Electron Configuration and Ionisation Energy — AQA A-level Chemistry

SPEC 3.1.1.3 - ATOMIC STRUCTURE

Electron Configuration & Ionisation Energies

Shells, sub-shells, orbital filling, and evidence for electron arrangement from ionisation energy data

Introduction

The arrangement of electrons determines how atoms bond and react. Understanding electron configuration is fundamental to chemistry.

Learning Objectives

Write electron configurations for atoms and ions up to Z = 36

Define and write equations for first and successive ionisation energies

Explain trends in ionisation energy across Period 3 and down Group 2

Use successive ionisation data as evidence for electron shells

Why This Matters

Ionisation energy trends provide direct experimental evidence for how electrons are arranged in atoms - connecting theory with observable data.

Electron Shells

Electrons are arranged in shells (also called energy levels or principal quantum levels) around the nucleus. Shells are numbered using the principal quantum number, n (n = 1, 2, 3, 4...).

Why Shells Exist

Electrons can only occupy specific, fixed energy levels - they cannot exist between shells. This is a fundamental principle of quantum mechanics. Electrons in shell 1 (closest to nucleus) have the lowest energy and are most strongly attracted to the positive nucleus.

Important

Electrons fill the lowest energy shell first (closest to nucleus) before moving to higher shells. This minimises the total energy of the atom - atoms are most stable in their lowest energy state.

Sub-shells and Energy Levels

Each shell contains sub-shells (s, p, d, f) with slightly different energies. Sub-shells are filled in order of increasing energy, not simply by shell number. Elements with more than 57 electrons also have an f sub-shell.

Why 4s Fills Before 3d

Although 3d is in shell 3 and 4s is in shell 4, the 4s sub-shell has lower energy than 3d. This is because 4s electrons penetrate closer to the nucleus than 3d electrons, experiencing stronger nuclear attraction. Electrons always fill the lowest available energy level first (Aufbau principle).

Sub-shellNumber of OrbitalsMax ElectronsShape

s12Spherical

p3 (px, py, pz)6Dumbbell

d510Cloverleaf

f714Complex

Energy Order of Sub-shells

Within the same shell, the energy of sub-shells increases in the order: s < p < d < f

The only exception is that the 3d sub-shell has slightly higher energy than 4s, so 4s fills before 3d.

Degenerate Orbitals

All orbitals in the same sub-shell have the same energy and are called degenerate. For example, the three 2p orbitals (2px, 2py, 2pz) are all equal in energy.

Energy Level Diagram

Sub-shells within each principal shell have different energies. Notice how 4s is lower in energy than 3d.

Filling Order

1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p

Worked Example01

Write the full electron configuration and draw the orbital box diagram for chlorine (Cl, Z = 17).

1

Count electrons: Z = 17, so 17 electrons to place

2

Fill in order: 1s² 2s² 2p⁶ 3s² 3p⁵

3

Check: 2 + 2 + 6 + 2 + 5 = 17 ✓

4

Shorthand: [Ne] 3s² 3p⁵

Orbital Box Diagram:

Orbital Shapes

An orbital is a region of space where there is a 95% probability of finding an electron. Electrons can only exist at specific energy levels, not in between them.

s orbital shape

The s orbitals are spherical. The size of s orbitals increases with shell number (e.g. 3s is bigger than 1s).

p orbital shape

The p orbitals have a dumbbell shape. Every shell has three p orbitals except the first one (n = 1). They occupy the x, y and z axes and are oriented perpendicular to one another. The lobes become larger and longer with increasing shell number.

Exam Note

The shape of d orbitals is NOT required. You need: s (spherical), p (dumbbell along x, y, z). d = 5 orbitals (10e⁻), f = 7 orbitals (14e⁻).

Exam Tip

The three p orbitals are labelled pₓ, p_y and p_z, but you do not need to include this in your electron configurations!

Periodic Table Blocks

The periodic table is organised into blocks based on which sub-shell is being filled:

Block Classification

Elements are classified based on the position of their outermost electron:

s block elements: Have their valence electron(s) in an s orbital

p block elements: Have their valence electron(s) in a p orbital

d block elements: Have their valence electron(s) in a d orbital

f block elements: Have their valence electron(s) in an f orbital

Using the Periodic Table

The block tells you which sub-shell contains the outer electrons:

Na (s-block) → 3s¹ | Cl (p-block) → 3p⁵ | Fe (d-block) → 3d⁶

Presenting the Electron Configuration

Electrons can be imagined as small spinning charges which rotate around their own axis in either a clockwise or anticlockwise direction. The spin of the electron is represented by its direction (↑ or ↓).

