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)`
A can be .

Options

A. `[Co(NH_(3))_(5)]CISO_(4)`
B. `[Co(NH_(3))_(5)SO_(4)]CI`
C. `[Co(NH_(3))_(5)CI]SO_(4)`
D. `[Co(NH_(3))_(4)SO_(4)]CI.NH_(3)`

Explanation

To solve the problem, we need to analyze the given information about the cationic complex and its isomers A and B.

### Step-by-Step Solution:

1. **Identify the Components of the Complex**:
The complex consists of:
- One Co^(3+) ion
- Five NH3 (ammonia) ligands
- One Cl^- (chloride) ion
- One SO4^(2-) (sulfate) ion

2. **Determine the Possible Structures**:
The two isomers A and B can be structured based on the arrangement of the ligands around the Co^(3+) ion. The two possible arrangements are:
- Isomer A: [Co(NH3)5Cl]SO4
- Isomer B: [Co(NH3)5SO4]Cl

3. **Analyze the Precipitation Reactions**:
- Isomer A gives a white precipitate with BaCl2. This indicates that the sulfate ion (SO4^(2-)) is present in the complex, as barium sulfate (BaSO4) is a white precipitate.
- Isomer B gives a white precipitate with AgNO3. This suggests that the chloride ion (Cl^-) is present in the complex, as silver chloride (AgCl) is a white precipitate.

4. **Conclude the Identity of Isomer A**:
Since Isomer A gives a white precipitate with BaCl2, it must contain the sulfate ion. Therefore, Isomer A can be identified as:
- A = [Co(NH3)5Cl]SO4

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

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