Among oxyacids of nitrogen `HNO_(2)` is unstabe, it can acts as

Options

A. Oxidising agent
B. Reducing agent
C. Oxidising agent and Reducing agent
D. Its solution in stable

Answer

To solve the problem regarding the stability of the oxyacid \( HNO_2 \) (nitrous acid) and its behavior as an oxidizing and reducing agent, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Oxidation State of Nitrogen in \( HNO_2 \)**:
- The formula for calculating the oxidation state is based on the overall charge of the molecule being zero.
- In \( HNO_2 \):
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- The equation can be set up as follows:
\[ +1 + x + 2(-2) = 0 \]
where \( x \) is the oxidation state of nitrogen (N).
- Simplifying gives:
\[ +1 + x - 4 = 0 \implies x - 3 = 0 \implies x = +3 \]
- Thus, the oxidation state of nitrogen in \( HNO_2 \) is +3.

2. **Analyze the Possible Oxidation States of Nitrogen**:
- The oxidation state of nitrogen can vary from -3 to +5.
- The possible oxidation states of nitrogen are:
- Maximum: +5 (in \( HNO_3 \))
- Minimum: -3 (in \( NH_3 \))
- Since \( HNO_2 \) has nitrogen in the +3 oxidation state, it is in an intermediate position.

3. **Determine the Behavior of \( HNO_2 \)**:
- Since nitrogen can exist in various oxidation states, \( HNO_2 \) can either:
- Be oxidized from +3 to +5 (acting as a reducing agent).
- Be reduced from +3 to -3 (acting as an oxidizing agent).
- Therefore, \( HNO_2 \) can act as both an oxidizing agent and a reducing agent.

4. **Conclusion**:
- The instability of \( HNO_2 \) allows it to participate in redox reactions, where it can either donate electrons (reducing agent) or accept electrons (oxidizing agent).
- Hence, the final answer is that \( HNO_2 \) can act as both an oxidizing agent and a reducing agent.

### Final Answer:
Among the oxyacids of nitrogen, \( HNO_2 \) is unstable and can act as both an oxidizing agent and a reducing agent.