# Radial part of polyelectronic atoms

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Posted (edited)

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!

Cheers

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I have a doubt

I have a doubt --> I have a question

which atom is going to be more diffuse (in this case the blue line) or penetrated (the red line).

which atom is going to be more diffuse (in this case the blue line) or penetrated (the red line). --> which electron is going to be more diffuse (in this case the blue line) or penetrating (the red line).

Not necessarily wrong, but IMO it's clearer this way.

Is this homework? 1s e- for S and Cl. You're dealing with very interior electrons here. So it looks like hydrogen-like electron states with a bigger Zeff and outer electrons pressing them harder against the nucleus.

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No it's not homework, it's something I found in the Internet. However I have a Chem exam (about this and more topics) in a few weeks.

Yeah well what I've deducted is that the radial wavefunction will be more spread out fore higher Z.

And thanks for correcting me!

Cheers

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it's something I found in the Internet.

it's something I found in the Internet --> it's something I found on the Internet

Your English is very good. I only need to fix minor details.

Yeah well what I've deducted is that the radial wavefunction will be more spread out fore higher Z.

[..] what I've deducted is that the radial wavefunction [...] --> [...] what I've deduced is that the radial wavefunction [...]

That's certainly true for outer electrons. Are you sure it is also for inner electrons?

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Hey, I study Chem. eng in Madrid (Spain), so first of all, my English might be awful, so I apologize.

Not sure why you need this for Chemical Engineering but good on you for wanting to know.

As a ChemE student you will be familiar with differential equations in general and the separation of variables in particular.

The good news is that because the potential energy of an electron in an atom is a function of distance from the nucleus only it is possible to separate the variables in the quantum differential wave equation (Schrodinger etc).
This means that we can express the wavefunction as the product of three functions, each one only involving one of the three coordinates

So we have in spherical coordinates

$\psi = R\left( e \right)\Theta \left( \theta \right)\Phi \left( \phi \right)$

So of the four quantum numbers, n, l, m and s

m is the magnetic quantum number and s the spin number and do not contribute to potential energy.

For an S shell electron l = 0 and the angular part is therefore constant.

So we are left with the radial part R(r)

The probability of finding an S electron in a small element of volume, dv at a distance between  r and (r + dr) is then [R(r)]2dv.

Also the probability of finding the electron anywhere between between r and (r+dr)
That is lying in a spherical shell of radius r and thickeness dr is given by replacing dv by the volume of the shell

volume  = 4πr2dr

So we have

The probability = 4π[R(r)]2 r2dr

and the function  4π[R(r)]2 ris the radial probability distribution function you have plotted.

Does this help?

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5 hours ago, joigus said:

That's certainly true for outer electrons. Are you sure it is also for inner electrons?

Well I don't know it for certain. All the exercises my teacher did were with outer electrons, except one, she did one with "inner electrons" (I at least consider 2s orbital as inner), and it was the same criterion, however, i'm not 100% sure.

4 hours ago, studiot said:

Not sure why you need this for Chemical Engineering but good on you for wanting to know.

Well it's not that I want to know it (I do, I find really triggering chemistry) but our teacher thinks this is extremely important (Schrödinger's equation and the angular&radial part), so important questions about this topic were 50% of the exam.

4 hours ago, studiot said:

Does this help?

Well, it is good to remember, but I already knew that, as I've said, this seems to be a very very important topic here in Spain (I have a friend who studies Pharmacy and she has been taught too the Schrödinger's equation). But it does not answer my question (I think).

I've been investigating and I've found that radial the wavefunction will be more spread out fore higher Z. @joigus helped me and told me that this is, without a shadow of doubt, true for outer electrons, but he's not sure about inner electrons.

You know, I could ask my teacher, but I have been asking her a lot of things (as you can imagine, virtual classes are by no means the same as "real" classes) and I'm a bit ashamed of asking her again (and in addition, I've already asked her about Schrödinger's equation, not this topic of polyelectronic atoms specifically, but a question about the angular part of an s orbital).

Thanks guys, you're great!

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Yeah well what I've deducted is that the radial wavefunction will be more spread out fore higher Z.

OMG, I don't know if this is correct, because I just seen another exercise made by my teacher which said that the radial wavefunction would be more spread out for the one with the lower Z. This is so frustrating

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Posted (edited)

Well it's not that I want to know it (I do, I find really triggering chemistry) but our teacher thinks this is extremely important (Schrödinger's equation and the angular&radial part), so important questions about this topic were 50% of the exam.

