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As far as I know, each $\ce{HF}$ molecule has two hydrogen bonds, one formed by its hydrogen atom and one which the $\ce{F}$-atom forms with hydrogen atom of a third $\ce{HF}$ molecule.

In other molecules, like $\ce{H2O}$, where both of oxygen's lone pairs are used in hydrogen bonding each water molecule forms as many as four hydrogen bonds. Why does fluorine, which has three lone pairs, not use all its lone pairs up to form three hydrogen bonds instead?

andselisk
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user1039203
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  • Then which part of these very electronegative atoms are the hydrogen atoms attracted to? – user1039203 Jul 03 '22 at 05:08
  • They're just attracted to the atom, not any part as such – Dhruv Kaushik Jul 03 '22 at 06:01
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    There are not enough H atoms to use all lone pairs. – Ivan Neretin Jul 03 '22 at 06:44
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    @Infinite (cc ananta) Thank you for doing editorial work, this has been really helpful. One nuance: we prefer to not use MathJax in the title field due to issues it gives rise to; see here for details. Also, it's a good practice to avoid chemical formulas and abbreviations in titles altogether whenever possible. – andselisk Jul 03 '22 at 14:01
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    https://chemistry.stackexchange.com/questions/20149/how-can-the-hydrogen-bonds-in-solid-hf-be-best-represented – Mithoron Jul 03 '22 at 17:06

3 Answers3

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Hydrogen fluoride has only one hydrogen atom per fluorine atom, making a branched hydrogen-bond network difficult to form. (This is one reason hydrogen fluoride also forms hydrogen-bonded structures in the gas phase, unlike water where hydrogen-bond branching is more likely to occur and drive condensation.)

Ammonium fluoride, $\ce{NH4F}$, is not so constrained. In this compound each fluorine can use all four of electron pairs to form bonds that are apparently both hydrogen bonds and ionic bonds:

Ammonium fluoride adopts the wurtzite crystal structure, in which both the ammonium cations and the fluoride anions are stacked in ABABAB... layers, each being tetrahedrally surrounded by four of the other. There are N−H···F hydrogen bonds between the anions and cations.[1] This structure is very similar to ice, and ammonium fluoride is the only substance which can form mixed crystals with water.[2]

Cited References

  1. A. F. Wells, Structural Inorganic Chemistry, 5th ed., Oxford University Press, Oxford, UK, 1984.

  2. Brill, R.; Zaromb, S. "Mixed Crystals of Ice and Ammonium Fluoride". Nature. 173 (4398): 316–317. https://doi.org/10.1038/173316a0.

Oscar Lanzi
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    I guess in the context of hydrogen bonds, we should also consider $\ce{2HF <=> [H2F]+ + F-}$. (You might want to correct how you spell hydrogen.) – Martin - マーチン Jul 03 '22 at 17:17
  • @martin I have a major problem typing on my device. Fingers are too fat for the tiny target offered by each letter. Thanks for the catch. – Oscar Lanzi Jul 03 '22 at 17:50
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    I do know exactly what you mean, I have somewhat similar issues sometimes. Anyway, I just noticed, in my last comment I've posted the wrong equilibrium. I meant of course: $\ce{2HF <=> [HF2]- + H+}$. Silly me... – Martin - マーチン Jul 03 '22 at 22:34
  • Right, but note that $\ce{HF2^-}$ does not force more than one hydrogen bond from fluorine. Hence my choice. – Oscar Lanzi Jul 03 '22 at 22:49
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    No, exactly. I agree with what you wrote. The equilibrium is posted last still would also support the notion, why a network is hard to form though. – Martin - マーチン Jul 03 '22 at 22:54
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In a true intermolecular hydrogen bond, the X-H---Y bond angle is approximately 180$^\circ$. Thus, each H atom can only participate in one H-bond.

In pure HF, the ratio of H:F is necessarily 1:1. Since each H-bond requires one H and one F and each H can only participate in one H-bond, the F atoms necessarily participate (on average) in only one H-bond each as well.

Oscar Lanzi
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Andrew
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  • So is the number of atoms participating in a hydrogen bond is proportional to the number of hydrogen atoms forming hydrogen bonds? – user1039203 Jul 03 '22 at 13:52
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    Yes, necessarily the number of H-bonds is determined by the number of H atoms that participate in H-bonds, since an H-bond must have an H. But when you are counting H atoms involved in H-bonds, remember that sometimes the number of H atoms is not limiting. For example, in NH3 the N has only one lone pair, so only 1/3 of the H atoms can participate in H bonds. – Andrew Jul 03 '22 at 14:12
  • Added "intermolecular". Our arguments apply in that case; intramolecular bonds may be constrained by overall molecular geometry. Then +1. – Oscar Lanzi Jul 03 '22 at 21:35
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$\ce{HF}$ can not form two hydrogen bonds since the two positively charged $\ce{H}$ atoms would repel each other.

HF wrong hydrogen bonding

Also, unlike it has been said in the comments, $\ce{H}$-bonds do involve orbital overlap and are not a consequence of electrostatic attraction alone. Although the overlap does not result in very strong bonds, as is the case with overlapping in covalent bonds, $\ce{H}$-bonds can be surprisingly stable ($\Delta G \approx 3\text{kJ/mol}$).

Never thought I would find a possible explanation while comparing neighbor interactions. If we assume that the the zig-zag structure for $\ce{HF}$ is correct, we can not rule out the repulsion caused by the lone pairs of fluorine. neighbor-neighbor repulsion in hydrogen fluoride

ananta
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    When there are sufficient H atoms available, F will happily bond with more than 2. For example, the solid crystal structure of NH4F is essentially identical to the ice-Ih form of solid H2O, with each F coordinated by 4 equivalent H atoms, just as O is coordinated by 4 H atoms in ice-Ih. – Andrew Jul 03 '22 at 14:14
  • @Andrew even in questions comments, the number of hydrogen atoms vs. the number of fluorine atoms has been pointed out. However, I do not think the limitation is the number of hydrogen atoms, actually it is energetic considerations. – ananta Jul 03 '22 at 14:40
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    If you have a source for that claim, do cite it. – Andrew Jul 03 '22 at 14:49
  • @Andrew You didn't cite a source for your claim either, don't you think it is unfair to ask me to do the same? An example of fluorine forming more than two bonds doesn't really answer the question if there can be more than three hydrogen bonds per $\ce{HF}$ molecule in an $\ce{HF}$ solid/liquid structure. – ananta Jul 03 '22 at 14:58
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    The claim for HF is based on simple arithmetic, and Oscar lanzi has provided citations for my statement regarding NH4F, so I'm not sure what other citation would be needed. – Andrew Jul 03 '22 at 15:53