Periodic Trends: Ionization Energy

Ionization energy is the energy needed to remove the most loosely held electron from a gaseous atom or ion, and the first ionization energy generally increases from left to right across a period and decreases down a group. In brief: removing one electron from sodium takes 496 kJ per mole, but chlorine needs 1,251 kJ/mol and neon 2,081 kJ/mol; for magnesium, the second ionization energy (1,451 kJ/mol) is about twice the first (738 kJ/mol), while the third (7,733 kJ/mol) is much larger, which shows that magnesium has two valence electrons.

Ionization energy tells how strongly an atom holds its electrons, and it determines whether an element forms positive ions. This article defines first and successive ionization energies, explains the trends and the dips that interrupt them, uses successive values to identify groups, shows data tables, works through calculations including unit conversion and photon wavelengths and gives some background on how the values are measured.

What is ionization energy?

The first ionization energy (IE₁) is the energy required to remove one mole of electrons from one mole of gaseous atoms: X(g) → X⁺(g) + e⁻. It is always positive, because energy has to be supplied to overcome the attraction of the nucleus. The second ionization energy (IE₂) is the energy to remove one electron from the gaseous 1+ ion: X⁺(g) → X²⁺(g) + e⁻, and so on. It is usually given in kilojoules per mole (kJ/mol) or in electron volts per atom (eV).

What are the trends?

• Across a period: IE₁ generally increases, because the effective nuclear charge increases and the atomic radius decreases, so the outer electron is held more strongly. The alkali metals have the lowest value and the noble gases the highest.
• Down a group: IE₁ decreases, because the outer electron is farther from the nucleus and is shielded by more inner shells, so it is easier to remove.

First ionization energies of periods 2 and 3

Why are there dips in the trend?

Two dips occur in each period of the main groups.

• Beryllium to boron (2s to 2p) and magnesium to aluminum: the outer electron of boron is in a 2p orbital, which is higher in energy and farther from the nucleus than 2s, and it is also shielded by the 2s electrons, so it is removed more easily despite the greater nuclear charge. IE₁ drops from 900 to 801 kJ/mol.
• Nitrogen to oxygen (and phosphorus to sulfur): in nitrogen, the three 2p electrons occupy separate orbitals. In oxygen, the fourth electron has to pair up with another electron in an orbital, and the repulsion between the two electrons makes it easier to remove. IE₁ drops from 1,402 to 1,314 kJ/mol.

What do successive ionization energies show?

Each successive electron is harder to remove, since the remaining electrons are held by the same nuclear charge. There is a very large jump when the removal starts from an inner shell. The size of this jump shows the number of valence electrons.

Worked examples

1. Identify the group. An element has successive ionization energies of 590, 1,145, 4,912 and 6,491 kJ/mol. The big jump comes after the second electron, so there are 2 valence electrons and the element is in group 2. The value of IE₁ and the group suggest calcium (590 kJ/mol).
2. Energy for a mass of sodium. The energy needed to ionize 1.00 g of sodium atoms: moles = 1.00 ÷ 22.99 = 0.0435 mol, and energy = 0.0435 × 496 = 21.6 kJ.
3. Conversion to electron volts. For sodium, 496 kJ/mol ÷ 96.485 kJ/mol per eV = 5.14 eV per atom.
4. Energy to form Mg²⁺. IE₁ + IE₂ = 738 + 1,451 = 2,189 kJ/mol. The energy to form Mg³⁺ would be 2,189 + 7,733 = 9,922 kJ/mol, which is too large, which is why magnesium forms 2+ ions only.
5. Which has the higher IE₁? Compare K and Ca. They are in the same period, and calcium is to the right, with the higher nuclear charge. Calcium (590) is higher than potassium (419). Compare Li and Cs: lithium is higher, since it is at the top of the group.
6. Photon needed to ionize hydrogen. The ionization energy of hydrogen is 13.6 eV. The wavelength is λ = hc ÷ E = 1,240 eV·nm ÷ 13.6 eV = 91.2 nm, which is ultraviolet light. Visible light (about 400 to 700 nm) does not have enough energy.

Why does ionization energy matter?

• Elements with low ionization energy (metals) lose electrons easily to form cations, and are good reducing agents.
• Elements with high ionization energy (nonmetals, noble gases) do not form positive ions easily.
• The values explain the oxidation states of elements and the type of bonding in their compounds.
• The photoelectric effect and photoelectron spectroscopy use ionization energies to study electron structure.
• The low values for alkali metals make them useful in photocells and the basis of ion sources in mass spectrometers.

How is ionization energy measured?

The first measurements in the early 20th century used the energy of electrons colliding with gas atoms. The Franck-Hertz experiment of 1914 demonstrated the discrete energy levels of atoms and gave experimental support for the Bohr model. Today, values are obtained by photoelectron spectroscopy, in which photons of known energy are directed at atoms and the energy of the electrons that come out is measured, as the difference between the photon energy and the electron’s kinetic energy gives the ionization energy. This technique shows the separate energy levels of all the electrons in an atom.

Which mistakes are common?

Mistake 1: Thinking that ionization energy is released

Energy must be supplied. Ionization is always endothermic.

Mistake 2: Ignoring the dips in the trend

Remember the drops from Be to B and from N to O, which can be explained by orbital types and electron pairing.

Mistake 3: Using atoms in the solid or liquid state

Ionization energy refers to isolated gaseous atoms.

Mistake 4: Expecting successive energies to increase evenly

They rise gradually while removing valence electrons, and then jump when an inner shell is reached.