To determine the oxidation state of sulfur (S) in the thiosulfate ion \( \text{S}_2\text{O}_3^{2-} \), we can follow these steps:
1. Identify the typical oxidation state of oxygen:
- Oxygen generally has an oxidation state of -2.
2. Let the oxidation state of each sulfur atom be \( x \).
3. Set up an equation involving the known oxidation states and the total charge of the ion:
- The thiosulfate ion has 2 sulfur atoms and 3 oxygen atoms.
- The overall charge of the thiosulfate ion is -2.
4. Write the equation using the sum of oxidation states:
[tex]\[
2 \cdot \text{(oxidation state of S)} + 3 \cdot \text{(oxidation state of O)} = \text{total charge}
\][/tex]
5. Substitute the known values:
[tex]\[
2x + 3(-2) = -2
\][/tex]
6. Simplify the equation:
[tex]\[
2x - 6 = -2
\][/tex]
7. Solve for \( x \):
\begin{align}
2x - 6 &= -2 \\
2x &= -2 + 6 \\
2x &= 4 \\
x &= \frac{4}{2} \\
x &= 2
\end{align}
Thus, the oxidation state of sulfur in \( \text{S}_2\text{O}_3^{2-} \) is \( +2 \).
Therefore, the correct answer is:
D. [tex]\( +2 \)[/tex]
