Question

The correct order of bond angle, $\angle XOX$ (X=H, O, or F) is:
 A.${H_2}O > {O_3} > {F_2}O$
 B.${O_3} > {H_2}O > {F_2}O$
 C.${F_2}O > {H_2}O > {O_3}$
 D.${O_3} > {F_2}O > {H_2}O$

Answer 1

Hint: To start with comparison of bond angle, we should know that there are certain factors to be taken into account which affects the bond angle:
-Hybridisation of central metal atoms.
-Electronegativity
-Repulsion of lone pairs and bond pairs
Now, we can start finding individual bond angles and then comparison.

Complete step by step answer:

We have to take into consideration the three factors above stated, and let us start with finding Hybridisation of each of these molecules. To find hybridisation, we need to know lone pairs and bond pairs of central atom
For \[{H_2}O\] Oxygen is central atom and to find hybridisation, we know, 2 bond pair with 2 hydrogen and 2 lone pair will be there, so total 4 hybrid orbital, means \[s{p^3}\] is hybridisation.
For \[{F_2}O\] oxygen is central atom, oxygen will form 2 bond pairs with 2 fluorine, and 2 lone pairs on oxygen will be present, thus 4 hybrid orbitals, means \[s{p^3}\] is hybridisation.
For \[{O_3}\] again oxygen is central atom, but this time we have one double bond with one atom of oxygen and central atom, and one coordinate bond (lone pair of oxygen shared) with another oxygen atom, so thus we have one coordinate bond, one sigma bond, and one lone pair on oxygen, thus 3 hybrid orbitals involved, means \[s{p^2}\] is hybridisation.
Now, in general, \[s{p^2}\] hybridisation has an angle of 120 degree, and \[s{p^3}\] hybridisation has an angle of 109.5 degree.
Thus \[{O_3}\] has highest angle, and now compare between \[{H_2}O\] and \[{F_2}O\]
To compare this, here comes the second factor of lone pair lone pair repulsion, the one in which lone pair and lone pair repulsion is there, so angle decreases.
But again there are 2 lone pairs in both molecules. So, here comes the role of electronegativity.
We know that fluorine has highest electronegativity, thus in fluorine, there is repulsion between lone pairs of oxygen electrons with fluorine atoms, thus the bond angle is decreased when compared with Hydrogen.
Thus, \[{H_2}O\] has greater bond angle than \[{F_2}O\]
Hence, the correct option is (B) ${O_3} > {H_2}O > {F_2}O$.

Note:
We should know the concept of hybridisation, and how to find it. Even after knowing hybridisation, we should know bond angle between molecules in general in any kind of hybridisation, and the effect of lone pair repulsion and electronegativity of atoms on bond angle.
One more thing to notice is bond angles are formed between bonds, and not with lone pairs.
Lone pair is changing angle and shape due to repulsion, but angles are not calculated in reference to lone pairs.