LIST-I LIST-II (P) F e C 2 O 4 (1) Only one of the radical decolorize aq. KMnO 4 (Q) F e S O 4 (2) Both the radicals decolorize aq. KMnO 4 (R) Z n ( N O 3 ) 2 (3) Can give a colorless gas wi

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LIST-I LIST-II (P) F e C 2 ​ O 4 ​ (1) Only one of the radical decolorize aq. KMnO 4 (Q) F e S O 4 ​ (2) Both the radicals decolorize aq. KMnO 4 (R) Z n ( N O 3 ​ ) 2 ​ (3) Can give a colorless gas with conc. H 2 SO 4 (S) ( N H 4 ​ ) 2 ​ S (4) Adding NaOH can produce a white precipitate

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P: FeC₂O₄ contains Fe²⁺ and C₂O₄²⁻. Both are reducing agents, decolorizing KMnO₄ (2). Fe²⁺ forms Fe(OH)₂ ppt with NaOH (4). C₂O₄²⁻ with conc. H₂SO₄ gives CO and CO₂ (3).

Q: FeSO₄ contains Fe²⁺ and SO₄²⁻. Fe²⁺ decolorizes KMnO₄ (1). Fe²⁺ forms Fe(OH)₂ ppt with NaOH (4). SO₄²⁻ with conc. H₂SO₄ gives SO₂ (3).

R: Zn(NO₃)₂ contains Zn²⁺ and NO₃⁻. Zn²⁺ forms Zn(OH)₂ ppt with NaOH (4).

S: (NH₄)₂S contains NH₄⁺ and S²⁻. S²⁻ decolorizes KMnO₄ (2). NH₄⁺ with NaOH gives NH₃ gas (3).

Final Matching: P -> 2,3,4; Q -> 1,3,4; R -> 4; S -> 3

LIST-I	LIST-II
(P) $F e C_{2} O_{4}$	(1) Only one of the radical decolorize aq. KMnO₄
(Q) $F e S O_{4}$	(2) Both the radicals decolorize aq. KMnO₄
(R) $Z n (N O_{3})_{2}$	(3) Can give a colorless gas with conc. H₂SO₄
(S) $(N H_{4})_{2} S$	(4) Adding NaOH can produce a white precipitate

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