alleryThe oxidation state of sulphur in SO₄²⁻ is:
A+2
B+4
C+6
D−2
Answer & Solution
Correct answer: C. +6
1. Let O.S. of S = x. Each O is −2; total charge of ion = −2.
2. x + 4(−2) = −2 ⇒ x − 8 = −2 ⇒ x = +6.
3. So S in SO₄²⁻ is +6 (its highest common oxidation state).
_Source: syllabus Chem Ch 8 §8.3 Rules for Oxidation Number_
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