Question

Compound $A$ is a light green crystalline solid. It gives the following tests
(i) It dissolves in dilute sulphuric acid. No gas is produced.
(ii) A drop of $KMnO_4$ is added to the above solution. The pink colour disappears.
(iii) Compound $A$ is heated strongly. Gases $B$ and $C$, with pungent smell, come out. A brown residue $D$ is left behind.
(iv) The gas mixture ($B$ and $C$) is passed into a dichromate solution. The solution turns green.
(v) The green solution from step (iv) gives a white precipitate $E$ with a
solution of barium nitrate.
(vi) Residue $D$ from step (iii) is heated on charcoal in a reducing flame. It gives a magnetic substance.
Name the compound $A, B, C, D$ and $E$.

Answer 1

Compound A is a light green crystalline solid, so it may be $FeSO_4.$
(i) $FeSO_4$ is a salt of strong acid and weak base, so it hydrolyses in dil. $H_2 SO_4$ but no gas is evolved.
(ii) $FeSO_4$ is a strong reducing agent, thus decolourises $KMnO_4$ solution :
$ 5Fe^{2+} + \underset {\text{Purple}}{MnO_4^-} + 8H^+ \longrightarrow 5Fe^{3+} + \underset {\text{Colourless}}{Mn^{2+}} + 4H_2 O$
(iii) $FeSO_4$ on strong heating gives both $SO_2$ (B) and $SO_3$ (C) gases along with a residue of $Fe_2O_3 (D).$
$ 2FeSO_4 \xrightarrow {\Delta} \, \underset {D}{Fe_2 O_3} + \underset {B}{SO_2} + \underset {C}{SO_3}$
(iv) The gaseous mixture reduced dichromate solution to green
solution and also gives $H_2SO_4$ in solution :
$ \, \, Cr_2 O_7^{2-} + 3SO_2 + 2H^+ \longrightarrow 3SO_4^{2-} + H_2 O + \underset {\text{Green}}{2Cr^{3+}}$
$ H_2 O + SO_3 \longrightarrow H_2 SO_4$
(v) The sulphuric acid (formed in previous step) gives white ppt.
with $Ba(NO_3)_2$ due to formation of $BaSO_4 (E)$ :
$ \, \, H_2 SO_4 + Ba(NO_3)_2 \longrightarrow \underset {\text{White } (E)}{BaSO_4 \downarrow} + 2HNO_3$
(vi) The residue $D$ when heated on charcoal in a reducing flame reduces to iron $(Fe)$ which is a magnetic substance.
Hence, $A = FeSO_4 , B = SO_2 , C = SO_3 , D = Fe_2 O_3$ and $ E = BaSO_4.$