Water has actually the formula $ceH2O$ and we can attract a Lewis structure with 2 lone pairs on the central oxygen. Together a physics student and not a chemist, ns think come myself, "Okay, there space two lone pairs, they will repel each other" and also so us arrive at the conclusion of bent shape.

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The worry in mine mind is that ns don"t watch why the lone electron pairs have to exist on the very same side of the atom. Also, wouldn"t the Schrödinger equation carry out an equally plausible structure for water through the lone pairs on the opposite side of the oxygen indigenous what we assume (imaging the electron on the height or top top the bottom of the oxygen in the Lewis structure)? If that were true, then there would certainly be a resonance structure in between the 2 states and also we would acquire a direct geometry. Clearly I"m running approximately in circles here, you re welcome someone enlighten me!

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I mean, there is a time and also place for VSEPR, and this is most likely as an excellent a time together any, since all beginning mmsanotherstage2019.com college student go through it. The actual version has already been explained multiple times, so i will just briefly say the according come this theory, over there are 4 pairs that electrons about the central oxygen. In order come minimise electron-electron repulsions, this pairs embrace a tetrahedral arrangement approximately the oxygen. The does not issue which two room lone pairs and also which two are connected to hydrogen atoms; the resulting form is always bent.

What"s worth bearing in mind (and hasn"t been explained really carefully for this reason far) is that VSEPR is a model the chemists use to suspect the shape of a molecule. The truth is the there is no real way to predict the shape of a molecule, personally from solving the Schrodinger equation, i m sorry is no analytically possible for water. Every little thing else is one approximation to the truth. Several of these approximations space pretty accurate, such as the use of thickness functional theory. Some of them are extremely crude, and VSEPR drops into this category: it essentially treats electrons as classical suggest charges, and also seeks come minimise the electrostatic repulsion in between these allude charges. As a physics student you should know much better than to execute this. Thus, while that predicts the correct result in this case, that is an ext in despite the of the version rather than due to the fact that of the model. And also you must not be surprised to hear the in some slightly more complex cases, VSEPR deserve to predict completely wrong outcomes.

As you learn an ext mmsanotherstage2019.com friend will uncover that there space increasingly advanced ways of explaining molecular geometry. Most revolve roughly molecular orbital theory. For a qualitative method, you have actually Walsh diagrams which have been defined at Why go bond angle decrease in the stimulate H2O, H2S, H2Se?. For a much more rigorous an approach you would certainly likely need to run some quantum mmsanotherstage2019.com computations, e.g. Are the lone pairs in water equivalent?. Of course, the border of this is the it becomes an ext and more an overwhelming to extract true chemical understanding native the numbers. Although the should also be stated that you can not extract any true chemical expertise from the VSEPR model.

What interests me an ext is the followup question:

Also, wouldn"t the Schrödinger equation carry out an same plausible structure for water through the lone pairs on the opposite side of the oxygen native what we assume (imaging the electron on the top or top top the bottom that the oxygen in the Lewis structure)?

Because the Hamiltonian the the water molecule is invariant upon rotation, this way that indeed, any orientation of the water molecule is same likely. However, this only refers to the orientation the the water molecule as a whole. The does not say anything around the internal degrees of freedom, such as the link angle.

In the absence of any kind of external force, the molecule is complimentary to bending in whichever direction that likes, and most water molecules without doubt do perform this as they to rise through space or swim in a lake. However it will always be bent.

If the were true, then there would certainly be a resonance structure in between the two states and also we would acquire a linear geometry.

If you to be to think that a solitary particle in a double-well potential, to speak something with

$$V = egincasesinfty & x b endcases$$

then because of the symmetry of your system, in every eigenstate of your system, the expectation value of $x$ would be $langle x angle = 0$. This is quite comparable to her argument. In the instance of water, let"s collection the oxygen nucleus come be at the origin. Because it can allude either increase or down, the expectation value of the hydrogen nucleus place along the up-down axis would certainly be specifically level with the oxygen atom, i.e. 0. In fact, don"t protect against there: it can suggest to the left or the right, and to the former or the back. For this reason the hydrogen nucleus has a place expectation worth of precisely $(0, 0, 0)$, 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 difficult for the particle to be in ~ $x = 0$, due to the fact that $V = infty$ there. If you to be to measure its position, you would never discover it in ~ $x = 0$; you would only uncover it in the left-hand next $<-b, -a>$, or the right-hand side $$. Just because the particle has an expectation value the $langle x angle = 0$ walk not average that it is physically there, or that $x = 0$ is somehow its equilibrium state. You"re confusing an expectation value with a actual eigenstate (which is what a resonance framework is).

See more: If The Denominator Is 0 Is It Undefined, What Happens When The Denominator Is 0

In exactly the very same way, if you ever were to measure the nature of water (and be afflicted with in mind that nearly every interaction with a water molecule is, in effect, a measurement), we would find that that is indeed always bent.