`2KI+HgI_(2) darr to K_(2)[HgI_(4)]`

Options

A. For coloured ppt./Black ppt
B. For coloured solution.
C. for clear/colourless solution
D. For white ppt.

Answer

To determine the nature of the product formed in the reaction \(2 \text{KI} + \text{HgI}_2 \rightarrow \text{K}_2[\text{HgI}_4\), we will analyze the reaction step by step.

### Step 1: Identify the Reactants
The reactants in this reaction are potassium iodide (KI) and mercury(II) iodide (HgI2).

### Step 2: Analyze the Reaction
When potassium iodide reacts with mercury(II) iodide, they form a complex compound, which is potassium tetraiodomercurate(II), represented as \( \text{K}_2[\text{HgI}_4] \).

### Step 3: Determine the Solubility of the Product
The product \( \text{K}_2[\text{HgI}_4] \) is a soluble complex. Since it is soluble, it will not precipitate out of the solution. Therefore, we can eliminate options that suggest the formation of a precipitate.

### Step 4: Identify the Color of the Product
The freshly prepared \( \text{K}_2[\text{HgI}_4] \) is colorless. Although it can form yellow crystals upon standing or reacting with water, the freshly prepared solution is clear and colorless.

### Conclusion
Based on the analysis, the reaction \(2 \text{KI} + \text{HgI}_2 \rightarrow \text{K}_2[\text{HgI}_4]\) results in a clear and colorless solution.

### Final Answer
The correct option is C: clear and colorless solution.

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