# Which of these is the best Lewis acid?

## A) ${\text{BPh}}_{3}$ B) ${\text{BPh}}_{4}^{-}$ C) "B"("C"_6"F"_5)_3 D) "B"("C"_6"F"_5)_4^(-)

Aug 15, 2017

Well, I would hope that $C$ is glaringly obvious, but if you aren't accustomed to looking for a bunch of fluorine atoms on a ligand as a sign, it may not be so clear-cut.

And in case you are not convinced, this paper briefly mentions that "B"("C"_6"F"_5)_3 is a stronger Lewis acid than "BPh"("C"_6"F"_5)_2.

These are all boron-class acids, i.e. they are derivatives of ${\text{BH}}_{3}$ (choices $A$ and $C$) or ${\text{BH}}_{4}^{-}$ (choices $B$ and $D$).

${\text{BH}}_{3}$ is known to be trigonal planar, and to have an empty $\boldsymbol{2 {p}_{z}}$ nonbonding orbital on the boron atom that is perpendicular to the plane of the molecule.

(the image depicts ${\text{NH}}_{3}$ acting as a Lewis base and ${\text{BH}}_{3}$ acting as a Lewis acid.)

That empty $2 {p}_{z}$ orbital on boron acts as an electron pair acceptor, making the molecule a Lewis acid.

The four-coordinate, tetrahedral boron compounds (having the maximum possible number of bonds boron can make) are not Lewis acids, because there is no empty orbital on the central atom to accept electron density anymore. So, we can eliminate $B$ and $D$.
However, in $C$, the massive number of fluorine atoms gives "B"("C"_6"F"_5)_3 three highly-electron-withdrawing $\boldsymbol{- {\text{C"_6"F}}_{5}}$ ligands (hexafluorophenyl).
These inductively withdraw electron density from the central atom, so that boron becomes extremely electropositive. That greatly increases the tendency of the central $2 {p}_{z}$ orbital to accept electron density (which is negative), making "B"("C"_6"F"_5)_3 the strongest Lewis acid (choice $C$).