An inorganic salt taken in a test tube was mixed with freshly prepared ferrous sulphate solution. Concentrated sulphuric acid was then added along the walls of the test tube. Predict the formula of the compound formed :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → [Fe(H2O)5NO]2+
The procedure described is the brown ring test for nitrates. In acid, \( \ce{Fe^{2+}} \) reduces the nitrate ion to nitric oxide, and NO then replaces one water ligand of the hexaaquairon(II) ion:
\( \displaystyle \ce{NO3^- + 3Fe^2+ + 4H^+ -> NO + 3Fe^3+ + 2H2O} \)
\( \displaystyle \ce{[Fe(H2O)6]^2+ + NO -> [Fe(H2O)5(NO)]^2+ + H2O} \)
The brown complex, pentaaquanitrosyliron(II), forms as a ring where the heavy concentrated \( \ce{H2SO4} \) layer meets the aqueous layer. It has one NO and five H2O ligands and a charge of 2+.
Why the other options are incorrect
Option (1) [Fe(NO)4(SO4)2]+ — Incorrect. The complex has only one NO ligand and no sulphate ligand; sulphate stays in solution as the counter-ion.
Option (2) [Fe(NO)5H2O]+ — Incorrect. The ratio is reversed; NO replaces only one of the six water molecules, and the charge is 2+, not 1+.
Option (3) [Fe(NO)2(H2O)2(SO4)2]+ — Incorrect. There is no such brown-ring complex; it has two NO groups and sulphate ligands, neither of which is present.
Note: 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.”
\( \displaystyle \ce{NO3^- + 3Fe^2+ + 4H^+ -> NO + 3Fe^3+ + 2H2O} \)
\( \displaystyle \ce{[Fe(H2O)6]^2+ + NO ->} \underset{\text{brown}}{\ce{[Fe(H2O)5(NO)]^2+}} + \ce{H2O} \)