Ionization energy is defined as the minimum amount of energy required to remove the electron from the outermost shell of an atom. When electrons are removed, the atom becomes a positively charged ion. In the periodic table, the ionization energy of atoms shows an increasing trend from left to right across a period, while it decreases down the group.
The ionization energy trends can be understood by studying the change in size and the effective nuclear charge. Along the period, the number of protons and electrons increases at each step shrinking the atomic radius. One can understand this shrinking phenomenon as a tug-of-war between protons and electrons. In other words, the greater number of participants on both sides, the lower the in-between distance.
Down the group, there is an addition of an extra shell on each atom, which makes the valence electron farther apart. As a result, less energy is required to remove this electron.
Exceptions in Ionization Energy Trends
These are some irregularities in the trends of ionization energy described below:
- In groups, II-A and III-A, boron has lower first ionization energy than beryllium. This can be explained by analyzing its electronic configuration. Boron has outer electrons in the ‘s’ orbital which is very close to the nucleus whereas, beryllium releases its first electron from the p-orbital which is comparatively away from the nucleus.
- Another exception of such trends in the periodic trends is found in Group V-A and VI-A. Nitrogen has three electrons in the p orbital while oxygen has four electrons. Nitrogen should have lower ionization energy but on contrary, it has lower first ionization energy than oxygen. This is so because of two main reasons: One is the symmetrical distribution of three electrons in the p orbital of a nitrogen atom, by Hund’s rule and the other is electron-electron repulsion.
- Another exception to the general ionization trend is found in the lanthanides and actinides, which have lower ionization energies due to the presence of an inner ‘f’ electron shell.
Calculation of Ionization energy
I = RH Zeff / n2
Where,
- I denote the ionization energy
- RH is Redberg’s constant
- Zeff is an effective nuclear charge
- n is the principal quantum number
Related Topics
Key Takeaways
At its simplest, the first ionization energy trends are summarized as:
- Increases from left to right in the period: Due to an increase in Zeff (effective nuclear charge).
- Decrease from top to bottom in a group: This happens due to a decrease in Zeff.
- Noble gases: They have very high ionization potential because of completely filled outermost shell
Note that, exceptions in the trends are due to the subshells and electron-electron repulsion.
Concepts Berg
What causes the ionization energy trend to decrease?
Ionization energy generally increases along the period but, anomalies in the outermost electronic configuration tend to decrease the ionization energy in some elements.
Does ionization energy increase down a group?
Ionization energy decreases down the group as the outermost shell is going far apart as we move down the group due to the addition of a new shell.
What influences the trend of ionization energy?
The major influencer to the trend of ionization energy in the period are given below
- Difference of Zeff of s and p orbital
- Symmetrical distribution of electrons by Hund’s rule
- Electron-electron pair repulsion
How can I find the ionization of energy of a doubly ionized lithium atom?
The 2nd ionization energy of the lithium atom can be found by removing two moles of an electron from one mole of the lithium atom.
Which ion has the maximum ionization energy, O- or S-?
Sulfite ion has a smaller size than peroxide ion so earlier has greater ionization potential.
What has higher first ionization energy, carbon or chlorine?
The size of chlorine is smaller than carbon. So, chlorine has a greater nuclear charge so it has higher first ionization energy.
