Which of the following compound show optical isomerism?

Options

A. cis-`[CrCl_(3)(NH_(3))_(3)]`
B. cis-`[Co(NH_(3))_(4)Cl_(2)]^(+)`
C. `[Co(en)_(3)]^(3+)`
D. trans-`[Co(en)_(2)Cl_(2)]^(+)`

Explanation

To determine which of the given compounds show optical isomerism, we need to analyze each compound based on the criteria for optical isomerism. Optical isomerism occurs when a compound can exist in two non-superimposable mirror images, known as enantiomers. This typically requires the compound to lack a plane of symmetry.

### Step-by-Step Solution:

1. **Understanding Optical Isomerism**:
- Optical isomerism arises in coordination compounds when they can form non-superimposable mirror images. This usually occurs when the compound lacks a plane of symmetry.

2. **Analyzing the First Compound (Cis form of Me3b3 complex)**:
- The cis form of the complex does not show optical isomerism because it has similar monodentate ligands that can be arranged in a way that allows for superimposition. Thus, it is eliminated as a candidate for optical activity.

3. **Analyzing the Second Compound (CONH3O4)**:
- This compound can exist in cis and trans forms. However, both forms are superimposable on their mirror images. Therefore, they do not exhibit optical isomerism and are considered optically inactive.

4. **Analyzing the Third Compound (MAA3 complex)**:
- The MAA3 complex, which includes ethylene diamine as a ligand, can show optical isomerism. The arrangement of ligands around the central metal ion (cobalt in this case) can lead to non-superimposable mirror images. Hence, this compound is optically active.

5. **Analyzing the Fourth Compound (Trans-COEN)**:
- The trans form of COEN also results in a configuration where the mirror image is superimposable. Therefore, it does not exhibit optical isomerism and is optically inactive.

6. **Conclusion**:
- Among the given compounds, the one that shows optical isomerism is the MAA3 complex.

### Final Answer:
The compound that shows optical isomerism is **MAA3 complex**.

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