# Why is AlCl3 a lewis acid ?

Due to the electronegativity differences between $\text{Cl}$ ($3.16$) and $\text{Al}$ ($1.61$), $\text{Cl}$ makes a good electron-withdrawing group in ${\text{AlCl}}_{3}$.
Furthermore, $\text{Al}$ must access both its $2 s$ and its three $2 p$ orbitals to bond, so it uses $s {p}^{3}$ hybridization (one $2 s$ and three $2 p$ orbitals), giving it four bonding orbitals (one of which is empty as ${\text{AlCl}}_{3}$). This allows it to form a fourth bond and acquire a tetrahedral structure as ${\text{AlCl}}_{4}^{-}$.
With one empty orbital and three electron-withdrawing $\text{Cl}$ atoms attached, the compound is thus an electron-acceptor at the $\text{Al}$ center. By definition, that is a Lewis Acid.
Here is an example of Friedel-Crafts Acylation that shows the Lewis Acid behavior of $\text{Al}$ in ${\text{AlCl}}_{3}$: