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What 'wills' certain atoms to covalently form?

  1. Sep 17, 2012 #1
    Let me explain. I recently wanted to work on self improvement and have started studying different things. I'm a mess... but I'm working on it.

    So what I really mean is... what is it that make atoms come together? Like something as simple as water. Why do two hydrogen atoms somehow 'decide' that they're meant for an oxygen atom (I know it's not a will, but I don't know how else to explain it)?

    Hope that makes sense. Maybe the question of why does this stuff work is too broad, but I gotta start somewhere. Thanks.
     
  2. jcsd
  3. Sep 17, 2012 #2
    This is very complicated. Chemical bonding is a hard topic and physicists nowadays don't like to talk about it so they leave it to chemists =)

    Here's my explanation which may be helpful, assuming you know a bit of quantum mechanics... it gets kind of long though so please be patient.

    Lets not start with H2O. That's pretty complicated. It has 10 electrons and 3 nuclei. Instead lets start off with H2. But that's still too complicated since it has 2 electrons that repel each other. Indeed, lets go with H2+ - 2 protons and an electron. That's the simplest possible molecule.

    First we apply the Born Oppenheimer Approximation - the nuclei are stationary. This seems obvious but its subtly not so - the electrons are after all negatively charged and should pull the positively charged hydrogen nuclei right? But since the nuclei are really big and heavy compared to the electron (2000x), we'll just say they don't move. In terms of quantum mechanics, we can separate the molecular Hamiltonian into a nuclear Hamiltonian and an electronic Hamiltonian.

    The dynamics of the molecule is determined by the electron motion alone, after all nuclei don't participate in chemical bonding. We write this as the electronic Hamiltonian He = -h^2/2m * ∇^2 - e^2/rA - 2^2/rB + e^2/R

    where ∇^2 is the Laplacian operator, R is the internuclear distance, rA is the distance between nucleus A and the electron, and rB is the distance between nucleus B and the electron.

    So now we write the Schrodinger equation: He ψ = Ee ψ where ψ is the electronic wavefunction. This can be solved, but its really complicated, so we use a trick to solve it approximately.

    We know for a fact that if R increases to infinity, the "molecule" actually becomes a hydrogen atom + a lone proton. So lets say the electron was closer to nucleus A.

    Then it would have the same wavefunction as a ground state hydrogenic electron:

    ψA = constants*e^rA/a0 where a0 is the Bohr radius and the constants are important but ultimately just constants so they don't impact the dynamics of the system.

    If the electron was closer to B, similarly:

    ψB = constants*e^rB/a0

    So as a first approximation lets just say that the molecular wavefunction is the summation of these 2 atomic wavefunctions: ψM = C1*ψA + C2*ψB where C1 and C2 are just random constants.

    Now, we could use the variational method... but that's too mathematical. Instead, lets use a trick: we know that the nuclei are identical, so the electron distribution must be symmetric! The only way for this to happen is the absolute values of C1 and C2 are the same, otherwise the electron would spend more time at one of the protons despite both protons being identical! So |C1| = |C2|.

    The probability distribution for the electrons are therefore (ψM)^2 = C1^2*ψA^2 + C2^2*ψB^2 + 2C1*C2*ψAψB. Note here that C1 and C2 are squared. That means it could be either positive OR negative and yield the same answer, so C1 actually is = +/- C2.

    So the molecular wavefunction is actually 2 wavefunctions:

    ψM+ = C1(ψA + ψB) and ψM- = C1(ψA - ψB) corresponding to the 2 cases I discussed above.

    Skipping some math - you should define an overlap integral S = ∫(ψAψB)dr ... yeah forget about it, thats not too important and it ends up being constants anyways.

    When you take the 2 wavefunctions and plot them on a graph, you'll find this:

    For ψM+ the wavefunction is very low far away from both nuclei and hit a peak at the nuclei. Between the nuclei, there is a residual "bridge" of wavefunction, so when ψM+ is squared, there is a nonzero electron probability distribution between the nuclei. That's the bonding orbital.

    For ψM- its still very low far away from both nuclei, but since the B nucleus has a negative coefficient, the wavefunction there is negative. Note, because the wavefunction at A is positive, and at B it is positive, it had to cross ZERO somewhere between A and B. Therefore, when you square it, there is a point in the graph that is ZERO and has NO ELECTRONS there. That is the anti-bonding orbital.

    Now note that we had 2 ground state atomic orbitals and it transformed into 2 molecular orbitals, one bonding and one anti bonding. Anti-bonding orbitals have higher energies than bonding orbitals. Each orbital can hold a maximum of 2 electrons, due to the Pauli Exclusion Principle.

