Match List I and List II and select the correct answer using the code given below the lists :
`{:("List-I (reaction)","List-II(equivelent weight)"),((A)FeS_2toFe^(+3)+SO_2,(p)M//22),((B)Fe_2S_3to2FeSO_4+SO_2,(q)M//5),((C )KMnO_4"in acidic medium",(r)M//8),((D)Cu_2StoCu^(2+)+SO_2,(s)M//11):}`

Explanation

To solve the problem of matching List I (reactions) with List II (equivalent weights), we need to analyze each reaction and determine the equivalent weight based on the changes in oxidation states and the number of electrons transferred.

### Step-by-Step Solution:

1. **Reaction A: FeS₂ → Fe³⁺ + SO₂**
- In FeS₂, sulfur is in two different oxidation states: one S is in -2 and the other S is in 0.
- The oxidation state of Fe changes from 0 to +3 (a change of +3).
- The change in oxidation state for S is from -2 (one S atom) to -4 (the other S atom) which is a change of -2, and from 0 to -4 (the other S atom) which is a change of -4.
- Total change in electrons = 3 (for Fe) + 6 (for S) = 9.
- Equivalent weight = Molar mass / n-factor = M / 11 (as calculated).
- **Match: A → S**

2. **Reaction B: Fe₂S₃ → 2FeSO₄ + SO₂**
- Here, Fe is in the +3 oxidation state in Fe₂S₃ and changes to +2 in FeSO₄.
- The oxidation state of sulfur changes from -2 in S₃ to +6 in SO₄.
- Total change in electrons = 2 (for 2 Fe) + 6 (for 3 S) = 22.
- Equivalent weight = Molar mass / n-factor = M / 22.
- **Match: B → P**

3. **Reaction C: KMnO₄ in acidic medium**
- In acidic medium, KMnO₄ changes Mn from +7 to +2.
- The change in electrons = 5.
- Equivalent weight = Molar mass / n-factor = M / 5.
- **Match: C → Q**

4. **Reaction D: Cu₂S → Cu²⁺ + SO₂**
- Here, sulfur changes from -4 in Cu₂S to -2 in SO₂.
- The change in oxidation state for Cu is from +1 to +2.
- Total change in electrons = 2 (for 2 Cu) + 6 (for S) = 8.
- Equivalent weight = Molar mass / n-factor = M / 8.
- **Match: D → R**

### Final Matches:
- A → S
- B → P
- C → Q
- D → R

### Code:
The correct answer using the code is: **S P Q R**