I was recently studying coordination compounds and i came across optical isomerism shown by complexes.I think that the complex i gave above have 4 geometrical isomers(I think this is where I am wrong, that it only has three geometrical isomers),but among those four geometrical isomers,how many of them are optically active?.According to what i studied only two of the three geometrical isomers should be optically active,other two having plane of symmetry(I am not too clear about the plane of symmetry concept),but all the solutions i looked up for the question said the complex only have 3 Geometrical isomers.Where am i wrong?.Is it that among the four geometrical isomers, two geometrical isomers are same?,if yes then how?

  • $\begingroup$ I suppose the easiest approach is to draw them all and then visually decide. The rules for being optical isomers are clear. They are mutually the mirror image for each other and at the same time they are not identical molecules. $\endgroup$
    – Poutnik
    Commented Dec 6, 2020 at 9:33

1 Answer 1


You have an octahedral complex where two of the adjacent positions are occupied by the ethylenediamine ligand. Let's make these equatorial for a clearer picture.


There are only two positions in the remaining four positions that are trans (C and D). You can choose to put two chlorides there, two ammines there, or one of each. Those are the 3 geometric isomers.

If the trans ligands are the same, then there is a plane of symmetry that goes through C-M-D and bisects the ethylene diamine ligand, so the complex will not be chiral. But if the two ligands are different in positions C and D, then there will not be a plane of symmetry and the complex will be chiral. You can draw them out to see what they are.

To summarize, there are three geometric isomers, one of which is chiral. If you have trouble with them types of problems, sometimes, it's helpful to just consider the cis/trans or mer/fac relationships between groups of ligands.


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