# Periodic trends in atomic properties

> IB Chemistry SL · Reactivity 1: Stoichiometry and Periodicity
> Source: https://www.owlsprep.com/study/ib-chemistry-sl-u4-periodic-trends-in-atomic-properties/

This module covers key periodic trends in atomic radius, ionization energy, electronegativity, and electron affinity. You will learn to explain these trends using effective nuclear charge and electron shielding, and predict relative properties.

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

## Learning objectives

- Describe and explain the trend in atomic radius across a period and down a group
- Explain general trends and exceptions in first ionization energy
- Describe trends in electronegativity and electron affinity
- Predict relative atomic properties from an element's periodic position

## Atomic Radius: Trends and Explanation

**Atomic radius** — Half the distance between the nuclei of two adjacent covalently bonded atoms of the same element

*Notation:* r

*Example:* For metallic elements, it is half the distance between two adjacent atoms in the metal lattice

Atomic radius trends are explained by two core factors: effective nuclear charge ($Z_{eff}$) and electron shielding (the ability of inner electrons to block nuclear attraction for valence electrons).

- Down a group: Number of occupied electron shells increases → shielding increases → valence electrons are further from the nucleus → atomic radius increases
- Across a period (left to right): Proton number increases → $Z_{eff}$ increases, while number of inner shells (and shielding) stays constant → valence electrons are pulled closer → atomic radius decreases

**Worked example:** Arrange the following elements in order of increasing atomic radius: Mg, Na, P, Rb, Al

1. Identify the position of each element in the periodic table:
2. $$\text{Na (Period 3, Group 1)}, \text{Mg (Period 3, Group 2)}, \text{Al (Period 3, Group 13)}, \text{P (Period 3, Group 15)}, \text{Rb (Period 5, Group 1)}$$
3. Down groups, atomic radius increases, so Rb (Period 5) is larger than all Period 3 elements here.
4. Across Period 3, atomic radius decreases left to right, so order from smallest to largest is: P < Al < Mg < Na
5. Combining these gives the final order of increasing atomic radius:
6. $$P < Al < Mg < Na < Rb$$

> **Exam tip:** Never confuse atomic number with atomic radius: increasing atomic number across a period does not mean increasing atomic radius, due to higher $Z_{eff}$.

## First Ionization Energy Trends

**First ionization energy** — Minimum energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous +1 ions

*Notation:* IE_1

*Example:* Na_{(g)} \rightarrow Na^+_{(g)} + e^- \quad \Delta H = IE_1

The general trend for first ionization energy follows $Z_{eff}$: higher $Z_{eff}$ holds electrons more tightly, so more energy is required to remove an electron.

- Down a group: Valence electrons are further from the nucleus, more shielded → attraction is weaker → $IE_1$ decreases
- Across a period: Increasing $Z_{eff}$ → stronger attraction for electrons → $IE_1$ generally increases

**Worked example:** Explain why the first ionization energy of aluminum is lower than that of magnesium

1. Write the full electron configurations for both elements:
2. $$Al: 1s^2 2s^2 2p^6 3s^2 3p^1 \quad Mg: 1s^2 2s^2 2p^6 3s^2$$
3. The outermost electron in aluminum is in the 3p orbital, which is higher in energy and experiences additional shielding from the 3s electrons, compared to magnesium's outermost 3s electrons.
4. Less energy is required to remove aluminum's 3p electron, so $IE_1(Al) < IE_1(Mg)$

## Electronegativity and Electron Affinity

**Electronegativity** — A measure of the ability of an atom in a covalent bond to attract shared bonding electrons to itself

Electron affinity is the energy change when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous -1 ions. A more negative electron affinity means the atom more readily accepts an electron.

Both properties follow the same general trend, driven by $Z_{eff}$ and atomic radius: they increase across a period left to right, and decrease down a group. Fluorine is the most electronegative element, while francium is the least electronegative.

