Question

What is the hybridization of bromine in each of the following: (a) BrF₂ ; (b) BrO₂ ; (c) BrF₅ ?

Short answer

The hybridization of bromine is given by

$\mathit{H}\mathbf{=}\frac{\mathbf{G}\mathbf{A}\mathbf{+}\mathbf{(}\mathbf{V}\mathbf{E}\mathbf{-}\mathbf{V}\mathbf{-}\mathbf{C}\mathbf{)}}{\mathbf{2}}$

Where H-hybridization of a central atom

GA- number of groups attached to a central atom

VE-valence electrons of a central atom

V-valency of a central atom

C-charge on molecule

Step-by-step solution

hybridization of bromine in BrF3

GA=3, VE=7, V=3, C=0

By substituting the value in the equation

$\begin{array}{rcl}\mathit{H}& \mathbf{=}& \frac{\mathbf{G}\mathbf{A}\mathbf{+}\mathbf{(}\mathbf{V}\mathbf{E}\mathbf{-}\mathbf{V}\mathbf{-}\mathbf{C}\mathbf{)}}{\mathbf{2}}\\ & \mathbf{=}& \frac{\mathbf{3}\mathbf{+}\mathbf{\left[}\mathbf{7}\mathbf{-}\mathbf{3}\mathbf{-}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{\right]}}{\mathbf{2}}\\ & \mathbf{=}& \mathbf{5}\mathbf{\left(}\mathit{s}{\mathit{p}}^{\mathbf{3}}\mathit{d}\mathbf{\right)}\end{array}$

H = Hence, the hybridization of chlorine in BeF₃is sp³d

hybridization of bromine in BrO2 

GA=2, VE=7, V=4, C=-1

By substituting the value in equation

$\begin{array}{rcl}\mathit{H}& \mathbf{=}& \frac{\mathbf{GA}\mathbf{+}\mathbf{(}\mathbf{VE}\mathbf{-}\mathbf{V}\mathbf{-}\mathbf{C}\mathbf{)}}{\mathbf{2}}\\ & \mathbf{=}& \frac{\mathbf{2}\mathbf{+}\mathbf{\left[}\mathbf{7}\mathbf{-}\mathbf{4}\mathbf{-}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{\right]}}{\mathbf{2}}\\ & \mathbf{=}& \mathbf{4}\mathbf{\left(}\mathit{s}{\mathit{p}}^{\mathbf{3}}\mathbf{\right)}\end{array}$

Hence, the hybridization of chlorine in BrO₂is sp³

Hybridization of bromine in BrF5 

GA=5, VE=7, V=5, C=0

By substituting the value in equation

$\begin{array}{rcl}\mathit{H}& \mathbf{=}& \frac{\mathbf{GA}\mathbf{+}\mathbf{(}\mathbf{VE}\mathbf{-}\mathbf{V}\mathbf{-}\mathbf{C}\mathbf{)}}{\mathbf{2}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{+}\mathbf{\left[}\mathbf{7}\mathbf{-}\mathbf{5}\mathbf{-}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{\right]}}{\mathbf{2}}\\ & \mathbf{=}& \mathbf{5}\mathbf{\left(}\mathit{s}{\mathit{p}}^{\mathbf{3}}{\mathit{d}}^{\mathbf{2}}\mathbf{\right)}\end{array}$

Hence, the hybridization of bromine in BrF₅is sp³D²