explain e- charge transfer in details in context with kmno4

							
Dear student,
Mn in the tetrahredral [MnO4]– anion is Mn(VII), so formally d(0), with no valence electrons.  The bonding can be described as primarily ionic, so we can think of the  filled orbitals as simply the electrons on the [O]2- groups surrounding  the Mn - the highest energy filled orbitals will be the O 2p atomic orbitals. Crystal field theory predicts a splitting of the empty d orbitals on Mn into two energy levels,  the lower two "e" orbitals and the upper three "t2" orbitals. There are  two transitions in the UV-vis spectrum of the [MnO4]– anion: one from  the filled O 2p to the lower-energy "e" Mn d orbitals, and one from the  filled O 2p to the higher-energy "t2" Mn d orbitals. Both are  ligand-to-metal charge transfer bands, because they formally move  electrons from the O to the Mn, which is why the colour is so intense --  CT bands have extremely high extinction coefficients.
 

































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