MnO4 - is a stronger oxidizing agent than ReO4 - . Both ions have charge-transfer (LMCT) bands; however, the charge-transfer band for ReO4 - is in the ultraviolet, whereas the corresponding band for MnO4 - is responsible for its intensely purple color. Are the relative positions of the charge-transfer absorptions consistent with the oxidizing abilities of these ions? Explain

The 5d orbitals of Re are higher in energy than 3 d orbitals of Mn. So an LMCT ligand to metal charge transfer excitation requires more energy of [tex]$ReO^-_4$[/tex].

Also, since the molecular orbitals are derived primarily from 3d orbitals of [tex]$MnO^-_4$[/tex] are lower in energy than the corresponding MO's of [tex]$ReO^-_4$[/tex], [tex]$MnO^-_4$[/tex] is better able to accept the electrons.

So it is a better oxidizing agent.

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-12-03T16:24:55+01:00

The ligand to metal charge transfer band in ReO4- occurs in the near UV region hence ReO4 - appears colorless.

The electron configuration of Re is Xe 4f14 5d5 6s2 and the electron configuration of Mn is [Ar] 3d5 4s2. We can see that Mn^7+ and Re^7+ have empty d orbitals.

The color of MnO4 - must result from ligand to metal charge transfer hence the purple color of MnO4 -. In the case of ReO4 -, the ligand to metal charge transfer occurs at a much higher energy owing to the fact that 5d orbitals are involved. This transition occurs in the near UV region hence ReO4 - appears colorless. The ligand to metal charge transfer in MnO4- involves lower energy 3d orbitals hence it occurs in the visible region of the spectrum.

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