Question

ii) Draw a simplified molecular orbital diagram (i.e. like those used for ligand field theory in lectures) to show the possible origin of the transition responsible for the purple colour of an aqueous KMnO₄ solution.

Answer 1

In kMnou the oxidation State of mnis +7 E means that there is no electron in d- orbital or Mn. so d.d transition is not possible because mn (U11) has d configura Hon. However KMnO4 shows a deep purple colour due to charge toansfer from a ligand to meral se from or to the central metal. This transfer is called ligand to metal charge transfer. Charge transfer are beyond Laporte & Spin selection rules so they always allowed giving the complex very high molar extinction coeblichents. Mn metal has 3d, 45 4 4p orbitals. f on the other side oxygen alom has twelve degenerate ligand -orbital. Molecular orbital diagram is as follows: Tetrahedral metal complex - - ettz e 11 11 11t a tett,tatz 14 14 42 tz 11 11 1 tz The electronic transition that causes the intense purple colour carrousponds to an excitation from t to e-t, should

always be the highest ligand orbital, while tetrahedral symmety typically impljer that Le lies lower than to on Higher the Central metals oxidation state, Tower its orbitals & closer it is to the ligands p-orbital. since the transition is from ligand Centered to metal-centered orbistan it is called as ligand to metal charge transfer ICLMCT: This transition is spin allowed & Laporte-rule is invalid in tetrahedral complexes as there is no centre of symmety. Therefore the observed colour is intense purple colour.