess an angle greater than ${90}^{\circ }$ ?

Answer 1

$XeO{F}_4$
Number of bond pairs $=5$
Number of lone pair $=1$
Hybridization $=5+1=6$
$\Rightarrow s{p}^3{d}^2$
Octahedral geometry and shape of $XeO{F}_4$ is given below.
Because of lone pair repulsions, all $F-Xe=O$ bond angles becomes less than ${90}^0$ and two $F-Xe-F$ bond angles make ${120}^0$ and the reamaining two $F-Xe-F$ bond anlgles make ${60}^0$