On adding KI solution in excess to a solution of `CuSO_(4)`, we get a precipitate 'P' and another liquor 'M'. Select the correct pair:

Options

A. P is CuI and M is `I_(2)` solution
B. P is `CuI_(2)` and M is `I_(2)` solution
C. P is CuI and M is `KI_(3)` solution
D. P is `CuI_(2)` and M is `KI_(3)` solution

Answer

To solve the question, we need to analyze the reaction that occurs when potassium iodide (KI) is added to copper sulfate (CuSO₄) in excess.

### Step-by-Step Solution:

**Step 1: Identify the Reactants**
- The reactants in this reaction are potassium iodide (KI) and copper sulfate (CuSO₄).

**Step 2: Write the Reaction**
- When KI is added to CuSO₄, a reaction occurs. The balanced chemical equation for this reaction is:
\[ 4 KI + 2 CuSO₄ \rightarrow 2 CuI + I₂ + 2 K₂SO₄ \]

**Step 3: Identify the Products**
- From the balanced equation, we can see that the products formed are:
- Cuprous iodide (CuI), which is a precipitate (denoted as 'P').
- Iodine (I₂), which remains in solution (denoted as 'M').

**Step 4: Determine the Precipitate and the Liquor**
- The precipitate 'P' formed in this reaction is CuI (cuprous iodide).
- The liquor 'M' consists of iodine (I₂) dissolved in the solution.

**Step 5: Select the Correct Pair**
- Therefore, the correct pair is:
- Precipitate 'P': CuI (cuprous iodide)
- Liquor 'M': I₂ (iodine)

### Final Answer:
The correct pair is:
- Precipitate 'P': CuI (cuprous iodide)
- Liquor 'M': I₂ (iodine)

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