The Moeller diagram o Rain method is a graphic and mnemonic method to learn Madelung's rule; that is, how to write the electron configuration of an element. It is characterized by drawing diagonals through the columns of the orbitals, and following the direction of the arrow, the appropriate order of the same is established for an atom.
In some parts of the world the Moeller diagram is also known as the rain method. Through this, an order is defined in the filling of the orbitals, which are also defined by the three quantum numbers n, l Y ml.
A simple Moeller diagram is shown in the image above. Each column corresponds to different orbitals: s, p, d and f, with their respective energy levels. The first arrow indicates that the filling of any atom must begin with the 1s orbital.
Thus, the next arrow must start from the 2s orbital, and then from the 2p through the 3s orbital. In this way, as if it were a rain, the orbitals and the number of electrons they contain (4l+two).
The Moeller diagram represents an introduction for those who study electron configurations..
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Since the Moeller diagram consists of a graphical representation of Madelung's rule, it is necessary to know how the latter works. The filling of the orbitals must obey the following two rules:
-The orbitals with the lowest values of n+l are filled first, being n the principal quantum number, and l the orbital angular momentum. For example, the 3d orbital corresponds to n= 3 and l= 2, therefore, n+l= 3 + 2 = 5; meanwhile, the 4s orbital corresponds to n= 4 and l= 0, and n+l= 4 + 0 = 4. From the above it is established that the electrons fill the 4s orbital first than the 3d one.
-If two orbitals have the same value of n+l, the electrons will occupy first the one with the lowest value of n. For example, the 3d orbital has a value of n+l= 5, like the 4p orbital (4 + 1 = 5); but since 3d has the smallest value of n, will fill up earlier than 4p.
From the two previous observations, the following order of filling of the orbitals can be reached: 1s 2s 2p 3s 3p 4s 3d 4p.
Following the same steps for different values of n+l for each orbital the electronic configurations of other atoms are obtained; which in turn can also be determined by the Moeller diagram graphically.
Madelung's rule states the formula n+l, with which the electronic configuration can be "armed". However, as mentioned, the Moeller diagram already graphically represents this; so just follow its columns and draw diagonals step by step.
How then do you start the electronic configuration of an atom? To do this, you must first know its atomic number Z, which by definition for a neutral atom is equal to the number of electrons.
Thus, with Z the number of electrons is obtained, and with this in mind we begin to draw diagonals through the Moeller diagram.
The s orbitals can accommodate two electrons (applying the formula 4l+2), the p six electrons, the d ten, and the f fourteen. It stops at the orbital where the last electron given by Z has been occupied.
For further clarification, below there are a series of solved exercises.
Using the periodic table, the element beryllium is located with a Z = 4; that is, its four electrons must be accommodated in the orbitals.
Starting then with the first arrow in the Moeller diagram, the 1s orbital occupies two electrons: 1stwo; followed by the 2s orbital, with two additional electrons to add 4 in total: 2stwo.
Therefore, the electron configuration of beryllium, expressed as [Be] is 1stwo2stwo. Note that the summation of the superscripts is equal to the number of total electrons.
The element phosphorus has a Z = 15, and therefore, it has 15 electrons in total which must occupy the orbitals. To advance the way, you start at once with the 1s settingtwo2stwo, which contains 4 electrons. Then 9 more electrons would be missing.
After the 2s orbital, the next arrow "enters" the 2p orbital, finally landing in the 3s orbital. Since the 2p orbitals can occupy 6 electrons, and the 3s 2 electrons, we have: 1stwo2stwo2 P63stwo.
There are still 3 more electrons missing, which occupy the following 3p orbital according to the Moeller diagram: 1stwo2stwo2 P63stwo3p3, electron configuration of phosphor [P].
The element zirconium has a Z = 40. Shortening path with 1s setuptwo2stwo2 P63stwo3p6, with 18 electrons (that of the noble gas argon), then 22 more electrons would be missing. After the 3p orbital, the next to fill according to the Moeller diagram are the 4s, 3d, 4p and 5s orbitals.
Filling them completely, that is, 4stwo, 3d10, 4p6 and 5stwo, a total of 20 electrons are added. The 2 remaining electrons are therefore housed in the following orbital: the 4d. Thus, the electron configuration of zirconium, [Zr] is: 1stwo2stwo2 P63stwo3p64stwo3d104p65stwo4dtwo.
Iridium has a Z = 77, so it has 37 additional electrons compared to zirconium. Starting from [Cd], that is, 1stwo2stwo2 P63stwo3p64stwo3d104p65stwo4d10, add 29 electrons with the following orbitals of the Moeller diagram.
Drawing new diagonals, the new orbitals are: 5p, 6s, 4f and 5d. Filling the first three orbitals completely, we have: 5p6, 6stwo and 4f14, to give a total of 22 electrons.
So 7 electrons are missing, which are in the 5d: 1s orbital.two2stwo2 P63stwo3p64stwo3d104p65stwo4d105 p66stwo4f145 d7.
The above is the electron configuration of iridium, [Go]. Note that the 6s orbitalstwo and 5d7 are highlighted in bold to indicate that they correspond properly to the valence layer of this metal.
There are many elements in the periodic table that do not obey what has just been explained. Their electron configurations differ experimentally from those predicted for quantum reasons..
Among the elements that present these discrepancies are: chromium (Z = 24), copper (Z = 29), silver (Z = 47), rhodium (Z = 45), cerium (Z = 58), niobium (Z = 41) and many more.
Exceptions are very frequent in the filling of the d and f orbitals. For example chrome should have a valence setting of 4stwo3d4 according to Moeller diagram and Madelung's rule, but it is actually 4s13d5.
Also, and finally, the silver valence setting should be 5stwo4d9; but it really is 5s14d10.
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