# How is "thiosulfate", S_2O_3^(2-), oxidized to "dithionate", S_4O_6^(2-)?

$2 {S}_{2} {O}_{3}^{2 -} \rightarrow {S}_{4} {O}_{6}^{2 -} + 2 {e}^{-}$
In thiosulfate the average sulfur oxidation state is $+ I I$. In dithionate, the average oxidation state is $+ I I \frac{1}{2}$ (the central sulfur linkage has oxidation state $0$, and the termini are $S \left(V\right)$.
We could assign individual sulfur oxidation numbers in dithionate as...""^(-)O(O=)stackrel(+V)S-stackrel(0)S-stackrel(0)S-stackrel(+V)S(=O)O^(-). The internal sulfurs are conceived to share the electrons in the $S - S$ bonds equally...and are thus formally zerovalent.