Choose the corret statement regarding following reaction
`HNO_(2) to HNO_(3) + NO uarr`

Options

A. It is an example of disproportionation reaction
B. Equivalent weight of `HNO_(2) =(3M)/(2)`
C. Equivalent weight of `HNO_(3)= 1M`
D. It is an example of intramolecular redox reaction

Answer

To analyze the reaction \( \text{HNO}_2 \rightarrow \text{HNO}_3 + \text{NO} \), we need to determine the oxidation states of nitrogen in the reactants and products, and identify the nature of the reaction.

### Step 1: Determine the oxidation state of nitrogen in HNO2
In \( \text{HNO}_2 \):
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of nitrogen (N) be \( x \).

The overall charge of the molecule is zero, so we can set up the equation:
\[ +1 + x + 2(-2) = 0 \]
This simplifies to:
\[ +1 + x - 4 = 0 \implies x - 3 = 0 \implies x = +3 \]
Thus, the oxidation state of nitrogen in \( \text{HNO}_2 \) is +3.

### Step 2: Determine the oxidation state of nitrogen in HNO3
In \( \text{HNO}_3 \):
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

Let the oxidation state of nitrogen (N) be \( y \). The equation becomes:
\[ +1 + y + 3(-2) = 0 \]
This simplifies to:
\[ +1 + y - 6 = 0 \implies y - 5 = 0 \implies y = +5 \]
Thus, the oxidation state of nitrogen in \( \text{HNO}_3 \) is +5.

### Step 3: Determine the oxidation state of nitrogen in NO
In \( \text{NO} \):
- Oxygen (O) has an oxidation state of -2.

Let the oxidation state of nitrogen (N) be \( z \). The equation becomes:
\[ z + (-2) = 0 \]
This simplifies to:
\[ z - 2 = 0 \implies z = +2 \]
Thus, the oxidation state of nitrogen in \( \text{NO} \) is +2.

### Step 4: Analyze the changes in oxidation states
- In \( \text{HNO}_2 \) (N = +3) to \( \text{HNO}_3 \) (N = +5), nitrogen is oxidized (increase in oxidation state).
- In \( \text{HNO}_2 \) (N = +3) to \( \text{NO} \) (N = +2), nitrogen is reduced (decrease in oxidation state).

### Step 5: Identify the type of reaction
Since one species (HNO2) is both oxidized and reduced in the reaction, this is classified as a **disproportionation reaction**.

### Step 6: Calculate the equivalent weight of HNO2
The equivalent weight is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n \text{ factor}} \]
For \( \text{HNO}_2 \), the molecular weight is approximately 63 g/mol. The n factor is determined by the number of electrons transferred. In this case, since nitrogen goes from +3 to +5 (2 electrons lost) and +3 to +2 (1 electron gained), the total n factor is 2.

Thus, the equivalent weight of \( \text{HNO}_2 \) is:
\[ \text{Equivalent Weight} = \frac{63 \text{ g/mol}}{2} = 31.5 \text{ g/equiv} \]

### Conclusion
The correct statement regarding the reaction \( \text{HNO}_2 \rightarrow \text{HNO}_3 + \text{NO} \) is that it is a **disproportionation reaction**.