The $M-\mathrm{O}-M$ bond angles in $\mathrm{M}_2 \mathrm{O}$ (where $M$ is halogen) is in the order

• Correct answer
  A

  $\mathrm{Br}_2 \mathrm{O}>\mathrm{Cl}_2 \mathrm{O}>\mathrm{F}_2 \mathrm{O}$
• B

  $\mathrm{F}_2 \mathrm{O}>\mathrm{Br}_2 \mathrm{O}>\mathrm{Cl}_2 \mathrm{O}$
• C

  $\mathrm{F}_2 \mathrm{O}>\mathrm{Cl}_2 \mathrm{O}>\mathrm{Br}_2 \mathrm{O}$
• D

  $\mathrm{Cl}_2 \mathrm{O}>\mathrm{F}_2 \mathrm{O}>\mathrm{Br}_2 \mathrm{O}$

Solution

In $\mathrm{OF}_2$ repulsion between lone pairs is greater than that between bond pair since electrons are away from $\mathrm{O}$ and nearer to $\mathrm{F}$. In $\mathrm{Cl}_2 \mathrm{O}$, bonding electrons are nearer to $\mathrm{O}$ than to $\mathrm{Cl}$, so the bond angle is greater than $109^{\circ} 28^{\prime}$. $\operatorname{In} \mathrm{Br}_2 \mathrm{O}$, the bonding electrons are more closer to oxygen than in $\mathrm{Cl}_2 \mathrm{O}$, so the bond angle is largest $\left(116^{\circ}\right)$.

Hence, the correct answer is option: 'A'.