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I have studied that according to Aufbau rule the energy of subshells is dependent on the sum of $ n $ and $l$ values. This would imply that the energy of subshells in a shell varies as

$$ ns \lt np \lt nd \lt nf $$

and this makes sense (intuitively from Bohr model) too as the closer the electron is to the nucleus the lesser energy it would have. And this idea corresponds to penetration power as well where the trend is $$ nf \lt nd \lt np \lt ns $$ as farther the subshell is from nucleus lesser it experiences its force and hence lesser penetration power. Everything made sense until I saw the graphs of Radial Probability Distribution.

enter image description here

and also the expression for average radius of subshell $$ 1/2(3n^2 - l(l+1)) $$ According to these both my conclusions about penetration power and energy levels should be wrong as for greater $l$ value average radius and even most probable radius both are smaller than corresponding values of smaller $l$ values. What is wrong with my reasoning?

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  • $\begingroup$ Just a point to note - When you say closer to the nucleus, the lesser the energy, it means the potential energy is more and more negative. $\endgroup$ May 30, 2023 at 8:09
  • $\begingroup$ @Proscionexium yes exactly my point, not talking about magnitude. $\endgroup$
    – bm27
    May 30, 2023 at 8:20
  • $\begingroup$ Note the energy E is proportional to -1/r, so it will help if the radial probability was weighted by multiplying by 1/r. Parts near a nucleus would become more prominent. Similarly, if the weighted average radius is calculated, it would be closed to nucleus, (to be formally equal to equivalent radius of circular orbit with the same el. energy). It would be more closer for orbitals with more significant near nucleus maxima. $\endgroup$
    – Poutnik
    May 30, 2023 at 8:33
  • $\begingroup$ @Poutnik could you explain that in simpler terms as I have only studied this topic at just a little more than high school level and don't have that much knowledge on the same. $\endgroup$
    – bm27
    May 30, 2023 at 8:39
  • $\begingroup$ The mean or most probable radius and the energy are directly comparable for circular orbits, what is not our case. Probabilities for being closer to a nucleus are more significant for the average energy. $\endgroup$
    – Poutnik
    May 30, 2023 at 8:54

3 Answers 3

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I have studied that according to Aufbau rule the energy of subshells is dependent on the sum of $n$ and $l$ values. This would imply that the energy of subshells in a shell varies as

$$ n_s<n_p<n_d<n_f $$

** Let us try a few examples, from the picture diagram. **

$ 1 S : n=1, l=0; n+l = 1$

$ 2 S : n=2, l=0; n+l = 2$

$ 2 P : n=2, l=1; n+l = 3$

$ 3 S : n=3, l=0; n+l = 3$

Here is a surprise, different from your understanding. For the $ 2 P $ and $ 3 S $ examples, the sum of $n+l$ is equal instead of the increasing of order that you had expected.

$ (n+1)_{2P} = 3 = (n+l)_{3S} $

And this makes sense (intuitively from Bohr model) too as the closer the electron is to the nucleus the lesser energy it would have. And this idea corresponds to penetration power as well where the trend is

$ n_f < n_d <n_p < n_s $

Well that quoted statement needs to be checked carefully. Unfortunately a little bit of background in Quantum Mechanics and the Schrödinger equation are needed for a more intuitive understanding of what is going on.

There are a few different forms of the Schrödinger equation. To fundamentally understand electron-ionization of atoms, the time-independent form is most useful 2 :

The simplest way to write the time--independent Schrödinger equation is $H\psi = E\psi$, however, with the radially-symmetric Hamiltonian operator expanded it becomes:

$\sum_i{(-\frac{\hbar^2}{2m_e} \frac{d^2 \psi_i}{dr_i^2})} + \sum_i{V_i\psi_i} = E_i\psi_i\\ $

The left-hand-most side of the equation is calculated from momentum contributions (of the wave-functions themselves) versus radius. In Quantum Mechanics, the electron can be found anywhere when it is ionized with some probability, but until it is ionized is is found no-where. There is just some probability that it it exists at any particular place to enable wave-function expected-energy calculations. The wave-function, in short, corresponds to a probability distribution for each $i_{th}$ electron.

