Step 1: Boron in BX3 has an empty 2p orbital, making it a Lewis acid.
Step 2: Halogens donate a lone pair via pi back bonding into this empty orbital, reducing Lewis acidity.
Step 3: Fluorine's 2p orbital matches boron's 2p size best, giving strongest back bonding and weakest acidity in BF3.
Step 4: Larger halogens (Cl, Br) overlap progressively worse, weakening back bonding.
Step 5: So acidity increases BF3 < BCl3 < BBr3.
\[\boxed{\text{BBr}_3 > \text{BCl}_3 > \text{BF}_3}\]