For calcium, that is the case, and the 3d orbitals are known to not be occupied, so one wrote 4s after 3p :. At scandium and past, the 3d orbitals dip below the 4s in energy, but some sloppy extensions of the Aufbau principle you may see out there that have a VERY specific, winding path through orbitals You can read more about that here.
It's just become a bad "habit" to write 4s before 3d. It really should be with the forethought of which orbital is next highest in energy , and that one goes next in the configuration. Hence, you really should write, for the first-row transition metals,. This indicates that the 4s orbital is higher in energy.
All data for orbital potential energies I reference in this answer are found here. See how the 3d orbitals are lower in energy than the 4s for the first-row transition metals here:. And you can further see how the Aufbau principle fails for the heavier transition metals, in that the n-1 d and ns orbital potential energies criss-cross rather unpredictably:. This is because the 4s sublevel has higher energy than the 3d sublevel.
Coote , Clemence Corminboeuf. Theory and practice of uncommon molecular electronic configurations. Historical Teaching of Atomic and Molecular Structure. Catalysis Letters , 1 , Mendl , Gero Friesecke. Efficient algorithm for asymptotics-based configuration-interaction methods and electronic structure of transition metal atoms. The Journal of Chemical Physics , 18 , Angewandte Chemie , 19 , Angewandte Chemie International Edition , 48 19 , Eugen Schwarz , Shu-Guang Wang.
Some solved problems of the periodic system of chemical elements. Foundations of Chemistry , 9 2 , Classification of Organotransition Metal Compounds. Wang , Y. Qiu , H. Fang , W. Chemistry - A European Journal , 12 15 , Autschbach , S. Siekierski , M. Seth , P. 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. 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. For convenience, [Ar] is used to represent 1s 2 2s 2 2p 6 3s 2 3p 6. 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.
Consider the electronic structure of neutral iron and iron III. The 4s electrons are lost first followed by one of the 3d electrons. This last bit about the formation of the ions is clearly unsatisfactory. When discussing ionization energies for these elements, you talk in terms of the 4s electrons as the outer electrons being shielded from the nucleus by the inner 3d levels.
We say that the first ionization energies do not change much across the transition series, because each additional 3d electron more or less screens the 4s electrons from the extra proton in the nucleus. The explanations around ionization energies are based on the 4s electrons having the higher energy, and so being removed first. The usual way of teaching this is an easy way of working out what the electronic structure of any atom is - with a few odd cases to learn like chromium or copper.
The problems arise when you try to take it too literally. It is way of working out structures - no more than that. The flaw lies in the diagram we started with Figure 1 and assuming that it applies to all atoms. In other words, we assume that the energies of the various levels are always going to be those we draw in this diagram.
If you stop and think about it, that has got to be wrong. As you move from element to element across the Periodic Table, protons are added to the nucleus and electrons surrounding the nucleus. The various attractions and repulsions in the atoms are bound to change as you do this - and it is those attractions and repulsions which govern the energies of the various orbitals.
That means that student must rethink this on the basis that what we drew above is not likely to look the same for all elements. So rather than working out the electronic structure of scandium by imagining that you just throw another electron into a calcium atom, with the electron going into a 3d orbital because the 4s is already full, you really need to look more carefully at it.
Remember that, in reality, for Sc through to Zn the 3d orbitals have the lower energy - not the 4s. So why is not the electronic configuration of scandium [Ar] 3d 3 rather than [Ar] 3d 1 4s 2? 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.
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] 3d 1. You might expect the next electron to go into a lower energy 3d orbital as well, to give [Ar] 3d 2. 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.
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