Question

Arrange the molecules in the increasing order of bond angle.

• A. $H_{2}O<O{F}_2<OC{l}_2$
• B. $OC{l}_2<H_{2}O<O{F}_2$
• C. $O{F}_2<OC{l}_2<H_{2}O$
• D. $O{F}_2<H_{2}O<OC{l}_2$

Answer 1

The correct option is D $O{F}_2<H_{2}O<OC{l}_2$

The correct order of increasing bond angle is:
$O{F}_2<H_{2}O<OC{l}_2$
Here $O{F}_2,O{H}_2$ and $OC{l}_2$ each have 2 lone pairs and 2 bond pairs. Now since the size of $Cl$ is comparatively large, thus due to the repulsion between the two $Cl$ atoms, bond angle of $OC{l}_2$ is highest.
Now again bond angle of $O{H}_2$ is greater than $O{F}_2$, because due to the greater electronegativity value of $F$, it will pull the bonding electron pairs towards itself. Hence bond pair-bond pair repulsion will be less in $O{F}_2$. Thus bond angle will be least in the case of $O{F}_2$