Optical isomerism in normal organic molecules can be found out by looking for chiral carbon, and the number of isomers can be found out by drawing different spacial orientations of the same.

But, I don't know how to check whether a given coordination compound is optically active or not, and how many optical isomers does it have. I don't know how to draw different spacial orientations with all those polydentate ligands.

I would like to learn the general guidelines for finding optical isomerism in co-ordination compounds. I would like to solve questions like how many no. of optical isomers are there for $\ce{[CoCl2(en)2]^+}$


2 Answers 2


Any 3rd-year inorganic chemistry textbook worth its weight in salt will provide some information about isomers in coordination compounds. An optically active compound contains no rotation-reflection symmetry element. Typically, this means that if a molecule has an inversion center or a mirror plane then it is not chiral.

It takes a systematic brute force and ignorance approach to determining the number of isomers a coordination compound has, and then a bit more visual thinking to determine if any of the isomers are enantiomers. The following figure is taken from Miessler, Fischer, Tarr 5th edition and is useful for identifying isomers of 6-coordinate complexes.

enter image description here

Note that a 6-coordinate complex with 6 unique ligands in an octahedral field would be enantiomers. In fact, there are 15 different isomers, each of which is chiral resulting in a total of 30 isomers!

The method of determining isomers this way is to formulate a structure of the type M<ab><cd><ef> where <ab> indicates two ligands that are trans to each other. Then, you start varying a, b, c, ... until you've come up with as many unique combinations as possible. For your case, you have 2 instances of a bidentate ligand. Bidentate ligands are typically designated with capital letters in this notation, so the formula for your molecule would be $\ce{M(AA)_2b_2}$. Further, since bidentate ligands are too small to coordinate in a trans conformation, you will decrease the number of possible isomers. One isomer, for example, would be M<Ab><A'b><A'A>, where I use A and A' to designate the different $\ce{(en)}$ ligands.

The answer to your question, then, is that there are two isomers, and they are enantiomers, since you can't rearrange the A, A' and b in the formula above to come up with another unique configuration.


There is an easy way of identifying optical isomers in coordination compounds.

  1. Roughly draw the structures of all possible isomers of the compound by interchanging the ligand positions.

  2. Cross out those structures whose mirror image overlaps with itself.

  3. The remaining are optical isomers of the given compound.

Once you do this a few times, you can draw this in your head using your imagination and figure it out faster.

To check if a coordination compound is optically active, just see if if the mirror image of the compound overlaps with itself (for the given isomer), if it doesn't, the compound is optically active.

  • 1
    $\begingroup$ This is correct, but if someone actually has significant problems with optical isomerism, then will stumble somewhere. $\endgroup$
    – Mithoron
    May 8, 2018 at 14:32
  • $\begingroup$ If you know the theory well , and have a basic understanding of chirality , enantiomers and optical activity in organic compounds , this should make perfect sense. :p $\endgroup$ May 8, 2018 at 18:25
  • 1
    $\begingroup$ Of course, but guys who know it all don't need help. BTW there shouldn't be any space before your commas. $\endgroup$
    – Mithoron
    May 8, 2018 at 18:28
  • $\begingroup$ @Mithoron my answer is way less complicated than the first ! , I used this in my 12th grade exams a month ago...! $\endgroup$ May 8, 2018 at 18:32
  • $\begingroup$ It's OK, maybe a bit like TL:DR, just saying that if you answer in the future, OP's may be nagging you about details etc., good luck though. $\endgroup$
    – Mithoron
    May 8, 2018 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.