Sorry if my answer was too basic and simple.

A detailed answer to both this question and your Moellier question will take some time to answer.

But do not expect formulae they can only give trends. Solution of Schrodinger for the exceptional cases can only be done by numerical methods which involve a best guess and refining this to fit observed data.

Edited by studiot

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24 minutes ago, studiot said:

But do not expect formulae they can only give trends

Don't understand this sorry hahaha

24 minutes ago, studiot said:

Extremely interesting, I knew Nitrogen and Oxygen were exceptions (and also P/S, Be/B and Mg/Al, but I didn't know about As and Se, I thought they weren't involved). I really want to thank you and @joigus, you're really talented guys and very helpful and kind.

26 minutes ago, studiot said:

Sorry if my answer was too basic and simple.

No worries!!

Thanks very ver very much

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Yeah well what I've deducted is that the radial wavefunction will be more spread out fore higher Z

Okay, so the guy that helped me out with this apparently got a bit confused and started to rethink what he told me (the radial wavefunction will be more spread out for higher Z). He has plotted this:

Where "The red curve is R20, the dark green is R30, and the bright green (which you can barely see the humps) is R40" As we can see, as Z gets higher, the first maximum gets nearer to the origin of the coordinates, and the peaks get lower. The first part (that the maximums are much closer to the y axis) seems good to me, however the second part (the peaks are lower) does not coincide with what my teacher told us. Do you have something to add @joigus and @studiot??

Thanks guys, you're saving  my semester

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Posted (edited)
5 hours ago, studiot said:

But do not expect formulae they can only give trends. Solution of Schrodinger for the exceptional cases can only be done by numerical methods which involve a best guess and refining this to fit observed data.

I agree with @studiot. Solving a high-Z atom is not like solving Schrödinger's equation for the hydrogen atom, substituting +e by +Ze and assuming that all the hydrogen-like "orbitals" are filled with electrons. There is the Hartree-Fock method, there are other methods that I can't remember now, and the problem is highly non-trivial.

You pointed to one of the clues yourself here:

On 6/15/2020 at 11:23 PM, cogujada said:

but things get difficult when we talk about atoms with more than one electron (due to the shielding effect between electrons)

It's even worse than that, AAMOF. You've got spin-spin effect, spin-orbit effects, the nucleus, London, and others I forget and would have to review. Highly complicated N-body problem mess.

But

intuition can guide you if you're interested in qualitative discussion. That's why I asked you if they were outer (valence) or inner electrons (1s). Outer electrons get more spread. But for very internal electrons my intuition (maybe my memory in part) tells me that it's the opposite. Electrons midway from both extremes are more difficult to predict.

I think it's no coincidence that they're giving you 1s electrons 1st principal quantum number and s-wave e-, so they don't stray very far from the nucleus-- for relatively high-Z atoms:

On 6/15/2020 at 11:23 PM, cogujada said:

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

Those e- will be very close to a +Ze charged nucleus. External electrons, although many of them will be in high-l orbital angular momentum and stray farther from the nucleus, will, on the average, act as a shell of negative charge outside the internal ones. First approx. to this is a shell of negative charge covering them would be a spherical shell of negative charge which has zero electric field inside (I'm using electrostatics as a clue). It's not a sphere, I know, but we're thinking schematically. And AAMOF some of them are s-wave spheres!! So my intuition is 1s electrons will get closer to the nucleus (more "penetrating") in the case of the higher-Z neutral atom, even though the valence electrons suffer the opposite trend and spread out for higher Z.

Don't take for granted anything I say here. Take my cue, see if it makes sense, discuss it with your classmate, and do your research and finally do your own thinking. It's going to teach you a lot more than anything anybody can do by telling you the answer.

And plan B is ask your teacher. It may be embarrassing, but you must get over it.  She won't mind, I'm sure. Quite the opposite. You can use my idea about internal electrons as a foil. I'm interested in the answer.

Edit: I can't see your picture.

Edited by joigus

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The red curve is R20, the dark green is R30, and the bright green (which you can barely see the humps) is R40

8 hours ago, joigus said:

I agree with @studiot. Solving a high-Z atom is not like solving Schrödinger's equation for the hydrogen atom, substituting +e by +Ze and assuming that all the hydrogen-like "orbitals" are filled with electrons. There is the Hartree-Fock method, there are other methods that I can't remember now, and the problem is highly non-trivial.

You pointed to one of the clues yourself here:

It's even worse than that, AAMOF. You've got spin-spin effect, spin-orbit effects, the nucleus, London, and others I forget and would have to review. Highly complicated N-body problem mess.

