Chemistry · Practical Inorganic chemistry (Qualitative Analysis)

JEE Main 2026 — 6 April, Evening Shift — Question 59

Treatment of a gas ' X ' with a freshly prepared ferrous sulphate solution gives a compound ' Y ' as a brown ring. The compounds X and Y are.

Option A:
NO and $[\mathrm{Fe}(\mathrm{NO})] \mathrm{SO}_{4}$
Correct
Option B:
$\mathrm{NO}_{2}$ and $\left[\mathrm{Fe}\left(\mathrm{NO}_{2}\right)\right] \mathrm{SO}_{4}$
Option C:
$\mathrm{N}_{2} \mathrm{O}$ and $\left[\mathrm{Fe}\left(\mathrm{N}_{2} \mathrm{O}\right)\right] \mathrm{SO}_{4}$
Option D:
$\mathrm{N}_{2} \mathrm{O}_{4}$ and $\left[\mathrm{Fe}\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\right] \mathrm{SO}_{4}$

Answer: A

$\mathrm{NO}+\mathrm{FeSO}_{4} \xrightarrow{\text { dil. } \mathrm{H}_{2} \mathrm{SO}_{4}}[\mathrm{Fe}(\mathrm{NO})] \mathrm{SO}_{4}$

Answer key and solution verified before publishing.

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