The Lewis acid character of halides of boron are as follows:

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

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

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

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

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Solution

The correct option is B
$B B r_{3} > B C l_{3} > B F_{3}$

Boron has an empty p orbital, which can accommodate a lone pair of electrons. In boron trihalides, depending on which period the halogen belongs to, there is a varied scope for back-bonding from the lone pairs of the halogen atoms.

Boron and fluorine belong to the same period and hence the ease with which back bonding happens is much greater than for halogens in other boron trihalides. As we go down the halogen group, the energy of the lone pairs increases and the overlapping efficacy of lone pairs via back bonding decreases. Hence, Lewis acidity follows the order:

$B F_{3} < B C l_{3} < B B r_{3} < B I_{3}$

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B F_{3}

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The correct order of Lewis acidity of $B X_{3}$ molecule is

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B - X

π

B B r_{3}, B C l_{3}

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