Which of the following reactions are disproportionation reactions?
`(A)" " Cu^(+)rarrCu^(2+)+Cu`
`(B)" "3 MnO_(4)^(2-)+4H^(+)rarr2MnO_(4)^(-)+MnO_(2)+2H_(2)O`
`(C )" "2KMnO_(4)rarrK_(2)MnO_(4)+MnO_(2)+O_(2)`
`(D)" "2MnO_(4)^(-)+3Mn^(2+)+2H_(2)O rarr5MnO_(2)+4H^(+)`

Options

A. (i)
B. (i),(ii) and (iii)
C. (ii), (iii) and (iv)
D. (i) and (iv)

Explanation

To determine which of the given reactions are disproportionation reactions, we need to understand the definition of a disproportionation reaction. A disproportionation reaction is one in which the same species is both oxidized and reduced in the same reaction.

Let's analyze each of the reactions step by step:

### Reaction A:
\[ \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu} \]

1. **Identify oxidation states**:
- Cu in Cu⁺: +1
- Cu in Cu²⁺: +2
- Cu in elemental Cu: 0

2. **Determine changes**:
- Cu⁺ is oxidized to Cu²⁺ (losing an electron).
- Cu⁺ is also reduced to elemental Cu (gaining an electron).

3. **Conclusion**:
- Since the same species (Cu) is both oxidized and reduced, this is a disproportionation reaction.

### Reaction B:
\[ 3 \text{MnO}_4^{2-} + 4 \text{H}^+ \rightarrow 2 \text{MnO}_4^{-} + \text{MnO}_2 + 2 \text{H}_2\text{O} \]

1. **Identify oxidation states**:
- Mn in MnO₄²⁻: +6
- Mn in MnO₄⁻: +7
- Mn in MnO₂: +4

2. **Determine changes**:
- Mn⁶⁺ (from MnO₄²⁻) is oxidized to Mn⁷⁺ (in MnO₄⁻).
- Mn⁶⁺ (from MnO₄²⁻) is reduced to Mn⁴⁺ (in MnO₂).

3. **Conclusion**:
- Since the same species (Mn) is both oxidized and reduced, this is also a disproportionation reaction.

### Reaction C:
\[ 2 \text{KMnO}_4 \rightarrow \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2 \]

1. **Identify oxidation states**:
- Mn in KMnO₄: +7
- Mn in K₂MnO₄: +6
- Mn in MnO₂: +4

2. **Determine changes**:
- Mn⁷⁺ (from KMnO₄) is reduced to Mn⁶⁺ (in K₂MnO₄).
- Mn⁷⁺ (from KMnO₄) is also reduced to Mn⁴⁺ (in MnO₂).

3. **Conclusion**:
- Since Mn is only being reduced and not oxidized, this is not a disproportionation reaction.

### Reaction D:
\[ 2 \text{MnO}_4^{-} + 3 \text{Mn}^{2+} + 2 \text{H}_2\text{O} \rightarrow 5 \text{MnO}_2 + 4 \text{H}^+ \]

1. **Identify oxidation states**:
- Mn in MnO₄⁻: +7
- Mn in Mn²⁺: +2
- Mn in MnO₂: +4

2. **Determine changes**:
- Mn⁷⁺ (from MnO₄⁻) is reduced to Mn⁴⁺ (in MnO₂).
- Mn²⁺ is oxidized to Mn⁴⁺ (in MnO₂).

3. **Conclusion**:
- Since Mn is being oxidized and reduced, this is a disproportionation reaction.

### Final Summary:
- **Disproportionation Reactions**: A and B
- **Not Disproportionation Reactions**: C and D

### Answer:
The disproportionation reactions among the given options are **A and B**.