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Our teacher told us that for coordination number 4 and tetrahedral shape $\ce{[M(AB)2]}$ doesn't have a plane of symmetry, so it shows optical isomerism.

What if we take a plane passing through the two $\ce{A}$'s so the $\ce{B's}$ will be the mirror image of each other? Am I wrong?

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    $\begingroup$ AB is implied to mean a bidentate ligand with a bridge between A and B. See where the bridge will go in the mirror. $\endgroup$ Jun 30 at 10:31
  • $\begingroup$ Do we see the bonds as well while considering the mirror image? $\endgroup$
    – Vega
    Jun 30 at 10:37
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    $\begingroup$ In all likeness, the bridges consist of quite a few atoms, not just bonds. $\endgroup$ Jun 30 at 10:43
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    $\begingroup$ Yes, you are right, I understood. $\endgroup$
    – Vega
    Jun 30 at 10:45
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Just to record the question as answered:

  • If you had separate A and B ligands with no ligand-ligand bonds, then the complex would have a plane (actually two planes) of symmetry and would not be chiral.

  • But when you have a bridge between each A ligand and a separate B ligand, the bridges do not conform with the above plane(s) of symmetry and so those symmetry planes are lost. Thereby the complex becomes chiral.

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