# Final Valence Question, I hope.

Besides, the fact that I don't understand why atoms with three shells can be stable even without filling their D orbitals, I have finally come to one final question.
Why do atoms gain a special stabilization, when the entire valence shell is filled?

## Answers and Replies

Astronuc
Staff Emeritus
With a full p-subshell - all the electrons are paired. For the details, one will have to immerse oneself in the Quantum (Wave) Mechanics of atomic (electron) orbitals.

Gokul43201
Staff Emeritus
Gold Member
Dual Op Amp said:
Why do atoms gain a special stabilization, when the entire valence shell is filled?

In case you didn't get this before: Only some atoms (the smaller ones, as explained in the other thread) gain additional stability by fully filling the valence shell. The bigger atoms attain highest stability by having 8 (not 2N^2) electrons in the valence shell. In these atoms, it is IMPOSSIBLE to fully fill the valence shell, because it is energetically advantageous to start filling a higher shell before completely filling the next inner one. This is because, the energy of an electron goes roughly like the square of (n+l).

So, the question should then read :
Why do atoms gain a special stabilization, when the valence shell is filled with exactly 8 electrons?

This is a very sensible question to ask, but unfortunately, is a little hard to satisfactorily (and classically) answer. I'll need some time to think about it, before I make an attempt. Perhaps Zz, Astronuc or someone else might find a good way to resolve this first.

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Gokul43201
Staff Emeritus
Gold Member
There are two rules that you will get used to with time (and I'll attempt to explain their origin, further down) :

Rule #1 : The energy of an electron increases with increasing n+l, where n is the shell number, and l is the sub-shell (for s, p, d, f, l = 0, 1, 2, 3 respectively) number. (Note : there can be conditions when this rule gets slightly altered)

Rule #2 : When two subshells have the same n+l value (like 2p and 3s), the one with the lower n has a lower energy.

From these above rules (which are outcomes - albeit approximate - of QM calculations), one can determine the order in which sub-shells are chosen for occupancy.

This is the order (verify using above rules, and correct me, if I've made a mistake) :

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < ...

Argument for the Origin of the above Rules :

Let's look at what happens to the energy of the atom/ion when you add electrons. There are two effects that dictate the changes in energy.

(i) Initially, adding more electrons to a given shell reduces the energy, because it increases the attractive force on the nucleus (which has a negative sign). Now, at the same time, this addition will also increase the repulsion (positive sign) between the electrons themselves, but this increase is found to be smaller than that with the nucleus. Since the reduction in energy (due to increased attraction from the nucleus) dominates the increase in energy (due to increased repulsion between electrons), the net effect is a reduction of energy - or an improvement of stability. However, as more electrons are added to the shell, the shell starts to get more crowded, so after a point, the repulsion between the electrons becomes more dominant. As a result, further addition of electrons beyond a particular number causes the energy to go up. Typically, this starts to happen whenever you've just filled (or sometimes, exactly half-filled) a sub-shell. Sometimes, this increase can be small, but other times it can be large.

(ii) Adding an electron into a new shell increases the energy too. Since this new (bigger) shell is farther away from the nucleus, the increase in attraction to the nucleus is smaller (remember, the electrostatic force is inversely proportional to the square of the distance between the charges) and so the repulsion is more dominant, causing the energy to go up (and usually by a significant amount).

So at certain points in the filling process (usually when a sub-shell has just been filled), you have to make a decision between filling the next sub-shell in the same shell (which increases the energy) and going to the next shell itself (this too increases the energy). The decision depends on which route causes a smaller increase in energy. Sometimes, it's better to go to the next subshell and at other times it's preferable to go to the next shell. Stated differently, it is possible for a lower subshell in a higher shell to have a lower energy than a higher subshell in the same shell. The energies of these sub-shells have been calculated using QM and the essence of the results of this calculation is conveyed by the 2 rules above.

(While this explains why the filling order follows some complex pattern, it does not explain why this pattern can be described (approximately) by the above rules. To understand this, you must know how to do the calculations.)

Let's now apply the rules to the elements to better understand how they work.

