Lesson Explainer: Ionization Energy Chemistry • Second Year of Secondary School

In this explainer, we will learn how to describe and explain the ionization energy of elements and ions.

The atoms of any one element have an overall neutral electrostatic charge because they contain an equal number of protons and electrons. Ions have a positive or negative electrostatic charge because they contain an unequal number of protons and electrons. Ions can be formed during chemical reactions, and they can also be formed when neutrally charged atoms are bombarded with ionizing radiation.

The first ionization energy is the energy required to remove an electron from a neutrally charged gas atom. It quantifies the amount of energy that is needed to make a charge cation from a neutral gas atom. The first ionization energy varies across the periodic table because it depends on the interplay of both easy-to-understand properties, like atom size (atomic radii) or effective nuclear charge, and more complex quantum effects like electron shielding.

Chemists usually represent first ionization energies with highly simplified chemical equations. The equations include terms for neutrally charged elements before they are ionized and terms for the positively charged ions and electrons that are made during the ionization process itself. The neutrally charged element and the positively charged ion terms are always written with gaseous state symbols because ionization energies are measured for gases rather than liquids or solids. The following equation shows the first ionization energy for gaseous-state sodium atoms:

First ionization energies are always positive numbers because we have to expend energy to remove outer-shell electrons from neutrally charged atoms.

The first ionization energy depends on physical properties like atom size (atomic radius) because electrostatic interaction strengths depend on the distance between oppositely charged particles. Relatively large atoms have a lot of distance between their protons and electrons, and they have weak forces of attraction acting on their highest-energy electrons. There is more distance between protons and valence electrons in large atoms, and this explains why first ionization energy values generally decrease in size down any one group of the periodic table. Atoms get larger down a group, and this means they have weaker forces of attraction acting on their highest-energy electrons.

The association between the first ionization energy and the atomic radius can be better understood by considering how first ionization energies change as we move down group 18 of the periodic table. The group 18 elements are usually called the noble gas elements. They include elements like helium, neon, and argon. Helium is the smallest noble gas, and it has the highest first ionization energy of any element in the periodic table. Neon is a bit larger than helium, and it has a slightly lower first ionization energy of 2‎ ‎081 kJ⋅mol⁻¹. Argon is larger still, and it has an even lower first ionization energy of 1‎ ‎521 kJ⋅mol⁻¹.

The following figure shows first ionization energies for elements with an atomic number between 1–20. It has data for helium, and it also has data for neon and argon elements. The figure shows quite clearly that first ionization energies tend to decrease down the group of noble gas elements. It also shows a more general relationship between the first ionization energy and the atomic radius because it has data for groups 1–2 and 13–17 as well. The graph shows that first ionization energies are linked with atomic radii in a fundamental way. First ionization energies decrease down all groups of the periodic table. Small atoms are shown to have a relatively high first ionization energy and larger atoms are shown to have a lower first ionization energy.

The first ionization energy also depends on effective nuclear charge. Electrons experience strong electrostatic attraction forces when they are affected by a high nuclear charge. This line of reasoning can be used to understand why noble gas elements have higher first ionization energies than other elements of the same period. Noble gas elements have the higher effective nuclear charges and there are incredibly strong electrostatic attraction forces between their highest-energy electrons and their positively charged protons. Helium has a higher first ionization energy than hydrogen primarily because helium has the higher effective nuclear charge. Neon has a higher first ionization energy than all of the other period two elements primarily because it has the higher effective nuclear charge. This line of reasoning could similarly be applied to understand why lithium has a first ionization of kJ⋅mol⁻¹, while beryllium has a much larger first ionization energy of kJ⋅mol⁻¹. Beryllium has the higher first ionization energy primarily because it has a greater number of protons and the higher effective nuclear charge.

The first ionization energy values are also affected by electron shielding effects. Electrons have weak electrostatic interactions with positively charged protons when there are lots of negatively charged subshells between them. The inner subshells screen electrostatic interactions, and they make the electrostatic interactions between positively charged protons and valence electrons less intense. This explains why boron has a lower first ionization energy than beryllium despite the fact that boron has five positively charged protons and beryllium only has four. Boron has its highest-energy electron in the 2p subshell, and beryllium has its highest-energy electron in the 2s subshell. The highest-energy electron in boron is screened from positively charged protons by a greater number of inner subshells. This line of reasoning can also be used to understand why aluminum has a lower first ionization energy than magnesium despite the fact that aluminum has thirteen positively charged protons and magnesium has twelve. Aluminum has its highest-energy electron in the 3p subshell, and magnesium has its highest-energy electron in the 3s subshell. The highest-energy electron of aluminum is screened from positively charged protons by a greater number of inner subshells.

