MnO 4-is of intense pink colour, though Mn is in +7 oxidation state. It is due to

Question

MnO 4-is of intense pink colour, 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) presence of unpaired electrons

Tardigrade Admin · Accepted Answer

Although in MnO 4-, the oxidation state of Mn is +7, ie, it contains no unpaired electrons, yet it is coloured. This is because oxygen gives its one electron to Mn and converts it into Mn ( VI ), which is coloured. Thus, the colour of MnO 4-is due to charge transfer.

Q. $M n O_{4}^{-}$ is of intense pink colour, though $M n$ is in $+ 7$ oxidation state. It is due to

oxygen gives colour to it

charge transfer when Mn gives its electron to oxygen

charge transfer when oxygen gives its electron to Mn making it Mn ( +VI)

presence of unpaired electrons

Although in $M n O_{4}^{-}$ , the oxidation state of $M n$ is $+ 7$ , ie, it contains no unpaired electrons, yet it is coloured. This is because oxygen gives its one electron to $M n$ and converts it into $M n (V I)$ , which is coloured. Thus, the colour of $M n O_{4}^{-}$ is due to charge transfer.

Q. MnO4−​is of intense pink colour, though Mn is in +7 oxidation state. It is due to

oxygen gives colour to it

charge transfer when Mn gives its electron to oxygen

charge transfer when oxygen gives its electron to Mn making it Mn ( +VI)

presence of unpaired electrons

Although in MnO4−​, the oxidation state of Mn is +7, ie, it contains no unpaired electrons, yet it is coloured. This is because oxygen gives its one electron to Mn and converts it into Mn(VI), which is coloured. Thus, the colour of MnO4−​is due to charge transfer.

Q. $M n O_{4}^{-}$ is of intense pink colour, though $M n$ is in $+ 7$ oxidation state. It is due to