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Finding angle between hydrogens and oxygen in water molecule

  1. Sep 24, 2006 #1
    Hi, I have a quick question. Ignoring quantum stuff, shouldn't I be able to predict the angle between the two hydrogens in a water molecule if I model the three atoms as point charges and assume that both hydrogens have a partial charge of +D and that the oxygen has a partial charge of -2D, and that both the hydrogens are the same distance L away from the oxygen molecule? If I assume these things then assume that the net force on each point charge is 0, shouldn't I be able to get the angle as a function of D and L?

    Or is this way hopeless for predicting the angle?
  2. jcsd
  3. Sep 24, 2006 #2
    Ignoring quantum stuff (like the asymmetry of the oxygen atom).. what angle do you think you'll need so that the force from the oxygen (on the first H+) is in the opposite direction to the force from the second H+?

    Of course, this is somewhat useful, but not as accurate as you hoped. Also, it is difficult classically to explain why L > 0..
  4. Sep 25, 2006 #3
    I may be mistaken, but I believe the quantum effects make a difference large enough for a non-quantum method to produce the 'wrong' answer. Unless I'm missing something obvious, there's nothing in classical theory which would account for the angles not being 180 degrees - the obvious state for the hydrogen atoms in a classical system is directly opposite one another - which obviously isnt the case.

    Unless you're simply thinking about the method and working.

    I'm thinking this is probably a quantum electrodynamics question, which is beyond me, I'm afraid.
  5. Sep 26, 2006 #4

    Besides the fact that classical physics cannot explain the stability of atoms and molecules, I would say that it would predict an angle of 180° (by symmetry an extremum of the potential energy should be there). But you would need some artificial internal forces to avoid the molecule to collapse on itself by electrostatic forces.

    The shape of the water molecule is completely quantum mechanical property.

  6. Sep 26, 2006 #5
    This reasoning is wrong. This "common sense" analysis yields a bond angle of 180 degrees, which is wrong. The actual bond angle is 109.5 degrees (techically, the bond angle is a little lower since there are two non-bonding electron pairs around the oxygen) and you DO need QM to explain it. It's not like the water molecule is "a little bent", but close to a 180 deg bond angle....no...it's not even close to 180.

    the valence S and P orbitals of the oxygen form SP3 hybrid orbitals when it bonds with hydrogen, where each orbital is 109.5 degrees apart. The SP3 hybrid orbitals form a tetrahedral structure around the oxygen atom. Two of the hydrid orbitals covalently bond with hydrogens. For a visual image of what the atomic orbitals and the molecular orbitals look like do a google search, or look inside any intro chemistry book.

    However, for a deep understanding of why the orbitals look and behave the way they do, you need QM. Classical physics doesn't cut it.
    Last edited: Sep 26, 2006
  7. Sep 26, 2006 #6
    Yes, I also get a bond angle of pi. Just wanted to see what basic classical calculations (Coulomb's etc.) would yield. I'm still in that stage of physical understanding where the joy is limited to the concrete and tangible... I have Newton and Lagrange hammers in my mind, and I am looking everywhere for nails.
    Last edited: Sep 26, 2006
  8. Sep 26, 2006 #7
    :smile: I said the classical analysis is somewhat useful, not that it is as accurate as someone might like. Yes, it suggests that if the atoms have any separation then the angle will be 180 degrees. A huge number of molecules do have just this type of arrangement (I think CO2 is one example.. H2Be also). Classical analysis is important because it gives you the first-order estimate of what to expect.

    The second-order approximation is to consider perhaps the simplest non-trivial quantum mechanical model, the hydrogen atom. From this you see that the possible electron distributions aren't all symmetric and, by assuming oxygen's bonding electrons behave according to corresponding levels in that hydrogen model, your second-order estimate is that H2O-like bonds will have the 109.5 degree angle.

    But after you satisfy yourself with higher order estimates, or more detailed simulations, you can perform a real experiment on H2O and measure an angle of perhaps 104.50 degrees. Yes, "180" is wrong. But "109.5" (or even "109.47" or whatever) is equally wrong. For that matter, if not for implying the uncertainty range, just saying "104.50" would also be equally wrong (as more precise measurements will show). In science (as should be emphasized in 1st year physics labs) you cannot compare "closeness" of numbers (two measurements can only be consistant or inconsistant). I also think it's meaningless to perform an experiment without some prior estimate (otherwise why attempt that particular experiment?), and looking at the cost/benefit of higher order estimates, I still think the first order one is generally most useful.
  9. Sep 26, 2006 #8
    Yes, molecules like CO2 are linear (180 degree bond angle) because the central atom (carbon) is SP1 hybridized. However, water is nothing like CO2, and the classical picture completely breaks down. I hardly call 180 degrees a good common sense approximation of the water molecule's bond angle.

    Short answer...you need QM.

    Your other points are irrelevent. I understand that 109.5 is an approximate, and the actual bond angle is around 104.5. My point was that trying to explain the bond angle of water with the classical theory is incorrect.
    Last edited: Sep 26, 2006
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