What is the oxidation number of S in $$ \mathrm{Na}_{2} \mathrm{SO}_{3} $$ ? You must include the + or - charge with the number.

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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.
The oxidation number of sulfur (S) in $$ \mathrm{Na}_{2}\mathrm{SO}_{3} $$ is **+4**.

What is the oxidation number of S in
? You must include the + or -
charge with the number.

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What is the oxidation number of S in ? You must include the + or - charge with the number.