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Which of the following does not show optical isomerism? (en = ethylenediamine)
A
$[Co(en)_3]^{3+}$
B
$[Co(en)_2Cl_2]^+$
C
$[Co(NH_3)_3Cl_3]^0$
D
$[Co(en)Cl_2(NH_3)_2]^+$
Detailed Solution
A complex is optically active if it is chiral, i.e., if it has no plane of symmetry and is not superimposable on its mirror image.
$[Co(en)_3]^{3+}$: three bidentate ligands give a propeller-like structure with no plane of symmetry; it shows optical isomerism.
$[Co(en)_2Cl_2]^+$: the cis isomer has no plane of symmetry and is optically active.
$[Co(en)Cl_2(NH_3)_2]^+$: its cis forms can be chiral and show optical isomerism.
$[Co(NH_3)_3Cl_3]$ is of the type $[MA_3B_3]$ with only monodentate ligands; it shows geometrical isomerism (fac and mer forms), but both forms have planes of symmetry, so neither is optically active.
Hence $[Co(NH_3)_3Cl_3]^0$ does not show optical isomerism.
$[Co(en)_3]^{3+}$: three bidentate ligands give a propeller-like structure with no plane of symmetry; it shows optical isomerism.
$[Co(en)_2Cl_2]^+$: the cis isomer has no plane of symmetry and is optically active.
$[Co(en)Cl_2(NH_3)_2]^+$: its cis forms can be chiral and show optical isomerism.
$[Co(NH_3)_3Cl_3]$ is of the type $[MA_3B_3]$ with only monodentate ligands; it shows geometrical isomerism (fac and mer forms), but both forms have planes of symmetry, so neither is optically active.
Hence $[Co(NH_3)_3Cl_3]^0$ does not show optical isomerism.