
When zinc reacts with very dilute nitric acid it produces:
(A) $\text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{3}}$
(B) $\text{NO}$
(C) $\text{N}{{\text{O}}_{2}}$
(D) ${{\text{H}}_{2}}$
Answer
530.8k+ views
Hint: Zinc is a d-block element with ${{\text{d}}^{10}}$ configuration. It exist in its $+2$ oxidative state as $\left( \text{Z}{{\text{n}}^{+2}} \right)$. $\text{HN}{{\text{O}}_{3}}$ is said to be strong oxidising reagent. Nitric acid is concentrated or dilute depending on its concentration. Moreover, it depends on the reactivity series too. In reaction with zinc, it gives nitrogenous compounds of ammonia.
Complete step by step solution:
Concentrated nitric acid is a strong oxidising agent and it attacks most metals. The products of oxidation depend upon the concentration of the acid, temperature and the nature of the compounds.
The reaction of zinc $\left( \text{Zn} \right)$ with very dilute nitric acid $\left( \text{HN}{{\text{O}}_{3}} \right)$ to produce ammonium nitrate $\left( \text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{3}} \right)$, zinc nitrate $\text{Zn}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}$ and water $\left( {{\text{H}}_{2}}\text{O} \right)$. The balanced chemical reaction is $4\text{Zn}+\text{10HN}{{\text{O}}_{3}}\to 4\text{Zn}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}+\text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{3}}+3{{\text{H}}_{2}}\text{O}$.
The oxidation state of nitrogen remains unchanged as it is in nitrate form only with +5 as its oxidation state.
In this reaction, the hydrogen atom of nitric acid is replaced by zinc. It is due to reactivity or electrochemical series. The oxidation potential of zinc is higher than that of hydrogen. So, the oxidation state of zinc changes from 0 to +2.
When zinc reacts with very dilute nitric acid it produces $\text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{3}}$.
The correct answer is option (A).
Additional Information:
Applications of ammonium nitrate:
(1) Ammonium nitrate is used as a fertilizer since it provides half of the nitrogen in the form of nitrate and half in the form of ammonium. Its advantage over urea is that it is more stable moreover, does not rapidly lose nitrogen to the atmosphere.
(2) Ammonium nitrate is an ingredient in explosives. Like, Ammonal is made of TNT and ammonium nitrate.
Note: The reaction of zinc with nitric acid in different conditions gives out different products. Like,
(1) Warm and dilute nitric acid: $4\text{Zn}+\text{10HN}{{\text{O}}_{3}}\to 4\text{Zn}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}+{{\text{N}}_{2}}\text{O}+\text{5}{{\text{H}}_{2}}\text{O}$
(2) Cold and very dilute nitric acid: $4\text{Zn}+\text{10HN}{{\text{O}}_{3}}\to 4\text{Zn}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}+\text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{3}}+3{{\text{H}}_{2}}\text{O}$
(3) Hot and concentrated nitric acid: $\text{Zn}+4\text{HN}{{\text{O}}_{3}}\to \text{Zn}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}+2\text{N}{{\text{O}}_{2}}+2{{\text{H}}_{2}}\text{O}$
Complete step by step solution:
Concentrated nitric acid is a strong oxidising agent and it attacks most metals. The products of oxidation depend upon the concentration of the acid, temperature and the nature of the compounds.
The reaction of zinc $\left( \text{Zn} \right)$ with very dilute nitric acid $\left( \text{HN}{{\text{O}}_{3}} \right)$ to produce ammonium nitrate $\left( \text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{3}} \right)$, zinc nitrate $\text{Zn}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}$ and water $\left( {{\text{H}}_{2}}\text{O} \right)$. The balanced chemical reaction is $4\text{Zn}+\text{10HN}{{\text{O}}_{3}}\to 4\text{Zn}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}+\text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{3}}+3{{\text{H}}_{2}}\text{O}$.
The oxidation state of nitrogen remains unchanged as it is in nitrate form only with +5 as its oxidation state.
In this reaction, the hydrogen atom of nitric acid is replaced by zinc. It is due to reactivity or electrochemical series. The oxidation potential of zinc is higher than that of hydrogen. So, the oxidation state of zinc changes from 0 to +2.
When zinc reacts with very dilute nitric acid it produces $\text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{3}}$.
The correct answer is option (A).
Additional Information:
Applications of ammonium nitrate:
(1) Ammonium nitrate is used as a fertilizer since it provides half of the nitrogen in the form of nitrate and half in the form of ammonium. Its advantage over urea is that it is more stable moreover, does not rapidly lose nitrogen to the atmosphere.
(2) Ammonium nitrate is an ingredient in explosives. Like, Ammonal is made of TNT and ammonium nitrate.
Note: The reaction of zinc with nitric acid in different conditions gives out different products. Like,
(1) Warm and dilute nitric acid: $4\text{Zn}+\text{10HN}{{\text{O}}_{3}}\to 4\text{Zn}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}+{{\text{N}}_{2}}\text{O}+\text{5}{{\text{H}}_{2}}\text{O}$
(2) Cold and very dilute nitric acid: $4\text{Zn}+\text{10HN}{{\text{O}}_{3}}\to 4\text{Zn}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}+\text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{3}}+3{{\text{H}}_{2}}\text{O}$
(3) Hot and concentrated nitric acid: $\text{Zn}+4\text{HN}{{\text{O}}_{3}}\to \text{Zn}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}+2\text{N}{{\text{O}}_{2}}+2{{\text{H}}_{2}}\text{O}$
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