What is the oxidation number of sulfur in Na2S2O3?

• A. +2
• B. +2 and +6
• C. 0
• D. +4

Answer: (B)

Verified step by step guidance

1. Identify the chemical formula: Na2S2O3 is sodium thiosulfate, which contains sodium (Na), sulfur (S), and oxygen (O).
2. Recall the typical oxidation states: Sodium (Na) is usually +1, oxygen (O) is usually -2, and sulfur (S) can have multiple oxidation states.
3. Set up an equation for the sum of oxidation numbers: The compound is neutral, so the sum of all oxidation numbers must be zero. Let the oxidation numbers of the two sulfur atoms be x and y (since they can differ). Then, the equation is: \$2(+1) + x + y + 3(-2) = 0$.
4. Use the structure of thiosulfate ion (S2O3^{2-}): In thiosulfate, one sulfur is typically in a different environment than the other, leading to different oxidation states. The overall charge on the thiosulfate ion is -2, so the sum of oxidation numbers of S and O must be -2.
5. Solve the system of equations to find the oxidation numbers of each sulfur atom: Use the known oxidation state of oxygen (-2) and the total charge (-2) to find the individual oxidation states of sulfur, which will be +2 and +6.