AQA A-Level Chemistry 7405

3.1.1 Atomic Structure

1. Specification Coverage

SPEC: 3.1.1 — Atomic structure

AQA requires students to understand how atomic structure, especially electron arrangement, determines chemical behaviour and links to the organisation of the Periodic Table.

This topic contains three official specification subsections:

• 3.1.1.1 Fundamental particles
• 3.1.1.2 Mass number and isotopes
• 3.1.1.3 Electron configuration

SPEC: 3.1.1.1 — Fundamental particles

You must know and apply:

• that understanding of atomic structure has evolved over time;
• the existence and properties of:
  • protons;
  • neutrons;
  • electrons;
• the relative charge of each fundamental particle;
• the relative mass of each fundamental particle;
• that an atom consists of:
  • a nucleus containing protons and neutrons;
  • electrons surrounding the nucleus.

SPEC: 3.1.1.2 — Mass number and isotopes

You must know and apply:

• mass number, (A);
• atomic number / proton number, (Z);
• determination of numbers of protons, neutrons and electrons in:
  • atoms;
  • positive ions;
  • negative ions;
• the existence and meaning of isotopes;
• the principles of a simple time-of-flight mass spectrometer, limited to:
  • ionisation;
  • acceleration so all ions have the same kinetic energy;
  • ion drift;
  • ion detection;
  • data analysis;
• that mass spectrometry provides:
  • accurate relative isotopic masses;
  • relative isotopic abundances;
  • identification of elements;
  • relative molecular mass;
• interpretation of simple mass spectra of elements;
• calculation of relative atomic mass from isotopic abundances;
• calculations are limited here to mononuclear ions.

SPEC: 3.1.1.3 — Electron configuration

You must know and apply:

• electron configurations of atoms and ions up to: [ Z=36 ]
• shells;
• subshells:
  • (s);
  • (p);
  • (d);
• orbitals;
• first ionisation energy;
• successive ionisation energies;
• equations for:
  • first ionisation energy;
  • successive ionisation energies;
• how first ionisation energies across Period 3, Na–Ar provide evidence for electron configuration in subshells;
• how first ionisation energies down Group 2, Be–Ba relate to electron structure;
• how successive ionisation-energy patterns provide evidence for electrons arranged in different shells.

2. Core Content — Extreme Detail

2.1 SPEC: 3.1.1.1 — Fundamental Particles

2.1.1 The structure of an atom

An atom contains a very small, dense central nucleus surrounded by electrons.

The nucleus contains:

• protons
• neutrons

Electrons occupy regions of space surrounding the nucleus.

DEFINITION — Proton
A positively charged subatomic particle found in the nucleus with relative charge (+1) and relative mass approximately (1).

DEFINITION — Neutron
An electrically neutral subatomic particle found in the nucleus with relative charge (0) and relative mass approximately (1).

DEFINITION — Electron
A negatively charged subatomic particle found outside the nucleus with relative charge (-1) and relative mass approximately (1/1836).

Required particle data

For most routine A-level calculations, the electron's mass is treated as negligible compared with the mass of a proton or neutron.

2.1.2 Nuclear charge

The total charge of the nucleus depends only on the number of protons because neutrons are uncharged.

If an atom contains (Z) protons, nuclear charge is:

[ +Ze ]

where (e) is the elementary charge.

The number of protons determines the identity of the element.

For example:

• 6 protons → carbon;
• 8 protons → oxygen;
• 11 protons → sodium.

Changing the number of neutrons does not change the element.

Changing the number of electrons produces an ion but does not change the element.

2.1.3 Neutral atoms

A neutral atom has equal numbers of protons and electrons.

For example, a neutral magnesium atom has:

[ 12\text{ protons} ]

and:

[ 12\text{ electrons} ]

because the positive and negative charges balance.

2.1.4 Development of atomic structure

AQA requires students to appreciate that scientific understanding of atomic structure has evolved over time.

