Which one of the following reagents can be used for differentiating the nitrates of `Cu^(2+)` and `Bi^(3+)`?

Options

A. Excess of water
B. `NH_4OH` (excess)
C. Excess of KI
D. All of above

Explanation

To differentiate between the nitrates of \( \text{Cu}^{2+} \) and \( \text{Bi}^{3+} \), we can use three different reagents: excess water, ammonium hydroxide, and excess potassium iodide (KI). Let's analyze the reactions step by step.

### Step 1: Reaction with Excess Water
1. **Copper(II) Nitrate (\( \text{Cu(NO}_3\text{)}_2 \))**: When treated with excess water, there is no observable change. The copper ions remain in solution.
2. **Bismuth Nitrate (\( \text{Bi(NO}_3\text{)}_3 \))**: In contrast, bismuth nitrate reacts with excess water to form a white precipitate of bismuth oxide nitrate (\( \text{BiO(NO}_3\text{)} \)).

**Conclusion**: Excess water can differentiate between \( \text{Cu}^{2+} \) (no change) and \( \text{Bi}^{3+} \) (white precipitate).

### Step 2: Reaction with Excess Ammonium Hydroxide
1. **Copper(II) Nitrate**: When treated with excess ammonium hydroxide, it forms a deep blue complex, specifically \( \text{Cu(NH}_3\text{)}_4^{2+} \).
- Reaction:
\[ \text{Cu}^{2+} + 4 \text{NH}_4\text{OH} \rightarrow \text{Cu(NH}_3\text{)}_4^{2+} + 4 \text{H}_2\text{O} \]
2. **Bismuth Nitrate**: On the other hand, bismuth reacts with ammonium hydroxide to form a white precipitate.
- Reaction:
\[ \text{Bi}^{3+} + 3 \text{NH}_4\text{OH} \rightarrow \text{Bi(OH)}_3 \text{ (white precipitate)} + 3 \text{NH}_4^{+} \]

**Conclusion**: Ammonium hydroxide can also differentiate between \( \text{Cu}^{2+} \) (deep blue solution) and \( \text{Bi}^{3+} \) (white precipitate).

### Step 3: Reaction with Excess Potassium Iodide (KI)
1. **Copper(II) Nitrate**: When treated with excess KI, it forms a white precipitate of cupric iodide (\( \text{Cu}_2\text{I}_2 \)).
- Reaction:
\[ \text{Cu}^{2+} + 4 \text{I}^{-} \rightarrow \text{Cu}_2\text{I}_2 \text{ (white precipitate)} + \text{I}_2 \]
2. **Bismuth Nitrate**: In contrast, bismuth reacts with KI to form a soluble orange complex \( \text{BiI}_4^{-} \).
- Reaction:
\[ \text{Bi}^{3+} + 4 \text{I}^{-} \rightarrow \text{BiI}_4^{-} \text{ (orange solution)} \]

**Conclusion**: Excess KI can also differentiate between \( \text{Cu}^{2+} \) (white precipitate) and \( \text{Bi}^{3+} \) (orange solution).

### Final Conclusion
Since all three reagents (excess water, ammonium hydroxide, and excess KI) can differentiate between \( \text{Cu}^{2+} \) and \( \text{Bi}^{3+} \), the correct answer is **all of the above**.