Potassium (K) and Rubidium (Rb) forms
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Superoxides
Alkali metals burn in excess air or oxygen, and the product depends on the size of the metal ion. A larger cation stabilises a larger anion, so the type of oxide changes down the group:
Lithium forms mainly the monoxide, \( \ce{Li2O} \) (O2–).
Sodium forms mainly the peroxide, \( \ce{Na2O2} \) (O22–).
Potassium, rubidium and caesium form superoxides, MO2 (O2–).
\( \ce{K + O2 -> KO2} \)
\( \ce{Rb + O2 -> RbO2} \)
The superoxide ion O2– has one unpaired electron, so these superoxides are paramagnetic and coloured.
Why the other options are incorrect
Option (1) Suboxides — Incorrect. Metal-rich oxides such as \( \ce{Rb6O} \) form only when the oxygen supply is very limited; they are not the product of burning K or Rb in air.
Option (2) Peroxides — Incorrect. The peroxide is the main product for sodium, whose smaller ion cannot stabilise the O2– ion as well.
Option (4) Dioxides — Incorrect. This name is used for covalent oxides with two oxide oxygens, such as \( \ce{CO2} \) or \( \ce{SO2} \), not for the ionic oxides of alkali metals.