Among the following , what is the total number of speices which are very good oxidising agents ?
a. `F_(2)`
b. `F^(ө)`
c. `Na`
d. `Na^(o+)`
e. `MnO_(4)^(ө)`

Explanation

To determine the total number of species that are very good oxidizing agents among the given options, we need to analyze the oxidation states of each species and their ability to accept electrons.

**Step 1: Analyze each species for oxidation state and oxidizing ability.**

1. **F₂ (Fluorine gas)**:
- Oxidation state of F in F₂ is 0.
- It can be reduced to F⁻ (oxidation state -1), meaning it can act as an oxidizing agent.
- **Conclusion**: F₂ is a good oxidizing agent.

2. **F⁻ (Fluoride ion)**:
- Oxidation state of F in F⁻ is -1.
- It cannot be reduced further (it can only be oxidized to F₂), so it does not act as a good oxidizing agent.
- **Conclusion**: F⁻ is not a good oxidizing agent.

3. **Na (Sodium)**:
- Oxidation state of Na is 0.
- It can be oxidized to Na⁺ (oxidation state +1), but it cannot be reduced, so it does not act as a good oxidizing agent.
- **Conclusion**: Na is not a good oxidizing agent.

4. **Na⁺ (Sodium ion)**:
- Oxidation state of Na is +1.
- It cannot be reduced further (it can only be reduced to Na), so it does not act as a good oxidizing agent.
- **Conclusion**: Na⁺ is not a good oxidizing agent.

5. **MnO₄⁻ (Permanganate ion)**:
- The oxidation state of Mn in MnO₄⁻ is +7.
- It can be reduced to lower oxidation states (such as +6, +4, +2), meaning it can accept electrons and act as an oxidizing agent.
- **Conclusion**: MnO₄⁻ is a very good oxidizing agent.

**Step 2: Count the good oxidizing agents identified.**

From the analysis:
- F₂ is a good oxidizing agent.
- MnO₄⁻ is a very good oxidizing agent.
- F⁻, Na, and Na⁺ are not good oxidizing agents.

Thus, the total number of species that are very good oxidizing agents is **2** (F₂ and MnO₄⁻).

**Final Answer: 2**

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