SATHEE: Qualitative Analysis - Result Question 76

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 $K M n O_{4}$ 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 $F e S O_{4}$ .

(i) $F e S O_{4}$ is a salt of strong acid and weak base, so it hydrolyses in dil. $H_{2} S O_{4}$ but no gas is evolved.

(ii) $F e S O_{4}$ is a strong reducing agent, thus decolourises $K M n O_{4}$ solution :

$5 F e^{2 +} + \underset{Purple}{M n O_{4}^{-}} + 8 H^{+} ⟶ 5 F e^{3 +} + \underset{Colourless}{M n^{2 +}} + 4 H_{2} O$

(iii) $F e S O_{4}$ on strong heating gives both $S O_{2} (B)$ and $S O_{3} (C)$ gases alongwith a residue of $F e_{2} O_{3} (D)$ .

$2 F e S O_{4} \overset{Δ}{⟶} \underset{D}{F e_{2} O_{3}} + \underset{B}{S O_{2}} + \underset{C}{S O_{3}}$

(iv) The gaseous mixture reduced dichromate solution to green solution and also gives $H_{2} S O_{4}$ in solution :

$\begin{matrix} C r_{2} O_{7}^{2 -} + 3 S O_{2} + 2 H^{+} ⟶ 3 S O_{4}^{2 -} + H_{2} O + \underset{Green}{2 C r^{3 +}} H_{2} O + S O_{3} ⟶ H_{2} S O_{4} \end{matrix}$