Chemistry · Practical Inorganic chemistry (Qualitative Analysis)

JEE Main 2024 — 4 April, Shift 2 — Question 61

When $\mathrm{MnO}_{2}$ and $\mathrm{H}_{2} \mathrm{SO}_{4}$ is added to a salt (A), the greenish yellow gas liberated as salt (A) is :

Option A:
NaBr
Option B:
$\mathrm{CaI}_{2}$
Option C:
$\mathrm{KNO}_{3}$
Option D:
$\mathrm{NH}_{4} \mathrm{Cl}$
Correct

Answer: D

Step-by-step solution

$2 \mathrm{NH}_{4} \mathrm{Cl}+\mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \xrightarrow{\Delta} \mathrm{MnSO}_{4}$ $+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}+2 \mathrm{H}_{2} \mathrm{O}+\underset{\substack{\text { grenish } \text { yellow } \text { gas }}}{\mathrm{Cl}_{2} \uparrow}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Chemistry
Chapter: Practical Inorganic chemistry (Qualitative Analysis)
Topic: Identification of Anions