Whenever I try to find the number of geometrical isomers (including optical isomers) of coordination compound, I got confused, and mostly miss few isomers. Is there any standard method to know the number of isomers of compounds such as $\ce{Ma2b2c2}$ or $\ce{[M(A-A)a2b2]}$.

where $\ce{M}$ is metal and $\ce{a,b}$ are monodentate ligands while $\ce{A-A}$ is a bidentate ligand.


Do it many times over, so as not to get confused.

Alternatively, there is a thing called Polya's formula, but you won't be able to use it anyway. In trivial cases like this, the said formula is about 100 times more complicated than counting isomers by hand. It is not before polysubstituted fullerenes that its use in chemistry starts making any sense.


For an easier method without going much into mathematics you can use simple logic to do find the number of geometrical isomers of [Ma2b2c2]. Though such logics are very much situation based, and I believe there's no such generic one.

Here you can start by choosing no. of pairs of same ligands to be in trans (i.e. 180 degrees to each other). After some thought, its clear that one can choose 0,1 or 3 such pairs, because choosing 2 will automatically force the 3rd pair to be in trans. For choosing 1 such pair there are 3 ways, namely (a,a),(b,b),(c,c); and only 1 way each for choosing none (0) or all (3) such pairs. Thus accounting for all 5 isomers. Take the following example of $\ce{[Co(NH3)2Cl2(NO2)2]-}$.

5 geometrical isomers of [Co(NH3)2Cl2(NO2)2]-

For further references you can visit this site.


For Ma2b2c2, you could go by:

  1. All cis (aCa) (bCb) (cCc)
  2. All trans (aTa) (bTb) (cTc)
  3. Only 1 trans (aTa)
  4. Only 1 trans (bTb)
  5. Only 1 trans (cTc)

Similar cases could be done for M(AA)b2c2, except of the fact that, (AA) will remain always cis [all cases EXCEPT where the 2 donor atoms are connected by NOT-A-VERY-LONG-CHAIN]

  • 1
    $\begingroup$ And how to know about optical activity $\endgroup$ – search Apr 22 '17 at 3:09

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