In the reaction, `M + CO_(2) to MO_(2)` (superoixde) the metal M is :

In the reaction, M+O_(2)to MO_(2) (superoxide), the metal M is

In the reaction M+O_(2) to MO_(2) (superoxide), the metal M is

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. Consider the following reaction : M+O_(2) to MO_(2) (M=alkali metal) (stable superoxide)

A metal carbonate decomposes according to the following reaction M_(2)CO_(3) (s) to M_(2)O (s) + CO_(2) (g) Percentage loss in mass on complete decomposition of M_(2)CO_(3) (s) (Atomic mass of M= 102 )

Formula of the carbonate of a metal M is M_(2)CO_(3) . Write the fomula of its chloride.

An active metal (M) gives its only oxide MO_(2) . Guess the metal.