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We have been taught to look for plane of symmetries in inorganic complexes. If there is a plane of symmetry then the complex is optically inactive, otherwise it is optically active. In this particular complex, I am unable to find the plane of symmetry. I have tried creating mirror images in my head and they were superimposable.

This was actually a test question and the answer was that this is optically inactive

Can someone help me find the plane of symmetry here? Or is this method not applicable to all compounds?

enter image description here

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2 Answers 2

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There is indeed a plane of symmetry. The symmetry axis passes through the top-most 'a', 'M', and 'c' (as you correctly surmised it must). Have a look at this diagram to see where it goes:

Symmetry axis

For octahedral complexes, it sometimes helps to visualise the structure from a different angle. If you 'look' at the complex with 'b', 'c'. and 'b' facing towards you, all three 'a' atoms will face away from you, and the symmetry axis might be easier to see:

enter image description here

You can see how to re-orientate the molecule by following the atom numbering.

Finally, the best way to visualise sterics is in 3D. This isn't much help for an exam, but to convince yourself of the symmetry this can be a great tool. This visualisation is in Avogadro, which you can download and play with for free. This first image is in the same orientation as your question:

enter image description here

And here it is re-orientated:

enter image description here

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Yes, there is an oblique plane of symmetry passing from between the two b ligands. Ergo, the compound is optically inactive.

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