For decolourisation of `1 "mol of" KMnO_(4)`, the moles of `H_(2)O_(2)` required is

Options

A. `1//2`
B. `3//2`
C. `5//2`
D. `7//2`

Explanation

To determine the moles of \( H_2O_2 \) required for the decolourisation of \( 1 \) mole of \( KMnO_4 \), we need to analyze the redox reaction involved.

### Step-by-Step Solution:

1. **Identify the Reduction of KMnO4:**
- \( KMnO_4 \) contains the permanganate ion \( MnO_4^- \). In this ion, the manganese (Mn) has an oxidation state of +7.
- During the reaction, \( MnO_4^- \) is reduced to \( Mn^{2+} \) (where Mn has an oxidation state of +2).

2. **Calculate the Electrons Gained:**
- The change in oxidation state from +7 to +2 indicates that \( Mn \) gains 5 electrons during the reduction process.
- Therefore, for the reduction of \( 1 \) mole of \( KMnO_4 \), \( 5 \) moles of electrons are needed.

3. **Identify the Oxidation of H2O2:**
- Hydrogen peroxide \( H_2O_2 \) acts as a reducing agent in this reaction. It gets oxidized to oxygen \( O_2 \).
- The oxidation state of oxygen in \( H_2O_2 \) is -1, and in \( O_2 \) it is 0.
- The oxidation of \( H_2O_2 \) to \( O_2 \) involves the loss of 2 electrons.

4. **Calculate the Moles of H2O2 Required:**
- Since \( 1 \) mole of \( H_2O_2 \) donates \( 2 \) electrons, to provide \( 5 \) moles of electrons, we can set up the following relationship:
\[ \text{Moles of } H_2O_2 = \frac{\text{Moles of electrons required}}{\text{Electrons donated by 1 mole of } H_2O_2} \]
\[ \text{Moles of } H_2O_2 = \frac{5 \text{ moles of electrons}}{2 \text{ electrons/mole}} = \frac{5}{2} \text{ moles} \]

5. **Final Answer:**
- Therefore, the moles of \( H_2O_2 \) required for the decolourisation of \( 1 \) mole of \( KMnO_4 \) is \( \frac{5}{2} \) or \( 2.5 \) moles.

### Conclusion:
The moles of \( H_2O_2 \) required for the decolourisation of \( 1 \) mole of \( KMnO_4 \) is \( \frac{5}{2} \).