A) \[{{H}_{2}}\]
B) \[{{N}_{2}}\]
C) \[{{O}_{2}}\]
D) \[C{{O}_{2}}\]
Correct Answer: C
Solution :
\[2N{{a}_{2}}{{O}_{2}}+2{{H}_{2}}O\to 4NaOH+{{O}_{2}}\]. In this reaction oxygen \[2NaCl+2{{H}_{2}}O\,\xrightarrow{\text{Electrolysis}}\] is formed.
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