Periodic Trends: Ionization Energy

Ionization energy (IE) is the amount of energy required to remove an electron from an atom in the gas phase:

A(g) → A⁺(g) + e⁻             ΔH ≡ IE

IE is usually expressed in kJ/mol of atoms. It is always positive because the removal of an electron always requires that energy be put in (i.e., it is endothermic). IE also shows periodic trends. As you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus. Thus,

as ↓PT, IE↓

However, as you go across the periodic table and the electrons get drawn closer in, it takes more energy to remove an electron; as a result, IE increases: as →PT, IE↑

Figure 1.1 “Ionization Energy on the Periodic Table” shows values of IE versus position on the periodic table. Again, the trend isn’t absolute, but the general trends going across and down the periodic table should be obvious.

IE also shows an interesting trend within a given atom. This is because more than one IE can be defined by removing successive electrons (if the atom has them to begin with):

A(g) → A⁺(g) + e⁻ IE₁
A⁺(g) → A²⁺(g) + e⁻ IE₂
A²⁺(g) → A³⁺(g) + e⁻ IE₃

and so forth.

Each successive IE is larger than the previous because an electron is being removed from an atom with a progressively larger positive charge. However, IE takes a large jump when a successive ionization goes down into a new shell. For example, the following are the first three IEs for Mg, whose electron configuration is 1s²2s²2p⁶3s²:

Mg(g) → Mg⁺(g) + e⁻ IE₁ = 738 kJ/mol
Mg⁺(g) → Mg²⁺(g) + e⁻ IE₂ = 1,450 kJ/mol
Mg²⁺(g) → Mg³⁺(g) + e⁻ IE₃ = 7,734 kJ/mol

The second IE is twice the first, which is not a surprise: the first IE involves removing an electron from a neutral atom, while the second one involves removing an electron from a positive ion. The third IE, however, is over five times the previous one. Why is it so much larger? Because the first two electrons are removed from the 3s subshell, but the third electron has to be removed from the n = 2 shell (specifically, the 2p subshell, which is lower in energy than the n = 3 shell). Thus, it takes much more energy than just overcoming a larger ionic charge would suggest. It is trends like this that demonstrate that electrons are organized in atoms in groups.