Question #aaf68

Sep 12, 2017

We gots sulfur $\left(V I +\right)$

Explanation:

And so we have $C {u}^{2 +}$ and $S {O}_{4}^{2 -}$. Now for sulfate, the SUM of the individual elemental oxidation numbers is equal to the charge on the ion.....

And thus ${S}_{\text{oxidation number"+4xxO_"oxidation number}} = - 2$

Now ${O}_{\text{oxidation number}}$ is usually $- I I$, and it is here, and clearly, ${S}_{\text{oxidation number}} = + V I$.

For $\text{thiosulfate anion}$, ${S}_{2} {O}_{3}^{2 -}$, the given approach would yield $S \left(+ I I\right)$. Here, however, I like to think that sulfur has replaced ONE of the oxygens in SULFATE, $S {O}_{4}^{2 -}$, and assumed its oxidation state, i.e. $S \left(- I I\right)$. The other sulfur is of course remains $S \left(V I +\right)$ by this formalism. The average oxidation state is still $S \left(+ I I\right)$, but of course both ideas are necessarily formalisms.

For $\text{sulfite anion}$, $S {O}_{3}^{2 -}$ we gots $S \left(I V +\right)$, and of course for $\text{sulfide dianion}$, ${S}^{2 -}$, we got $S \left(- I I\right)$.

So here is a question.....we can add elemental sulfur (possibly ${S}_{8}$) to a sulfide solution as shown.....

${S}_{8} \left(s\right) + {S}^{2 -} \rightarrow {S}_{10}^{2 -}$

The sulfur goes up to give polysulfide chains of various length. What is the oxidation state of the sulfur here?