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

Jan 18, 2016

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 $\left[N e\right] 3 {s}^{2} 3 {p}^{1}$. It thus has an empty $3 p$ atomic orbital (AO) as an atom.

So, aluminum then has an empty $\setminus m a t h b f \left(3 p\right)$ valence unhybridized orbital after it needs to bond with three $\text{Cl}$ atoms.

Since ${\text{AlCl}}_{3}$ has an empty $3 p$ valence AO, that AO can accept electrons, making ${\text{AlCl}}_{3}$ a (good) Lewis acid.

Accepting electrons in this way is still favorable. These electrons can be donated from ${\text{Cl}}^{-}$ (such as when reacting with $\text{HCl}$) in this example, thus forming ${\text{AlCl}}_{4}^{-}$.

At this point, the molecular geometry around aluminum has changed from trigonal planar to tetrahedral. In other words, the orbital hybridization changed from $s {p}^{2}$ to $s {p}^{3}$.