In the brown ring test, nitrate is first reduced to NO by Fe^2+ in acid, and the NO then combines with further Fe^2+ to form [Fe(H2O)5NO]^2+. For each mole of nitrate converted fully into the ring complex, the moles of Fe^2+ consumed and the moles of electrons gained by nitrogen are respectively:

JEE Advanced levelSingle correct optionDifficulty 3 of 3qa155

In the brown ring test, nitrate is first reduced to NO by $Fe X^{2 +}$ in acid, and the NO then combines with further $Fe X^{2 +}$ to form $[Fe (H X_{2} O) X_{5} NO] X^{2 +}$ . For each mole of nitrate converted fully into the ring complex, the moles of $Fe X^{2 +}$ consumed and the moles of electrons gained by nitrogen are respectively:

3 Fe X^{2 +} + NO X_{3} X^{-} + 4 H X^{+} 3 Fe X^{3 +} + NO + 2 H X_{2} O

uses 3 mol

Fe X^{2 +}

and N goes from +5 to +2 (3 electrons). The NO then needs one more

Fe X^{2 +}

to form the complex, so the total is 4 mol

Fe X^{2 +}

per mole of nitrate.