InSO3−− x+3(−2)=−2;x=+4 In S2O4−−2x+4(−2)=−22x−8=−22x=6;x=+3 In S2O62−2x+6(−2)=−22x=10;x=+5 hence the correct order is S2O4−−<SO3−−<S2O6−−
JEE Main 2013 — Chemistry Inorganic Chemistry
Oxidation state of sulphur in anions SO32−,S2O42− and S2O62− increases in the orders :
Held on 22 Apr 2013 · Verified 6 Jul 2026.
S2O62−<S2O42−<SO32−
SO62−<S2O42−<S2O62−
S2O42−<SO32−<S2O62−
S2O42−<S2O62−<SO32−
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