Which of the following aqueous solution of cation(s) give(s) white ppt. with NaOH and `NH_(4)OH` solution and formed ppt. is/are further completely dissolved in one of the excess reagent?

Options

A. `Cd^(2+)`
B. `Cr^(3+)`
C. `Sn^(2+)`
D. `Bi^(3+)`

Explanation

To solve the question, we need to analyze the behavior of the given cations (Cadmium \( \text{Cd}^{2+} \), Chromium \( \text{Cr}^{3+} \), Tin \( \text{Sn}^{2+} \), and Bismuth \( \text{Bi}^{3+} \)) when they react with sodium hydroxide (NaOH) and ammonium hydroxide (\( \text{NH}_4\text{OH} \)). We will look for those that form a white precipitate and can dissolve in one of the excess reagents.

### Step-by-Step Solution:

1. **Cadmium \( \text{Cd}^{2+} \)**:
- When \( \text{Cd}^{2+} \) reacts with NaOH, it forms a white precipitate of cadmium hydroxide \( \text{Cd(OH)}_2 \).
- In excess NaOH, \( \text{Cd(OH)}_2 \) is **not soluble**.
- In excess \( \text{NH}_4\text{OH} \), \( \text{Cd(OH)}_2 \) is **soluble** and forms the complex \( \text{Cd(NH}_3\text{)}_4^{2+} \).
- **Conclusion**: Cadmium \( \text{Cd}^{2+} \) gives a white precipitate and is soluble in excess \( \text{NH}_4\text{OH} \).

2. **Chromium \( \text{Cr}^{3+} \)**:
- When \( \text{Cr}^{3+} \) reacts with NaOH, it forms a green precipitate of chromium hydroxide \( \text{Cr(OH)}_3 \).
- In excess NaOH, \( \text{Cr(OH)}_3 \) is **soluble** and forms the complex \( \text{NaCr(OH)}_4 \).
- In excess \( \text{NH}_4\text{OH} \), it forms \( \text{Cr(NH}_3\text{)}_6^{3+} \) which is also **soluble**.
- **Conclusion**: Chromium \( \text{Cr}^{3+} \) does not give a white precipitate.

3. **Tin \( \text{Sn}^{2+} \)**:
- When \( \text{Sn}^{2+} \) reacts with NaOH, it forms a white precipitate of tin hydroxide \( \text{Sn(OH)}_2 \).
- In excess NaOH, \( \text{Sn(OH)}_2 \) is **soluble** and forms the complex \( \text{Na}_2\text{Sn(OH)}_4 \).
- In excess \( \text{NH}_4\text{OH} \), it is **not soluble**.
- **Conclusion**: Tin \( \text{Sn}^{2+} \) gives a white precipitate and is soluble in excess NaOH.

4. **Bismuth \( \text{Bi}^{3+} \)**:
- When \( \text{Bi}^{3+} \) reacts with NaOH, it forms a white precipitate of bismuth hydroxide \( \text{Bi(OH)}_3 \).
- In excess NaOH, \( \text{Bi(OH)}_3 \) is **not soluble**.
- In excess \( \text{NH}_4\text{OH} \), it is also **not soluble**.
- **Conclusion**: Bismuth \( \text{Bi}^{3+} \) gives a white precipitate but is not soluble in either reagent.

### Final Answer:
The cations that give a white precipitate with NaOH and \( \text{NH}_4\text{OH} \) and are soluble in one of the excess reagents are:
- **Cadmium \( \text{Cd}^{2+} \) (soluble in excess \( \text{NH}_4\text{OH} \))**
- **Tin \( \text{Sn}^{2+} \) (soluble in excess NaOH)**

Thus, the correct options are **A (Cadmium \( \text{Cd}^{2+} \)) and C (Tin \( \text{Sn}^{2+} \))**.