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    Can someone describe the selection rule that controls the value of the extinction coefficients in octahedral complexs please

    Oh and why are the values for the extinction coefficient higher in tetrahedral complexes?


    Assuming I have got the 'right end of the stick' the selection rules are as follows:

    Transitions (of electrons) between like orbitals i.e p to p, d to d etc are formally forbidden (laporte forbidden). Transition between different orbitals e.g. s to p, p to d etc are allowed.

    Chaning the spin of an electron in a transition is formally forbidden as well.

    I hope from my memory I have quoted those right.

    The extinction coefficient in tetrahedral complexes would be higher because the gap between tetrahedral homo and lumo (t2g to eg) are less than that of octahedrals, thus electrons are able to be excited into higher orbitals easier and create a visiable spectrum.

    Does that make sence?

    Spin selection rule and Laporte selection rule.

    - The spin-selection rule can be overcome by spin-orbit coupling. This causes a mixing of states which partially allows transitions. This is weak for 1st row transition metals therefore absorbtions will also be weak. For heavier metals, spin orbit coupling is stronger.

    -The Laporte selection rule can be overcome by vibronic coupling. This causes the loss of a centre of symmetry in your octahedral complex which allows a mixing of p and d orbitals. This then leads to P - D transitions which is partially allowed (since delta l = +/- 1 ).

    Tetrahedral complexes have a higher coffecient because they are non-centrosymmetric . As a result, mixing of p and d orbitals occurs to a greater extent and therefore d-d transitions are more probable.
    Also tetradehedral complex are capable of charge transfer (either ligand to metal /metal to ligand) which are intense absorbtions.

    I'm studying this **** as we speak...exam tomorrow ****!
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