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Hydrogen bonding and Oxygen electron count ? (how)

  1. Aug 8, 2011 #1
    Sorry to butt in on your form.

    Please can someone help me with this – I’m sure it’s simple to those who know but I have been thinking about these 2 problems for ages (and reading up on every website I can find) and still no answer. I am trying to get a basic understanding of this and seem to be getting nowhere

    Water Molecules = 2 hydrogen atoms with 1 electron each. Plus 1 Oxygen atom with 8 electrons (with the 8 electrons being put – for want of a better word, into 4 pairs ;-). 2 of these pairs will form covalent bonds between the H and the O. 1 pair will be the Lone pair (to bond with another H in a dimer) and the other pair doesn’t bond. Is that right ??

    If it is right how does the hydrogen bond take place been each molecule to form a Tetrahedral shape (you'd need 4 pairs and you only have 3 ??)– is there just 1 electron needed for a hydrogen bond and 2 for a covalent bond ?? If it’s 2 where do the other 2 come from ??

    2nd (this may be a silly question but I have thinking about this for a long time ;-)
    If the angle (of the hydrogen bond) is 104.5degrees, how can 6 molecules form a hexagon with 720 degrees ??

    Thanks for your time in reading my (hopefully simple to you) questions.
  2. jcsd
  3. Aug 8, 2011 #2
    Last edited by a moderator: Apr 26, 2017
  4. Aug 8, 2011 #3
    The standard water monomer has TWO lone pairs of electrons, not one. Moreover, the bonding angles are not rigid, they flex (libration). For a very readable discourse on water structures try: Martin Chapin: Water Structures: Introduction;
    http://www.lsbu.ac.uk/water/intro.html His bibliography is worth a visit all by itself.
  5. Aug 8, 2011 #4
    But I am reading a book (by David Lind) that says (page 31) that "of the oxygen electrons, 2 electrons compose an inside shell and do not react with other electrons of other atoms ", so only six electrons form the form bonds.
    Now I'm lost ;-(
    Oh, I "may" have it.
    Do only 2 electrons from the O form the covalent bond (1 for each ???) with the H and this would leave 4 more (2 pairs) to form the hydrogen bonds with other molicules to make the Tetrahedral shape which need the O to be "linked" to 4 H

    Sorry to ask this but to my 52 year old mind this is all a little ..... well lets just say I am having fun learning

    But 6 X 104.5 = 627 which must leave a "lot" of flex to make a hexagon of 720 degrees (over 15 per angle)
    Last edited: Aug 8, 2011
  6. Aug 8, 2011 #5
    the inner shell electrons (or inside shell) do not play any part in any interactions.

    and while there are 4 electrons to form 2 lone paired electrons (not sure how you got 3), this is not how hydrogen bonding arises. you may have been thinking of lone pair donation instead.

    Hydrogen bonding is the attractive interaction of a hydrogen atom with an electronegative atom (like nitrogen, oxygen or fluorine) that comes from another molecule or chemical group.
    You may want to read this for more info: http://en.wikipedia.org/wiki/Hydrogen_bond#Hydrogen_bonds_in_water

    also, the hexagon is not a flat plane hexagon, but 3 dimension. that's how the angles are accommodated.
  7. Aug 8, 2011 #6
    Thank for that - I have just read it and it is a wonderful link that I had not found. Thanks

    But (sorry about this, I'll go away soon ;-)
    O has 8 electrons
    2 don't interreact with other atoms
    4 form lone pairs
    this leave 2 - does this mean that the bond between the O and the H (in the H2O molicule) needs only 1 electron for each bond ??
    Also I thought Hydrogen was electronegative and O was electropositive ???
    Last edited: Aug 8, 2011
  8. Aug 8, 2011 #7
    haha, no worries, just ask away.

    both hydrogen and oxygen are non-metals. so what they will form is a covalent bond between them, which is very basically the sharing of electrons between the O and H atoms. so there isn't any negative or positive to speak of (yet, as later there would be partial positive and partial negative when talking of H-bonding).

    also, electronegative and electropositive talking about something else. (related, but not quite)

    so in the case of water, it would be:
    O __H

    where each line represents 2 electrons between the atoms.

    you can also have others, like


    where 4 electrons are shared between 2 oxygen atoms

    and more links:
  9. Aug 8, 2011 #8
    Bingo! You've got it! The two hydrogen electrons complete the ring.

    My mind is over 80 years old, and I'm still having fun learning.

    Water structures are not planar. They are three-dimensional. Therefore you cannot project the true angles onto a plane. You can think of tetrahedrons and icosahedrons as tiny "bucky" balls. Chapin has some excellent illustrations on his web site, some of them animated.
  10. Aug 9, 2011 #9
    Just wanted to say a big thank you to everyone for your help.

    I have just bought this book on physics and I really want to be able to understand it - so I may be back for more clarification, hope you don't mind.:smile:

    So thanks to everyone
  11. Aug 30, 2011 #10
    Thank you both for your advice. As a result of this advice, and a private message from ### (I don't know if I'm allowed to say the name - don't want to be shouted at by the forum moderators, again :grumpy:, I have now change my study to include this text and graphic (the picture took my "a long time to draw")

    Thank you all again for your expert advice.


    Notes regarding the angles of molecular bonds in an ice (Ih) lattice.
    Almost all graphics (including mine) displaying H2O molecules will show a standard hexagon shape Fig.12; with an angle 120o between each molecule, this is a downwards view. In reality the bonds are warped; with internal angles of 109.5o Fig.13 & Fig.14. This warped formation allows molecules to bond with others to form complex sturdy lattices Fig15.
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