    So we have 1 electron. It goes into the bonding orbital. That's great. But why do we see H2? Well, H2 has 2 electrons in the bonding orbital, which makes the chemical bond more stable, but none in the anti-bonding orbital. If we add a 3rd electron, it destabilizes the chemical bond a bit. But since there's electrostatic repulsion in H2-, the 3rd electron can never be added, and since there's electrons just floating around, in practice H2+ always will get the 2nd electron, only the most stable configuration is seen in nature, though H2+ can be produced artificially.

    So in water the same thing applies, oxygen has 2 free "electron slots" in the p2 orbital, the 2 hydrogens donate an electron to oxygen, and since the anti-bonding molecular orbitals aren't filled, and no more bonding orbitals can be filled, H2O is stable.

    References if you want to look further: "Quantum Chemistry" Ira Levine
     
    Last edited: Sep 17, 2012
  4. Sep 17, 2012 #3

    DrDu

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    Let me try a somewhat simpler explanation. Bonding is a quantum mechanical effect and a rough understanding can be gained from the Heisenberg uncertainty relation [itex]\Delta x \Delta p >h[/itex]. That means if you enclose an electron into a "box" of size Δx, its mean absolute momentum will be about [itex] \Delta p\approx h/\Delta x[/itex] and its kinetic energy [itex] E_\mathrm{kin}\approx (\Delta p)^2/2m\approx h^2/(2m (\Delta x)^2)[/itex].
    If you think about the size of the box as being due to the effective range of the attractive nuclear potential in an atom, this effective range can be increased for an electron if two atoms are brought close to each other so that their nuclear potentials overlapp. This increased size of the box will lead to a reduction of kinetic energy and thus to bonding. However, due to the Pauli principle, a bond cannot contain more than two electrons.
     
  5. Sep 17, 2012 #4
    Wow, this is quite a bit to work with, but it's what I wanted. Thanks! I'll get back to you both with, surprise, more questions. What a handy forum.
     
  6. Sep 17, 2012 #5
    Now both of those explanations looked like physicists' explanations. Let me try a chemist's explanation on you.

    First let's not go all the way to H2+. An ordinary hydrogen molecule will do just fine.

    Now, let us also forget quantum mechanics for the minute with the understanding that the almost classical explanation I am trying to give will also work in the slightly more complicated maths of quantum theory.

    A hydrogen atom has a single proton and a single electron. Why do those two particles stick together? Because the positively charged proton has an electrostatic attraction for the negatively charged electron. The attraction can be seen either as a force bringing the two particles together or as a potential energy advantage that the two particles get by being close together -- a lower potential energy situation.

    In a hydrogen molecule, four particles come together -- two protons and two electrons. The electrons are light and fast-moving. The protons are heavier and much more sluggish in any movements they make.

    In two separate hydrogen atoms, there are two forces of attraction. If the atoms come together, there are four forces of attraction (electron-proton interactions) and two forces of repulsion (an electron-electron interaction, and a proton-proton interaction).

    If the electrons a) keep out of one another's way as they zip around, and b) spend most of their time in the space that is more or less in between the two protons, we will arrive at a situation where each of the attractive forces is much larger in magnitude than either of the two repulsive forces, and the sum of four large attractive interactions and two small repulsive interactions will be more attractive than the sum of two ordinary attractive interactions.

    In this way, a hydrogen molecule arrangement obtains an electrostatic potential energy advantage over two separate hydrogen atoms -- meaning, in other words, that a chemical bond forms.
     
  7. Sep 17, 2012 #6
    Your explanation is better. I just parroted what my physical chemistry teacher taught me.

    The semiclassical model actually works really nicely: it accounts for rotational and vibrational spectra, chemical bonding and alot of important stuff.
     
  8. Sep 18, 2012 #7

    chemisttree

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    hmmm... all this time I thought it was MY will!

    Here's one for you. If two hydrogen atoms meet but nobody is around to observe them, do they bond?
     
  9. Sep 18, 2012 #8
    This is now becoming a philosophical question, that has no real place in a scientific forum, except to indicate the following:

    Because of certain phenomena that arise in particle physics and cosmology, most physics-based philosophers feel compelled to adopt a philosophy that is described as "anti-realist", that is, that does not believe that there is a world that can exist and have properties independent of any observer. Many working physicists who are not well versed in philosophical debates find themselves pushed into a discredited philosophical position designated "positivism".

