Transition metals form coloured ions or compounds due to partially filled d orbitals. In the presence of solvent molecules (in solutions), ligands (in complexes) or counter ions (in crystals), the d orbitals split into two sets. An electron in a transition metal ion that occupies the set of d orbitals of lower energy can be excited to the set of higher energy by absorbing energy from visible light. Since the energy difference (ΔE) between the two sets of d orbitals is small, only light of the visible region is absorbed during this excitation. The colour of a transition metal ion is due to this d–d excitation (d–d transition) of the electron. During d–d excitation the electron absorbs one colour of visible light, and the ion appears in the complementary colour of the absorbed light. Complementary colours can be identified from the colour wheel shown below (complementary colours lie opposite each other).
The number of electrons undergoing d–d transition and the energy difference between the two sets of orbitals decide the colour.
Acidified potassium permanganate solution is added to a green solution of iron(II) sulphate. Which of the following gives the species formed from the iron(II) ion, with its colour in aqueous solution, and the species formed from the permanganate ion, respectively?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \ce{Fe^3+} \) (yellow) and \( \ce{Mn^2+} \)
Permanganate is a strong oxidising agent. In acid solution manganese is reduced from +7 to +2, and the iron(II) ion is oxidised to iron(III):
\( \displaystyle \ce{MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O} \)
\( \displaystyle \ce{Fe^2+ -> Fe^3+ + e^-} \)
Five iron(II) ions supply the five electrons taken up by one permanganate ion, which gives the full reaction:
\( \displaystyle \ce{5Fe^2+ + MnO4^- + 8H+ -> Mn^2+ + 4H2O + 5Fe^3+} \)
The colour of the iron solution changes because the d-electron count changes: green Fe2+ (3d6) becomes yellow Fe3+ (3d5).
Why the other options are incorrect
Option (1) \( \ce{Fe^3+} \) (yellow) and \( \ce{MnO2} \) — Incorrect. Permanganate gives MnO2 in neutral or faintly alkaline solution, as in the oxidation of iodide to iodate. In acid solution it is reduced to Mn2+.
Option (2) \( \ce{Fe^3+} \) (green) and \( \ce{Mn^2+} \) — Incorrect. Green is the colour of the starting Fe2+ ion. Aquated Fe3+ is yellow.
Option (4) \( \ce{FeO4^2-} \) (ferrate) and \( \ce{Mn^2+} \) — Incorrect. Ferrates(VI) are formed in alkaline media, and even there they readily decompose to Fe2O3 and O2. Acidified permanganate takes iron(II) only to iron(III).
NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “1. In acid solutions:” … “(b) Fe2+ ion (green) is converted to Fe3+ (yellow): 5Fe2+ + MnO4– + 8H+ → Mn2+ + 4H2O + 5Fe3+”