Whenever a reaction between an oxidising agent and reducing agent is carried out

Options

A. A compound of lower oxidation state is formed if the reducing agent is in excess
B. A compound of higher oxidation state is formed if the oxidising agent is in excess
C. both(1)(2)
D. none of the above

Answer

To solve the question regarding the reaction between an oxidizing agent and a reducing agent, we can break down the process step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- An oxidizing agent (e.g., \( F_2 \)) and a reducing agent (e.g., \( P_4 \)) are involved in the reaction.

2. **Understand the Role of Each Agent**:
- The oxidizing agent gains electrons and is reduced, leading to an increase in its oxidation state.
- The reducing agent loses electrons and is oxidized, leading to a decrease in its oxidation state.

3. **Determine the Oxidation States**:
- In the case of \( P_4 \) reacting with \( F_2 \):
- When \( P_4 \) acts as a reducing agent, it can form \( PF_3 \) where phosphorus has an oxidation state of +3.
- When \( F_2 \) is in excess (acting as the oxidizing agent), it can form \( PF_5 \) where phosphorus has an oxidation state of +5.

4. **Analyze the Outcomes Based on the Excess Agent**:
- If the reducing agent is in excess, the product will have a lower oxidation state (e.g., \( PF_3 \)).
- If the oxidizing agent is in excess, the product will have a higher oxidation state (e.g., \( PF_5 \)).

5. **Conclusion**:
- The key takeaway is that the oxidation state of the product depends on which agent is in excess. The reducing agent leads to lower oxidation states, while the oxidizing agent leads to higher oxidation states.