**Worked example:** Which of C, O, N, Si has the highest electronegativity? Explain your answer.

1. Identify positions: C, N, O are all Period 2, Si is Group 14 Period 3 (below C).
2. Electronegativity increases across a period (left to right) and decreases down a group.
3. Order across Period 2: C < N < O, and Si is less electronegative than C.
4. Oxygen therefore has the highest electronegativity of the four elements.

## Common Exceptions to General Trends

IB exams frequently test exceptions to the general ionization energy trend. There are two consistent drops across every period: between Group 2 and Group 13, and between Group 15 and Group 16.

Half-filled and full subshells have extra stability due to symmetric electron distribution and minimal electron-electron repulsion. This stability changes the expected ionization energy.

**Worked example:** Explain why the first ionization energy of oxygen is lower than that of nitrogen

1. Write the valence electron configurations for both elements:
2. $$N: 2s^2 2p^3 \quad O: 2s^2 2p^4$$
3. Nitrogen has a half-filled 2p subshell, which has extra stability due to all electrons being unpaired with symmetric distribution.
4. Oxygen has one paired electron in its 2p subshell; this paired electron experiences extra repulsion from its partner, so it requires less energy to remove it.
5. Therefore $IE_1(O) < IE_1(N)$, even though oxygen has a higher atomic number than nitrogen.

> **Exam tip:** Always mention the extra stability of half-filled or full subshells when explaining these exceptions—examiners actively look for this point.

## Common pitfalls

- **Wrong:** Claiming atomic radius increases across a period because atomic number increases
  - Why it fails: Increasing proton count raises effective nuclear charge, which pulls electrons closer, overriding the increase in number of electrons
  - Correct: State that atomic radius decreases across a period due to increasing effective nuclear charge with constant shielding
- **Wrong:** Confusing ionization energy with electron affinity
  - Why it fails: Ionization energy describes energy to remove an electron, electron affinity describes energy change when adding an electron
  - Correct: Memorize: Ionization = Remove, Affinity = Add
- **Wrong:** Claiming fluorine is not the most electronegative element
  - Why it fails: Electronegativity increases up and to the right, and noble gases do not form covalent bonds so they are not counted
  - Correct: Remember that fluorine is the most electronegative element, francium is the least
- **Wrong:** Explaining trends only by total nuclear charge, ignoring shielding
  - Why it fails: Down a group, total nuclear charge increases, but increased shielding and distance of valence electrons dominates
  - Correct: Always explain trends using a combination of effective nuclear charge, shielding, and electron distance from the nucleus
- **Wrong:** Forgetting to mention subshell stability when explaining IE exceptions
  - Why it fails: Examiners specifically award marks for recognizing the extra stability of half-filled and full subshells
  - Correct: Always explicitly state the extra stability of half-filled/full subshells when explaining drops in ionization energy

## Cheatsheet

| Property | Trend across period (left → right) | Trend down a group |
| --- | --- | --- |
| Atomic radius | Decreases | Increases |
| First ionization energy | Generally increases | Decreases |
| Electronegativity | Increases | Decreases |
| Magnitude of negative electron affinity | Generally increases | Generally decreases |
| Effective nuclear charge ($Z_{eff}$) | Increases | Approximately constant |

## What's next

Understanding periodic trends in atomic properties is the foundation for explaining all chemical reactivity, bonding, and compound properties in IB Chemistry. These trends are used consistently to predict element behavior, why elements form specific ions, and how bonds form between different atoms. This topic is a core building block for all subsequent content, and regularly appears in both paper 1 and paper 2 exam questions. You will now apply these ideas to bonding, then to group reactivity trends.

- [S- and p-block element chemistry](https://www.owlsprep.com/study/ib-chemistry-sl-u4-s-and-p-block-element/)
- [Reactivity 2: Energetics and Kinetics](https://www.owlsprep.com/study/ib-chemistry-sl-u5-overview/)
- [Temperature change and heat capacity](https://www.owlsprep.com/study/ib-chemistry-sl-u5-temperature-change-and-heat-capacity/)

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