acetylmandarin

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Oct 20, 2014
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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?
 
May 13, 2016
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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.
 

JPS398

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Dec 10, 2013
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So H2O is still the highest boiling point here? Even if HF Is the most EN??
 
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acetylmandarin

acetylmandarin

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Oct 20, 2014
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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?
 
May 13, 2016
77
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Pre-Medical
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.
 
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acetylmandarin

acetylmandarin

5+ Year Member
Oct 20, 2014
1,109
213
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Pre-Health (Field Undecided)
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
 
Oct 30, 2017
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Pre-Health (Field Undecided)
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" ?