`NaI+H_(2)SO_(4)(Conc.) to Na_(2)SO_(4)+SO_(2)uarr+I_(2)uarr`

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 analyze the reaction \( \text{NaI} + \text{H}_2\text{SO}_4 \text{(Conc.)} \rightarrow \text{Na}_2\text{SO}_4 + \text{SO}_2 \uparrow + \text{I}_2 \uparrow \) and determine its type, we will follow these steps:

### Step 1: Identify the Reactants and Products
The reactants are sodium iodide (\( \text{NaI} \)) and concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)). The products are sodium sulfate (\( \text{Na}_2\text{SO}_4 \)), sulfur dioxide (\( \text{SO}_2 \)), and iodine (\( \text{I}_2 \)).

### Step 2: Assign Oxidation States
- For \( \text{NaI} \):
- Sodium (\( \text{Na} \)): +1
- Iodine (\( \text{I} \)): -1

- For \( \text{H}_2\text{SO}_4 \):
- Hydrogen (\( \text{H} \)): +1 (2 atoms contribute +2)
- Sulfur (\( \text{S} \)): +6 (calculated as follows: \( 2 + x - 8 = 0 \) → \( x = +6 \))
- Oxygen (\( \text{O} \)): -2 (4 atoms contribute -8)

- For \( \text{Na}_2\text{SO}_4 \):
- Sodium (\( \text{Na} \)): +1 (2 atoms contribute +2)
- Sulfur (\( \text{S} \)): +6 (same as above)
- Oxygen (\( \text{O} \)): -2 (4 atoms contribute -8)

- For \( \text{SO}_2 \):
- Sulfur (\( \text{S} \)): +4 (calculated as follows: \( x - 4 = 0 \) → \( x = +4 \))
- Oxygen (\( \text{O} \)): -2 (2 atoms contribute -4)

- For \( \text{I}_2 \):
- Iodine (\( \text{I} \)): 0 (as it is in elemental form)

### Step 3: Determine Changes in Oxidation States
- Iodine changes from -1 in \( \text{NaI} \) to 0 in \( \text{I}_2 \) (oxidation).
- Sulfur changes from +6 in \( \text{H}_2\text{SO}_4 \) to +4 in \( \text{SO}_2 \) (reduction).

### Step 4: Identify the Type of Reaction
Since one element (iodine) is oxidized (losing electrons) and another element (sulfur) is reduced (gaining electrons), this reaction is a redox reaction.

### Step 5: Determine if it is Disproportionation
A disproportionation reaction involves a single element undergoing both oxidation and reduction. In this case, iodine is oxidized, and sulfur is reduced, but they are different elements. Therefore, it is not a disproportionation reaction.

### Step 6: Conclusion
The reaction is classified as an **intermolecular redox reaction** because it involves the transfer of electrons between different species (iodine and sulfur).

### Final Answer
The type of reaction is **intermolecular redox reaction**.

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