Which of the following octahedral complexes do not show geometrical isomerism ?

Options

A. `[Co(NH_(3))_(3)Cl_(3)]`
B. `[PtCl_(2)(NH_(3))_(4)]`
C. `[Cr(H_(2)O)_(5)Cl]^(2+)`
D. `[Co(en)_(3)]^(3+)`

Answer

To determine which of the given octahedral complexes do not show geometrical isomerism, we need to analyze the ligands present in each complex and apply the conditions for geometrical isomerism.

### Step-by-Step Solution:

1. **Identify the complexes**: The complexes given are:
- Cu(NH3)3Cl3
- PtCr2I3C4
- Cr(H2O)5Cl2
- Cu(NH3)3O3

2. **Understand geometrical isomerism**: Geometrical isomerism in octahedral complexes typically occurs when:
- There are at least two different types of ligands.
- At least one type of ligand must be present in a quantity of at least two.

3. **Analyze each complex**:
- **Cu(NH3)3Cl3**: This complex has 3 ammonia (NH3) ligands and 3 chloride (Cl) ligands. Since there are two different ligands (NH3 and Cl) and Cl is present in a quantity of 3, this complex can show geometrical isomerism.
- **PtCr2I3C4**: This complex has 3 iodine (I) ligands and 4 carbon (C) ligands. Here, there are two different ligands (I and C), and I is present in a quantity of 3. Therefore, this complex can also show geometrical isomerism.
- **Cr(H2O)5Cl2**: This complex has 5 water (H2O) ligands and 2 chloride (Cl) ligands. There are two different ligands (H2O and Cl), and Cl is present in a quantity of 2. Thus, this complex can show geometrical isomerism.
- **Cu(NH3)3O3**: This complex has 3 ammonia (NH3) ligands and 3 oxide (O) ligands. Here, there are only one type of ligand (O) and one type of ligand (NH3) in equal quantity. Therefore, this complex does not have two different ligands and cannot show geometrical isomerism.

4. **Conclusion**: The complex that does not show geometrical isomerism is **Cu(NH3)3O3**.

### Final Answer:
The octahedral complex that does not show geometrical isomerism is **Cu(NH3)3O3**.