The correct option is C +2, +2.5
charge on oxygen atom is =-2
charge on sodium atom is =+1
(i) Let oxidation number of S-atom in Na2S2O3 is x.
∴(+1)×2+(x)×2+(−2)×3=0
⇒x=+2
(ii) Let oxidation number of S-atom in Na2S4O6 is x
∴(+1)×2+(x)×4+(−2)×6=0⇒x=+2.5
It is important to note here that Na2S2O3 have two S-atoms and there are four S-atoms in Na2S4O6. However none of the sulphur atoms in both the compounds have +2 or + 2.5 oxidation number, it is the average of oxidation number, which reside on each sulphur atom.