Lesson Explainer: Ionization Energy Chemistry

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 1+ 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.

Definition: First Ionization Energy

The first ionization is the amount of energy required to remove the most loosely bound electron completely from an isolated gaseous atom.

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: Na()Na()+erstionizationenergykJmolgg+=+496

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.

Example 1: Determining the Relationship Between the Atomic Radius and the Ionization Energy

How does the atomic radius affect the ionization energy?

  1. Decreasing the atomic radius leads to a higher ionization energy.
  2. Increasing the atomic radius leads to a higher ionization energy.
  3. Decreasing the atomic radius leads to a lower ionization energy.
  4. The atomic radius does not affect ionization energy at all.

Answer

There is less distance between the outer-shell electrons and the positively charged protons when the atomic radius is small compared with the situation when the atomic radius is relatively large. The electrostatic attraction forces will be bigger when the atoms are small and there is not much distance between the oppositely charged protons and electrons. The (first) ionization energies will be larger when the atoms are relatively small and there are stronger electrostatic attraction forces between the electrons and the positively charged protons. As such, the correct answer is option A.

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 +520 kJ⋅mol−1, while beryllium has a much larger first ionization energy of +900 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.

Example 2: Determining Which Element Has the Higher First Ionization Energy

Which element in the pair of adjacent elements, Mg and Na, has a higher first ionization energy?

  1. Na
  2. Mg

Answer

Electrostatic attraction forces depend on the number of positively charged protons and not just the distance between the positively charged protons and the outer-shell electrons. There is a stronger electrostatic interaction between negatively charged outer-shell electrons and positively charged atomic nuclei when the nuclei contain lots of positively charged protons. Magnesium has one more positively charged proton than sodium, and the valence electrons in magnesium experience a stronger electrostatic attraction force than the valence electrons in sodium. It takes more energy to remove an outer shell electron from a magnesium atom that has twelve protons than from a sodium atom that has just eleven protons. As such, the answer is option B, Mg (magnesium).

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 2+ and 3+ positively charged ions. The second electrons are difficult to remove because they are interacting with 1+ positively charged ions, and the third electrons are even more difficult to remove because they are interacting with 2+ positively charged ions.

The following equation shows the successive removal of electrons from one mole of helium atoms to make one mole of 2+ helium ions: He()He()+erstionizationenergykJmolHe()He()+esecondionizationenergykJmolgggg++2+=+2372=+5250

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

Definition: Second Ionization Energy

The second ionization energy is the energy required to remove an electron from a gaseous ion that has the 1+ charge state.

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 12ss 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 (𝑛=2), and the second electron is taken from the first electron shell (𝑛=1): Li()Li()+erstionizationenergykJmolLi()Li()+esecondionizationenergykJmolgggg++2+=+520=+7298

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.

Definition: Successive Ionization Energies

Successive ionization energies quantify how much energy is needed to consecutively remove electrons from one gaseous-state 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 1+ charge-state potassium ion because it is in an altogether different electron shell. It is in the third electron shell (𝑛=3), whereas the first electron came from the fourth electron shell (𝑛=4).

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 (𝑛=3), while the first two come from the much higher energy fourth energy shell (𝑛=4).

Ionization Energy (kJ/mol)
Element1st2nd3rd4th5th
Potassium (K)4193‎ ‎0514‎ ‎4195‎ ‎8787‎ ‎975
Calcium (Ca)5901‎ ‎1454‎ ‎9126‎ ‎4918‎ ‎140

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?

Element1st2nd3rd4th5th6th
17901‎ ‎6003‎ ‎2004‎ ‎40016‎ ‎10019‎ ‎800
25004‎ ‎6006‎ ‎9009‎ ‎50013‎ ‎40016‎ ‎600
35901‎ ‎1004‎ ‎9006‎ ‎5008‎ ‎10010‎ ‎500
45801‎ ‎8002‎ ‎70011‎ ‎60014‎ ‎80018‎ ‎400
58701‎ ‎8002‎ ‎7003‎ ‎6005‎ ‎7006‎ ‎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 12233sspsp 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 (𝑛=2), which has much lower energy and is much closer to the positively charged atomic nucleus. Taken as a whole, element 4 is the answer.

