Discussion
The oxidation number of oxygen in O2, O2F2 and RbO2 are respectively
Answer: C
The oxidation numbers are 0, +1, -1/2.
In \(\mathrm{O_2}\) oxygen is an element, so 0. In \(\mathrm{O_2F_2}\) fluorine, the most electronegative element, is always -1, so the two oxygens share +2: each is +1. In \(\mathrm{RbO_2}\), a superoxide, Rb is +1, so the two oxygens share -1: each is -1/2.
\[ \begin{aligned} \mathrm{O_2F_2}: 2x + 2(-1) &= 0 \Rightarrow x = +1 \\ \mathrm{RbO_2}: +1 + 2x &= 0 \Rightarrow x = -\tfrac{1}{2} \end{aligned} \]
The option with +2 uses \(\mathrm{OF_2}\) instead of \(\mathrm{O_2F_2}\); -1 is the value in peroxides, not superoxides.
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