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Why does HF have a lower boiling point that H2O?


Science Student
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I checked the electronegativities of the intermolecular forces, and the difference between fluorine and hydrogen is 1.8, much higher than oxygen and hydrogen which is 1.2. So there must be another reason other than hydrogen bonding why H2O has such a higher boiling point than HF, which is only 19.5c.

 

Also, why isn't HF called hydrogen monoflouride?

Edited by Science Student
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What are the shapes of the molecules and which ones have lone pairs?

HF has 3 lone pairs and H2O has 2 lone pairs. H2O is a bent tetrahedral, and HF has only a single bond, so I am not sure what to call it. This question is in my textbook for grade 11 chemistry; it wants me to know why, but I can't find anything in the chapter that explains it.

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So how many hydrogen bonds must be 'broken' to release a molecule from the liquid to the gas in each case?

Knowing the number of hydrogen bonds for these polar molecules is not something that is covered in the material before the question.

 

However, I see that there would probably be more hydrogen bonds in HF per molecule than H2O because there are more lone pairs on F for hydrogen from the other hydrogen fluorides to bond to. This makes HF's boiling point even more counterintuitive to me.

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H2O

 

H1F

 

How many hydrogen bonds (involving hydrogen) can each molecule engage in?

 

:)

 

I think the answer is 4 for water and 2 for hydrogen flouride

 

Note also that boiling point increases with increasing molecular weight, in the absence of molecular interactions.

 

Water is (16 + 2) = 18 : BP = 100C

 

Hydrogen Flouride is (19 + 1) = 20 : BP = 20C

 

Ammonia is (14 + 3) = 17 : BP = -33C

 

Methane is (12 + 4) = 16 : BP = -182C

 

So water and hydrogen flouride are anomalous, going across the periodic table, as are hydrogen sulphide and hydrogen chloride in the next line, for the same reasons.

 

 

 

 

Edited by studiot
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H2O

 

H1F

 

How many hydrogen bonds (involving hydrogen) can each molecule engage in?

 

:)

 

I think the answer is 4 for water and 2 for hydrogen flouride

 

Note also that boiling point increases with increasing molecular weight, in the absence of molecular interactions.

 

Water is (16 + 2) = 18 : BP = 100C

 

Hydrogen Flouride is (19 + 1) = 20 : BP = 20C

 

Ammonia is (14 + 3) = 17 : BP = -33C

 

Methane is (12 + 4) = 16 : BP = -182C

 

So water and hydrogen flouride are anomalous, going across the periodic table, as are hydrogen sulphide and hydrogen chloride in the next line, for the same reasons.

 

 

 

 

Thanks for your help, but it's just that my textbook doesn't seem to get into enough details for this question. I found an answer that seems correct at http://www.chemguide.co.uk/atoms/bonding/hbond.html about a third of the way down.

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Notice that each water molecule can potentially form four hydrogen bonds with surrounding water molecules. There are exactly the right numbers of delta.GIF+ 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.

In hydrogen fluoride, the problem is a shortage of hydrogens. In water, there are exactly the right number of each. Water could be considered as the "perfect" hydrogen bonded system.

 

Extracted from your link.

 

Yes that's exactly what I was trying to steer you towards.

 

There is not one factor involved but a combination of them.

 

The number of hydrogen atoms

The number of lone pairs

The geometry which is the result of hybridisation and allows attractions between lone pairs and the Hd+

 

all play their part.

Note that the 4 and 2 are the max numbers of hydrogen bonds, not all may be established.

 

Keep asking searching questions and you will go far. +1 for a good question

 

;)

Edited by studiot
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Extracted from your link.

 

Yes that's exactly what I was trying to steer you towards.

 

There is not one factor involved but a combination of them.

 

The number of hydrogen atoms

The number of lone pairs

The geometry which is the result of hybridisation and allows attractions between lone pairs and the Hd+

 

all play their part.

Note that the 4 and 2 are the max numbers of hydrogen bonds, not all may be established.

 

Keep asking searching questions and you will go far. +1 for a good question

 

;)

Thank-you, and thank-you for giving your opinion about the link; that is helpful too. :)

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