Dear Yasmin,

Please read the text in the following link:

http://chemwiki.ucdavis.edu/Core/Inorganic_Chemistry/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/The_Order_of_Filling_3d_and_4s_Orbitals

The Order of Filling Orbitals

The aufbau principle explains how electrons fill low energy orbitals (closer to the nucleus) before they fill higher energy ones. Where there is a choice between orbitals of equal energy, they fill the orbitals singly as far as possible (Hunds rules). The diagram (not to scale) summarizes the energies of the orbitals up to the 4p level.

The oddity is the position of the 3d orbitals, which are shown at a slightly higher level than the 4s. This means that the 4s orbital which will fill first, followed by all the 3d orbitals and then the 4p orbitals. Similar confusion occurs at higher levels, with so much overlap between the energy levels that the 4f orbitals do not fill until after the 6s, for example. 

Everything is straightforward up to this point, but the 3-level orbitals are not all full - the 3d levels have not been used yet. But if you refer back to the energies of the orbitals, you will see that the next lowest energy orbital is the 4s - so that fills first.

K

1s22s22p63s23p64s1

Ca

1s22s22p63s23p64s2

d-block elements are thought of as elements in which the last electron to be added to the atom is in a d orbital (actually, that turns out not to be true! We will come back to that in detail later.) The electronic structures of the d-block elements are shown in the table below. Each additional electron usually goes into a 3d orbital.

d-block ions

This is probably the most unsatisfactory thing about this approach to the electronic structures of the d-block elements In all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. The reversed order of the 3d and 4s orbitals only seems to apply to building the atom up in the first place. In all other respects, the 4s electrons are always the electrons you need to think about first.

The d-block elements: For reasons which are too complicated to go into at this level, once you get to scandium, the energy of the 3d orbitals becomes slightly less than that of the 4s, and that remains true across the rest of the transition series.

Example 2: Scandium

So why is not the electronic configuration of scandium [Ar] 3d3 rather than [Ar] 3d14s2?

SOLUTION

Making Sc3+

Imagine you are building a scandium atom from boxes of protons, neutrons and electrons. You have built the nucleus from 21 protons and 24 neutrons, and are now adding electrons around the outside. So far you have added 18 electrons to fill all the levels up as far as 3p. Essentially you have made the ion Sc3+.

Making Sc2+

Now you are going to add the next electron to make Sc2+. Where will the electron go? The 3d orbitals at scandium have a lower energy than the 4s, and so the next electron will go into a 3d orbital. The structure is [Ar] 3d1.

Making Sc+

You might expect the next electron to go into a lower energy 3d orbital as well, to give [Ar] 3d2. But it doesn't. You have something else to think about here as well. If you add another electron to any atom, you are bound to increase the amount of repulsion. Repulsion raises the energy of the system, making it less energetically stable. It obviously helps if this effect can be kept to a minimum.

The 3d orbitals are quite compactly arranged around the nucleus. Introducing a second electron into a 3d orbital produces more repulsion than if the next electron went into the 4s orbital. There is not a very big gap between the energies of the 3d and 4s orbitals. The reduction in repulsion more than compensates for the energy needed to do this.

The energetically most stable structure for Sc+ is therefore [Ar] 3d14s1.

Making Sc:

Putting the final electron in, to make a neutral scandium atom, needs the same sort of discussion. In this case, the lowest energy solution is the one where the last electron also goes into the 4s level, to give the familiar [Ar] 3d14s2 structure.

In conclusion

The current method of teaching students to work out electronic structures is fine as long as you realize that that is all it is - a way of working out the overall electronic structures, but not the order of filling.

You can say that for potassium and calcium, the 3d orbitals have a higher energy than the 4s, and so for these elements, the 4s levels fill before than the 3d. That, of course, is entirely true!

Then you can say that, looking at the structures of the next 10 elements of the transition series, the 3d orbitals gradually fill with electrons (with some complications like chromium and copper). That is also true.

What is not right is to imply that the 3d levels across these 10 elements have higher energies than the 4s. That is definitely not true, and causes the sort of problems we have been discussing. 

