[Francis]

Hi, as an atom's Z number increases its size different energy levels of electrons appears. As the electrons are filled in an atom we see that the 4s orbital is being filled before the 3d orbital. Then when we take away electrons from transition metals, the 4s orbital electrons goes away first then the 3d's are removed.

 In my attempt to understand this weird phenomenon I have come across an article by Eric R. Scerri who is a chemist at the University of Purdue. This criticism of Atkins textbooks states that even though at a basic level we learn in general chemistry that the reason why 4s is filled before 3d is because 4s is lower in energy than in 3d this is infact false because 4s never is lower in energy than 3d (http://www.chem.ucla.edu/dept/Faculty/scerri/pdf/Atkins_critique.pdf - beginning of page 6).

So now my two questions are why does the 4s orbital is being filled before the 3d one and why are the 4s electrons are removed before the 3d one.

If this article is wrong and 4s is indeed really lower in energy than 3d then the first answer is easily answer with Hund's rule but then the second question takes all it's importance, if on the other hand the article is right about the 4s and 3d energy level then the first question becomes hard to answer and the second one is answer by a '' because the 4s electrons have more energy than the 3d''.

Anyhow ignore the last paragraph if it confused you

Thanks a lot for your help,
Francis

By the way, my knowledge of Schrodinger Equation's and Wavefunction is fairly low because I never learned that in class yet, I only know these from layman book's Ive read about quantum mechanics so if you can give me a good explanation that doesn't require the mathematical parts I'll be happy and wait till I learn everything to understand it mathematically!

[Yggdrasil]

The way that I have learned it that the average distance from the nucleus in the 4s and 3d orbitals is about the same, and therefore, the average energy is roughly the same (with the 4s orbital having slightly lower energy).  Because it has lower energy, electrons will occupy the 4s orbital before the 3d orbital.  However, when electrons are removed, they are removed from the most easily accessible orbital (i.e. the orbital with the most electron density farthest away from the nucleus).  It turns out, because the 4s orbital has various radial nodes at intermediate distances from the nucleus, the 4s orbital has a greater electron density away from the nucleus than the 3d orbital (whose electron density is packed more closely to the mean distance from the nucleus).

[Francis]

Thanks a lot for your answer, I was starting to look into spin orbit coupling and the implication when two electrons exchange place in an atom to understand this since I couldn't find a more basic explanation, anyhow I'll look more into your answer!

Thanks a lot I really appreciate that you have took the time to answer my question.
Francs

[FeLiXe]

If you want a quantitative answer you can use: Slater's rules

You know that the energy of an orbital in a hydrogenic atom is proportional to -Z^2/n^2. In a multi-electron atom the nuclear charge Z is lowered because of the other electrons. The remaining effective charge is calculated by Slater's rules.

Now you have two adverse effects on the energy of the orbital. You want a high effective nuclear charge and a low principle quantum number.

High Z means low energy. It is achieved if screening is little. Screening for d electrons, as calculated by slaters rules, is very high. Therefore s electrons are preferred.

But with 4s electrons the principle quantum number is higher than for 3d electrons.

If you consider a neutral atom the first thing is more important and s-levels are populated. If you add a positive charge then differences in effective charge will be lower and the second part is more important.

I tried the calculation for V(0) and V(II) and it worked. You can see that here

I hope it was kind of understandable. It was interesting to think about it.