Spin-Pair Repulsion

Electrons with similar spin repel each other - this is called spin-pair repulsion.

To minimise this repulsion, electrons will occupy separate orbitals in the same sub-shell where possible, and have their spin in the same direction.

Example: 3 Electrons in a p Sub-shell

If there are three electrons in a p sub-shell, one electron will go into each pₓ, p_y and p_z orbital - all with the same spin direction.

Electron configuration: three electrons in a p sub-shell

When Do Electrons Pair?

Electrons are only paired when there are no more empty orbitals available within a sub-shell. When pairing occurs, the spins are opposite to minimise repulsion.

Example: 4 Electrons in a p Sub-shell

If there are four electrons in a p sub-shell: the first 3 electrons fill up the empty p orbitals one at a time, and then the 4th one pairs up in the pₓ orbital with opposite spin.

Electron configuration: four electrons in a p sub-shell

Writing Electron Configurations

The electron configuration shows the number of electrons occupying a sub-shell in a specific shell.

The electron configuration shows the number of electrons occupying a subshell in a specific shell

Full Notation

Write sub-shells in order with superscript numbers showing electron count:1s² 2s² 2p⁶ 3s² 3p⁶This is the configuration for Argon (Ar, Z = 18)

Noble Gas Core Shorthand

Use the previous noble gas symbol in brackets to simplify:

Sodium: 1s² 2s² 2p⁶ 3s¹ = [Ne] 3s¹

Calcium: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² = [Ar] 4s²

Orbital Box Diagrams

Each box = one orbital. Arrows (↑↓) show electron spin. Pauli: max 2e⁻ per orbital, opposite spins. Hund's: fill singly first, same spin. Aufbau: lowest energy first.

Exam Tip

You can use full headed arrows (↑↓) or half headed arrows to represent electrons in box notation - both are acceptable.

Filling Rules & Ground State

Ground State

The ground state is the most stable electronic configuration of an atom, with the lowest amount of energy. This is achieved by filling the lowest energy sub-shells first.

Three Rules for Filling Orbitals

1. Aufbau Principle: Electrons fill orbitals starting from the lowest energy level first.

2. Hund's Rule: Orbitals of the same energy fill singly before pairing (minimises repulsion).

3. Pauli Exclusion: Each orbital holds max 2 electrons with opposite spins (↑↓).

Common Mistake

Remember: 4s fills BEFORE 3d because 4s is lower in energy. But when forming ions, electrons are removed from 4s first!

02Worked Example

Draw the orbital box diagram (electrons in boxes notation) for:

(a) Phosphorus (P, Z = 15)

(b) Titanium (Ti, Z = 22)

Phosphorus (P, Z = 15):

Titanium (Ti, Z = 22):

Transition Metals and Ions

For transition metals (Z = 21-30), the 3d sub-shell fills after 4s:

ElementZExpectedActual Configuration

Scandium (Sc)21—[Ar] 4s² 3d¹

Chromium (Cr)24[Ar] 4s² 3d⁴[Ar] 4s¹ 3d⁵ ⚠️

Iron (Fe)26—[Ar] 4s² 3d⁶

Copper (Cu)29[Ar] 4s² 3d⁹[Ar] 4s¹ 3d¹⁰ ⚠️

Zinc (Zn)30—[Ar] 4s² 3d¹⁰

Cr and Cu Exceptions - MUST KNOW

Chromium: [Ar] 4s¹ 3d⁵ (NOT 4s² 3d⁴) — half-filled d sub-shell is more stable

Copper: [Ar] 4s¹ 3d¹⁰ (NOT 4s² 3d⁹) — fully-filled d sub-shell is more stable

Ions: Remove from 4s First

When forming positive ions, remove electrons from 4s before 3d:

Fe: [Ar] 4s² 3d⁶ → Fe²⁺: [Ar] 3d⁶ → Fe³⁺: [Ar] 3d⁵

Worked Example03

Write the electron configurations for: (a) Bromine (Br, Z = 35), (b) Fe²⁺, (c) Cl⁻

a)

Bromine (Z = 35): Fill orbitals in order until 35 electrons are placed.

