Does $\ce{[Co(H2O)3Br3]}$ show geometrical and optical isomerism?

According to me $\ce{Co}$ should be $\mathrm{sp^3d^2}$ hybridised giving an octahedral geometry. It is of the form $\ce{[Ma3b3]}$ type. I know it can show geometrical isomerism, namely fac and mer isomers. But can it show optical isomerism?

  • $\begingroup$ Is there any way you could position the ligands such that it has a non-superimposable mirror image? $\endgroup$ Feb 4, 2016 at 11:38
  • $\begingroup$ @DGS yeah I think so. When similar ligands are all adjacent to each other. But I think they will have a centre of symmetry. $\endgroup$ Feb 4, 2016 at 11:54
  • $\begingroup$ I don't believe either fac or mer isomers result in non-superimposable mirror images. $\endgroup$ Feb 4, 2016 at 11:56
  • $\begingroup$ This very much looks like a homework question. Therefore I have given it the corresponding tag. As per our homework policy, please show your thoughts on this or the question may get closed. Note that you need to consider both fac and mer separately whether they can (or can not) exhibit optical isomerism. Hint: It may help to draw the isomers on paper. Also note that you should remove hybridisation from your mind for transition metal coordination compounds; they are much better described without it. $\endgroup$
    – Jan
    Feb 4, 2016 at 16:39
  • $\begingroup$ @Jan No, this is not a homework question. I am asking if anyone of the fac or mer isomers can show optical activity. $\endgroup$ Feb 4, 2016 at 17:06

1 Answer 1


According to me $\ce{Co}$ should be $\mathrm{sp^3d^2}$ hybridised giving an octahedral geometry.

No, please don’t do this. The most accurate description of cobalt here would be $\mathrm{d^5}$, maybe $\mathrm{d^5s}$. But don’t extend the hybridisation concept from main group elements to the transition metals; it doesn’t belong and it doesn’t help here. Rather, there are six ligands around cobalt, so you have an extremely high chance for octahedral geometry. Which is also the case.

I know it can show geometrical isomerism, fac and mer.

That’s a good start.

But can it show optical isomerism?

When asking about optical isomerism, ask yourself, if the compound is superimposable on its mirror image. One method to check this is to draw the compound, draw its mirror image and rotate either until they overlap — or until you think they cannot. The second method is much more efficient though: Still draw them, but check whether you can find a centre of inversion symmetry, a plane of symmetry or an axis of improper rotation (turn by $\frac{360^\circ}{n}$, where $n$ is a natural number and then mirror the result through a plane perpendicular to the rotation axis). If you can find any of these, the compound is automatically identical with its mirror image, and if there are none of these, then it displays optical isomerism. Remember to consider 3D structures.

It should be rather simple to determine this for the mer isomer. For the fac, consider a view where three ligands of the same type are above the paper plane, the central cobalt occupies the paper plane and the three ligands of the other type are below it. For both considerations, you can ignore the two hydrogens on each oxygen.

The results are given in a spoiler (mouse over to see).

The mer isomer has planes of symmetry that transect three meridional atoms of one kind and the middle atom of the other three.

The fac isomer has planes of symmetry that transect the cobalt, one ligand of each kind and the middle between the other two ligands of each type.

Therefore, neither displays optical isomerism.

  • $\begingroup$ pls explain how do you highlight certain part of question while answering. It will be helpful for me while answering questions so that it will be clear for the viewer. $\endgroup$ Jul 12, 2021 at 12:08

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