JEE Chemistry 2025 Question with Solution

Given: A salt treated with sodium hydroxide solution gives gas X. Passing gas X through reagent Y gives a brown coloured precipitate.

Find: The correct identities of X and Y.

Step 1: Understand the reaction with sodium hydroxide When an ammonium salt such as ammonium chloride reacts with sodium hydroxide, ammonia gas is evolved.

Thus, gas X is $\text{NH}_3$.

Step 2: Analyze reagent Y Ammonia reacts with potassium tetraiodomercurate(II) in the presence of potassium hydroxide to give a brown precipitate. The solution states that the brown precipitate formed is mercury(I) iodide.

Therefore, reagent Y is $\text{K}_2\text{HgI}_4 + \text{KOH}$.

Therefore, the correct option is B.

Given: Gas X is produced when a salt is treated with sodium hydroxide, and this gas gives a brown coloured precipitate with reagent Y.

Find: Which pair of X and Y matches the observation.

Ammonium salts characteristically liberate ammonia on treatment with a strong base such as sodium hydroxide. Hence the gas released is ammonia, $\text{NH}_3$.

The confirmatory test mentioned is Nessler's reagent, represented here as $\text{K}_2\text{HgI}_4 + \text{KOH}$. Ammonia gives a brown coloured precipitate with this reagent.

So,

• X = $\text{NH}_3$
• Y = $\text{K}_2\text{HgI}_4 + \text{KOH}$

Hence, the correct option is B.