Question

Match List-I with the List-II

List-I Reaction	List-II Type of redox reaction
(A) N2(g) + O2(g) → 2NO(g)	(I) Decomposition
(B) 2Pb (NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g)	(II) Displacement
(C) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)	(III) Disproportionation
(D) 2NO2(g) + 2OH-(aq) → NO2-(aq) + NO3-(aq) + H2O(l)	(IV) Combination

Choose the correct answer from the options given below:

• A. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
• B. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
• C. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
• D. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Answer 1

The Correct Option is D

\(\textbf{(A) $N_2(g) + O_2(g) \rightarrow 2NO(g)$:}\) This reaction is a combination reaction because two reactants combine to form a single product. Thus, $A \rightarrow (IV)$. 
\(\textbf{(B) $2Pb(NO_3)_2(s) \rightarrow 2PbO(s) + 4NO_2(g) + O_2(g)$:} \\\) This is a decomposition reaction because a single compound breaks down into multiple products. Thus, $B \rightarrow (I)$. 
\(\textbf{(C) $2Na(s) + 2H_2O \rightarrow 2NaOH(aq) + H_2(g)$:} \\\) This is a displacement reaction as sodium displaces hydrogen from water. Thus, $C \rightarrow (II)$.
\(\textbf{(D) $2NO_2(g) + 2OH^-(aq) \rightarrow NO^-_2(aq) + NO^-_3(aq) + H_2O(l)$:} \\\) This reaction involves disproportionation because the same species ($NO_2$) is oxidized and reduced simultaneously. Thus, $D \rightarrow (III)$.