Question

In the disproportionation reaction of nitrous acid, X is formed along with nitric acid and water. X can also be obtained by the reaction of:

• A. Zn + dil. HNO\(_3\)
• B. Zn + Conc. HNO\(_3\)
• C. Cu + dil. HNO\(_3\)
• D. Cu + Conc. HNO\(_3\)

Hint:

Remember the reactions of metals with dilute and concentrated nitric acid. The products formed depend on the metal and the concentration of the acid.

Answer 1

The Correct Option is C

1. Disproportionation of nitrous acid (HNO\(_2\)):

\( 3 \text{HNO}_2 \rightarrow \text{HNO}_3 + 2 \text{NO} + \text{H}_2\text{O} \)

Thus, X is NO (nitric oxide).

2. Reaction of metals with nitric acid: Zinc (Zn):

\( \text{Zn} + \text{dil. HNO}_3 \rightarrow \text{Zn(NO}_3)_2 + \text{NH}_4\text{NO}_3 + \text{H}_2\text{O} \) (Ammonium nitrate is formed)

\( \text{Zn} + \text{Conc. HNO}_3 \rightarrow \text{Zn(NO}_3)_2 + \text{NO}_2 + \text{H}_2\text{O} \) (Nitrogen dioxide is formed)

Copper (Cu):

\( 3 \text{Cu} + 8 \text{dil. HNO}_3 \rightarrow 3 \text{Cu(NO}_3)_2 + 2 \text{NO} + 4 \text{H}_2\text{O} \) (Nitric oxide is formed)

\( \text{Cu} + 4 \text{Conc. HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2 \text{NO}_2 + 2 \text{H}_2\text{O} \) (Nitrogen dioxide is formed)

From the above reactions, we can see that:

Cu + dil. HNO\(_3\) produces NO (nitric oxide), which is X.

Therefore, X can also be obtained by the reaction of Cu + dil. HNO\(_3\).

Final Answer:
Cu + dil. HNO\(_3\).