You do not need to memorise a long historical essay, but you should understand the principle:

Scientific models change when new experimental evidence cannot be explained adequately by an existing model.

A concise development sequence is:

1. early atomic ideas treated atoms as indivisible particles;
2. discovery of the electron showed atoms contained smaller components;
3. scattering evidence showed positive charge and most mass were concentrated in a tiny nucleus;
4. later work established protons and neutrons in the nucleus;
5. quantum theory led to modern electron-shell, subshell and orbital models.

Exam language

“The atomic model was modified because new experimental evidence was inconsistent with the previous model.”

Avoid:

“Scientists changed their minds.”

The marks are gained for evidence leading to model refinement.

2.2 SPEC: 3.1.1.2 — Mass Number and Isotopes

2.2.1 Atomic number / proton number

DEFINITION — Atomic number, (Z)
The number of protons in the nucleus of an atom.

Atomic number uniquely identifies the element.

For a neutral atom:

[ \text{number of electrons}=Z ]

2.2.2 Mass number

DEFINITION — Mass number, (A)
The total number of protons and neutrons in the nucleus.

Therefore:

[ \boxed{ A=\text{protons}+\text{neutrons} } ]

So:

[ \boxed{ \text{neutrons}=A-Z } ]

2.2.3 Nuclide notation

An isotope may be written:

[ {}^{A}_{Z}X ]

where:

• (X) = element symbol;
• (A) = mass number;
• (Z) = atomic number.

Example:

[ {}^{23}_{11}\text{Na} ]

contains:

• 11 protons;
• (23-11=12) neutrons;
• 11 electrons if neutral.

2.2.4 Ions

When an atom forms an ion, only its number of electrons changes.

Positive ion

A positive ion has lost electrons.

Example:

[ \text{Mg}^{2+} ]

Magnesium has atomic number 12.

Therefore:

• protons = 12;
• electrons = (12-2=10).

Negative ion

A negative ion has gained electrons.

Example:

[ \text{Cl}^{-} ]

Chlorine has atomic number 17.

Therefore:

• protons = 17;
• electrons = (17+1=18).

Universal rule

For an ion:

[ \text{charge}

\text{number of protons}

\text{number of electrons} ]

when charge is expressed in units of the elementary charge.

2.2.5 Worked particle-count example

For:

[ {}^{56}_{26}\text{Fe}^{3+} ]

Protons:

[ 26 ]

Neutrons:

[ 56-26=30 ]

Electrons:

[ 26-3=23 ]

So:

[ \boxed{ 26\text{ p},\ 30\text{ n},\ 23\text{ e}^- } ]

2.3 Isotopes

DEFINITION — Isotopes
Atoms of the same element with the same number of protons but different numbers of neutrons.

Because isotopes belong to the same element, they have the same:

[ Z ]

but different:

[ A ]

Example: chlorine

[ {}^{35}_{17}\text{Cl} ]

and:

[ {}^{37}_{17}\text{Cl} ]

Both contain:

[ 17\text{ protons} ]

but:

[ {}^{35}\text{Cl}:18\text{ neutrons} ]

[ {}^{37}\text{Cl}:20\text{ neutrons} ]

2.3.1 Why isotopes have similar chemical properties

Chemical behaviour mainly depends on:

• electron configuration;
• especially outer-shell electrons.

Neutral isotopes of the same element contain the same number of electrons.

Therefore, they have essentially the same electron configuration and very similar chemical properties.

High-mark wording

“Isotopes of an element have the same number and arrangement of electrons, so they undergo the same types of chemical reaction.”

Avoid saying simply:

“They are the same element.”

That does not explain why their chemistry is similar.

2.3.2 Why isotopes can have different physical properties

Isotopes have different numbers of neutrons, so their masses differ.

Therefore mass-dependent physical properties can differ.

The chemical properties remain similar because electron arrangement remains the same.

2.4 Mass Spectrometry

AQA requires the principles of a simple time-of-flight mass spectrometer.

The examinable sequence is:

[ \boxed{ \text{ionisation} \rightarrow \text{acceleration} \rightarrow \text{ion drift} \rightarrow \text{detection} \rightarrow \text{data analysis} } ]

Students frequently lose marks by naming stages without explaining what occurs during each stage.

2.5 Stage 1 — Ionisation

DEFINITION — Ionisation
The formation of charged particles from atoms or molecules.

In a mass spectrometer, the sample must be converted into ions so that electric fields can accelerate and detect the particles.

For a simple atomic sample:

[ X(g)\rightarrow X^{+}(g)+e^- ]

The ions commonly considered in AQA calculations are singly positively charged:

[ X^+ ]

Why positive ions are needed

Neutral atoms do not experience the required electrostatic acceleration in an electric field.

2.6 Stage 2 — Acceleration

Positive ions are accelerated by an electric field.

AQA specifies that ions are accelerated so that they have the same kinetic energy.

For an ion of charge (q) accelerated through potential difference (V):

[ \boxed{ E_k=qV } ]

and:

[ E_k=\frac12mv^2 ]

Therefore:

[ qV=\frac12mv^2 ]

Rearranging:

[ v=\sqrt{\frac{2qV}{m}} ]

For ions with the same charge and accelerated through the same potential difference:

[ v\propto\frac{1}{\sqrt{m}} ]

Therefore lighter ions travel faster.

2.6.1 A* derivation

From:

[ \frac12mv^2=qV ]

multiply by 2:

[ mv^2=2qV ]

so:

[ v^2=\frac{2qV}{m} ]

and:

[ \boxed{ v=\sqrt{\frac{2qV}{m}} } ]

If (q) and (V) are fixed:

[ v\propto m^{-1/2} ]

Thus a lower (m/z) ion reaches the detector sooner.

2.7 Stage 3 — Ion Drift

After acceleration, ions enter a drift region.

This region is essentially field-free, so ions move at approximately constant velocity.

The flight time is:

[ t=\frac{d}{v} ]

where:

• (d) = drift length;
• (v) = ion speed.

Substitute:

[ v=\sqrt{\frac{2qV}{m}} ]

to give:

[ t

d\sqrt{\frac{m}{2qV}} ]

Therefore:

[ \boxed{ t\propto\sqrt{\frac{m}{q}} } ]

or in mass-spectrometry language:

[ \boxed{ t\propto\sqrt{\frac{m}{z}} } ]

for equivalent charge notation.

Key implication

For singly charged ions:

• lower mass → higher speed → shorter time of flight;
• higher mass → lower speed → longer time of flight.

2.8 Stage 4 — Detection

When ions reach the detector, they generate an electrical signal.

The detector response allows the instrument to determine:

• arrival time;
• relative number of ions arriving.

The time of arrival is used to determine:

[ m/z ]

The signal size is related to abundance.

Exam wording

“The detector produces a signal when ions arrive; the arrival time is used to determine (m/z), while signal intensity is related to the number/abundance of ions.”

2.9 Stage 5 — Data Analysis

The instrument produces a mass spectrum.

A simple mass spectrum plots:

• horizontal axis: [ m/z ]
• vertical axis: relative abundance / relative intensity.

DEFINITION — Mass spectrum
A plot showing the relative abundance or intensity of ions as a function of their mass-to-charge ratio, (m/z).

For singly charged mononuclear ions:

[ z=1 ]

so:

[ m/z ]

numerically corresponds to the ion's relative isotopic mass.

2.10 Interpreting Elemental Mass Spectra

Suppose an element gives peaks at:

[ m/z=35 ]

and:

[ m/z=37 ]

This indicates two isotopes with relative isotopic masses approximately 35 and 37, assuming singly charged mononuclear ions.

The relative heights/areas of the peaks indicate their relative abundances.

