Question

In $B{F}_3$ the $B-F$ bond length is $1.30\mathring{A}$, when $B{F}_3$ is allowed to be treated with $M{e}_{3}N$ it forms an adduct $M{e}_{3}N\to B{F}_3$ the bond length of $B-F$ in the adduct is :

• A. greater than $1.30\mathring{A}$
• B. smaller than $1.30\mathring{A}$
• C. equal to $1.30\mathring{A}$
• D. None of these

Answer 1

The correct option is A greater than $1.30\mathring{A}$

In $B{F}_3$, three electronegative Fluorine atoms bonded to single Boron and then ${}^{\prime }{3}^{\prime }{}^{\prime }{F}^{\prime }$ atoms pull electrons towards it.

So the bond length is less. But in $M{e}_{3}N\to B{F}_3$, the electrons are shared from ${}^{\prime }{N}^{\prime }$ and the scarcity of electrons is reduced. So they are free and bond length increases.

So the bond length will be greater than $1.30\mathring{A}$.

Hence, the correct option is A.