The ease with which electrons can be removed from an atom or ion has a major impact on chemical behavior. The ionization energy of an atom or ion is the minimum energy required to remove an electron from the ground state of the isolated gaseous atom or ion. The first ionization energy,I1, is the energy needed to remove the first electron from a neutral atom. For example, the first ionization energy for the sodium atom is the energy required for the process

The second ionization energy, I2, is the energy needed to remove the second electron, and so forth, for successive removals of additional electrons. Thus, I2 for the sodium atom is the energy associated with the process

The greater the ionization energy, the more difficult it is to remove an electron.

Variations in Successive Ionization Energies

Notice in TABLE 7.2 that ionization energies for a given element increase as successive electrons are removed: I1I2 < I3, and so forth. This trend exists because with each successive removal, an electron is being pulled away from an increasingly more positive ion, requiring increasingly more energy.


Light can be used to ionize atoms and ions. Which of the two processes shown in Equations 7.2 and 7.3 requires shorter-wavelength radiation?

A second important feature shown in Table 7.2 is the sharp increase in ionization energy that occurs when an inner-shell electron is removed. For example, consider silicon, 1s2s2p63s23p2. The ionization energies increase steadily from 786 kJ/mol to 4356 kJ/mol for the four electrons in the 3s and 3p subshells. Removal of the fifth electron, which comes from the 2p subshell, requires a great deal more energy: 16,091 kJ/mol. The large increase occurs because the 2p electron is much more likely to be found close to the nucleus than are the four n = 3 electrons and, therefore, the 2p electron experiences a much greater effective nuclear charge than do the 3s and 3p electrons.

TABLE 7.2 • Successive Values of Ionization Energies, I, for the Elements Sodium through Argon (kJ/mol)


Which would you expect to be greater, I1 for a boron atom or I2 for a carbon atom?

Every element exhibits a large increase in ionization energy when one of its inner electrons is removed. This observation supports the idea that only the outermost electrons are involved in the sharing and transfer of electrons that give rise to chemical bonding and reactions. The inner electrons are too tightly bound to the nucleus to be lost from the atom or even shared with another atom.

SAMPLE EXERCISE 7.5 Trends in Ionization Energy

Three elements are indicated in the periodic table in the margin. Which one has the largest second ionization energy?


Analyze and Plan The locations of the elements in the periodic table allow us to predict the electron configurations. The greatest ionization energies involve removal of core electrons. Thus, we should look first for an element with only one electron in the outermost occupied shell.

Solve The red box represents Na, which has one valence electron. The second ionization energy of this element is associated, therefore, with the removal of a core electron. The other elements indicated, S (green) and Ca (blue), have two or more valence electrons. Thus, Na should have the largest second ionization energy.

Check A chemistry handbook gives these I2 values: Ca 1145 kJ/mol, S 2252 kJ/mol, Na 4562 kJ/mol.


Which has the greater third ionization energy, Ca or S?

Answer: Ca

Periodic Trends in First Ionization Energies

FIGURE 7.9 shows, for the first 54 elements, the trends we observe in first ionization energy as we move from one element to another in the periodic table. The important trends are as follows:

1. I1 generally increases as we move across a period. The alkali metals show the lowest ionization energy in each period, and the noble gases show the highest. There are slight irregularities in this trend that we will discuss shortly.

2. I1 generally decreases as we move down any column in the periodic table. For example, the ionization energies of the noble gases follow the order He > Ne > Ar > Kr > Xe.

3. The s- and p-block elements show a larger range of I1 values than do the transition-metal elements. Generally, the ionization energies of the transition metals increase slowly from left to right in a period. The/-block metals (not shown in Figure 7.9) also show only a small variation in the values of I1.

In general, smaller atoms have higher ionization energies. The same factors that influence atomic size also influence ionization energies. The energy needed to remove an electron from the outermost occupied shell depends on both the effective nuclear charge and the average distance of the electron from the nucleus. Either increasing the effective nuclear charge or decreasing the distance from the nucleus increases the attraction between the electron and the nucleus. As this attraction increases, it becomes more difficult to remove the electron and, thus, the ionization energy increases. As we move across a period, there is both an increase in effective nuclear charge and a decrease in atomic radius, causing the ionization energy to increase. As we move down a column, the atomic radius increases while the effective nuclear charge increases rather gradually. Thus, the attraction between the nucleus and the electron decreases, causing the ionization energy to decrease.


Which has a larger first ionization energy, Ar or As? Why?

FIGURE 7.9 Trends in first ionization energies of the elements.

