Question

What are the oxidation numbers in the ion `SO_3^(2-)`?

Answer 1

`"Sulfur = (+4), Oxygen = (-2)"`

Explanation:

To assign `"oxidation numbers"`, I would advise you to take a good look at this really nice website, Rules for Assigning Oxidation Numbers.

For this particular problem, the `"sulfite"` anion is a `"polyatomic ion"` which bears a `(-2)` charge. `"Oxygen"` here will have a `(-2)` for oxidation number and since there are `3`, we multiply the oxidation number by it. We let `"x"` represent `"sulfur"` since no rules assign an oxidation number. We let the whole equation equal to the charge of the polyatomic ion.

`color(blue)"Step 1: Set ion equal to charge:"`
`SO_3 = (-2)`

`color(blue)"Step 2: Solve"`

`"Sulfur"color(white)(aa)"Oxygen"`
`darrcolor(white)(aaaaa)darr`
`x+(-2)(3) = -2`
`x +(-6) = -2`
`x = -2+6`
`x = +4`

`"Answer: Sulfur = (+4), Oxygen = (-2)"`