    Most chemists, on the other hand, believe (on the basis of the observations they make and the science that they do) that there is an external world that is objective and independent of observers. They adopt a position known as "naïve realism". If they read any philosophical debates they have to concede that this position is also flawed, and they shift either to other versions of realism, or to anti-realist positions like that of van Fraassen
     
  10. Sep 18, 2012 #9

    DrDu

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    I am a chemist, you?

    Alas, this explanation is not quite correct. Increased electron density between the nuclei means reduced density near the nuclei, which would result in increased potential energy.
    The driving force behind bond formation is reduction of kinetic energy.
    If you really want to know how bonding works you really ought to read:
    http://onlinelibrary.wiley.com/doi/10.1002/anie.197305461/abstract
     
  11. Sep 18, 2012 #10
    very true. i didn't think about that clearly and should not have agreed. i still have much learning to do.

    increased electron density between nuclei reduces that near the nuclei which is incorrect because if the internuclear distance R increases, then in that situation by symmetry the electrons will stay in the middle which conflicts with the atomic orbital approximation, and the atomic orbital approximation is correct since if R diverges it will indeed be proton + neutral hydrogen atom.

    however regarding your article, i was taught in physical chemistry that it was symmetry effects that created the chemical bond. is this incorrect? also, in classical mechanics, there's a gravitational analog for the potential well in molecules, so couldn't the covalent bond also be partially explained that way?
     
    Last edited: Sep 18, 2012
  12. Sep 18, 2012 #11
    I am also a chemist. So l think you are misinterpreting the results of this Angewandte Chemie article, and that my original formulation is correct. So let me translate my post into quantum mechanical terms:

    Suppose that two hydrogen atoms start to approach one another. The initial effect will be that the electron on one of the atoms "sees" the attractive field of the other nucleus and the repulsive field of the other electron. There will also be another repulsive effect because of the operation of the Pauli exclusion principle and the fact that the two electrons are starting to intrude on each other's space.(***EDIT -- no there will not be such a term because the two electrons are in opposite spin orientations***). Meanwhile, because the geometry is 3-dimensional rather than 1-dimensional, the electrons can move some of their density towards the other nucleus and away from the internuclear line in such a way as -- not to diminish the attractive potential of each in the field of "its own" nucleus more than it increases the attractive potential in the field of "the other" nucleus, and to ensure that the repulsive field between the two electron distributions is no greater than any of the four attractive fields.

    It is also the case that in quantum theory kinetic and potential energy are not in eigenstates for fixed total energy "stationary states" of the molecule, so that attempts to distinguish between potential and kinetic effects are somewhat illusory.

    However, there are readily available average or "expectation" values for individual kinetic and potential energy terms arising from various factors, and if there is a serious argument that I should be looking at kinetic rather than potential factors, then I will accept it for purposes of "the truth". But the art of theoretical chemistry, and quantum chemistry in particular, is in drawing caricatures that can be extended by analogy to more complicated systems, and that provide real explanations without complicated mathematics, and I certainly will not withdraw from my model explanation as one that will provide a valid understanding of the underlying principles of the OP's question.
     
  13. Sep 18, 2012 #12

    DrDu

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    Away from the internuclear line? In molecules, the density is increased on the line joining the two atoms. I still don't get your picture.
     
  14. Sep 18, 2012 #13
    All right then -- to analyze this in simple terms, let me retreat to the hydrogen molecular ion, and let us consider it in terms of two stationary protons and a single mobile electron.

    For a hydrogen atom

    EH = <ψ(1s)|H|ψ(1s)> -- exact value of total energy
    <V> = <ψ(1s)|V|ψ(1s)> = 2 * EH -- average value of potential energy via virial theorem
    <T> = – EH -- via virial theorem or subtraction

    Suppose, then, that we take ø = ψ1(1s) + ψ2(1s) as a trial wavefunction for the molecular ion when the nuclei have separation r.

    Then

    <V> = [itex]\frac{4*E_{H} + 2 V_{r}}{2 + 2 S_{r}}[/itex] + [itex]\frac{e^{2}}{4 π \epsilon r}[/itex]

    <V> = [itex]\frac{2*E_{H} + V_{r}}{1 + S_{r}}[/itex] + [itex]\frac{e^{2}}{4 π \epsilon r}[/itex]

    It is quite clear that for a distributed charge, the potential function is steeper for the closer part of the distribution, and therefore that the electrostatic attraction potential for the nearside of the distribution will dominate to make the attraction potential greater in magnitude than the fixed point charge repulsion potential between the two nuclei. It is also quite clear that the increase in the denominator of the attractive term can be made sufficiently small at sufficiently large R that this factor will not reverse the trend embodied in the previous inequality.

