Bond angle order of SF2, OF2, HOF

Question

Arrange the following componds in the increasing order of bond angle. $$\ce{HOF,~OF2,~SF2}$$

I know that bond angle of $\ce{SF2}$ will be less than the bond angle of $\ce{OF2}$ because of larger size of sulfur atom.

How can I compare $\ce{HOF}$ with the other two? Since $\ce{H}$ is less electronegative than $\ce{O}$, the bond pair of $\ce{O - H}$ bond will be more closer to $\ce{O}$. The opposite happens for $\ce{O - F}$ bond. In $\ce{SF2}$ both bond pairs are near $\ce{F}$ so the bonds can come closer to each other.

Answer 1

When considering bond angles of molecules of main group ($\ce{\angle A-X-B}$) there are several different factors to consider:

1. The row of the the central atom $\ce{X}$ determines the energy difference between the valence s and p orbital, which determines the extent of mixing (or hybridisation if you wish)
2. The sizes of A and B, because steric repulsion, A and B would want to stay away from each other the bigger they get.
3. Electronegativities of the A and B. More electronegative atoms would prefer to have a bond with higher p-character. (This is explained in VSEPR model with the bond pair being drawn away from the central atom)
4. Other forces betwen A and B

The bond angles of the molecules mentioned in question are: $$\begin{array} {c|c}\hline & \text{bond angle} \\ \hline \ce{OF2} & 103.2^\circ \\ \hline \ce{SF2} & 98^\circ \\ \hline \ce{HOF} & 97.2^\circ \\ \hline \end{array}$$

The difference between $\ce{OF2}$ and $\ce{SF2}$ would be usually explained with the fact that oxygen is in 2nd period so, the energy difference between $\mathrm{2s}$ and $\mathrm{2p}$ is low which means it is $\mathrm{sp^3}$ hybridised. (Or at least close to $\mathrm{sp^3}$). Whereas sulfur is in the 3rd period so, the energy difference between $\text{3s}$ and $\text{3p}$ is high which means it is almost unhybridised and the bonds are formed by almost pure $\text{3p}$ (which is why the angle is closer to $90^\circ$).

The bond angle of $\ce{HOF}$ is however, unusally small. This is difficult to explain and is usually attributed to the electrostatic attraction between $\delta+\;$ of H and $\delta-\;$ of F, which pulls the H and F close together.

From Chemistry of the Elements ^[1],

Spectroscopic data establish a nonlinear structure with $\ce{H-O}$ $\pu{96.4 pm}$, $\ce{O-F}$ $\pu{144.2 pm}$, and bond angle $\ce{H-O-F}$ $97.2^\circ$: this is the smallest known bond angle at an unrestricted O atom (cf. $\ce{H-O-H}$ $104.7^\circ$, $\ce{F-O-F}$ $103.2^\circ$). It has been suggested that this arises in part from eletrostatic attraction of the 2 terminal atoms, since NMR data lead to a charge of $\sim+0.5e$ on H and $\sim-0.5e$ on F.

So, the unusually small bond angle is likely from the electrostatic attraction. This type of effect is not just limited to $\ce{HOF}$, it is also found it other molecules. For instance, in hydrogen peroxide $\ce{H-O-O-H}$, the $\ce{H-O-O}$ angle is $94.8^\circ$ which is quite lower than the ideal tetrahedral angle, and has been attributed the electrostatic attraction between $\ce{O}$ and $\ce{H}$.^[2]

As an aside, it should be noted that the above values of bond angles were measured in the gas phase. In solid phase, $\ce{HOF}$ has a slightly higher bond angle, at $101^\circ$.

References-

1. N. N. Greenwood, A Earnshaw, Chemistry of the Elements, 2nd ed., Butterworth-Heinemann, Oxford, 1998
2. E. Wiberg, A. F. Holleman, N. Wiberg, Inorganic Chemistry, Academic Press, 2001

Answer 2

I looked up the net and found the following order-

HOF(97.2°)<SF₂(98°)<OF₂(103°)

(Of course, the values might not be exact but this is what I found on the net.)

Now, my POV-

Yes, your analogy that bond angle of OF₂> SF₂ is correct. It's a direct result of Drago's Rule.

The electron cloud in OF₂ will be shifted much towards the Fluorine atom. Bond pair-Bond pair repulsions are larger if the electron cloud is nearer to the central atom. But the correct reason is the distance between the electron clouds, not the "near the central atom, away from it" analogy.

In the HOF molecule, the cloud is more towards the central atom Oxygen in the H-O bond and towards F in the O-F bond. So the distance(very misleading term tho) between the two electron clouds is a bit larger than in the case of OF₂. A larger distance means lesser repulsions and a smaller bond angle.

But the catch here is that the bond angle of SF₂ is slightly bigger than that of HOF. This is because the effect of electron cloud repulsions ( both electron clouds in both S-F bonds shifted towards Fluorine) and the size of sulfur atom ( Drago's rule) work against each other. Also, Drago's Rule doesn't apply much when the Electronegativity of the surrounding atom exceeds 2.5 which is the case with Fluorine ( 4- on the Pauling Scale).

You should also not forget the weak but existent back bonding in the S-F bond due to the presence of vacant d orbital in the Sulphur. The double-bondish character leads to a bit stronger bond pair-bond pair repulsions. Why did I call the back bonding weak? It's because the Sulphur atom in SF₂ has two lone pairs that repel the electrons flowing in due to the back bond. Also adding on to that point, the 2p-3d back bond has appreciable overlap (more than 2p-3p) but it's still weak.

Drago's Rule is explained by the unfavorability of hybridization - the energy required for hybridization is much larger than what will be obtained after hybridization. In more EN atoms(>2.5) the net charge on the central atom is more positive which reduces the size of the central atom. Thus, the sizes ( read energies) of s, p, d orbitals become similar and hybridization is favorable only when atomic orbitals involved have similar energies.

Answer 3

$\ce{H-O-F}$ has a lower bond angle than both $\ce{F-S-F}$ and $\ce{F-O-F}$. The reason is probably that the lower electron density around $\ce{H}$ results in significantly less steric interaction between $\ce{H}$ and the lone pairs of the $\ce{F}$ atom. In both $\ce{SF2}$ and$\ce{ OF2}$, the two $\ce{F}$ atoms with high electron densities would tend to repel each other, thus resulting in a greater bond angle. Hence the bond angle order comes out to be- $$\ce{F-O-F \gt F-S-F \gt H-O-F}$$ Hope this helps!