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