A solutionf of a metal ion when treated with KI gives a red precipitate which dissolves in excess KI to give a colourless solution.Moreover, the solution of metal ion on treatment with a solution of cobalt (II) thiocyanate gives rise to a deep-blue crystalline precipitate. The metal ions is

Options

A. `Pb^(2+)`
B. `Hg^(2+)`
C. `Cu^(2+)`
D. `CO^(2+)`

Explanation

To determine the metal ion based on the given reactions, we can break down the information step by step.

### Step 1: Identify the Red Precipitate with KI
When a solution of a metal ion is treated with potassium iodide (KI), it forms a red precipitate. This is characteristic of the mercurous ion (Hg2²⁺), which reacts with iodide ions to form mercurous iodide (HgI2), which appears red.

**Hint:** Look for metal ions that form colored precipitates with iodide ions.

### Step 2: Dissolution in Excess KI
The red precipitate formed (HgI2) dissolves in excess KI to form a colorless solution. This indicates the formation of a soluble complex, specifically K2HgI4 (Nessler's reagent). This further confirms that the metal ion is indeed mercury (Hg).

**Hint:** Consider the behavior of precipitates in the presence of excess reagents to identify soluble complexes.

### Step 3: Reaction with Cobalt Thiocyanate
Next, the solution of the metal ion is treated with cobalt (II) thiocyanate (Co(SCN)2), resulting in a deep blue crystalline precipitate. This deep blue color is characteristic of the mercuric thiocyanate complex formed when mercuric ions (Hg²⁺) react with thiocyanate ions (SCN⁻).

**Hint:** Identify the color changes and precipitate formations that are characteristic of specific metal ions.

### Conclusion
Based on the above steps, we conclude that the metal ion in question is mercuric ion (Hg²⁺).

**Final Answer:** The metal ion is **mercury (Hg²⁺)**.