Question

The tendency of $X$ in $B{X}_3$ $(X=F,Cl,OMe,NMe)$ to form a$\pi$ -bond with boron follows the order

• A. $BC{l}_3<B{F}_3<B(OMe{)}_3<B(NM{e}_2{)}_3$
• B. $B{F}_3<BC{l}_3<B(OMe{)}_3<B(NM{e}_2{)}_3$
• C. $BC{l}_3<B(NM{e}_2{)}_3<B(OMe{)}_3<B{F}_3$
• D. $BC{l}_3<B{F}_3<B(NM{e}_2{)}_3<B(OMe{)}_3$

Answer 1

Answer: (A)

Boron atom has an incomplete octet. It can accept a lone pair from $X$ given in the question and can also donate electron to it, thus it can form a $\pi$-bond also known as back bonding. The extent of back bonding depends upon the electronegativity of $X$ and on the size of the valence shell. Lesser is the electronegativity of $X$ and similar is the size of valence shell of $X$ and $B$, more will be the extend of back bonding.
$Cl$ and $F$ are more electronegative than $OMe$ and $NMe$ group, so there extent of back bonding will be least. Among $BC{l}_3$ and $B{F}_3,Cl$ will have least tendency to form $\pi$ - bond with boron, because of large size of its $3p$-orbital.
Among $B(OMe{)}_3$ and $B(NM{e}_2{)}_3,(NMe)$ will have the maximum tendency to form $\pi$-bond because $N$ in $NMe$ is less electronegative than $O$ in $OMe$ group. Thus, the tendency to form $\pi$ -bond with boron follows the order. $BC{l}_3<B{F}_3<B(OMe{)}_3<B(NM{e}_2{)}_3$.