Describe the change in hybridisation (if any) of the Al atom in the following reaction.
` AlCl_3 +Cl^(-) to AlCl_(4)^(-)`

Explanation

Valence electronic configuration of `Al(Z=13)` in the excited state is: `3s^(1)3p_(x)^(1)3p_(y)^(1)3p_(z)^(0)`.
Therefore, `AlCl_(3)` undergoes `sp^(2)` hybridisation to give it planar triangular shape. For the formation of `AlCl_(4)^(ө)`, the empty `3p_(z)` orbital accepts one lp of electrons from `Cl^(ө)` ion and forms coordinate bond, In this process empty `3p_(z)` orbital is also involved in the hybridisation and the hybridisation is `sp^(3)` with tetrahedral shape `(TH)`.