Presence of nitrate ions can be easily detected in a solution by adding to it a freshly prepared \(FeSO_4\) solution. Subsequently few drops of concentrated sulphuric acid are added along the walls of the test-tube. This results in the formation of a brown ring at the interface (Brown Ring Test) having the formula \([Fe(H_2O)_5(NO)]^{2+}\).
The oxidation state of Fe in this complex is _____.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (+1)
In the brown ring test \( \ce{Fe^2+} \) reduces nitrate to nitric oxide, and NO then combines with excess \( \ce{Fe^2+} \):
\( \displaystyle \ce{NO3^- + 3Fe^2+ + 4H+ -> NO + 3Fe^3+ + 2H2O} \)
\( \displaystyle \ce{[Fe(H2O)6]^2+ + NO -> [Fe(H2O)5(NO)]^2+ + H2O} \)
On binding, NO gives its odd electron to iron and is held as the nitrosonium ion, NO+. With \( \ce{H2O} \) neutral, the oxidation state x of Fe follows from the charge of the complex:
\( \displaystyle x + 5(0) + (+1) = +2 \;\Rightarrow\; x = +1 \)
The measured magnetic moment of the brown complex (about 3.9 BM, three unpaired electrons) fits Fe(I), 3d7, in high spin.
Why the other options are incorrect
Option (2) (+3) — Incorrect. +3 would need NO to carry a −1 charge (\( \ce{NO^{-}} \)), which does not fit the electron transfer from NO to iron.
Option (3) (+2) — Incorrect. +2 treats NO as a neutral ligand; in the complex NO is bound as \( \ce{NO^{+}} \), which lowers iron to +1.
Option (4) zero — Incorrect. Zero would need a total ligand charge of +2, but five neutral water molecules and one \( \ce{NO^{+}} \) give only +1.
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. NO3– + 3Fe2+ + 4H+ → NO + 3Fe3+ + 2H2O; [Fe(H2O)6]2+ + NO → [Fe(H2O)5(NO)]2+ + H2O”