In the above mentioned complex, there is a chiral centre at the metal, and hence it seems the compound must be optically active.

But my course notes say that the ligands keep on changing their their positions at a very fast rate and as a result such complexes get converted into their mirror images and are hence optically inactive.

I am unable to reconcile my understanding of the stereochemistry with this statement in the notes. How can such complexes be optically inactive?

  • $\begingroup$ What's unclear about it? Yes, technically the molecule is optically active, and would have (R) and (S) isomers. But you can't separate them, because (R) changes to (S) and back pretty quickly (or not so quickly, depends on the metal, but whatever). $\endgroup$ Apr 14, 2016 at 10:22
  • $\begingroup$ Please see my edits & change anything if I've misrepresented the situation or your confusion. $\endgroup$
    – hBy2Py
    Apr 14, 2016 at 11:07

1 Answer 1


In a labile complex, ligands are detaching and attaching to the metal centre at extremely high rates. This means the solution will racemise quickly. Therefore, even if the complex is truly optically active, you have equal proportions of each enantiomer and thus equal rotations of the polarised light plane - in opposite directions.
The end result is that rotation cancels and so none is observed.


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