Question

The CORRECT order of stability of the given metal oxides is

• A. LiO\(_2\) > NaO\(_2\) > KO\(_2\) > RbO\(_2\)
• B. LiO\(_2\) < NaO\(_2\) < KO\(_2\) < RbO\(_2\)
• C. LiO\(_2\)<NaO\(_2\) > KO\(_2\) > RbO\(_2\)
• D. LiO\(_2\) > NaO\(_2\) < KO\(_2\) < RbO\(_2\)

Hint:

Stability of alkali metal oxides decreases as the size of the metal ion increases. The smaller the cation, the more stable the oxide.

Answer 1

The Correct Option is A

The stability of alkali metal oxides generally decreases as the size of the metal ion increases. This is because the charge density of the metal ion decreases as the ion gets larger, making the oxide less stable. The oxide of lithium (LiO\(_2\)) is the most stable, followed by sodium (NaO\(_2\)), and the stability decreases further with potassium (KO\(_2\)) and rubidium (RbO\(_2\)).
- LiO\(_2\) has the highest stability due to the small size of Li\(^+\).
- NaO\(_2\) is next in stability as Na\(^+\) is larger than Li\(^+\).
- KO\(_2\) and RbO\(_2\) are progressively less stable due to their larger cations.
Thus, the correct order of stability is: \[ \boxed{\text{LiO}_2>\text{NaO}_2>\text{KO}_2>\text{RbO}_2} \]