A) KO2
B) BaO2
C) MnO2
D) NO2
Correct Answer: B
Solution :
| [b] Key Idea In peroxides, the oxidation state of O is 1 and they give \[{{H}_{2}}{{O}_{2}},\]with dilute acids, and have peroxide linkage. |
| In \[K{{O}_{2}}\], \[+1+(x\times 2)=0\] |
| \[x=-\frac{1}{2}\](thus, it is a superoxide, not a peroxide.) |
| In \[Ba{{O}_{2}},\] \[+2+(x\times 2)=0\] |
| \[x=-1\] |
| Thus, it is a peroxide. Only it gives \[{{H}_{2}}{{O}_{2}}\] when reacts with dilute acids and has peroxide linkage as\[\underset{peroxide\,linkage}{\mathop{B{{a}^{2+}}{{[O-O]}^{2-}}}}\,\] |
| In \[\text{Mn}{{\text{O}}_{\text{2}}}\] and \[\text{N}{{\text{O}}_{\text{2}}},\]Mn and N exhibit variable oxidation states, thus, the oxidation state of O in these is - 2. Hence, these are not peroxides. |
| Thus, it is clear, that among the given molecules only \[\text{Ba}{{\text{O}}_{\text{2}}}\] is a peroxide. |
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