Step 1: Understanding the Concept:
In complex ions like tetrathionate (\(\text{S}_4\text{O}_6^{2-}\)), atoms of the same element can have different oxidation states depending on their local chemical environment and the atoms they are bonded to. Step 2: Key Formula or Approach:
The oxidation state of an atom is calculated by assigning valence electrons to the more electronegative atom in a bond. Bonds between atoms of the same element do not contribute to a change in oxidation state. Step 3: Detailed Explanation:
Looking at the structure \([\text{O}_3\text{S}^{(1)}\text{-S}^{(2)}\text{-S}^{(3)}\text{-S}^{(4)}\text{O}_3]^{2-}\):
- Sulphur atoms 2 and 3: These are central and are bonded only to other sulphur atoms. Since they have no bonds to more electronegative atoms like oxygen, their oxidation state is $0$.
- Sulphur atoms 1 and 4: Each of these terminal sulphurs is bonded to three oxygen atoms and one sulphur atom. In the \(\text{S-O}\) bonds, oxygen is more electronegative.
Counting bonds for S(1) or S(4): Two double bonds to oxygen (\(2 \times 2 = 4\)) and one single coordinate bond or ionic oxygen bond (\(1 \times 1 = 1\)). The total is $+5$.
The bond between \(\text{S}^{(1)}\text{-S}^{(2)}\) is between identical atoms, so it adds $0$ to the count.
Thus, the states are: \(\text{S}^{(1)} = +5\), \(\text{S}^{(2)} = 0\), \(\text{S}^{(3)} = 0\), \(\text{S}^{(4)} = +5\). Step 4: Final Answer:
The sequence is \(+5, 0, 0, +5\).