Start with no electrons, and keep adding one at a time, putting the first electron in 1s. The addition of the second electron to 1s reduces the energy as explained in (i), so He (=1s^2) is more stable than H(=1s^1). Now the next electron must go to 2s, causing the energy to increase (by quite a large number actually). Thus, Li (=1s^2 2s^1) is more unstable than He. So, He sits at something like a stability peak (Duet/Fully Filled Shell Rule). Adding a second electron to 2s makes Be (=1s^2 2s^2) more stable than Li. But adding the first electron to 2p causes the energy to increase, though by a small enough amount that this is still preferable to 3s. So B (=1s^2 2s^2 2p^1) is marginally more unstable than Be. Adding more electrons to 2p makes C (=1s^2 2s^2 2p^2), N (=1s^2 2s^2 2p^3),..., Ne (=1s^2 2s^2 2p^6) more and more stable. The only deviation from this trend is that O (=1s^2 2s^2 2p^4) is actually a little bit more unstable than N, because N has an exactly half-filled 3p sub-shell (the reason for this increase in energy beyond half-filling has to do with the spins of the electrons). Now, as expected, adding an electron to Ne involves starting a new shell, making Na much more unstable than Ne. So, Ne occupies the second major stability peak. But keep in mind that, along the way, Be and N also occupied little peaks themselves.

(Check the bar graph at this link to confirm that the above analysis is indeed correct. The ionization energy is a measure of stability)

Now finally, we can show why the Octet Rule works, based upon the ordering of sub-shells as determined by the above two rules. Let's list the order again :
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < ...

First 1s is filled (Duet/Fully Filled Shell Rule) .

Then 2s and 2p are filled completing the second shell ((Octet/ Fully Filled Shell Rule)

Next, 3s and 3p are filled. But the next electron goes to 4s (a new shell), making a large drop in stability. So, Ar (=1s^2 2s^2 2p^6 3s^2 3p^6) sits at a stability peak with 8 electrons in the valence (n=3) shell (Octet Rule).

Similarly, the filling of 4s, 3d (which is more energetic than 4s, but better than 4p) and 4p gives us Kr (=1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6). But since the next electron goes to 5s (a new shell), Kr sits at another peak with 8 electrons in the valence (n=4) shell (Octet Rule).

Checking the others in the list shows that the Octet Rule applies to all atoms beyond He.

Further Insights :

This pattern (which is a result of the two rules) suggests that the d sub-shells are energetically expensive, because whenever some {N}p (n+l =N+1) sub-shell gets filled, the next electron goes to {N+1}s (n+l = N+1) rather than {N}d (n+l = N+2), since clearly N+1 < N+2. This is, in fact true. The shape of the d sub-shells makes them relatively unfavorable.

Conclusion :

So, in short, the "lower n" rule (#2) ensures that the valence shell gets an Octet (since {N+1}s will not be filled before {N}p, as long as {N}p exists), and the "n+l" rule (#1) ensures that you won't get more than an Octet (since {N+1}s will get filled before {N}d).

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dextercioby
Homework Helper
Thanks,Gokul,for providing this great insight to Hund's rules and how they help building the periodic table.(I my school,in the atom and molecule phyiscs course,they were both missing,though i can remember discussing the periodic table... :grumpy: )
I would have done it myself,had i been able to come up with an english text,but the only one i could find among my documents was in German. And translating it to English would have taken me a couple of days. :yuck:

Daniel.

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Gokul43201
Staff Emeritus
Gold Member
It took me all of the two hours following the first post to think and write up the second one. <tired>

Well, I greatly appreciate it. I finally understand, this also explains the other post on titanium. That was an EXTREMELY useful and descriptive post, I thank you very much. This clears up all my questions. You must be a physics teacher!
Amp.
I do have one question, though, how does adding electrons to the atom increase the attractive force to the nucleus?

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Anyone here?

Gokul43201
Staff Emeritus
Gold Member
Dual, I think I clarified in my PM to you, that adding electrons does not increase the net force, but each new attractive force vector is associated with a new (negative) energy term (due to the attractive interaction between the nucleus and the electron).

