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  4. Oxidation state of oxygen in compounds OF 2 and Na 2 O respectively are

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Q. Oxidation state of oxygen in compounds $OF _{2}$ and $Na _{2} O$ respectively are

Classification of Elements and Periodicity in Properties

Solution:

$OF _{2}$, here EN of fluorine is more than oxygen.
So, oxidation state of fluorine $=-1$
and oxidation state of oxygen $=+2$
In $Na _{2} O$, oxygen has more electronegative value than sodium,
So, oxidation state of oxygen $=-2$
and oxidation state of $Na =+1$