Example

If peak intensity ratio is:

[ 35:37 = 3:1 ]

then approximately:

[ 75% ]

of atoms are isotope 35 and:

[ 25% ]

are isotope 37.

2.11 Relative Isotopic Mass

DEFINITION — Relative isotopic mass
The mass of an atom of an isotope compared with one-twelfth of the mass of an atom of carbon-12.

Relative isotopic mass has no unit because it is a ratio.

It is not always an exact whole number.

For high-precision mass spectrometry, the measured isotopic mass reflects effects such as nuclear binding energy and exact particle masses.

2.12 Relative Atomic Mass

DEFINITION — Relative atomic mass, (A_r)
The weighted mean mass of an atom of an element compared with one-twelfth of the mass of an atom of carbon-12.

Again:

[ A_r ]

has no unit.

The word weighted is essential because naturally occurring isotopes are not necessarily equally abundant.

2.13 Calculating Relative Atomic Mass

General formula:

[ \boxed{ A_r

\frac{\sum(\text{isotopic mass}\times\text{isotopic abundance})} {\sum(\text{isotopic abundance})} } ]

If abundances are percentages:

[ \boxed{ A_r

\frac{\sum(\text{isotopic mass}\times%\text{ abundance})}{100} } ]

2.13.1 Example

An element contains:

• isotope (24): 79.0%;
• isotope (25): 10.0%;
• isotope (26): 11.0%.

Then:

[ A_r

\frac{(24\times79.0)+(25\times10.0)+(26\times11.0)}{100} ]

[

\frac{1896+250+286}{100} ]

[

24.32 ]

Therefore:

[ \boxed{A_r=24.32} ]

No unit.

2.13.2 Using non-percentage intensity data

Suppose intensities are:

• mass 10: intensity 20;
• mass 11: intensity 80.

Then:

[ A_r

\frac{(10\times20)+(11\times80)}{20+80} ]

[

\frac{1080}{100} ]

[ \boxed{ A_r=10.8 } ]

The denominator is the total abundance, not automatically 100 unless the values are percentages.

2.14 Identifying an Element from a Mass Spectrum

A mass spectrum can identify an element using:

• isotopic (m/z) values;
• isotopic abundance pattern;
• calculated (A_r).

Example:

Peaks near:

[ 35,\ 37 ]

with approximate abundance ratio:

[ 3:1 ]

strongly indicate chlorine.

The calculated relative atomic mass is approximately:

[ 35.5 ]

which agrees with the Periodic Table value.

2.15 Determining Relative Molecular Mass

Mass spectrometry can also determine relative molecular mass, (M_r).

For a molecular species, the molecular ion peak may correspond to the intact molecule with a single positive charge.

For:

[ M^+ ]

with charge (+1):

[ m/z=M_r ]

when considering nominal integer masses.

This becomes much more important later in Organic Chemistry mass spectrometry.

2.16 SPEC: 3.1.1.3 — Electron Configuration

Electron arrangement is central to A-level Chemistry because it explains:

• periodicity;
• ion formation;
• bonding;
• ionisation energies;
• transition-metal chemistry.

2.17 Shells

Electrons occupy principal energy levels called shells.

DEFINITION — Electron shell
A principal energy level around the nucleus occupied by electrons.

Shells are identified by principal quantum number:

[ n=1,2,3,4,\ldots ]

The shell with lower (n) is generally closer to the nucleus and lower in energy.

2.18 Subshells

Each shell contains one or more subshells.

At AQA A-level, the relevant subshell types are:

[ s,\ p,\ d ]

The subshells required for configurations up to (Z=36) include:

• (1s)
• (2s)
• (2p)
• (3s)
• (3p)
• (4s)
• (3d)
• (4p)

2.19 Orbitals

DEFINITION — Atomic orbital
A region of space around the nucleus that can hold up to two electrons with opposite spins.

This precise distinction matters:

• a subshell contains one or more orbitals;
• an orbital holds a maximum of two electrons.

