Question

Which of the following is the correct representation when $B{F_3}$ reacts with ammonia?

1. 
2. 
   (A) 1 is incorrect and 2 is correct
   (B) 1 is correct and 2 is incorrect
   (C) Both 1 and 2 are correct
   (D) Both 1 and 2 are incorrect

Answer 1

To answer the question we must know about the concept of Lewis acid and Lewis base. Both $B{F_3}$and $N{H_3}$ are Lewis acid and Lewis base respectively. And we have to know the octet rule and the concept of stable compounds. A detailed discussion is shown below.

Complete step-by-step answer: Lewis acid: In the Lewis hypothesis of acid-base reactions, bases give sets of electrons and acids acknowledge sets of electrons. A Lewis acid is any substance that can acknowledge a couple of nonbonding electrons. As such, a Lewis acid is an electron-pair acceptor.
One favorable position of the Lewis hypothesis is the manner in which it supplements the model of oxidation-reduction reactions. Oxidation-reduction reactions include an exchange of electrons starting with one particle then onto the next, with a net change in the oxidation number of at least one atom.
Lewis base: A Lewis base is a substance which can donate at least a pair of nonbonding electrons. That means it can be called an electron pair donor.

In the above question, two molecules are $N{H_3}$and$B{F_3}$. In $N{H_3}$there are $5$ electrons in the outer shell of $N$ and $3$electrons are involved in the bond formation with $H$ atom and rest of the electrons remains as lone pair that means nonbonding electron-pair. So, we can conclude $N{H_3}$ is a Lewis base.
And in $B{F_3}$ there are 3 electrons in the outer shell of $B$and all are involved in bond formation so there is no nonbonding electron left in $B{F_3}$and we can consider $B{F_3}$ as a Lewis acid.
So the correct representation for the reaction of $B{F_3}$ with $N{H_3}$ will be:

Hence the correct option is (A).

Note: In the reaction it is basically an acid base reaction. When $B{F_3}$ reacts with $N{H_3}$ an adduct is formed shown in the above diagram. And in that adduct $N$ will donate electrons to $B$ centre.