Consider #"sulfate ion"#, i.e. #SO_4^(2-)#; we have #"sulfur(+VI)"# and #"oxygen(-II)"#. As always, the weighted sum of the oxidation numbers is equal to the charge on the ion, i.e. #4xx-II+VI=-2#, if you will forgive me mixing Roman and Arabic numerals (of course you will).
As regards #"thiosulfate ion"#, #S_2O_3^(2-)#, I have always liked to think that one of the oxygens has been replaced by a #"sulfide ion"# with PRECISELY the same oxidation state as the oxygen, i.e. #stackrel(-II)S#. And with this formalism, we gots #stackrel(-II)S+stackrel(+VI)S+3xxstackrel(-II)O=-2#, the charge on the ion as required. (And note that both sulfur and oxygen are Group 16 chalcogens, and they should sometimes have the same oxidation number!)
And so FORMALLY we gots #stackrel(-II)S#, AND #stackrel(+VI)S#; and we may continue the formalism, and assign an AVERAGE oxidation number of #(-II+VI)/2=+II# for both sulfurs.
Are you happy with this treatment?
And another way of looking at this is to consider the #S^(2-)# ion....which, I hope you will agree has a formal #-II# oxidation state. Now it is a FACT that sulfur flowers, i.e. the yellow powder, that we can find in a bottle, is mighty insoluble stuff......and this is probably the #S_8# molecule, i.e. an 8-membered sulfur ring. So if you have got a slurry of sulfur in water......you can add some sulfide salt, and get the sulfur up into solution......
#S_8+S^(2-) rarr S_9^(2-)# where the sulfido anion is present as some form of ring.....the formal oxidation state of #S# is #-2/9#; of course, FORMALLY we got #8xxstackrel(0)S+stackrel(-II)S# OR #2xxstackrel(-I)S+6xxstackrel(0)S#. These sulfido rings are strongly oxidizing.
