# Why is [M(AB)2] a geometrical isomer?

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?

• AB is implied to mean a bidentate ligand with a bridge between A and B. See where the bridge will go in the mirror. Jun 30, 2021 at 10:31
• Do we see the bonds as well while considering the mirror image?
– Vega
Jun 30, 2021 at 10:37
• In all likeness, the bridges consist of quite a few atoms, not just bonds. Jun 30, 2021 at 10:43
• Yes, you are right, I understood.
– Vega
Jun 30, 2021 at 10:45

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.