LIST-I contains compounds and LIST-II contains reaction

LIST-I	LIST-II
(I) H$$_2$$O$$_2$$	(P) Mg(HCO$$_3$$)$$_2$$ + Ca(OH)$$_2$$ $$\to$$
(II) Mg(OH)$$_2$$	(Q) BaO$$_2$$ + H$$_2$$SO$$_4$$ $$\to$$
(III) BaCl$$_2$$	(R) Ca(OH)$$_2$$ + MgCl$$_2$$ $$\to$$
(IV) CaCO$$_3$$	(S) BaO$$_2$$ + HCl $$\to$$
	(T) Ca(HCO$$_3$$)$$_2$$ + Ca(OH)$$_2$$ $$\to$$

Match each compound in LIST-I with its formation reaction(s) in LIST-II, and choose the correct option

The compounds in LIST-I are I $$H_2O_2$$, II $$Mg(OH)_2$$, III $$BaCl_2$$ and IV $$CaCO_3$$.

For each reaction given in LIST-II, first write the balanced equation and then inspect the products that precipitate or remain in solution:

$$BaO_2 + H_2SO_4 \rightarrow BaSO_4 \downarrow + H_2O_2$$
Barium sulphate precipitates, while $$H_2O_2$$ is produced in solution. Hence reaction Q forms compound I.

$$Ca(OH)_2 + MgCl_2 \rightarrow Mg(OH)_2 \downarrow + CaCl_2$$
Magnesium hydroxide precipitates, so reaction R forms compound II.

$$BaO_2 + 2\,HCl \rightarrow BaCl_2 + H_2O_2$$
Barium chloride is obtained (no precipitate because $$BaCl_2$$ is soluble). Thus reaction S forms compound III.

$$Mg(HCO_3)_2 + 2\,Ca(OH)_2 \rightarrow Mg(OH)_2 \downarrow + 2\,CaCO_3 \downarrow + 2\,H_2O$$
Calcium carbonate precipitates; therefore reaction P forms compound IV. (Reaction T, with $$Ca(HCO_3)_2$$, would also give $$CaCO_3$$, but T does not appear in any option together with the other correct matches.)

Collecting the correct pairs:
I $$H_2O_2$$  →  Q
II $$Mg(OH)_2$$  →  R
III $$BaCl_2$$  →  S
IV $$CaCO_3$$  →  P

The option that contains this exact matching is: Option D which is: I → Q; II → R; III → S; IV → P.