Which of the following can't show isomerism?

Options

A. `[Ni(NH_3)_2Cl_2]`
B. `[Ni(en)_3]`
C. `[Pt(NH_3)_2Cl_2]`
D. `[Ni(NH_3)_4(H_2O)_2]`

Answer

To determine which of the given complexes cannot show isomerism, we will analyze the geometry of each complex. Isomerism can be classified into geometrical isomerism and optical isomerism, and tetrahedral complexes do not exhibit either type of isomerism.

### Step-by-Step Solution:

1. **Identify the complexes:**
- Option 1: Ni(NH3)2Cl2
- Option 2: Ni(en)3 (where en = ethylenediamine)
- Option 3: Pt(NH3)2Cl2
- Option 4: Ni(NH3)4(H2O)2

2. **Analyze Option 1: Ni(NH3)2Cl2**
- This complex has a coordination number of 4 (two NH3 ligands and two Cl ligands).
- The hybridization is sp³, leading to a **tetrahedral geometry**.
- Since it is tetrahedral, it **cannot show isomerism**.

3. **Analyze Option 2: Ni(en)3**
- Ethylenediamine (en) is a bidentate ligand, and with three bidentate ligands, the coordination number is 6.
- The hybridization is d²sp³, leading to an **octahedral geometry**.
- Octahedral complexes can show geometrical and optical isomerism.

4. **Analyze Option 3: Pt(NH3)2Cl2**
- The coordination number is 4 (two NH3 ligands and two Cl ligands).
- The hybridization is dsp², leading to a **square planar geometry**.
- Square planar complexes can show geometrical isomerism.

5. **Analyze Option 4: Ni(NH3)4(H2O)2**
- This complex has a coordination number of 6 (four NH3 ligands and two H2O ligands).
- The hybridization is d²sp³, leading to an **octahedral geometry**.
- Octahedral complexes can show geometrical and optical isomerism.

### Conclusion:
The complex that cannot show isomerism is **Option 1: Ni(NH3)2Cl2**, as it has tetrahedral geometry.