Question

The colour of ${\text{KMn}}{{\text{O}}_{\text{4}}}$ is due to:
A) ${\text{M}} \to {\text{L}}$ charge transfer transition
B) ${\text{d}} - {\text{d}}$ transition
C) ${\text{L}} \to {\text{M}}$ charge transfer transition
D) $\sigma - {\sigma ^*}$ transition

Answer 1

We know that ${\text{KMn}}{{\text{O}}_{\text{4}}}$ is known as potassium permanganate. Potassium is deep purple in colour. In potassium permanganate, the central metal atom is manganese i.e. ${\text{Mn}}$ and the ligands are oxygen i.e. ${{\text{O}}^{2 - }}$. We can see the coloured species when the atoms release energy and return to initial state and the energy has wavelength in the visible region.

Complete solution:
We know that the atoms absorb energy in the form of photons and get excited. When the atoms release energy and return to the ground state and the energy has wavelength in a visible region we can see coloured compounds.
In potassium permanganate i.e. ${\text{KMn}}{{\text{O}}_{\text{4}}}$, the central metal atom is manganese i.e. ${\text{Mn}}$ and the ligands are oxygen i.e. ${{\text{O}}^{2 - }}$. One electron in ${\text{Mn}} - {\text{O}}$ bond absorbs energy and gets excited to a higher orbital i.e. to the vacant d-orbital of manganese.
The excited electron then comes back to the ${\text{Mn}} - {\text{O}}$ bond orbital i.e. the electron comes back to the lower energy. During this, energy is released in the form of a wave which has its wavelength in a visible region.
The wavelength of the emitted energy or light is in the region where purple colour is visible. Thus the colour in ${\text{KMn}}{{\text{O}}_{\text{4}}}$ is due to the charge transfer transition. The ligand oxygen atom transfers electrons from its orbitals to vacant d-orbitals of manganese. Thus, the charge transfer is from ligand to metal. This is known as LMCT.
Thus, the colour of ${\text{KMn}}{{\text{O}}_{\text{4}}}$ is due to ${\text{L}} \to {\text{M}}$ charge transfer transition.

Thus, the correct option is (C) ${\text{L}} \to {\text{M}}$ charge transfer transition.

Note: Remember the colour of ${\text{KMn}}{{\text{O}}_{\text{4}}}$ is not due to d-d transition. Many d-block metals show colour due to d-d transition. This is not possible for ${\text{KMn}}{{\text{O}}_{\text{4}}}$ because there are no electrons in the d-orbital of ${\text{KMn}}{{\text{O}}_{\text{4}}}$. Thus, ${\text{KMn}}{{\text{O}}_{\text{4}}}$ shows colour due to ligand to metal charge transfer transition.