`Na_(2)C_(2)O_(4)+H_(2)SO_(4)(Conc.) to Na_(2)SO_(4)+CO+CO_(2)`

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 solve the question regarding the reaction of sodium oxalate (Na₂C₂O₄) with concentrated sulfuric acid (H₂SO₄), we will analyze the reaction step by step, focusing on the oxidation states of the elements involved and identifying the type of reaction.

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction given is:
\[ \text{Na}_2\text{C}_2\text{O}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{CO} + \text{CO}_2 \]

2. **Determine Oxidation States**:
- For Na₂C₂O₄:
- Sodium (Na) = +1 (each)
- Oxygen (O) = -2 (each)
- Let the oxidation state of Carbon (C) be \( x \).
- The equation for oxidation states in Na₂C₂O₄ is:
\[ 2(+1) + 2x + 4(-2) = 0 \implies 2 + 2x - 8 = 0 \implies 2x - 6 = 0 \implies x = +3 \]
- Therefore, in Na₂C₂O₄, Carbon is in the +3 oxidation state.

- For H₂SO₄:
- Hydrogen (H) = +1 (each)
- Sulfur (S) = +6
- Oxygen (O) = -2 (each)

- For Na₂SO₄:
- Sodium (Na) = +1 (each)
- Sulfur (S) = +6
- Oxygen (O) = -2 (each)

- For CO:
- Oxygen (O) = -2
- Therefore, Carbon (C) = +2.

- For CO₂:
- Oxygen (O) = -2
- Therefore, Carbon (C) = +4.

3. **Identify Changes in Oxidation States**:
- In the reaction:
- Carbon in Na₂C₂O₄ changes from +3 to +2 in CO (reduction).
- Carbon in Na₂C₂O₄ also changes from +3 to +4 in CO₂ (oxidation).

4. **Classify the Reaction**:
Since a single species (C in Na₂C₂O₄) is both oxidized (to CO₂) and reduced (to CO), this type of reaction is classified as a **disproportionation reaction**.

### Conclusion:
The reaction of sodium oxalate with concentrated sulfuric acid is a **disproportionation reaction**.