# How to assign oxidation number of sulfur in thiocyanate anion?

Recently I was trying to balance a redox chemical equation:

$$\ce{CuSCN + KIO3 + HCl <=> CuSO4 + KCl + HCN + ICl + H2O}$$

The equation is in principle simple to solve; one element ($$\ce{I}$$) is reduced and two elements ($$\ce{Cu}$$, $$\ce{S}$$) are oxidized. The number of exchanged electrons in both cases must be the same. However, my problem is that I don't know how to assign the oxidation number to sulfur in a thiocyanate anion.

Later, I solved the equation using the matrix method and via obtained reaction coefficients ($$4$$, $$7$$, $$14$$, $$4$$, $$7$$, $$4$$, $$7$$, $$5$$) and concluded that the oxidation state of sulfur in thiocyanate is zero. This result is very surprising, because the oxidation number zero in heteroatomic molecules is not common.

My questions are:

• How to assign the oxidation number in such cases?
• How is it possible that oxidation number of an atom bound to a more eletropositive atom is zero?
• If you assume that carbon isn't reduced that you'll get such result, but it's wrong assumption. – Mithoron Mar 3 '16 at 20:04
• Thanks for the idea! Only now oxidation states in relation to atom's valency make more sense. In $SCN^{-}$ I assigned oxidation states as S=-2, C=4, N=-3 and in $CN^{-}$ as C=2, N=-3 – Rok Narobe Mar 3 '16 at 20:39

Clearly the iodine in the iodate ion is the oxidizing agent, dropping from $$+5$$ to $$+1$$. But copper, sulfur, and even carbon are all being oxidized or reduced from the cuprous thiocyanate. We have multiple parts of the $$\ce{CuSCN}$$ participating in the electron exchange.
Why not then identify the whole $$\ce{CuSCN}$$ molecule as the reducing agent, thus covering all the reducing agent atoms in their proper proportions? With this approach you see that the sum of oxidation states for this reductant goes from $$0$$ to $$+7$$ ($$+2$$ for $$\ce{Cu}$$ in the copper sulfate, $$+6$$ for the sulfur in the same compound, and $$-1$$ for $$\ce{CN}$$ in the hydrogen cyanide). Then $$\ce{CuSCN}$$ gives $$7$$ electrons, iodate takes $$4$$ electrons, and thus you get $$\ce{4 CuSCN + 7 KIO3}$$, just like with the matrix method.