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Boron trihalides ($\ce {BX3}$) is a Lewis acid, because it can accept an electron pair, as Gilbert N. Lewis definition. Now my question is, please arrange ($\ce {BF3}$), ($\ce {BCl3}$), ($\ce {BBr3}$), and ($\ce {BI3}$) from the weakest to the strongest acid properties.

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The following order is experimentally found:

$$\ce{BF3}<\ce{BCl3}<\ce{BBr3}<\ce{BI3}$$

This stands in contrast to what is expected when the electronegativity of the halides is considered. A more electronegative halide should be able to stabilize the negative charge at B in the Lewis acid-base complex better, and this would suggest that $\ce{BF3}$ would be the strongest Lewis acid.

However, the nature of the B-X bond needs to be taken into account. The change of hybridization from $sp^2$ in $\ce{BX3}$ to $sp^3$ in $\ce{[BX4]^-}$ involves the loss of partial $\pi$ bonding which results from the overlap of the $p_{z}$ orbitals of B and X. The $p_{z}$ orbitals of B and F are of similar size, because both elements are members of the same period. Therefore, they have the best overlap and form the strongest $\pi$ bond. Because the reaction with a Lewis base would require some energy to break this bond, $\ce{BF3}$ is a comparably weak Lewis acid. The larger $p_{z}$ orbitals of the heavier halides have increasingly less overlap with the smaller boron $p_{z}$ orbital. Therefore, the $\pi$ bonding is weaker and formation of the tetrahedral Lewis acid-base adduct via rehybridization becomes energetically more favorable.

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Relative Lewis acidities parallel exothermicities of a Lewis base adduct-forming reaction. The adduct is a dative bond tetrahedral boron anion. The worst pi-bonding substituent has the highest energy versus product valley, maximizing exotherm difference. Order radii of the participating pi-orbitals by their worst matches.

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