The reaction which proceeds in the backward direction is :

F e_{3} O_{4} + 6 H C l = 2 F e C l_{3} + 3 H_{2} O

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N H_{3} + H_{2} O + N a C l = N H_{4} C l + N a O H

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S n C l_{4} + H g_{2} C l = S n C l_{2} + 2 H g C l_{2}

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2 C u I + I_{2} + 4 K^{\oplus} = 2 C u^{2 +} + 4 K I

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Solution

The correct options are
B $2 C u I + I_{2} + 4 K^{\oplus} = 2 C u^{2 +} + 4 K I$
C $S n C l_{4} + H g_{2} C l = S n C l_{2} + 2 H g C l_{2}$
D $N H_{3} + H_{2} O + N a C l = N H_{4} C l + N a O H$
According to Le Chatelier's Principle, when a stress is applied to a system at equilibrium the system will adjust to relieve the stress.
In the given reactions
$F e_{2} O_{3} + 6 H C l \to 2 F e C l_{3} + 3 H_{2} O$
The reaction is acid base neutralization reaction will proceed in forward direction.
$N H_{3} + H_{2} O + N a C l \to N H_{4} C l + N a O H$
The reaction will proceed in backward direction as the product side has strong base reacting with salt.
$S n C l_{4} + H g_{2} C l_{2} \to S n C l_{2} + 2 H g C l_{2}$
The reaction will proceed in backward direction as the $S n^{+ 4}$ and $H g^{+ 2}$ are more stable oxidation states than $H g^{+}$ and $S n^{2 +}$
$2 C u I + I_{2} + 4 K^{+} \to 2 C u^{2 +} + 4 K I$
The reaction will proceed in backward direction.

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Similar questions

2 F e C l_{3} + S n C l_{2} \to 2 F e C l_{2} + S n C l_{4}

2 F e C l_{3} + S n C l_{2} \to 2 F e C l_{2} + S n C l_{4}

2 F e C l_{3} + S n C l_{2} \to 2 F e C l_{2} + S n C l_{4}

S n C l_{2} + 2 F e C l_{3} \to S n C l_{4} + 2 F e C l_{2}

A

N H_{4} C l

D

C O_{2}

B

C

+ N a O H \to N a C l + N H_{3} + H_{2} O

N H_{3} + C O_{2} + H_{2} O \to (B)

+ N a C l \to (C) + N H_{4} C l

\to N a_{2} C O_{3} + H_{2} O + (D)

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