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵

(2 + 2 + 6 + 2 + 6 + 2 + 10 + 5 = 35 ✓)

b)

Fe²⁺: Start with Fe (Z = 26): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶

Remove 2 electrons from 4s (highest energy level first):

Fe²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶

c)

Cl⁻: Cl (Z = 17) gains 1 electron: 1s² 2s² 2p⁶ 3s² 3p⁵ + e⁻

Cl⁻: 1s² 2s² 2p⁶ 3s² 3p⁶ (same as Argon - isoelectronic)

First Ionisation Energy

Definition (MUST MEMORISE)

First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.

X(g) \to X^{+} (g) + e^{-}

Standard Conditions & Units

Ionisation energies are measured under standard conditions: 298 K (25°C) and 100 kPa.

Units: kJ mol⁻¹ (kilojoules per mole)

Why These Conditions?

Gaseous atoms (g): We measure IE for isolated atoms in the gas phase because this eliminates intermolecular forces. In solids or liquids, atoms are bonded together, which would affect the energy measurement.

One mole: We use moles to give a measurable, comparable quantity. Removing one electron from one atom requires a tiny amount of energy (measured in eV); using moles gives values in kJ that are easier to work with.

Always endothermic: Energy must always be supplied to overcome the electrostatic attraction between the negatively charged electron and the positively charged nucleus. Electrons don't leave atoms spontaneously.

Exam Marking Points

The definition must include: (1) "mole of electrons", (2) "gaseous atoms", (3) "gaseous ions". Missing any loses marks. State symbols (g) are essential in the equation.

Factors Affecting Ionisation Energy

Four factors determine how strongly an electron is held by the nucleus, and therefore how much energy is needed to remove it:

FactorEffect on IEExplanation

Nuclear charge↑ protons → ↑ IEMore protons = stronger attraction to outer electrons

Atomic radius (distance)↑ radius → ↓ IEOuter electrons further from nucleus = weaker attraction (Coulomb's law: force ∝ 1/r²)

Electron shielding↑ shielding → ↓ IEInner shell electrons repel outer electrons, reducing the effective nuclear charge they experience

Spin-pair repulsionPaired e⁻ → ↓ IETwo electrons in the same orbital repel each other, making one easier to remove

Group Trends

As expected from their electron configurations, Group 1 metals have relatively low ionisation energies (one outer electron, easy to remove), whereas the noble gases have very high ionisation energies (full outer shells, very stable).

Understanding Shielding

Inner electrons "shield" outer electrons from the full positive charge of the nucleus. An outer electron doesn't experience the full nuclear charge (+Z) because inner electrons partially cancel it out. The effective nuclear charge felt by the outer electron is less than the actual nuclear charge. More inner shells = more shielding = weaker attraction to outer electrons.

How to Explain IE Trends

When explaining any IE trend in an exam, always identify which factor dominates and explain why. For example: "IE increases because nuclear charge increases while shielding stays the same, so outer electrons are more strongly attracted."

Worked Example04

Write equations, including state symbols, for the first, second, and third ionisation energies of magnesium.

1

First ionisation energy: Remove 1 electron from neutral atom

Mg(g) \to Mg^{+} (g) + e^{-}

2

Second ionisation energy: Remove 1 electron from 1+ ion

Mg^{+} (g) \to Mg^{2 +} (g) + e^{-}

3

Third ionisation energy: Remove 1 electron from 2+ ion

Mg^{2 +} (g) \to Mg^{3 +} (g) + e^{-}

Why does IE increase with each successive removal? Each time you remove an electron, the remaining electrons experience less repulsion AND the same nuclear charge is now shared among fewer electrons, so each is held more tightly.

Trend Across Period 3

Ionisation energies show periodicity - a repeating trend across a period of the Periodic Table. First ionisation energy generally increases from left to right across Period 3 (Na → Ar) and decreases down a group.

Why IE Increases Across a Period

Nuclear charge increases: Each element has one more proton than the previous one (Na has 11 protons, Mg has 12, Al has 13, etc.).

Shielding stays approximately constant: Electrons are being added to the same outer shell (shell 3). Inner shell shielding doesn't change significantly because no new inner shells are added.

