One cationic complex has to isomers `A` and `B` Each has one `Co^(3+)` five `NH_(3)` one `CI^(Θ)` and one `SO_(4)^(2-)` stoichiometically A give white precipitate with `BaCI_(2)` white `B` gives white precipitate with `AgNO_(3)`
` (B)` can be .

Options

A. `[Co(NH_(3))_(5)SO_(4)]CI`
B. `[CO(NH_(3))_(5)CI]SO_(4)`
C. `[Co(NH_(3))_(3)CI(SO_(4))]2NH_(3)`
D. None of these

Explanation

To solve the problem, we need to analyze the given information about the two isomers, A and B, of the cationic complex containing Co^(3+), NH3, Cl^-, and SO4^(2-).

### Step-by-Step Solution:

1. **Identify the Composition of the Complex**:
- Each isomer has the following components:
- 1 Co^(3+)
- 5 NH3 (ammonia)
- 1 Cl^- (chloride)
- 1 SO4^(2-) (sulfate)

2. **Determine the Possible Structures**:
- The complex can be represented as [Co(NH3)5X]Y, where X can be Cl^- or SO4^(2-) and Y is the counter ion.
- Given that we have two isomers, one is likely to have Cl^- as the outer ligand and the other with SO4^(2-) as the outer ligand.

3. **Analyze the Precipitation Reactions**:
- Isomer A gives a white precipitate with BaCl2. This suggests that it contains the sulfate ion (SO4^(2-)), as BaSO4 is insoluble and would precipitate out.
- Isomer B gives a white precipitate with AgNO3. This indicates that it contains chloride ions (Cl^-), since AgCl is also insoluble and would precipitate out.

4. **Assign the Isomers**:
- Based on the precipitation reactions:
- Isomer A must be [Co(NH3)5(SO4)]Cl, where SO4^(2-) is the inner complex and Cl^- is the counter ion.
- Isomer B must be [Co(NH3)5(Cl)]SO4, where Cl^- is the inner complex and SO4^(2-) is the counter ion.

5. **Conclusion**:
- Therefore, the isomer B can be represented as [Co(NH3)5(Cl)]SO4.

### Final Answer:
Isomer B can be [Co(NH3)5(Cl)]SO4.