In the following reaction using isotopic `.^(18)O` in `H_(2)O_(2), 2MnO_(4)^(-)+3H_(2)O_(2)^(18)rarr 2MnO_(2)+3O_(2)+2H_(2)O+2OH^(-)` isotopic oxygen goes,

Options

A. with `O_(2)`
B. with `MnO_(2)`
C. with `OH^(-)`
D. one with `O_(2)` and one with `MnO_(2)`

Answer

To solve the question regarding the reaction involving isotopic oxygen \(^{18}O\) in \(H_2O_2\), we will analyze the reaction step by step.

### Step 1: Write down the reaction
The given reaction is:
\[ 2MnO_4^- + 3H_2O_2^{(18)} \rightarrow 2MnO_2 + 3O_2 + 2H_2O + 2OH^- \]

### Step 2: Identify the components
In this reaction, we have:
- \(MnO_4^-\) (permanganate ion)
- \(H_2O_2^{(18)}\) (hydrogen peroxide with isotopic oxygen)
- Products: \(MnO_2\), \(O_2\), \(H_2O\), and \(OH^-\)

### Step 3: Analyze the role of \(H_2O_2^{(18)}\)
In the reaction, \(H_2O_2\) decomposes in a basic medium to produce water and dioxygen. The isotopic oxygen \(^{18}O\) is present in the \(H_2O_2\) molecule.

### Step 4: Determine where the isotopic oxygen goes
When \(H_2O_2\) decomposes, it produces \(O_2\) and \(H_2O\). The isotopic oxygen \(^{18}O\) can either end up in the water or in the dioxygen produced.

### Step 5: Identify the final products
The products of the reaction include:
- \(O_2\) (dioxygen)
- \(H_2O\) (water)

### Step 6: Conclusion on the fate of isotopic oxygen
Since \(H_2O_2\) decomposes to produce \(O_2\) and some of its oxygen atoms are released as dioxygen, the isotopic oxygen \(^{18}O\) from \(H_2O_2\) will primarily go to form \(O_2\).

### Final Answer
The isotopic oxygen \(^{18}O\) goes with the dioxygen \(O_2\).

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