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In `"thiosulfate"`, `S_2O_3^(2-)`, I like to think that ONE of the oxygen atoms of sulfate has BEEN REPLACED by one sulfur as sulfide.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`.