![]() Introducing a second electron into a 3d orbital produces more repulsion than if the next electron went into the 4s orbital. The 3d orbitals are quite compactly arranged around the nucleus. It obviously helps if this effect can be kept to a minimum. Repulsion raises the energy of the system, making it less energetically stable. If you add another electron to any atom, you are bound to increase the amount of repulsion. You have something else to think about here as well. You might expect the next electron to go into a lower energy 3d orbital as well, to give 3d 2. 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. Now you are going to add the next electron to make Sc 2 +. Essentially you have made the ion Sc 3 +. ![]() So far you have added 18 electrons to fill all the levels up as far as 3p. You have built the nucleus from 21 protons and 24 neutrons, and are now adding electrons around the outside. Imagine you are building a scandium atom from boxes of protons, neutrons and electrons. 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. 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. If you stop and think about it, that has got to be wrong.Īs you move from element to element across the Periodic Table, protons are added to the nucleus and electrons surrounding the nucleus. In other words, we assume that the energies of the various levels are always going to be those we draw in this diagram. The flaw lies in the diagram we started with (Figure 1) and assuming that it applies to all atoms. It is way of working out structures - no more than that. The problems arise when you try to take it too literally. 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 explanations around ionization energies are based on the 4s electrons having the higher energy, and so being removed first. 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. 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. Those statements are directly opposed to each other and cannot both be right. So the 4s orbital must have a higher energy than the 3d orbitals. The electrons lost first will come from the highest energy level, furthest from the influence of the nucleus.
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