Question

The correct order of decreasing $X - O - X$ bond angle is $(X = H,F$ 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

Hint:We know that the bond angle will be affected by the presence of lone pairs of electrons present on the central atom. Also, we will take into account the electronegativity of $X$ here. Electronegativity is the tendency of the atom to attract electrons towards itself in a chemical bond. The most electronegative atom in the periodic table is fluorine and the least electronegative or the most electropositive atom francium.

Complete answer:
We already know that ${H_2}O,{F_2}O\, \& \, C{l_2}O$ are all similar as the oxygen atom forms a single bond with the other two atoms. Hence all the three molecules are tetrahedral in geometry.
We determine the bond angle of a molecule by the repulsion caused by the electron pairs hence the strength of repulsion will decrease as:
Lone pair-lone pair repulsion $ > $ lone pair-bond pair repulsion $ > $ bond pair-bond pair repulsion
Here the bond angle will be determined by the bond pair-lone pair repulsion. Hence the bond angle will increase if the bond pair-lone pair repulsion will increase.
Also, we know that as hydrogen has a half filled $1s$ orbital and needs only one electron to complete its octet, therefore hydrogen has a high electronegativity but oxygen is more electronegative than hydrogen hence it will pull the electrons of hydrogen towards itself and will cause high electron density near oxygen. This will tend to increase the repulsion with the lone pairs of oxygen and will cause bond pair-lone pair repulsion, therefore will increase the bond angle.
Also, we know that chlorine is more electronegative than oxygen so the difference of electronegativity is also low between oxygen and chlorine so the electron density will decrease and will decrease the bond angle.
On the other hand, as we know fluorine is the most electronegative atom and the difference of electron between oxygen and fluorine is least so the electron density of fluorine near oxygen will be less than the hydrogen and oxygen bond.
So, the correct order of decreasing bond angle will be ${H_2}O > C{l_2}O > {F_2}O$

Hence the correct answer is Option A. ${H_2}O > C{l_2}O > {F_2}O$

Note:
The electronegativity of an atom increases on moving from left to right in a period and decreases on moving down the period. Here tetrahedral geometry means that the hybridization is $s{p^3}$ . The covalent bonds between two species of different electronegativities are polar in nature hence ${H_2}O,{F_2}O \,and\,C{l_2}O$ are all polar molecules.