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Topic: Radial part of polyelectronic atoms  (Read 342 times)

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Offline cogujada

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Radial part of polyelectronic atoms
« on: June 15, 2020, 05:29:02 PM »
Hey, I study Chem. eng in Madrid (Spain), so first of all, my English might be awful, so I apologize.

I have a doubt about the radial part of polyelectronic atoms (you know, Schrödinger's equation etc). The radial part for Hydrogen is quite simple, but things get difficult when we talk about atoms with more than one electron (due to the shielding effect between electrons). My question is the following: How does the radial part of Schrödinger's equation change for polyelectronic atoms? I know there are a few ways of estimating the new radial part (SCF or STOs) but I don't need that. What I need is how does the graph change.

For example, here it is represented the 2s orbital of F and Li. How can I know which one is one in this case?

What really bothers me is that I want to know which factors influence the graph and if there is a general rule to know which atom is going to be more diffuse (in this case the blue line) or penetrated (the red line). For example, between two atoms, analyzing the same orbital (for example 2s and 2s, or 3p and 3p etc) the one with the highest Z is the most penetrated, and the one with the lowest Z is the most diffuse (obviously I don't know if this is true hahah, it is an example). You know, something like that so I can answer these type of questions:

Plot the graph of the radial part of S and Cl (1s orbital)

In order to answer this, what I need is to know which one of them is more penetrated.

I don't know if you understand me. Again, sorry, my English is awful...

Thank you guys, if there's something you don't understand please tell me and I will answer shortly. By the way, feel free to correct me in any grammatical, spelling error please, I really want to boost my level of English!


Offline Enthalpy

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Re: Radial part of polyelectronic atoms
« Reply #1 on: June 17, 2020, 03:38:06 PM »
¡Bienvenido(a), Cogujada! No te preocupes por tu inglés, está muy bien.

a0 is a constant independent of the element, usually and here supposedly too. The most important effect is that more protons attract all electrons nearer to the nucleus. The rest, electrostatic repulsion AND fermion nature, is only a correction.

So, I would say too: F has a smaller 2d orbital than Li.


You mentioned shielding. If wanting such a detailed description (which is necessary to get meaningful figures), the idea of "orbitals" should be refined. Shielding is in fact only an attempt to keep the relatively manageable orbitals of the hydrogen atom, with one electron. This attempt tries to describe electrons as almost independent, a bit as if an inner electron had a distribution little altered by the outer ones, and as if an inner electron acted globally on an outer electron. Such attempts are useful, but they are formally wrong, and they need much unjustified numerical tinkering before they give meaningful figures.

Take the electrostatic repulsion. An electron has the same charge magnitude as a proton, and the distances are about an atomic radius too. So the interactions are very strong. If you find one electron in a small volume within an atom, finding an other electron in that small volume is very improbable. That is, if p1 and p2 are individual probabilities for electrons 1 and 2, taken as mean values over the positions of the other electrons, then the probability of finding both electrons in the same small volume is less than p1×p2. This means that ψ(r1, r2) can't be written as ψ(r1)×ψ(r2). ψ(r1, r2, ...) should be solved at once, but this is rarely done, and only for few electrons, since no algebraic solution exists, and computers stop at very few electrons. So, simplified methods are used.

I wrote "if" an electron is in a small volume, not "when". By definition of an orbital, it's independent of time. ψ(r1, r2, ...) has a component as exp(iωt), and everything else is static. |ψ| is completely independent of time. This is a bit abstract, some meditation is useful here.

ψ(r1, r2) ≠ ψ(r1)×ψ(r2) is called quantum entanglement
it's the usual situation for nearly all particles on Earth.

There is more. Fermions, including electrons, have by nature an antisymmetric wave function ψ(r1, r2, ...). One consequence is "two electrons with antiparallel spin per orbital", but the consequences reach much farther. The wave function is antisymmetric for all electrons, including on different orbitals, which constraints the shape of the orbitals. Have a look there
with two electrons on different orbitals of a helium atom, the energy levels depend on the alignment of the spins, because the spin is a component of antisymmetry, so it changes the shape of the possible wave functions. The effect on energy is not at all explainable by magnetic interaction.

Offline wildfyr

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Re: Radial part of polyelectronic atoms
« Reply #2 on: June 18, 2020, 02:05:42 PM »
Too far over my head, but your English is excellent, don't be down on yourself about that!

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