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(B) For a complex to exhibit optical isomerism,it must be chiral,meaning it should not have a plane of symmetry or a center of symmetry.$1$. $[Pd(en)_

Vedclass · Accepted Answer

$[Co(H_2O)_4Cl_2]Cl \cdot 2H_2O$

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(B) For a complex to exhibit optical isomerism,it must be chiral,meaning it should not have a plane of symmetry or a center of symmetry.$1$. $[Pd(en)_2]^{2+}$ and $[Pt(en)_2]^{2+}$ are square planar complexes. Square planar complexes with bidentate ligands like $en$ (ethylenediamine) are planar and possess a plane of symmetry,hence they are optically inactive.$2$. $[Co(H_2O)_4Cl_2]Cl \cdot 2H_2O$ is an octahedral complex. The isomer $trans-[Co(H_2O)_4Cl_2]^+$ has a plane of symmetry,while the $cis$ isomer can exist in chiral forms if the ligands are arranged appropriately,but generally,simple octahedral complexes of the type $[MA_4B_2]$ are not optically active unless the ligands are chelating.$3$. However,looking at the options provided,there is a common error in the question structure. Usually,$[M(en)_3]^{n+}$ or $[M(en)_2X_2]^{n+}$ (cis-form) are the standard examples for optical activity. Given the options,if we consider $[Co(H_2O)_4Cl_2]Cl \cdot 2H_2O$,it is the only one that can potentially form a $cis$ isomer which lacks a plane of symmetry if the geometry allows. But strictly,none of these are classic examples of optical activity. Re-evaluating: $[Pt(en)_2]^{2+}$ is square planar (inactive). $[Pd(en)_2]^{2+}$ is square planar (inactive). $[ZnCl_4]^{2-}$ is tetrahedral (achiral). The question likely intended to include a complex like $[Co(en)_2Cl_2]^+$. Given the options,$B$ is the most complex structure capable of geometric isomerism,but none are optically active. If forced to choose,the question is flawed.

Which of the following complexes or complex ions can have an optically active isomer?

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