Q11.9P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the hybridization of chlorine in each of the following: (a) ${{\mathbf{CO}}_{\mathbf{2}}}^{\mathbf{-}}$; (b) ${{\mathbf{ClO}}_{\mathbf{3}}}^{\mathbf{-}}$ ; (c) ${{\mathbf{ClO}}_{\mathbf{4}}}^{\mathbf{-}}$ ?

Step-by-Step Solution

Answer

The hybridization of chlorine is given by

$\mathbf{H}\mathbf{=}\frac{\mathbf{GA}\mathbf{+}\mathbf{(}\mathbf{VE}\mathbf{-}\mathbf{V}\mathbf{-}\mathbf{C}\mathbf{)}}{\mathbf{2}}$

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

Step1: Hybridization of chlorine in ClO 2 -

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

By substituting the value in the equation

$$\begin{gathered} \mathbf{H}\mathbf{=}\frac{\mathbf{GA}\mathbf{+}\mathbf{(}\mathbf{VE}\mathbf{-}\mathbf{V}\mathbf{-}\mathbf{C}\mathbf{)}}{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{2}\mathbf{+}\mathbf{[}\mathbf{7}\mathbf{-}\mathbf{4}\mathbf{-}\left(-1\right)\mathbf{]}}{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{4}\mathbf{\left(}{\mathbf{Sp}}^{\mathbf{3}}\mathbf{\right)} \end{gathered}$$

Hence, the hybridization of chlorine in ${{\mathbf{CO}}_{\mathbf{2}}}^{\mathbf{-}}$ is ${\mathbf{Sp}}^{\mathbf{3}}$

Step2: hybridization of chlorine in ClO 3 -

GA=3, VE=7, V=6, C=-1

By substituting the value in equation

$$\begin{gathered} \mathbf{H}\mathbf{=}\frac{\mathbf{GA}\mathbf{+}\mathbf{(}\mathbf{VE}\mathbf{-}\mathbf{V}\mathbf{-}\mathbf{C}\mathbf{)}}{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{2}\mathbf{+}\mathbf{[}\mathbf{7}\mathbf{-}\mathbf{6}\mathbf{-}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{]}}{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{4}\mathbf{\left(}{\mathbf{Sp}}^{\mathbf{3}}\mathbf{\right)} \end{gathered}$$

Hence, the hybridization of chlorine in ${{\mathbf{ClO}}_{\mathbf{3}}}^{\mathbf{-}}$ is ${\mathbf{Sp}}^{\mathbf{3}}$

Step3: hybridization of chlorine in ClO 4 -

GA=4, VE=7, V=8, C=-1

By substituting the value in equation

$$\begin{gathered} \mathbf{H}\mathbf{=}\frac{\mathbf{GA}\mathbf{+}\mathbf{(}\mathbf{VE}\mathbf{-}\mathbf{V}\mathbf{-}\mathbf{C}\mathbf{)}}{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{2}\mathbf{+}\mathbf{[}\mathbf{7}\mathbf{-}\mathbf{8}\mathbf{-}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{]}}{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{4}\mathbf{\left(}{\mathbf{Sp}}^{\mathbf{3}}\mathbf{\right)} \end{gathered}$$

As a result, chlorine hybridization in ${{\mathbf{ClO}}_{\mathbf{4}}}^{\mathbf{-}}$ is ${\mathbf{Sp}}^{\mathbf{3}}$