# What is the lewis structure for SO_4 with 2- charge?

Jun 11, 2016

This is the tetrahedral sulfate ion, $S {O}_{4}^{2 -}$

#### Explanation:

Number of valence electrons

$=$ $6 \left(S\right) + 4 \times 6 \left(O\right) + 2 {e}^{-} = 32 \text{ electrons}$, to be distributed across 5 centres, with a central $S$ atom.

$S {\left(= O\right)}_{2} {\left(- {O}^{-}\right)}_{2}$, i.e. 16 electron pairs distributed around the 5 centres.

In the actual molecule, all the oxygen atoms are equivalent.

The sulfate ion derives from sulfuric acid, ${H}_{2} S {O}_{4}$, i.e.
$S {\left(= O\right)}_{2} {\left(- O H\right)}_{2}$. The oxidation state of sulfur here is $V I +$.