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(D) The given complex is a square planar complex of the type $[Pt(AB)_2]$,where $AB$ represents an unsymmetrical bidentate ligand. Due to the presence

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It is optically active because it does not have a plane of symmetry

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(D) The given complex is a square planar complex of the type $[Pt(AB)_2]$,where $AB$ represents an unsymmetrical bidentate ligand. Due to the presence of different substituents on the two ligands (one side has $C_6H_5$ groups,the other has $CH_3$ groups),the molecule lacks a plane of symmetry. Since the complex does not contain any element of symmetry,the configuration is asymmetric,and therefore,it is optically active.

List-$I$ (Coordination compounds)	List-$II$ (Number of unpaired electrons)
$A$. $[Cr(CN)_6]^{3-}$	$I$. $0$
$B$. $[Fe(H_2O)_6]^{2+}$	$II$. $3$
$C$. $[Co(NH_3)_6]^{3+}$	$III$. $2$
$D$. $[Ni(NH_3)_6]^{2+}$	$IV$. $4$

Select the correct statement about the given square planar complex.

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