Question

Boron forms $B{X}_3$ type of halides. The correct increasing order of Lewis-acid strength of these halides is :

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

Answer 1

The correct option is B $B{I}_3$>$BB{r}_3$>$BC{l}_3$>$B{F}_3$

The correct increasing order of Lewis-acid strength of these halides is $B{I}_3$>$BB{r}_3$>$BC{l}_3$>$B{F}_3$.
This can be explained on the basis of tendency of halogen atom to back donate its lone pair of electrons to the empty $p$-orbital of $B$ atom through $p\pi -p\pi$ bonding. As this tenedency is higher in case of $F$ (due to identical size of 2 $p$- orbitals of $B$ and $F$) the electron deficiency of $B$ decreases and thus $B{F}_3$ behaves as weakest Lewis acid. As the size of halogen atom increases from $Cl$ to $I$, the tendency for back donation decreases and consequently the electron deficiency of $B$ increases and Lewis acid character increases accordingly from $B{F}_3$ to $B{I}_3$.