Question

$O_2F_2$ is an unstable yellow orange solid and ${{{H}}_2}{{{O}}_2}$ is a colorless liquid. Both have ${{O - O}}$ bond and ${{O - O}}$ bond length in ${{{H}}_2}{{{O}}_2}$ and ${{{O}}_2}{{{F}}_2}$ respectively is:
A. $1.22{{{A}}^ \circ },1.48{{{A}}^ \circ }$
B. $1.48{{{A}}^ \circ },1.22{{{A}}^ \circ }$
C. $1.22{{{A}}^ \circ },1.22{{{A}}^ \circ }$
D. $1.48{{{A}}^ \circ },1.48{{{A}}^ \circ }$

Answer 1

This question is solved using a rule called Bent’s rule. Structures of several molecules are determined using this rule. This rule was stated by Henry Bent. It explains that there is a relationship between hybridizations of the central atoms and electronegativity.

Complete step by step answer:
The given compounds are ${{{O}}_2}{{{F}}_2}$ and ${{{H}}_2}{{{O}}_2}$. By looking into the compounds itself, we would understand there is a great difference in the electronegativity between oxygen and hydrogen in ${{{H}}_2}{{{O}}_2}$ and oxygen and hydrogen in ${{{H}}_2}{{{O}}_2}$.
Both of them have the same structures. But the bond angle is different for ${{{H}}_2}{{{O}}_2}$. In ${{{H}}_2}{{{O}}_2}$, the bond angle of ${{H}} - {{O}} - {{O}}$ is ${101.9^ \circ }$. While in ${{{O}}_2}{{{F}}_2}$, the bond angle between oxygen and fluorine is ${109.5^ \circ }$. The bond length of ${{O - O}}$ in ${{{O}}_2}{{{F}}_2}$ is very shorter than that in ${{{H}}_2}{{{O}}_2}$. This is because of its electronegativity. The electron density is attracted by fluorine towards itself. Thus the bond length is reduced. But in ${{{H}}_2}{{{O}}_2}$, oxygen is not as electronegative as fluorine. Thus the pulling of electron density is not as intense as in ${{{O}}_2}{{{F}}_2}$. Thus the bond length of ${{O - O}}$ in ${{{O}}_2}{{{F}}_2}$ is less than that in ${{{H}}_2}{{{O}}_2}$. So the possible option will be B.
Hence the bond lengths are $1.48{{{A}}^ \circ },1.22{{{A}}^ \circ }$.

Hence, the correct option is B.

Note:
A p character is more in ${{O - F}}$ bond than ${{O}} - {{H}}$ bond in ${{{H}}_2}{{{O}}_2}$. Thus s character is increased in ${{O - O}}$ bond in ${{{O}}_2}{{{F}}_2}$. When there is a more electronegative element, then it will have less s character. Thus it will have a smaller bond angle. Less s character also denotes that there is more p character.