On the left-side is also a sum of the wave-function multiplied by a potential function that accounts for the energy that the electron has because of the protons at the nucleus or because of other electrons.

To get to understand this potential (labeled as $V$ in the equation), there are a couple of interactions that are important, for the moment I will describe for Helium, for you to generalize to other atoms:

  • The potential of the inner-electron due to the two protons in the center of the nucleus and its probability density. (This well is so deep that the wave-function of the inner electron is not much influenced by the outer-electron. For a good first calculation of ionization energy, this inner-electron wave-function can be assumed to not change at all.)
  • The potential of the outer-electron including the shielding of the nucleus by going to a value of $e^{2+}$ to $e^{1+}$ because the inner electron has a shielding value of $e^{1-}$. Ironically, for a basic good calculation, the outer-shielded-electron wave-function does not change very much either, and for a first start, this change can also be entirely neglected.
  • The interaction of the inner electron and outer electron as a coulomb correction to the Schroedinger model for Hydrogen, which accounts only for a point-source nucleus versus a spread-out expectation of the electric field and the expectation of the outer electron charge density integrated (on the entire volume).
  • For this Coulomb interaction of the inner electron and outer-electron, the inner-electron wave-function is that from $e_{2+} hydrogen$ and the outer-electron wave-function is that from $e_{1+} hydrogen$.

Thus regarding:

as farther the sub-shell is from nucleus lesser it experiences its force and hence lesser penetration power. Everything made sense until I saw the graphs of Radial Probability Distribution.

The component of the Schrödinger equation that deals with the nuclear attraction is the fully-shielded atom, because the outer-electron that is ejected for the first-ionization is always out-most. However, the near-field Coulomb energy contribution changes quite a bit as the electrons are pushed further and further from each-other, as is shown in the below diagram.

Outer-most Ionization Energies:

far-field component.

Like you stated, we can expect that for higher element-numbers in the Periodic-Table, as a general trend, the outer electron ionization energy is lower because the inner-electron-shielding is more complete.

I have not yet plotted out your diagrams, but instead I think this reference about Hydrogen Wave Functions might be more-complete.

One of the illustrations is that reference 6 is very much easier to understand:

Hydrogen Wave Function Pictures

Also, I found for you this reference websites.umich.edu/~chem461/QMChap7.pdf. It also has the linear plots in it with a more detailed explanation about how to calculate the probability distributions which should be more insightful than the previous diagrams you referenced.

From the reference Hydrogen Electron Density Versus Radius R, there is a plot on page 19:

Some radial distribution functions for Hydrogen-Wave-Functions

It does seem that the expected radius for the $3 S$ - wave-function is greater than the expected-radius for the $3 P$ - wave-function, which is also apparently greater than the expected-radius for the $3 D$ - wave-function. So the expected trend from your plots seem somewhat confirmed, in terms of expected radius. The problem with your understanding is that expected radius for Hydrogen is not the only thing determining the first ionization potential. Indeed for Hydrogen, only the quantum number $n$ determines potential ionization energy, independent of $l$ and $s$ quantum numbers. The plots you have seem to be similar (in trends) to those for pure hydrogen wave-functions, all with expected energy just from $n=3$ and nothing else, making the ionization energies for the three first ionization energies exactly the same because of other effects, such as distributions versus spherical coordinates (aside from the radial coordinate $r$).