Atomic radius decreases: The increased nuclear charge pulls all electrons closer to the nucleus, so the outer electrons are actually closer despite being in the same shell.

Result: Outer electrons are held more tightly → more energy needed to remove them → IE increases.

Period 3 First Ionisation Energies (kJ mol⁻¹)

Na: 496 → Mg: 738 → Al: 578 → Si: 789 → P: 1012 → S: 1000 → Cl: 1251 → Ar: 1521

Notice the Anomalies

The general trend increases, but Al < Mg and S < P. These "dips" are explained by electron configuration and are commonly asked in exams.

NaMgAlSiPSClAr

4967385787891012100012511521

Period 3 Anomalies

There are two decreases in the general trend that need explaining:

Anomaly 1: Al (578) lower than Mg (738)

Anomaly 2: S (1000) lower than P (1012)

Memory Aid

Al: 3p is higher energy than 3s | S: Paired electrons experience repulsion

Worked Example05

Explain why the first ionisation energy of aluminium (578 kJ mol⁻¹) is lower than that of magnesium (738 kJ mol⁻¹).

1

State the electron configurations:

Mg: 1s² 2s² 2p⁶ 3s²

Al: 1s² 2s² 2p⁶ 3s² 3p¹

2

Identify the key difference: The outer electron in Al is in a 3p sub-shell, while in Mg it is in a 3s sub-shell.

3

Explain the energy difference: The 3p sub-shell is higher in energy than the 3s sub-shell, meaning the electron is further from the nucleus and more shielded by the 3s electrons.

4

Conclude: Therefore, less energy is required to remove the outer electron from aluminium than from magnesium.

Trend Down Group 2

First ionisation energy decreases down Group 2 (Be → Ba). This is the opposite of the trend across a period, and understanding why is essential for exam success.

Why IE Decreases Down a Group

1. Atomic radius increases significantly: Each element has an additional electron shell compared to the one above. The outer electron is progressively further from the nucleus. According to Coulomb's law, the force of attraction decreases with the square of the distance (F ∝ 1/r²).

2. Shielding increases: Each element has more inner shells of electrons. These inner electrons repel the outer electrons and shield them from the full positive charge of the nucleus. The outer electron experiences a lower effective nuclear charge.

3. Nuclear charge also increases: Yes, there are more protons down the group, BUT this effect is outweighed by the increased distance and shielding. The extra protons are "cancelled out" by the extra shielding electrons.

ElementOuter Shell1st IE / kJ mol⁻¹

Beryllium (Be)n = 2900

Magnesium (Mg)n = 3738

Calcium (Ca)n = 4590

Strontium (Sr)n = 5550

Barium (Ba)n = 6503

Successive Ionisation Energies

Definition

Successive ionisation energies are the energies required to remove electrons one at a time from an atom. The 1st IE removes one electron from the neutral atom, the 2nd IE removes one electron from the 1+ ion, the 3rd IE from the 2+ ion, and so on.

Why Each Successive IE is Higher

1. Removing from a more positive ion: After removing the first electron, you have a 1+ ion. The remaining electrons now experience the same nuclear charge but with less electron-electron repulsion. Each electron is held more tightly.

2. Greater effective nuclear charge: With fewer electrons, each remaining electron experiences a greater share of the positive nuclear charge. Think of it as the same positive charge being "spread over" fewer electrons.

3. Ionic contraction: The ion becomes slightly smaller as electrons are removed because the remaining electrons are pulled closer to the nucleus. This shorter distance means stronger attraction.

Key Pattern to Spot

A very large jump between successive IEs indicates that the next electron is being removed from a new inner shell. This provides direct evidence for electron shell structure.

Evidence for Shell Structure

Successive ionisation energies provide direct experimental evidence for the existence of electron shells. By analysing where the big jumps occur, we can determine how many electrons are in each shell.

Example: Sodium (Na, Z = 11)

Successive IE values (kJ mol⁻¹): 496, 4563, 6913, 9544, 13352, 16611, 20115, 25491, 28934, 141367, 159079

Analysis: There is a massive jump (×9) between the 1st and 2nd IE. This tells us the 1st electron is easy to remove (outer shell) but the 2nd electron requires much more energy (inner shell). Another big jump occurs after the 9th electron, showing the 10th electron is from an even closer inner shell.

