alleryIn KO₂ (potassium superoxide), the oxidation state of oxygen is:
A0.00
B−0.5
C−1.0
D−2.0
Answer & Solution
Correct answer: B. −0.5
1. K is +1. The compound is neutral. Let each O be x.
2. (+1) + 2 x = 0 ⇒ x = −1/2.
3. Each O in the superoxide ion O₂⁻ carries a fractional oxidation state of −1/2.
4. Other unusual O states: peroxide −1 (H₂O₂), normal oxide −2, OF₂ +2.
_Source: syllabus Chem Ch 8 §8.3 Examples of Fractional Oxidation Numbers_
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