Which of the following complex will give white ppt. with `BaCl_(2)` solution ?

Options

A. `[Co(NH_(3))_(5) SO_(4)]Br`
B. `[Co(NH_(3))_(5)Br]SO_(4)`
C. `[Co(NH_(3))_(4) (SO_(4))_(2) ]Br`
D. `[Co(NH_(3))_(4) Br(SO_(4))]`

Explanation

To determine which complex will give a white precipitate with BaCl₂ solution, we need to analyze the options provided and understand the chemistry involved.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- Barium chloride (BaCl₂) can react with sulfate ions (SO₄²⁻) to form barium sulfate (BaSO₄), which is a white precipitate. The reaction can be represented as:
\[ \text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 (s) \]

2. **Identifying the Complexes**:
- We need to examine the complexes provided in the options to see if they contain sulfate ions (SO₄²⁻) or can release sulfate ions upon dissociation.

3. **Analyzing Option B**:
- The complex given in option B is [Co(NH₃)₅Br]SO₄.
- When this complex dissociates, it releases the sulfate ion (SO₄²⁻) along with the cobalt complex ion [Co(NH₃)₅Br]²⁺.

4. **Reaction with BaCl₂**:
- The released sulfate ions from the dissociation of the complex can react with BaCl₂ to form barium sulfate:
\[ \text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 (s) \]
- This confirms that option B will indeed produce a white precipitate when reacted with BaCl₂.

5. **Conclusion**:
- Based on the analysis, the correct answer is option B: [Co(NH₃)₅Br]SO₄, as it produces a white precipitate with BaCl₂ due to the formation of barium sulfate.

### Final Answer:
The complex that will give a white precipitate with BaCl₂ solution is **Option B: [Co(NH₃)₅Br]SO₄**.