What is the oxidation state of S in S_2O_3^-2S2O23?

3 Answers
Apr 16, 2018

+2+2

Explanation:

We can see that the net charge of the molecule is -22.
To start of we find the oxidation state of oxygen, in the molecule.
Oxygen has a oxidation state of -22 since we have 3 oxygens this would give a total oxidation state of -66 for oxygen. But since the net charge of the molecule is -22 we have to all 2 the the total charge of oxygen, thereby giving -44. Now we can see that the sulfur must have a charge of +2+2.

Apr 16, 2018

Formally we gots stackrel(VI+)SVI+S and stackrel(-II)SIIS...

Explanation:

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

"Thiosulfate ion"Thiosulfate ion is an interesting customer in terms of sulfur oxidation state. If we look at sulfate, SO_4^(2-)SO24, CLEARLY we got S^(VI+)SVI+ and 4xxO^(-II)4×OII..and as usual, the weighted sum of the oxidation numbers, 6-8=-268=2, i.e. the charge on the ion.

In "thiosulfate"thiosulfate, S_2O_3^(2-)S2O23, I like to think that ONE of the oxygen atoms of sulfate has BEEN REPLACED by one sulfur as sulfide.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+VI, and the terminal sulfur is S^(-II)SII, 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+2

Explanation:

We got the thiosulfate ion S_2O_3^(2-)S2O23.

Since oxygen is more electronegative than sulfur, then oxygen will have its usual -22 oxidation state. There are three oxygen atoms, and so the total charge of the oxygens is -2*3=-623=6.

Let xx be the total charge of two sulfur atoms.

We got:

x-6=-2x6=2

x=-2+6x=2+6

x=4x=4

So, the sum of the oxidation numbers of the sulfur atoms is +4+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+42=+2.