Chemistry · Practical Inorganic chemistry (Qualitative Analysis)

JEE Main 2024 — 29 January, Shift 2 — Question 67

On passing a gas, ' $X$ ', through Nessler's reagent, a brown precipitate is obtained. The gas ' X ' is

Option A:
$\mathrm{H}_{2} \mathrm{~S}$
Option B:
$\mathrm{CO}_{2}$
Option C:
$\mathrm{NH}_{3}$
Correct
Option D:
$\mathrm{Cl}_{2}$

Answer: C

Step-by-step solution

Nessler's reagent is an alkaline solution of potassium tetraiodomercurate(II) (K₂[HgI₄]). It is used as a sensitive test for ammonia (NH₃). When NH₃ gas is passed through Nessler's reagent, it forms a brown precipitate. The brown precipitate is Millon's base iodide, HgO·Hg(NH₂)I. The reaction is: $2K_2[HgI_4] + NH_3 + 3KOH \rightarrow HgO·Hg(NH_2)I + 7KI + 2H_2O$ . H₂S gives a black precipitate with Nessler's reagent. CO₂ does not give a precipitate with Nessler's reagent. Cl₂ may bleach or oxidize the reagent but does not give a characteristic brown precipitate. Thus, gas X is NH₃.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Chemistry
Chapter: Practical Inorganic chemistry (Qualitative Analysis)
Topic: Other Misc Reactions/ Tests

Answer: C

Step-by-step solution

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