
Chlorine reacts with excess ammonia to form
(A) $N{{H}_{4}}Cl$
(B) ${{N}_{2}}+HCl$
(C) ${{N}_{2}}+N{{H}_{4}}Cl$
(D) ${{N}_{2}}+HC{{l}_{3}}$
Answer
542.4k+ views
Hint: When a halogen reacts with any electron rich species which are capable of donating its electrons, the halogen gets reduced, as they need only one electron to attain a stable configuration of the outermost orbital.
In this process the electron rich species would act as a Lewis base, and the halogen would act as a Lewis acid.
Complete step-by-step answer:In the given reaction mentioned in the question, we get that the chlorine is present along with excess of ammonia. In order to predict the products of the following reaction we need to know the chemical formulas of these molecules. So the chlorine is represented by the symbol $Cl$ and ammonia is represented by $N{{H}_{3}}$. Now, in the reaction, as we know that chlorine is a halogen and needs one electron in order to complete its outermost electronic configuration hence it has a tendency of acting as a Lewis acid. By the definition of Lewis acid, the substances which have a tendency to accept electrons, are known as the Lewis acids. Similarly in case of Lewis bases, the substances which have a tendency to donate its electrons, are called Lewis bases. So in case of ammonia we know that it has a lone pair of electrons, hence it could act as a base.
So the chlorine will get reduced in the reaction, and the nitrogen will get oxidised, giving the equation,
$C{{l}_{2}}+N{{H}_{3}}\to {{N}_{2}}+2C{{l}^{-}}$
And now, it is given in the question that the ammonia is present in excess, so this excess ammonia will combine with the chlorine ion to form ammonium chloride.
$C{{l}_{2}}+N{{H}_{3}}\to {{N}_{2}}+N{{H}_{4}}Cl$
Hence the product formed are nitrogen gas and ammonium chloride. Now if we consider all the options, only option C matches this combination of products so it would be the correct option.
So the correct answer would be option (C).
Note: The reaction of chlorine with excess of ammonia results in the formation of ammonium chloride along with nitrogen gas.
The chlorine acts as a Lewis acid, because of its tendency to accept electrons and the ammonium acts as a Lewis base, because of its tendency to donate the electrons.
In this process the electron rich species would act as a Lewis base, and the halogen would act as a Lewis acid.
Complete step-by-step answer:In the given reaction mentioned in the question, we get that the chlorine is present along with excess of ammonia. In order to predict the products of the following reaction we need to know the chemical formulas of these molecules. So the chlorine is represented by the symbol $Cl$ and ammonia is represented by $N{{H}_{3}}$. Now, in the reaction, as we know that chlorine is a halogen and needs one electron in order to complete its outermost electronic configuration hence it has a tendency of acting as a Lewis acid. By the definition of Lewis acid, the substances which have a tendency to accept electrons, are known as the Lewis acids. Similarly in case of Lewis bases, the substances which have a tendency to donate its electrons, are called Lewis bases. So in case of ammonia we know that it has a lone pair of electrons, hence it could act as a base.
So the chlorine will get reduced in the reaction, and the nitrogen will get oxidised, giving the equation,
$C{{l}_{2}}+N{{H}_{3}}\to {{N}_{2}}+2C{{l}^{-}}$
And now, it is given in the question that the ammonia is present in excess, so this excess ammonia will combine with the chlorine ion to form ammonium chloride.
$C{{l}_{2}}+N{{H}_{3}}\to {{N}_{2}}+N{{H}_{4}}Cl$
Hence the product formed are nitrogen gas and ammonium chloride. Now if we consider all the options, only option C matches this combination of products so it would be the correct option.
So the correct answer would be option (C).
Note: The reaction of chlorine with excess of ammonia results in the formation of ammonium chloride along with nitrogen gas.
The chlorine acts as a Lewis acid, because of its tendency to accept electrons and the ammonium acts as a Lewis base, because of its tendency to donate the electrons.
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