In BF_3,the B-F bond length is 1.3 mathring{ | vedclass

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(A) In $BF_3$,the boron atom is $sp^2$ hybridized and has an incomplete octet,which allows for $p\pi-p\pi$ back-bonding from fluorine to boron.
This

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Greater than $1.3 \ \mathring{A}$

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(A) In $BF_3$,the boron atom is $sp^2$ hybridized and has an incomplete octet,which allows for $p\pi-p\pi$ back-bonding from fluorine to boron.
This back-bonding gives the $B-F$ bond some double bond character,shortening it to $1.3 \ \mathring{A}$.
When $BF_3$ reacts with $Me_3N$ (a Lewis base),it forms an adduct $Me_3N 	o BF_3$.
In this adduct,the boron atom becomes $sp^3$ hybridized and completes its octet by accepting a lone pair from nitrogen.
Due to the formation of the $N 	o B$ coordinate bond,the $p\pi-p\pi$ back-bonding from fluorine to boron is no longer possible.
As a result,the $B-F$ bond loses its double bond character and becomes a pure single bond,which is longer than the original $B-F$ bond in $BF_3$.
Therefore,the $B-F$ bond length increases and becomes greater than $1.3 \ \mathring{A}$.

In $BF_3$,the $B-F$ bond length is $1.3 \ \mathring{A}$. When $BF_3$ reacts with $Me_3N$,it forms an adduct $Me_3N \to BF_3$. In this adduct,the $B-F$ bond length will be ..........

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