The number of chloride ions which would be precipated when one mole of the complex `PtCl_(4).4NH_(3)` is treated with silver nitrate is : (here coordination number of platinum is 6).

Options

A. four
B. one
C. three
D. two

Explanation

To solve the problem, we need to analyze the complex \( \text{PtCl}_4 \cdot 4\text{NH}_3 \) and understand how many chloride ions can be precipitated when it reacts with silver nitrate (\( \text{AgNO}_3 \)).

### Step-by-Step Solution:

1. **Identify the Coordination Complex**:
The complex given is \( \text{PtCl}_4 \cdot 4\text{NH}_3 \). Here, platinum (Pt) has a coordination number of 6, which means it can coordinate with 6 ligands.

2. **Determine the Ligands**:
In the complex, there are 4 ammonia (\( \text{NH}_3 \)) ligands and 4 chloride (\( \text{Cl}^- \)) ions. However, since the coordination number is 6, we need to determine how many of these chloride ions are ionizable.

3. **Distribution of Chloride Ions**:
In coordination complexes, some ligands are bound within the coordination sphere (non-ionizable) while others are outside (ionizable). Given that the coordination number is 6 and there are 4 ammonia ligands, this means that 2 chloride ions must be outside the coordination sphere as ionizable ions.

4. **Reaction with Silver Nitrate**:
When the complex \( \text{PtCl}_4 \cdot 4\text{NH}_3 \) is treated with silver nitrate, the ionizable chloride ions will react with silver ions (\( \text{Ag}^+ \)) to form silver chloride (\( \text{AgCl} \)), which is a precipitate.

5. **Calculate the Precipitate**:
Since there are 2 ionizable chloride ions in the complex, when treated with silver nitrate, 2 moles of \( \text{AgCl} \) will precipitate.

6. **Conclusion**:
Therefore, the number of chloride ions that would be precipitated when one mole of the complex \( \text{PtCl}_4 \cdot 4\text{NH}_3 \) is treated with silver nitrate is **2**.

### Final Answer:
**2 chloride ions will be precipitated.**

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