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According to the values of boiling points that I found on internet the order is as follows:

$\ce{H2O}$ > $\ce{HF}$ > $\ce{NH3}$

I was expecting $\ce{HF}$ to have highest boiling point because F is the most electronegative atom hence should form stronger H Bonds.

Now, I think the factor here might be number of H Bond rather than strength of H Bond.

  • $\ce{H2O}$ have 2 lone pair and 2 H atoms, hence it can form 4 H Bonds with surrounding molecules.

  • Similarly $\ce{NH3}$ is also capable of forming 4 H Bonds with surrounding molecules.

  • $\ce{HF}$ on the other hand is only capable of forming 2 H Bonds with surrounding molecules.

Still the above hypothesis doesn’t explain why the boiling point of $\ce{HF}$ is greater than $\ce{NH3}$. I’m not sure what am I missing here.

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Though ammonia has three H atoms, it is not capable of forming four H-bonds

An important difference in terms of hydrogen bonding between ammonia and water, lies in the ratio between how many partial positive hydrogen atoms and how many lone pairs of electrons each have.

A water molecule has two partial positive hydrogen atoms and two lone pairs of electrons located on the oxygen atom. This implies that each water molecule can potentially have both its hydrogen atoms and both its lone pairs involved in hydrogen bonding.

But in ammonia, there is a shortage of lone pairs on the central nitrogen atom. Hence it is capable of forming only two H-bonds.

And the difference in electronegativity is higher in $\ce{HF}$ than in $\ce{NH3}$.

Hence boiling point order is $\ce{H2O} > \ce{HF} > \ce{NH3}$

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