A black mineral (P) on heating in presence of air gives a gas (Q). The mineral (P) on reaction with dilute H₂SO₄ gives a gas (R) and a solution of a compound (S). On passing the gas (R) into an aqueous solution of (Q) a white turbidity is obtained. The aqueous solution of (S) on reaction with postassium ferricyanide gives a blue compound (T). Then

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$(P) = CuS, (Q) = {SO}_{2}, (R) = H_{2} S; (S) = {CuSO}_{4} and (T) = {Cu}_{3} {[Fe {(CN)}_{6}]}_{2}$

$(P) = PbS, (Q) = {SO}_{2}, (R) = H_{2} S; (S) = {PbSO}_{4} and (T) = {Cu}_{3} {[Fe {(CN)}_{6}]}_{2}$

$(P) = FeS, (Q) = {SO}_{2}, (R) = H_{2} S; (S) = {FeSO}_{4} and (T) = {Fe}_{4} {[Fe {(CN)}_{6}]}_{3}$

$(P) = FeS, (Q) = {SO}_{2}, (R) = H_{2} S; (S) = {Fe}_{2} {({SO}_{4})}_{3} and (T) = Fe [Fe {(CN)}_{6}]$

answer is C.

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Detailed Solution

$\underset{(P)}{2 FeS} + 3 O_{2} \to 2 FeO + \underset{(Q)}{2 {SO}_{2}} FeS + H_{2} {SO}_{4} \to \underset{(S)}{{FeSO}_{4}} + \underset{(R)}{H_{2} S} H_{2} S + {SO}_{2} \to \underset{(Milky white)}{3 S} ↓ + 2 H_{2} O {FeSO}_{4} + K_{3} [Fe {(CN)}_{6}] \to \underset{\underset{(T)}{(Blue ppt)}}{{Fe}_{4} {[Fe {(CN)}_{6}]}_{3}}$