Question

The $M - O - M$ bond angles in $M_2O$ (where $M$ is halogen) is in the order

• A. $Br _{2} O > Cl _{2} O > F _{2} O$
• B. $F _{2} O > Br _{2} O > Cl _{2} O$
• C. $F _{2} O > Cl _{2} O > Br _{2} O$
• D. $Cl _{2} O > F _{2} O > Br _{2} O$

Answer 1

Answer: (A)

$Br _{2} O > Cl _{2} O > F _{2} O$

Answer 2

In $OF _{2}$ repulsion between lone pairs is greater than that between bond pair since electrons are away from $O$ and nearer to $F$. In $Cl _{2} O$, bonding electrons are nearer to $O$ than to $Cl$, so the bond angle is greater than $109^{\circ} 28' .$ In $Br _{2} O$, the bonding electrons are more closer to oxygen than in $Cl _{2} O$, so the bond angle is largest $\left(116^{\circ}\right)$.