A hydrated metallic salt (A) , light green in colour gives a white anhydrous residue (B) after being heated gradually. (B) is soluble in water and its aqueous solution reacts with NO to give a dark brown compound (C). (B) on strong heating gives a brown residue (D) and a mixture of two gases (E) and (F). The gaseous mixture , when passed through acidified permanganate, discharges the pink colour and when passed through acidified BaCl2 solution, gives a white precipitate. Compounds (A), (B), (C) , (D) , (E) and (F) are identified as: (A) : $F e S O_{4} .7 H_{2} 0$ (B) : $F e S O_{4}$ ,(C) : $F e S O_{4} . N O$ or $[F e {(H_{2} 0)}_{6} N 0] S O_{4}$ (D) : $F e_{2} 0_{3}$ ,(E) : $S O_{2}$ and (F) : $S O_{3}$ If true enter 1, else enter 0.

Question

A hydrated metallic salt (A) , light green in colour gives a white anhydrous residue (B) after being heated gradually. (B) is soluble in water and its aqueous solution reacts with NO to give a dark brown compound (C). (B) on strong heating gives a brown residue (D) and a mixture of two gases (E) and (F). The gaseous mixture , when passed through acidified permanganate, discharges the pink colour and when passed through acidified BaCl2 solution, gives a white precipitate. Compounds (A), (B), (C) , (D) , (E) and (F) are identified as: (A) : $F e S O_{4} .7 H_{2} 0$ (B) : $F e S O_{4}$ ,(C) : $F e S O_{4} . N O$ or $[F e {(H_{2} 0)}_{6} N 0] S O_{4}$ (D) : $F e_{2} 0_{3}$ ,(E) : $S O_{2}$ and (F) : $S O_{3}$ If true enter 1, else enter 0.

Vedantu Content Team · Accepted Answer

Hint: An inorganic hydrated salt is an ionic compound in which water molecules are attracted by the ions and therefore get enclosed in the lattice crystal of the salt. There  are so many examples of hydrated salts which are,  ferrous sulfate heptahydrate, calcium chloride hexahydrate etc. Complete step by step solution:The light green colour salt is ferrous sulfate heptahydrate. : (A) : $$FeS{O_4}.7{H_2}0$$. After being heated gradually, a light green in colour gives a white anhydrous residue (B) $$FeS{O_4}$$. This anhydrous residue is soluble in water and its aqueous solution reacts with NO to give a dark brown compound (C) :$$FeS{O_4}.NO$$ or $$\left[ {Fe{{\left( {{H_2}0} \right)}_6}N0} \right]S{O_4}$$. The reaction is shown below.$$   FeS{O_{4}} + NO \to FeS{O_{4}}.NO\;   \\  or,FeS{O_4} + 6{H_2}O + NO \to Fe{({H_{2}}O)_6}NO]S{O_{4}}]   \\   $$ On strong heating the anhydrous residue (B) gives a brown residue (D) :$$F{e_2}{0_3}$$ and a mixture of two gases (E) $$S{O_2}$$ and (F) : $$S{O_3}$$. The reaction is shown below.$$2FeS{O_{4}}\xrightarrow{{heat\;strongly}}F{e_{2}}{O_3} + S{O_{2}} \uparrow  + S{O_3} \uparrow $$ The gaseous mixture , when passed through acidified permanganate, discharged the pink colour and when passed through acidified $$BaC{l_{2}}$$ solution, $$BaC{l_{2}}$$ reacts with $${\text{S}}{{\text{O}}_3}$$ and gives a white precipitate of barium sulfate. The reaction is shown below.$$BaC{l_{2}} + {H_2}O + S{O_{3}} \to BaS{O_{4}} + 2HCl$$ Note: $$\left[ {Fe{{\left( {{H_2}O} \right)}_5}NO} \right]S{O_4}$$ this is a brown ring complex. This complex is formed in brown ring test. The brown ring test is basically to determine the presence of nitrate ion in the solution. It is performed by iron(II) sulfate to a solution of a nitrate. After that sulfuric acid is added slowly, which forms a layer below  the aqueous solution as a ring.

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