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p-Block Elements - I
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The tendency of $X$ in $B X_{3}(X= F , Cl$, OMe, NMe $)$ to form a $\pi$-bond with boron follows the order
A$BCl _{3}\,<\, BF _{3}\,< \,B ( OMe )_{3}\,<\, B \left( NMe _{2}\right)_{3}$
B$BF _{3}\,<\, BCl _{3}\,<\, B ( OMe )_{3}\,< \,B \left( NMe _{2}\right)_{3}$
C$BCl _{3}\,<\, B \left( NMe _{2}\right)_{3}\,<\, B ( OMe )_{3}\,< \,BF _{3}$
D$BCl _{3}\,<\, BF _{3}\,<\, B \left( NMe _{2}\right)_{3}\,<\, B ( OMe )_{3}$
(KVPY-2018)
Solution
(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 $BCl _{3}$ and $BE _{3}, Cl$ will have least tendency to form $\pi$ – bond with boron, because of large size of its $3 p$-orbital.
Among $B ( OMe )_{3}$ and $B \left( NMe _{2}\right)_{3},( NMe )$ will have the maximum tendency to form $\pi$-bond because $N$ in NMe is less electronegative than $O$ in $OM$ group. Thus, the tendency to form $\pi$-bond with boron follows the order.
$BCl _{3}\,<\, BF _{3}\,< \,B ( OMe )_{3}\,<\, B \left( NMe _{2}\right)_{3}$
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 $BE _{3}, Cl$ will have least tendency to form $\pi$ – bond with boron, because of large size of its $3 p$-orbital.
Among $B ( OMe )_{3}$ and $B \left( NMe _{2}\right)_{3},( NMe )$ will have the maximum tendency to form $\pi$-bond because $N$ in NMe is less electronegative than $O$ in $OM$ group. Thus, the tendency to form $\pi$-bond with boron follows the order.
$BCl _{3}\,<\, BF _{3}\,< \,B ( OMe )_{3}\,<\, B \left( NMe _{2}\right)_{3}$
Standard 11
Chemistry
