Comparing H bonding in H2O, HF, NH3, and CH4

[deleted647690]

So H2O has the highest boiling point out of these four compounds because it can potentially form 4 H bonds. Why does HF have a higher BP than NH3? Doesn't NH3 form more H bonds than HF? Same goes for CH4. Is it due to the increased EN of fluorine over N and subsequently over C? So that the H bond formed by HF is stronger than NH3's bonds which are stronger than CH4's?

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[kryptonxenon]

HF and NH3 can both form 2 Hydrogen bonds. But like you mentioned since Fluorine is more EN than Nitrogen, HF should have stronger IMFs and thus a higher boiling point.

CH4 actually can't form any hydrogen bonds. Remember that hydrogen bonding requires a H bonded to an EN atom like F, O, or N.

 

[deleted585480]

So H2O is still the highest boiling point here? Even if HF Is the most EN??

 

[kryptonxenon]

So H2O is still the highest boiling point here? Even if HF Is the most EN??

Yes because it forms more hydrogen bonds than HF does.

 

[goodtime]

Yes because it forms more hydrogen bonds than HF does.

why does it form more hydrogen bonds, i thought i would only form 2?

 

[deleted647690]

HF and NH3 can both form 2 Hydrogen bonds. But like you mentioned since Fluorine is more EN than Nitrogen, HF should have stronger IMFs and thus a higher boiling point.

CH4 actually can't form any hydrogen bonds. Remember that hydrogen bonding requires a H bonded to an EN atom like F, O, or N.

So two of the hydrogens on NH3 are not considered H bonded to N?

 

[kryptonxenon]

So two of the hydrogens on NH3 are not considered H bonded to N?

Remember IMFs are between 2 different molecules of NH3 and the hydrogen bonds we are talking about are between 2 different molecules. Here's a trick I learned in my TPR class to calculate the number of hydrogen bonds a compound can form:
1. Identify number of hydrogen and lone pairs
2. To calculate the number of possible hydrogen bonds take the lowest number from step 1 and multiply this by 2.

 

[deleted647690]

Remember IMFs are between 2 different molecules of NH3 and the hydrogen bonds we are talking about are between 2 different molecules. Here's a trick I learned in my TPR class to calculate the number of hydrogen bonds a compound can form:
1. Identify number of hydrogen and lone pairs
2. To calculate the number of possible hydrogen bonds take the lowest number from step 1 and multiply this by 2.

Ah, I think I forgot what an H bond is. haha whoops

 

[Songyeon 17]

Remember IMFs are between 2 different molecules of NH3 and the hydrogen bonds we are talking about are between 2 different molecules. Here's a trick I learned in my TPR class to calculate the number of hydrogen bonds a compound can form:
1. Identify number of hydrogen and lone pairs
2. To calculate the number of possible hydrogen bonds take the lowest number from step 1 and multiply this by 2.

What do you mean by "take the lowest number from step 1" ?