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  • $\begingroup$ thanks again for your answer but I think you misunderstood my question (or maybe I haven't explained it well), I'm not bothered with the size comparison of orbitals of different shells (K,L,M...) , my confusion arises from the fact that according the average radius formula mentioned in my question and also the plots (even in the reference you shared) depict that the $s$ subshells is larger than $d$ for the same $n$ value, despite the fact that it has less energy and less penetration power. Is there a reason for that or the information provided to me is false? $\endgroup$
    – bm27
    Jun 1, 2023 at 17:08
  • $\begingroup$ If the S-shell radius is larger than the D-shell radius, that would put the S-shell-energy conceptually lower. It is the nucleus that binds the electron; so if the distance is expected to be further from the nucleus, then the nucleus's effect is weaker. The energy you are talking about is the binding energy to the nucleus. Far away from the nucleus, this energy is Hydrogen-like, corresponding to $e^{+1}$. Closer to the nucleus this binding energy is greater because of the lower shielding of the other electrons. Please keep on asking be questions if this is not clear. $\endgroup$ Jun 6, 2023 at 12:00
  • $\begingroup$ How would the energy of S-shell be lower if its radius is greater than that of D-shell, wouldn't it be greater as further from the nucleus the magnitude of Columbic attractive force would decrease and hence the energy would decrease, as closer to the nucleus more negative the energy is and hence smaller. I'm not sure the energy we are talking about is the same but I think they have the same magnitude, as the energy the electron has in an orbital should be the same as its binding/ionization energy. $\endgroup$
    – bm27
    Jun 7, 2023 at 4:17
  • $\begingroup$ @Banaj Mahajan : please see my updated answer to your question. The ionization-energy picture is not clear from just the radial-coordinate plot. Plots versus the 3 spherical coordinates are necessary to get a more complete picture about contributions to the energy aside from the radial-distribution contribution to the ionization-energy. I suspect that a good part of the periodic table can be calculated straight-forwardly, but I have not had the time yet to take on this expanded task versus just the Helium calculations that were very successful. $\endgroup$ Jun 8, 2023 at 9:29
  • $\begingroup$ So considering all the factors including the spherical coordinates what would be the size comparison for the subshells of the 3rd shell? $\endgroup$
    – bm27
    Jun 8, 2023 at 16:50
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The Aufbau Principle reference is summarized with the following picture:

energy ordering picture

Keep in mind that the Aufbau principle is founded on 1) the energy levels of Hydrogen, which are only determined by the quantum number $n$. But that is not the entire picture. Because of the shielding effect as well as near-nucleus penetration, there is an additional perturbation that comes in to play.

I have calculated this for Helium, and I discovered that the binding energy taking into account this addition perturbation could be calculated with around $3$ $Percent$ $accuracy$ which I considered a great success. There are two electrons in this model, an inner electron and an outer electron. For Helium, the inner electron is so closer to the nucleus, and even as the radius approaches infinity, it is still inner, meaning that it bears the $2e^+$ charge attraction. And the outer electron, even as the radius approaches infinity is still outer, meaning that it bears $1e^+$, being shielded by the inner electron.

The two spins are opposite for Helium for the inner and outer electron, so there is no major problem with spin interaction. The perturbation comes almost entirely from the effect of the outer electron penetrating into the shielding of the inner at the near field, close to the nucleus.

So the Aufbau Principle also considers this shielding perturbation of energies to arrive at an accurate ordering of shells, which makes it different than the pure Hydrogen calculations that just depend on the quantum number $n$.

There are a couple of other considerations. First of all, when the Aufbau Principle references the $2$ $P$ $Orbital$, it is really a hybridized orbital, which is some combination of $2$ $S$ and $2$ $P$ orbitals (which for idealized Hydrogen have instead equal energy levels). This hybridization pushes the $S-P$ orbital outward, with more focus. This same hybridization effect increases with $S-P-D$ hybridized orbitals and so forth.

Taking account with the VESPR model, which includes this Hybridization-focusing effect of electron bonds, one can see that as that from the order of $1S$ which is spherically symmetrical, going on to $2S, 2P$ which has 4-fold symmetry, that the electrons of that level are able to distance themselves out further from one another in the near field to the nucleus. The Aufbau Principle is a kind of short-hand notation, labeling $2P$ as what is really a $2S-2P$ hybridized orbital that allows for this additional separation. The additional separation in the near-field to the nucleus of the electrons allow for lower interaction energy there.

