The oxidation state of chromium in the final product formed by the reaction between Kl and acidified potassium dichromate solution is

Options

A. `+3`
B. `+2`
C. `+6`
D. `+4`

Answer

To determine the oxidation state of chromium in the final product formed by the reaction between potassium iodide (KI) and acidified potassium dichromate (K2Cr2O7), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are potassium dichromate (K2Cr2O7) and potassium iodide (KI) in an acidic medium (usually sulfuric acid, H2SO4).

### Step 2: Write the Reaction
The overall reaction can be represented as:
\[ \text{K}_2\text{Cr}_2\text{O}_7 + \text{KI} + \text{H}_2\text{SO}_4 \rightarrow \text{Cr}_2(\text{SO}_4)_3 + \text{K}_2\text{SO}_4 + \text{I}_2 + \text{H}_2\text{O} \]

### Step 3: Determine the Oxidation States in Reactants
- In potassium dichromate (K2Cr2O7), the oxidation state of chromium (Cr) is +6.
- In potassium iodide (KI), the oxidation state of iodine (I) is -1.

### Step 4: Identify the Products
The products of the reaction include chromium sulfate (Cr2(SO4)3), potassium sulfate (K2SO4), iodine (I2), and water (H2O).

### Step 5: Determine the Oxidation States in Products
- In chromium sulfate (Cr2(SO4)3), the oxidation state of chromium (Cr) is +3.
- In iodine (I2), the oxidation state of iodine (I) is 0 (elemental state).

### Step 6: Analyze the Change in Oxidation States
- The oxidation state of chromium changes from +6 in K2Cr2O7 to +3 in Cr2(SO4)3, indicating a reduction of 3 units.
- The oxidation state of iodine changes from -1 in KI to 0 in I2, indicating an increase of 1 unit for each iodine atom. Since there are 2 iodine atoms produced, the total increase is 2 units.

### Step 7: Conclusion
Since the oxidation state of chromium decreases from +6 to +3, we conclude that in the final product (Cr2(SO4)3), the oxidation state of chromium is +3.

### Final Answer
The oxidation state of chromium in the final product formed by the reaction is **+3**.

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