An aqueous solution of a metal ion (A) on reaction with Kl gives a black brownish ppt (B) and this aqueous suspension on treatment with excess Kl gives orange yellowish solution. Then A is

Options

A. `pb^(2+)`
B. `Bl^(3+)`
C. `Hg_(2)^(2+)`
D. `Hg^(2+)`

Explanation

To solve the question step by step, we need to identify the metal ion (A) based on the reactions described.

### Step 1: Analyze the Reaction with KI
We start with an aqueous solution of a metal ion (A) that reacts with potassium iodide (KI) to produce a blackish-brown precipitate (B).

**Hint:** Think about common metal ions that form colored precipitates with iodide ions.

### Step 2: Identify Possible Metal Ions
Let's consider some metal ions that could react with KI:
- **Lead (Pb²⁺)**: Forms a yellow precipitate (PbI₂).
- **Bismuth (Bi³⁺)**: Forms a blackish-brown precipitate (BiI₃).
- **Mercurous (Hg₂²⁺)**: Forms a yellow precipitate (HgI).
- **Mercuric (Hg²⁺)**: Forms a scarlet red precipitate (HgI₂).

**Hint:** Focus on the color of the precipitate formed. We are looking for a blackish-brown precipitate.

### Step 3: Evaluate the Precipitate Color
From the options:
- **Pb²⁺** gives a yellow precipitate, which does not match.
- **Bi³⁺** gives a blackish-brown precipitate, which matches.
- **Hg₂²⁺** gives a yellow precipitate, which does not match.
- **Hg²⁺** gives a scarlet red precipitate, which does not match.

**Hint:** Eliminate options that do not produce a blackish-brown precipitate.

### Step 4: Reaction with Excess KI
Next, we consider the reaction of the precipitate (B) with excess KI. The blackish-brown precipitate (BiI₃) formed from bismuth will react with excess KI to form a soluble complex, BiI₄⁻, which is known to give an orange-yellow solution.

**Hint:** Understand the concept of complexation and how excess reagents can change the solubility and color of precipitates.

### Step 5: Confirm the Identity of A
Since Bi³⁺ produces a blackish-brown precipitate (BiI₃) and forms an orange-yellow solution (BiI₄⁻) upon treatment with excess KI, we can conclude that the metal ion (A) is indeed bismuth (Bi³⁺).

**Final Conclusion:** The metal ion (A) is **Bismuth (Bi³⁺)**.

### Summary
- Aqueous solution of metal ion A reacts with KI to form a blackish-brown precipitate (B).
- Upon treatment with excess KI, the precipitate forms an orange-yellow solution.
- The correct identification of A is **Bismuth (Bi³⁺)**.