Question

The Bond Angle In ${\mathrm{OF}}_2\mathrm{Out}\mathrm{Of}{\mathrm{OF}}_2{\mathrm{Cl}}_2\mathrm{O}{\mathrm{Br}}_2\mathrm{O}$ Is Minimum. It Is Because In the Case Of ${\mathrm{OF}}_2$?

• A. Lone Pair-lone Pair Repulsion Decreases Bond Angle
• B. Electrons Are Nearer To Fluorine Due To High Electronegativity Of $\mathrm{F}$Compared To $\mathrm{Cl}\mathrm{To}\mathrm{Br}$
• C. Both (A) And (B) Are Correct
• D. None Of The Above Is Correct

Answer 1

The correct option is B

Electrons Are Nearer To Fluorine Due To High Electronegativity Of $\mathrm{F}$Compared To $\mathrm{Cl}\mathrm{To}\mathrm{Br}$

Explanation for the correct option:

Option(B): Electrons Are Nearer To Fluorine Due To High Electronegativity Of $\mathrm{F}$Compared To $\mathrm{Cl}\mathrm{To}\mathrm{Br}$

• In the case of ${\mathrm{OF}}_2$ , the electrons are nearer to Fluorine due to the high electronegativity of $\mathrm{F}$Compared To $\mathrm{Cl}\mathrm{To}\mathrm{Br}$.
• This results in the decrease of the repulsion between electrons, which in turn helps in decreasing the bond angle in ${\mathrm{OF}}_2$.

Explanation for the incorrect option:

Option(A): Lone Pair-lone Pair Repulsion Decreases Bond Angle

• Lone Pair-lone Pair Repulsion does not explain the decreases in the bond angle of ${\mathrm{OF}}_2$.
• Hence, the given option is the incorrect one.

Thus, the Bond Angle In ${\mathrm{OF}}_2\mathrm{Out}\mathrm{Of}{\mathrm{OF}}_2{\mathrm{Cl}}_2\mathrm{O}{\mathrm{Br}}_2\mathrm{O}$ Is Minimum Is Because In the Case Of ${\mathrm{OF}}_2$Electrons Are Nearer To Fluorine Due To High Electronegativity Of $\mathrm{F}$Compared To $\mathrm{Cl}\mathrm{To}\mathrm{Br}$.