What is the average oxidation number of sulfur in S_4O_6^(2-)?

Nov 6, 2015

+2.5 (No, this isn't an anomaly. Let me explain first.)

Explanation:

First we need to calculate the oxidation state of $S$ atom the usual way.

${S}_{4} {O}_{6}^{\text{2-}}$ : overall oxidation state is -2

[oxidation state of $S$ x 4] + [oxidation state of $O$ atom x 6] = -2

The most common oxidation state of oxygen is -2. Thus,

[oxidation state of $S$ x 4] + [(-2) (6)] = -2

Let $\textcolor{red}{y}$ be the oxidation state of $S$. Therefore, we can rewrite the equation as

[($\textcolor{red}{y}$ ) (4)] + [(-2) (6)] = -2

[($\textcolor{red}{y}$ ) (4)] + (-12) = -2

[($\textcolor{red}{y}$ ) (4)] = -2 + (+12)

[($\textcolor{red}{y}$ ) (4)] = +10

$\textcolor{red}{y}$ = $\frac{+ 10}{4}$

$\textcolor{red}{y}$ = + 2.5

Why the decimal place? Because the individual partial charges of the four $S$ atoms are not equal to each other. You need to consider the Lewis structure of this ion.

[Notice how the formal charges are given as $+ 2$ on the central sulfurs ($\textcolor{red}{red}$) and $0$ on the bridging sulfurs ($\textcolor{b l u e}{b l u e}$). These are not oxidation states!]

The oxidation state +2.5 is just the average oxidation state for the $S$ atom.

According to the structure, the symmetry suggests a $- 1$ on each bridging sulfur ($\textcolor{b l u e}{b l u e}$) (just like the bridging $O$ atoms in a peroxide), and a $+ 6$ ($\textcolor{red}{red}$) on each central sulfur (like in sulfate).

Indeed, $\frac{+ 6 - 1}{2} = 2.5$. Hence, the fractional oxidation state.