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Complexes of type $\ce{MA3B3}$ have two geometrical isomers, namely:

  1. fac-isomer
  2. mer-isomer

If we look closely at the mer-isomer, it has a plane of symmetry, so it is optically inactive. But the fac-isomer has no plane of symmetry. Still, it is optically inactive, why?

Being a 12th grader, I wonder where am I wrong in my understanding of optical isomerism in coordination complexes.

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    $\begingroup$ Because of three planes of symmetry, to begin with. $\endgroup$ Mar 1 at 7:55
  • $\begingroup$ There is no plane of symmetry in fac-isomer of type ${MA_3B_3}$ $\endgroup$
    – Harsh
    Mar 1 at 8:23
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    $\begingroup$ Your assertiveness is not quite justified. $\endgroup$ Mar 1 at 8:30
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    $\begingroup$ there is a plane of symmetry in the fac- along the axis AMB and between two A and two B atoms. see the image here courses.lumenlearning.com/introchem/chapter/…. $\endgroup$
    – porphyrin
    Mar 1 at 8:31
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    $\begingroup$ chemtube3d.com/symfecl3br3fac $\endgroup$
    – S R Maiti
    Mar 2 at 17:13
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Orient the fac- isomer so one A atom is facing towards you and a space between two B atoms is below it. Then the left half of this view is a reflection of the right half so you are actually looking down a mirror plane.

Because you could have chosen any of three A atoms in your view there are three mirror planes, which pass through a common rotation axis. The axis passes through the face that has all A atoms at its corners and the opposite face which has all B atoms. This symmetry is called $C_{\mathrm{3v}}$ in the language of point groups.

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