# What is the oxidation state of each element in SO_4^(2-)?

Mar 4, 2017

S = +6 O = -2

#### Explanation:

The sulfate ion $S {O}_{4}^{-} 2$ has a charge of -2

so $S + 4 \times O = - 2$

Oxygen is almost always -2 in charge in order to achieve the stable electron configuration of Neon substituting -2 for O gives

$S + 4 \times \left(- 2\right) = - 2$ solving for S by anding +8 to both sides gives
$S + \left(- 8\right) + \left(+ 8\right) = - 2 + \left(+ 8\right)$ adding the integers results in
# S = +6