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

Apr 16, 2018

$+ 2$

Explanation:

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 $+ 2$.

Apr 16, 2018

Formally we gots $\stackrel{V I +}{S}$ and $\stackrel{- I I}{S}$...

Explanation:

And thus ${S}_{\text{average oxidation state}} = \frac{V I + \left(- I I\right)}{2} = + I I$.

$\text{Thiosulfate ion}$ is an interesting customer in terms of sulfur oxidation state. If we look at sulfate, $S {O}_{4}^{2 -}$, CLEARLY we got ${S}^{V I +}$ and $4 \times {O}^{- I I}$..and as usual, the weighted sum of the oxidation numbers, $6 - 8 = - 2$, i.e. the charge on the ion.

In $\text{thiosulfate}$, ${S}_{2} {O}_{3}^{2 -}$, I like to think that ONE of the oxygen atoms of sulfate has BEEN REPLACED by one sulfur as sulfide.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 $+ V I$, and the terminal sulfur is ${S}^{- I I}$, 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?

Apr 17, 2018

$+ 2$

Explanation:

We got the thiosulfate ion ${S}_{2} {O}_{3}^{2 -}$.

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 $- 2 \cdot 3 = - 6$.

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 $\frac{+ 4}{2} = + 2$.