Question

Which of the following is correct representation of the reaction when $B{F}_3$ reacts with ammonia?

• A. Both are correct
• B. $\left(i\right)$ is correct $\left(ii\right)$ is incorrect
• C. $\left(ii\right)$ is correct $\left(i\right)$ is incorrect
• D. No reaction takes place

Answer 1

The correct option is B $\left(i\right)$ is correct $\left(ii\right)$ is incorrect

As $B{F}_3$ is a Lewis acid and $N{H}_3$ is a Lewis base,so, $N{H}_3$ donates its lone pair to $B{F}_3$ and forms a dative bond.
In $\left(i\right)$ $N{H}_3$ donates its lone pair to $B{F}_3$, whereas in $\left(ii\right)$ $B{F}_3$ donates its electron to $N{H}_3$.
So, $\left(i\right)$ is correct.