A) Chlorine
B) \[Ba{{O}_{2}}\]
C) \[{{H}_{2}}{{O}_{2}}\]
D) \[Mn{{O}_{2}}\]
Correct Answer: C
Solution :
\[PbO+\underset{(black)}{\mathop{{{H}_{2}}S}}\,\to PbS+{{H}_{2}}O\] \[PbS+4{{H}_{2}}{{O}_{2}}\to \underset{(white)}{\mathop{PbS{{O}_{4}}}}\,+4{{H}_{2}}O\] When blackened statues are treated with \[{{H}_{2}}{{O}_{2}},\] the \[PbS\]is oxidised to \[PbS{{O}_{4}}\] which is colourless (white).
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