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The issue in my mind is that I don"t watch why the lone electron pairs need to exist on the very same side of the atom. Also, wouldn"t the Schrödinger equation carry out an equally plausible framework for water with the lone pairs on the oppowebsite 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 framework between the two says and also we would gain a direct geometry. Clearly on I"m running around in circles below, please someone enlighten me!
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edited Apr 14 "20 at 15:13
Martin - マーチン♦
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asked Apr 14 "20 at 13:28
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I expect, tright here is a time and also place for VSEPR, and also this is more than likely as great a time as any, bereason all beginning chemisattempt students go through it. The actual model has currently been described multiple times, so I will certainly just briefly say that according to this theory, there are 4 pairs of electrons approximately the central oxygen. In order to minimise electron-electron repulsions, these pairs take on a tetrahedral arrangement approximately the oxygen. It does not matter which two are lone pairs and also which 2 are connected to hydrogen atoms; the resulting shape is always bent.
What"s worth bearing in mind (and also hasn"t been defined exceptionally closely so far) is that VSEPR is a model that chemists use to predict the form of a molecule. The reality is that there is no actual means to predict the form of a molecule, acomponent from resolving the Schrodinger equation, which is not analytically feasible for water. Everything else is an approximation to the fact. A few of these approximations are pretty accurate, such as the usage of density sensible concept. Some of them are incredibly crude, and also VSEPR falls into this category: it essentially treats electrons as classical point charges, and also seeks to minimise the electrostatic repulsion between these point charges. As a physics student you should understand much better than to execute this. Therefore, while it predicts the correct lead to this instance, it is more in spite of the version fairly than because of the version. And you should not be surprised to hear that in some slightly more complex instances, VSEPR have the right to predict entirely wrong outcomes.
As you learn even more chemisattempt you will find that there are significantly sophisticated methods of explaining molecular geometry. Many revolve around molecular orbital concept. For a qualitative technique, you have Walsh diagrams which have been defined at Why does bond angle decrease in the order H2O, H2S, H2Se?. For an extra rigorous technique you would most 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 even more and more difficult to extract true chemical understanding from the numbers. Although it need to additionally be sassist that you cannot extract any true chemical knowledge from the VSEPR model.
What interests me even more is the followup question:
Also, wouldn"t the Schrödinger equation carry out an equally plausible framework for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the height or on the bottom of the oxygen in the Lewis structure)?
Since the Hamiltonian of the water molecule is invariant upon rotation, this means that indeed, any orientation of the water molecule is equally likely. However before, this just describes the orientation of the water molecule as a whole. It does not say anything around the internal degrees of freedom, such as the bond angle.
In the absence of any outside pressure, the molecule is free to bend in whichever direction it likes, and also most water molecules indeed do do this as they float via space or swim in a lake. But it will always be bent.
If that were true, then tright here would be a resonance framework in between the 2 claims and also we would certainly get a straight geometry.
If you were to think of a solitary pwrite-up in a double-well potential, say somepoint with
$$V = egincasesinfty & x b endcases$$
then because of the symmeattempt of your system, in every eigenstate of your system, the expectation worth of $x$ would be $langle x angle = 0$. This is quite comparable to your debate. In the situation of water, let"s collection the oxygen nucleus to be at the beginning. Because it can allude either up or down, the expectation worth of the hydrogen nucleus position along the up-down axis would certainly be precisely level with the oxygen atom, i.e. 0. In reality, do not soptimal there: it have the right to suggest to the left or the best, and also to the front or the ago. So the hydrogen nucleus has actually a place expectation value of specifically $(0, 0, 0)$, i.e. appropriate inside the oxygen nucleus.
Does that suppose it"s actually tright here, though? In our contrived double-well device, it"s patently impossible for the pshort article to be at $x = 0$, bereason $V = infty$ there. If you were to meacertain its place, you would never before uncover it at $x = 0$; you would certainly just uncover it in the left-hand side $<-b, -a>$, or the right-hand side $$. Just because the ppost has actually an expectation value of $langle x
angle = 0$ does not suppose that it is physically there, or that $x = 0$ is somehow its equilibrium state. You"re confutilizing an expectation value via a genuine eigenstate (which is what a resonance framework is).
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In specifically the very same way, if you ever before were to meacertain the properties of water (and also bear in mind that practically eextremely interactivity with a water molecule is, in result, a measurement), we would find that it is indeed always bent.