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Question: Answered & Verified by Expert
The $M-\mathrm{O}-M$ bond angles in $\mathrm{M}_2 \mathrm{O}$ (where $M$ is halogen) is in the order
ChemistryChemical Bonding and Molecular StructureAIIMSAIIMS 2009
Options:
  • 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:
1756 Upvotes Verified Answer
The correct answer is: $\mathrm{Br}_2 \mathrm{O}>\mathrm{Cl}_2 \mathrm{O}>\mathrm{F}_2 \mathrm{O}$


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 $\left(116^{\circ}\right)$.

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