In “Ionization Energy,” we defined the term “First Ionization Energy.” There is also second ionization energy, third ionization energy, etc.
Note that we will sometimes abbreviate ionization energy as “I.E.”
The nᵗʰ ionization energy is the energy required to remove the nᵗʰ electron from an atom
1st I.E. – the energy required to remove the first electron from an atom.
Ex. Na + 1st I.E. → Na⁺ + e⁻
2nd I.E. – the energy required to remove the second electron from an atom.
Ex. Na⁺ + 2nd I.E. → Na²⁺ + e⁻
3rd I.E. – the energy required to remove the third electron from an atom.
Ex. Na²⁺ + 3rd I.E. → Na³⁺ + e⁻
And you get the idea. The 4th I.E. is the energy required to remove the fourth electron, and so on.
As you remove more electrons, the ionization energy increases, so 1st I.E. < 2nd I.E. < 3rd I.E. < 4th I.E… This is because as you remove electrons, the electron-electron repulsion decreases. Valence electrons are closer to the nucleus, so they experience more attraction. So, more ionization energy is required to remove electrons.
There is a large jump in ionization energy when an electron is removed from a lower principal energy level. This large jump is often indicated by a “<<”.
For example, let’s look at magnesium. The 1st I.E. of magnesium is removing its 3s² electron. The 2nd I.E. is greater, removing its 3s¹ electron. The 3rd I.E., however, is much greater, removing the 2p⁶ electron. The 3rd I.E. is at a lower principal energy level than the 1st and 2nd I.E. The 2p⁶ electron is much closer to the nucleus, so it requires much more energy to remove. The 4th I.E. removes the 2p⁵ electron, so it is again a little greater than the 3rd I.E.
So if we were to describe the successive ionization energies of Mg, we would write: 1st I.E. < 2nd I.E. << 3rd I.E. < 4th I.E.
We can also use this idea to determine an element from its successive ionization energies.
Example Problem
Identify this element in the third period of the periodic table based on its successive ionization energies ↴
1st I.E. < 2nd I.E. < 3rd I.E. << 4th I.E < 5th I.E.
Answer
Based on the successive ionization energies, this element has 3 valence electrons. After you remove the third electron, there is a large increase in ionization energy, indicating that the 4th electron is in a lower principal energy level. The problem also told us that the element is in the 3rd period. The element in the 3rd period with 3 valence electrons is aluminum.
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