A) \[N{{a}_{2}}S{{O}_{3}}\]
B) \[KI\]
C) \[PbS\]
D) \[{{O}_{3}}\]
Correct Answer: D
Solution :
: \[{{H}_{2}}{{O}_{2}}\]oxidises to \[N{{a}_{2}}S{{O}_{3}}\], \[KI\] to 12, \[PbS\] 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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