Suppose we have a complex of type $\ce{[Mabcdef]}$ then we know it’ll show geometrical isomerism. According to my textbook (Cengage Inorganic Chemistry II, Page 7.37) there are 15 geometrical isomers possible.

I tried calculating it but ended up with different answers every time.

I think we have to use the knowledge of permutations and combinations for this calculation.


3 Answers 3

  1. Assign "a" to one of the six equivalent positions.

  2. Note that there are now 5 choices (b,c,d,e,f) for the opposing position.

  3. You are left with four equivalent equatorial positions. Once a ligand is assigned to one of these four, there are 3 ligands remaining to select for the opposing equatorial position.

  4. Finally, there are 2 ways to assign the last two positions. The two possibilities are mirror images of each other.

So $5\times3\times2= 30$

30 isomers (15 pairs of enantiomers).

  • $\begingroup$ Great answer, but I need a clarification. In step 2, if I chose group b anti to group a, I will get four equivalent equatorial positions. However, if I choose group b cis to group a, I'm unable to see how that gives me four equivalent equatorial positions, as you've said. Could you please clarify? Thanks! $\endgroup$ Apr 20, 2018 at 14:33
  • 4
    $\begingroup$ @GaurangTandon In step 2, you are choosing which element to put across from group a, so if you are following this procedure you don't have the option of putting anything in the cis positions yet. $\endgroup$
    – Tyberius
    Apr 20, 2018 at 14:55

Every geometrical isomer would have 3 unique pairs of ligands that would be opposite (or trans) to eachother.

To divide 6 different objects into 3 pairs: $$\frac{6!}{2!^3*3!}$$

In case you don't know this method of division into groups, the answer provided by Dave would be easier.

But if you do want to get into the math part, then the formula can be worked out as:

Permutations of 6 different objects. $$=6!$$

But, two objects belong to the same group, which makes them "similar". So, divide by 2!. Thrice, because 3 groups. $$=\frac{6!}{2!^3}$$

The order of the three pairs is not significant to geometrical isomerism. So further divide by 3!. $$=\frac{6!}{2!^3*3!}$$


Theoretically, there are 15 possible geometrical isomers for 'Mabcdef' type complex. Possible geometrical isomers (in terms of combinations three ligand pairs) are as follows :

  1. ab cd ef
  2. ab ce df
  3. ab cf de
  4. ac bd ef
  5. ac be df
  6. ac bf de
  7. ad bc ef
  8. ad be cf
  9. ad bf ce
  10. ae bc df
  11. ae bd cf
  12. ae bf cd
  13. af bc de
  14. af bd ce
  15. af be cd

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