Which of the following mixtures cann be separated by using excess `NH_(3) ` solution?

Options

A. `Bi_((aq.))^(3+) and Al_((aq.))^(3+)`
B. `Al_((aq.))^(3+) and Zn_((aq.))^(2+)`
C. `Hg_((aq.))^(2+) and Pb_((aq.))^(2+)`
D. `Cu_((aq.))^(2+) and Cd_((aq.))^(2+)`

Explanation

To solve the question of which mixtures can be separated using excess ammonia solution, we will analyze the behavior of aluminum ions (Al³⁺) and zinc ions (Zn²⁺) in the presence of ammonia.

### Step-by-Step Solution:

1. **Identify the Ions**: The ions we are considering for separation are Al³⁺ (aluminum ions) and Zn²⁺ (zinc ions).

2. **Reaction of Aluminum Ions with Ammonia**:
- When aluminum ions (Al³⁺) are treated with ammonia (NH₃) in water, they react to form aluminum hydroxide (Al(OH)₃):
\[ \text{Al}^{3+} + 3 \text{NH}_3 + 3 \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 \downarrow + 3 \text{NH}_4^+ \]
- Aluminum hydroxide precipitates out as a solid (insoluble in excess ammonia).

3. **Reaction of Zinc Ions with Ammonia**:
- When zinc ions (Zn²⁺) are treated with ammonia, they initially form zinc hydroxide (Zn(OH)₂):
\[ \text{Zn}^{2+} + 2 \text{NH}_3 + 2 \text{H}_2\text{O} \rightarrow \text{Zn(OH)}_2 \downarrow + 2 \text{NH}_4^+ \]
- Zinc hydroxide is also a precipitate, but when excess ammonia is added, it dissolves to form a soluble complex:
\[ \text{Zn(OH)}_2 + 4 \text{NH}_3 \rightarrow \text{[Zn(NH}_3\text{)}_4]^{2+} \]

4. **Separation of the Mixture**:
- After the initial reaction with ammonia, we have:
- Al(OH)₃ as an insoluble precipitate.
- [Zn(NH₃)₄]²⁺ as a soluble complex in solution.
- By filtering the mixture, we can separate the insoluble aluminum hydroxide from the soluble zinc complex.

5. **Conclusion**:
- Therefore, the mixture of Al³⁺ and Zn²⁺ can be effectively separated using excess ammonia solution.

### Final Answer:
The mixture of aluminum ions (Al³⁺) and zinc ions (Zn²⁺) can be separated using excess ammonia solution.

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