Question:
When a salt is treated with sodium hydroxide solution, it gives gas X. On passing gas X through reagent Y, a brown coloured precipitate is formed. X and Y respectively, are:
When a salt is treated with sodium hydroxide solution, it gives gas X. On passing gas X through reagent Y, a brown coloured precipitate is formed. X and Y respectively, are:
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Ammonia (\( \text{NH}_3 \)) reacts with potassium tetraiodomercurate(II) and potassium hydroxide to form a brown precipitate of mercury(I) iodide (\( \text{Hg}_2\text{I}_2 \)).
Updated On: Jan 14, 2026
- X = NH₃ and Y = HgO
- X = NH₃ and Y = K₂HgI₄ + KOH
- X = NH₄Cl and Y = KOH
- X = HCl and Y = NH₄Cl
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The Correct Option is B
Approach Solution - 1
Step 1: Understand the reaction with sodium hydroxide
When a salt, such as ammonium chloride (\( \text{NH}_4\text{Cl} \)), reacts with sodium hydroxide (\( \text{NaOH} \)), ammonia gas (\( \text{NH}_3 \)) is evolved: \[ \text{NH}_4\text{Cl} + \text{NaOH} \rightarrow \text{NH}_3 + \text{NaCl} + \text{H}_2\text{O} \]
Thus, gas X is ammonia (\( \text{NH}_3 \)).
Step 2: Analyze the reagent Y
When ammonia gas is passed through potassium tetraiodomercurate(II) (\( \text{K}_2\text{HgI}_4 \)) in the presence of potassium hydroxide (KOH), a brown precipitate of mercury(I) iodide (\( \text{Hg}_2\text{I}_2 \)) is formed: \[ \text{NH}_3 + \text{K}_2\text{HgI}_4 + \text{KOH} \rightarrow \text{Hg}_2\text{I}_2 + \text{KCl} + \text{H}_2\text{O} \] The brown precipitate is mercury(I) iodide (\( \text{Hg}_2\text{I}_2 \)).
Thus, reagent Y is \( \text{K}_2\text{HgI}_4 + \text{KOH} \).
When a salt, such as ammonium chloride (\( \text{NH}_4\text{Cl} \)), reacts with sodium hydroxide (\( \text{NaOH} \)), ammonia gas (\( \text{NH}_3 \)) is evolved: \[ \text{NH}_4\text{Cl} + \text{NaOH} \rightarrow \text{NH}_3 + \text{NaCl} + \text{H}_2\text{O} \]
Thus, gas X is ammonia (\( \text{NH}_3 \)).
Step 2: Analyze the reagent Y
When ammonia gas is passed through potassium tetraiodomercurate(II) (\( \text{K}_2\text{HgI}_4 \)) in the presence of potassium hydroxide (KOH), a brown precipitate of mercury(I) iodide (\( \text{Hg}_2\text{I}_2 \)) is formed: \[ \text{NH}_3 + \text{K}_2\text{HgI}_4 + \text{KOH} \rightarrow \text{Hg}_2\text{I}_2 + \text{KCl} + \text{H}_2\text{O} \] The brown precipitate is mercury(I) iodide (\( \text{Hg}_2\text{I}_2 \)).
Thus, reagent Y is \( \text{K}_2\text{HgI}_4 + \text{KOH} \).
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Approach Solution -2
Step 1:
The reaction described in the question involves the formation of gas X upon treatment with sodium hydroxide. This is typical of ammonia (NH₃), which is released when ammonium salts (like NH₄Cl) react with sodium hydroxide.
Step 2:
Now, gas X (NH₃) reacts with reagent Y. The formation of a brown colored precipitate suggests the reaction with a mercury(I) compound, which is typically represented as K₂HgI₄ (also known as potassium tetraiodomercurate(I)). When ammonia is passed through K₂HgI₄ in the presence of KOH, a brown precipitate of mercury(I) iodide (Hg₂I₂) is formed.
Step 3:
Thus, the gas X is ammonia (NH₃), and the reagent Y is potassium tetraiodomercurate(I) (K₂HgI₄), which, when treated with ammonia, forms a brown precipitate of Hg₂I₂.
Final Answer:
\[ \boxed{X = \text{NH}_3 \text{ and } Y = \text{K}_2\text{HgI}_4 + \text{KOH}} \]
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