Oxidation-Reduction Reactions

AP Chemistry· difficulty 3/5

The oxidation state of S in SO₄²⁻ is

A
$+4$
B
$-2$
C
$+2$
D
$+6$
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Explanation

O is -2 each (4 \times -2 = -8). Charge of ion = -2. So S = -2 + 8 = +6.

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