Chemistry Q63 – JEE-MAIN 2025

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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

Clue 1 of 3: Reaction of salts with bases

Consider what type of gas is typically evolved when a salt reacts with a strong base like sodium hydroxide.

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Ninja Strategy•Identify Gas and Specific Reagent

First, identify that the reaction of a salt with NaOH producing a gas implies an ammonium salt yielding ammonia. Then, recall the specific reagent (Nessler's) that gives a brown precipitate with ammonia.

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Step 1: Identify Gas X✦ Active

When an ammonium salt reacts with a strong base like sodium hydroxide, ammonia gas ( $N H_{3}$ ) is evolved. For example, ${NH}_{4} Cl + NaOH \to NaCl + H_{2} O + {NH}_{3}$ . Thus, gas X is ammonia ( $N H_{3}$ ). This eliminates options 1 and 2.

Step 2: Identify Reagent Y for Ammonia Detection○ Expand

Ammonia gas ( $N H_{3}$ ) gives a characteristic brown precipitate when passed through Nessler's reagent. This is a standard qualitative test for ammonia.

Step 3: Determine the Composition of Nessler's Reagent○ Expand

Nessler's reagent is an alkaline solution of potassium tetraiodomercurate(II), which is $K_{2} {HgI}_{4}$ in $KOH$ . The reaction with ammonia forms a brown precipitate of Millon's base iodide ( ${I-Hg-O-Hg-NH}_{2}$ ). Therefore, X is ${NH}_{3}$ and Y is $K_{2} {HgI}_{4} + KOH$ , which corresponds to option 4. Option 3 is incorrect because $HgO$ alone is not Nessler's reagent.

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Chemistry Question 63 – JEE-MAIN 2025

Clue 1 of 3: Reaction of salts with bases

🧠 Full Solution Path