There is change in hybridization from $sp^2$ → $sp^3$. In $AlCl_3$, Al is $sp^2$ hybridised. $AlCl_3$ thus has three $sp^2$ hybrid orbitals and a vacant unhybrid p-orbital. When $AlCl_4^{–}$ is formed, Al undergoes $sp^3$ hybridisation and thus $Cl^{–}$ overlaps with the fourth orbital.