The tendency of X in BX_{3} (X = F, Cl, OMe, | vedclass

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Question

(A) The boron atom in $BX_{3}$ has an incomplete octet and can accept a lone pair from the substituent $X$ to form a $\pi$-bond,commonly known as back

Vedclass · Accepted Answer

$BCl_{3} < BF_{3} < B(OMe)_{3} < B(NMe_{2})_{3}$

Vedclass · Answer

(A) The boron atom in $BX_{3}$ has an incomplete octet and can accept a lone pair from the substituent $X$ to form a $\pi$-bond,commonly known as back bonding.The extent of back bonding depends on the electronegativity of $X$ and the energy match between the orbitals of $X$ and $B$. Lower electronegativity of $X$ facilitates better donation of the lone pair.Comparing the substituents: $Cl$ and $F$ are more electronegative than $OMe$ and $NMe_{2}$,so their back bonding tendency is lower.Between $BCl_{3}$ and $BF_{3}$,$Cl$ has a lower tendency to form a $\pi$-bond due to the size mismatch between its $3p$-orbital and boron's $2p$-orbital.Between $B(OMe)_{3}$ and $B(NMe_{2})_{3}$,the $NMe_{2}$ group has a higher tendency to form a $\pi$-bond because nitrogen is less electronegative than oxygen,making its lone pair more available for donation.Thus,the correct order of the tendency to form a $\pi$-bond with boron is $BCl_{3} < BF_{3} < B(OMe)_{3} < B(NMe_{2})_{3}$.

The tendency of $X$ in $BX_{3}$ $(X = F, Cl, OMe, NMe_{2})$ to form a $\pi$-bond with boron follows the order:

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