# How do you find the oxidation number of SO_4^(-2)?

Jun 22, 2016

Sulfate is a covalent ion. So it's more realistic to use formal charges, not oxidation states, to describe its atoms.

It's OK to give sulfate an overall oxidation state, but it doesn't make sense to do so for sulfur and oxygen in sulfate.

OXIDATION STATE APPROACH: IDEAL IONIC BONDS

Oxidation states are hypothetical charges for a compound that has ideal ionic bonds, or would form ideal ionic bonds, i.e. complete transfer of the electron pair to only one atom in the bond.

You wrote ${\text{SO}}_{4}^{2 -}$, which is the sulfate polyatomic ion. You wrote that its total charge was $- 2$.

Since sulfate (${\text{SO}}_{4}^{2 -}$) is already ionized, its charge is its oxidation state. So it's kind of pointless to find the "oxidation state" for ${\text{SO}}_{4}^{2 -}$ because it's already given.

The oxidation states for sulfur and oxygen atom are determined from:

• electronegativities
• total charge

${\text{SO}}_{4}^{2 -}$ has a total charge of $- 2$, and oxygen is more electronegative than sulfur. So, oxygen polarizes the electron density towards itself more, and thus has the more negative oxidation state here.

Since oxygen is on column 16, its expected oxidation state is $\textcolor{red}{- 2}$. Therefore:

$- 2 \times 4 + x = - 2$

$\implies$$x$ gives the oxidation state of sulfur, which is $\textcolor{red}{+ 6}$.

$\implies \textcolor{h i g h l i g h t}{\stackrel{\textcolor{red}{+ 6}}{{\text{S")stackrel(color(red)(-2))("O}}_{4}^{2 -}}}$

But keep in mind that this is unrealistic. If you ever do this for redox balancing, it's only an accounting scheme, and nothing more.

FORMAL CHARGE APPROACH: IDEAL COVALENT BONDS

${\text{SO}}_{4}^{2 -}$ is primarily a covalent ion. The electronegativity differences are only $1.0$. So, it's more practical to find the formal charges.

Formal charges are the hypothetical partial charges for if the compound has ideal covalent bonds, i.e. perfectly even sharing of electron pairs.

When we cleave each bond in half, we then assume that one electron goes to each atom, and we can assign formal charges that way.

Then:

• The "owned" electrons are the ones the atom has upon this cleavage
• The "valence" electrons are the ones that the atom typically has as a neutral atom.

So...

Each 6-electron oxygen has a formal charge of:

$\text{valence" - "owned} = 6 - 6 = 0$, of which there are 2 oxygens.

Each 7-electron oxygen has a formal charge of:

$\text{valence" - "owned} = 6 - 7 = \textcolor{red}{- 1}$, of which there are 2 oxygens.

The 6-electron sulfur has a formal charge of:

$\text{valence" - "owned} = 6 - 6 = 0$

That accounts for a total charge of $2 \times \left(- 1\right) = \textcolor{red}{- 2}$.