Question

If the $B-F$ bond length in $B{F}_3$ is 1.30 $\stackrel{\circ }{A}$, then the $B-F$ bond length in $M{e}_{3}N.B{F}_3$ is:

• A. 1.30 $\stackrel{\circ }{A}$
• B. 1.28 $\stackrel{\circ }{A}$
• C. 1.25 $\stackrel{\circ }{A}$
• D. 1.35 $\stackrel{\circ }{A}$

Answer 1

The correct option is D 1.35 $\stackrel{\circ }{A}$

In $M{e}_{3}N\to B{F}_3$, the electron deficiency of boron is compensated by the lone pair of electrons donated by nitrogen atom by forming the sigma coordinate bond. Whereas in $B{F}_3$, it is compensated by backbonding with $F$ atom. Back bonding is delocalised. Thus, each $B-F$ bond has a partial double bond character. Hence, the bond length in $M{e}_{3}N\to B{F}_3$ will be greater than that in $B{F}_3$.