explain e- charge transfer in details in context with kmno4

aryan kashyap, 12 years ago

Grade:12

1 Answers

SAGAR SINGH - IIT DELHI

879 Points

12 years ago

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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