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As per your syllabus, the AVERAGE oxidation number of S in S₄O₆²⁻ (tetrathionate) is which?
A+2.0
B+2.5
C+0.0
D+5.0
Answer & Solution
Correct answer: B. +2.5
syllabus box: 4S + 6(−2) = −2 ⇒ 4S = +10 ⇒ S = +2.5 (avg).
Structurally two terminal S are +5, two middle S are 0.
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