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 \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\text{,}\]when treated with dilute acids. Among the given compounds, \[\text{N}{{\text{a}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]is a true peroxide, so gives \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\] with dilute acid. \[\underset{\begin{smallmatrix} sodium \\ peroxide \end{smallmatrix}}{\mathop{N{{a}_{2}}{{O}_{2}}}}\,+2HCl\xrightarrow{{}}2NaCl+\underset{\begin{smallmatrix} hydrogen \\ peroxide \end{smallmatrix}}{\mathop{{{H}_{2}}{{O}_{2}}}}\,\]
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