Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hydrogen bonding for Fluorine

  1. Jul 15, 2011 #1
    Every water molecule is H-bonded with up to four other molecules (two through its two lone pairs, and two through its two hydrogen atoms.

    These 4 intermolecular hydrogen bonds are what contribute to water's high boiling point..

    My textbook compares this with Fluorine, which it says can only form two bonds because it has 3 lone pairs on the F atom but only one H atom....


    This makes no sense to me. Why would Fluorine, which has 3 lone pairs as opposed to water's 2 lone pairs, not form just as many bonds than water?

    Can't each lone pair attract it's own hydrogen just like each lone pair of water attract's each own hydrogen?
  2. jcsd
  3. Jul 15, 2011 #2
    when there are lone pairs on the central atom, the central atom develops a delta negative charge, therefore to keep the charge of the molecule neutral, the other atoms develop a delta positive charge.

    in water, every hydrogen atom develops a delta positive charge and forms a hydrogen bond with the oxygen atom in another molecule (which has a delta negative charge), so there are 4 h-bonds in 1 molecule.

    in HF, fluorine develops a delta negative charge (which is obviously greater than the oxygen atom has in water) and hydrogen develops a delta positive (which is greater than the hydrogen atom in water, due to the greater delta negative charge) charge.

    in water, the delta positive charges acquired by the hydrogen atoms are shared equally. so, for example, lets say the delta positive charge acquired by 1 hydrogen is +c. therefore the oxygen atom has 2(-c) charge.
    so think about it, when 2 hydrogen atoms form a bond with water, this happens:
    2(-c) + (+c) + (+c) = 0

    lets use the same logic in the case of HF. fluorine acquires a delta negative charge of 3(-c). so, hydrogen acquires a delta positive charge of 3(+c). so, when the fluorine atom forms a h-bond with a hydrogen atom:
    3(-c) + 3(+c) = 0

    according to what you say,
    3(-c) + 3(+c) + 3(+c) + 3(+c) = 6(+c) [hence unstable]
  4. Jul 15, 2011 #3


    User Avatar
    Science Advisor

    Your description is qualitatively reasonable, and I think the end result is the correct one. However, the physical basis of your reasoning can't be the whole story, since H-bonding interactions are not purely electrostatic in nature, but have partial covalent character as well, and thus are directional.

    Having said that, your explanation does capture the essence of the answer, which is that while an HF molecule can in principle *accept* 3 H-bonds, it can only donate one. Thus in pure HF, each molecule is (on average) donating one H-bond and accepting one H-bond. At any given time, a particular HF molecule might be accepting more than one H-bond, but over time, the average value of accepted H-bonds must be no larger than one.

    On the other hand, water can donate two, and accept two ... so it is natural to find it with a tetrahedral arrangement of H-bonds around it (two donated and two accepted).
  5. Jul 15, 2011 #4
    Yes I think spatial is part of the key, but hydrogen bonding does occur in HF.

    The two lone pairs in water are on the opposite side to the two bonded hydrogens.

    This cannot happen if there are three lone pairs.

    Further the ratio of hydrogen to oxygen is 2:1 whereas the ratio of hydrogen to flourine is 1:1

    2:1 allows cross linking - 1:1 limits the linking to chains

    See this extract from WP


    Attached Files:

    Last edited: Jul 15, 2011
  6. Jul 16, 2011 #5
    (I learnt chemical bonding recently, I don't know whether my diagram is correct)
    from the diagram, it's pretty clear that 3 hydrogen bonds are not going to form at right angles (because it is obvious that while forming a covalent bond with the hydrogen atom, the pair of shared electrons will be farther from the nucleus of fluorine atom as compared to the lone pairs. stronger repulsion between the lone pairs as compared to the shared electrons will not allow the formation of a right angle).

    therefore, accepting 3 and donating 1 cannot be possible.

    Attached Files:

  7. Jul 16, 2011 #6


    User Avatar
    Science Advisor

    Well, in this case your reasoning is not quite right. There is nothing wrong in principle with accepting 3 and donating 1. You are right that the bond angles will not be 90º ... what you will get is a tetrahedral arrangement of H-atoms around the F-atom. However, as I said, while this may represent an instantaneous configuration of molecules in pure HF, it is much more probable to find each HF molecule donating 1 and accepting one H-bond.
  8. Jul 16, 2011 #7
    What does hydrogen bonding have to do with accepting/donating electrons in bonding either in pairing or full transfer?
  9. Jul 16, 2011 #8


    User Avatar
    Science Advisor

    Sorry, but what are you talking about? This whole thread is about H-bonding ...

    [EDIT] Just to clarify .. H-bonds are formed between H-atoms and electron pairs, therefore, the number of H-bonds an atom can accept (in principle) is equal to the number of lone pairs that it has. There are other kinds of H-bonds which are much weaker, but can still be observed for low temperature gas-phase clusters. For example, HF can form an H-bond with the pi-electrons of acetylene or benzene. As Audioflux said, the H-atom has a partial positive charge, therefore when forming intermolecular interactions, it "wants" to associate itself with regions of (relatively) high electron-density. Lone pairs are the best choice, but pi-bonds will do in a pinch.
    Last edited: Jul 16, 2011
  10. Jul 16, 2011 #9
    I forgot that water has a bent structure :(
    anyway, I kept thinking that a molecule with non-right angled bonding could not form a crystal structure, my bad.
  11. Jul 17, 2011 #10
    I had always understood that the basis of hydrogen bonding was interaction between the nonbonding electrons (ie the lone pairs) and protons (ie hydrogen nuclei) and that the bonding electrons took no part in this process.

    I thought the thread was lurching towards the opposite. My apologies to any who did not mean this.
    Last edited: Jul 17, 2011
  12. Jul 17, 2011 #11


    User Avatar
    Science Advisor

    Ok .. I guess I see where the confusion arose. However, in the community of people who study H-bonding, it is common to speak of "donors" of H-bonds (the molecules to which the H-atom in the H-bond is covalently attached), and "acceptors" of H-bonds (the molecules which "own" the lone pairs to which the H-atoms bond). There is no implication the the electron pairs in questions are somehow transferred between molecules.

    However, it is worth noting that the H-bonds *can* be transferred between molecules. This is the reason why the diffusion of protons through water is about 6 orders of magnitude faster than diffusion of the lightest heavy ions. Protons diffuse not by the physical motion of nuclei, but rather by the exchange of H-bonds by the Grotthuis mechanism (http://en.wikipedia.org/wiki/Grotthuss_mechanism). This has been directly observed experimentally by Erik Nibbering at the Max Born Institute, and has been modelled theoretically. You can find better animations for proton-hopping for larger clusters than the one shown on the wiki if you are interested.
  13. Jul 17, 2011 #12
    Not convinced this is a good nomnenclature especially for beginners, who have 'just learned bonding',
    But thank you for the explanation.

    I also think that the simple point that in water there are twice as many protons available for H-bonding action as compared to hydrogen flouride needs bringing out.

    go well
  14. Jul 17, 2011 #13


    User Avatar
    Science Advisor

    I disagree, particularly since I was always careful to be explicit that I was talking about donating and accepting H-bonds in my posts. Furthermore, it is the most clear nomenclature for addressing the question that was asked.

    You are right that the fact water has two H-atoms is significant (I pointed this out in my first post) ... that means that it can donate two H-bonds, while the fact that it has two electron pairs means that it can accept two H-bonds (see how useful that nomenclature is?).

    The equivalence between the number of potential H-bonds that can be accepted or donated by water molecules leads to the extraordinarily high degree of H-bonding observed in the liquid and solid phases. This in turn is the primary feature that makes water such a unique solvent, contributing to its high surface-tension, and the fact that it has such a high boiling point relative to other molecules of comparable molecular weights (like HF), and also leads to the solid phase (hexagonal ice) having a lower density than the liquid.
  15. Jul 17, 2011 #14
    I'm sorry I have to disagree with this statement as the basis of the answer.

    There are two conditions for the creation of a hydrogen bond and your statement fulfils only one of them.

    The OP asked why (in your terms) HF has 3 acceptor sites but cannot or does not enter into so many H-bonds as water which only has 2

    The answer lies in the second condition viz there not only has to be the acceptor site but there also have to be sufficient protons to enter them.

    So I repeat the key is that there are twice as many protons in water as HF.
  16. Jul 17, 2011 #15
    In fact, thinking about your statement and the octet

    There is more to the 'equivalence' that I first thought.

    Considering some general molecule [itex]\Omega[/itex]-Hn it can have

    0,4 ; 1,3 ; 2,2 ; 3,1 ; 4,0 lone pairs and covalently bonded protons

    of your acceptor and donor sites.

    So in cases 1 and 5 you cannot have any H bonding since you either have no pairs or no protons.

    In cases 2 and 4 you can have only one H bond since you either have only one pair or one proton,

    But case 3 (oxygen) you can have 2 since you have 2 of each. So this is the maximum.

    But it would still be the maximum if you were able to find an [itex]\Omega[/itex] with only 1 pair and one covalent proton.
  17. Jul 17, 2011 #16


    User Avatar
    Science Advisor

    I have no idea what distinction you are trying to draw .. it appears that you just want to argue semantics. Everything in your post was mentioned in my own post. Read my post again more carefully. Furthermore, your answer just repeats the answer I already gave in an earlier post, so I have no idea what you are getting at.
  18. Jul 17, 2011 #17
    So why are you complaining about someone you say is agreeing with you?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook