Question

The hybrid orbitals used by bromine atom in $Br{F}_3$ are

• A. $s{p}^2$
• B. $s{p}^3$
• C. $s{p}^{3}d$
• D. $s{p}^3{d}^3$

Answer 1

Answer: (C)

$Br{F}_3$.
Total $e^{-}=7+7\times 3=28$
$=8\times 3+4$
no. of hybridised orbitals $=\frac{8\times 3}{8}+\frac{4}{2}$
$=3+2=5$
$\therefore s{p}^{3}d$ hybridization