Question

Why is `"AlCl"_3` a Lewis acid? What happens to its geometry after bonding with another atom?

Answer 1

Having only three valence electrons, aluminum is (along with Boron, for instance) special in that it doesn't need an octet to become stable.

Its electron configuration is `[Ne] 3s^2 3p^1`. It thus has an empty `3p` atomic orbital (AO) as an atom.

So, aluminum then has an empty `\mathbf(3p)` valence unhybridized orbital after it needs to bond with three `"Cl"` atoms.

Since `"AlCl"_3` has an empty `3p` valence AO, that AO can accept electrons, making `"AlCl"_3` a (good) Lewis acid.

Accepting electrons in this way is still favorable. These electrons can be donated from `"Cl"^(-)` (such as when reacting with `"HCl"`) in this example, thus forming `"AlCl"_4^(-)`.

At this point, the molecular geometry around aluminum has changed from trigonal planar to tetrahedral. In other words, the orbital hybridization changed from `sp^2` to `sp^3`.