Incorrect order of solubility product `(K_(sp))` of given precipitated compound is:

Options

A. `AgCl lt PbCl_(2)`
B. `Al(OH)_(3) lt Zn(OH)_(2)`
C. `BaCO_(3) lt MgCO_(3)`
D. `MnS lt Ag_(2)S`

Explanation

To determine the incorrect order of solubility products (Ksp) for the given precipitated compounds, we will analyze each pair of compounds step by step.

### Step 1: Analyze AgCl and PbCl2
- **Dissociation**:
- AgCl dissociates into Ag⁺ and Cl⁻:
\[ \text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \]
- Let the solubility of AgCl be \( S \). Therefore, \( K_{sp} = S^2 \).

- PbCl2 dissociates into Pb²⁺ and 2Cl⁻:
\[ \text{PbCl}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{Cl}^- \]
- Let the solubility of PbCl2 be \( S' \). Therefore, \( K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = S' \cdot (2S')^2 = 4S'^3 \).

- **Comparison**:
- Since \( K_{sp} \) for PbCl2 is greater than that for AgCl, this order is correct:
\[ K_{sp}(\text{PbCl}_2) > K_{sp}(\text{AgCl}) \]

### Step 2: Analyze Al(OH)3 and Zn(OH)2
- **Dissociation**:
- Al(OH)3 dissociates into Al³⁺ and 3OH⁻:
\[ \text{Al(OH)}_3 \rightleftharpoons \text{Al}^{3+} + 3\text{OH}^- \]
- Let the solubility of Al(OH)3 be \( S \). Therefore, \( K_{sp} = S \cdot (3S)^3 = 27S^4 \).

- Zn(OH)2 dissociates into Zn²⁺ and 2OH⁻:
\[ \text{Zn(OH)}_2 \rightleftharpoons \text{Zn}^{2+} + 2\text{OH}^- \]
- Let the solubility of Zn(OH)2 be \( S' \). Therefore, \( K_{sp} = S' \cdot (2S')^2 = 4S'^3 \).

- **Comparison**:
- Al(OH)3 precipitates first due to lower OH⁻ concentration, thus having a lower Ksp:
\[ K_{sp}(\text{Al(OH)}_3) < K_{sp}(\text{Zn(OH)}_2) \]
- This order is correct.

### Step 3: Analyze BaCO3 and MgCO3
- **Dissociation**:
- BaCO3 dissociates into Ba²⁺ and CO3²⁻:
\[ \text{BaCO}_3 \rightleftharpoons \text{Ba}^{2+} + \text{CO}_3^{2-} \]

- MgCO3 dissociates into Mg²⁺ and CO3²⁻:
\[ \text{MgCO}_3 \rightleftharpoons \text{Mg}^{2+} + \text{CO}_3^{2-} \]

- **Comparison**:
- As we move down the group in the periodic table, the solubility decreases. Therefore, MgCO3 has a higher solubility than BaCO3:
\[ K_{sp}(\text{MgCO}_3) > K_{sp}(\text{BaCO}_3) \]
- This order is correct.

### Step 4: Analyze Ag2S and MnS
- **Dissociation**:
- Ag2S dissociates into 2Ag⁺ and S²⁻:
\[ \text{Ag}_2\text{S} \rightleftharpoons 2\text{Ag}^+ + \text{S}^{2-} \]
- Let the solubility of Ag2S be \( S \). Therefore, \( K_{sp} = (2S)^2 \cdot S = 4S^3 \).

- MnS dissociates into Mn²⁺ and S²⁻:
\[ \text{MnS} \rightleftharpoons \text{Mn}^{2+} + \text{S}^{2-} \]
- Let the solubility of MnS be \( S' \). Therefore, \( K_{sp} = S' \cdot S' = S'^2 \).

- **Comparison**:
- Ag2S precipitates first (lower Ksp) compared to MnS:
\[ K_{sp}(\text{Ag}_2\text{S}) < K_{sp}(\text{MnS}) \]
- The order given (Ksp of Ag2S > Ksp of MnS) is incorrect.

### Conclusion
The incorrect order of solubility products (Ksp) is:
\[ K_{sp}(\text{Ag}_2\text{S}) > K_{sp}(\text{MnS}) \]

### Final Answer
The correct answer is option 4.

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