Difference in Oxidation Numbers of Sulphur in Sodium Tetrathionate

Sodium tetrathionate has the chemical formula $\text{Na}_2\text{S}_4\text{O}_6$, which dissociates to form two sodium cations ($2\text{Na}^+$) and one tetrathionate anion ($\text{S}_4\text{O}_6^{2-}$).

The structure of the tetrathionate ion, $\text{S}_4\text{O}_6^{2-}$, is represented as:
$[\text{O}_3\text{S}-\text{S}-\text{S}-\text{SO}_3]^{2-}$

In this ion, there are two distinct types of sulphur atoms:

1. Terminal Sulphur Atoms:

   • There are two terminal sulphur atoms ($\text{S}^*$) in the structure:
     $\text{O}_3\text{S}^*-\text{S}-\text{S}-\text{S}^*\text{O}_3^{2-}$
   • Each terminal sulphur atom is bonded to three oxygen atoms and one central sulphur atom.
   • The three bonds to oxygen atoms (two $\text{S}=\text{O}$ double bonds and one $\text{S}-\text{O}^-$ single bond) contribute $+5$ to the oxidation state of the terminal sulphur atom due to the higher electronegativity of oxygen compared to sulphur.
   • The homonuclear $\text{S}^*-\text{S}$ single bond contributes $0$ to the oxidation state as both atoms have the same electronegativity.
   • Therefore, the oxidation number of each terminal sulphur atom is $+5$.

2. Central Sulphur Atoms:

   • There are two central sulphur atoms in the middle of the chain ($\text{S}-\text{S}-\text{S}$).
   • Each central sulphur atom is bonded exclusively to other sulphur atoms ($\text{S}-\text{S}$ bonds).
   • Since there is no electronegativity difference between identical sulphur atoms, these bonds do not contribute to any formal oxidation state.
   • Therefore, the oxidation number of each central sulphur atom is $0$.

The difference in the oxidation numbers of the two types of sulphur atoms is:
$\text{Difference} = |+5 - 0| = 5$