But

intuition can guide you if you're interested in qualitative discussion. That's why I asked you if they were outer (valence) or inner electrons (1s). Outer electrons get more spread. But for very internal electrons my intuition (maybe my memory in part) tells me that it's the opposite. Electrons midway from both extremes are more difficult to predict.

I think it's no coincidence that they're giving you 1s electrons 1st principal quantum number and s-wave e-, so they don't stray very far from the nucleus-- for relatively high-Z atoms:

Those e- will be very close to a +Ze charged nucleus. External electrons, although many of them will be in high-l orbital angular momentum and stray farther from the nucleus, will, on the average, act as a shell of negative charge outside the internal ones. First approx. to this is a shell of negative charge covering them would be a spherical shell of negative charge which has zero electric field inside (I'm using electrostatics as a clue). It's not a sphere, I know, but we're thinking schematically. And AAMOF some of them are s-wave spheres!! So my intuition is 1s electrons will get closer to the nucleus (more "penetrating") in the case of the higher-Z neutral atom, even though the valence electrons suffer the opposite trend and spread out for higher Z.

Don't take for granted anything I say here. Take my cue, see if it makes sense, discuss it with your classmate, and do your research and finally do your own thinking. It's going to teach you a lot more than anything anybody can do by telling you the answer.

And plan B is ask your teacher. It may be embarrassing, but you must get over it.  She won't mind, I'm sure. Quite the opposite. You can use my idea about internal electrons as a foil. I'm interested in the answer.

Edit: I can't see your picture.

yeah I'm afraid I need to ask her, however, what about the peaks? We've seen plotting the graphs (I think that if you click the hyperlink you can see it) that for higher Z, the graph is less spread out and has lower peaks, whereas our teacher told us that the peaks were higher (which makes sense, because the area under the graph has to be 1, and if you make it more narrow you'll have to make it higher no?).

On the other hand, I just saw an exercise about the 3p orbital of Cl and S, and it follows the same tendency (for higher Z, narrow graph and higher peaks). Do you consider them as outter electrons??

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whereas our teacher told us that the peaks were higher (which makes sense, because the area under the graph has to be 1, and if you make it more narrow you'll have to make it higher no?).

Yes, that's correct. Narrower --> Higher peaks. Conservation of probability.

On the other hand, I just saw an exercise about the 3p orbital of Cl and S, and it follows the same tendency (for higher Z, narrow graph and higher peaks). Do you consider them as outter electrons??

Yes, for S (Z=16) and Cl (Z=17) the valence electrons are in 3p orbital, so those are outer electrons.

(I think that if you click the hyperlink you can see it)

I'm busy now. I'll get back to you in maybe 6+ hours...

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1 hour ago, joigus said:

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1 hour ago, joigus said:

Yes, for S (Z=16) and Cl (Z=17) the valence electrons are in 3p orbital, so those are outer electrons.

Yeah, and it stills follow the criterion of Higher Z, narrower graph and higher peaks (at least according to my teacher).

1 hour ago, joigus said:

Yes, that's correct. Narrower --> Higher peaks. Conservation of probability.

Yeah, seems logical to me, but now that studiot posted the photo (thanks by the way), you can see that for higher Z, the graph is narrower but the peaks are lower (which seems non-sense).

1 hour ago, joigus said:

I'm busy now. I'll get back to you in maybe 6+ hours...

No problem, my exam is the 15 of July, so I have still plenty of time haha

Thanks to both of you!

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Posted (edited)

3 hours ago, studiot said:

<image>

Thank you. +1.

Edited by joigus
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Posted (edited)
21 hours ago, studiot said:

A detailed answer to both this question and your Moellier question will take some time to answer.

I said it would take a while to answer but first a word of gentle friendly advice.

This must be a very small part of your first year curriculum. I have never heard of Chemical Engineers requiring such detail.
So don't be diverted from the main thrust of your subject, there is so much introduced in the first year, from fluid mechanics to science of materials to advanced mathematics to thermodynamics to transport phenomena......
Remember that this is only about atoms. So think about how few times a CE will deal with pure atoms as opposed to the great many times he will deal with molecules?

So to proceed with atoms. I will try to clear up a few things you may know or may only partly know on the way.

Firstly the Hamiltonian in a system (H) relates the total energy of a system to some controlling parameter.
In the case of wave mechanics the Schrodinger equation  this is the wave function.
We say that the Hamiltonian operates on the wave fucntion to ouput the total energy.

