Dow's process of extraction of `Mg` involves extraction of `Mg` from sea water. Sea water is concentrated in sun light and is then treated with skaked lime. Magnesium hydroxide is heated in a stream of `HCI` to give `MhCi` which is electrolysed to dischate `Mg`. the mixture is in the ratio `35% MgCI_(2) +50% NaCI + 15% CaCI_(2). NaCI_(2)` and `CaCI_(2)` are added to lower the fusion temperature and to increase the conductance.
`Mg^(2+) +Ca(OH)_(2) rarr Mg (OH)_(2) +Ca^(2+)`
`Mg(OH)_(2) + 2HCI rarr MgCI_(2) +2H_(2)O` (liquid)
Electrolysis of fused `MgCI_(2) overset("Anode")rarr 2CI^(-) rarr 2_(-I)e^(@)`
`overset("Cathode")rarr Mg^(2+) +2_(-I)e^(@) rarr Mg`
Mg electrolysed is protected from atmospheric oxidation by a blanket of inert gases.
`Mg^(2+) + Ca(OH)_(2) rarr Mg(OH)_(2)darr +Ca^(2+)` This reaction indicates:

Options

A. `Mg (OH)_(2)` is weaker base than `Ca(OH)_(2)`
B. Solubility prodcucts of `Mg(OH)_(2)` is less than that of `Ca(OH)_(2)`
C. Polarising power of `Mg^(2+)` is more than that of `Ca^(2+)` ion
D. Both (B) and (C)

Answer

To solve the question regarding the reaction:

\[ \text{Mg}^{2+} + \text{Ca(OH)}_2 \rightarrow \text{Mg(OH)}_2 \downarrow + \text{Ca}^{2+} \]

we need to analyze the implications of this reaction step by step.

### Step 1: Understanding the Reaction
The reaction shows that magnesium ions (\(\text{Mg}^{2+}\)) react with calcium hydroxide (\(\text{Ca(OH)}_2\)) to form magnesium hydroxide (\(\text{Mg(OH)}_2\)), which precipitates out of the solution, and calcium ions (\(\text{Ca}^{2+}\)) remain in the solution.

**Hint:** Identify the reactants and products in the reaction to understand what is happening.

### Step 2: Identifying the Precipitate
The formation of \(\text{Mg(OH)}_2\) as a precipitate indicates that it is less soluble in water compared to the other ions present. This suggests that the solubility product (\(K_{sp}\)) of \(\text{Mg(OH)}_2\) is lower than that of \(\text{Ca(OH)}_2\).

**Hint:** Recall the concept of solubility product and how it relates to the formation of precipitates.

### Step 3: Comparing Bases
We can compare the strengths of the bases involved. Since \(\text{Mg(OH)}_2\) precipitates out, it indicates that it is a weaker base than \(\text{Ca(OH)}_2\). This is because stronger bases tend to remain in solution and do not precipitate.

**Hint:** Consider the definitions of strong and weak bases in terms of their solubility in water.

### Step 4: Polarizing Power of Ions
The polarizing power of an ion is related to its charge and size. \(\text{Mg}^{2+}\) has a smaller ionic radius compared to \(\text{Ca}^{2+}\), which means it has a higher polarizing power. This higher polarizing power can lead to stronger interactions with water molecules and other ions, facilitating the formation of precipitates.

**Hint:** Remember how ionic size and charge affect polarizing power.

### Step 5: Conclusion
Based on the analysis, we can conclude that:
- The solubility product of \(\text{Mg(OH)}_2\) is less than that of \(\text{Ca(OH)}_2\).
- \(\text{Mg(OH)}_2\) is a weaker base than \(\text{Ca(OH)}_2\).
- The polarizing power of \(\text{Mg}^{2+}\) is greater than that of \(\text{Ca}^{2+}\).

Thus, the correct answer is that both statements regarding the solubility product and the polarizing power are correct.

**Final Answer:** The correct option is D: Both B and C are correct.