Answer
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Hint: Whenever any atom or species absorbs energy and gets excited, after this process the atom then gets back to the initial state and releases energy and if this energy is having wavelength in visible region, then we can see them as colored.
Complete step by step solution:
We know that all atoms and molecules can absorb and release energy in the form of photons. The amount of energy absorbed or released is the difference between the energies of two quantum states.
- In case of \[KMn{{O}_{4}}\], initially one of the electron in Mn-O bond absorbs energy from photon and gets excited to higher orbital, here it gets excited to vacant d-orbitals on manganese. This process is called excitation.
- Then, the excited electron comes back to the Mn-O bond orbital from vacant d-orbital, so it comes to the orbital having lower energy from the orbital having high energy, hence it releases the excess energy in the form of waves which have its wavelength in visible region.
- The wavelength of the light emitted is in such a region that we can see it purple in color.
- The perception of light is directed by three types of color receptors in the eye which are sensitive to different ranges of wavelength within the band.
- So, we can say that the color property of \[KMn{{O}_{4}}\] arises due to charge transfer transitions. Here, the ligand oxygen atom transfers electrons from its orbitals to vacant orbitals of manganese, hence this type of charge transfer is called LMCT (Ligand to metal charge transfer).
Therefore, we can conclude that the purple color of \[KMn{{O}_{4}}\] is due to charge transfer.
So, correct option is (C) Charge transfer
Note: Remember that color properties of \[KMn{{O}_{4}}\] is not due to d-d transition. However many d-block metal shown colored complexes due to d-d transition but in case of \[KMn{{O}_{4}}\], it does not occur because there is no electrons present in the d-orbital in the compound \[KMn{{O}_{4}}\]. That is the reason why d-d transitions do not occur in \[KMn{{O}_{4}}\].
Complete step by step solution:
We know that all atoms and molecules can absorb and release energy in the form of photons. The amount of energy absorbed or released is the difference between the energies of two quantum states.
- In case of \[KMn{{O}_{4}}\], initially one of the electron in Mn-O bond absorbs energy from photon and gets excited to higher orbital, here it gets excited to vacant d-orbitals on manganese. This process is called excitation.
- Then, the excited electron comes back to the Mn-O bond orbital from vacant d-orbital, so it comes to the orbital having lower energy from the orbital having high energy, hence it releases the excess energy in the form of waves which have its wavelength in visible region.
- The wavelength of the light emitted is in such a region that we can see it purple in color.
- The perception of light is directed by three types of color receptors in the eye which are sensitive to different ranges of wavelength within the band.
- So, we can say that the color property of \[KMn{{O}_{4}}\] arises due to charge transfer transitions. Here, the ligand oxygen atom transfers electrons from its orbitals to vacant orbitals of manganese, hence this type of charge transfer is called LMCT (Ligand to metal charge transfer).
Therefore, we can conclude that the purple color of \[KMn{{O}_{4}}\] is due to charge transfer.
So, correct option is (C) Charge transfer
Note: Remember that color properties of \[KMn{{O}_{4}}\] is not due to d-d transition. However many d-block metal shown colored complexes due to d-d transition but in case of \[KMn{{O}_{4}}\], it does not occur because there is no electrons present in the d-orbital in the compound \[KMn{{O}_{4}}\]. That is the reason why d-d transitions do not occur in \[KMn{{O}_{4}}\].
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