`NH_(4)Cl+NaNO_(2) overset(Delta)toN_(2)uarr+NaCl+H_(2)O`

Options

A. For disproportionation reaction.
B. For comproportionation reaction.
C. For either intermolecular redox reaction or displacement reaction
D. For either thermal combination redox reaction or thermal decomposition redox reaction.

Answer

To determine the type of reaction represented by the equation:

\[ \text{NH}_4\text{Cl} + \text{NaNO}_2 \xrightarrow{\Delta} \text{N}_2 + \text{NaCl} + \text{H}_2\text{O} \]

we will analyze the oxidation states of the elements involved in the reactants and products.

### Step 1: Identify the oxidation states in the reactants

1. **For NH4Cl**:
- NH4+ has a charge of +1.
- Chlorine (Cl) has an oxidation state of -1.
- Let the oxidation state of nitrogen in NH4+ be \( x \).
- The equation for NH4+ is:
\[ x + 4(-1) = +1 \implies x - 4 = 1 \implies x = +3 \]
- Therefore, nitrogen in NH4Cl is in the -3 oxidation state.

2. **For NaNO2**:
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of nitrogen in NaNO2 be \( y \).
- The equation for NaNO2 is:
\[ +1 + y + 2(-2) = 0 \implies y - 3 = 0 \implies y = +3 \]
- Therefore, nitrogen in NaNO2 is in the +3 oxidation state.

### Step 2: Identify the oxidation states in the products

1. **For N2**:
- Nitrogen in its elemental form (N2) has an oxidation state of 0.

2. **For NaCl**:
- Sodium (Na) is +1 and Chlorine (Cl) is -1.

3. **For H2O**:
- Hydrogen (H) is +1 and Oxygen (O) is -2.

### Step 3: Analyze the changes in oxidation states

- In the reactants:
- Nitrogen in NH4Cl is -3.
- Nitrogen in NaNO2 is +3.

- In the products:
- Nitrogen in N2 is 0.

### Step 4: Determine the type of reaction

- One nitrogen atom is oxidized from -3 to 0 (loss of electrons).
- Another nitrogen atom is reduced from +3 to 0 (gain of electrons).

This indicates that we have a situation where a single species (nitrogen) is present in two different oxidation states in the reactants, which is characteristic of a **comproportionation reaction**.

### Conclusion

The type of reaction represented by the equation is a **comproportionation reaction**.

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