In boron trihalides ($\text{BX}_3$), the central boron atom has an empty $2p$ orbital and the halogen atoms have lone pairs in their $p$ orbitals. The lone pair of electrons from the halogen can be donated to the empty $2p$ orbital of boron, forming a $p\pi-p\pi$ back bond. This reduces the electron deficiency of boron, decreasing its Lewis acid character. The effectiveness of back bonding is maximum in $\text{BF}_3$ because both boron and fluorine have $2p$ orbitals of similar size, leading to strong orbital overlap. As the size of the halogen atom increases from $\text{F}$ to $\text{Br}$ ($\text{F}: 2p$, $\text{Cl}: 3p$, $\text{Br}: 4p$), the size mismatch between the overlapping orbitals increases, making back bonding progressively weaker. Consequently, the electron deficiency of boron remains higher for larger halogens, increasing their Lewis acid strength. Thus, the Lewis acid strength increases in the order $\text{BF}_3 < \text{BCl}_3 < \text{BBr}_3$. Therefore, the decreasing sequence is $\text{BBr}_3 > \text{BCl}_3 > \text{BF}_3$.