This is my understanding of how all of this works. I am welcome to additional insight and comments, as I have not yet gone through the calculations for Lithium and so forth, which I estimate are not hard calculations, just tedious ones.

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  • $\begingroup$ Hello. The figures in the OP just refer to the radial distribution function in the H-atom (multiplied strangely by $4\pi$), no need to analyze $\ce{Be}$, an so on. I saw the link to "calculated". You wrote a book about the helium atom? $\endgroup$ May 30, 2023 at 13:01
  • $\begingroup$ @Metal Storm : Yes (pdf free), I went through the symbolic math for Helium using this inner-electron, outer-electron model. Because all the integrals are powers of radius and an exponential, the symbolic solution is straight-forward. The inner electron is just like Hydrogen in this model, except it faces a nucleus charge of $ 2e^+ $. The outer electron's wave-function is a polynomial times an exponential that in the far-field is bound by a shielded $ 1 e^+ $ field. One can calculate the interaction energy using the expectation for the Electric field (it is after all spherically symmetric). $\endgroup$ May 30, 2023 at 13:51
  • $\begingroup$ Never heard of the inner-electron, outer-electron model. So you were able to obtain an analytical (symbolic in you terms) expression for the $2$-electron wavefunction $\Psi(\mathbf{r}_1,\mathbf{r}_2)$? $\endgroup$ May 30, 2023 at 14:10
  • $\begingroup$ @StephenElliott thanks for your answer, but it didn't really clear my doubts, additionally I haven't studied such in depth didn't even know aside from penetration and shielding there was something called perturbation. $\endgroup$
    – bm27
    May 30, 2023 at 15:51
  • $\begingroup$ @Metal Storm : Introductory College Chemistry courses talk about the inner electron and the outer electron. But it is hard to believe the texts or the teachers! There is actually an inner electron for Helium. When the outer electron is ionized, the inner electron does not change its distribution very much. There is not really $ \Psi(r_1,r_2) $ but rather $ \Psi_1(r_1) $ and $ \Psi_2(r_2) $ which are somewhat independent wave functions for electron one and electron two. Except the Energy of the system with these two electrons is different than with Hydrogen. $\endgroup$ May 31, 2023 at 13:57
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Where did you get this diagrams from?There is something definitely suspicious with your diagrams.The more the wavefunction of a particle changes over r the more energy the particle has.From your diagrams obviously the "3s" wavefunction has more energy than the other two because it has more valleys and troughs which according to the Aufbau principle cannot be true.

Also if you look at some of your diagrams at some points the wavefunction $\Psi(r)$ takes a value $>1$ which isnt possible as well.

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  • $\begingroup$ I got this from JD LEE but I have checked several other sources and they too have a similar curve $\endgroup$
    – bm27
    May 30, 2023 at 7:54
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    $\begingroup$ the figures appear to be correct they show $4\pi r^2 R_{n\ell}(r)^2$ $\endgroup$
    – porphyrin
    May 30, 2023 at 7:56
  • $\begingroup$ @porphyrin even if the x axis is $r^2$ diagram A still has more valleys and troughs which indicates a bigger energy of the wavefunction.Remember the more the wavefunction changes the bigger the energy of the particle which is described by that wavefunction $\endgroup$
    – Volpina
    May 30, 2023 at 8:05
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    $\begingroup$ @BanajMahajan can you site all your sources of the diagrams?Thanks.I could be wrong and if I am wrong I want to learn. $\endgroup$
    – Volpina
    May 30, 2023 at 8:56
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    $\begingroup$ @Volpina The radial probability diagrams are correct. but are misleading due non-homogeneous potential gradient. $\endgroup$
    – Poutnik
    May 30, 2023 at 11:52

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