Question

Describe the change in hybridisation (if any) of the $Al$ atom in the following reaction: $AlC{l}_3+C{l}^{-}\to AlC{l}_4^{-}$.

Answer 1

The formation of $AlC{l}_3$ can be explained as:

Electronic configuration of ${}_{13}Al=1s^{2}2s^{2}2p^{6}3s^{2}3p^1$

Electronic configuration of ${}_{17}Cl=1s^{2}2s^{2}2p^{6}3s^{2}3p^5$

In excited state, one electron in $Al$ gets promoted to $p-$orbital. It can be shown as:
Thus it has three unpaired electrons. Chlorine needs one electron to complete its octet. Three atoms hybridise with $Al$ to undergo $s{p}^2$ hybridisation to give a trigonal planar arrangement in $AlC{l}_3$. But in the formation of $AlC{l}_4^{-}$ the empty $3p_z$ orbital while also be involved in the hybridisation and thus the hybridization changes from $sp2$ to $sp3$. As a result, the shapes gets changed to tetrahedral.