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I came across a statement saying,

"Consider a Octahedral coordination compound of type [MA2X2]. Then, if the compound is in a cis form, it shows optical activity as there is no plane of symmetry and hence is non-superimposable on its mirror image. On the other hand, a trans form is symmetric and superimposable on its mirror image and so is optically inactive." enter image description here

I do not see why the cis form is not superimposable on its mirror image. If all the ligands lie on the same plane, then isn't there a line of symmetry passing diagonally? Does that not make it superimposable on its mirror image?

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    $\begingroup$ $\ce{[CoCl2(en)2]+}$ isn't square planar. Please add a reference for the quoted part. $\endgroup$
    – andselisk ♦
    Oct 28, 2019 at 5:17
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    $\begingroup$ Point 1. The compound is not planar(it is octahedral). Point 2. Line/Axis of symmetry doesn't affect chirality. Only POS,COS,alternating Axis of symmetry does. But for any planar molecule, molecular plane is itself a plane of symmetry so planar compounds are always achiral $\endgroup$
    – user600016
    Oct 28, 2019 at 5:17
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    $\begingroup$ ah duly noted. Turns out I got the structures messed up. I will edit my question for future viewers. $\endgroup$
    – The Jade Reaper
    Oct 29, 2019 at 1:40

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It is not always as easy to see from a standard octahedral depiction. Optical activity in octahedral complexes can be understood better, if you draw the octahedron as looking onto a triangular face rather than along an edge; this results in a hexagonal arrangement of ligands around the central atom with ligands alternately pointing into and out of the paper plane.

If we do that with $\ce{[Co(en)2Cl2]}$, we get the following two enantiomers which you cannot convert into each other:

both enantiomers of the [Co(en)2Cl2] complex

The complex is chiral because the $\ce{en}$ ligands necessarily cross the paper plane and can do so either in a right-turn screw motion or in a left-turn screw motion.

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