Q.Potassium burns in excess oxygen to form a superoxide rather than a normal oxide or a peroxide. Write the balanced equation and explain why the superoxide is the stable product.

Potassium, being a still larger cation than sodium, burns in excess dry oxygen to give the superoxide rather than the normal oxide or the peroxide:

The choice among normal oxide ($\text{O}^{2-}$), peroxide ($\text{O}_2^{2-}$) and superoxide ($\text{O}_2^{-}$) as the stable product for a given alkali metal is decided by a balance of lattice energies: a small cation gives the highest lattice energy (most stable product) when paired with the small oxide ion, while a large cation gives its highest lattice energy when paired with a correspondingly large anion. As the cation grows from $\text{Li}^+$ through $\text{Na}^+$ to $\text{K}^+$ (and further to $\text{Rb}^+$, $\text{Cs}^+$), the anion that best matches it in size grows too -- from oxide, to peroxide, to the still larger superoxide ion. For potassium (and likewise rubidium and caesium), the superoxide, $\text{O}_2^{-}$, is the best size match, so $\text{KO}_2$ is the thermodynamically stable product of burning potassium in excess oxygen. …

Apply the same cation-anion size-matching (lattice energy) rule used for Li and Na, extended to the largest common …

Writing K2O2 (the peroxide) instead of KO2 (the actual major product for K, Rb, Cs); forgetting to balance K:O as 1:2 in KO2 itself (only one K per formula unit …