Which of the following complex compound have highest number of geometrical isomers?

Question

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The number of geometrical isomers depends on the coordination number and ligand types.

A. $[PtC l_{2} B r_{2}]^{2 -}$ : Square planar, MA₂B₂ type, 2 geometrical isomers (cis, trans).

B. $[C o C l_{2} B r_{2}]^{2 -}$ : Tetrahedral, MA₂B₂ type, 0 geometrical isomers.

C. $[Pt (N H_{3}) (Gl y) B r]$ : Square planar, MABCD type, 3 geometrical isomers.

D. $[C o (N H_{3})_{2} (Gl y)_{2}]^{+}$ : Octahedral, MA₂B₂ type, 3 geometrical isomers (cis-cis, cis-trans, trans-cis).

The question asks for the highest number of geometrical isomers.

For option D, the complex is $[C o (N H_{3})_{2} (Gl y)_{2}]^{+}$ . Glycine (Gly) is a bidentate ligand.

This is an octahedral complex of the type MA₂B₂ (where A = NH₃ and B = Gly).

For MA₂B₂ type octahedral complexes, there are 3 geometrical isomers: cis-cis, cis-trans, and trans-cis.

Therefore, option D has the highest number of geometrical isomers.

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