Boron trihalides acidity
The following order is experimentally found:

$$\ce{BF3}&lt;\ce{BCl3}&lt;\ce{BBr3}&lt;\ce{BI3}$$

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 $\ce{B}$ 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.