Example 4: Determining Chemical Formula from Successive Ionization Energies

The following table shows the data of successive ionization energies of a metal, M. What is the likely formula of the sulfide compound formed in the reaction between sulfur and metal M?

Successive Ionization Energies of Element M (kJ/mol)
5901104‎ ‎9006‎ ‎5008‎ ‎10010‎ ‎50012‎ ‎30014‎ ‎200
  1. MS23
  2. MS
  3. MS2
  4. MS2

Answer

Consecutive ionization energy patterns can be used to determine the identity of an unknown element. Group 1 elements have one pattern of successive ionization energies. Group 2 elements have an altogether different pattern of successive ionization energies. Group 13 elements have their own pattern of successive ionization energies. Group 14 elements have their own pattern as well.

The pattern of successive ionization energies is relatively simple for group 1 elements. They have a first ionization that is relatively low. This is followed by a second ionization energy that is much larger. There is a very large energy difference between the first and second group 1 ionization energies. There is a less substantial difference between their second and third ionization energies. The pattern of successive ionization energies is slightly different for group 2 elements. They have first and second ionization energies that are relatively low. They have a third ionization that is significantly larger. The pattern of successive ionization energies is interesting for the group 13 elements. They have a first ionization energy that is relatively low. This is followed by a second ionization energy that is much larger. They also have a fourth ionization energy that is significantly larger than their first three. The pattern of successive ionization energies is different for the group 14 elements. They have first and second ionization energies that are relatively low. These are followed by a third that is significantly larger. There is a less substantial energy difference between their third and fourth ionization energies.

The table shows successive ionization energies for the unknown metal. The ionization energy series matches the pattern for group 2 elements. It has first and second ionization energies that are relatively low. This is followed by a third ionization energy that is significantly larger. The unknown metal must be a group 2 element.

Sulfur is a group 16 element. Sulfur gets a negative oxidation state when it reacts with metals. It gets the 2 oxidation state. Metals (M) get a positive oxidation state when they react with sulfur. They get whatever oxidation state gives the product MS compound no overall electrostatic charge.

The MS compound must contain metals with the +2 oxidation state. It is without charge if it has one +2 oxidation-state metal for each 2 oxidation-state sulfur. The MS compound has to contain group 2 metals. The MS2 compound must contain metals with the +1 oxidation state. It is without charge if it has two +1 oxidation-state metals for each 2 oxidation-state sulfur. The MS2 compound has to contain group 2 metals. The MS23 compound must contain metals with the +3 oxidation state. It is without charge if it has two +3 oxidation-state metals for every three 2 oxidation-state sulfurs. The MS23 compound has to contain group 13 metals. The MS2 compound must contain metals with the +4 oxidation state. It is without charge if it has one +4 oxidation-state metal for every two 2 oxidation-state sulfurs. The MS2 compound has to contain group 14 metals.

Group 2 metals get the +2 oxidation state when they react with a nonmetal element like sulfur. Metal M must form the MS compound when it reacts with sulfur. MS compounds have no overall electrostatic charge if they have one +2 oxidation-state metal for each 2 oxidation-state sulfur. This line of reasoning can be used to determine that option B is the correct answer for this question.

Key Points

  • Chemists can use energy to remove outer shell electrons from neutrally charged atoms to make positively charged ions.
  • The first ionization energy is the energy required to remove the highest-energy electron from a neutrally charged gas atom.
  • Ionization energies depend on electron shielding, effective nuclear charge, and atomic radii.
  • Ionization energies increase with effective nuclear charge across any one period of the periodic table.
  • Ionization energies decrease with increasing atomic radii down any one group of the periodic table.
  • Noble gas elements have a higher first ionization energy than other elements of the same period.
  • Successive ionization energies quantify how much energy is needed to consecutively remove electrons from one gaseous-state element.
  • Successive ionization energy values can be used to determine the group and the electron configuration of an unknown element.

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