Question
What is the oxidation number of S in \( \mathrm{Na}_{2} \mathrm{SO}_{3} \) ? You must include the + or - charge with the number.
Ask by Vargas Olson. in the United States
Jan 27,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The oxidation number of sulfur (S) in \( \mathrm{Na}_{2}\mathrm{SO}_{3} \) is **+4**.
Solution
In the compound \( \mathrm{Na}_{2}\mathrm{SO}_{3} \) (sodium sulfite), you can determine the oxidation number of sulfur (S) by following these steps:
1. **Assign known oxidation states:**
- Sodium (Na) typically has an oxidation state of **+1**.
- Oxygen (O) typically has an oxidation state of **-2**.
2. **Set up the equation:**
- The compound is neutral, so the sum of all oxidation states should be **0**.
- Let the oxidation state of sulfur (S) be **x**.
\[
2(\text{Na}) + \text{S} + 3(\text{O}) = 0
\]
\[
2(+1) + x + 3(-2) = 0
\]
3. **Solve for x:**
\[
2 + x - 6 = 0
\]
\[
x - 4 = 0
\]
\[
x = +4
\]
**Therefore, the oxidation number of sulfur in \( \mathrm{Na}_{2}\mathrm{SO}_{3} \) is \( +4 \).**
Answer: The oxidation number of S in Na₂SO₃ is +4.
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Simplify this solution Bonus Knowledge
To find the oxidation number of sulfur (S) in \( \mathrm{Na}_{2} \mathrm{SO}_{3} \), we can use the rules of oxidation states. Sodium (Na) has an oxidation state of +1. Since there are two sodium atoms, their contribution is +2. The overall charge of the compound is neutral, so we have: \[ +2 + (\text{oxidation number of S}) + 3 \times (-2) = 0 \] The oxidation state of oxygen (O) is -2. There are three oxygen atoms, giving us a total of -6 from oxygen. Now we can plug in this information: \[ +2 + x - 6 = 0 \] Solving for x (the oxidation state of sulfur): \[ x - 4 = 0 \] \[ x = +4 \] Thus, the oxidation number of sulfur in \( \mathrm{Na}_{2} \mathrm{SO}_{3} \) is +4.
