In an acidified solution of `K_2Cr_2O_7, H_2O_2` is added. Then

Options

A. solution turns green due to formation of `Cr_2O_3`
B. solution turns yellow due to formation of `K_2CrO_4`
C. a blue coloured compound `CrO(O_2)_(2)` is formed
D. solution gives green precipitate of `Cr(OH)_(3)`

Answer

To solve the question regarding the reaction that occurs when hydrogen peroxide (H2O2) is added to an acidified solution of potassium dichromate (K2Cr2O7), we can follow these steps:

### Step 1: Identify the Reactants
In this reaction, the reactants are potassium dichromate (K2Cr2O7) and hydrogen peroxide (H2O2) in an acidified solution.

### Step 2: Understand the Reaction Conditions
The solution is acidified, which means that an acid (like sulfuric acid, H2SO4) is present. This acidic environment is crucial for the reaction to proceed correctly.

### Step 3: Determine the Products
When H2O2 is added to an acidified solution of K2Cr2O7, a redox reaction occurs. The chromium in K2Cr2O7 is in the +6 oxidation state and gets reduced, while H2O2 acts as a reducing agent.

### Step 4: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
\[ K_2Cr_2O_7 + H_2O_2 + H_2SO_4 \rightarrow CrO_2 \cdot H_2O + K_2SO_4 + H_2O \]
In this reaction, the chromium is reduced to a lower oxidation state, forming a deep blue compound known as chromium diperoxide (CrO2).

### Step 5: Identify the Color Change
The formation of chromium diperoxide (CrO2) results in a deep blue color in the solution. This is a significant indicator of the reaction taking place.

### Conclusion
Based on the above steps, the correct answer to the question is that the solution turns blue due to the formation of CrO2, which corresponds to option C.