Qualitative Analysis - Result Question 76
79. 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 $E$.
(1980,4 M)
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Solution:
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 :
$5 Fe^{2+}+\underset{\text { Purple }}{MnO _4^{-}}+8 H^{+} \longrightarrow 5 Fe^{3+}+\underset{\text { Colourless }}{Mn^{2+}}+4 H _2 O$
(iii) $FeSO _4$ on strong heating gives both $SO _2(B)$ and $SO _3(C)$ gases alongwith a residue of $Fe _2 O _3(D)$.
$2 FeSO _4 \stackrel{\Delta}{\longrightarrow} \underset{D}{Fe _2 O _3}+\underset{B}{SO _2}+\underset{C}{SO _3}$
(iv) The gaseous mixture reduced dichromate solution to green solution and also gives $H _2 SO _4$ in solution :
$\begin{gathered} Cr _2 O _7^{2-}+3 SO _2+2 H^{+} \longrightarrow 3 SO _4^{2-}+H _2 O+\underset{\text { Green }}{2 Cr^{3+}} \\ H _2 O+SO _3 \longrightarrow H _2 SO _4 \end{gathered}$
(v) The sulphuric acid (formed in previous step) gives white ppt. with $Ba\left(NO _3\right) _2$ due to formation of $BaSO _4(E)$ :
$H _2 SO _4+Ba\left(NO _3\right) _2 \longrightarrow \underset{\text { White (E)}}{BaSO _4 \downarrow}+2 HNO _3$
(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 $E=BaSO _4$
BETA
