The fact that both of the electrons in the 2p subshell have the same spin quantum number can be shown by representing an electron for which s = +1/2 with an
arrow pointing up and an electron for which s = -1/2 with an arrow pointing down.
The electrons in the 2p orbitals on carbon can therefore be represented as follows.
When we get to N (Z = 7), we have to put one electron into each of the three degenerate 2p orbitals.
N (Z = 7):1s2 2s2 2p3
Because each orbital in this subshell now contains one electron, the next electron added to the subshell must have the opposite spin quantum number, thereby filling one of the 2p orbitals.
The ninth electron fills a second orbital in this subshell.
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F (Z = 9): 1s2 2s2 2p5
The tenth electron completes the 2p subshell.
There is something unusually stable about atoms, such as He and Ne, that have electron configurations with filled shells of orbitals. By convention, we therefore write abbreviated electron configurations in terms of the number of electrons beyond the previous element with a filled-shell electron configuration. Electron configurations of the next two elements in the periodic table, for example, could be written as follows.

Na (Z = 11): [Ne] 3s1

Mg (Z = 12): [Ne] 3s2

Practice Problem 8:
Predict the electron configuration for a neutral tin atom (Sn, Z = 50).
The aufbau process can be used to predict the electron configuration for an element. The actual configuration used by the element has to be determined experimentally. The experimentally determined electron configurations for the elements in the first four rows of the periodic table are given in the table in the following section.

(1st, 2nd, 3rd, and 4th Row Elements)

Atomic NumberSymbol Electron Configuration
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
1H1s1
2He1s2 = [He]
3Li[He] 2s1
4Be[He] 2s2
5B[He] 2s2 2p1
6C[He] 2s2 2p2
7N[He] 2s2 2p3
8O[He] 2s2 2p4
9F[He] 2s2 2p5
10Ne[He] 2s2 2p6 = [Ne]
11Na[Ne] 3s1
12Mg[Ne] 3s2
13Al[Ne] 3s2 3p1
14Si[Ne] 3s2 3p2
15P[Ne] 3s2 3p3
16S[Ne] 3s2 3p4
17Cl[Ne] 3s2 3p5
18Ar[Ne] 3s2 3p6 = [Ar]
19K[Ar] 4s1
20Ca[Ar] 4s2
21Sc[Ar] 4s2 3d1
22Ti[Ar] 4s2 3d2
23V[Ar] 4s2 3d3
24Cr[Ar] 4s1 3d5
25Mn[Ar] 4s2 3d5
26Fe[Ar] 4s2 3d6
27Co[Ar] 4s2 3d7
28Ni[Ar] 4s2 3d8
29Cu[Ar] 4s1 3d10
30Zn[Ar] 4s2 3d10
31Ga[Ar] 4s2 3d10 4p1
32Ge[Ar] 4s2 3d10 4p2
33As[Ar] 4s2 3d10 4p3
34Se[Ar] 4s2 3d10 4p4
35Br[Ar] 4s2 3d10 4p5
36Kr[Ar] 4s2 3d10 4p6 = [Kr]
There are several patterns in the electron configurations listed in the table in the previous section. One of the most striking is the remarkable level of agreement between these configurations and the configurations we would predict. There are only two exceptions among the first 40 elements: chromium and copper.
Strict adherence to the rules of the aufbau process would predict the following electron configurations for chromium and copper.
predicted electron configurations:Cr (Z = 24): [Ar] 4s2 3d4
Cu (Z = 29): [Ar] 4s2 3d9
The experimentally determined electron configurations for these elements are slightly different.
actual electron configurations:Cr (Z = 24): [Ar] 4s1 3d5
Cu (Z = 29): [Ar] 4s1 3d10
In each case, one electron has been transferred from the 4s orbital to a 3d orbital, even though the 3d orbitals are supposed to be at a higher level than the 4s orbital.
Once we get beyond atomic number 40, the difference between the energies of adjacent orbitals is small enough that it becomes much easier to transfer an electron from one orbital to another. Most of the exceptions to the electron configuration predicted from the aufbau diagram shown earlier therefore occur among elements with atomic numbers larger than 40. Although it is tempting to focus attention on the handful of elements that have electron configurations that differ from those predicted with the aufbau diagram, the amazing thing is that this simple diagram works for so many elements.
When electron configuration data are arranged so that we can compare elements in one of the horizontal rows of the periodic table, we find that these rows typically correspond to the filling of a shell of orbitals. The second row, for example, contains elements in which the orbitals in the n = 2 shell are filled.
Li (Z = 3):[He] 2s1
Be (Z = 4):[He] 2s2
B (Z = 5):[He] 2s2 2p1
C (Z = 6):[He] 2s2 2p2
N (Z = 7): [He] 2s2 2p3
O (Z = 8):[He] 2s2 2p4
F (Z = 9):[He] 2s2 2p5
Ne (Z = 10): [He] 2s2 2p6
There is an obvious pattern within the vertical columns, or groups, of the periodic table as well. The elements in a group have similar configurations for their outermost electrons. This relationship can be seen by looking at the electron configurations of elements in columns on either side of the periodic table.
Group IAGroup VIIA
H 1s1
Li [He] 2s1F[He] 2s2 2p5
Na [Ne] 3s1Cl[Ne] 3s2 3p5
K[Ar] 4s1Br[Ar] 4s2 3d10 4p5
Rb[Kr] 5s1I[Kr] 5s2 4d10 5p5
Cs[Xe] 6s1At[Xe] 6s2 4f14 5d10 6p5

Quantum Calculation

The figure below shows the relationship between the periodic table and the orbitals being filled during the aufbau process. The two columns on the left side of the periodic table correspond to the filling of an s orbital. The next 10 columns include elements in which the five orbitals in a d subshell are filled. The six columns on the right represent the filling of the three orbitals in a p subshell. Finally, the 14 columns at the bottom of the table correspond to the filling of the seven orbitals in an f subshell.
Practice Problem 9:
Predict the electron configuration for calcium (Z = 20) and zinc (Z = 30) from their positions in the periodic table.

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