# What are the oxidation numbers of each element in the salt MgSO_4?

Mar 27, 2017

A metal oxidation state of $+ I I$ is expressed.

#### Explanation:

The oxidation number is the charge left on the central atom when all the bonding pairs of electrons are removed with the charge assigned to the most electronegative atom.

This is a mouthful, but for ionic compounds, such as $M g S {O}_{4}$, which is composed of $M {g}^{2 +}$ and $S {O}_{4}^{2 -}$ ions, the oxidation number is generally the charge on the ion. And thus for $M {g}^{2 +}$, the oxidation number is $+ I I$.

For the anion, $S {O}_{4}^{2 -}$, the sum of the oxidation numbers equals the charge on the ion, here $- 2$. Oxygen generally assumes an oxidation number of $- I I$ in its compounds, and it does here, and thus ${S}_{\text{oxidation number}} + 4 \times \left(- 2\right) = - 2$, i.e. ${S}_{\text{oxidation number}} = + V I$.

What is the oxidation number of sulfur in $S {O}_{3}^{2 -}$; what about ${S}_{2} {O}_{3}^{2 -}$?

Mar 27, 2017

$+ 2$

#### Explanation:

The sum of oxidation numbers in a neutral compound must be equal to 0.

We know that $S {O}_{4}$has a charge of -2. i.e. $S {O}_{4}^{2 -}$

Knowing that, we can set up an equation.

Let the charge of $M g$ be $x$

$M g + S {O}_{4}^{-} 2 = 0$

$x + \left(- 2\right) = 0$

$x - 2 = 0$

$x = + 2$

Therefore, the oxidation state of $M g$ in $M g S {O}_{4}$ is $+ 2$