Question

Which of the following statements is/are correct for the 1 : 1 complex of trimethyl amine and boron trifluoride?

• A. The $B—F$ bond length in the complex is longer than
  that in $B{F}_3$.
• B. The $N$ is pyramidal with $s{p}^3$ hybridisation and $B$ is planar with $s{p}^2$ hybridisation.
• C. The coordination geometry of both $N$ and $B$ are tetrahedral with $s{p}^3$ hybridisation.
• D. Both $\left(a\right)$ and $\left(c\right)$

Answer 1

The correct option is D Both $\left(a\right)$ and $\left(c\right)$

In $B{F}_3$, the $B–F$ bond length is shorter than the normal bond character on the account of $p\pi –p\pi$ back bonding $\left(1.30\stackrel{\circ }{A}\right)$.

In 1 : 1 complex there is no $p\pi –p\pi$ back bonding and hence, the $B–F$ bond length is longer $\left(1.35\stackrel{\circ }{A}\right)$.
In this complex, the steric number of both $N$ and $B$ are 4.
Hence, both $N$ and $B$ have tetrahedral geometry with $s{p}^3$ hybridisation.