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Standard Electrode Potentials

A light bluis...

Question

A light bluish green crystalline compound responds to the following tests:
(i) Its aqueous solution gives a brown precipitate or colouration with alkaline $K_{2} [{H g I}_{4}]$ solution.
(ii) Its aqueous solution gives a blue colour with $K_{3} [{F e (C N)}_{6}]$ solution.
(iii) Its solution in hydrochloric acid gives a white precipitate with ${B a C l}_{2}$ solution. The molecular formula of the compound is $F e S O_{4} . (N H_{4})_{2} S O_{4} . 6 H_{2} O$ (Mohr's salt)

If the given statement is true enter 1 else enter 0.

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Solution

The light bluish green crystalline compound is $F e S O_{4} . (N H_{4})_{2} S O_{4} . 6 H_{2} O$ (Mohr's salt).
(i) Its aqueous solution gives a brown precipitate or colouration with alkaline $K_{2} [{H g I}_{4}]$ solution. This is due to presence of ammonium ions.
$2 K_{2} H g I_{4} + 4 K O H + N H_{4} \to H g_{2} O N H_{2} I ↓ + 7 K I + C l^{-} + 3 H_{2} O$
$i o d i d e o f m i l l o n^{'} s b a s e, b r o w n p p t$
(ii) Its aqueous solution gives a blue colour with $K_{3} [{F e (C N)}_{6}]$
solution. This is due to presence of Fe(I) ions.
$F e S O_{4} (C) + K_{3} [F e (C N)_{6}] ⟶ K F e^{I I} [F e^{I I I} (C N)_{6}] (H), T u r n b u l l^{'} s b l u e$
(iii) Its solution in hydrochloric acid gives a white precipitate with ${B a C l}_{2}$ solution.
This is due to presence of sulphate ions.
$B a C l_{2} + S O_{4}^{2 -} \to B a S O_{4} ↓ (w h i t e) + 2 C l^{-}$

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