Chemistry · Practical Inorganic chemistry (Qualitative Analysis)

JEE Main 2025 — 7 April, Morning Shift — Question 2

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

Option A:
$\mathrm{X}=\mathrm{NH}_{3}$ and $\mathrm{Y}=\mathrm{K}_{2} \mathrm{Hgl}_{4}+\mathrm{KOH}$
Correct
Option B:
$X=\mathrm{NH}_{3}$ and $Y=\mathrm{HgO}$
Option C:
$\mathrm{X}=\mathrm{HCl}$ and $\mathrm{Y}=\mathrm{NH}_{4} \mathrm{Cl}$
Option D:
$\mathrm{X}=\mathrm{NH}_{4} \mathrm{Cl}$ and $\mathrm{Y}=\mathrm{KOH}$

Answer: A

Step-by-step solution

lodide of Millon's base (brown ppt)

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Chemistry
Chapter: Practical Inorganic chemistry (Qualitative Analysis)
Topic: Other Misc Reactions/ Tests