Boron forms $B X_{3}$ type of halides. The correct decreasing order of Lewis-acid strength of these halides is:

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

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

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

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

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Solution

The correct option is B $B I_{3} > B B r_{3} > B C l_{3} > B F_{3}$
The decreasing order of the Lewis acidity of boron halides is $B I_{3} > B B r_{3} > B C l_{3} > B F_{3}$ .
This order is due to the relative tendency of the halogen atom to back donate its unutilized electrons to the vacant p-orbitals of boron atom.
Due to the back donation of electrons from fluorine to boron, the electron deficiency of boron is reduced and hence, Lewis acidity is decreased.
The tendency for the formation of back bonding is maximum in boron trifluoride and decreases very rapidly from boron trifluoride to boron triodide.

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Similar questions

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

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

B F_{3}

B C l_{3}

B B r_{3}

B I_{3}

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

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

B F_{3}

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