Which is the correct order of decreasing acidity of lewis acids?

The correct option is BBr₃ > BCl₃ > BF₃.

To understand the decreasing order of Lewis acidity among boron trihalides, we need to analyze their electron-accepting tendencies and the concept of back-bonding (or pπ-pπ back-bonding).
A Lewis acid is a chemical species that can accept a pair of electrons. Boron in boron trihalides ($B{X}_3$) is $s{p}^2$ hybridized and has an empty, unhybridized 2p orbital, making it electron-deficient (containing only 6 valence electrons).

Halogen atoms have lone pairs of electrons in their p orbitals that can be donated into the empty 2p orbital of the central boron atom. This coordinate covalent bonding is known as back-bonding.
The efficiency of this back-bonding depends on the size match between the orbital of boron (which is 2p) and the orbital of the halogen (2p for F, 3p for Cl, and 4p for Br):

1. In $B{F}_3$, the back-bonding occurs between the 2p orbital of boron and the 2p orbital of fluorine. Since both orbitals belong to the same energy level (2p-2p overlap), the overlap is highly efficient and strong. This satisfies the electron deficiency of boron to a large extent, making $B{F}_3$ the weakest Lewis acid.
2. In $BC{l}_3$, the overlap is between the 2p orbital of boron and the larger 3p orbital of chlorine. Because of the size mismatch, the pπ-pπ back-bonding is weaker than in $B{F}_3$.
3. In $BB{r}_3$, the overlap involves the 2p orbital of boron and the even larger 4p orbital of bromine. This overlap is extremely poor, resulting in negligible back-bonding. Consequently, the boron atom in $BB{r}_3$ remains highly electron-deficient and is the most eager to accept an electron pair from an external donor.

Therefore, as the strength of pπ-pπ back-bonding decreases in the order:
$B{F}_3>BC{l}_3>BB{r}_3$
The Lewis acidity (tendency to accept external electrons) increases in the opposite direction:
$BB{r}_3>BC{l}_3>B{F}_3$