If you've done some basic electrostatics (usually covered late in high school), you'll understand what I mean by this. If not I can find you some links to look up.

I'm not following you.

Gokul43201 said:
There are two rules that you will get used to with time (and I'll attempt to explain their origin, further down) :

Rule #1 : The energy of an electron increases with increasing n+l, where n is the shell number, and l is the sub-shell (for s, p, d, f, l = 0, 1, 2, 3 respectively) number. (Note : there can be conditions when this rule gets slightly altered)

Rule #2 : When two subshells have the same n+l value (like 2p and 3s), the one with the lower n has a lower energy.

Argument for the Origin of the above Rules :
But no good arguments for the origins of these rules were given...

That Rule #1 is not really correct. In hydrogen or hydrogen-like atoms, the orbital energy is independent of $$l$$.

As to the origin of Rule #2: The outer electrons do not feel a pure $$1/r^2$$ force, but the nuclear charge is partly screened by the inner electrons. Orbitals with high angular momentum are like classical circular orbitals. They do not penetrate this screening charge cloud, and they do not feel as much of the nuclear attraction as the the elliptical orbitals with lower $$l$$.

In particular, s-orbitals have a non-zero electron density near the nucleus. That is why the 4s-orbital is lower in energy than the 3d. (At least in the Aufbau; when you look at core binding energies of say silver, the 3d binding energy is higher than that of 4s.)
http://www.webelements.com/webelements/elements/text/Ag/bind.html

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Gokul43201
Staff Emeritus
Gold Member
Pieter Kuiper said:
But no good arguments for the origins of these rules were given...
Yes, I should have called them "arguments for the existence of the rules". I thought going any deeper would have been befyond the scope of the OP.

That Rule #1 is not really correct. In hydrogen or hydrogen-like atoms, the orbital energy is independent of $$l$$.

Thanks for the critique. Yes, for Hydrogen-like atoms, E(n) ~ 1/n^2. That includes H, He+, Li++..., where there are no electron-electron interactions. Among these only H is a stable atom.

As to the origin of Rule #2: The outer electrons do not feel a pure $$1/r^2$$ force, but the nuclear charge is partly screened by the inner electrons. Orbitals with high angular momentum are like classical circular orbitals. They do not penetrate this screening charge cloud, and they do not feel as much of the nuclear attraction as the the elliptical orbitals with lower $$l$$.

In particular, s-orbitals have a non-zero electron density near the nucleus. That is why the 4s-orbital is lower in energy than the 3d. (At least in the Aufbau; when you look at core binding energies of say silver, the 3d binding energy is higher than that of 4s.)
http://www.webelements.com/webelements/elements/text/Ag/bind.html
I tried to avoid raising screening (or shielding), at this point, because I couldn't think of a good way to explain it (quickly) to a 9/10 grade student (who, I think, has not yet had instruction in electrostatics, and probably has never heard the term orbital angular momentum). In retrospect, I realize this may be an omission on my part to not mention it.

On the other hand, you don't really have to go to Ag or Au (where the screening effect is pronounced) to see screening. The reason that 2p is higher than 2s (in non H-like atoms) is itself partially due to a screening effect.

Gokul43201 said:
On the other hand, you don't really have to go to Ag or Au (where the screening effect is pronounced) to see screening. The reason that 2p is higher than 2s (in non H-like atoms) is itself partially due to a screening effect.
I just mentioned silver to show that the quantitative details cannot be explained by simple arguments. Screening always results in 3d being higher than 3p, but to know whether 3d is below 4s (as in silver) or not (as in potassium, calcium, iron...) requires numerical computer calculations.

My impression of th OP's questions is that he wants to "see" a physical explanation of why all these rather arbitrary-looking rules are what they are. But yes, it would be easier to explain things sitting beside him, or making sketches on the blackboard.

Gokul43201
Staff Emeritus
Gold Member
My original intention was to only point out the fallacy in the OP's understanding that a fully filled valence shell created stability. But what started out as a 5-line post, went through successive edits to try and convey more explanation, without invoking too many different effects, as well as trying to avoid mentioning words like 'probability', 'charge density' or others such, that would only result in an explosion of more questions.

