Identify the optically active compounds from the following

Options

A. `[Co(en)_(3)]^(3+)`
B. trans`- [Co(en)_(2) Cl_(2)]^(+)`
C. cis`- [Co(en)_(2)Cl_(2)]^(+)`
D. `[Cr(NH_(3))_(5)Cl]`

Explanation

To identify the optically active compounds from the given options, we will analyze each compound based on its symmetry and the presence of non-superimposable mirror images. Here’s the step-by-step solution:

### Step 1: Analyze the First Compound
- **Compound**: [Co(en)3]³⁺
- **Structure**: This is a cobalt complex with three bidentate ethylene diamine (en) ligands.
- **Symmetry**: The arrangement of the ligands in an octahedral geometry does not exhibit any symmetry.
- **Optical Activity**: Since it lacks a plane of symmetry, it has non-superimposable mirror images, making it optically active.

### Step 2: Analyze the Second Compound
- **Compound**: [Co(en)2Cl2] (trans)
- **Structure**: This is an octahedral complex with two bidentate ethylene diamine ligands and two trans chlorine ligands.
- **Symmetry**: The trans arrangement creates a plane of symmetry.
- **Optical Activity**: Because it has a plane of symmetry, it is optically inactive.

### Step 3: Analyze the Third Compound
- **Compound**: [Co(en)2Cl2] (cis)
- **Structure**: This is an octahedral complex with two bidentate ethylene diamine ligands and two cis chlorine ligands.
- **Symmetry**: The cis arrangement does not have a plane of symmetry.
- **Optical Activity**: It has non-superimposable mirror images, making it optically active.

### Step 4: Analyze the Fourth Compound
- **Compound**: [CrCl(NH3)5]²⁺
- **Structure**: This is a chromium complex with five ammonia ligands and one chloride ligand.
- **Symmetry**: The arrangement has a plane of symmetry due to the presence of identical ammonia ligands.
- **Optical Activity**: Because it has a plane of symmetry, it is optically inactive.

### Conclusion
Based on the analysis:
- **Optically Active Compounds**: First compound [Co(en)3]³⁺ and Third compound [Co(en)2Cl2] (cis).
- **Optically Inactive Compounds**: Second compound [Co(en)2Cl2] (trans) and Fourth compound [CrCl(NH3)5]²⁺.

Thus, the answer is that the optically active compounds are the **first and third options**.

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