$H\left( \psi \right) = {E_\psi }$

Now this can be broken down to measurable quantities such as momentum, charge, mass etc

For a single electron atom Ie a hydrogen-like atom this becomes

$H = - \frac{{{\hbar ^2}}}{{2\mu }}{\nabla ^2} - Z\frac{{{e^2}}}{r}$

Where Z is the atomic number that is the number of protons in the nucleus.
So the energy depends upon the number of protons in the nucleus.
Further this equation has been solved analytically.

Some notes here are in order.
A hydrogen-like atom has one outer electron over a core of fully paired, if any, electrons so includes lithium, sodium, potassium etc.
This electron will inhabit an S orbital,  generally of higher energy than any orbital in the core. More of this later.

Your diagram refers to S orbitals only, and further more is proportional to the square of slutions of the Schrodinger equation.
Since it is a squared term it is non negative ie positve or zero.
Plots of the wavefunctions themselves show positive and negative regions.

The various humps in the square (probability plot can be seen forming in these plots)
Further the S orbital is the only one that is not zero at the origin (I have only show p but this is also true of d and f orbitals).

The second point is as already made; the actual energy value depends upon Z.
So the actual energies of 'the same' orbital in the atom of one element will be different from those in the atoms of other hydrogen-like atoms.
Note carefully on this chart

Such variation of levels becomes even more complicated when we consider atoms with two or more outer electrons

Looking back at the Hamiltonian this is because we need to add more terms due to the interaction of the other (Z-1) electrons.

$H = - \frac{{{\hbar ^2}}}{{2\mu }}{\nabla ^2} - Z\frac{{{e^2}}}{r}$

Now the electrostatic energy is increased by the addition of this interaction energy.
Unfortunately even this simple addition cannot be solved analytically.

This accounts for the majority of the effects and the reason why all the building up , aufbau, Hund, Slater diagram, Mollier etc rules have exceptions.

And all this effort is only for less than 0.01% of the substances a CE will be dealing with.

For  molecules and chemical bonding several more terms involving cross products of the Z numbers for the second and subsequent nuclei in the molecule are needed.

This is why the build up of atomic orbitals, which was greatly studied in the first half of the 20th century, faded in the second half, when the subsidiary Science of Computational Chemistry was born.

How are we doing?

Edited by studiot

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Posted (edited)

Yeah, and it stills follow the criterion of Higher Z, narrower graph and higher peaks (at least according to my teacher).

For what kind of n?

I don't think that's universal. Take a look at @studiot's graphs. +1.

Electrons in 3d, 4d, or 4f start going a bit nuts. I think by the way that's when new orbital types are being inaugurated! Not sure now, but it may have to do with exceptions in Moeller's diagram. Aufbau is more solid, if I remember correctly.

Edited by joigus

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29 minutes ago, joigus said:

I don't think that's universal. Take a look at @studiot's graphs. +1.

Sadly I see I have pasted in the same equation twice.
Hhere is the correct one for many electroned atoms.

$H = - \frac{{{\hbar ^2}}}{{2m}}\sum\limits_i {\nabla _i^2} - \sum\limits_i {Z\frac{{{e^2}}}{{{r_i}}}} + \sum\limits_{i > j} {\frac{{{e^2}}}{{{r_{ij}}}}}$

My apologies to all.

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8 hours ago, studiot said:

This must be a very small part of your first year curriculum. I have never heard of Chemical Engineers requiring such detail

Well, in Spain we are told that "el saber no ocupa lugar". @joigus can translate it to you hahah

2 hours ago, joigus said:

I don't think that's universal.

Yeah well I'm afraid I've been talking with other first year students (like me) but from other colleges (UAM, UB and UPM) and they don't even know about Schrödinger's equation. I think that she told us that to avoid messing our minds hahaha. I dunno really.

8 hours ago, studiot said:

How are we doing?

Very very good. Thanks, I mean, you've both done more than any other person for me in my academic career I think

I know understand much better, not only Schrödinger's equation, but how difficult it gets chemistry when talking about insignificant electrons, what the hell man.

Thanks guys

2 hours ago, studiot said:

My apologies to all.

hahahahahahah, mate you've been killing it out there explaining it all (along with joigus)... Hopefully, someday, my errors will be as insignificant as yours!

Best regards boys!

know

omg, now wtf is going on in my head

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omg, now wtf is going on in my head

Todavía no es juernes, cogujada.  Take it easy.

I'll dig out something for you about Zeff that I think will help you. Give me some time.

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