Question

The correct order of strengths of back-bonding and lewis acidic strengths in Boron trihalides respectively is:

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

Answer 1

The correct option is C $B{F}_3>BC{l}_3>BB{r}_3$ and $B{F}_3<BC{l}_3<BB{r}_3$

Lewis acids are electron pair acceptors.
In $B{F}_3$, though the partial positive charge on boron atom is more due to the high electronegativity of flourine, the tendency to accept a lone pair is reduced because of the ability of the fluorine atoms to donate their own lone pair to the empty boron orbitals forming $p\left(\pi \right)-p\left(\pi \right)$ back bonding. flourine forms the strongest back bond due to the effective overlap between $2p\left(F\right)$ and $2p\left(B\right)$ orbitals. So, the electron deficiency of boron gets compensated and the lewis acidic character decreases.
But as the size of the orbital increases, the extent of back-bonding decreases. This is due to the ineffective overlap as in case of $3pand4p$ obitals of $ClandBr$ respectively with that of $2p$ orbital of $B$. So, the extent of back bonding is $B{F}_3>BC{l}_3>BB{r}_3$ and hence that of lewis acidity is $B{F}_3<BC{l}_3<BB{r}_3$.