A) \[N{{a}_{2}}S{{O}_{3}}\]
B) KI
C) PbS
D) \[{{O}_{3}}\]
Correct Answer: D
Solution :
[d]: \[{{H}_{2}}{{O}_{2}}\] oxidises \[N{{a}_{2}}S{{O}_{3}}\]to \[N{{a}_{2}}S{{O}_{4}},KI\] to \[{{I}_{2}},PbS\] to \[PbS{{O}_{4}}\] and reduces \[{{O}_{3}}\]to \[{{O}_{2}}\]. \[{{H}_{2}}{{O}_{2}}+{{O}_{3}}\xrightarrow[{}]{{}}{{H}_{2}}O+2{{O}_{2}}\]
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