Ionisation Energy Explained Revision Notes | A Level AQA Chemistry

1. Why Ionisation Energy Matters

Picture trying to pull a magnet off a fridge. The closer it is, the harder you have to pull. Electrons behave the same way around a nucleus — and ionisation energy is the measure of exactly how hard that pull is.

Ionisation energy sits at the heart of atomic structure, periodicity, and bonding. Examiners use it to test whether you can connect electronic configuration to observable data — so getting this right unlocks marks across multiple topics.

2. Defining First Ionisation Energy

Every word in that definition earns marks. The atom must be gaseous (no intermolecular forces to overcome), in its ground state (lowest energy configuration), and the process produces a gaseous 1+ ion. Miss any of these conditions and you lose the mark.

The equation for first ionisation energy always follows the same pattern: one neutral gaseous atom loses one electron to form a 1+ gaseous ion. The electron released is written as $e^-$, and all species carry the state symbol (g).

3. Successive Ionisation Energies and Their Equations

Once you have removed the first electron, you can keep going — removing a second, third, fourth, and so on. Each of these is called a successive ionisation energy, and each one is always larger than the one before it.

There are two reasons successive ionisation energies always increase. First, each removal leaves fewer electrons shielding the nucleus, so the remaining electrons experience a greater effective nuclear charge. Second, the ion becomes progressively more positive, making it harder to pull away a negatively charged electron against that increasing attraction.

• First ionisation energy ($IE_1$): $\text{X}(g) \rightarrow \text{X}^+(g) + e^-$
• Second ionisation energy ($IE_2$): $\text{X}^+(g) \rightarrow \text{X}^{2+}(g) + e^-$
• Third ionisation energy ($IE_3$): $\text{X}^{2+}(g) \rightarrow \text{X}^{3+}(g) + e^-$
• Fourth ionisation energy ($IE_4$): $\text{X}^{3+}(g) \rightarrow \text{X}^{4+}(g) + e^-$

Notice the pattern: the charge on the ion on the left always equals the number of electrons already removed, and the charge on the right is one higher. State symbols (g) are required throughout — every species in these equations is gaseous.

4. Factors Affecting Ionisation Energy

Three factors control how large an ionisation energy is: nuclear charge, atomic radius, and electron shielding. These three always work together, and exam questions frequently ask you to use all three in a single explanation.

When comparing ionisation energies across a period, nuclear charge increases but shielding stays roughly constant (electrons are added to the same shell). This means effective nuclear charge rises across a period, so ionisation energy generally increases from left to right.

Going down a group, each new period adds a full inner shell of electrons. The shielding increases significantly and the atomic radius grows, so the outer electron is both further away and more shielded — ionisation energy decreases down a group despite the higher nuclear charge.

5. Successive Ionisation Energy Data and Shell Evidence

Here is where ionisation energy becomes genuinely powerful as a tool. When you plot successive ionisation energies for a single element, the graph does not rise smoothly — it rises gradually within a shell, then shows a large jump when you move between shells.

That large jump is the key signal. It tells you that the next electron being removed is in a shell closer to the nucleus, experiencing far less shielding and much greater effective nuclear charge. The position of the jump reveals how many electrons are in the outer shell of that element.

Look at the sodium data above. There is a large jump between $IE_1$ and $IE_2$: the first electron is in the third shell (outer, well-shielded), but the second electron is in the second shell (much closer to the nucleus, far less shielded). This confirms sodium has one electron in its outer shell — consistent with its position in Group 1.

A second large jump appears between $IE_9$ and $IE_{10}$, marking the transition from the second shell to the first shell (just two electrons in the 1s sub-shell). In absolute terms, this jump (from 28,932 to 141,362 kJ mol⁻¹) is even larger than the first. The graph therefore provides direct experimental evidence for the existence of distinct electron shells.

Worked example: Reading a Successive Ionisation Energy Graph

The successive ionisation energies (in kJ mol⁻¹) for an unknown element are: 738, 1451, 7733, 10540, 13630, 17995, 21703. Identify the group of this element and suggest its identity.

1. Values: 738, 1451, 7733, 10540... Jump between $IE_2$ (1451) and $IE_3$ (7733): ratio = 7733 ÷ 1451 ≈ 5.3 — A jump of this magnitude (more than five times larger) signals a change of electron shell. All smaller increases are within the same shell.
2. The large jump occurs after removing 2 electrons, so 2 electrons were in the outer shell before the jump. — Electrons removed before the large jump belong to the outermost shell. Here, 2 electrons were removed relatively easily, then the third was much harder to remove.
3. 2 outer-shell electrons → Group 2 — The number of outer-shell electrons equals the group number for main-group elements.
4. $IE_1 = 738$ kJ mol$^{-1}$, which matches magnesium (literature value: 738 kJ mol$^{-1}$). — Cross-referencing the first ionisation energy value with data book values confirms the element. Magnesium is in Period 3, Group 2, with configuration [Ne] 3s².

Answer: The element is in Group 2. The first ionisation energy of 738 kJ mol⁻¹ identifies it as magnesium (Mg).

This method — locating the large jump, counting the electrons removed before it, and matching to a group — is a standard AQA exam task. Practise it until it feels automatic.

6. Periodic Trends: Across a Period and Down a Group

Now that the underlying factors are clear, the periodic trends follow directly. Across Period 3 (Na to Ar), nuclear charge increases from 11 to 18 protons while all electrons are added to the same third shell — shielding barely changes. The result is a general increase in first ionisation energy across the period.

Two dips in this trend are worth knowing precisely. The first ionisation energy of aluminium (Group 3) is lower than that of magnesium (Group 2), because the outer electron in aluminium is in a 3p sub-shell rather than a 3s sub-shell. The 3p electron is slightly higher in energy and experiences slightly more shielding from the filled 3s sub-shell, making it easier to remove despite the higher nuclear charge.

The second dip occurs between phosphorus (Group 5) and sulphur (Group 6). Phosphorus has three unpaired electrons in its 3p sub-shell (one in each 3p orbital). Sulphur has four 3p electrons, meaning one orbital contains a pair. That paired electron experiences extra electron–electron repulsion within the same orbital, making it easier to remove than expected — so sulphur has a lower first ionisation energy than phosphorus.

The chart makes both dips visible: Al is lower than Mg, and S is lower than P. These are the two anomalies examiners test most frequently in Period 3 ionisation energy questions.

7. Exam Technique and Common Mistakes

Ionisation energy questions appear in multiple formats: writing equations, explaining trends, reading graphs, and identifying elements from data. Each format has its own mark-winning habits and its own traps.

Every spec requirement for ionisation energy connects back to one central idea: the balance between nuclear charge, shielding, and distance determines how strongly an electron is held. Master that triangle and the definitions, equations, trends, and graph-reading all fall into place.