Step-by-step Derivations:

1. Analyze the reference complex [Pt(NH₃)₂Br₂] to determine the type of isomerism it exhibits.

   • The central metal ion is Pt²⁺. Pt(II) complexes with a coordination number of 4 are typically square planar.
   • The complex has the general formula [MA₂B₂], where M = Pt, A = NH₃, and B = Br⁻.
   • Square planar complexes of the type [MA₂B₂] can exist as two different geometrical isomers: cis and trans.
     • In the cis-isomer, the two identical ligands (e.g., NH₃) are adjacent to each other (at 90°).
     • In the trans-isomer, the two identical ligands are opposite to each other (at 180°).
   • Therefore, [Pt(NH₃)₂Br₂] exhibits geometrical (cis-trans) isomerism.
   • The task is to identify which of the given options also exhibit this type of isomerism.

2. Evaluate each option:

   • A: [Pt(en)(SCN)₂]

     • This is a Pt(II) complex, so it is square planar.
     • It contains a bidentate ligand, ethylenediamine (en), and two monodentate ligands, thiocyanate (SCN⁻).
     • The formula type is [M(AA)B₂] where (AA) is the bidentate ligand.
     • In a square planar complex, the bidentate ligand en must occupy two adjacent positions. Consequently, the two SCN⁻ ligands must also occupy the remaining two adjacent positions. There is only one possible spatial arrangement, so this complex does not show geometrical isomerism.
     • However, SCN⁻ is an ambidentate ligand and can coordinate through either S (thiocyanato) or N (isothiocyanato), leading to linkage isomerism. This is a different type of isomerism. Thus, A is incorrect.

   • B: [Zn(NH₃)₂Cl₂]

     • The central metal ion is Zn²⁺, which has a d¹⁰ electronic configuration.
     • Zn(II) complexes with coordination number 4 are tetrahedral.
     • In a tetrahedral geometry, all four ligand positions are equivalent with respect to each other (all angles are 109.5°). Therefore, changing the positions of ligands does not create a new isomer. Tetrahedral complexes of the type [MA₂B₂] do not exhibit geometrical isomerism.
     • Thus, B is incorrect.

   • C: [Pt(NH₃)₂Cl₄]

     • The central metal ion is Pt⁴⁺. Pt(IV) complexes with a coordination number of 6 are octahedral.
     • The complex has the general formula [MA₂B₄] where M = Pt, A = NH₃, and B = Cl⁻.
     • Octahedral complexes of this type exhibit geometrical isomerism based on the relative positions of the two 'A' ligands.
       • cis-isomer: The two NH₃ ligands are adjacent (90° apart).
       • trans-isomer: The two NH₃ ligands are opposite (180° apart).
     • Since this complex shows geometrical isomerism, C is correct.

   • D: [Cr(en)₂(H₂O)(SO₄)]⁺

     • The central metal ion is Cr³⁺. Cr(III) complexes with a coordination number of 6 are octahedral.
     • The ligands are two bidentate en, one monodentate H₂O, and one monodentate SO₄²⁻. The coordination number is 2×2 + 1 + 1 = 6.
     • The complex has the general formula [M(AA)₂BC].
     • Octahedral complexes of this type exhibit geometrical isomerism. The two en ligands can be arranged cis or trans to each other. This arrangement dictates the relative positions of the monodentate ligands H₂O and SO₄²⁻.
       • cis-isomer: The two en ligands are adjacent. The remaining two positions are also adjacent, so H₂O and SO₄²⁻ are cis to each other.
       • trans-isomer: The four N donor atoms from the two en ligands lie in a plane. The H₂O and SO₄²⁻ ligands occupy the axial positions, making them trans to each other.
     • Since the complex can exist as cis and trans isomers, it exhibits geometrical isomerism. Thus, D is correct.

Conclusion:

Both complexes [Pt(NH₃)₂Cl₄] and [Cr(en)₂(H₂O)(SO₄)]⁺ exhibit geometrical isomerism, which is the same type of isomerism shown by the reference complex [Pt(NH₃)₂Br₂]. Therefore, options C and D are the correct answers.