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Redox Reactions question

2003 · Q128
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Redox Reactions question

2003 · Q128

NEETChemistryRedox ReactionsMCQ+4 / −1
The oxidation states of sulphur in the anions

SO332−{_3^{2 - }}32−​, S2O42−{_4^{2 - }}42−​ and S2O62−{_6^{2 - }}62−​ follow the order
  1. A
    S2O42−{_4^{2 - }}42−​ < SO32−{_3^{2 - }}32−​ < S2O62−{_6^{2 - }}62−​
  2. B
    SO32−{_3^{2 - }}32−​ < S2O42−{_4^{2 - }}42−​ < S2O62−{_6^{2 - }}62−​
  3. C
    S2O42−{_4^{2 - }}42−​ < S2O62−{_6^{2 - }}62−​ < SO32−{_3^{2 - }}32−​
  4. D
    S2O62−{_6^{2 - }}62−​ < S2O42−{_4^{2 - }}42−​ < SO32−{_3^{2 - }}32−​
View written solutionFree

Correct answer: A

SO32– : oxidation state of ‘S’ is +4

S2O42– : oxidation state of ‘S’ is +3.

S2O62– : oxidation state of ‘S’ is +5.

So, the order is S2O42−{_4^{2 - }}42−​ < SO32−{_3^{2 - }}32−​ < S2O62−{_6^{2 - }}62−​.

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