A brown ring is formed in the ring test for NO_3^- ion. It is due to…

The correct answer is Option (2) → \([Fe(H_2O)_5(NO)]^{2+}\)

In the brown ring test, \( \ce{Fe^{2+}} \) reduces the nitrate to nitric oxide, and the NO then combines with more \( \ce{Fe^{2+}} \) to give the brown complex at the junction of the two layers:

    \( \ce{NO3^{-}} \) + \( \ce{3Fe^{2+}} \) + \( \ce{4H^{+}} \) → NO + \( \ce{3Fe^{3+}} \) + \( \ce{2H2O} \)

    \( \ce{[Fe(H2O)6]^{2+}} \) + NO → [Fe(H₂O)₅(NO)]²⁺ (brown) + \( \ce{H2O} \)

Why the other options are incorrect

• Option (1) [Fe(H₂O)₄(NO)₂]²⁺ — Incorrect. Only one NO molecule enters the complex, in place of one water molecule, so the brown complex has five water molecules and one NO.

• Option (3) FeSO₄.NO₂ — Incorrect. Fe²⁺ reduces the nitrate to NO (nitric oxide), not to \( \ce{NO2} \); the brown species is the complex ion [Fe(H₂O)₅(NO)]²⁺.

• Option (4) FeSO₄.HNO₃ — Incorrect. Nitric acid does not bind to iron; the nitrate is first reduced to NO by Fe²⁺, and it is this NO that bonds to iron.

Note 1: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Brown Ring Test – “The familiar brown ring test for nitrates depends on the ability of Fe²⁺ to reduce nitrates to nitric oxide, which reacts with Fe²⁺ to form a brown coloured complex. The test is usually carried out by adding dilute ferrous sulphate solution to an aqueous solution containing nitrate ion, and then carefully adding concentrated sulphuric acid along the sides of the test tube. A brown ring at the interface between the solution and sulphuric acid layers indicate the presence of nitrate ion in solution.”

Note 2: The same/similar question has been asked in different years by NTA as per the details given below: