Question

Match the List-l with List-II.

	List-I (Reactions)		List-II (Oxides of nitrogen)
(P)	$NaNO_2 + FeSO_4 + H_2 SO_4 \rightarrow$	(1)	$N_2O$
(Q)	$HNO_3 + P_4O_{10} \rightarrow$	(2)	$NO_2$
(R)	$Cu + HNO_3(conc.) \rightarrow$	(3)	$NO$
(S)	$NH_4NO_3 \stackrel{\Delta}{\rightarrow}$	(4)	$N_2O_5$
		(5)	$N_2O_3$

• A. P - 1, Q - 2, R - 3, S - 4
• B. P - 3, Q - 4, R - 2, S - 1
• C. P - 5, Q - 4, R - 3, S - 2
• D. P - 2, Q - 3, R - 4, S - 5

Answer 1

Answer: (B)

P-3, Q-4, R-2, S-1

Answer 2

The solution involves identifying the product oxides of nitrogen for each given reaction.

(P) $NaNO_2 + FeSO_4 + H_2 SO_4 \rightarrow$
This reaction is part of the brown ring test. In an acidic medium, ferrous ions ($Fe^{2+}$) reduce nitrite ions ($NO_2^-$) to nitric oxide ($NO$).
The ionic reaction is: $NO_2^- + Fe^{2+} + 2H^+ \rightarrow NO + Fe^{3+} + H_2O$
Thus, the oxide of nitrogen produced is $NO$.
(P) matches with (3) $NO$.

(Q) $HNO_3 + P_4O_{10} \rightarrow$
Phosphorus pentoxide ($P_4O_{10}$) is a powerful dehydrating agent. It dehydrates nitric acid ($HNO_3$) to form dinitrogen pentoxide ($N_2O_5$).
The reaction is: $4HNO_3 + P_4O_{10} \rightarrow 2N_2O_5 + 4HPO_3$
Thus, the oxide of nitrogen produced is $N_2O_5$.
(Q) matches with (4) $N_2O_5$.

(R) $Cu + HNO_3(conc.) \rightarrow$
Copper reacts with concentrated nitric acid. Concentrated nitric acid acts as a strong oxidizing agent and is reduced to nitrogen dioxide ($NO_2$).
The reaction is: $Cu + 4HNO_3(conc.) \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$
Thus, the oxide of nitrogen produced is $NO_2$.
(R) matches with (2) $NO_2$.

(S) $NH_4NO_3 \stackrel{\Delta}{\rightarrow}$
Ammonium nitrate ($NH_4NO_3$) decomposes on heating to produce nitrous oxide ($N_2O$) and water. This is a redox reaction where nitrogen in the ammonium ion (oxidation state -3) is oxidized and nitrogen in the nitrate ion (oxidation state +5) is reduced.
The reaction is: $NH_4NO_3 \stackrel{\Delta}{\rightarrow} N_2O + 2H_2O$
Thus, the oxide of nitrogen produced is $N_2O$.
(S) matches with (1) $N_2O$.

Final Matching:
(P) - (3)
(Q) - (4)
(R) - (2)
(S) - (1)

Explanation of the solution:

• (P) $NaNO_2 + FeSO_4 + H_2SO_4 \rightarrow NO$: Nitrite is reduced by $Fe^{2+}$ in acidic medium to $NO$.
• (Q) $HNO_3 + P_4O_{10} \rightarrow N_2O_5$: $P_4O_{10}$ dehydrates $HNO_3$ to form $N_2O_5$.
• (R) $Cu + HNO_3(conc.) \rightarrow NO_2$: Concentrated $HNO_3$ oxidizes $Cu$ and is itself reduced to $NO_2$.
• (S) $NH_4NO_3 \stackrel{\Delta}{\rightarrow} N_2O$: Thermal decomposition of ammonium nitrate yields $N_2O$.