The reaction that proceeds in the forward direction is:

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

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

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{M n}^{2 +} + 2 H_{2} O + C l_{2} ⟶ M n O_{2} + 4 H^{+} + 2 C l^{-}

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S_{4} O_{6}^{2 -} + 2 I^{-} ⟶ 2 S_{2} O_{3}^{2 -} + I_{2}

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Solution

The correct option is B $N H_{4} C l + N a O H ⟶ H_{2} O + N H_{3} + N a C l$
For a reaction to proceed in forward direction, it has to be spontaneous.
A spontaneous reaction is characterized by increase in entropy.

From the given reactions only option B shows an increase in entropy, because the number of molecules in the product is higher than that of the reactants.

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2 S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{⊖}

I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{-}

2 S_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -}

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I_{2 (a q)} + 2 S_{2} O_{3}^{2 -} (a q) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

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S_{2} O_{3}^{2 -} (a q) + 2 {B r}_{2} (l) + 5 H_{2} O (l) ⟶ 2 {S O}_{4}^{2 -} (a q) + 4 {B r}^{-} (a q) + 10 H^{+} (a q)

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