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

Hint: 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.