Describe the change in hybridisation (if any) of the $Al$ atom in the following reaction.
$AlCl_{3} + Cl^{-} \longrightarrow AlCl_{4}^{-}$

(N/A) In $AlCl_{3}$,the aluminium atom is in the $sp^{2}$ hybridised state,resulting in a trigonal planar geometry.
When $AlCl_{3}$ reacts with $Cl^{-}$ to form $AlCl_{4}^{-}$,the lone pair from the chloride ion is donated into the empty $3p_{z}$ orbital of the aluminium atom.
Consequently,the hybridisation of the $Al$ atom changes from $sp^{2}$ to $sp^{3}$,and the geometry changes from trigonal planar to tetrahedral.