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The oxidation states of sulphur in the anions SO32–, S2O42– and S2O62– follow the order [2003]
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The oxidation states of sulphur in the anions SO32–, S2O42&ndash...
The correct answer is option B
Oxidation state of S2​O42−
2(x) + 4(−2) = −2
 2x = 8 − 2
 2x = 6
  x = 3
Oxidation state of SO32−
x + 3(−2) = −2
 x = 6 − 2
 x = 4
Oxidation state of S2​O62−
2(x) + 6(−2) = −2
 2x = 12 − 2
 2x = 10
   x = 5
So the oxidation state of sulphur in the anions S2​O42−​, S2​O42−​ and S2​O62−​ follows the order.S2​O42−​ < SO32−​ < S2​O62−​.
 
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The oxidation states of sulphur in the anions SO32–, S2O42&ndash...
SO3^2- has o.s=+4
S2O4^2- has o.s=+3
S2O6^2- has o.s=+5
so correct order is option B
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