`Cr_(2)O_(7)^(2-)+H^(+)+SO_(3)^(2-) to Cr^(3+)(aq.)+SO_(4)^(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 analyze the reaction \( \text{Cr}_2\text{O}_7^{2-} + \text{H}^+ + \text{SO}_3^{2-} \rightarrow \text{Cr}^{3+} + \text{SO}_4^{2-} \), we will follow these steps:

### Step 1: Determine the oxidation states of the elements in the reactants.
- **Chromium in \( \text{Cr}_2\text{O}_7^{2-} \)**:
- Let the oxidation state of Cr be \( x \).
- The total charge from the two Cr atoms is \( 2x \).
- The total contribution from the seven O atoms is \( 7 \times (-2) = -14 \).
- The overall charge of the dichromate ion is \( -2 \).

Therefore, the equation is:
\[ 2x - 14 = -2 \implies 2x = 12 \implies x = +6 \]
So, the oxidation state of Cr in \( \text{Cr}_2\text{O}_7^{2-} \) is +6.

- **Hydrogen in \( \text{H}^+ \)**:
- The oxidation state of H is +1.

- **Sulfur in \( \text{SO}_3^{2-} \)**:
- Let the oxidation state of S be \( y \).
- The total contribution from the three O atoms is \( 3 \times (-2) = -6 \).
- The overall charge of the sulfite ion is \( -2 \).

Therefore, the equation is:
\[ y - 6 = -2 \implies y = +4 \]
So, the oxidation state of S in \( \text{SO}_3^{2-} \) is +4.

### Step 2: Determine the oxidation states of the elements in the products.
- **Chromium in \( \text{Cr}^{3+} \)**:
- The oxidation state of Cr is +3.

- **Sulfur in \( \text{SO}_4^{2-} \)**:
- Let the oxidation state of S be \( z \).
- The total contribution from the four O atoms is \( 4 \times (-2) = -8 \).
- The overall charge of the sulfate ion is \( -2 \).

Therefore, the equation is:
\[ z - 8 = -2 \implies z = +6 \]
So, the oxidation state of S in \( \text{SO}_4^{2-} \) is +6.

### Step 3: Analyze the changes in oxidation states.
- **Chromium**: Changes from +6 (in \( \text{Cr}_2\text{O}_7^{2-} \)) to +3 (in \( \text{Cr}^{3+} \)), indicating a reduction.
- **Sulfur**: Changes from +4 (in \( \text{SO}_3^{2-} \)) to +6 (in \( \text{SO}_4^{2-} \)), indicating an oxidation.

### Step 4: Identify the type of reaction.
Since there is a simultaneous oxidation (of sulfur) and reduction (of chromium), this is classified as a redox reaction. Furthermore, since the reaction involves different species undergoing oxidation and reduction, it is classified as an intermolecular redox reaction.

### Final Answer:
The reaction is an **intermolecular redox reaction**.

---