It is interesting to note that oxygen has a lower first ionization energy than nitrogen. This seems to contradict what we have learned so far. Oxygen has the higher atomic number, and it does not have its highest-energy electron in one subshell, while nitrogen has its highest-energy electron in an altogether different subshell. Scientists can explain the anomalously low first ionization energy of oxygen by considering the spin state of its electrons. The following figure uses upward-facing arrows (↑) and downward-facing arrows (↓) to represent the electron configurations of comparative nitrogen and oxygen elements. It shows that nitrogen has its three highest-energy electrons singly occupying the three atomic orbitals of the 2p subshell. The electrons are all in the spin-up state, and there is no pairing of electrons in any one orbital of the 2p subshell. The figure also shows that oxygen has its four highest-energy electrons in the 2p subshell as well, but one of the 2p orbitals is doubly occupied with electrons. One of the orbitals contains a spin-up state electron and a spin-down state electron. There is repulsion between these two paired electrons, and it is relatively easy to remove one of them from the 2p subshell. Scientists use this line of reasoning to explain why oxygen has a lower first ionization energy than nitrogen. This line of reasoning could similarly be used to explain the anonymously low first ionization energy of sulfur compared to that of phosphorus.

The following figure compares the electron configuration of comparative sulfur and phosphorus elements. It shows that phosphorus has its three highest-energy electrons in the three atomic orbitals of the 3p subshell. The electrons are all in the spin-up state, and there is no pairing of electrons in any one orbital of the 3p subshell. The figure also shows that sulfur has its four highest-energy electrons in the 3p subshell as well, but one of the 3p orbitals is doubly occupied with electrons. There is repulsion between these two paired electrons, and it is relatively easy to remove one of them from the 3p subshell.

It usually takes a lot of energy to remove a single electron from a neutrally charged atom, but it takes even more energy to remove the second and third electrons to make and positively charged ions. The second electrons are difficult to remove because they are interacting with positively charged ions, and the third electrons are even more difficult to remove because they are interacting with positively charged ions.

The following equation shows the successive removal of electrons from one mole of helium atoms to make one mole of helium ions:

The second ionization energy is more than twice as large as the first ionization energy because the second electron is being removed from the ion instead of a neutrally charged helium atom.

There can be an even bigger difference between successive ionization energies when we go from removing an electron from one electron shell to removing an electron from an altogether different energy level. Lithium has the electron configuration, and it has first and second ionization energy values of 520 kJ⋅mol⁻¹ and 7‎ ‎298 kJ⋅mol⁻¹. The second ionization energy is so much larger than the first because the first electron is taken from the second electron shell , and the second electron is taken from the first electron shell :

The successive changes in ionization energies can be used to determine the identity of an unknown element because each periodic table group has its own pattern of successive ionization-energy values. Group 1 elements have first ionization energies that are much smaller than their second ionization energies, and group 2 elements have second ionization energies that are much smaller than their third ionization energies. Group 3 elements have second ionization energies that are noticeably larger than their first ionization energies, and they have fourth ionization energies that are significantly larger than their first three ionization energy values. The series of successive ionization energies is usually enough to determine the group number of an unknown element.

The following table compares the successive ionization energies of potassium and calcium elements. Potassium is a group 1 element, and it has a first ionization energy that is much smaller than its second ionization energy. The first electron is relatively easy to remove from potassium because it is in the 4s subshell. The second electron is much more difficult to remove from the charge-state potassium ion because it is in an altogether different electron shell. It is in the third electron shell , whereas the first electron came from the fourth electron shell .

Calcium is a group 2 element, and the table shows that it has a second ionization energy that is much smaller than its third. It is relatively easy to remove two electrons from calcium because the first two electrons come from the fourth electron shell. It takes a lot of energy to remove another electron from calcium because the third electron comes from an altogether different electron shell. The third electron comes from the third electron shell , while the first two come from the much higher energy fourth energy shell .

Example 3: Understanding How We Can Identify an Element from a Series of Its Successive Ionization Energy Values

Given the data of successive ionization energies (in kJ/mol) in the table shown, which of the following elements is most likely to be in group III of the periodic table?

Element	1st	2nd	3rd	4th	5th	6th
1	790	1‎ ‎600	3‎ ‎200	4‎ ‎400	16‎ ‎100	19‎ ‎800
2	500	4‎ ‎600	6‎ ‎900	9‎ ‎500	13‎ ‎400	16‎ ‎600
3	590	1‎ ‎100	4‎ ‎900	6‎ ‎500	8‎ ‎100	10‎ ‎500
4	580	1‎ ‎800	2‎ ‎700	11‎ ‎600	14‎ ‎800	18‎ ‎400
5	870	1‎ ‎800	2‎ ‎700	3‎ ‎600	5‎ ‎700	6‎ ‎700

Answer

Group 13 (3) elements have noticeably different first and second ionization energies because the first and second ionized electrons come from different electron subshells. The first ionized electron is from a p subshell, and the second ionized electron is from an s subshell. There is an even larger difference between the third and fourth ionization energies because the third and fourth ionized electrons come from entirely differently energy levels.

Aluminum is a group 13 (3) metal that has the electron configuration. It can be used as a representative example to understand the ionization energies of group 13 (3) elements. The following figure shows the electron configuration of aluminum. The aluminum electrons are represented as a series of upward-facing arrows (↑) and downward-facing arrows (↓). The electrons are ordered from left-to-right in terms of increasing orbital energy. This explains why the rightmost electrons are ionized first.

The figure shows that the first ionized electron is removed from the highest-energy 3p subshell, and the second and third ionized electrons come from the slightly lower-energy 3s subshell. The fourth ionized electron is much more challenging to remove because it comes from the 2p orbital. The 2p orbital is in the second electron shell , which has much lower energy and is much closer to the positively charged atomic nucleus. Taken as a whole, element 4 is the answer.