In the following reaction `2MnO_(4^(-))+5H_(2)O_(2)^(18)+6H^(+)rarr2Mn^(2+) + 8H_(2)O+5O_(2)`
The radioactive oxygen will appear in :

Options

A. `H_(2)O`
B. `O_(2)`
C. both
D. above reaction does not take place

Explanation

To solve the problem, we need to analyze the given chemical reaction and identify where the radioactive oxygen (O-18) appears in the products.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
The reaction provided is:
\[ 2 \text{MnO}_4^{-} + 5 \text{H}_2\text{O}_2^{(18)} + 6 \text{H}^{+} \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} + 5 \text{O}_2 \]
Here, \(\text{H}_2\text{O}_2^{(18)}\) indicates that we have hydrogen peroxide with radioactive oxygen (O-18).

2. **Understand the Role of H2O2**:
Hydrogen peroxide (H2O2) acts as a reducing agent in this reaction. It gets oxidized to form oxygen gas (O2) and water (H2O).

3. **Analyze the Oxidation Process**:
During the oxidation of H2O2, the O-O bond in H2O2 does not break. Instead, the oxygen atoms from H2O2 are released as molecular oxygen (O2). This means that the oxygen in the O2 produced comes from the H2O2.

4. **Identify the Source of Oxygen in O2**:
Since the H2O2 used in the reaction is labeled as H2O2^(18), the oxygen atoms in the O2 produced must also contain the radioactive isotope O-18. Therefore, the O2 produced in the reaction will contain the radioactive oxygen.

5. **Conclusion**:
The radioactive oxygen (O-18) will appear in the molecular oxygen (O2) produced as a result of the reaction.

### Final Answer:
The radioactive oxygen will appear in **O2**.

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