Re: Why does the 4s orbital come before the 3d? I don't understand the exact reason why 3d comes before 4s when writing the electron configuration. Post by Jennifer Tuell 1B » Sat Apr 28, am 3d is at a lower energy level than 4s therefore it comes before it in the electron configuration because it going from lowest to highest energy level. You do not have the required permissions to view the files attached to this post. Quick links. Email Link. 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.
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. 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] 3d 1 4s 2 structure.
In each of these cases we have looked at, the 3d orbitals have the lowest energy, but as we add electrons, repulsion can push some of them out into the higher energy 4s level. The difficulty with this approach is that you cannot use it to predict the structures of the rest of the elements in the transition series.
In fact, what you have to do is to look at the actual electronic structure of a particular element and its ions, and then work out what must be happening in terms of the energy gap between the 3d and 4s orbitals and the repulsions between the electrons. The common way of teaching this based on the wrong order of filling of the 3d and 4s orbitals for transition metals gives a method which lets you predict the electronic structure of an atom correctly most of the time.
Raising l raises orbital energy. Higher l values result in orbitals with more nuclear nodes a node being a place where the probability of finding the electron is zero. We say high- l orbitals are "less penetrating" because their electrons have a lower probability of being found at or near the nucleus.
That gives high- l orbitals like d orbitals more energy than low- l orbitals like s orbitals within the same shell. This effect causes 4s orbitals to have lower energy than 3d orbitals for elements lighter than copper. Although for hydrogen, the unoccupied 4s and 3d orbitals have nearly identical energies.
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