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$\ce{F}$ has more unshared electron pairs and is very electronegative, so $\ce{H}$ of another $\ce{HF}$ molecule can $\ce{H}$-bond with it.

$\ce{HF}$ has normal boiling point of $\pu{19.5^oC}$ while $\ce{H2O}$, as you know, has normal boiling of $\pu{100^oC}$.

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I think there are a variety of qualitative ways of looking at this:

  1. Perhaps the most obvious is that $\ce{H2O}$ can form a greater number of hydrogen bonds due to having an equal numbers of hydrogen bond acceptors and donors. Each of the hydrogen atoms can be hydrogen bond acceptors; each of the lone pairs on the oxygen can be donors. In $\ce{HF}$ however there is only one hydrogen bond acceptor and theoretically three hydrogen bond donors. This imbalanced ratio inevitably leaves some acceptors without a donor. Imagine this as two dance parties; one party has two men and two women but the other has one man and three women. Obviously fewer bonds can be successfully formed in the latter, assuming that everything is monogamous.
  2. As mentioned by the other poster, both O and F are very electronegative elements. F however is more EN than O, which means that F stabilizes electrons better than O. As a result, electrons localized on F are weaker hydrogen bond donors; they are already fairly well stabilized. Remember, electrons "want" to be stabilized by nuclei. If they already are fairly well stabilized then they will "feel" less of a need to be associated (and therefore stabilized) with (by) other nuclei. This suggests that the hydrogen bonding present among $\ce{HF}$ molecules might be weaker than the hydrogen bonds present among $\ce{H2O}$ molecules.

Experimental results are the gold standard, of course, as opposed to qualitative reasoning, and qualitative reasoning here can also lead us in the opposite direction; one might argue that by having three lone pairs, F has a lot on its plate, so to speak; it might be able to stabilize one lone pair of electrons very well but three lone pairs is a bigger ordeal, and perhaps big enough that the hydrogen bonds among $\ce{HF}$ molecules are stronger than those among water molecules. Again, this is all qualitative, but this is the sort of reasoning introductory chemistry teachers desire.

If we limit our thinking to just electrostatics then we might surmise that $\ce{HF}$ should have the stronger hydrogen bonds because F is more electron withdrawing and therefore hydrogen should be more positively polarized in $\ce{HF}$ as opposed to hydrogen in water. However, hydrogen bonding is more than just electrostatics. Hydrogen bonding actually has a covalent component; this, however is usually ignored by introductory treatments of chemistry. The bond angle of elements involved in a hydrogen bond is critical. The closer the elements involved in a hydrogen bond are to 180 degrees, the stronger the bond (this specific angle is the case with hydrogen bonding in water; not necessarily other molecules). If hydrogen bonding were purely electrostatic then this would not be the case; angles would not matter - only distance would.

Other issues must be explored as well, such as the number of viable hydrogen bonds and the electron-donating/releasing tendencies of the involved elements.

After scouring the web some common "explanations" that would not explain why water has a higher boiling point than $\ce{HF}$ would be:

  1. water can form 4 per molecule while HF can only form 2.

Incorrect because if we only look at hydrogen bonding as involving the polarity of atoms, then how can water form four hydrogen bonds per molecule? It has two positively charged hydrogens, and a negatively charged oxygen. Seems like it should only form three hydrogen bonds. One must understand that lone pairs can each be hydrogen bond donors.

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    $\begingroup$ So, should we accept the reasoning that goes with the experiment and disregard the others? Well, I am asking this because I am wondering what percentage of my high school book is true! $\endgroup$ – Mockingbird May 12 '17 at 7:49
  • $\begingroup$ It doesn't really detract from your point, but you have acceptor and donor the wrong way round. $\endgroup$ – orthocresol Jun 24 '17 at 13:00
  • $\begingroup$ @orthocresol It is known that HF shows hydrogen bonding in gaseous phase too. It simply means that while going from liquid to gaseous phase, the hydrogen bonds don't have to be broken in the case of HF while in the case of water, the hydrogen bonds need to be broken. This simply explains why boiling point of water is greater despite HF making stronger hydrogen bonds. Is this reasoning wrong because it hasn't been discussed in the answer? $\endgroup$ – Arishta Aug 7 '17 at 13:42
  • $\begingroup$ We don't only have to explain why the boiling point of water is higher than that of HF, we also have to explain this "huge difference" in their boiling points. The factors discussed above cannot alone account for this. $\endgroup$ – Arishta Aug 7 '17 at 13:48
  • $\begingroup$ Water can form 2 hydrogen bonds, as oxygen has a 2δ− partial charge, therefore it attracts to hydrogen atoms with a δ+ bond to it. What you found on the internet wasn't entirely incorrect because they may have meant that water has 2 covalent bonds and can have 2 hydrogen bonds (4 bonds). $\endgroup$ – Dastur Aug 8 '18 at 2:52
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The answer lies in hydrogen bonding.

