Describe the change in hybridisation of the $A l$ atom in the following reaction :

$A l C l_{3} + C l^{-} \to {A l C l_{4}}^{-}$

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Solution

The hybridization of Al atom changes from $s p^{2}$ in $A l C l_{3}$ to $s p^{3}$ in ${A l C l_{4}}^{-}$ .

In $A l C l_{3}$ , $A l$ contains 3 bond pairs of electrons and zero lone pair of electrons.

Hence, it is $s p^{2}$ hybridized. In ${A l C l_{4}}^{-}$ , $A l$ atom contains 4 bond pairs of electrons and zero lone pairs of electrons.

Hence, it is $s p^{3}$ hybridized.

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Similar questions

Describe the change in hybridisation (if any) of the Al atom in the following reaction.

$A l C l_{3} + C l^{-} \to A l C l_{4}^{-}$

Describe the change in hybridisation (if any) of the Al atom in the following reaction.

A l C l_{3} + C l^{-} \to A l C l_{4}^{-}, A l C l_{3}

A l C l_{3} + C l^{-} \to [A l C l_{4}]^{-}, A l C l_{3}

A l {C l}_{3} + {C l}^{-} ⟶ {[A l {C l}_{4}]}^{-}

A l {C l}_{3}

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