Question

The $M-O-M$ bond angles in $M_{2}O$ (where $M$ is halogen) is in the order

• A. $B{r}_{2}O>C{l}_{2}O>F_{2}O$
• B. $F_{2}O>B{r}_{2}O>C{l}_{2}O$
• C. $F_{2}O>C{l}_{2}O>B{r}_{2}O$
• D. $C{l}_{2}O>F_{2}O>B{r}_{2}O$

Answer 1

Answer: (A)

In $O{F}_2$ repulsion between lone pairs is greater than that between bond pair since electrons are away from $O$ and nearer to $F$. In $C{l}_{2}O$, bonding electrons are nearer to $O$ than to $Cl$, so the bond angle is greater than $109^{\circ }28^{\prime }.$ In $B{r}_{2}O$, the bonding electrons are more closer to oxygen than in $C{l}_{2}O$, so the bond angle is largest $\left(116^{\circ }\right)$.