Hydrogen bond energy depends on the electronegativity of a highly electronegative atom which is bonded to hydrogen. Electronegativity of hydrogen is $2.2$, for oxygen it's $3.44$ and it's $4$ for fluorine.

Difference in electronegativity between $\ce{F}$ and $\ce{H}$ is $1.8$ and between $\ce{O}$ and $\ce{H}$ is $1.24$. Hydrogen bond energy of $\ce{H-F}$ is $\pu{41.83 kJ/mole}$ and that of $\ce{O-H}$ is $\pu{23 kJ/mole}$.

$\ce{H-F}$ bond is stronger in comparison to $\ce{O-H}$ bond. In the case of $\ce{H-F}$, there exists hydrogen bonding even in vapour state, 4 to 7 $\ce{HF}$ molecules together form one unit in vapour state. However, in the case of water, there is no hydrogen bonding in vapour state; each water molecule exists independently.

So, to boil liquid water, all hydrogen bonds have to be broken and it requires a large amount of energy. This isn't the case in $\ce{HF}$; all hydrogen bonds need not to be broken, and therefore a lesser amount of energy is required. So $\ce{HF}$ boils at a much lower temperature as compared to water.

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    $\begingroup$ Hydrogen fluoride is known to have a way stronger hydrogen bond than water; but, the question asks why the seemingly opposite view is true. $\endgroup$ – M.A.R. Aug 22 '15 at 7:57
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Both fluoride and oxygen are very electronegative. When they bond with hydrogen the hydrogen becomes slightly positive and the electronegative atom slightly negative. Because of this, attraction occurs between the negative atoms and hydrogen in different molecules, called hydrogen bonding.

In water there are two hydrogen making more charge dipoles for stronger and more numerous hydrogen bonds.

Another factor is water disassociates (i.e. liquid water is partially made up of $\ce{H3O+}$ (three hydrogen and an oxygen, which is positively charged) and some $\ce{OH-}$ (one hydrogen and one oxygen, negatively charged). This increases inter-molecule interaction, making the boiling point higher.

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    $\begingroup$ Well, [HF2]-, the bifluoride anion, exists in (concentrated) HF solutions. Wouldn't this have some effect on intermolecular interactions as well? Also HF definitely ionizes just like water and actually HF ionizes to a greater extent than water since HF is a stronger acid than water. $\endgroup$ – Dissenter Aug 22 '15 at 5:41
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Intermolecular forces (IMFs) are the key here. IMFs are directly related to boiling point.

Both $\ce{HF}$ and $\ce{H2O}$ have hydrogen bonds ($\ce{H}$ attached to $\ce{N}$, $\ce{O}$ or $\ce{F}$). But $\ce{H2O}$ has two hydrogen bonds whereas $\ce{HF}$ only has one. Thus, $\ce{H2O}$ should have a much higher boiling point.

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  • $\begingroup$ please note that there are exactly zero hydrogen bonds in H₂O or HF $\endgroup$ – mykhal May 12 '17 at 12:44
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I think we may have been focusing too much on $\ce{HF}$ and $\ce{H2O}$. If we look at the bigger picture of the boiling points of the hydrides of the elements of their respective groups, we can see that group 17 hydrides really have lower boiling points than that of group 16 hydrides. I still don't know why is the trend this way within each period, but the figure definitely tells us that:

  • the deviation is more probably due to the differences between the groups
  • the H-bond concept still works the same way for HF; it really has higher BPs than the other hydrogen halides

So I think now the main question is: why do hydrogen chalcogenides have higher BPs than hydrogen halides (and is actually the group of halides that have the highest BPs in each period)?

(figure is from http://www.vias.org/genchem/kinetic_12450_08.html)

Boiling points of Groups 14, 15, 16, and 17 hydrides

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    $\begingroup$ In addition, HF and H2O both can make the maximum number of H-bonds per molecule. HF has three H-bond acceptors (three lone pairs of F) and one H-bond donor (from H of HF) while H2O has 2 H-bond acceptors (2 lone pairs of O) and two H-bond donors (from 2 H of H2O). $\endgroup$ – Acnologia Sep 15 '17 at 20:19

protected by orthocresol Mar 19 '17 at 13:28

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