# Why is FeBr_3 a lewis acid?

If you look at $F e B {r}_{3}$, the first thing that should stand out is the fact that you've got a transition metal, $F e$, bonded to a highly electronegative element, $B r$.
This difference in electronegativity creates a partial positive charge on the $F e$, which in turns allows it to accept an electron pair. Remember that transition metals are capable of expanding their octets in order to accomodate more electrons, so a good rule of thumb is that compounds formed by transition metals paired with highly electronegative elements will most likely be Lewis acids.
$F e B {r}_{3}$ is used as a catalyst in the bromination of benzene because of its strong Lewis acid character.
Without going into detail, the bromine ($B {r}_{2}$) molecule reacts with $F e B {r}_{3}$ and donates a pair of electrons to it. This polarizes $B {r}_{2}$, making it a better electrophile (more attracted to electrons).