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Because this outermost shell is the most important shell, it is given the name “valence shell,” and the electrons in that shell are called “valence electrons.” The word “valence” means “importance.” The valence electrons are the most important electrons in each atom.

Observation 3: successive ionization energies of the atom

There is another more direct way for us to observe the number of valence electrons in each atom and to show that they are all in about the same shell. We can attempt to remove each electron in the valence shell, one after the other, increasing the charge on the ion to higher and higher values. We have discussed the first ionization energy, which is the energy to remove one electron. The second ionization energy is the energy needed to remove an electron from the positive ion to form an ion with a +2 charge. There are also third and fourth ionization energies and beyond:

First ionization energy IE 1 A (g) → A + (g) + e - (g)

Second ionization energy IE 2 A + (g) → A 2+ (g) + e - (g)

Third ionization energy IE 3 A 2+ (g) → A 3+ (g) + e - (g)

Fourth ionization energy IE 4 A 3+ (g) → A 4+ (g) + e - (g)

The experimental data observed when doing these “successive” ionizations is shown in [link] :

Successive ionization energies of the atoms (kj/mol)
Na Mg Al Si P S Cl Ar
IE 1 496 738 578 787 1012 1000 1251 1520
IE 2 4562 1451 1817 1577 1903 2251 2297 2665
IE 3 6912 7733 2745 3231 2912 3361 3822 3931
IE 4 9543 10540 11575 4356 4956 4564 5158 5770
IE 5 13353 13630 14830 16091 6273 7013 6542 7238
IE 6 16610 17995 18376 19784 22233 8495 9458 8781
IE 7 20114 21703 23293 23783 25397 27106 11020 11995

Let’s analyze this data, looking for evidence of the valence shell and the number of valence electrons. First, for every atom listed, IE 2 is always greater than IE 1 , IE 3 is greater than IE 2 , and so forth. This makes sense when we remember that the negatively charged electrons in an atom repel one another. Once an electron has been removed from an atom, the remaining electrons will have lower energy and be harder to remove.

Looking more closely, though, the increases in the ionization energies are not very constant. In Na, IE 2 is greater than IE 1 by a factor of almost nine, but IE 3 is greater than IE 2 by a fraction, and the same is true with the higher ionization energies. This means that the first electron in Na is fairly easy to remove, but the second one is much harder. This means that the first electron removed from Na is in the valence shell, far from the nucleus, but the second electron removed is closer in and more strongly attracted. The third electron removed is harder still to remove, but not much. This means that the third electron removed is in the same shell as the second one. Therefore, Na has only one valence electron.

Now look at Mg. The first electron doesn’t require much energy to remove, although it is more than in Na. The second electron is harder again, but the real change is when we try to remove the third electron. Suddenly we see an increase of a factor of 5 in the ionization energy. This means that Mg has two relatively easily removed electrons, and therefore Mg has two valence electrons. Looking at the data in [link] , we can simply count the number of valence electrons in each atom. (It is hard to do this for Cl and Ar, because it is hard to remove this many electrons from a single atom.)

The shell model of the atom tells us how the electrons are arranged in each atom, but it does not tell us why. We don’t know why the electrons can’t all be added to a single shell, because we don’t know why a shell seems to “fill up.” It is clear from the data that the number of valence electrons in each noble gas is the number needed to fill the valence shell. But we don’t have a reason why this is true. These questions will require further observations and reasoning.

Review and discussion questions

  1. Provide the experimental evidence that reveals the electrons in an atom are grouped into a valence shell and inner shell electrons.
  2. State and explain the evidence that reveals the outer shell of each inert gas is full.
  3. Why does the ionization energy for each successive ionization increase for every atom? Why is the increase from IE 4 to IE 5 in Si much larger than any of the other increases for Si?

By John S. Hutchinson, Rice University, 2011

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Source:  OpenStax, Concept development studies in chemistry 2012. OpenStax CNX. Aug 16, 2012 Download for free at http://legacy.cnx.org/content/col11444/1.4
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