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Q.
The tendency of $X$ in $BX_3$ $(X = F, Cl, OMe, NMe)$ to
form a$\pi$ -bond with boron follows the order
KVPYKVPY 2018
Solution:
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 $BCl_3$ and $BF_3, Cl$ will have least tendency to form $\pi$ - bond with boron, because of large size of its $3p$-orbital.
Among $B(OMe)_3$ and $B(NMe_2)_3 , (NMe)$ 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.
$BCl_3 < BF_3 < B(OMe)_3 < B(N Me_2)_3$.