JEE Advanced 2022 Paper 1 · Q16 · s-Block Elements

Question

LIST-I contains compounds and LIST-II contains reactions.

LIST-I:
(I) H$_2$O$_2$
(II) Mg(OH)$_2$
(III) BaCl$_2$
(IV) CaCO$_3$

LIST-II:
(P) Mg(HCO$_3$)$_2$ + Ca(OH)$_2$ $\rightarrow$
(Q) BaO$_2$ + H$_2$SO$_4$ $\rightarrow$
(R) Ca(OH)$_2$ + MgCl$_2$ $\rightarrow$
(S) BaO$_2$ + HCl $\rightarrow$
(T) Ca(HCO$_3$)$_2$ + Ca(OH)$_2$ $\rightarrow$

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

SolveKar AI · Accepted Answer

Work out the product of each reaction in LIST-II:

(P) Mg(HCO$_3$)$_2$ + 2 Ca(OH)$_2 \rightarrow$ Mg(OH)$_2$$\downarrow$ + 2 CaCO$_3$$\downarrow$ + 2 H$_2$O. Forms Mg(OH)$_2$ (II) and CaCO$_3$ (IV).

(Q) BaO$_2$ + H$_2$SO$_4 \rightarrow$ BaSO$_4$$\downarrow$ + H$_2$O$_2$. Forms H$_2$O$_2$ (I) (this is the standard laboratory preparation of H$_2$O$_2$ from barium peroxide).

(R) Ca(OH)$_2$ + MgCl$_2 \rightarrow$ Mg(OH)$_2$$\downarrow$ + CaCl$_2$. Forms Mg(OH)$_2$ (II).

(S) BaO$_2$ + 2 HCl $\rightarrow$ BaCl$_2$ + H$_2$O$_2$. Forms BaCl$_2$ (III) and H$_2$O$_2$ (I).

(T) Ca(HCO$_3$)$_2$ + Ca(OH)$_2 \rightarrow$ 2 CaCO$_3$$\downarrow$ + 2 H$_2$O. Forms CaCO$_3$ (IV).

Now choose a unique mapping where each LIST-I compound is matched to a LIST-II reaction that produces it:
• I (H$_2$O$_2$) from Q or S — choose Q.
• II (Mg(OH)$_2$) from P or R — choose R.
• III (BaCl$_2$) from S — choose S.
• IV (CaCO$_3$) from P or T — choose P.

Mapping: I $\rightarrow$ Q; II $\rightarrow$ R; III $\rightarrow$ S; IV $\rightarrow$ P. This is option (D).

JEE Advanced 2022 Paper 1 · Q16 · s-Block Elements

Solution