A) \[N{{a}_{2}}O\]
B) \[SnC{{l}_{2}}\]
C) \[N{{a}_{2}}{{O}_{2}}\]
D) \[NaN{{O}_{2}}\]
Correct Answer: D
Solution :
\[NaN{{O}_{2}}\] (sodium nitrite) act both as oxidising as well as reducing agent because in it N atom is in +3 oxidation state (intermediate oxidation state) Oxidising property \[2NaN{{O}_{2}}+2KI+2{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}N{{a}_{2}}S{{O}_{2}}\] \[+{{K}_{2}}S{{O}_{4}}+2NO+2{{H}_{2}}O+{{I}_{2}}\] Reducing property \[{{H}_{2}}{{O}_{2}}+NaN{{O}_{2}}\xrightarrow{{}}NaN{{O}_{3}}+{{H}_{2}}O\]
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