Identification of Metal Peroxide or Superoxide Liberating Oxygen on Hydrolysis

To determine which compound liberates dioxygen gas ($\text{O}_2$) upon hydrolysis, let us analyze the reaction of each given species with water ($\text{H}_2\text{O}$):

1. Potassium Superoxide ($\text{KO}_2$): $\text{KO}_2$ contains the superoxide ion ($\text{O}_2^-$). Superoxides are strong oxidizing agents and undergo hydrolysis with water to yield potassium hydroxide, hydrogen peroxide, and dioxygen gas: $2\text{KO}_2(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{KOH}(aq) + \text{H}_2\text{O}_2(aq) + \text{O}_2(g) \uparrow$ Hence, $\text{KO}_2$ liberates $\text{O}_2$ gas.

2. Sodium Peroxide ($\text{Na}_2\text{O}_2$) and Lithium Peroxide ($\text{Li}_2\text{O}_2$): These compounds contain the peroxide ion ($\text{O}_2^{2-}$). Hydrolysis of alkali metal peroxides produces metal hydroxide and hydrogen peroxide ($\text{H}_2\text{O}_2$) without liberating oxygen gas under normal conditions: $\text{Na}_2\text{O}_2(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2\text{O}_2(aq)$ $\text{Li}_2\text{O}_2(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{LiOH}(aq) + \text{H}_2\text{O}_2(aq)$

3. Red Lead ($\text{Pb}_3\text{O}_4$): $\text{Pb}_3\text{O}_4$ is a mixed oxide ($2\text{PbO} \cdot \text{PbO}_2$) and is insoluble in water. It does not undergo hydrolysis to liberate oxygen gas.

Therefore, only $\text{KO}_2$ liberates $\text{O}_2$ gas upon hydrolysis.

Correct Option: (B)