The blue colour produced on adding `H_2O_2` to acidified `K_2Cr_2O_7` is due to the formation of

Options

A. `CrO_(5)`
B. `Cr_(2)O_(3)`
C. `CrO_(4)^(2-)`
D. `CrO_(3)`

Answer

To solve the question regarding the blue color produced when adding H₂O₂ to acidified K₂Cr₂O₇, we can follow these steps:

### Step 1: Understand the Reactants
We start with potassium dichromate (K₂Cr₂O₇) in an acidic medium, typically with sulfuric acid (H₂SO₄). The dichromate ion (Cr₂O₇²⁻) is known for its orange color.

**Hint:** Identify the reactants and their states in the reaction.

### Step 2: Identify the Reaction
When hydrogen peroxide (H₂O₂) is added to the acidified potassium dichromate, a redox reaction occurs. The H₂O₂ acts as a reducing agent and reduces the dichromate ion.

**Hint:** Consider the roles of H₂O₂ and K₂Cr₂O₇ in the reaction.

### Step 3: Determine the Products
The reduction of the dichromate ion leads to the formation of chromium species. In this case, one of the products formed is CrO₅ (chromium pentoxide), which is responsible for the blue color observed.

**Hint:** Think about the oxidation states of chromium before and after the reaction.

### Step 4: Conclude the Color Change
The blue color observed in the solution is due to the formation of CrO₅. This is a characteristic color change that indicates the presence of this specific chromium compound.

**Hint:** Relate the color change to the specific chromium compound formed.

### Final Answer
The blue color produced on adding H₂O₂ to acidified K₂Cr₂O₇ is due to the formation of CrO₅.

### Summary of Steps:
1. Identify the reactants (K₂Cr₂O₇ and H₂O₂).
2. Recognize the redox reaction taking place.
3. Determine that CrO₅ is formed as a product.
4. Conclude that the blue color is due to CrO₅.