Based on VSEPR theory, the number of $90°$ $\mathrm{F}-\mathrm{Br}-\mathrm{F}$ angles in ${\mathrm{BrF}}_5$ is______________

According to VESPER theory valence ${\mathrm{e}}^{-}$ & surrounding atom  repel each other, and due to this adopt structure that minimizes the repulsion. All four planar bonds ( $\mathrm{F}-\mathrm{Br}-\mathrm{F}$ ) will reduce from $90°$ to $84.8°$ after lone pair - bond pair repulsion.

Hence, there are no 90$°$ $\mathrm{F}-\mathrm{Br}-\mathrm{F}$ angles in ${\mathrm{BrF}}_5$.