In which of the following, the oxidation number of oxygen has been arranged in increasing order?

BaO₂ < KO₂ < O₃ < OF₂

OF₂ < KO₂ < BaO₂ <O₃

BaO₂ < O₃ < OF₂ < KO₂

KO₂ < OF₂ < O₃ < BaO₂

Solution

BaO₂ < KO₂ < O₃ < OF₂

Oxidation of oxygen are as follows,

$\underset{(- 1)}{{BaO}_{2}}, \underset{\frac{- 1}{2}}{{KO}_{2}}, \underset{zero}{O_{3}}, \underset{+ 2}{{OF}_{2}}$

Thus, the correct order is

BaO₂ <KO₂ < O₃ < OF₂

Some More Questions From The p-Block Elements Chapter

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