The oxidation state of each oxygen atom in $KO_{2}$ is:

Question

Accepted Answer

The oxidation state of each oxygen atom in $KO_{2}$ is indeed -1/2, as indicated by Option 3. Let's delve into the reasoning behind this:

1. **Explanation for Option 3 (-1/2):**
   - Potassium superoxide ($KO_{2}$) is a compound where potassium (K) is combined with the superoxide ion ($O_{2}^{-}$). In this compound, potassium typically has an oxidation state of +1, as it is an alkali metal and tends to lose one electron.
   - The superoxide ion ($O_{2}^{-}$) consists of two oxygen atoms sharing a total charge of -1. To find the oxidation state of each oxygen atom in the superoxide ion, we divide the total charge by the number of oxygen atoms: 
     $$
     \text{Oxidation state of each O} = \frac{-1}{2} = -\frac{1}{2}
     $$
   - Therefore, each oxygen atom in $KO_{2}$ has an oxidation state of -1/2, which confirms that Option 3 is correct.

2. **Explanation for why other options are incorrect:**
   - **Option 1 (-2):** This oxidation state is typical for oxygen in most compounds, such as oxides (e.g., $H_2O$, $CO_2$). However, in the superoxide ion, the oxygen atoms do not achieve the -2 oxidation state because the total charge of the ion is only -1, not -2.
   - **Option 2 (-1):** This oxidation state is characteristic of the peroxide ion ($O_{2}^{2-}$), where each oxygen atom has an oxidation state of -1. However, in the superoxide ion ($O_{2}^{-}$), the total charge is -1, not -2, so each oxygen atom cannot have an oxidation state of -1.
   - **Option 4 (0):** An oxidation state of 0 would imply that the oxygen atoms are in their elemental form, such as in $O_2$ gas. In $KO_{2}$, the oxygen atoms are part of the superoxide ion, which carries a charge, so their oxidation state cannot be 0.

In conclusion, the correct oxidation state of each oxygen atom in $KO_{2}$ is -1/2, as supported by the structure and charge of the superoxide ion. This validates Option 3 as the correct answer.