Question

Describe the change in hybridization of the $Al$ atom in the following reaction:
$AlC{l_3} + C{l^ - } \to AlCl_4^ - $

Answer 1

Hint: Whenever the number of bond pairs increase in a structure, there is a change in the hybridization of the compound. In the case of $AlC{l_3}$ there are 3 bond pairs and zero lone pairs, thus the structure will be triangular planar but as soon as a chloride ion attacks it, it acquires a tetrahedral geometry.

Complete answer:
Let us understand the hybridization, geometry and structure of the aluminium chloride. It contains three bond pairs and zero lone pairs. The aluminium atom has a vacant orbital which tries to acquire a lone pair from a Lewis base. It has a $s{p^2}$ hybridization and a trigonal planar geometry.
Now let us discuss the attack of a $C{l^ - }$ ion which acts as a Lewis base, on the vacant orbital of the aluminium atom. The reaction is as follows:
$AlC{l_3} + C{l^ - } \to AlCl_4^ - $
The aluminium tetrachloride ion has a $s{p^3}$ hybridization and is tetrahedral in structure. The number of bond pairs is 4 and the number of lone pairs is zero.
Thus, we can deduce from the above reaction that there is a change of hybridization from $s{p^2}$ to $s{p^3}$ when a Lewis base like $C{l^ - }$ attacks a Lewis acid like $AlC{l_3}$.

Note:
Whenever there is a change in the hybridization of a compound from $s{p^2}$ to $s{p^3}$ , the bonds become free and the bond angles decrease but move in different planes in the space or orientation and the planar structure is lost.