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Question

A black coloured compound (A) on reaction with dilute H2SO4 gives a gas (B) which on passing in a solution of an acid (C) gives a white turbidity (D). Gas (B) when passed in an acidified solution of a compound (E) gives a precipitate (F) soluble in dil. HNO3. After boiling this solution when an excess of NH4OH is added, a blue coloured compound (G) is formed. To this solution on addition of acetic acid and aqueous K4[Fe(CN)6] a chocolate precipitate (H) is obtained. On addition of an aqueous soluion of BaCl2 to an aqueous solution of (E), a white precipitate insoluble in HNO3 is obtained. Identify (G) and (H).

A
(G)−[Cu(NH3)4(NO3)2(H)−Cu2[Fe(CN)6]
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B
(G)−[Cu(NH3)3(NO3)2(H)−Cu2[Fe(CN)6]
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C
(G)−[Cu(NH3)4(NO3)2(H)−Cu[Fe(CN)6]
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D
(G)−[Cu(NH3)4(NO3)(H)−Cu2[Fe(CN)6]
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Solution

The correct option is A (G)−[Cu(NH3)4(NO3)2(H)−Cu2[Fe(CN)6]
Option (A) is correct. The sequence of the formed compounds are : (A)−FeS(B)−H2S(C)−HNO3(D)−S(E)−CuSO4(F)−CuS(G)−[Cu(NH3)4(NO3)2(H)−Cu2[Fe(CN)6]
FeS + H2SO4 → Fe2O3 + H2S
H2S + HNO3 → H2O + N2 +S
H2S + CuSO4 → CuS + H2SO4
CuS + NH4OH → [Cu(NH3)4(NO3)2
[Cu(NH3)4(NO3)2 + CH3COOH +K4[Fe(CN)6] → Cu2[Fe(CN)6]
CuSO4 + BaCl2 → BaSO4 + CuCl2

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