2
$\begingroup$

Boiling point of water is 100 degrees Celsius, while boiling point of ammonia is minus 33 degrees Celsius, which makes 133 degrees difference. Now when we discuss value of boiling point, we also say that it depends on inter-molecular forces, and in case of both water and ammonia I can see two such forces: London dispersion forces and hydrogen bonds. Relative molecular mass of water is 18 amu, and relative molecular mass of ammonia is 17 amu, so unlikely we can explain the difference in boiling points by difference in London dispersion forces: as they depend on mass, the LDF should be about the same for two compounds. Speaking about hydrogen bonds, it looks as each molecule (water or ammonia) is able to form 4 hydrogen bonds. And in this way amount of hydrogen bonds again should be about the same. So what is an explanation for the difference in boiling point between water and ammonia?

UPDATE: I found some sort of explanation, which partially seems OK, but I see at least one problem with it though:

Notice that each water molecule can potentially form four hydrogen bonds with surrounding water molecules. There are exactly the right numbers of δ+ hydrogens and lone pairs so that every one of them can be involved in hydrogen bonding. This is why the boiling point of water is higher than that of ammonia or hydrogen fluoride. In the case of ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen only has one lone pair. In a group of ammonia molecules, there aren't enough lone pairs to go around to satisfy all the hydrogens. That means that on average each ammonia molecule can form one hydrogen bond using its lone pair and one involving one of its δ+ hydrogens. The other hydrogens are wasted. In hydrogen fluoride, the problem is a shortage of hydrogens. On average, then, each molecule can only form one hydrogen bond using its δ+ hydrogen and one involving one of its lone pairs. The other lone pairs are essentially wasted. In water, there are exactly the right number of each. Water could be considered as the "perfect" hydrogen bonded system.

Source: http://www.chemguide.co.uk/atoms/bonding/hbond.html

OK, so each ammonia molecule forms 2 hydrogen bonds and so does each hydrogen fluoride molecule. Their masses are 17 amu and 20 amu respectively (difference is very small). Boiling point of fluoride is 19.5 degrees Celsius, while boiling point of ammonia is minus 33 degrees Celsius, which makes 53.5 degrees difference. So what is an explanation for the difference in boiling point between hydrogen fluoride and ammonia?

$\endgroup$
  • $\begingroup$ Man, you could really use some formatting. Now, you've pretty much answered your titular question yourself. As for the difference between $\ce{HF}$ and $\ce{NH3}$, the former is much more polar, hence its hydrogen bonds are stronger. $\endgroup$ – Ivan Neretin Sep 27 '16 at 9:11
  • $\begingroup$ It’s not related to you actual question, but please note that the quantity “relative molecular mass” $M_\mathrm r$ is a dimensionless quantity (to be precise: a quantity of dimension one). Thus, the atomic mass unit $(\mathrm{amu})$ is not a unit of relative molecular mass. Please also note that the unit $\mathrm{amu}$ is obsolete (in Chemistry since 1961). $\endgroup$ – Loong Sep 27 '16 at 11:31
  • $\begingroup$ I would like to add another relevant question with very good answer here: chemistry.stackexchange.com/q/64191/25519 $\endgroup$ – Sleepy Hollow Dec 10 '16 at 4:53
2
$\begingroup$

If I interpret your edited question correctly, it now boils down to why is the difference between the boiling points of $\ce{HF}$ and $\ce{NH3}$ so large, even though they have almost the same molecular masses ($M_\mathrm{r}(\ce{NH3}) = 17, M_\mathrm{r}(\ce{HF}) = 20$).

Ammonia can self-ionise much like water can according to the following equation $(1)$. And much like water, the process is important to determine a $\mathrm{pH}$ value, but rather neglegible when it comes to determining attractive forces between ammonia molecules where ‘traditional’ hydrogen bonding predominates.

$$\ce{2 NH3 <<=> NH4+ + NH2-}\tag{1}$$

On the other hand, self-ionisation in hydrogen fluoride, especially in the liquid and solid states, is much more important and leads to the two fragments shown in equation $(2)$.

$$\ce{3 HF <=> H-F-H+ + F-H-F-}\tag{2}$$

These two linear fragments both feature de facto four-electron-three-centre bonds and are thus much more prevalent than the corresponding ammonia-derived ones. (Indeed, something analogous to $\ce{FHF-}$ doesn’t exist for ammonia.) Since these also introduce ionic interactions into the equation, the intermolecular forces between $\ce{HF}$ molecules are much stronger than those between ammonia molecules contributing to $\ce{HF}$’s higher boiling point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.