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In cases where a metal ion like, Fe(III), Cr(III), Pt(II) is made to react with a ligand species like oxalate, CN-, CO etc and it forms a complex then, How do we determine whether 6 ligands will be coordinated to the central atom(i.e. in case of octahedral) or 4 ligands will be coordinated (as in tetrahedral or square planar arrangement) ?

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From your question, I suppose you do not know about crystal field theory. To properly understand how and why certain complexes are octahedral and other ones are tetrahedral you need to fully understand this theory. You can find information in Greenwood and Earnshaw, Chemistry of the Elements. A more detailed info about the theory is in Theory of Groups in Chemical Applications of Group Theory of Cotton.

I will explain the basis concept here. The isolated ion has a spherical symmetry, then all the microstates for a specific electronic configuration have the same energy. The ligands can be imagined as a source of electric potential. This field alters the symmetry, than the microstates have not all the same energy. In particular, the 5d orbitals will not have the same energy in presence of the ligands because they are shifted up and down. A degeneracy remotion, as always in QM, will alter the symmetry of the system, that is the structure will change. Then the geometry of a complex depends upon:

  • the nature of the metallic center
  • the oxidation state of the metallic center
  • the nature of the ligands ($\pi$-basic, $\pi$-acid, $\sigma$-donor and so on).
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    $\begingroup$ Thanks a lot, I do know the basics of cft. One of the questions I came across asked the reaction of Fe3+ with oxalate ligand in aq. sol containing KOH. Although octahedral geometry felt right to me I wasn't able to rule out the reason why it is not tetrahedral for instance. Is it this something based on experimentation ? Or is there any logic using which If given a pair of metal and ligand I can figure out what the geometry will be. $\endgroup$ – Ashish Mar 24 at 11:58
  • $\begingroup$ Fe(III) has a d$^5$ electronic configuration and oxalate is a $\sigma$-donor ligand, then it will be an high spin complex, no net gain in energy. Anyway, remember for other cases that the octahedral field is always stronger than the tetrahedral one. In this specific case it is more convenient to form six ligands because there is not so much steric hinderance. But, for other metals other factors can play a role like the Jahn-Teller distorsions in Cu(II) ammonia complexes $\endgroup$ – TheOldJonny Mar 24 at 12:41

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