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:

• A. X = NH₃ and Y = HgO
• B. X = NH₃ and Y = K₂HgI₄ + KOH
• C. X = NH₄Cl and Y = KOH
• D. X = HCl and Y = NH₄Cl

Hint:

Ammonia (\( \text{NH}_3 \)) reacts with potassium tetraiodomercurate(II) and potassium hydroxide to form a brown precipitate of mercury(I) iodide (\( \text{Hg}_2\text{I}_2 \)).

Answer 1

The Correct Option is B

Step 1: Identify the reaction with sodium hydroxide
Ammonium chloride (\( \text{NH}_4\text{Cl} \)) reacts with sodium hydroxide (\( \text{NaOH} \)) to produce ammonia gas (\( \text{NH}_3 \)):\[\text{NH}_4\text{Cl} + \text{NaOH} \rightarrow \text{NH}_3 + \text{NaCl} + \text{H}_2\text{O}\]
Therefore, gas X is ammonia (\( \text{NH}_3 \)).
Step 2: Analyze reagent Y
When ammonia gas is passed through potassium tetraiodomercurate(II) (\( \text{K}_2\text{HgI}_4 \)) with potassium hydroxide (KOH), a brown precipitate of mercury(I) iodide (\( \text{Hg}_2\text{I}_2 \)) forms:\[\text{NH}_3 + \text{K}_2\text{HgI}_4 + \text{KOH} \rightarrow \text{Hg}_2\text{I}_2 + \text{KCl} + \text{H}_2\text{O}\]The brown precipitate is mercury(I) iodide (\( \text{Hg}_2\text{I}_2 \)).
Thus, reagent Y consists of \( \text{K}_2\text{HgI}_4 + \text{KOH} \).