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, fac and mer. But can it show optical isomerism?
Answer
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.
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