The relative Lewis acid strengths of boron trihalides are in the order:

B B r_{3} > B C l_{3} > B F_{3}

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B C l_{3} > B F_{3} > B B r_{3}

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B F_{3} > B C l_{3} > B B r_{3}

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B F_{3} > B B r_{3} > B C l_{3}

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Solution

The correct option is A $B B r_{3} > B C l_{3} > B F_{3}$
In boron trihalides the B atom is $s p^{2}$ hybridised and has its vacant 2p orbital perpendicular to the plane of the molecule. F atoms have filled 2p orbitals perpendicular to the plane of the molecule. These filled orbitals donates the electron density back to the B atom through pi-back bonding. This reduces the electron deficiency on B and thus the Lewis acidity of $B F_{3}$ is the least among the boron trihalides. But in case of higher halogens where the ligand orbital is 3p, 4p etc. the extent of back bonding decreases. Thus, the Lewis acidity increases in the order $B F_{3} < B C l_{3} < B B r_{3} < B I_{3}$ .

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