Based on VSEPR theory, the number of 90 degree F-Br-F angles in $B r F_{5}$ is

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Solution

The correct option is A 0
BrF₅ has square bipyramidal geometry with one lone pair of electron. Thereby forms a square pyramidal structure. Due to lone pair-bond pair repulsion the bond angle of axial lone pair and equatorial F gets distorted and results in less than 90° bond angle.
Therefore none of the F - Br - F bond angle is of 90°.

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