Match List-I with List-II and chose the correct answer from the code
`{:("List-I Electrolyte","List-I Solubility product"),(a.Bi_(2)S_(3),i.4s^(3)),(b.CdS,ii. 27s^(2)),(c.AI(OH)_(3),iii.108s^(5)),(d.CaF_(2),iv.s^(2)):}`

Options

A. iii, iv, ii, i
B. I, ii, iii, iv
C. iv, I, ii, iii
D. ii, iii, I, iv

Answer

To solve the problem of matching the electrolytes from List-I with their corresponding solubility products from List-II, we will analyze each electrolyte step by step.

### Step 1: Analyze Bi2S3
- **Dissociation**: Bi2S3 dissociates into 2 Bi³⁺ and 3 S²⁻ ions.
- **Solubility**: Let the solubility be 's'.
- Concentration of Bi³⁺ = 2s
- Concentration of S²⁻ = 3s
- **Ksp Expression**:
\[ K_{sp} = [Bi^{3+}]^2 \times [S^{2-}]^3 = (2s)^2 \times (3s)^3 = 4s^2 \times 27s^3 = 108s^5 \]
- **Match**: Bi2S3 corresponds to **iii. 108s⁵**.

### Step 2: Analyze CdS
- **Dissociation**: CdS dissociates into Cd²⁺ and S²⁻ ions.
- **Solubility**: Let the solubility be 's'.
- Concentration of Cd²⁺ = s
- Concentration of S²⁻ = s
- **Ksp Expression**:
\[ K_{sp} = [Cd^{2+}] \times [S^{2-}] = s \times s = s^2 \]
- **Match**: CdS corresponds to **iv. s²**.

### Step 3: Analyze Al(OH)₃
- **Dissociation**: Al(OH)₃ dissociates into Al³⁺ and 3 OH⁻ ions.
- **Solubility**: Let the solubility be 's'.
- Concentration of Al³⁺ = s
- Concentration of OH⁻ = 3s
- **Ksp Expression**:
\[ K_{sp} = [Al^{3+}] \times [OH^{-}]^3 = s \times (3s)^3 = s \times 27s^3 = 27s^4 \]
- **Match**: Al(OH)₃ corresponds to **ii. 27s⁴**.

### Step 4: Analyze CaF₂
- **Dissociation**: CaF₂ dissociates into Ca²⁺ and 2 F⁻ ions.
- **Solubility**: Let the solubility be 's'.
- Concentration of Ca²⁺ = s
- Concentration of F⁻ = 2s
- **Ksp Expression**:
\[ K_{sp} = [Ca^{2+}] \times [F^{-}]^2 = s \times (2s)^2 = s \times 4s^2 = 4s^3 \]
- **Match**: CaF₂ corresponds to **i. 4s³**.

### Final Matching
Now, we can summarize the matches:
- a. Bi₂S₃ → iii. 108s⁵
- b. CdS → iv. s²
- c. Al(OH)₃ → ii. 27s⁴
- d. CaF₂ → i. 4s³

### Conclusion
The correct matching is:
- (a) → iii
- (b) → iv
- (c) → ii
- (d) → i

Thus, the answer code is **(iii, iv, ii, i)**.