Oxidation number of potassium in K2O, K2O2 and KO2, respectively, is:

Key rule: Potassium is an alkali metal and always has oxidation state +1 in its compounds.

Step 1 — $\mathrm{K_2O}$ (potassium oxide): $2(+1) + x = 0 \Rightarrow x_{\mathrm{O}} = -2$ K = +1 ✓ (O is $-2$, normal oxide)

Step 2 — $\mathrm{K_2O_2}$ (potassium peroxide): $2(+1) + 2x = 0 \Rightarrow x_{\mathrm{O}} = -1$ K = +1 ✓ (O is $-1$, peroxide)

Step 3 — $\mathrm{KO_2}$ (potassium superoxide): $(+1) + 2x = 0 \Rightarrow x_{\mathrm{O}} = -\frac{1}{2}$ K = +1 ✓ (O is $-\frac{1}{2}$, superoxide)

Note: The oxidation state of O changes (+2, -1, -½) but K remains +1 throughout.

Answer: Option (2) — +1, +1 and +1