What is the oxidation state of S in #S_2O_3^-2#?

We can see that the net charge of the molecule is #-2#.
To start of we find the oxidation state of oxygen, in the molecule.
Oxygen has a oxidation state of #-2# since we have 3 oxygens this would give a total oxidation state of #-6# for oxygen. But since the net charge of the molecule is #-2# we have to all 2 the the total charge of oxygen, thereby giving #-4#. Now we can see that the sulfur must have a charge of #+2#.

And thus #S_"average oxidation state"=(VI+(-II))/2=+II#.

#"Thiosulfate ion"# is an interesting customer in terms of sulfur oxidation state. If we look at sulfate, #SO_4^(2-)#, CLEARLY we got #S^(VI+)# and #4xxO^(-II)#..and as usual, the weighted sum of the oxidation numbers, #6-8=-2#, i.e. the charge on the ion.

In #"thiosulfate"#, #S_2O_3^(2-)#, I like to think that ONE of the oxygen atoms of sulfate has BEEN REPLACED by one sulfur as sulfide. And thus the central sulfur is #+VI#, and the terminal sulfur is #S^(-II)#, i.e. its oxidation state is PRECISELY the same as its Group 16 congener, oxygen. Of course, this is a formalism, but so is the whole concept of oxidation state and oxidation number.

Claro?

We got the thiosulfate ion #S_2O_3^(2-)#.

Since oxygen is more electronegative than sulfur, then oxygen will have its usual #-2# oxidation state. There are three oxygen atoms, and so the total charge of the oxygens is #-2*3=-6#.

Let #x# be the total charge of two sulfur atoms.

We got:

#x-6=-2#

#x=-2+6#

#x=4#

So, the sum of the oxidation numbers of the sulfur atoms is #+4#. If we consider both sulfur atoms to have the same oxidation number, then each sulfur will have an oxidation number of #(+4)/2=+2#.