Identify the compound

The compound is $KO_2$.

Potassium is an alkali metal, so it almost always has oxidation state: $K = +1$

Find the charge on the oxygen species

Since the compound is neutral: $+1 + (\text{charge on } O_2) = 0$ $\Rightarrow \text{charge on } O_2 = -1$

So the oxygen species present is: $O_2^-$

Identify the oxygen species

Common oxygen species are:

• Oxide: $O^{2-}$
• Peroxide: $O_2^{2-}$
• Superoxide: $O_2^-$

Since in $KO_2$ the anion is $O_2^-$, it is a superoxide.

Find oxidation state of each oxygen atom

Let the oxidation state of each oxygen atom be $x$.

In $O_2^-$: $2x = -1$ $x = -\frac{1}{2}$

So each oxygen atom has oxidation state: $-\frac{1}{2}$

Check options

• A: Oxide and $-2$ $\rightarrow$ Incorrect
• B: Superoxide and $-1/2$ $\rightarrow$ Correct
• C: Peroxide and $-1/2$ $\rightarrow$ Incorrect
• D: Superoxide and $-1$ $\rightarrow$ Incorrect

Final answer

The correct option is B: Superoxide and $-\frac{1}{2}$.