A) \[N{{a}_{2}}S{{O}_{3}}\]
B) \[PbS\]
C) KI
D) \[{{O}_{3}}\]
Correct Answer: D
Solution :
\[N{{a}_{2}}S{{O}_{3}}\] is oxidised by \[{{H}_{2}}{{O}_{2}}\] to \[N{{a}_{2}}S{{O}_{4}}\] PbS is oxidised by \[{{H}_{2}}{{O}_{2}}\] to \[PbS{{O}_{4}}\] KI is oxidised by \[{{H}_{2}}{{O}_{2}}\] to \[{{I}_{2}}\] \[{{O}_{3}}\] cannot be oxidised by \[{{H}_{2}}{{O}_{2}}\] but it is reduced to \[{{O}_{2}}\] by \[{{H}_{2}}{{O}_{2}}\] \[{{H}_{2}}{{O}_{2}}+{{O}_{3}}\xrightarrow{{}}{{H}_{2}}O+2{{O}_{2}}\]
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