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

1 Answer

Mar 4, 2017

S = +6 O = -2

Explanation:

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

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