Cobalt (III) chloride forms several octahedral complexes with amonia. Which of the following will not give test for chloride ions with silver nitrate at `25^@C`?

Options

A. `CoCl_(3)cdot3NH_(3)`
B. `CoCl_(3)cdot4NH_(3)`
C. `CoCl_(3)cdot5NH_(3)`
D. `CoCl_(3)cdot6NH_(3)`

Explanation

To solve the problem, we need to analyze the cobalt (III) chloride complexes with ammonia and determine which of them will not give a test for chloride ions when reacted with silver nitrate.

### Step-by-Step Solution:

1. **Understand the Coordination Complexes**: Cobalt (III) chloride can form octahedral complexes with ammonia (NH₃). The general formula for these complexes can be represented as [Co(NH₃)₆]Cl₃, [Co(NH₃)₄Cl₂], [Co(NH₃)₅Cl], and so on.

2. **Identify the Ionizable Chloride Ions**: The test for chloride ions with silver nitrate (AgNO₃) involves the formation of a white precipitate of silver chloride (AgCl). For chloride ions to be detected, they must be free in the solution, meaning they must be ionizable.

3. **Analyze Each Complex**:
- **Option A: [Co(NH₃)₆]Cl₃**: Here, all three chloride ions are ionizable. Therefore, it will give a positive test for chloride ions.
- **Option B: [Co(NH₃)₄Cl₂]**: This complex has two ionizable chloride ions. It will also give a positive test for chloride ions.
- **Option C: [Co(NH₃)₅Cl]**: This complex has one ionizable chloride ion. It will give a positive test for chloride ions.
- **Option D: [Co(NH₃)₆]**: In this case, there are no chloride ions present outside the coordination sphere. Thus, there are no ionizable chloride ions available in the solution.

4. **Conclusion**: The complex that will not give a test for chloride ions with silver nitrate is **[Co(NH₃)₆]** because it has no ionizable chloride ions.

### Final Answer:
The complex that will not give a test for chloride ions with silver nitrate at 25°C is **[Co(NH₃)₆]**.