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The S-Block Elements

Question
CBSEENCH11007087

Lithium forms normal oxide, sodium forms peroxides while K, Rb and Cs form superoxides. Explain. 

Solution

Lithium forms normal oxide left parenthesis 4 Li space plus space straight O subscript 2 space rightwards arrow space space 2 Li subscript 2 straight O right parenthesis. Lithium-ion with small size has a strong positive field around it. On combination with oxide anion, the positive field of lithium ion restricts the spread of negative charge towards another oxygen atom and thus prevents the formation of a higher oxide. 
Sodium reacts with dioxygen to form sodium peroxide left parenthesis 2 Na space plus space straight O subscript 2 space rightwards arrow space space Na subscript 2 straight O subscript 2 right parenthesis. Sodium ion with a larger size than lithium ion has weaker positive field than lithium ion. This positive field is so weak that it cannot prevent the conversion of the oxide anion left parenthesis straight O to the power of 2 minus end exponent right parenthesis into a peroxide ion left parenthesis straight O subscript 2 superscript 2 minus end superscript right parenthesis., However, it is strong enough to prevent further oxidation of peroxide to superoxide. 
Potassium, rubidium and caesium react with dioxygen to form superoxide left parenthesis straight K space plus straight O subscript 2 space rightwards arrow space space KO subscript 2 right parenthesis.
Potassium, rubidium and caesium ions are large sized and thus have a very weak positive field around them. The positive field around these ions is so weak that it cannot prevent the conversion of peroxide left parenthesis straight O subscript 2 superscript 2 plus end superscript right parenthesis anion to superoxide anion left parenthesis straight O subscript 2 superscript minus right parenthesis.