Question : Which of the following is not a disproprotionation reaction?

Options

A. `KO_(2)+H_(2)O+CO_(2)rarrKHCO_(3)+O_(2)`
B. `KClO_(3)rarr KClO_(4)+KCl`
C. `PbO_(2)+H_(2)OrarrPbO+H_(2)O_(2)`
D. `OHC-COOHoverset(overset(ө)OH)rarrHOH_(2)C-COOH+overset(ө)OOC-COO^(ө)`

Explanation

To determine which of the following reactions is not a disproportionation reaction, we need to understand what a disproportionation reaction is. In a disproportionation reaction, the same element undergoes both oxidation (increase in oxidation state) and reduction (decrease in oxidation state) simultaneously.

Let's analyze each option step by step.

### Step 1: Analyze the first option
**Reaction:** \( KO_2 + H_2O + CO_2 \rightarrow KHCO_3 + O_2 \)

1. Determine the oxidation states:
- In \( KO_2 \): K = +1, O = -1 (for each O in \( O_2^{2-} \))
- In \( H_2O \): H = +1, O = -2
- In \( CO_2 \): C = +4, O = -2
- In \( KHCO_3 \): K = +1, H = +1, C = +4, O = -2
- In \( O_2 \): O = 0 (elemental state)

2. Changes in oxidation states:
- K goes from +1 to +1 (no change)
- O in \( KO_2 \) goes from -1 to 0 (oxidation)
- C remains +4 (no change)

Since there is no simultaneous oxidation and reduction of the same element, this is **not a disproportionation reaction**.

### Step 2: Analyze the second option
**Reaction:** \( KClO_3 \rightarrow KClO_4 + KCl \)

1. Determine the oxidation states:
- In \( KClO_3 \): K = +1, Cl = +5, O = -2
- In \( KClO_4 \): K = +1, Cl = +7, O = -2
- In \( KCl \): K = +1, Cl = -1

2. Changes in oxidation states:
- Cl goes from +5 (in \( KClO_3 \)) to +7 (in \( KClO_4 \)) and to -1 (in \( KCl \)).

Since Cl is both oxidized and reduced, this is a **disproportionation reaction**.

### Step 3: Analyze the third option
**Reaction:** \( PbO_2 + H_2O \rightarrow PbO + H_2O_2 \)

1. Determine the oxidation states:
- In \( PbO_2 \): Pb = +4, O = -2
- In \( H_2O \): H = +1, O = -2
- In \( PbO \): Pb = +2, O = -2
- In \( H_2O_2 \): H = +1, O = -1

2. Changes in oxidation states:
- Pb goes from +4 (in \( PbO_2 \)) to +2 (in \( PbO \)) (reduction).
- O in \( H_2O_2 \) goes from -2 to -1 (oxidation).

Since there is no simultaneous oxidation and reduction of the same element, this is **not a disproportionation reaction**.

### Step 4: Analyze the fourth option
**Reaction:** \( CHO + COOH \rightarrow CH_2OH + COO^- \)

1. Determine the oxidation states:
- In \( CHO \): C = +1, O = -2, H = +1
- In \( COOH \): C = +3, O = -2, H = +1
- In \( CH_2OH \): C = +1, O = -2, H = +1
- In \( COO^- \): C = +3, O = -2

2. Changes in oxidation states:
- C remains +1 in \( CHO \) and +1 in \( CH_2OH \) (no change).
- C remains +3 in both \( COOH \) and \( COO^- \) (no change).

Since there is no increase or decrease in oxidation state, this is also **not a disproportionation reaction**.

### Conclusion
The reactions that are **not** disproportionation reactions are:
1. \( KO_2 + H_2O + CO_2 \rightarrow KHCO_3 + O_2 \)
2. \( PbO_2 + H_2O \rightarrow PbO + H_2O_2 \)
3. \( CHO + COOH \rightarrow CH_2OH + COO^- \)

### Final Answer
The reaction that is not a disproportionation reaction is the first option, the third option, and the fourth option.