Which one out of $\ce{H2O}, \ce{Cl2O}, \&\ \ce{F2O}$ will have largest bond angle?

I think it should be $\ce{H2O}$ because oxygen is most electronegative in this case so electrons will be more towards oxygen hence increasing bond angle, but the answer is $\ce{Cl2O}$. Why?


2 Answers 2


There is practically only one effect that determines the angle, but there is an underlying mechanism that you also should understand.

Ideally, a central atom such as oxygen would want to use its p-orbitals exclusively for bonding and have the s-orbital electrons sitting comfortably as a lone pair close to its nucleus. However, that requires a $90^\circ$ angle which can create rather strained molecules — remember that each atom is surrounded by an electron cloud with negative charge that repulses other atoms’ electron clouds. There are now two possibilities of how to relieve this steric strain:

  • enlarge the bond lengths, so that different substituents are further away from each other while keeping a bond angle as perfect as possible;
  • enlarge the bond angles so the substituents move further apart from each other while retaining a close distance to their central atom with a more perfect bond length.

In practice, almost all systems stabilise themselves by modifying the angles. Increasing bond lengths decreases bond strengths and thus energy gained much more strongly than tinkering around with angles, because the overlap will inevitably get significantly smaller in absolute value number no matter which way you look at it.

An average bond length can be thought of as the distance which is just large enough for those orbitals that interact to do so favourably while minimising nonfavourable interactions. For certain atom pairs, bond length can essentially be considered almost a constant.

All of this taken together means that the very small hydrogen atoms have a much easier time of crowding together closer to the $90^\circ$ angle than the larger fluorine or huge (in this context) chlorine atoms. Even though the latters’ bond lengths are longer in absolute values their larger radii makes the effective bond lengths shorter and thus makes them require more space. Since chlorine is the largest of the lot, the $\ce{OCl2}$ bond angle is largest.

  • 3
    $\begingroup$ Even though this is the the accepted answer, this is partially wrong because it does not consider the effect of electronegativity difference between atoms. The last part about "hydrogen atoms having easier time of crowding together closer to the 90 degree angle" is completely wrong: it isn't supported by experimental data. H2O has bond angle 104.45 degree and F2O has a bond angle of 103 degree. So clearly hydrogen is not happy to stay closer to 90 degrees. $\endgroup$
    – S R Maiti
    Commented May 30, 2021 at 21:05

There are a few factors which affect bond angle which you must consider. Loan pairs typically occupy more space than bonding pairs, so you should expect the two loan pairs of oxygen to repel the bonding pairs inward such that the angle between them is less than that of a perfect tetrahedral (109.5°). This holds true for $\ce{H2O}$ and $\ce{F2O}$, but as for $\ce{Cl2O}$, the opposite is seen ($\ce{Cl2O}$ bond angle is 110.9°). This can be attributed to a concept known as Ligand Close-Packing. The ligands (atoms attached to the central atom, in this case oxygen) have a fixed radius, and the lone pairs of the central atom (oxygen) will repel the bonding pairs (between the ligand and the central atom) until the ligands "just touch". The bond angle is dependent on how far the lone pairs can repel the bonding pairs before the ligands "just touch" (and therefore is also dependent on the atomic radius of the ligand and bond length between the ligand and the central atom). This explains why $\ce{H2O}$ and $\ce{F2O}$ have a smaller bond angle than tetrahedral, while $\ce{Cl2O}$ has a larger bond angle.


  • $\begingroup$ Is this possible that oxygen has 2 lone pairs right, so it makes another bond with one of the chlorine to minimise lone pair lone pair repulsion and hence its bond angle is most $\endgroup$
    – user354545
    Commented Aug 20, 2016 at 3:37

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