The irregularities in a given period are subtle but still readily explained. For example, the decrease in ionization energy from beryllium ([He]2s2) to boron ([He]2s22p1), shown in Figure 7.9, occurs because the third valence electron of B must occupy the 2p subshell, which is empty for Be. Recall that the 2p subshell is at a higher energy than the 2s subshell (Figure 6.24). The decrease in ionization energy when moving from nitrogen ([He]2s22p3) to oxygen ([He]2s22p4) is because of the repulsion of paired electrons in the p4 configuration (FIGURE 7.10). Remember that according to Hund's rule, each electron in the p3 configuration resides in a different p orbital, which minimizes the electron-electron repulsion among the three 2p electrons. (Section 6.8)

SAMPLE EXERCISE 7.6 Periodic Trends in Ionization Energy

Referring to a periodic table, arrange the atoms Ne, Na, P, Ar, K in order of increasing first ionization energy.


Analyze and Plan We are given the chemical symbols for five elements. To rank them according to increasing first ionization energy, we need to locate each element in the periodic table. We can then use their relative positions and the trends in first ionization energies to predict their order.

Solve Ionization energy increases as we move left to right across a period and decreases as we move down a group. Because Na, P, and Ar are in the same period, we expect I1 to vary in the order Na < P < Ar. Because Ne is above Ar in group 8A, we expect Ar < Ne. Similarly, K is directly below Na in group 1A, and so we expect K < Na.

From these observations, we conclude that the ionization energies follow the order

K < Na < P < Ar < Ne

Check The values shown in Figure 7.9 confirm this prediction.


Which has the lowest first ionization energy, B, Al, C, or Si? Which has the highest?

Answer: Al lowest, C highest


Explain why it is easier to remove a 2p electron from an oxygen atom than from a nitrogen atom.

FIGURE 7.10 2p orbital filling in nitrogen and oxygen.

Electron Configurations of Ions

When electrons are removed from an atom to form a cation, they are always removed first from the occupied orbitals having the largest principal quantum number, n. For example, when one electron is removed from a lithium atom (1s22s1), it is the 2s1 electron:

Likewise, when two electrons are removed from Fe ([Ar]3d64s2), the 4s2 electrons are the ones removed:

If an additional electron is removed, forming Fe3+, it comes from a 3d orbital because all the orbitals with n = 4 are empty:

It may seem odd that 4s electrons are removed before 3d electrons in forming transition-metal cations. After all, in writing electron configurations, we added the 4s electrons before the 3d ones. In writing electron configurations for atoms, however, we are going through an imaginary process in which we move through the periodic table from one element to another. In doing so, we are adding both an electron to an orbital and a proton to the nucleus to change the identity of the element. In ionization, we do not reverse this process because no protons are being removed.

If there is more than one occupied subshell for a given value of n, the electrons are first removed from the orbital with the highest value of l. For example, a tin atom loses its 5p electrons before it loses its 5s electrons:

Electrons added to an atom to form an anion are added to the empty or partially filled orbital having the lowest value of n. For example, an electron added to a fluorine atom to form the F ion goes into the one remaining vacancy in the 2p subshell:


Do Cr3+ and V2+ have the same or different electron configurations?

SAMPLE EXERCISE 7.7 Electron Configurations of Ions

Write the electron configuration for (a) Ca2+, (b) Co3+, and (c) S2−.


Analyze and Plan We are asked to write electron configurations for three ions. To do so, we first write the electron configuration of each parent atom, then remove or add electrons to form the ions. Electrons are first removed from the orbitals having the highest value of n. They are added to the empty or partially filled orbitals having the lowest value of n.


(a) Calcium (atomic number 20) has the electron configuration [Ar]4s2. To form a 2+ ion, the two outer electrons must be removed, giving an ion that is isoelectronic with Ar:

Ca2+: [Ar]

(b) Cobalt (atomic number 27) has the electron configuration [Ar]3d7s2. To form a 3+ ion, three electrons must be removed. As discussed in the text, the 4s electrons are removed before the 3d electrons. Consequently, the electron configuration for Co3+ is

Co3+: [Ar]3d6

(c) Sulfur (atomic number 16) has the electron configuration [Ne]3s23p4. To form a 2− ion, two electrons must be added. There is room for two additional electrons in the 3p orbitals. Thus, the S2− electron configuration is

S2−: [Ne]3s2 3p6 = [Ar]

Comment Remember that many of the common ions of the s- and p-block elements, such as Ca2+ and S2−, have the same number of electrons as the closest noble gas. (Section 2.7)


Write the electron configuration for (a) Ga3+, (b) Cr3+, and (c) Br.

Answers: (a) [Ar]3d10, (b) [Ar]3d3, (c) [Ar]3d104s24p6 = [Kr]