Atomic structure

i still don't understand why when the 3d sublevel is half-filled or completly filled we get : 3d5 4s1 or 3d10 4s1..

I need a real expert to clarify it for me..
Thanks Joe

Those configurations occur when the 3d orbital is almost half-filled or full, meaning 3d4 or 3d9. Because of the extra stability associated with a half-filled or full 3d orbital, an electron from the 4s orbital jumps into the 3d to achieve that state.

and one more thing, why instead of numbering: 4s1 3d5 (for example)
we number: 3d5 4s1.
the energy levels are changed???

Gokul43201
Staff Emeritus
Gold Member
I can answer this (the original question) in part for now, but will need to think more before I venture further.

The reason why the $4s^1~3d^5$ is prefered to $4s^2~3d^4$ comes from Hund's First Rule (rather, a variant of it). The 3d and 4s subshells are energetically very close to each other. The energy cost from moving (raising) the second 4s electron to the 3d subshell is smaller than the gain from now having to have no spin pairing.

Why is spin pairing costly ? Two words : coulomb repulsion (if you are familiar with the Hubbard Hamiltonian, you will remember a term that looks like $U \sum (n_{ \uparrow } ~ n_{ \downarrow })$ for the coulomb interaction; this is similar). It is energetically expensive to put two electrons in the same orbital and that is the origin of Hund's first rule.

Furthermore, I believe there may be an additional stabilization coming from half/full filling. I need to think about this (I can not explain the $4s^1~3d^{10}$ config this way...or can I ?)

Last edited:
Gokul43201
Staff Emeritus
Gold Member
A_I_ said:
and one more thing, why instead of numbering: 4s1 3d5 (for example)
we number: 3d5 4s1.
the energy levels are changed???
This is actually a far deeper question than it appears to be.

The reason that the 4s subshell is sometimes written after the 3d subshell is that it has a lower "ionization energy" (hence 4s electrons are removed before 3d electrons : for instance, Mn(2+) is 3d5 4s0). This leads to the obvious follow up question : "How can the 4s subshell have a lower "ionization energy" when it is lower in energy than the 3d subshell ?" The answer to this is that ionization must take into account both the energy of the system before (system = atom) ionization as well as after (system = ion + free electron). This calculation has been performed for various elements in the 30s and 40s and verified experimentally.

Gokul43201
Staff Emeritus
Gold Member
Even more elusive a question to answer is simply "why does a fully filled subshell have extra stability ?"

A_I_ : It would help for us to know what your educational background is. I should have asked this question earlier, as I now suspect I may have be addressing the issue on a different level than is required.

r u asking me a question??
i guess it is because it realises it's octet rule :S
i really didnt understand what is ur primary goal from ur ultime post

Gokul43201
Staff Emeritus
Gold Member
A_I_ said:
r u asking me a question??
i guess it is because it realises it's octet rule :S
i really didnt understand what is ur primary goal from ur ultime post
No, I'm not asking you a question; nevermind.

I wanted to know what your educational background is (high school, college chemistry major, graduate student,... something else) so I can answer your question accordingly.

oh, that will be surprising.
I'm senior (two month until graduation from high school)
I'm in a french system school here in Lebanon
but positively moving to English education
Now i will be writing in the 7th of May the SAT2 subjects test in maths2, physics, chemistry

The french system is very different from the american one or maybe yours;
since high school they start separating students those in the litterature field and those i nthe scientific field.
this occurs in the pre-senior year.
in our senior year, there is always another partition where scientific students are separated into two other different fields: the experimental sciences field (biology most studied) and the theoretical sciences filed (math and physics) while chem,istry plays nearly the same role in both.
Next year will be my first year in university; my chosen major is "Earth and Planetary sciences"; i got accepted in a US university; i am still waiting for the decision of the canadian one.

You got a pretty good background!

Basically a 4s orbital is lower in energy as compared to a 3d orbital. This is due to the effective nuclear charge experienced by the electron from the nucleus; the degree to which the electron is able to penetrate the nucleus.

Generally due to the rate of penetration and shielding of electrons, the order of energies of many-electron atoms is s<p<d<f where s orbitals are most penetrating and f orbitals are least penetrating. Think of an s orbital electron as a tennis ball and an f orbital electron as a sponge ball hitting paper. Which will fly right through? Not only that, as you go across and down the periodic table, the atom gains more subshells and the sponge ball has to fly through not only one paper, but a stack of it to get through the nucleus which is pretty much impossible.

Secondly, the reason why atomic orbitals have half filled shells is answered by Hund's rule - "For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximised." This is based on the fact that electron-electron repulsion exists in the same orbital (opp spins +1/2 and -1/2), so by placing them in separate orbitals of parallel spins, this minimises the repulsions since the electrons are spaced apart from one another. Therefore the atom is able to achieve its lowest energy.

Congrats! All the best to your studies!