Question

$Mn{O}_4^{-}$ is of intense pinl color, though Mn is in (+7) oxidation state, It is due to:

• A. Oxygen gives colour to it
• B. Charge transfer when Mn gives its electron to oxygen
• C. Charge transfer when oxygen gives its electron to Mn making it Mn (+VI)
• D. None is correct

Answer 1

The correct option is C Charge transfer when oxygen gives its electron to Mn making it Mn (+VI)

It is because of charge transfer, when oxygen gives its electron to $Mn$ making it $Mn(+VI)$, in $+VI$ oxidation state it has one unpair electron, so it is colored due to presence of unpair electron. (The color in $KMn{O}_4$ arises from an electronic transition, but it is actually not a d-d transition, since the $Mn$ in this compound has no d electrons. It arises from a charge transfer reaction within the molecule, in which photons promote an electron from the highest energy molecular orbital in one of the $Mn-O$ bonds to an empty d orbital on the manganese. This promotion is equivalent to the energy of a yellow photon, so yellow light is absorbed leaving us to see purple, the complementary colour.)