A) \[Pb{{O}_{2}}\]
B) \[Ti{{O}_{2}}\]
C) \[Mn{{O}_{2}}\]
D) \[N{{a}_{2}}{{O}_{2}}\]
Correct Answer: D
Solution :
Only true peroxides gives \[{{H}_{2}}{{O}_{2}}\] when treated with dilute acids. Among the given compounds, \[N{{a}_{2}}{{O}_{2}}\] is a true peroxide, so gives \[{{H}_{2}}{{O}_{2}}\] with dilute acid. \[\underset{\begin{smallmatrix} sodium \\ peroxide \end{smallmatrix}}{\mathop{N{{a}_{2}}{{O}_{2}}}}\,+2HCl\xrightarrow{{}}2\,NaCl+\underset{\begin{smallmatrix} hydrogen \\ peroxide \end{smallmatrix}}{\mathop{{{H}_{2}}{{O}_{2}}}}\,\]
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