Complexes Exhibiting Same Type of Isomerism

To determine which of the given complexes exhibit the same type of isomerism as $[\text{Pt}(\text{NH}_3)_2\text{Br}_2]$, let us first analyze the isomerism shown by $[\text{Pt}(\text{NH}_3)_2\text{Br}_2]$.

1. Analysis of $[\text{Pt}(\text{NH}_3)_2\text{Br}_2]$

• Platinum(II) is a $d^8$ transition metal ion that forms square planar complexes with a coordination number of 4.
• The complex is of the general form $[\text{MA}_2\text{B}_2]$, where $\text{A} = \text{NH}_3$ and $\text{B} = \text{Br}^-$.
• Since all four positions lie in a single plane, the complex exhibits geometrical isomerism ($cis$-$trans$ isomerism):
  • $cis$-isomer: The two $\text{NH}_3$ ligands are at an angle of $90^\circ$ to each other.
  • $trans$-isomer: The two $\text{NH}_3$ ligands are at an angle of $180^\circ$ to each other.

Therefore, we need to find the complexes that can exhibit geometrical isomerism.

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2. Evaluation of Options

• Option (A): $[\text{Pt}(\text{en})(\text{SCN})_2]$

  • $\text{Pt}^{2+}$ forms a square planar complex (coordination number 4).
  • Ethylenediamine ($\text{en}$) is a bidentate chelating ligand that can only coordinate in a $cis$ configuration ($90^\circ$ bite angle) to avoid ring strain.
  • Consequently, the two $\text{SCN}^-$ ligands are forced into the remaining adjacent ($cis$) positions.
  • Thus, no $trans$-isomer exists, and the complex cannot exhibit geometrical isomerism.

• Option (B): $[\text{Zn}(\text{NH}_3)_2\text{Cl}_2]$

  • $\text{Zn}^{2+}$ is a $d^{10}$ metal ion and forms a tetrahedral complex ($sp^3$ hybridized, coordination number 4).
  • In a regular tetrahedral geometry, all four ligand positions are symmetrically equivalent and adjacent to each other.
  • Therefore, tetrahedral complexes of the type $[\text{MA}_2\text{B}_2]$ do not exhibit geometrical isomerism.

• Option (C): $[\text{Pt}(\text{NH}_3)_2\text{Cl}_4]$

  • $\text{Pt}^{4+}$ is a $d^6$ metal ion that forms an octahedral complex ($d^2sp^3$ hybridized, coordination number 6).
  • This is an octahedral complex of the type $[\text{MA}_2\text{B}_4]$.
  • It exhibits geometrical isomerism:
    • $cis$-isomer: The two $\text{NH}_3$ ligands are $90^\circ$ apart.
    • $trans$-isomer: The two $\text{NH}_3$ ligands are $180^\circ$ apart.
  • Hence, this complex exhibits geometrical isomerism.

• Option (D): $[\text{Cr}(\text{en})_2(\text{H}_2\text{O})(\text{SO}_4)]^+$

  • $\text{Cr}^{3+}$ forms an octahedral complex with a coordination number of 6.
  • This complex is of the type $[\text{M}(\text{AA})_2\text{BC}]$, where $\text{AA} = \text{en}$, $\text{B} = \text{H}_2\text{O}$, and $\text{C} = \text{SO}_4^{2-}$.
  • It exhibits geometrical isomerism:
    • $cis$-isomer: The monodentate/unidendate ligands $\text{H}_2\text{O}$ and $\text{SO}_4^{2-}$ are at $90^\circ$ to each other.
    • $trans$-isomer: The monodentate ligands $\text{H}_2\text{O}$ and $\text{SO}_4^{2-}$ are at $180^\circ$ to each other.
  • Hence, this complex exhibits geometrical isomerism.

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Conclusion

The complexes that exhibit geometrical isomerism are given by options (C) and (D).