You are watching: What causes water molecules to have a bent shape according to vsepr theory
The concern in my mind is that ns don"t view why the lone electron pairs have to exist on the same side the the atom. Also, wouldn"t the Schrödinger equation provide an equally plausible framework for water through the lone bag on the opposite next of the oxygen from what we assume (imaging the electron on the height or ~ above the bottom of the oxygen in the Lewis structure)? If that were true, climate there would be a resonance structure in between the two states and also we would gain a linear geometry. Clearly I"m running about in circles here, you re welcome someone enlighten me!
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I mean, there is a time and also place because that VSEPR, and this is most likely as an excellent a time as any, because all start stclairdrake.net student go through it. The actual version has already been explained multiple times, so ns will only briefly say that according come this theory, over there are four pairs the electrons roughly the central oxygen. In order come minimise electron-electron repulsions, this pairs take on a tetrahedral arrangement roughly the oxygen. That does not issue which two space lone pairs and which two are linked to hydrogen atoms; the resulting form is always bent.
What"s worth bearing in psychic (and hasn"t been explained really carefully so far) is that VSEPR is a model that chemists usage to guess the shape of a molecule. The reality is the there is no real means to suspect the form of a molecule, apart from resolving the Schrodinger equation, i beg your pardon is no analytically feasible for water. Every little thing else is an approximation to the truth. Some of these approximations room pretty accurate, such together the usage of density functional theory. Few of them are extremely crude, and also VSEPR drops into this category: it essentially treats electrons together classical point charges, and also seeks to minimise the electrostatic repulsion between these suggest charges. As a physics student you need to know better than to perform this. Thus, while it predicts the correct result in this case, the is much more in despite the of the version rather than due to the fact that of the model. And you must not it is in surprised come hear that in some slightly more complicated cases, VSEPR can predict totally wrong outcomes.
As friend learn an ext stclairdrake.net you will find that there space increasingly sophisticated ways that explaining molecular geometry. Most revolve around molecular orbital theory. Because that a qualitative method, you have actually Walsh diagrams which have actually been defined at Why walk bond edge decrease in the order H2O, H2S, H2Se?. For a more rigorous an approach you would likely have to run some quantum stclairdrake.net computations, e.g. Space the lone bag in water equivalent?. The course, the limit of this is that it becomes more and more complicated to extract true stclairdrake.net understanding native the numbers. Although it should also be stated that you can not extract any type of true chemical understanding from the VSEPR model.
What understand me more is the followup question:
Also, wouldn"t the Schrödinger equation carry out an same plausible framework for water through the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or top top the bottom that the oxygen in the Lewis structure)?
Because the Hamiltonian of the water molecule is invariant upon rotation, this method that indeed, any orientation that the water molecule is equally likely. However, this just refers come the orientation the the water molecule as a whole. The does not say anything around the internal levels of freedom, such as the bond angle.
In the lack of any type of external force, the molecule is free to bending in whichever direction it likes, and most water molecules undoubtedly do carry out this as they to rise through space or swimming in a lake. However it will constantly be bent.
If that were true, climate there would be a resonance structure between the two states and also we would get a linear geometry.
If you were to think of a single particle in a double-well potential, say something with
$$V = egincasesinfty & x b endcases$$
then due to the fact that of the the contrary of her system, in every eigenstate of her system, the expectation worth of $x$ would certainly be $langle x angle = 0$. This is quite similar to your argument. In the case of water, let"s set the oxygen nucleus come be at the origin. Due to the fact that it can allude either increase or down, the expectation worth of the hydrogen nucleus position along the up-down axis would certainly be specifically level through the oxygen atom, i.e. 0. In fact, don"t prevent there: the can point to the left or the right, and also to the front or the back. For this reason the hydrogen nucleus has a position expectation worth of specifically $(0, 0, 0)$, i.e. Best inside the oxygen nucleus.
Does that median it"s in reality there, though? In our contrived double-well system, it"s patently impossible for the bit to be at $x = 0$, because $V = infty$ there. If you were to measure up its position, you would never discover it in ~ $x = 0$; you would only find it in the left-hand side $<-b, -a>$, or the right-hand next $$. Just since the particle has an expectation value that $langle x
angle = 0$ does not typical that that is physically there, or the $x = 0$ is somehow its equilibrium state. You"re confusing an expectation value through a actual eigenstate (which is what a resonance framework is).
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In exactly the same way, if you ever before were to measure the nature of water (and be afflicted with in psychic that practically every communication with a water molecule is, in effect, a measurement), we would find that that is indeed always bent.