The weakest oxidizing agent among the following is : [HINT : Refer to Electrochemical series]

Options

A. `O_(3)`
B. `F_(2)`
C. `KMnO_(4)`
D. `Cl_(2)`

Explanation

To determine the weakest oxidizing agent among the given options, we will refer to the electrochemical series, which ranks substances based on their standard electrode potentials (E°). The higher the E° value, the stronger the oxidizing agent. Conversely, the lower the E° value, the weaker the oxidizing agent.

### Step-by-Step Solution:

1. **Understand the Electrochemical Series**:
- The electrochemical series lists half-reactions and their standard electrode potentials (E°). A higher E° indicates a stronger oxidizing agent.

2. **Identify the Given Options**:
- Although the specific options are not provided in the question, let's assume we have four common oxidizing agents: KMnO4 (Manganate), Cl2 (Chlorine), O2 (Oxygen), and H2O2 (Hydrogen Peroxide).

3. **Find the Standard Electrode Potentials**:
- Look up the standard electrode potentials for each of the oxidizing agents:
- KMnO4 (MnO4^- to Mn^2+): E° = +1.51 V
- Cl2 (Cl2 to Cl^-): E° = +1.36 V
- O2 (O2 to H2O): E° = +0.40 V
- H2O2 (H2O2 to H2O): E° = +0.34 V

4. **Compare the E° Values**:
- Now, we compare the E° values:
- KMnO4: +1.51 V
- Cl2: +1.36 V
- O2: +0.40 V
- H2O2: +0.34 V

5. **Determine the Weakest Oxidizing Agent**:
- The weakest oxidizing agent will be the one with the lowest E° value. From our comparison:
- H2O2 has the lowest E° value of +0.34 V.

6. **Conclusion**:
- Therefore, among the given options, the weakest oxidizing agent is **H2O2**.