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)

$B{{I}_{3}}>BB{{r}_{3}}>BC{{l}_{3}}>B{{F}_{3}}$

Answer 2

$BI _{3}> BBr _{3}> BCl _{3}> BF _{3}$ 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 $2 p$-orbital of $B$ and the $2 p$-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, $BF _{3}$ 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 $BF _{3}$ is the weakest Lewis acid. As the size of the halogen atom increases from $Cl$ to $I$, the extent of overlap between $2 p$ orbital of $B$ and a bigger $p$-orbital of halogen (3p in Cl, $4 p$ in Br and $5 p$ in I) decreases and consequently the electron deficiency of B increases and thus the Lewis acid character increases accordingly from $BF _{3}$ to $BI _{3} .$ Thus, the relative acid strength of the boron trihalides follows the sequence : $BI _{3}> BBr _{3}> BCl _{3}> BF _{3}$