Reason: Effective $p\pi - p\pi$ Back Bonding • In $BF_3$ , the filled $2p$ orbital of $F$ overlaps effectively with the empty $2p$ orbital of $B$ (similar size). This donates electron density to Boron, partially satisfying its electron deficiency. • As the halogen size increases ( $Cl \rightarrow I$ ), the orbital mismatch ( $2p-3p$ , $2p-5p$ ) makes back bonding negligible, leaving Boron more electron-deficient and thus a stronger acid.