Question

Number of ${90}^{\circ }$ $F-Br-F$ bond angles in $Br{F}_5$ is:

• A. \N
• B. 8
• C. 6
• D. 4

Answer 1

The correct option is A \N

Geometry of $Br{F}_5$:
Steric number of $Br{F}_5$ = $\frac{1}{2}(7+5)=6$
So, the number of bond pairs are 5 and the number of lone pairs is 1. The hybridisation is $s{p}^3{d}^2$. It assumes a square pyramidal structure. Hence, due to the distortion caused by the lone pair, none of the bonds will be ${90}^{\circ }$ to each other.