A boron trihalide ($\ce {BX3}$) is a Lewis acid, because it can accept an electron pair, as per Gilbert N. Lewis' definition.

Now my question is-

Arrange $\ce {BF3}$, $\ce {BCl3}$, $\ce {BBr3}$ and $\ce {BI3}$ from the weakest to the strongest acid properties.

  • $\begingroup$ 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. $\endgroup$ – Uncle Al Apr 27 '14 at 17:09
  • $\begingroup$ hint; most important factor in BX3 is the bonding between X and B which is going to determine the acidic strength , and here we can clearly see that in BF3 there is the partial double bond character between atoms which is due to back bonding (2pi-2pi), and gradually back bonding weakens and lewis acidic character increases so your final order will be. $$\ce{BF3}<\ce{BCl3}<\ce{BBr3}<\ce{BI3}$$ $\endgroup$ – Jack Rod May 18 at 9:46

The following order is experimentally found:


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 $\ce{B}$ 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 $\ce{B-X}$ bond needs to be taken into account. The change of hybridization from $\ce{sp^2}$ in $\ce{BX3}$ to $\ce{sp^3}$ in $\ce{[BX4]-}$ involves the loss of partial $\pi$ bonding which results from the overlap of the $\ce{p_z}$ orbitals of $\ce{B}$ and $\ce{X}$. The $\ce{p_z}$ orbitals of $\ce{B}$ and $\ce{F}$ 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 $\ce{p_z}$ orbitals of the heavier halides have increasingly less overlap with the smaller boron $\ce{p_z}$ 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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  • $\begingroup$ d if uou can add back bonding concept answer will look more good $\endgroup$ – Jack Rod May 18 at 9:26
  • $\begingroup$ order is same as that of backbonding $\endgroup$ – Jack Rod May 18 at 9:34

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