Question

Which of the following statement is incorrect regarding $O_2{F}_2$?

• A. $O-F$ bond length in $O_2{F}_2$ is longer than $O-F$ bond length in $O{F}_2$
• B. The oxidation state of oxygen in $O_2{F}_2$ is $+1$
• C. The $O-O$ bond length in $O_2{F}_2$ is shorter than $O-O$ bond length in $H_2{O}_2$
• D. None of these

Answer 1

The correct option is D None of these

$\left(a\right)$ In the case of $O{F}_2,$ a single oxygen atom is surrounded by two fluorine atoms. Hence the shared pairs of electrons from both sides of bond will be attracted more towards fluorine, creating a partial positive charge on oxygen from both sides. In the case of $O_2{F}_2,$ again the electron pairs get more attracted towards $F$ and hence, the partial positive charge does come on both oxygen atoms, but it is distributed and hence lesser in comparison to $O{F}_2.$ Due to more attraction in $O{F}_2,$ the bond length of $O-F$ is less in $O{F}_2$ than of that in $O_2{F}_2.$

$\left(b\right)$ Since fluorine is more electronegative than oxygen and the oxidation state of fluorine is $-1$ in all its compounds. So, the oxidation state of oxygen will be $+1$ as calculated.
$\stackrel{-1}{F}-\stackrel{+1}{O}-\stackrel{+1}{O}-\stackrel{-1}{F}$
Oxidation number of $O=+1$

$\left(c\right)$ The $O-O$ bond length in $O_2{F}_2$ is shorter than $O-O$ bond length in $H_2{O}_2.$
We know that, $F$ is more electronegative than $H$. The difference in electronegativity between $O$ and $F$ is lesser than that between $O$ and $H$. Hence, the magnitude of partial positive charge on $O$ atoms in $O_2{F}_2$ is lesser than the magnitude of partial negative charge on $O$ atoms in $H_2{O}_2.$ Due to this, the bond pair of electrons of $O-O$ bond in $O_2{F}_2$ has less repulsion between them than the bond pair of electrons of $O-O$ bond in $H_2{O}_2$.As a result, $O-O$ bond length is shorter in the case of $O_2{F}_2$