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Water has the formula $\ce{H2O}$ and we can draw a Lewis structure with two lone pairs on the central oxygen. As a physics student and not a chemist, I think to myself, "Okay, there are two lone pairs, they will repel each other" and so we arrive at the conclusion of bent shape.

The issue in my mind is that I don't see why the lone electron pairs have to exist on the same side of the atom. Also, wouldn't the Schrödinger equation provide an equally plausible structure for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or on the bottom of the oxygen in the Lewis structure)? If that were true, then there would be a resonance structure between the two states and we would get a linear geometry. Clearly I'm running around in circles here, please someone enlighten me!

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  • $\begingroup$ The lone pairs repel each other, but you should also throw the bonding electrons of the O-H bonds into the mix if you're approaching the problem from just electron repulsion. $\endgroup$ – Zhe Apr 14 at 13:36
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    $\begingroup$ You must account for repulsion between bonding electrons and non bonding electrons too (Edit: just as @zhe said). Therefore, each pair are repelling every other pair. So, every pair of electrons must have farthest possible distance from each other, which results in tetrahedral shape. And because lone pair repulsions are greater, it is distorted tetrahedral. $\endgroup$ – B.Anshuman Apr 14 at 13:55
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    $\begingroup$ You could have a look at the answers here chemistry.stackexchange.com/questions/14981/… which deal with Walsh diagrams. $\endgroup$ – porphyrin Apr 14 at 14:50
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    $\begingroup$ Does this answer your question? Are the lone pairs in water equivalent? $\endgroup$ – Mithoron Apr 14 at 16:37
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    $\begingroup$ a minor side point on terminology - in chemistry the term "resonance structures" is not used to describe structures in which atoms occupy different spatial positions (such as your example of water molecules with H atoms on opposite sides from each other). Resonance structures are electronic states. Crucially, molecules do not alternate between resonance structures. All resonance structures simultaneously contribute to the single electronic structure of the molecule. $\endgroup$ – Andrew Apr 14 at 17:52
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I mean, there is a time and place for VSEPR, and this is probably as good a time as any, because all beginning chemistry students go through it. The actual model has already been explained multiple times, so I will only briefly say that according to this theory, there are four pairs of electrons around the central oxygen. In order to minimise electron-electron repulsions, these pairs adopt a tetrahedral arrangement around the oxygen. It does not matter which two are lone pairs and which two are connected to hydrogen atoms; the resulting shape is always bent.

What's worth bearing in mind (and hasn't been explained very carefully so far) is that VSEPR is a model that chemists use to predict the shape of a molecule. The truth is that there is no real way to predict the shape of a molecule, apart from solving the Schrodinger equation, which is not analytically possible for water. Everything else is an approximation to the truth. Some of these approximations are pretty accurate, such as the use of density functional theory. Some of them are extremely crude, and VSEPR falls into this category: it essentially treats electrons as classical point charges, and seeks to minimise the electrostatic repulsion between these point charges. As a physics student you should know better than to do this. Thus, while it predicts the correct result in this case, it is more in spite of the model rather than because of the model. And you should not be surprised to hear that in some slightly more complicated cases, VSEPR can predict entirely wrong outcomes.

As you learn more chemistry you will find that there are increasingly sophisticated ways of explaining molecular geometry. Most revolve around molecular orbital theory. For a qualitative method, you have Walsh diagrams which have been explained at Why does bond angle decrease in the order H2O, H2S, H2Se?. For a more rigorous method you would likely have to run some quantum chemical computations, e.g. Are the lone pairs in water equivalent?. Of course, the drawback of this is that it becomes more and more difficult to extract true chemical understanding from the numbers. Although it should also be said that you cannot extract any true chemical understanding from the VSEPR model.

What interests me more is the followup question:

Also, wouldn't the Schrödinger equation provide an equally plausible structure for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or on the bottom of the oxygen in the Lewis structure)?

Because the Hamiltonian of the water molecule is invariant upon rotation, this means that indeed, any orientation of the water molecule is equally likely. However, this only refers to the orientation of the water molecule as a whole. It does not say anything about the internal degrees of freedom, such as the bond angle.

In the absence of any external force, the molecule is free to bend in whichever direction it likes, and most water molecules indeed do do this as they float through space or swim in a lake. But it will always be bent.

If that were true, then there would be a resonance structure between the two states and we would get a linear geometry.

