Ionization energy (IE)
First, we have to know what is ionization energy before we could discuss its periodic trends.
Ionization energy is the energy required to remove one electron from a particle in gaseous state under standard conditions.
We could use a general chemical equation to represent this process.
\[ {\color{red}{1}}\ce{X^{n+}_{\color{red}{\left(g\right)}}} \rightarrow \ce{X^{\left(n+1\right)+}_{\color{red}{\left(g\right)}}} + {\color{red}{1}} \ce{e^{-}} \]
In the chemical equation above, there are two things that we highlighted with red.
- Particles must be in the gaseous state, regardless of their natural forms;
- Only one electron is removed from one particle.
When both criteria are met, the amount of energy required for that process would be the ionization energy. However, if multiple electrons are removed, or if electrons are removed from a solid, that's not the ionization energy.
Now let's take a closer look using boron (B) as an example. As boron is a solid, we'll have to first vaporize it to obtain boron atoms in the gaseous state. This would allow us to measure the energy required to remove an electron from an individual boron atom, without the interference of other boron atoms because in a gaseous state, boron atoms are separated by such large distances that they don't interact with each other.
Once we've got gaseous boron atoms, we'll use a beam of high speed electrons to knock out one of the electrons from the boron atom. This would give us a chemical equation as following:
\[ \ce{B_{(g)}} \rightarrow \ce{B^{+}_{\left(g\right)}} + \ce{e^{-}} \]
Since the electron being removed in this step is the first electron that's being removed, we say the ionization energy for this step is the first ionization energy, \( \text{IE}_{1} \).
Subsequently, we'll remove one more electron from the \( \ce{B^{+}_{(g)}} \) cation. This is the second electron that's being removed, thus giving rise to second ionization energy, \( \text{IE}_{2} \).
\[ \ce{B^{+}_{\left(g\right)}} \rightarrow \ce{B^{2+}_{\left(g\right)}} + \ce{e^{-}} \]
This process could continue until we've removed all the electrons. Therefore, in a general equation, we could write:
\[ \ce{B^{\left(n-1\right)+}_{\left(g\right)}} \rightarrow \ce{B^{n+}_{\left(g\right)}} + \ce{e^{-}} \]
to represent the process accompanied by nth ionization energy, \( \text{IE}_{n} \).
We would like to emphasize again that each ionization energy is associated with only ONE electron being removed.
So, can you write an equation to show the 7th ionization energy of chlorine?
Periodic trends of first ionization energy
Down the group
As we go down the group, atoms have a greater number of energy levels, and thus a weaker interaction between the nucleus and the valence electrons. It would be easier to remove one of the valence electrons. Therefore, we could say that first ionization energy decreases down the group.
Across the period
On the other hand, first ionization energy increases across the period. This is because atoms in the same period have the same number of energy level but a smaller radius from the left to the right of the periodic table. As electrons are getting closer to the nucleus due to the smaller radius, the attraction between the nucleus and the valence electrons would be increasing, and resulting in a higher IE1.
At the same time, atoms in the same period would have an increasing atomic number from the left to the right of the period. A greater number of protons within the nucleus would give rise to a stronger attraction between the nucleus and the valence electrons. This would further increase the IE1 of the atom.
It is noteworthy that IE1 of transition elements may not show a very significant trend across the period. This is due to the fact that electrons in the d and f orbitals would have impact on the shielding effects. We'll limit our discussion to main group elements instead.
Next, we'll talk about some of the exceptions to this general trend of increasing IE1 across the period.