Why an atom is more stable when only sublevels s and p are full?

Question

Supposedly when explaining electronegativity and stability of an element, they tell you that it is more stable if the last level is full. That works up to the third period, but after transition elements, they are full when the d sublevels are full, not when the p sublevels are full. Why is this? Is there any simple way to justify when one is teaching this to a high school level student?

Answer 1

Supposedly when explaining electronegativity and stability of an element, they tell you that it is more stable if the last level is full.

No "supposedly" about it. That concept is definitely taught.

That works up to the third period, but after transition elements, they are full when the d sublevels are full, not when the p sublevels are full. Why is this?

You seem to be defining "level" as all the orbitals with a given primary quantum number. Perhaps that comes from relying too heavily on the Bohr model of the (hydrogen) atom, which gets less and less adequate for elements with higher and higher atomic numbers. In any case, that is not the definition of "level" that applies to the concept.

The definition of "level" (a.k.a. "shell") that applies is more along the lines of "all the orbitals populated in the ground state of any neutral element in the n^th period but not in the ground state of any neutral element in the (n-1)^th period". I can appreciate that you may not find that definition very satisfying, but the whole concept is somewhat empirical.

In my experience, usually no descriptive definition is taught at all. Instead, students are directly taught the pattern of which orbitals appear in which levels. One could use this as an opportunity to point out the limitations of the Bohr model.

Is there any simple way to justify when one is teaching this to a highschool level student?

The fact that some levels comprise orbitals with a variety of primary quantum numbers is typically explained in terms of the relative energy of electrons in the various elements' various orbitals. That is, for elements whose nd orbitals are occupied in the ground state, those are close in energy to the (n+1)s and (n+1)p, but comparatively far in energy from the ns and np. An explanation of why the relative orbital energies follow the pattern they do would not be accessible to most high school chemistry students, plus it would constitute a diversion that available instruction time typically does not permit.

Answer 2

What you wrote might be a bit confused and you didn't include sources for what you said.. You may have misinterpreted your own material.

There is a pattern that breaks and i'll address that

In basic chemistry they can teach electronic configurations up to and including Calcium and stop there. And not go to Scandium. And they'd say(rightly) Calcium's electronic configuration is 2,8,8,2.

Things go funny after that because in basic chemistry they might say pretend that the third shell can take only 8 electrons. Really it can take 18 but pretend it can take 8. So then you get as far as calcium(atomic number 20). You mention transition metals. Scandium(atomic number 21) is the first transition metal, where in basic chemistry they might say let's stop just before Scandium. Because the pattern up after 2,8,8,2 then changes.

What happens at Scandium is after the third shell gets 8 electrons, the 4th shell has to get two, (hence 2,8,8,2), but then, the third shell can start filling. And after the third shell is complete. 2,8,18,2 then the fourth shell can be added to e.g. 2,8,18,3 e.t.c. Often that format of electronic configuration is used in basic chemistry and up to Calcium. This online periodic table shows the electronic configuration in that format for each element. https://ptable.com/#Properties/Series

It then becomes easier to use a model that is more complex, but says, there's each shell, 1,2,3,4,5,6 (maybe the order there is proximity to the nucleus). And each shell has subshells.

But the factor that controls the order that they fill up is "energy level". And the order of energy level goes a bit interesting around 3d(a subshell of the third shell), and 4s(a subshell of the fourth shell). And it's all simply subshells filling up.

They don't teach about the subshells at very basic chemistry.

The order of subshells filling up is, which is in ascending order of energy level, low to high, is 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f ...

There is a diagram to help work it out by hand, and a graph showing energy levels too, and the periodic table helps.

So you see 3p fills up, then 4s, then 3d, then some more 4.. So it's one subshell that fills up and then the next and then the next, in order of their energy levels.

It's not one shell that fills up and then the next e.t.c. because the shells don't divide completely by energy level as e.g. part of shell 4 could be higher than parts of shell 5. It's that one subshell that fills up and then the next e.t.c.

And Where they teach that if the outter shell is full then it's more stable, I can't comment on whether that's true even pre calcium. But you can adapt that to the highest energy subshell. i.e. the last subshell to be filled.

You mention s,p,d.. so you might be aware that each subshell e.g. s,p,d,f.. contains orbitals and each orbital can contain a maximum of two electrons. They tend to not teach the subshells and those electronic configurations in basic level chemistry but you seem to have researched it somewhat.

You can look at a periodic table where they block parts off, as 1s, 2s, 2p, e.t.c. 4s precedes 3d. And that is helpful for predicting or determining electronic configurations. And there are "exceptions" like copper and chromium in the 3d section. And even more exceptions in 4d, as you see from that periodic table showing electron configuration https://ptable.com/#Properties/Series

One sees from a diagram showing energy levels of orbitals, https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/07%3A_The_Quantum-Mechanical_Model_of_the_Atom/7.06%3A_The_Shape_of_Atomic_Orbitals "the energy levels become closer and closer together as the value of n increases". If orbitals are close, there can be some competition.

An oddity re this is that in academia there seems to be some discussion that the order of orbitals in order of energy differs a bit, from element to element, and debate on whether 4s<3d in the case of scandium. Eric Scerri argues that actually 3d<4s. Whereas Geoffrey Neuss https://www.thinkib.net/chemistry/page/37492/the-electron-configuration-of-scandium mentions this paper of his freely available on springer. And says that even though recent papers have argued that 3d<4s for scandium, he argues in response to those recent papers, that we don't know if 3d<4s for scandium.

At a very high level they understand some further things about orbitals and energy levels. https://pubs.acs.org/doi/pdf/10.1021/ed200673w

There is no dispute as to the electronic configurations of the elements. So you can see e.g. for potassium,calcium,scandium. A list of electron configurations of elements is here, in the notation that shows the subshells. https://sciencenotes.org/list-of-electron-configurations-of-elements/

Answer 3

No. Transition metals without any electrons in the 4p or 5p orbital are very reactive just like atoms without 2p or 3p orbitals fully filled.

The reason is quantum mechanical .Without any details if you solve the time independent Schrodinger equations for atoms which do not have fully filled the p orbitals by combining them with other orbitals of atoms the energy is reduced.