A) \[N{{a}_{2}}O\]
B) \[N{{a}_{2}}{{O}_{2}}\]
C) \[NaN{{O}_{3}}\]
D) \[NaN{{O}_{2}}\]
Correct Answer: D
Solution :
| [d] \[NaN{{O}_{2}},\] is the only compound among the given options which can acts as an oxidizing as well as reducing agent due to the reason that in\[NaN{{O}_{2}},\] N is in its intermediate oxidation state +3. |
| It oxidises K.I to \[{{I}_{2}}\] as follows: |
| \[2NaN{{O}_{2}}+2KI+4C{{H}_{3}}COOH\xrightarrow[{}]{{}}\] |
| \[2NO+{{I}_{2}}+2C{{H}_{3}}COOK+2C{{H}_{3}}COONa+2{{H}_{2}}O\]Similarly, it reduces \[{{H}_{2}}{{O}_{2}}\] to \[{{H}_{2}}O\] |
| \[NaN{{O}_{2}}+{{H}_{2}}{{O}_{2}}\xrightarrow[{}]{{}}NaN{{O}_{3}}+{{H}_{2}}O\] |
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