Question

Deduce the structure of $Br{F}_5$ on the basis of VSEPR theory.

Answer 1

Valence electrons in $Br{F}_5$ = $7+5\times 7=42$. The distribution of these electrons in $Br{F}_5$ is given as follows in image. There are six lone pairs of electrons around Br. Hence, the arrangement of these electrons around Br will be octahedral. The hybridization involved is $s{p}^3{d}^2$.