Which of the following reactions represents disproportionation ?

Options

A. `CrO_(5) rarr Cr^(3+)+O_(2)`
B. `IO_(3)^(-)+I^(-)+H^(+)rarr I_(2)`
C. `CrO_(2)Cl_(2)+NaOHrarr Na_(2)CrO_(4)+NaCl+H_(2)O`
D. `Na_(2)S_(2)O_(3)+H_(2)SO_(2)rarr Na_(2)SO_(4)+SO_(2)+S_(8)+H_(2)O`

Explanation

To determine which of the given reactions represents a disproportionation reaction, we need to follow these steps:

### Step 1: Understand Disproportionation Reaction
A disproportionation reaction is a type of redox reaction in which a single substance is both oxidized and reduced, resulting in two different products. This means that one part of the substance loses electrons (is oxidized) while another part gains electrons (is reduced).

### Step 2: Analyze Each Reaction
We will analyze each of the given reactions (A, B, C, D) to check if they involve both oxidation and reduction of the same species.

#### Reaction A:
**CrO5 → Cr3+ + O2**
- Oxidation states:
- Cr in CrO5: +6
- Cr in Cr3+: +3
- O2: 0
- Here, Cr is reduced from +6 to +3, but there is no species being oxidized. Thus, this is **not** a disproportionation reaction.

#### Reaction B:
**I2 → I- + I5+**
- Oxidation states:
- I2: 0
- I-: -1
- I5+: +5
- Iodine is being reduced to -1 and oxidized to +5. However, this is not a classic disproportionation since the reactants are different. Thus, this is **not** a disproportionation reaction.

#### Reaction C:
**CrO2Cl2 + NaOH → Na2CrO4 + NaCl + H2O**
- Oxidation states:
- Cr in CrO2Cl2: +6
- Cr in Na2CrO4: +6
- There is no change in the oxidation state of Cr. Thus, this is **not** a disproportionation reaction.

#### Reaction D:
**Na2S2O3 + H2SO2 → Na2SO4 + SO2 + S8 + H2O**
- Oxidation states:
- S in Na2S2O3: +2
- S in SO2: +4
- S in S8: 0
- Here, sulfur goes from +2 to +4 (oxidized) and to 0 (reduced). Thus, this is a **disproportionation reaction**.

### Conclusion
The reaction that represents a disproportionation reaction is **Reaction D**.

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