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