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Tetrahedral and square planar crystal fields?

  1. Mar 1, 2012 #1
    This is two questions really in crystal-field theory.

    For tetrahedral crystal fields, why are the 't' and 'e' orbital sets switched in energy with the case in octahedral crystal fields?

    For square planar crystal fields, why do we discard the ligands along the z-axis? Why not discard the ligands along the x-axis or the y-axis?


  2. jcsd
  3. Mar 2, 2012 #2


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    That's not true, in general. You have to distinguish between the formal symmetry label of the complex and the concrete arrangement of the ligands. T_d is a subgroup of O_h, so there are complexes of tetrahedral symmetry with the same splitting pattern as octahedral ones. Think of eight ligands sitting at the corners of a cube (this arrangement has O_h symmetry). The field of this configuration will lead to the same splitting pattern as that of charges sitting on the corners of a tetrahedron, which is obtained by removing every second ligand.
    So the real question is why the splitting is reversed when going from octahedral to cubic complexes. This is easy to understand: The potential due to the charges has a minimum at the cubic corners in case of the octahedral coordination and maximum at the centers of the faces. In the case of cubic coordination it is just the other way round. So the effective potential seen by the orbitals is just inversed when going from cubic to octahedral complexation.
  4. Mar 2, 2012 #3
    Thank you! What about for square planar fields?

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