Periodic trends in atomic properties

IB Chemistry SL· 45 min read

1. Atomic Radius: Trends and Explanation★★☆☆☆⏱ 10 min

📘 Definition

Atomic radius

r

Half the distance between the nuclei of two adjacent covalently bonded atoms of the same element

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 ( $Z_{e f f}$ ) 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 → $Z_{e f f}$ 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 $Z_{e f f}$ .

2. First Ionization Energy Trends★★★☆☆⏱ 15 min

📘 Definition

First ionization energy

IE_1

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

Example:

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

The general trend for first ionization energy follows $Z_{e f f}$ : higher $Z_{e f f}$ 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 → $I E_{1}$ decreases
Across a period: Increasing $Z_{e f f}$ → stronger attraction for electrons → $I E_{1}$ 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 $I E_{1} (A l) < I E_{1} (M g)$

3. Electronegativity and Electron Affinity★★★☆☆⏱ 12 min

📘 Definition

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 $Z_{e f f}$ 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.

4. Common Exceptions to General Trends★★★★☆⏱ 10 min

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 $I E_{1} (O) < I E_{1} (N)$ , 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.

5. Common Pitfalls

Wrong move:

Claiming atomic radius increases across a period because atomic number increases

Why:

Increasing proton count raises effective nuclear charge, which pulls electrons closer, overriding the increase in number of electrons

Correct move:

State that atomic radius decreases across a period due to increasing effective nuclear charge with constant shielding

Wrong move:

Confusing ionization energy with electron affinity

Why:

Ionization energy describes energy to remove an electron, electron affinity describes energy change when adding an electron

Correct move:

Memorize: Ionization = Remove, Affinity = Add

Wrong move:

Claiming fluorine is not the most electronegative element

Why:

Electronegativity increases up and to the right, and noble gases do not form covalent bonds so they are not counted

Correct move:

Remember that fluorine is the most electronegative element, francium is the least

Wrong move:

Explaining trends only by total nuclear charge, ignoring shielding

Why:

Down a group, total nuclear charge increases, but increased shielding and distance of valence electrons dominates

Correct move:

Always explain trends using a combination of effective nuclear charge, shielding, and electron distance from the nucleus

Wrong move:

Forgetting to mention subshell stability when explaining IE exceptions

Why:

Examiners specifically award marks for recognizing the extra stability of half-filled and full subshells

Correct move:

Always explicitly state the extra stability of half-filled/full subshells when explaining drops in ionization energy

6. Quick Reference 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 ( $Z_{e f f}$ )	Increases	Approximately constant

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 in atomic radius across a period
2023 · 2
Explain IE exception between N and O
2021 · 1
Compare electronegativity of three elements

Going deeper

Interactive toolInteractive Periodic Table of TrendsGreat for visualizing all trends

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.