STATEMENT - 1 : All `Pb^(+2)` form white ppt with dil. HCl in estimation.
STATEMENT - 2 : `Al^(+3)` form white ppt with `NH_(4)OH//NH_(4)Cl`.
STATEMENT - 3 : `Fe(OH)_(3)` is soluble in excess `NaOH`.

Options

A. TTF
B. TTT
C. FTT
D. FTF

Answer

To analyze the statements provided, we will evaluate each statement one by one.

### Step 1: Evaluate Statement 1
**Statement 1:** All \( \text{Pb}^{2+} \) form white ppt with dilute HCl in estimation.

- When \( \text{Pb}^{2+} \) ions react with dilute HCl, they do not form a white precipitate. Instead, they can form a black precipitate when reacted with H2S in the presence of dilute HCl, which indicates that this statement is false.

**Conclusion for Statement 1:** False

### Step 2: Evaluate Statement 2
**Statement 2:** \( \text{Al}^{3+} \) forms white ppt with \( \text{NH}_4\text{OH} \) / \( \text{NH}_4\text{Cl} \).

- When \( \text{Al}^{3+} \) ions react with \( \text{NH}_4\text{OH} \), they form \( \text{Al(OH)}_3 \), which is a white precipitate. Therefore, this statement is true.

**Conclusion for Statement 2:** True

### Step 3: Evaluate Statement 3
**Statement 3:** \( \text{Fe(OH)}_3 \) is soluble in excess \( \text{NaOH} \).

- When \( \text{Fe(OH)}_3 \) is treated with excess \( \text{NaOH} \), it forms a soluble complex \( \text{Na}_3[\text{Fe(OH)}_6] \). Hence, this statement is also true.

**Conclusion for Statement 3:** True

### Final Conclusion
- Statement 1: False
- Statement 2: True
- Statement 3: True

The correct option based on the evaluation of the statements is that Statement 2 and Statement 3 are true, while Statement 1 is false.