Explain optical isomerism in six-coordinated complex compounds or octahedral complexes.

(N/A) Octahedral complexes are optically active if:
$(i)$ The plane of symmetry is absent.
$(ii)$ Their mirror images are non-superimposable on each other.
Hence,these are called enantiomers. The two forms of enantiomers are called $\text{dextro} (d)$ and $\text{laevo} (l)$ depending upon the direction they rotate the plane-polarized light in a polarimeter. The $\text{dextro}$ molecule rotates it to the right and the $\text{laevo}$ rotatory molecule rotates it to the left.
Optical isomerism in octahedral complexes is common if they contain bidentate ligands.
$(i)$ Complexes of type $\left[M(AA)_{3}\right]^{n \pm}$:
$AA$ : Symmetrical bidentate ligand.
For example: $\left[Co(en)_{3}\right]^{3+}$.
$(ii)$ Complexes of type $\left[M(AA)_{2}X_{2}\right]^{n \pm}$:
$AA$ : Symmetrical bidentate ligands.
$X$ : Monodentate ligand.
For example: $\left[PtCl_{2}(en)_{2}\right]^{2+}$.