Q.

Oxidation states of S in anions SO32-, S2O42- and S2O62- follow the order

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a

S2O62<S2O42<SO32

b

S2O42<SO32<S2O62

c

SO32<S2O42<S2O62

d

S2O42<S2O62<S2O32

answer is A.

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Detailed Solution

S2O4-2  sulphur oxidation number = 2(x) +4(-2) =-2   therefore x= +3                                                                                                                           

SO3-2   sulphur oxidation number =x +3(-2) =-2         therefore x = +4                                                                                                                  

S2O6-2   sulphur oxidation number = 2(x) +6(-2) = -2  therefore x = +5

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