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 one by one and derive their solubility product expressions.

### Step 1: Analyze Bi₂S₃
Bi₂S₃ dissociates in water as follows:
\[ \text{Bi}_2\text{S}_3 (s) \rightleftharpoons 2 \text{Bi}^{3+} (aq) + 3 \text{S}^{2-} (aq) \]

Let the solubility of Bi₂S₃ be \( S \). Then at equilibrium:
- The concentration of \(\text{Bi}^{3+}\) will be \( 2S \)
- The concentration of \(\text{S}^{2-}\) will be \( 3S \)

The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Bi}^{3+}]^2 [\text{S}^{2-}]^3 = (2S)^2 (3S)^3 \]
\[ K_{sp} = 4S^2 \cdot 27S^3 = 108S^5 \]

### Step 2: Analyze CdS
CdS dissociates as follows:
\[ \text{CdS} (s) \rightleftharpoons \text{Cd}^{2+} (aq) + \text{S}^{2-} (aq) \]

Let the solubility of CdS be \( S \). Then at equilibrium:
- The concentration of \(\text{Cd}^{2+}\) will be \( S \)
- The concentration of \(\text{S}^{2-}\) will also be \( S \)

The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Cd}^{2+}][\text{S}^{2-}] = S \cdot S = S^2 \]

### Step 3: Analyze Al(OH)₃
Al(OH)₃ dissociates as follows:
\[ \text{Al(OH)}_3 (s) \rightleftharpoons \text{Al}^{3+} (aq) + 3 \text{OH}^{-} (aq) \]

Let the solubility of Al(OH)₃ be \( S \). Then at equilibrium:
- The concentration of \(\text{Al}^{3+}\) will be \( S \)
- The concentration of \(\text{OH}^{-}\) will be \( 3S \)

The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Al}^{3+}][\text{OH}^{-}]^3 = S \cdot (3S)^3 \]
\[ K_{sp} = S \cdot 27S^3 = 27S^4 \]

### Step 4: Analyze CaF₂
CaF₂ dissociates as follows:
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{F}^{-} (aq) \]

Let the solubility of CaF₂ be \( S \). Then at equilibrium:
- The concentration of \(\text{Ca}^{2+}\) will be \( S \)
- The concentration of \(\text{F}^{-}\) will be \( 2S \)

The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2 = S \cdot (2S)^2 \]
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

### Summary of Results
Now we can summarize the results:
- A. Bi₂S₃ → \( K_{sp} = 108S^5 \) (matches with iii)
- B. CdS → \( K_{sp} = S^2 \) (matches with iv)
- C. Al(OH)₃ → \( K_{sp} = 27S^4 \) (matches with ii)
- D. CaF₂ → \( K_{sp} = 4S^3 \) (matches with i)

### Final Matching
- A → iii
- B → iv
- C → ii
- D → i

### Correct Answer
The correct option is (1): A-iii, B-iv, C-ii, D-i.