And so we have Cu^(2+) and SO_4^(2-). Now for sulfate, the SUM of the individual elemental oxidation numbers is equal to the charge on the ion.....
And thus S_"oxidation number"+4xxO_"oxidation number"=-2
Now O_"oxidation number" is usually -II, and it is here, and clearly, S_"oxidation number"=+VI.
For "thiosulfate anion", S_2O_3^(2-), the given approach would yield S(+II). Here, however, I like to think that sulfur has replaced ONE of the oxygens in SULFATE, SO_4^(2-), and assumed its oxidation state, i.e. S(-II). The other sulfur is of course remains S(VI+) by this formalism. The average oxidation state is still S(+II), but of course both ideas are necessarily formalisms.
For "sulfite anion", SO_3^(2-) we gots S(IV+), and of course for "sulfide dianion", S^(2-), we got S(-II).
So here is a question.....we can add elemental sulfur (possibly S_8) to a sulfide solution as shown.....
S_8(s)+S^(2-) rarr S_10^(2-)
The sulfur goes up to give polysulfide chains of various length. What is the oxidation state of the sulfur here?
