# First ionization energy trends

> IB Chemistry SL · IB Chemistry SL 2025
> Source: https://www.owlsprep.com/study/ib-chemistry-sl-u2-first-ionization-energy-trends/

This module covers general trends of first ionization energy across periods and down groups, explains common exceptions to these trends, and links all patterns to underlying atomic structure and electron configuration.

**Prerequisites:** [Electron configuration](https://www.owlsprep.com/study/ib-chemistry-sl-u2-electron-configuration/); [Atomic structure basics](https://www.owlsprep.com/study/ib-chemistry-sl-u1-atomic-structure/)

## Learning objectives

- Define first ionization energy correctly according to IB syllabus requirements
- Explain the general trends of first ionization energy across periods and down groups
- Identify and explain common exceptions to the general trends
- Link trends to underlying factors: nuclear charge, shielding, and electron configuration

## Definition and Key Influencing Factors

**First ionization energy** — 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.

*Notation:* IE_1

*Example:* For sodium, the process is written as $Na(g) \rightarrow Na^+(g) + e^-$ with $IE_1 = +496$ kJ mol⁻¹

Three core factors control the magnitude of $IE_1$ for any element:

- **Nuclear charge**: More protons in the nucleus increase attraction for outer electrons, which increases $IE_1$.
- **Shielding (screening)**: Inner electron shells block attraction from the nucleus. More inner shells increase shielding, which decreases $IE_1$.
- **Distance from nucleus**: Outer electrons further from the nucleus experience weaker attraction, which decreases $IE_1$.

> **info**
>
> Ionization is always endothermic, so all $IE_1$ values are positive.

**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. 2. Add correct state symbols for all species:
3. $$Ca(g) \rightarrow Ca^+(g) + e^-$$

## Trend of First Ionization Energy Down Groups

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 $IE_1$.
- 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. 2. Potassium has higher nuclear charge than sodium, but increased shielding means outer electrons are further from the nucleus.
3. 3. The increase in shielding and distance outweighs the higher nuclear charge, so net attraction for the outer electron is lower.
4. 4. Less energy is required to remove the outer electron from potassium, so its $IE_1$ is lower than sodium.

## Trend Across Periods and Common Exceptions

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 $IE_1$ 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. 2. 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. 3. Less energy is required to remove the higher energy 3p electron from aluminum, even though aluminum has a higher nuclear charge, so $IE_1$ 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. 2. In phosphorus, all three 3p electrons are unpaired in separate orbitals. In sulfur, the fourth 3p electron is paired in one 3p orbital.
5. 3. Paired electrons in the same orbital experience increased electron-electron repulsion, which makes it easier to remove the paired electron from sulfur.
6. 4. This repulsion effect outweighs sulfur's higher nuclear charge, so $IE_1$ of S is lower than P.

> **tip**
>
> The two exception patterns repeat across every period: 1) Group 13 has lower $IE_1$ than Group 2, 2) Group 16 has lower $IE_1$ than Group 15.

## Common pitfalls

- **Wrong:** Stating that $IE_1$ decreases down a group only because nuclear charge increases
  - Why it fails: Nuclear charge does increase down a group, but this effect is outweighed by shielding and distance. The explanation is incomplete and loses marks.
  - Correct: Explain that increased shielding and greater distance of outer electrons from the nucleus outweigh higher nuclear charge, leading to lower $IE_1$ down a group.
- **Wrong:** Claiming that $IE_1$ increases without exception across an entire period
  - Why it fails: Examiners regularly test understanding of the two common exceptions, so omitting them loses marks.
  - Correct: State the general increasing trend, then outline the two exceptions and their causes linked to electron configuration.
- **Wrong:** Using solid or aqueous state symbols in ionization energy equations
  - Why it fails: First ionization energy is defined specifically for gaseous atoms, so incorrect state symbols are penalized.
  - Correct: Always use (g) state symbols for both the neutral atom and the product 1+ ion.
- **Wrong:** Explaining the S/P $IE_1$ exception by claiming sulfur has more shielding
  - Why it fails: Both elements are in the same period, so they have the same number of inner shells and identical shielding.
  - Correct: Explain the exception using increased electron-electron repulsion between paired electrons in the same 3p orbital of sulfur.

## 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 |

## 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.

- [Structure 3: Chemical Bonding](https://www.owlsprep.com/study/ib-chemistry-sl-u3-overview/)
- [Ionic bonding and structure](https://www.owlsprep.com/study/ib-chemistry-sl-u3-ionic-bonding-and-structure/)
- [Covalent bonding](https://www.owlsprep.com/study/ib-chemistry-sl-u3-covalent-bonding/)

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