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Q.
Based on VSEPR theory, the number of $90^{\circ} \,F - Br - F$ bond angles in $BrF_3$ is
Solution:
In $BrF_{3}$, hybridisation is $sp^{3}d$ [two lone pair and $3\sigma$ bond]
$\Rightarrow $ Due to lone pair-bond pair repulsion, no $F-Br-F$ bond angle in it is equal to $90^{\circ}$
