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A metal on combustion in excess air forms X, X upon hydrolysis with water yields ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ and ${{\text{O}}_{\text{2}}}$ along with another product. The product is:

Last updated date: 13th Jun 2024
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Hint: Group I elements react with oxygen to form oxides, peroxides and superoxides. Group II elements form peroxides and simple oxides with oxygen from air.

Complete Step-by-step solution:
The given options consist of both alkali metals and alkaline earth metals. Let us get to know the reactions of the given elements upon reaction with excess of air. Reaction with excess of air is nothing but the reaction of any given element with the oxygen present in the air.
Group I elements i.e. alkali earth metals in the given question are Sodium (Na), Rubidium (Rb), lithium (Li). These elements are very reactive towards oxygen to form different types of oxides.
-Sodium upon reaction with excess air more likely forms sodium peroxide ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{O}}_{\text{2}}}$. When sodium peroxide is reacted with water it undergoes hydrolysis to form hydrogen peroxide and sodium hydroxide. ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{O}}_{\text{2}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}} \to {{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}{\text{ + 2NaOH}}$.
-Lithium upon reaction with excess air usually forms lithium oxide but lithium peroxide is also possible to form. But we know that peroxides upon hydrolysis give hydrogen peroxide and metal hydroxide. The reaction between lithium and oxygen is ${\text{4Li + }}{{\text{O}}_{\text{2}}} \to {\text{2L}}{{\text{i}}_{\text{2}}}{\text{O}}$. Rubidium upon reaction with excess air gives a superoxide. Superoxide ion upon hydrolysis forms hydrogen peroxide, metal hydroxide and oxygen. The reaction is ${\text{2Ru}}{{\text{O}}_{\text{2}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}} \to {{\text{H}}_2}{{\text{O}}_2}{\text{ + 2RuOH + }}{{\text{O}}_2}$.
Magnesium forms white powder of magnesium oxide (MgO) upon heating in excess air.

So the answer for the question is option (A) Rb- Rubidium.

Note: Superoxides are the oxides of metals with general formula ${\text{M}}{{\text{O}}_{\text{2}}}$. The superoxide ion is \[{{\text{O}}_2}^{\text{ - }}\] in which oxygen is with 0.5 oxidation number. They give peroxide and oxygen upon hydrolysis.