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Polar Molecules

Compound A is...

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 $K M n O_{4}$ is added to the above solution. the pink colour disappears.
(iii) Compound (A) is heated strongly. gases (B) and (C) with pungent smell came 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) The residue (D) from (v) is heated on charcoal in reducing the flame. It gives a magnetic substance.
Compound (A) to (E) are identified as:
(A): $F e S O_{4}$ , (B): $S O_{2},$ , (C): $S O_{3}$ , (D): $F e_{2} O_{3}$ , (E): $B a S O_{4}$
If true enter 1, else enter 0.

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Solution

Compounds $^{'} A^{'}$ is a light green crystalline solid, so it may be ferrous sulphate.

(i) $F e S O_{4}$ is a salt of strong acid weak base, so it hydrolysed in dil

$H_{2} S O_{4}$ but no gas is produced.

(ii) Ferrous supphate is a strong reducing agent, thus decolourises the

$K M n O_{4}$ solution

$5 F e^{2 +} + M n O_{4}^{-} (p i n k) + 8 H^{+} \to 5 F e^{3 +} + M n^{2 +} (c o l o u r l e s s) + 4 H_{2} O$

(iii) Ferrous sulphate on strong heating gives both $S O_{2}$ 'B' and $S O_{3}$ 'C' alongwith a residue of $F e_{2} O_{3}$ 'D'

$2 F e S O_{4} \to F e_{2} O_{3} + S O_{2} + S O_{3}$

(iv) This gaseous mixture dichromate to green solution and also give sulphuric acid in the solution.

$C r_{2} O_{7}^{2 -} + 3 S O_{2} + 2 H^{+} \to 3 K_{2} S O_{4} + H_{2} O + 2 C r^{3 +} g r e e n$
$H_{2} O + S O_{3} \to H_{2} S O_{4}$

(v) This sulphuric acid (formed in step(iv)) gives white ppt. with a solution of

$B a (N O_{3})_{2}$ due to formation of $B a S O_{4}^{'} E^{'}$

$H_{2} S O_{4} + B a (N O_{3})_{2} \to 2 H N O_{3} + B a S O_{4} w h i t e ↓$

(vi) The residue 'D' when heated on charcoal in a reducing flame, reduce to iron $(F e)$ which is a magnetic substance.

Hence, $A, B, C . D$ and $E$ are respectively $F e S O_{4}, S O_{2}, S O_{3}, F e_{2} O_{3}$ and $B a S O_{4}$

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If true enter 1, else enter 0.

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If true enter 1, else enter 0.

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