Answer:
Let \[x\] be the oxidation state of Na
in\[N{{a}_{2}}{{O}_{2}}\]. \[O_{2}^{2-}\]is a peroxo linkage in which O has
oxidation state (-1).
\[\therefore \] \[2x+2\times
(-1)=0\]
or \[x=+1\]
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