Question

Potassium when heated strongly in air, it forms:
A.${{K}_{2}}O$
B.$K{{O}_{2}}$
C.${{K}_{2}}{{O}_{2}}$
D.$K{{O}_{3}}$

Answer 1

Hint: The concept of formation of oxides, peroxide and superoxide when alkali metals are burnt in air is to be used in this question. The formation of oxides is dependent on whether the metal is heated firmly in limited supply of air or heated strongly in excess oxygen.

Complete answer:
In order to answer this question, we need to learn about the properties of alkali metals. Alkali metals are silvery white in colour and are generally soft and light metals. Because of the large size of the atoms, softness of alkali metals is due to weak metallic bonding in them. Hence, softness increases as we move down the group and metallic bonding weakens . Potassium is softer than sodium. The densities of alkali metals are low and increase down the group. Potassium, however, is lighter than sodium. Lithium can be said as the lightest metal having a density of $0.534g\,c{{m}^{-3}}$. Because it floats on the surface, it cannot be stored in kerosene oil, rather, it is kept wrapped in paraffin wax. The fact that alkali metals have low density is associated with their large atomic size and weak metallic bond. However, going down the group, atomic size as well as atomic mass increases but the corresponding increase in atomic mass is not neutralised by the increase in atomic volume : As a result, the ratio, mass/volume, i.e., density gradually increases. Alkali metals have low melting and boiling point. The low melting point is the consequence to their larger atomic size, which results in the binding energies of their atoms in the crystal lattice are low. As we move down the group, atomic size increases and the strength of metallic bonds decreases so the melting points decrease. The boiling points almost follow the same order for the above reasons.
Alkali metals when combusted in air form different kinds of oxides, for example, the alkali metals on reaction with restricted quantities of oxygen the common or normal oxides of the formula,${{M}_{2}}O$. However, when heated with excess of air lithium forms normal oxide, ($L{{i}_{2}}O$) and some peroxide $L{{i}_{2}}{{O}_{2}}$ sodium forms peroxide, $N{{a}_{2}}{{O}_{2}}$ and some superoxide $Na{{O}_{2}}$, whereas potassium, rubidium and caesium form superoxide having general formula. The reaction between potassium and oxygen of air can be represented as:
\[K+{{O}_{2}}+\Delta \to K{{O}_{2}}\]
Superoxide is formed. So our correct option is option B.

NOTE: The temperature required for the reaction to take place decreases on descending the group. As the size of the alkali metal increases, the stability of peroxide or superoxide increases. This is due to stabilization of large anions by large cations through lattice energy effects.