Gokul, I understand the basics, it's just the way you're wording that last part. Here this is an E-mail I wrote up for a friend, I've been searching the answer to covelant bonding for some time now.
Electrons are attracted to protons, but repell electrons. So, instead of all the electrons being bunched up right next to the nucleas, they orbit around the nucleas in shells. These shells can sometimes contain sub-shells. For example, the first shell contains only one sub-shell. As an electron gets further away from it's atom, it must have more "quantum energy." Electrons want to get as close to the nucleas as possible, but according to quantum physics, no to electrons can have the same "quantum energy." So, they orbit in shells. The electrons orbit in orbitals. The sub-shells have orbitals. For example, the 1 shell has an S orbital. Because it's an s orbital and it's the first shell it's labelled 1S. For 1-First shell-, S-S orbital. An S orbital has the shape of a sphere. An orbital wants to fill it's self. Alright, so why would the atom want to have 8 electrons in it's outer most shell, good question. The second shell has two sub-shells. One sub-shell has an S orbital, and the second has three P orbitals. The reason it has three is because they can arrange themselves according to X,Y,Z. Each orbital has only two electrons, because no two electrons can have the same "quantum energy." So, for the valence shell of an atom with two shells, one S orbital and three P orbitals. Two electrons an orbital adds to...8. Hydogen, on the other hand, only has one shell. So, to fill it's valence shell, it only needs two electrons. It already has one - Hydogen = one proton, one electron - so, it only needs to bond with one atom to fill itself. Carbon, on the other hand, has two shells, so it needs 8 to fill it's valence shell. So....

H
H C H Methane!!! CH4.
H

If you were to count it up everyone's filled. The carbon atom has 6 electrons. 2 in it's first shell, and 4 in it's valence shell. It needs 8 in it's valence shell. So, it shares one with hydrogen, and the hydrogen shares one of the carbons. This gives the carbon an extra electron, and the hydrogen it's desired two. The carbon, then, bonds with three more to add to 8.

HOH Water!!! H20. Oxygen has six valence electrons, meaning it needs 2 to gain, which it does with 2 hydrogen molecules.

O=O Oxygen!!! O2.

You're probably wondering, why is there an equals sign between the Oxygen molecules?
This indicated a double bond. Oxygen has six valence electrons, when it bonds with another oxygen, it gets 7. That's not the desired 8. So, it makes a double bond, and they share two electrons each. Which adds to 8.

O
O O Ozone!!! O3. Each one of these atoms share with each other, making 8.

That's covelant bonding!!!
This "quantum energy I told you about is somewhat true. What's really true is that there are four "quantum numbers" that cannot match.
The first is N.
N is the energy of an electron. For example, an electron in the first shell would have an N of 1. An electron in the second shell would have an N of 2. An electron in the third shell would have an N of 3.
N=1, means it's in the first shell.
The second is L. It's actually a greek cursive L kind of like this. l. Okay. This sign is the orbital. L = N - 1. That's the equasion. So, if N = 1, then, L = 0. 0 is an S orbital.
If N = 2, L can equal either 0 or 1. If it is 1, that's a P orbital. If N = 3, then that can be either 0,1 or 2. An S,P or...a D orbital.
Now, the third quantum number is M. It is the orientation of the orbitals, you know XYZ.
M can equal anything between -L and +L. For example if L is 1, then M can equal -1,0,1.
This is 3 different ways of arranging the P orbital.
Now the final one is Ms. For Spin. The spin of the electron can equal - 1/2 or 1/2.

Okay, so let's look at the possible arrangements of some electrons.

N L M Ms
1 0 0 -1/2
1 0 0 1/2 First shell, only can have two electrons.

2 0 0 -1/2
2 0 0 1/2
2 1 -1 -1/2
2 1 -1 1/2
2 1 0 -1/2
2 1 0 1/2
2 1 1 -1/2
2 1 1 1/2 Second shell, eight electrons, but none of them, nor the one's in the first shell have the same 4 quantum numbers.

