In the test for iodine, when $I_{2}$ is treated with sodium thiosulphate $N a_{2} S_{2} O_{3}$ .

$N a_{2} S_{2} O_{3} + I_{2} \to N a I + . . . .$

N a_{2} S_{4} O_{6}

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N a_{2} S O_{4}

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N a_{2} S

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N A_{3} I S O_{4}

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Solution

The correct option is A $N a_{2} S_{4} O_{6}$
When Iodine is treated with sodium thiosulphate $N a_{2} S_{2} O_{3}$ , iodine solution is decolourised by sodium thiosulphate as sodium tetrathionate $(N a_{2} S_{4} O_{6})$ and sodium iodide are formed. Both are colourless and soluble.

$2 N a_{2} S_{2} O_{3} + I_{2} \to 2 N a I + N a_{2} S_{4} O_{6}$

Hence the correct option is $(A)$ .

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Similar questions

I_{2}

(N a_{2} S_{2} O_{3})

(N a_{2} S_{2} O_{3}) + I_{2} ⟶ N a I + \dots \dots

(N a_{2} S_{2} O_{3} \cdot 5 H_{2} O)

2 N a_{2} S_{2} O_{3} + I_{2} \to 2 N a I + N a_{2} S_{4} O_{6}

I_{2} + N a_{2} S_{2} O_{3} \to N a_{2} S_{4} O_{6} + N a I

N a_{2} S_{2} O_{3}

I_{2}

N a_{2} S_{4} O_{6}

N a I

Balance the following equations:
1. FeCl₃ + NH₄OH → Fe(OH)₃ + NH₄Cl
2. Na₂S₂O₃ + I₂ → Na₂S₄O₆ + NaI

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