Lithium only forms monoxide when heated in oxygen. Sodium forms monoxide and peroxide in excess of oxygen. Other alkali metals form superoxide metals superoxide with oxygen i.e., `MO_(2)`. The abnormal behaviour of lithium is due to small size. The larger size of higher metals also decides the role in formation of superoxide. All the three anions abstract proton from water. The three anions are raleted to each other as follows:
`underset("oxide ion")(O_(2)^(2-)overset((1)/(2)O_(2))(to)underset("peroxide ion")(O_(2)^(2-))overset(O_(2))(to)underset("superoxide ion")(2O_(2)^(-))`
Q. Which compounds will liberate oxygen when reacts with ice cold water?

Options

A. `Na_(2)O_(2)`
B. `KO_(2)`
C. `Na_(2)O`
D. `Cs_(2)O_(2)`

Answer

B

Explanation

All superoxide liberate oxygen with water:
`2O_(2)^(-)+2H_(2)O(l) to 2OH^(-)(aq)+H_(2)O(l)+O_(2)(g)`