Question

The relative Lewis acid character of boron trihalides is in the order:

• A. $B{I}_3>BB{r}_3>B{F}_3>BC{l}_3$
• B. $B{I}_3>BB{r}_3>BC{l}_3>B{F}_3$
• C. $B{F}_3>BC{l}_3>BB{r}_3>B{I}_3$
• D. $BC{l}_3>B{F}_3>B{I}_3>BB{r}_3$

Answer 1

Answer: (B)

$B{I}_3>BB{r}_3>BC{l}_3>B{F}_3$
This order can be easily explained on the basis of the tendency of the halogen atom to back donate its lone pair of electrons to the empty p-orbital of the boron atom through $p\pi -p\pi$ bonding.
Since the size of the vacant $2p$-orbital of $B$ and the $2p$-orbital of $F$ containing a lone pair of electrons are almost identical, therefore, the lone pair of electrons on $F$ is donated towards the B atoms. Further due to back donation by three $F$ atoms, $B{F}_3$ can be represented as a resonance hybrid of the three structures.

As a result of $p\pi -p\pi$ back donation and resonance, the electron deficiency of $B$ decreases and thus $B{F}_3$ is the weakest Lewis acid. As the size of the halogen atom increases from $Cl$ to $I$, the extent of overlap between $2p$ orbital of $B$ and a bigger $p$-orbital of halogen (3p in Cl, $4p$ in Br and $5p$ in I) decreases and consequently the electron deficiency of B increases and thus the Lewis acid character increases accordingly from $B{F}_3$ to $B{I}_3.$ Thus, the relative acid strength of the boron trihalides follows the sequence :
$B{I}_3>BB{r}_3>BC{l}_3>B{F}_3$