Which of the following liberates $$O_2$$ upon hydrolysis?
Correct option is B. $$KO_2$$
(A) $$Pb_3O_4+H_2O\rightarrow $$ No reaction
(B) $$2KO_2+2H_2O\rightarrow 2KOH +H_2O_2+O_2$$
(C) $$Na_2O_2+2H_2O\rightarrow 2NaOH + H_2O_2$$
(D) $$Li_2O_2+2H_2O\rightarrow 2LiOH + H_2O_2$$
So, option B is the correct answer.