Which one of the following complex is not expected to exhibit isomerism?

Options

A. `[Ni(NH_(3))_(4)(H_(2)O_(2)]^(2+)`
B. `[Pt(NH_(3))_(2)CI_(2)]`
C. `[Ni(NH_(3))_(2)CI_(2)]`
D. `[Ni(en)_(3)]^(2+)`

Explanation

To determine which complex is not expected to exhibit isomerism, we need to analyze the geometry and ligand arrangement of each complex provided in the options.

### Step-by-Step Solution:

1. **Identify the Complexes**:
- We have four complexes to analyze:
1. Ni(NH3)4(H2O)2^2+
2. Pt(NH3)2Cl2
3. Ni(NH3)2Cl2
4. Ni(en)3^2+ (where en = ethylenediamine)

2. **Determine the Oxidation States**:
- For Ni(NH3)4(H2O)2^2+: Nickel is in +2 oxidation state (both NH3 and H2O are neutral ligands).
- For Pt(NH3)2Cl2: Platinum is also in +2 oxidation state (NH3 is neutral, Cl is -1).
- For Ni(NH3)2Cl2: Nickel is in +2 oxidation state (NH3 is neutral, Cl is -1).
- For Ni(en)3^2+: Nickel is in +2 oxidation state (en is a neutral bidentate ligand).

3. **Identify the Electron Configuration**:
- For Ni^2+: The electron configuration is 3d^8.
- For Pt^2+: The electron configuration is 5d^8.

4. **Determine the Geometry**:
- Ni(NH3)4(H2O)2: This is an octahedral complex (sp3d2 hybridization).
- Pt(NH3)2Cl2: This is a square planar complex (DSP2 hybridization).
- Ni(NH3)2Cl2: This is a tetrahedral complex (sp3 hybridization).
- Ni(en)3^2+: This is an octahedral complex (due to bidentate ligands).

5. **Analyze Isomerism**:
- Octahedral complexes (like Ni(NH3)4(H2O)2 and Ni(en)3^2+) can exhibit geometrical (cis-trans) and optical isomerism.
- Square planar complexes (like Pt(NH3)2Cl2) can exhibit geometrical isomerism (cis-trans).
- Tetrahedral complexes (like Ni(NH3)2Cl2) do not exhibit geometrical isomerism due to their symmetry.

6. **Conclusion**:
- Among the complexes analyzed, **Ni(NH3)2Cl2** is a tetrahedral complex and is not expected to exhibit isomerism.

### Final Answer:
**Ni(NH3)2Cl2** is the complex that is not expected to exhibit isomerism.