HOPE YOU UNDERSTAND. IT TOOK ME A WHILE TO WRITE, I'D HATE TO LOSE IT AT THE LAST MOMENT, LIKE THE POWER SHUT DOWN OR SOMETHING. IF YOU UNDERSTAND THIS, YOU WILL UNDERSTAND THE REST.
HERE'S SOME SITES.

http://chemed.chem.purdue.edu/gench...h6/quantum.html [Broken]

http://lectureonline.cl.msu.edu/~mm...od/electron.htm [Broken]
So, it's not going to result in an explosion of questions.

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Okay, I think I know what you're saying, but correct me if I'm wrong.

The protons are attracted to the other atom's electrons, and visa versa. The electrons are attracted to the other atom's protons, and visa versa.
As the atom gets closer, the repullsiveness of the electrons would grow
just as much as the attractiveness, cancelling out the force. But, the
electrons may spin in opposite direction, reducing the repullsive force.
Meaning, that the attractiveness is larger than the repullsiveness, but
eventually, the repullsiveness catches up.
The reason the electrons are in orbitals, is because that's the least
repullsive form.
Correct?
Even in the ball park?

Am I right?

Gokul43201
Staff Emeritus
Gold Member
Actually, in the discussion till now we only talk about a single atom, and not the effect of bringing atoms together. This gives rise to behavior that really can not be explained classically. Band formation and ligand field effects are quite complex (not to mention, Quantum Mechanical effects), and we have not involved them in the picture so far.

Electrons are in orbitals
(i) to reduce the repulsion, and
(ii) because of certain quantum mechanical rules that they are forced to obey (Exclusion)

Uhh...
Geeze, why is this so hard?
Do you have any links?
Okay, I understand it all. Only one part I don't understand.
One second.
Okay.
Initially, adding more electrons to a given shell reduces the energy, because it increases the attractive force on the nucleus (which has a negative sign).

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Gokul43201
Staff Emeritus
Gold Member
So hard ? Hardly !

The real problem is way harder than any of us have described here. It involves very complicated math, and a radically counter-intuitive framework of physics that defies visualization. In fact, calculating the energies of the various levels in a single atoms becomes incredibly difficult if you have more than just a couple of electrons. It takes powerful computers to solve for such simple atoms as Li, Be, and B.

To address your second question, one needs to know a little about your current level of physics education. Have you been tought electrostatics ? Do you know the expressions for force (coulomb's law), and work, and how to use them ?

Gokul43201 said:
It takes powerful computers to solve for such simple atoms as Li, Be, and B.
The computer you are using to read this should be sufficiently powerful for any atom.

(I agree with the rest of what you wrote.)

Yes, coulumb's law is the fact that opposites attract and likes repell, an electron repells an electron, but is attracted to a proton. I understand, I guess, the basics of quantum mechanics. I understand electrostatics, mainly, because I've been taught electronics from my Grandfather. I took electronics to the sub-atomic level. I understand the essentials. Quarks, orbitals, gluons. Is that enough.
Wow, you got here fast.

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Gokul43201
Staff Emeritus
Gold Member
I guess I'm not nearly up to speed on the computational side. I thought there were difficulties in incorporating relativistic effects. And I thought it was quite computationally intensive to do DFT calculations.

Need to do some catching up, I guess.

I'm not sure how it takes advanced math, just to compute the energy of an atom, but, I need to go to college before I get deep into this stuff, anyway last couple of posts is an answer to what you wanted.

To address your second question, one needs to know a little about your current level of physics education. Have you been tought electrostatics ? Do you know the expressions for force (coulomb's law), and work, and how to use them ?
Yes, coulumb's law is the fact that opposites attract and likes repell, an electron repells an electron, but is attracted to a proton. I understand, I guess, the basics of quantum mechanics. I understand electrostatics, mainly, because I've been taught electronics from my Grandfather. I took electronics to the sub-atomic level. I understand the essentials. Quarks, orbitals, gluons. Is that enough.
Wow, you got here fast.