Question

The correct order of increasing $X-O-X$ bond angle is $(X=H,F\text{or}Cl)$:

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

Bond angle is generally affected by the presence of lone pairs of electrons and the electronegativity of the central atom and side atom in different cases.

In order to find the order of bond angle, we can see that since the central atom is same in all cases with the same number of lone pairs on it but here, the surrounding atoms are different in each case.

Generally in these cases, the order of bond angle is inversely proportional to the electronegativity of surrounding atoms. Since $F$ has the highest EN and hence, $F_{2}O$ will have the lowest bond angle and $H$ has the lowest EN among all the given compounds and hence $H_{2}O$ must have the highest bond angle. But due to the large size of $Cl$, there occurs high repulsion between the two $Cl$ atoms', the lone pair in the case of $C{l}_{2}O$ and hence both $Cl$ atoms will go far away and have the highest bond angle among all.

Therefore $C{l}_{2}O>H_{2}O>F_{2}O$