This is a matching-type question where we need to pair compounds from LIST-I with the reactions in LIST-II that form them. We will analyze each reaction in LIST-II to determine its products and then match them with the compounds in LIST-I.

Step 1: Analyze the reactions in LIST-II

1. (P) $\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}+\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow$ This reaction is used in Clark's method for removing temporary hardness of water caused by magnesium bicarbonate. The balanced chemical equation is: $\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2} + 2\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{Mg}(\mathrm{OH})_{2} \downarrow + 2\mathrm{CaCO}_{3} \downarrow + 2\mathrm{H}_{2}\mathrm{O}$ The products formed are Magnesium Hydroxide ($\mathrm{Mg}(\mathrm{OH})_{2}$) and Calcium Carbonate ($\mathrm{CaCO}_{3}$). So, reaction (P) can be matched with compound (II) or (IV).

2. (Q) $\mathrm{BaO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow$ This is a standard laboratory method for the preparation of hydrogen peroxide. Barium peroxide reacts with dilute sulfuric acid. $\mathrm{BaO}_{2} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{BaSO}_{4} \downarrow + \mathrm{H}_{2} \mathrm{O}_{2}$ The product of interest is Hydrogen Peroxide ($\mathrm{H}_{2} \mathrm{O}_{2}$). So, reaction (Q) matches with compound (I).

3. (R) $\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{MgCl}_{2} \rightarrow$ This reaction involves the treatment of permanent hardness (due to $\mathrm{MgCl}_{2}$) with slaked lime. It's a precipitation reaction. $\mathrm{Ca}(\mathrm{OH})_{2} + \mathrm{MgCl}_{2} \rightarrow \mathrm{Mg}(\mathrm{OH})_{2} \downarrow + \mathrm{CaCl}_{2}$ The precipitate formed is Magnesium Hydroxide ($\mathrm{Mg}(\mathrm{OH})_{2}$). So, reaction (R) matches with compound (II).

4. (S) $\mathrm{BaO}_{2}+\mathrm{HCl} \rightarrow$ This is another reaction for preparing hydrogen peroxide. The balanced equation is: $\mathrm{BaO}_{2} + 2\mathrm{HCl} \rightarrow \mathrm{BaCl}_{2} + \mathrm{H}_{2} \mathrm{O}_{2}$ The products are Barium Chloride ($\mathrm{BaCl}_{2}$) and Hydrogen Peroxide ($\mathrm{H}_{2} \mathrm{O}_{2}$). So, reaction (S) can be matched with compound (III) or (I).

5. (T) $\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}+\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow$ This is Clark's method for removing temporary hardness due to calcium bicarbonate. $\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2} + \mathrm{Ca}(\mathrm{OH})_{2} \rightarrow 2\mathrm{CaCO}_{3} \downarrow + 2\mathrm{H}_{2}\mathrm{O}$ The product is Calcium Carbonate ($\mathrm{CaCO}_{3}$). So, reaction (T) matches with compound (IV).

Step 2: Determine the correct set of matches

Let's summarize the possible matches:

• (I) $\mathrm{H}_{2} \mathrm{O}_{2}$ $\rightarrow$ (Q), (S)
• (II) $\mathrm{Mg}(\mathrm{OH})_{2}$ $\rightarrow$ (P), (R)
• (III) $\mathrm{BaCl}_{2}$ $\rightarrow$ (S)
• (IV) $\mathrm{CaCO}_{3}$ $\rightarrow$ (P), (T)

We proceed by identifying the most constrained matches first.

• Compound (III), $\mathrm{BaCl}_{2}$, can only be formed by reaction (S). So, the match III $\rightarrow$ S is certain.
• Since reaction (S) is matched with (III), we must find another match for compound (I), $\mathrm{H}_{2} \mathrm{O}_{2}$. The only other option is reaction (Q). So, the match I $\rightarrow$ Q is also determined.

Now we have I $\rightarrow$ Q and III $\rightarrow$ S. Let's examine the options:

• A: I $\rightarrow$ Q; ... III $\rightarrow$ S; ... (Possible)
• B: I $\rightarrow$ T; ... (Incorrect)
• C: I $\rightarrow$ T; ... (Incorrect)
• D: I $\rightarrow$ Q; ... III $\rightarrow$ S; ... (Possible)

We are left with options A and D. Let's compare them:

• Option A: I $\rightarrow$ Q; II $\rightarrow$ P; III $\rightarrow$ S; IV $\rightarrow$ R
• Option D: I $\rightarrow$ Q; II $\rightarrow$ R; III $\rightarrow$ S; IV $\rightarrow$ P

Let's check the match IV $\rightarrow$ R from option A. Compound (IV) is $\mathrm{CaCO}_{3}$. Reaction (R) produces $\mathrm{Mg}(\mathrm{OH})_{2}$, not $\mathrm{CaCO}_{3}$. Therefore, option A is incorrect.

This leaves option D as the only possibility. Let's verify all matches in option D:

• I $\rightarrow$ Q: Correct, $\mathrm{BaO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4}$ produces $\mathrm{H}_{2} \mathrm{O}_{2}$.
• II $\rightarrow$ R: Correct, $\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{MgCl}_{2}$ produces $\mathrm{Mg}(\mathrm{OH})_{2}$.
• III $\rightarrow$ S: Correct, $\mathrm{BaO}_{2}+\mathrm{HCl}$ produces $\mathrm{BaCl}_{2}$.
• IV $\rightarrow$ P: Correct, $\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}+\mathrm{Ca}(\mathrm{OH})_{2}$ produces $\mathrm{CaCO}_{3}$.

All matches in option D are chemically correct. Thus, it is the correct answer.