The change in hybridization (if any) of the A | vedclass

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(B) In $AlCl_3$,the $Al$ atom is bonded to $3$ chlorine atoms with $3$ $\sigma$ bonds and has no lone pair. The steric number is $3$,which corresponds

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$sp^2$ to $sp^3$

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(B) In $AlCl_3$,the $Al$ atom is bonded to $3$ chlorine atoms with $3$ $\sigma$ bonds and has no lone pair. The steric number is $3$,which corresponds to $sp^2$ hybridization.In $AlCl_4^{-}$,the $Al$ atom is bonded to $4$ chlorine atoms with $4$ $\sigma$ bonds and has no lone pair. The steric number is $4$,which corresponds to $sp^3$ hybridization.Therefore,the hybridization of the $Al$ atom changes from $sp^2$ to $sp^3$.

The change in hybridization (if any) of the $Al$ atom in the following reaction is $AlCl_3 + Cl^{-} \rightarrow AlCl_4^{-}$

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