Conclusion: Sodium has 1 electron in the outer shell, 8 in the second shell, and 2 in the innermost shell. Configuration: 2, 8, 1.

Exam Strategy

To identify an element's group from successive IE data: find the biggest jump. The number of electrons removed before the big jump equals the number of outer shell electrons, which equals the group number.

06Worked Example

The successive ionisation energies (in kJ mol⁻¹) of an element X are: 578, 1817, 2745, 11577, 14842...

Identify the group of element X and explain your reasoning.

1

Look for the largest jump between consecutive IEs:

1st → 2nd: 1817 - 578 = 1239 kJ mol⁻¹

2nd → 3rd: 2745 - 1817 = 928 kJ mol⁻¹

3rd → 4th: 11577 - 2745 = 8832 kJ mol⁻¹ ← Largest jump!

4th → 5th: 14842 - 11577 = 3265 kJ mol⁻¹

2

The largest jump occurs between the 3rd and 4th ionisation energies.

3

This means the 4th electron is removed from an inner shell, while the first 3 electrons were in the outer shell.

Answer: Element X is in Group 3 (has 3 outer shell electrons)

Worked Example07

Explain why the second ionisation energy of sodium is much greater than the second ionisation energy of magnesium.

1

Write the electron configurations of the 1+ ions:

Na⁺: 1s² 2s² 2p⁶ (2nd electron removed from 2p)

Mg⁺: 1s² 2s² 2p⁶ 3s¹ (2nd electron removed from 3s)

2

For Na⁺, the 2nd electron is removed from the 2nd shell (2p orbital).

For Mg⁺, the 2nd electron is removed from the 3rd shell (3s orbital).

3

The 2nd shell electron in Na⁺ is:

Closer to the nucleus

Less shielded (only 1s² shielding it)

More strongly attracted to the nucleus

AnswerMuch more energy is needed to remove an electron from an inner shell (Na⁺) than from an outer shell (Mg⁺).

Practice

Question 1: Write the full electron configuration for:

(a) Vanadium (V, Z = 23)

(b) V³⁺

(c) Se²⁻ (Z = 34)

a)

V (Z = 23): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³

b)

V³⁺: Remove 3 electrons (2 from 4s, then 1 from 3d)

1s² 2s² 2p⁶ 3s² 3p⁶ 3d²

c)

Se²⁻: Se (Z = 34) + 2 electrons

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ (same as Kr)

Practice

Question 2: Explain why the first ionisation energy of sulfur (1000 kJ mol⁻¹) is lower than that of phosphorus (1012 kJ mol⁻¹), despite sulfur having a higher nuclear charge.

1

Phosphorus has configuration: [Ne] 3s² 3p³ — three unpaired 3p electrons

2

Sulfur has configuration: [Ne] 3s² 3p⁴ — one paired 3p orbital

3

The paired electrons in sulfur's 3p orbital experience electron-electron repulsion, which destabilises the electron and makes it easier to remove.

4

This repulsion effect outweighs the small increase in nuclear charge from P to S.

Practice

Question 3: The first seven successive ionisation energies (kJ mol⁻¹) of element Y are:

786, 1577, 3232, 4356, 16091, 19785, 23786

(a) Identify the group of element Y.

(b) Write an equation for the 4th ionisation energy of Y.

a)

Find largest jump: 4th → 5th = 16091 - 4356 = 11735 kJ mol⁻¹

This is the biggest jump, so 4 electrons are in the outer shell.

Element Y is in Group 4

b)

4th ionisation energy removes 4th electron:

Y^{3 +} (g) \to Y^{4 +} (g) + e^{-}

Summary

Key Concepts

ConceptKey Points

Electron ConfigurationFill 1s→2s→2p→3s→3p→4s→3d→4p; ions lose from highest shell first

1st Ionisation EnergyX(g) → X⁺(g) + e⁻; energy to remove 1 mol e⁻ from 1 mol gaseous atoms

Period 3 TrendGenerally increases; drops at Al (3p vs 3s) and S (paired e⁻ repulsion)

Group 2 TrendDecreases down group; increased radius and shielding

Successive IEsLarge jumps indicate electron removed from inner shell

Exam Essentials

Always include state symbols (g) in ionisation energy equations

For Period 3 anomalies: Al = 3p higher energy; S = paired electron repulsion

Position of largest jump in successive IEs = group number