    Therefore the expectation value of potential energy is lower -- more negative -- for a simple linear sum of hydrogen wavefunctions centred on the two nuclei than for the isolated hydrogen atom, and these functions are not even the optimised stationary wavefunctions for the molecular ion.

    In fact if we use another model that the physicists love, and consider the He+ ion, and start to separate its two protons, we can easily see that the expectation value of potential energy of the symmetric sum of 1s atomic orbitals, even though it is increasing from that of the helium ion, will start to fall below that of the hydrogen atom when the nuclei obtain a separation somewhat greater than the average proton/electron separation
     
  15. Sep 18, 2012 #14

    DrDu

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    Although I am trying hard to do so and without bad will, I don't understand your explanation. Which distribution? What is the closer part and what is the nearside of it?
    I also don't see how the denominator can be made sufficiently small - after all, it can't get smaller than 1.
     
  16. Sep 18, 2012 #15
    Do not worry at this stage about ill will. The misunderstanding is entirely due to my trying to shortcut things rather than present complete working, and to my poor expression.

    The "distribution" I was referring to is the distribution of electron density in atomic orbital ψ1(1s) when considering an attractive interaction between this orbital and proton(2) -- that is the nucleus on which it is not centred. A part of this distribution will be closer to proton(2) than the internuclear distance, and for this part of the distribution the attractive interaction will be greater than the internuclear repulsion. The part of the distribution farther from proton(2) will have an attractive interaction smaller than the internuclear repulsion. But because the potential gradient is steeper when closer to proton(2) the integrated interaction will have the closer part outweigh the farther part, and therefore the attraction between the distributed electron wavefunction ψ1(1s) and proton(2) will have a larger magnitude than that of the internuclear repulsion.

    I did not really mean to say a sufficiently small denominator, but rather a denominator sufficiently close to 1, or a sufficiently small overlap (non-orthogonality) between the two functions ψ1(1s) and ψ2(1s). My apologies.
     
  17. Sep 19, 2012 #16

    DrDu

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    Ok, I'll try to clean up this a little bit:
    The trial wavefunction is
    [tex]\psi=\frac{1}{\sqrt{2}(1+S)}(|1\rangle +|2\rangle)[/tex]
    [tex]H=H^0_1+V_2=H^0_2+V_1[/tex]
    where [itex]H^0_1[/itex] is the Hamiltonian of hydrogen atom 1 and [itex]V_2[/itex] is the potential of atom 2 and vice versa.
    Then the energy of the [itex] H_2^+[/itex] molecule is
    [tex]
    E=\frac{E_\mathrm{H}+\langle 1| V_2|1 \rangle +\langle 1|V_2|2\rangle}{1+S}+\frac{e^2}{4\pi \epsilon_0 R}
    [/tex]
    The next step is to derive the R-dependence of all terms which involves calculation of some integrals.
    Unless maybe for a genius it is hard to say beforehand which terms will dominate at what distance.
     
  18. Sep 19, 2012 #17
    Au contraire, mon ami!

    It is quite clear, without calculating any integrals, that <1|V2|1> must be greater than the internuclear repulsion term +e2/4πϵ0R at all distances comparable to the size of the atomic wavefunctions.

    The integral can be separately considered in two regions of space. For the region where |1> is farther from nucleus 2 than nucleus 1 is, the repulsion will be greater than the attractive contribution; in the other region where |1> is closer to nucleus 2, the repulsion will be less than the attractive contribution. The effect of the latter region of space will dominate over the effect of the former region of space because the gradient is steeper there. In other words, the advantage obtained by integration over the latter region must more than outweigh the disadvantage of having to integrate over the former region. This is the point I was trying to make in my previous post.
     
  19. Sep 19, 2012 #18

    DrDu

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    Good!
    Looks as if you could squeeze your argument into a hard inequality!
     
  20. Sep 19, 2012 #19

    DrDu

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    [tex]\langle 1| V_2|1 \rangle =-\frac{e^2}{4\pi \epsilon_0 R}[/tex]
    [tex] \langle 1| V_2|2\rangle=E_\mathrm{H}S-\langle 1|T|2\rangle [/tex]
    [tex] E_\mathrm{binding}=E-E_\mathrm{H}=\frac{-\langle 1|T|2\rangle}{1+S}+S \frac{e^2}{4\pi \epsilon_0 R(1+S)}[/tex]
     
  21. Sep 20, 2012 #20

    chemisttree

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    You guys have completely gone off the rails.
     
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