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

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Answer 1 · 2017-03-04T04:19:25

S = +6 O = -2

The sulfate ion $SO_4^-2$ has a charge of -2

so $S + 4xx O = -2$

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

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

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