If you were to think of a single particle in a double-well potential, say something with

$$V = \begin{cases} \infty & x < -b \\ 0 & -b \le x \le -a \\ \infty & -a < x < a \\ 0 & a \le x \le b \\ \infty & x > b \end{cases}$$

then because of the symmetry of your system, in every eigenstate of your system, the expectation value of $x$ would be $\langle x \rangle = 0$. This is quite similar to your argument. In the case of water, let's set the oxygen nucleus to be at the origin. Because it can point either up or down, the expectation value of the hydrogen nucleus position along the up-down axis would be exactly level with the oxygen atom, i.e. 0. In fact, don't stop there: it can point to the left or the right, and to the front or the back. So the hydrogen nucleus has a position expectation value of exactly $(0, 0, 0)$, i.e. right inside the oxygen nucleus.

Does that mean it's actually there, though? In our contrived double-well system, it's patently impossible for the particle to be at $x = 0$, because $V = \infty$ there. If you were to measure its position, you would never find it at $x = 0$; you would only find it in the left-hand side $[-b, -a]$, or the right-hand side $[a,b]$. Just because the particle has an expectation value of $\langle x \rangle = 0$ does not mean that it is physically there, or that $x = 0$ is somehow its equilibrium state. You're confusing an expectation value with a genuine eigenstate (which is what a resonance structure is).

In exactly the same way, if you ever were to measure the properties of water (and bear in mind that practically every interaction with a water molecule is, in effect, a measurement), we would find that it is indeed always bent.

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  • $\begingroup$ "Solving the Schrödinger equation" is of course also just an approximation to the truth. In a very basic sense one could even argue that its one that is not in principle any better than Lewis + VSEPR. It is in fact a misunderstanding to imply a truth behind a model. This is in my point of view the cause of almost all missunderstandings in Theiretical Chemistry. A model shall predict the phenomena. It does not make sense to say a chemical model (that is a" theory" really) is wrong. A theory is always correct. The question is only its range of validity. $\endgroup$ – Rudi_Birnbaum Apr 15 at 17:38
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    $\begingroup$ @Rudi_Birnbaum "A theory is always correct." -- I have a theory. My theory supposes that my theory is incorrect. $\endgroup$ – LordStryker Apr 17 at 18:35
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    $\begingroup$ I like your placing of ideas into a spectrum of accuracy. However, your discussion of the "good" approximations are slightly off. DFT, as a theory, is exact. It is the approximation of the unknown universal functional that makes its implementation inexact. Also, "solving the Schrödinger equation" and DFT are typically associated with electronic structure theory, unless you are treating atoms quantum mechanically as well, in which case you can only predict the expectation value of position. Thus, a pin-point description of atomic locations necessitates an approximation (Born-Oppenheimer). $\endgroup$ – jezzo Apr 17 at 19:53
  • $\begingroup$ @LordStryker: That's not a theory in the sense of epistemology. It could be a hypotheses. $\endgroup$ – Rudi_Birnbaum Apr 17 at 22:04
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    $\begingroup$ @jezzo thanks for your comment! I’m aware of your points, but am slightly loathe to include them because I feel like I already digressed too much. But the comment can and should stay as a useful reminder to other readers about technical accuracy :) $\endgroup$ – orthocresol Apr 18 at 5:25
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Lewis structures are dots around atoms on 2-dimensional paper. Although H-O-H is planar, when you throw in the lone pairs, you have to think three-dimensionally. How would oxygen be hybridized, in readiness to accept two incoming hydrogens, each with an electron?

Oxygen could hybridize as sp2p, with 120 degree angles between bonds to hydrogen and a 90 degree angle between the p2 lone pair and the sp2 lone pair. Or as sp3, with 109 degrees between all the bonds. In water, we have 104.5 degrees https://en.wikibooks.org/wiki/Structural_Biochemistry/Water. 104.5 is between 90 and 109, so there must be some accommodation between the repulsion of lone pairs, which would push toward sp3 hybridization, and perhaps some repulsion from the electrons in the O-H bonds, which would tend to spread the H-O-H angle toward 120 degrees (sp2) and push one of the lone pairs more completely into the p2 orbital.

I just can't decide which is the more powerful force; perhaps the water molecule can't either, so it just compromises.

