. Which of the following does not show geometrical isomerism?

Options

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

Answer

To determine which of the given compounds does not show geometrical isomerism, we need to analyze the geometries of each compound mentioned. Here’s a step-by-step solution:

### Step 1: Understand Geometrical Isomerism
Geometrical isomerism occurs when compounds have the same molecular formula but differ in the spatial arrangement of their atoms. This is often seen in coordination compounds with specific geometries, such as square planar and octahedral.

### Step 2: Analyze Each Compound
1. **Ni(CO)4**:
- Geometry: Tetrahedral
- Explanation: In a tetrahedral geometry, all four positions are equivalent. Therefore, regardless of how the CO ligands are arranged, the compound remains the same. Thus, it does not show geometrical isomerism.

2. **[Pt(NH3)2Cl2]**:
- Geometry: Square planar
- Explanation: In square planar geometry, we can have different arrangements of ligands (cis and trans). Thus, this compound can show geometrical isomerism.

3. **[Co(NH3)4Cl2]**:
- Geometry: Octahedral
- Explanation: In octahedral geometry, there are multiple arrangements (facial and meridional) possible for the ligands. Thus, this compound can also show geometrical isomerism.

4. **[Co(NH3)3Cl3]**:
- Geometry: Octahedral
- Explanation: Similar to the previous compound, this can also show geometrical isomerism due to different arrangements of ligands.

### Step 3: Conclusion
From the analysis, **Ni(CO)4** is the only compound that does not show geometrical isomerism due to its tetrahedral geometry where all ligand positions are equivalent.

### Final Answer
**Ni(CO)4** does not show geometrical isomerism.

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