I have a question concerning metal ligand complexes.

In every chemistry text book I have read which treats metal ligand complexes, following situation is discussed: When 6 ligands (in octahedral coordination) approaches a metal ion with d-orbitals, 2 of the three orbitals are energetically raised because these orbitals have their lobs in direction of the ligands.

But why is this always true? Why does always point the lobes of the z² and x²-y² orbital towards the ligands coming from "above" and "below"? Isn't it possible, that the z-axis of orbital system of the metal is a little bit inclined with respect to the octahedral ligand system surrounding the middle metal ion?

The only thing I can think of, is that this is because of energetic reasons. That in total, it's energetically more favourable when 3 of the 5 d-orbitals are arranged, so that electron occupying these orbitals have the lowest possible energies.

But up to now, I haven't found anything discussing this question.

Thanks in advance, Jürgen

  • $\begingroup$ See also: chemistry.stackexchange.com/questions/17284/… $\endgroup$ – Geoff Hutchison Oct 21 '14 at 1:36
  • $\begingroup$ This is not a duplicate question. The other question asks why does the energy of three orbitals ever decrease, because there exists repulsion. This question asks why does ligands always approach z² and x²-y² orbitals; why not z² and xy for example? $\endgroup$ – fz_salam Apr 26 '17 at 17:57

The answer lies in symmetry. When placed in an octahedral field, the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals are directed at the surrounding ligands because they have $e_g$ symmetry. The remaining three orbitals have $t_{2g}$ symmetry and are directed between the ligands. This explanation can be found in an undergraduate inorganic chemistry textbook such as that by Miessler, Fischer and Tarr. (In the 5th edition, that's section 10.2.1 - Crystal Field Theory).

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