# Question 6c70c

Jan 29, 2017

I think you might mean tetrathionite, ${S}_{4} {O}_{6}^{2 -}$?

#### Explanation:

Well, here the $\text{average oxidation number of sulfur}$ $=$ $+ \frac{5}{2}$.

As you know, the oxidation number is the charge left on the central atom when all the bonding pairs of electrons are removed, with the charge devolved to the central. In practice, the sum of the oxidation numbers equals the charge on the ion, here $- 2$.

And thus $4 \times {S}_{\text{oxidation number")+6xxO_("oxidation number}} = - 2$.

Given that the usual oxidation number of $O$ is $- I I$ (and it is this value here, then $4 \times {S}_{\text{oxidation number}} + 6 \times \left(- 2\right) = - 2$, then $4 \times {S}_{\text{oxidation number}} = + 10$, i.e. the $\text{average oxidation number of sulfur}$ $=$ $\frac{5}{2}$.

But we can refine this further, and make some more pronouncements.

The structure of the ion is ""^(-)O(O=)_2S-S-S-S(=O)_2O^-.

Given that we can break up the sulfur chain in radicals (i.e. the 2 electrons are shared by the sulfur atoms in a $S - S$ bond in that they have the same electronegativity), each TERMINAL sulfur has an oxidation state of $+ V$, and the inner sulfur atoms are zerovalent, i.e. ${S}^{0}$.

Do you see from where I am coming?

And thus ""^(-)O(O=)_2S^(+V)-S^(0)-S^(0)-^(+V)S(=O)_2O^-#

Of course, $\text{tetrathionite}$ could be oxidized up to $S {O}_{4}^{2 -}$. What is the oxidation state of sulfur here?

${S}_{4} {O}_{6}^{2 -} + 10 {H}_{2} O \rightarrow 4 S {O}_{4}^{2 -} + 20 {H}^{+} + 14 {e}^{-}$

I think mass and charge are balanced, as required.