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    $\begingroup$ This reasoning is backwards. Hybridisation is a model of description, hence it always follows the molecular structure. For more see: Are the lone pairs in water equivalent? As an additional note, while the phrasing 'oxygen is hybridised' is unfortunately very common, it is incorrect. Atoms cannot be hybridised, only orbitals can be constructed as hybrid orbitals. The repulsion of these lone pairs is also a lot more involved than what is presented here. $\endgroup$ – Martin - マーチン Apr 14 at 15:21
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Yet another way to think about the fact that $\ce{H2O}$ does not have the full symmetry theoretically possible, is that the number of electrons does not properly match the highest possible symmetry point group (which has the odd name $D_{\infty\mathrm{h}}$. In this sense it can be seen as simple example of symmetry breaking (its however not a simple first order Jahn-Teller distortion).

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  • $\begingroup$ Why is $D_\mathrm{\infty h}$ an odd name? Please use the chemical construct $\ce{H2O}$ instead of the concoction H$_2$O which may have plenty of unwanted side effects. If you want to know more about MathJax, please have a look here and here. $\endgroup$ – Martin - マーチン Apr 16 at 22:11
  • $\begingroup$ I am writing to a Physicist which possibly never came in contact to the Schönflies notation. They usually prefer other ones like Hermann-Mauguin or more mathematically oriented ones like Coxeter. $\endgroup$ – Rudi_Birnbaum Apr 17 at 5:15
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I don't see why the lone electron pairs have to exist on the same side of the atom.

Nobody says they exist "on the same side of the atom". In the picture showing Lewis structures of waters, the lone pairs are shown on opposite sides in the left panel , and on the same side on the right.

enter image description here

These two structures are identical. A Lewis structure does not make a statement about the geometry of a molecule.

If you look at a 3D model of water where the lone pairs are shown according to sp3-hybridization in the valence bond view of things, it depends on the orientation of the molecule whether it looks like "on opposite sides" or "on the same side" (the elongated shape - bunny ears - of the lone pairs is exaggerated; they should add up to a roughly spherical electron density). (Source: https://www.biotopics.co.uk/jsmol/watersingle.html)

enter image description hereenter image description here

In the molecular orbital view (where the molecular orbitals share the symmetry of the molecule) the two lone pairs have distinct shapes (each panel shows one lone pair, orientation of the molecule distinct from figure above).

enter image description hereenter image description here

This was calculated using molcalc.org. As these two molecular orbitals are of similar energy, you could make linear combinations of them to arrive at orbitals similar to the valence bond picture.

So what is the real picture? Looking at hydrogen bond geometries, either one describes the directions from which hydrogen acceptors on other molecules would be located, so both models are consistent with experimental data.

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There are eight valence electrons on the oxygen that have similar energies then there is a gap and two electrons very close to oxygen core follow. These eight electrons form four spin-pairs and those point in directions that more or less minimise the repulsion. These directions are called "tetrahedral" in Chemistry language, since they point from the center (= oxygen atom core) to the vertices of an tetrahedron. Now the two protons are connected with two of those "electron pairs" and two are "alone". Hence you obtain the bent shape of H$_2$O.

(This is kind of trivialized version of the VSEPR model.)

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    $\begingroup$ This kind of trivialised version of VSEPR is unfortunately as wrong as the VSEPR explanation for water itself. Water is indeed one of the most popular examples for when VSEPR breaks down. $\endgroup$ – Martin - マーチン Apr 14 at 15:23
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    $\begingroup$ @Martin-マーチン to add: I would make a distinction between when the VSEPR algorithm breaks down, and when the premises of the VSEPR model fail. The algorithm itself works surprisingly well on H2O: if one assumes that the model is correct (i.e. two equivalent LP + two equivalent BP + LP repel more strongly than BP), it predicts a bond angle slightly smaller than 109.5°, in line with experiment. The algorithm fails for H2S, even if one makes the necessary assumptions. The premises fail for H2O. But then again it is arguable as to whether the premises are ever true... $\endgroup$ – orthocresol Apr 14 at 17:36
  • $\begingroup$ Not sure about which premises you are talking. Obviously VSEPR works fine, you can even predict a compression of the tetrahedral angle if you include the refined rule the the sp3 LP requires more space than the sp3 bond. The VSEPR model in that sense is one of chemistries most important models at all I would say, since it has a huge predictive power. In the sense of the amount of data you need and the amount of molecular structures you get proprerly described by that. Its paper and pencil and the next better one is the MO and that you have to buy with a huge complexity. $\endgroup$ – Rudi_Birnbaum Apr 15 at 17:19

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