Question

The correct order of decreasing $X-O-X$ bond angle is:

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

Answer 1

The correct option is A $C{l}_{2}O>H_{2}O>F_{2}O$

Bond angle is affected by the presence of lone pair of electrons on the central atom. Since oxygen is the central atom and it forms bonds with hydrogen, chlorine and fluorine, in all three cases, the geometry will be tetrahedral with two lone pairs of electrons.

The strength of repulsions decrease in the following order:
$\text{l.p−l.p > l.p−b.p > b.p−b.p}$
$\text{l.p - lone pair}$
$\text{b.p - bond pair}$

The degree of this type of repulsion depends on the electronegativity of $X$. Hydrogen is the least EN atom. Thus, bond pairs of $O-H$ bonds will be closer to $O$, resulting in higher electron density near $O$, at the site of the bond. This increases repulsion with the lone pairs in the outer shell of oxygen, thus increasing bond angle.

Fluorine is the most electronegative element. Thus, electron density of the $O-F$ bond near $O$ is less than that of $O-H$.
In $C{l}_{2}O$, due to non bonding electron pair repulsion (NBEPR), the bond angle is increased to ${110.9}^o$.