The final products formed on the addition of KI to copper sulphate solution are:

Options

A. `K_(2)SO_(4), I_(2)`
B. `K_(2)SO_(4),Cu_(2)I_(2)` and `I_(2)`
C. `K_(2)SO_(4) and Cu_(2)O`
D. `K_(2)SO_(4),CuO and I_(2)`

Explanation

To determine the final products formed when potassium iodide (KI) is added to copper sulfate (CuSO₄) solution, we can follow these steps:

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

### Step 2: Identify the Reaction
When KI is added to CuSO₄, a double displacement reaction occurs. The iodide ions (I⁻) will react with copper ions (Cu²⁺) to form copper(I) iodide (CuI) and potassium sulfate (K₂SO₄).

### Step 3: Write the Chemical Equation
The unbalanced chemical equation for the reaction is:
\[ \text{CuSO}_4 + 2 \text{KI} \rightarrow \text{CuI} + \text{K}_2\text{SO}_4 + \text{I}_2 \]

### Step 4: Balance the Chemical Equation
To balance the equation, we need to ensure that the number of atoms for each element is the same on both sides:
- Copper (Cu): 2 on the left (from CuSO₄) and 2 on the right (from CuI).
- Iodine (I): 2 from 2KI on the left and 2 from I₂ on the right.
- Potassium (K): 2 from 2KI on the left and 2 from K₂SO₄ on the right.
- Sulfate (SO₄): 1 from CuSO₄ on the left and 1 from K₂SO₄ on the right.

The balanced equation is:
\[ \text{CuSO}_4 + 2 \text{KI} \rightarrow \text{Cu}_2\text{I}_2 + \text{K}_2\text{SO}_4 + \text{I}_2 \]

### Step 5: Identify the Final Products
From the balanced equation, the final products formed are:
1. Copper(I) iodide (Cu₂I₂)
2. Potassium sulfate (K₂SO₄)
3. Iodine (I₂)

### Conclusion
The final products formed on the addition of KI to copper sulfate solution are Cu₂I₂, K₂SO₄, and I₂.