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Byju's Answer

Standard XII

Chemistry

Multiple Bond Formation

A yellow colo...

Question

(A) (yellow coloured solution) changes to light green coloured solution (B) on passing $H_{2} S$ gas. (A) and (B) both give white precipitate with $B a {C l}_{2}$ solution, insoluble in conc. $H C l$ . A gives blue coloured ppt (C) with $K_{4} [{F e (C N)}_{6}]$ , (B) does not. Here,
(A) : ${F e C l}_{3}$
(B) : ${F e C l}_{2}$
(C) : $F e_{4} [F e (C N_{6})]_{3}$
If true enter 1, else enter 0.

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Solution

The yellow coloured solution (A) contains $F e C l_{3}$ .
The light green coloured solution (B) contains $F e C l_{2}$ .
$H_{2} S$ reduces Fe(III) ions to Fe(II) ions, therefore, the solution becomes light green in colour on passing $H_{2} S$ .
$2 F e^{3 +} + S^{2 -} \to 2 F e^{2 +} + S$
(B) reacts with $K_{4} [{F e (C N)}_{6}]$ to form white precipitate.
$2 F e C l_{2} + K_{4} [{F e (C N)}_{6}] \to F e_{2} [F e (C N)_{6}] w h i t e p r e c i p i t a t e + 4 K C l$
(A) reacts with $K_{4} [{F e (C N)}_{6}]$ to form blue precipitate.
$4 F e C l_{3} + 3 K_{4} [{F e (C N)}_{6}] \to F e_{4} [F e (C N)_{6}]_{3} b l u e p r e c i p i t a t e + 12 K C l$

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