The unique behaviour of CU, having a positive `E^(@)` (reduction potential) accounts for its inability to liberate `H_(2)` from acids,. Only oxidising acids (nitric acid and hot concentrated sulphuric acid) react with Cu. The high energy of transform Cu(s) to `Cu^(2+)(aq.)` is not balanced by its hydration enthalpy.
On the other hand, All Cu(II) halides are known except iodide. in this case, `Cu^(2+)` oxidises `I^(-)` to `I_(2)`:
`2Cu^(2+)+4I^(-) to 2CuI(s)+I_(2)`
However, copper (I) compounds are unstable in aqueous solution annd undergo disproportionation.
`2Cu^(+)(aq.)toCu^(2+)(aq.)+Cu`
The stability of `Cu^(2+)(aq.)` rather than `Cu^(+)(aq)` is due to the much more negative `Delta_(Hyd)` of `Cu^(2+)(aq.)` than `Cu^(+)(aq.)`
Q. Consider the following transformation:
`CuSO_(4)(aq.)+KI(excess)to`product
Select the correct statement:

Options

A. Product contains `[Cu(H_(2)O_(4))]^(2+)` ion.
B. Presence of brown colouration in product is due to `I_(3)^(-)` ion
C. Oxidation state of sulphur in reactant and product is different
D. white ppt. of `CuI_(2)` is observed in product

Explanation

To solve the given question, we need to analyze the reaction between copper(II) sulfate (CuSO₄) and potassium iodide (KI) in excess. Let's break down the steps to arrive at the correct answer.

### Step 1: Write the Reaction
The reaction between CuSO₄ and KI can be written as:
\[ \text{CuSO}_4(aq) + 2 \text{KI}(aq) \rightarrow \text{CuI}(s) + \text{I}_2(aq) + \text{K}_2\text{SO}_4(aq) \]

### Step 2: Identify the Products
From the reaction, we can identify the products:
- Copper(I) iodide (CuI) which is a white precipitate.
- Iodine (I₂) which is responsible for the brown coloration in the solution.
- Potassium sulfate (K₂SO₄) remains in the solution.

### Step 3: Analyze the Statements
Now, let's analyze the statements provided in the question:

1. **Product contains CuH₂O₄**: This is incorrect. The product does not contain CuH₂O₄.

2. **Presence of brown coloration in product is due to I₃⁻**: This is correct. The brown color is due to the formation of I₂ from the oxidation of iodide ions (I⁻).

3. **Oxidation state of sulfur in reactant and product is different**: This is incorrect. The oxidation state of sulfur in both CuSO₄ and K₂SO₄ remains +6.

4. **PPT of CuI₂ is observed in the product**: This is incorrect. The precipitate formed is CuI, not CuI₂.

### Conclusion
Based on the analysis:
- The correct statement is: "Presence of brown coloration in product is due to I₃⁻."

### Final Answer
The correct option is the one that states the presence of brown coloration due to I₃⁻.

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