Which of the following complex is optically inactive

Options

A. `[RhCl(CO)(PPh_(3))(NH_(3))]`
B. `[Fe(en)_(2)Cl_(2)]`
C. `[Fe(en)_(2)Cl_(2)]`
D. `[Pd(en)_(2)Cl_(2)]`

Explanation

To determine which of the given complexes is optically inactive, we need to analyze the structures and geometries of each complex.

### Step-by-Step Solution:

1. **Identify the Complexes**: We have four complexes to analyze. Let's denote them as:
- Complex A
- Complex B
- Complex C
- Complex D

2. **Analyze Complex A**:
- Complex A has a square planar geometry.
- In square planar complexes, the central metal atom is surrounded by four ligands arranged at the corners of a square.
- Since square planar complexes have a plane of symmetry, they are optically inactive.

3. **Analyze Complex B**:
- Complex B consists of Fe with bis-ethylenediamine and dichloride ligands.
- This complex can exhibit geometrical isomerism (cis and trans forms).
- The cis isomers will be optically active due to their lack of symmetry, while the trans isomer will be a meso form and thus optically inactive.

4. **Analyze Complex C**:
- Complex C is similar to Complex B (as stated in the problem).
- Therefore, it will have the same properties: the cis forms are optically active, and the trans form is optically inactive.

5. **Analyze Complex D**:
- Complex D has a palladium center with bis-ethylenediamine and dichloride ligands.
- Similar to Complex B, it can also exhibit cis and trans isomerism.
- The cis forms will be optically active, while the trans form will be optically inactive.

6. **Conclusion**:
- Among the complexes analyzed, **Complex A** is optically inactive due to its square planar geometry, which inherently possesses a plane of symmetry.
- Complexes B, C, and D can have optically active forms depending on their isomerism.

### Final Answer:
**Complex A is optically inactive.**