Ionisation energiesAQA A-Level Chemistry: Revision notes

The first ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous atoms of an element to form one mole of gaseous 1+ ions. It is measured in kJ mol⁻¹.

The equation must have state symbols, e.g. for magnesium:

Mg(g) → Mg⁺(g) + e⁻

The values depend on three factors: the nuclear charge, the distance of the outer electron from the nucleus, and the shielding by inner electrons.

The second ionisation energy is the energy needed to remove one electron from each ion in one mole of gaseous 1+ ions to form one mole of gaseous 2+ ions, and so on.

• First: Mg(g) → Mg⁺(g) + e⁻
• Second: Mg⁺(g) → Mg²⁺(g) + e⁻
• Third: Mg²⁺(g) → Mg³⁺(g) + e⁻

Successive ionisation energies always increase, because each electron is removed from an increasingly positive ion, where the same number of protons attract fewer electrons, which are held more strongly.

A large jump in successive ionisation energies shows that the next electron is being removed from a new, inner shell. The inner shell is closer to the nucleus and less shielded, so it is held much more strongly.

The number of electrons removed before the big jump equals the number of electrons in the outer shell, so it gives the group.

Worked example. Aluminium: 578, 1817, 2745, 11577, 14842 kJ mol⁻¹. The big jump comes after the third electron, so three electrons are in the outer shell and the element is in Group 3 (configuration 1s² 2s² 2p⁶ 3s² 3p¹).

First ionisation energy generally increases across Period 3. The nuclear charge increases while the electrons are added to the same shell with similar shielding, so the atomic radius decreases and the outer electron is held more strongly.

There are two exceptions, and they give evidence for sub-shells.

• Mg to Al decreases. The outer electron of Al is in a 3p sub-shell, which is higher in energy and shielded by the 3s electrons, while Mg's outer electrons are 3s. This is evidence that the 3p sub-shell is higher in energy than 3s.
• P to S decreases. In P (3p³) each 3p electron is in a separate orbital, but in S (3p⁴) two electrons share an orbital. The repulsion between them makes one easier to remove.

First ionisation energy decreases down Group 2. Going down the group, the atoms have more shells, so the outer electron is further from the nucleus and there is more shielding by inner electrons. This outweighs the increased nuclear charge, so the outer electron is held less strongly and less energy is needed to remove it.

This trend is why the metals become more reactive down the group: they lose their outer electrons more easily.