First ionization energy trends

IB Chemistry SL· Structure 2.2 Electron Configuration· 15 min read

1. Definition and Key Influencing Factors★★☆☆☆⏱ 5 min

📘 Definition

First ionization energy

IE_1

The minimum energy needed to remove 1 mole of electrons from 1 mole of neutral gaseous atoms, producing 1 mole of gaseous +1 ions. Units are typically kJ mol⁻¹.

Example:

For sodium, the process is written as $N a (g) \to N a^{+} (g) + e^{-}$ with $I E_{1} = + 496$ kJ mol⁻¹

Three core factors control the magnitude of $I E_{1}$ for any element:

Nuclear charge: More protons in the nucleus increase attraction for outer electrons, which increases $I E_{1}$ .
Shielding (screening): Inner electron shells block attraction from the nucleus. More inner shells increase shielding, which decreases $I E_{1}$ .
Distance from nucleus: Outer electrons further from the nucleus experience weaker attraction, which decreases $I E_{1}$ .

📐 Worked Example

Write the correct chemical equation for the first ionization energy of calcium.

1
1. Recall the definition: start with 1 mole of gaseous neutral calcium, remove 1 electron to form 1 mole of gaseous 1+ calcium ions.
2
1. Add correct state symbols for all species:
3
$Ca(g) \rightarrow Ca^+(g) + e^-$

2. Trend of First Ionization Energy Down Groups★★☆☆☆⏱ 5 min

The general trend moving down any group of the periodic table is that first ionization energy decreases. This trend is explained by the change in the three key factors listed above:

Nuclear charge increases moving down a group, which would tend to increase $I E_{1}$ .
However, each element down a group has one additional full inner electron shell, so shielding increases significantly, and outer electrons are further from the nucleus.
The effects of increased shielding and greater distance outweigh the higher nuclear charge, leading to a net decrease in attraction between the nucleus and outer electrons.

📐 Worked Example

Explain why the first ionization energy of potassium is lower than that of sodium.

1
1. Potassium is below sodium in group 1, so it has one more full inner electron shell than sodium.
2
1. Potassium has higher nuclear charge than sodium, but increased shielding means outer electrons are further from the nucleus.
3
1. The increase in shielding and distance outweighs the higher nuclear charge, so net attraction for the outer electron is lower.
4
1. Less energy is required to remove the outer electron from potassium, so its $I E_{1}$ is lower than sodium.

3. Trend Across Periods and Common Exceptions★★★☆☆⏱ 8 min

The general trend moving left to right across a period is that first ionization energy increases. Across a period, all elements have the same number of inner electron shells, so shielding is roughly constant. Nuclear charge increases by one proton per element, so effective nuclear charge increases. This pulls outer electrons closer to the nucleus, increasing attraction, so more energy is needed to remove an electron, hence $I E_{1}$ increases.

There are two common exceptions to this trend across every period. We illustrate them with examples from period 3 below:

📐 Worked Example

Explain why aluminum has a lower first ionization energy than magnesium.

1
1. Write the full electron configurations for both elements:
2
$Mg: 1s^2 2s^2 2p^6 3s^2$
3
$Al: 1s^2 2s^2 2p^6 3s^2 3p^1$
4
1. The outer electron of aluminum is in the higher energy 3p sub-shell, which is further from the nucleus than the 3s sub-shell that holds magnesium's outer electrons.
5
1. Less energy is required to remove the higher energy 3p electron from aluminum, even though aluminum has a higher nuclear charge, so $I E_{1}$ of Al is lower than Mg.

📐 Worked Example

Explain why sulfur has a lower first ionization energy than phosphorus.

1
1. Write the valence electron configurations for both elements:
2
$P: 3s^2 3p^3$
3
$S: 3s^2 3p^4$
4
1. In phosphorus, all three 3p electrons are unpaired in separate orbitals. In sulfur, the fourth 3p electron is paired in one 3p orbital.
5
1. Paired electrons in the same orbital experience increased electron-electron repulsion, which makes it easier to remove the paired electron from sulfur.
6
1. This repulsion effect outweighs sulfur's higher nuclear charge, so $I E_{1}$ of S is lower than P.

4. Common Pitfalls

Wrong move:

Stating that $I E_{1}$ decreases down a group only because nuclear charge increases

Why:

Nuclear charge does increase down a group, but this effect is outweighed by shielding and distance. The explanation is incomplete and loses marks.

Correct move:

Explain that increased shielding and greater distance of outer electrons from the nucleus outweigh higher nuclear charge, leading to lower $I E_{1}$ down a group.

Wrong move:

Claiming that $I E_{1}$ increases without exception across an entire period

Why:

Examiners regularly test understanding of the two common exceptions, so omitting them loses marks.

Correct move:

State the general increasing trend, then outline the two exceptions and their causes linked to electron configuration.

Wrong move:

Using solid or aqueous state symbols in ionization energy equations

Why:

First ionization energy is defined specifically for gaseous atoms, so incorrect state symbols are penalized.

Correct move:

Always use (g) state symbols for both the neutral atom and the product 1+ ion.

Wrong move:

Explaining the S/P $I E_{1}$ exception by claiming sulfur has more shielding

Why:

Both elements are in the same period, so they have the same number of inner shells and identical shielding.

Correct move:

Explain the exception using increased electron-electron repulsion between paired electrons in the same 3p orbital of sulfur.

5. Quick Reference Cheatsheet

Trend Type	General Direction	Core Cause	Common Exceptions
Down a group	Decreases	Increased shielding + greater outer electron distance outweigh higher nuclear charge	None
Across period (left → right)	Increases	Constant shielding, increasing effective nuclear charge	Group 13 < Group 2, Group 16 < Group 15

6. Frequently Asked

Why do first ionization energies not always increase across a period?

Exceptions arise from differences in sub-shell energy and electron-electron repulsion. For example, Al has a lower $I E_{1}$ than Mg because its outer electron occupies the higher energy 3p sub-shell, rather than the lower energy 3s.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Trend across period 3
2023 · 2
Explain Mg/Al IE exception
2021 · 1
Trend down group 1

Going deeper

official syllabusIB Chemistry SL 2025 SyllabusOfficial syllabus reference

What's Next

First ionization energy trends are core to understanding periodicity, and the observed exceptions provide direct experimental evidence for the existence of electron sub-shells, confirming the electron configuration model you learned earlier. This concept underpins all other periodic trends, including atomic radius, electronegativity, and metallic character, all of which are regularly tested in both Paper 1 and Paper 2 of IB Chemistry SL. It also connects directly to successive ionization energies, which are used to predict electron configuration from experimental data.