# Which is considered a stronger lewis acid, BF3 or BCl3?

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Dec 24, 2015

#### Answer:

${\text{BCl}}_{3}$ is the stronger Lewis acid.

#### Explanation:

We would expect ${\text{BF}}_{3}$ to be stronger, because $\text{F}$ is more electronegative than $\text{Cl}$.

Chemists explain this unexpected result by an electronic argument and a steric argument.

The electronic argument — backbonding

The boron atom in ${\text{BF}}_{3}$ is $s {p}^{2}$ hybridized, with a vacant $2 p$ orbital.

The $\text{F}$ atoms can also be $s {p}^{2}$ hybridised, with lone pairs in their $2 p$ orbitals.

These $\text{F}$ orbitals can overlap with the orbital on $\text{B}$, thereby increasing the electron density on the boron atom and making it less acidic.

This effect is called backbonding, because electron density is leaving the more electronegative atom.

In ${\text{BCl}}_{3}$, the $3 p$ orbitals on $\text{Cl}$ are bigger than the $2 p$ orbital on $\text{B}$, so orbital overlap is less efficient, and backbonding is less important.

Hence, the greater backbonding in ${\text{BF}}_{3}$ makes it a weaker Lewis acid.

The steric argument — ligand close-packing (LCP)

The LCP model is based on the observation that the $\text{X}$ atoms (ligands) in ${\text{AX}}_{n}$ systems are always the same distance from each other.

For example, the distance between the $\text{F}$ atoms in ${\text{BF}}_{3}$ and ${\text{BF}}_{4}^{-}$ is 226 pm, despite the longer $\text{B-F}$ distance in the tetrahedral structure.

(from alpha.chem.umb.edu)

It is as if the $\text{F}$ atoms are closest-packed (like in a crystal), with the $\text{F}$ atoms having a ligand radius of 113 pm.

When the ${\text{BF}}_{3}$ forms a Lewis complex, the $\text{F}$ atoms remain close-packed, but the $\text{B-F}$ bonds must become longer in the new tetrahedral geometry.

It takes more energy to lengthen the short, strong $\text{B-F}$ bonds than the longer, weaker $\text{B-Cl}$ bonds.

Hence ${\text{BF}}_{3}$ is a weaker Lewis acid than ${\text{BCl}}_{3}$.

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