Colour of Manganate ion in KMnO_4 is due to:

The correct answer is Option (1) → \(d-d\) transition

Note 1: The given answer is as per the official NTA Answer Key. However, the correct answer should be Option (2) Charge transfer spectra, as explained below.

The ion present in \( \ce{KMnO4} \) is the permanganate ion, \( \ce{MnO4-} \). The oxidation state of manganese is: +1 + x + 4(−2) = 0, so x = +7. Mn (Z = 25) is [Ar] 3d⁵4s²; removing all seven outer electrons gives Mn(VII), [Ar] 3d⁰. Permanganate is diamagnetic because it has no unpaired electron, and a d–d transition needs at least one d electron to move between d orbitals.

The intense purple colour arises from charge transfer: on absorbing visible light, an electron moves from an oxygen-based orbital into an empty d orbital of Mn(VII). Such transitions are very intense, which is why even dilute \( \ce{KMnO4} \) solution is deep purple.

• Option (1) d–d transition — The answer given by NTA, but incorrect. Mn(VII) is 3d⁰, so there is no d electron for a d–d jump.

• Option (2) Charge transfer spectra — Correct. Electron transfer from the oxide ligands to the empty d orbitals of Mn(VII) gives the colour.

• Option (3) f–f transition — Incorrect. f–f transitions occur in lanthanoid and actinoid ions with partly filled f orbitals; manganese has no f electrons.

• Option (4) its diamagnetic nature — Incorrect. Diamagnetism is a magnetic property; it does not cause colour (many diamagnetic compounds, such as \( \ce{Zn^2+} \) salts, are colourless).

Note 2: It seems that there is a mistake in the original NTA paper, which prints “Manganate ion in KMnO₄” when it should read “permanganate ion”.

NCERT (2026-27 edition):
Oxides and Oxoanions of Metals – …