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So there is this question which asks whether the following complex

$\ce{[Co(NH3)4Br2]+}$

will show optical isomerism or not.

According to my knowledge it will show optical isomerism in cis form only and not in trans form so overall the compound shouldn’t be optically active.enter image description here

This what I hav been taught. And moreover shouldn’t it be optically active in every scenario to be called so?

But the ans says that it shows optical isomerism. I don’t understand why.. is my understanding wrong?

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    $\begingroup$ OK, you seem to agree that the trans form does not show optical isomerism. Now what about cis form? Why would it be chiral, when it has an obvious plane of symmetry? $\endgroup$ – Ivan Neretin Jan 30 at 9:42
  • $\begingroup$ Alright I agree so then is the answer wrong? Cause if I remember correctly this question came in JEE exam (a prestigious exam in India)@IvanNeretin $\endgroup$ – Muskaan Jan 30 at 10:16
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    $\begingroup$ Yes, the answer is wrong. None of the forms is optically active. $\endgroup$ – Ivan Neretin Jan 30 at 11:50
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As already commented by Ivan Neretin, the cis form also does not show optical isomerism. Elaborating a bit further on the comments:

Complexes show optical isomerism when they are asymmetric (also known as chiral). They should not have any elements of symmetry, such as planes/axes/centres of symmetry. The trans form has a clearly visible plane of symmetry. It's a bit trickier to find the plane for the cis form. Here's a diagram showing the planes: enter image description here (source: Chemistry libretexts)

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  • $\begingroup$ In the first diagram, the blue plane is not the plane of symmetry. They are rotating the molecules in the left and right panel so that in either case, the two B ligands are in the blue plane. That makes it easier to compare, and to distinguish cis (90 degree angle BMB) vs trans (180 degree angle BMB). It is fine to use this diagram, but you have to specify the location of the mirror planes (there are two for the cis isomer, and five for the trans isomer). $\endgroup$ – Karsten Theis Jan 30 at 17:34

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