$(1992,1 \mathrm{M})$
(a) $\mathrm{PbO}_2$
(b) $\mathrm{H}_2 \mathrm{O}_2$
(c) $\mathrm{SrO}_2$
(d) $\mathrm{BaO}_2$
Correct Answer: 30. (a)
Solution: In $\mathrm{PbO}_2, \mathrm{~Pb}$ is in +4 oxidation state and oxygen is in -2 oxidation state. In all other case, peroxide ion $\left(\mathrm{O}_2^{2-}\right)$ is present.
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