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(C) For square planar complexes of the type $MABCD$,there are three possible geometrical isomers.These isomers arise from the different relative posit

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$MABCD$

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(C) For square planar complexes of the type $MABCD$,there are three possible geometrical isomers.These isomers arise from the different relative positions of the ligands $A, B, C,$ and $D$ around the central metal atom $M$.Specifically,one can fix the position of $A$ and then arrange the other ligands $B, C,$ and $D$ in different positions relative to $A$,leading to three distinct isomers.In contrast,complexes of the type $MA_3B$,$MA_4$,and $M(AA)_2$ (where $AA$ is a symmetrical bidentate ligand) do not exhibit geometrical isomerism because all positions are equivalent or the arrangement is fixed by the symmetry of the ligands.

Which one of the following will be able to show $cis-trans$ isomerism?

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