References

Dr Eric Scerri provided me with copies of a number of useful papers, and helped me to get my ideas about it sorted out.http://ericscerri.blogspot.com/2012/06/trouble-with-using-aufbau-to-find.html

R. N. Keller: Textbook errors, 38: Energy Level Diagrams and Extranuclear Building of the Elements: J. Chem. Educ., 1962, 39 (6), p 289, published June 1962

W. H. Eugen Schwarz: The Full Story of the Electron Configurations of the Transition Elements: Journal of Chemical Education, Vol. 87 No. 4 April 2010.

Hoping this will be helpful,

Rafik

Rafik, let us consider the complex ScF3. The formal oxidation number of Sc is +3, therefore, the d^n count is (oxidation # - group #) = 3-3 = 0. When calculating the number of d electrons we are assuming all the electrons are in the d orbital and not it the s orbital, does this contradict what is explained? 

Dear Yasmin,

Scandium forms one ion, Sc3+, with an electron configuration of 1s2 2s2 2p6 3s2 3p6. This is the main ion, and it has only empty d orbitals, and so scandium is excluded from the list of transition metals.

Absolutely right. There is no contradiction.

Rafik

Apologies, that was a poor example. Let's consider the [Co(NH3)6]3+ complex. The d^n count is 6, therefore, 6 electrons are in the d orbitals. My question what causes these electrons to occupy the d orbitals and not the (supposedly) 'lower' 4s orbital (as governed by the Aufbau principle)?

Dear Yasmin,

The Electron Configuration of Transition-Metal Ions

The relationship between the electron configurations of transition-metal

elements and their ions is complex. Example: Let's consider the chemistry of cobalt which forms complexes that contain either Co2+ or Co3+ ions.

The electron configuration of a neutral cobalt atom is written as follows.

Co: [Ar] 4s2 3d7 The discussion of the relative energies of the atomic orbitals suggests that the 4s orbital has a lower energy than the 3d orbitals. Thus, we might expect cobalt to lose electrons from the higher energy 3d orbitals, but this is not what is observed. The Co2+ and Co3+ ions have the following electron configurations.

Co2+: [Ar] 3d7

Co3+: [Ar] 3d6

In general, electrons are removed from the valence-shell s orbitals before

they are removed from valence d orbitals when transition metals are ionized.

A fast way of working out the d-electronic configuration of a transition metal ion is to subtract the oxidation number of the ion away from its group number.

The same treatment should be done with [Co(NH3)6]3+ 

Hoping this will be helpful,

Rafik

Dear Rafik, 

I am fully aware of the fact that when transition metal complexes are formed the d orbital becomes lower in energy than the s orbital. However, my question is regarding the explanation behind this, so, what is responsible for the lowering of energy of the 3d with respect to the 4s orbital?

Dear Yasmin,

I refer you to the following link which describes the mentioned phenomenon in transition metals (please see answer 2):

http://chemistry.stackexchange.com/questions/33297/why-do-3d-orbitals-have-lesser-energy-than-4s-orbitals-in-transition-metals

Hoping this will be helpful,

Rafik

Dear Yasmin

The electronic configuration of transition elements should start from [Ar]. Thus for atomic number 21, it is [Ar] + 3 proton + 3 electrons = [Ar]3d1 4s2, and so forth for the next atoms. Thus the increasing atomic number (protons in nucleus) should responsible to this. This in line with the PES data.

NOTE: aufbau should be defined as that by IUPAC. DO NOT confuse by aufbau of (n+l) values proposed by Madelung resulting in diagonal lines of filling electrons.

Thanks

Rafik Karaman

Dear Dr. Karaman,

Why would electrons be removed from valence-shell d orbital instead of s orbital for transition metals? If 3d is more compacted around nucleus than 4s is, and thus electrons should preferably be placed in 4s instead of 3d, then why would in Copper, the opposite is true? Why would the electron configuration [Ar] 4s13d10 is in reality more preferred than the predicted [Ar] 4s23d9?

Thank you,

Han