`PbCl_(2)+Klrarr underset("Yellow ppt")([A])+KCl`
`underset("Yellow ppt")([A])+underset("excess")(Kl) rarr underset("Soluble")([B])`
Compound [A] and [B] are

Options

A. `Pbl_(4) and K_(2)[Pbl_(4)]` respectively
B. `K_(2)[Pbl_(4)] and Pbl_(4)` respectively
C. `Pbl_(2) and K_(2)[Pbl_(4)]` respectively
D. `Pbl_(2) and K_(2)[Pbl_(2)]` respectively

Answer

To solve the question, we will analyze the reactions step by step.

### Step 1: Identify the first reaction
The first reaction given is:
\[ \text{PbCl}_2 + \text{KI} \rightarrow \text{[A]} + \text{KCl} \]

In this reaction, lead(II) chloride (PbCl₂) reacts with potassium iodide (KI). The product formed is a yellow precipitate.

### Step 2: Determine the yellow precipitate
The yellow precipitate formed in this reaction is lead(II) iodide (PbI₂). Thus, we can identify:
\[ \text{[A]} = \text{PbI}_2 \]

### Step 3: Identify the second reaction
The second part of the question states:
\[ \text{[A]} + \text{excess KI} \rightarrow \text{[B]} \]

Here, the yellow precipitate (PbI₂) reacts with excess potassium iodide (KI).

### Step 4: Determine the soluble compound
When lead(II) iodide (PbI₂) reacts with excess potassium iodide (KI), it forms a soluble complex:
\[ \text{PbI}_2 + 2\text{KI} \rightarrow \text{K}_2\text{PbI}_4 \]

Thus, we can identify:
\[ \text{[B]} = \text{K}_2\text{PbI}_4 \]

### Conclusion
From the above steps, we conclude that:
- Compound [A] is lead(II) iodide (PbI₂).
- Compound [B] is potassium tetraiodoplumbate(II) (K₂PbI₄).

### Final Answer
- [A] = PbI₂
- [B] = K₂PbI₄

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