JEE Advance - Chemistry (2020 - Paper 1 Offline - No. 2)

Which of the following liberates O2 upon hydrolysis?
Pb3O4
KO2
Na2O2
Li2O2

Explanation

(a) $$P{b_3}{O_4} + {H_2}O\buildrel {} \over \longrightarrow $$ No reaction Pb3O4 is insoluble in water or do not react with water.

(b) $$K{O_2} + 2{H_2}O\buildrel {} \over \longrightarrow KOH + {H_2}{O_2} + 1/2{O_2}$$

Potassium superoxide is a strong oxidant, able to convert oxides into peroxides or molecular oxygen. Hydrolysis gives oxygen gas, hydrogen peroxide and potassium hydroxide.

(c) $$N{a_2}{O_2} + 2{H_2}O\buildrel {} \over \longrightarrow 2NaOH + {H_2}{O_2}$$

When sodium peroxide dissolves in water, it is hydrolysed and forms sodium hydroxide and hydrogen peroxide. The reaction is highly exothermic.

(d) $$L{i_2}{O_2} + 2{H_2}O\buildrel {} \over \longrightarrow 2LiOH + {H_2}{O_2}$$

The reactivity of Li2O2 toward water differs from LiO2, in Li2O2 results in H2O2 as a product.

Hence, the correct option is (b).

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