Question

The magnitude of difference in oxidation number of ${}^{\prime }{O}^{\prime }$ in $O{F}_2$ and ​ $O{}_2{F}_2$ respectively is

Answer 1

$O{F}_2:$​
Let $y$ be the oxidation state of "$O$" in $O{F}_2$​.
Since the overall charge of a neutral molecule is zero. The sum of oxidation states of all elements in it should be equal to 0.
Therefore, $y+2(-1)=0y=+2$

Hence, the oxidation state of "$O$ "in $O{F}_2$​ is $+2.$

$O_2{F}_2$ :
Since fluorine is more electronegative than oxygen so, oxidation state of fluorine is $-1$ in all its compounds. Let oxidation number of $O$ atom in $O_2{F}_2$ is $x.$
$2\times x+1\times (-1)=0x=+1$

$\stackrel{-1}{F-}\stackrel{+1}{O-}\stackrel{+1}{O-}\stackrel{-1}{F}$
So, oxidation number of oxygen will be +1 on $O_2{F}_2$
Difference : $(+2)-(+1)=1$