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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-$I$	List-$II$
$I. [FeF_6]^{3-}$	$A. 1.73 \ BM$
$II. [Ti(H_2O)_6]^{3+}$	$B. 5.93 \ BM$
$III. [Cr(NH_3)_6]^{3+}$	$C. 0.00 \ BM$
$IV. [Ni(H_2O)_6]^{2+}$	$D. 2.83 \ BM$
$V. [Fe(CN)_6]^{4-}$	$E. 3.88 \ BM$

Select the correct statement about the given square planar complex.

Similar Questions

Provide evidence that $[Co(NH_3)_5Cl]SO_4$ and $[Co(NH_3)_5SO_4]Cl$ are ionization isomers.

Match List-$I$ with List-$II$ and select the correct answer using the codes given below:
List-$I$ List-$II$
$I. [FeF_6]^{3-}$ $A. 1.73 \ BM$
$II. [Ti(H_2O)_6]^{3+}$ $B. 5.93 \ BM$
$III. [Cr(NH_3)_6]^{3+}$ $C. 0.00 \ BM$
$IV. [Ni(H_2O)_6]^{2+}$ $D. 2.83 \ BM$
$V. [Fe(CN)_6]^{4-}$ $E. 3.88 \ BM$

Draw the structures of the geometrical isomers of $[Fe(NH_3)_2(CN)_4]^-$.

Which of the following coordination compounds would exhibit optical isomerism?

The complex compounds $[Co(NH_3)_5 SO_4] Br$ and $[Co(NH_3)_5 Br] SO_4$ are

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