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Q.

Sodium thiosulphate reacts with iodine to give iodide and tetrathionate. In this reaction S2O32- undergoes

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a

Oxidation

b

Reduction

c

Disproportionation

d

Comproportionation

answer is A.

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Detailed Solution

\large \mathop {2{S_2}}\limits^{ + 2} O_3^{ - 2}\; + \;{I_2}\; \to \;\mathop {{S_4}}\limits^{ + 2.5} O_6^{ - 2}\; + \;2{I^ - }
Oxidation state of 'S' in \large {S_2}O_3^{ - 2} increase from +2 to +2.5
\large \therefore \;{S_2}O_3^{ - 2}\; undergoes Oxidation

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