With the given anions, the correct stability order of tetrahaloborates is:

B C l_{4}^{-} > B B r_{4} > B I_{4}^{-}

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B I_{4}^{-} > B B r_{4}^{-} > B C I_{4}^{-}

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B C l_{4}^{-} = B B r_{4}^{-} > B I_{4}^{-}

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B C l_{4}^{-} = B B r_{4}^{-} = B I_{4}^{-}

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Solution

The correct option is B $B I_{4}^{-} > B B r_{4}^{-} > B C I_{4}^{-}$
In $B X_{3}, p_{π} - p_{π}$ back bonding takes place. The back bonding is stronger in case of $B F_{3}$ and reduces progressively down the series. This makes the formation of the coordinate bond $X \to B X_{3}$ more difficult.

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