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

Apr 30, 2017

$\text{Sulfur = (+4), Oxygen = (-2)}$

#### Explanation:

To assign $\text{oxidation numbers}$, I would advise you to take a good look at this really nice website, Rules for Assigning Oxidation Numbers.

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

$\textcolor{b l u e}{\text{Step 1: Set ion equal to charge:}}$
$S {O}_{3} = \left(- 2\right)$

$\textcolor{b l u e}{\text{Step 2: Solve}}$

$\text{Sulfur"color(white)(aa)"Oxygen}$
$\downarrow \textcolor{w h i t e}{a a a a a} \downarrow$
$x + \left(- 2\right) \left(3\right) = - 2$
$x + \left(- 6\right) = - 2$
$x = - 2 + 6$
$x = + 4$

$\text{Answer: Sulfur = (+4), Oxygen = (-2)}$