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The VO43-, CrO42- and MnO4- ions exhibit intense ligand to metal charge transfer transition. The wavelengths of this transition follow the order

  • A.\(\mathrm{CrO_{4}^{2-}} \lt \mathrm{VO_{4}^{3-}} \lt \mathrm{MnO_{4}^{-}}\)
  • B.\(\mathrm{MnO_{4}^{-}} \lt \mathrm{VO_{4}^{3-}} \lt \mathrm{CrO_{4}^{2-}}\)
  • C.\(\mathrm{VO_{4}^{3-}} \lt \mathrm{CrO_{4}^{2-}} \lt \mathrm{MnO_{4}^{-}}\)
  • D.\(\mathrm{CrO_{4}^{2-}} \lt \mathrm{MnO_{4}^{-}} \lt \mathrm{VO_{4}^{3-}}\)

Answer: C

The wavelengths follow the order VO43- \(\lt\) CrO42- \(\lt\) MnO4-.

All three are d0 tetrahedral oxo-anions, and their intense colour (or its absence) comes from charge transfer from oxygen to the metal. The higher the metal's oxidation state and the more easily it is reduced, the lower the energy of this transfer and the longer its wavelength: V(V) \(\lt\) Cr(VI) \(\lt\) Mn(VII).

That matches the colours: vanadate is colourless (its band is in the ultraviolet), chromate is yellow (absorbing in the violet-blue), and permanganate is deep purple (absorbing green light, the longest wavelength of the three). The other orders put chromate or permanganate in the wrong place, for example making permanganate the shortest-wavelength absorber.

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