Thiosulfate ion contains two sulfur atoms, but they have different oxidation numbers, 0 and +4. Sulfur's maximum oxidation number is +6. Therefore thiosulfate ion can be oxidized or reduced. Therefore thiosulfate can be converted into sulfur or sulfate ion in reactions.
There is a double bond between two sulfur atoms(S=S) and middle sulfur atom and one oxygen atom. (S=O). Electronegativity of oxygen is higher than sulfur. Therefore oxygen has negative oxidation number. Oxidation number of all oxygen atoms are -2. One sulfur atom has +4 oxidation number and other has 0.
One sulfur aton is the central atom in the thiosulfate ion. There are four sigma bonds and zero lone pairs around the central atom. Therefore molecular shape of thiosulfate ion is tetrahedral.
Reaction of rhombic form of sulfur and sulfite ion will give thiosulfate ion.
SO32- + S → S2O32-
Add little bit of farina to iodine solution. The solution will take blue colour.
Then add sodium sulphite (Na2S2O3) to that blue colour solution. You can see blue color is disappearing from the solution due to loss of I2.
A white precipitate (Ag2S2O3) is formed.
Add more S2O32- to Ag2S2O3. Precipitate is started to dissolve and give colourless solution.
But Ag2S2O3 is unstable. Therefore it turns black silver sulphide (Ag2S).
Potassium permanganate (KMnO4) is purple colour solution. KMnO4 reacts with S2O32- ion in two ways.
Ferric chloride (FeCl3) gives a deep violet colour with thiosulfate ion solutions. The colour quickly disappears. it is probaly due to the formation of ferric thiosulfates (or a corresponding complex ). The colour fades owing to reduction of ferric ion to ferrouss ion.
This reaction give SO2 and S(white precipitate) as products.
White precipitate, lead thiosulfate (PbS2O3) forms. But this PbS2O3 is unstable. When it boils, lead sulfide (PbS) black precipitate is formed.
In this section, we are going to find oxidation number of sulfur atoms in S2O32- ion using the equation. We assume oxidation numbers of sulfur atoms are same. Oxidation number of oxygen atoms are taken as -2.
Oxidation number of s is taken as x.
2*x + (-2)*3 = -6
x = +3
But, by calculating oxidation number from drawing the molecular shape, we got oxidation numbers of sulfur atoms are +4 and 0. Therefore we see, the oxidation number found by the formula is not much accurate.