# How does SO_4 have a charge of 2-?

Jul 3, 2016

We are agreed that sulfate anion derives from ${H}_{2} S {O}_{4}$, sulfuric acid, or from sodium sulfate, $N {a}_{2} S {O}_{4}$, which are both manifestly neutral entities.

#### Explanation:

A typical (if outdated) Lewis structure of sulfuric acid is:

${\left(H O -\right)}_{2} S {\left(= O\right)}_{2}$

This Lewis structure is equivalent to:

${\left(H O -\right)}_{2} {S}^{2 +} {\left(- {O}^{-}\right)}_{2}$

For a neutral chalcogen atom ($\text{chalcogen = S or O}$), there must be 6 valence electrons. In the representation ${\left(H O -\right)}_{2} S {\left(= O\right)}_{2}$ there are certainly 6 electrons associated with each sulfur or oxygen. Lone pairs are owned by the atom, and thus on neutral oxygen there are 2 electrons from the double bond, and 4 electrons in the lone pairs).

Now of course both ${H}_{2} S {O}_{4}$ and $H S {O}_{4}^{-}$ are strong acids, and undergoes almost complete ionization in water:

${H}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right) \rightarrow S {O}_{4}^{2 -} + 2 {H}_{3} {O}^{+}$

Conservation of charge demands that the sulfate ion has 2 formal negative charges.

Nitric acid has an even more problematic representation: $\left(O =\right) {N}^{+} \left(- {O}^{-}\right) \left(- O H\right)$, where there is formal charge separation in even the neutral acid (6 electrons around nitrogen rather than 7; 9 electrons around oxygen rather than 8 ).

See here for another example that assigns formal charge.