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1.4 Ionisation energy

Syllabus
9701–2028–2029
Topic
1.4
Level
AS

Learning objectives

1.4.0Scope note• In 1.4 each atom or ion described will be in the ground state. Only the elements hydrogen to krypton will be assessed.1.4.1Terms: first ionisation energy, IE• Define/use: first ionisation energy, IE1.4.2Equations to represent first, second• Construct equations to represent first, second and subsequent ionisation energies1.4.3The trends in ionisation energies across• Identify/explain: the trends in ionisation energies across a period and down a group of the Periodic Table1.4.4The variation in successive ionisation• Identify/explain: the variation in successive ionisation energies of an element1.4.5Ionisation energies are due to the attraction• Understand: ionisation energies are due to the attraction between the nucleus and the outer electron1.4.6Factors influencing the ionisation energies• Explain the factors influencing the ionisation energies of elements in terms of nuclear charge, atomic/ionic radius, shielding by inner shells and sub-shells and spin-pair repulsion1.4.7The electronic configurations of elements• Deduce the electronic configurations of elements using successive ionisation energy data1.4.8Periodic Table position from IE data• Deduce the position of an element in the Periodic Table using successive ionisation energy data

The assessed ionisation cases are ground-state species from H to Kr

Throughout section 1.4, every atom or ion is in its ground state, and the assessed elements run from hydrogen, H, to krypton, Kr. Use ground-state configurations when explaining or interpreting ionisation-energy data.

This is an assessment boundary, not a separate ionisation rule. Excited-state configurations and elements beyond krypton are outside the cases required here; definitions, equations and trends are taught in the following objectives.

First ionisation energy removes one mole of gaseous atoms' electrons

The first ionisation energy, IE₁, is the energy required to remove one mole of electrons from one mole of gaseous atoms, forming one mole of gaseous 1+ ions. Its unit is kJ mol⁻¹.

X(g)→X+(g)+e−\mathrm{X(g) \rightarrow X^+(g) + e^-}

All three details matter: the atoms are gaseous, exactly one electron is removed from each atom, and gaseous 1+ ions form. Removing an electron requires energy because attraction between the nucleus and electron must be overcome.

Each ionisation equation removes one electron from a gaseous species

Construct successive ionisation equations one step at a time. The reactant for each step is the gaseous ion produced by the preceding step, and its charge increases by one.

X(g)→X+(g)+e−\mathrm{X(g) \rightarrow X^+(g) + e^-}

X+(g)→X2+(g)+e−\mathrm{X^+(g) \rightarrow X^{2+}(g) + e^-}

X(n−1)+(g)→Xn+(g)+e−\mathrm{X^{(n-1)+}(g) \rightarrow X^{n+}(g) + e^-}

Do not combine several removals into one equation. For example, the second ionisation must start with X⁺(g), not X(g), and every atom or ion in the equation must carry the gaseous state symbol.

Ionisation energy rises across a period and falls down a group

Direction General IE₁ trend Main explanation
Across a period increases nuclear charge increases while shielding is similar, so radius decreases and attraction to the outer electron strengthens
Down a group decreases the outer electron is in a higher shell, farther from the nucleus and more shielded, so attraction weakens

The across-period rise is not perfectly smooth. Moving from an s to a higher-energy p sub-shell can lower IE₁, and pairing two electrons in one p orbital adds repulsion, making one easier to remove.

Explain the trend by following the electron being removed. Nuclear charge alone is insufficient when distance, shielding, sub-shell energy or spin-pair repulsion changes.

Successive ionisation energies rise and reveal shell changes

Successive ionisation energies increase because each electron is removed from an increasingly positive ion. The remaining electrons experience stronger attraction to the unchanged nuclear charge.

Calcium ionisation IE₁ IE₂ IE₃ IE₄
IE / kJ mol⁻¹ 590 1150 4940 6480

The large jump between IE₂ and IE₃ shows that two outer-shell electrons have been removed and the third electron comes from an inner shell. Inner-shell electrons are closer to the nucleus and less shielded from it, so much more energy is required.

Every successive value may be larger, but only a clear change of scale is evidence for moving to a new shell. Smaller changes can reflect different sub-shells or electron pairing within the same shell.

Ionisation energy measures nucleus–outer-electron attraction

An outer electron is electrostatically attracted to the positively charged nucleus. Ionisation energy is the energy needed to overcome this attraction and separate the electron from the gaseous atom or ion.

Stronger attraction gives a larger ionisation energy; weaker attraction gives a smaller one. The attraction changes with nuclear charge and with how far the electron is from the nucleus and how strongly inner electrons shield it.

The nucleus does not lose protons during ionisation. The positive charge increases because electrons are removed, so the remaining electrons are generally held more strongly.

Use the actual electron removed to compare ionisation energies

Factor Effect on attraction and ionisation energy
Greater nuclear charge stronger attraction; IE tends to increase
Greater atomic or ionic radius electron is farther away; IE tends to decrease
More inner-shell shielding lower effective attraction; IE tends to decrease
Higher-energy, more shielded sub-shell electron is easier to remove; IE tends to decrease
Spin-pair repulsion a paired electron is easier to remove; IE tends to decrease

First identify the shell, sub-shell and pairing of the electron removed. Then compare the factors and state which change dominates; do not merely list them.

This explains common dips across a period: Al loses a higher-energy 3p electron whereas Mg loses a 3s electron; S has a paired 3p electron whereas P has three singly occupied 3p orbitals.

Use successive ionisation data to deduce an electronic configuration

Locate the largest jump in successive ionisation energies. The number of electrons removed before that jump is the number in the outer shell. Combine this evidence with the atomic number or other supplied information to build the complete ground-state configuration.

Ionisation IE₁ IE₂ IE₃ IE₄
IE / kJ mol⁻¹ 577 1820 2740 11600

The jump after three removals shows three outer-shell electrons. With atomic number 13, the configuration is 1s² 2s² 2p⁶ 3s² 3p¹: three electrons occupy the third shell before removal begins from the second shell.

1s2 2s2 2p6 3s2 3p1\mathrm{1s^2\,2s^2\,2p^6\,3s^2\,3p^1}

The jump gives the outer-shell count, not the full configuration by itself. Use the other supplied evidence to determine how many inner electrons and occupied shells are present.

Use successive ionisation data to locate an element in the Periodic Table

For an assessed s- or p-block element, the first large jump reveals the number of outer-shell electrons and therefore its group pattern. A jump after two removals supports Group 2; a jump after seven supports Group 17.

Unknown Period 3 element IE₁ IE₂ IE₃ IE₄
IE / kJ mol⁻¹ 736 1450 7740 10500

The large jump between IE₂ and IE₃ shows two outer electrons, so the element follows the Group 2 pattern. The supplied Period 3 information fixes the row; Period 3 and Group 2 identify magnesium.

Successive values alone do not always reveal the period. Use the stated period, atomic number, identity or deduced configuration alongside the jump before claiming a complete Periodic Table position.