2.19.1 s subshell

An (s) subshell contains:

[ 1\text{ orbital} ]

Maximum electrons:

[ \boxed{2} ]

2.19.2 p subshell

A (p) subshell contains:

[ 3\text{ orbitals} ]

Maximum electrons:

[ \boxed{6} ]

2.19.3 d subshell

A (d) subshell contains:

[ 5\text{ orbitals} ]

Maximum electrons:

[ \boxed{10} ]

2.20 Filling Orbitals

Electrons occupy available orbitals starting with lower-energy orbitals.

Relevant filling order:

[ \boxed{ 1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d \rightarrow 4p } ]

For elements up to (Z=36), this order is sufficient.

2.20.1 Maximum occupancies

2.20.2 Electron spin and pairing

Each orbital holds a maximum of two electrons.

The two electrons in the same orbital must have opposite spins.

Within a set of equal-energy orbitals such as the three (p) orbitals, electrons occupy orbitals singly before pairing.

This helps explain details of electronic structure and later chemical behaviour.

2.21 Writing Electron Configurations

Hydrogen, (Z=1)

[ \boxed{1s^1} ]

Helium, (Z=2)

[ \boxed{1s^2} ]

Carbon, (Z=6)

[ \boxed{1s^2,2s^2,2p^2} ]

Oxygen, (Z=8)

[ \boxed{1s^2,2s^2,2p^4} ]

Sodium, (Z=11)

[ \boxed{1s^2,2s^2,2p^6,3s^1} ]

Chlorine, (Z=17)

[ \boxed{1s^2,2s^2,2p^6,3s^2,3p^5} ]

Potassium, (Z=19)

[ \boxed{ 1s^2,2s^2,2p^6,3s^2,3p^6,4s^1 } ]

Calcium, (Z=20)

[ \boxed{ 1s^2,2s^2,2p^6,3s^2,3p^6,4s^2 } ]

2.22 4s and 3d

For neutral atoms, (4s) fills before (3d).

Example:

Sc:

[ [\text{Ar}],4s^2,3d^1 ]

Fe:

[ [\text{Ar}],4s^2,3d^6 ]

Zn:

[ [\text{Ar}],4s^2,3d^{10} ]

Crucial ion rule

When transition-metal atoms form positive ions, electrons are removed from the 4s subshell before the 3d subshell.

Example:

[ \text{Fe}

[\text{Ar}],4s^2,3d^6 ]

[ \text{Fe}^{2+}

[\text{Ar}],3d^6 ]

[ \text{Fe}^{3+}

[\text{Ar}],3d^5 ]

This is a classic lost-mark area.

2.23 Chromium and Copper Exceptions

For completeness in configurations up to (Z=36), chromium and copper are commonly represented as:

[ \boxed{ \text{Cr}=[\text{Ar}],3d^5,4s^1 } ]

rather than the simple predicted:

[ 3d^4,4s^2 ]

and:

[ \boxed{ \text{Cu}=[\text{Ar}],3d^{10},4s^1 } ]

rather than:

[ 3d^9,4s^2 ]

These configurations are associated with the stability of half-filled and fully filled (d) subshell arrangements.

For ions, remove (4s) electrons first.

Example:

[ \text{Cu}^{2+}

[\text{Ar}],3d^9 ]

2.24 Ions and Electron Configuration

Sodium ion

Neutral Na:

[ 1s^2,2s^2,2p^6,3s^1 ]

Na loses one electron:

[ \boxed{ \text{Na}^+

1s^2,2s^2,2p^6 } ]

Oxide ion

Neutral O:

[ 1s^2,2s^2,2p^4 ]

O gains two electrons:

[ \boxed{ \text{O}^{2-}

1s^2,2s^2,2p^6 } ]

Calcium ion

Ca:

[ [\text{Ar}],4s^2 ]

Therefore:

[ \boxed{ \text{Ca}^{2+}

[\text{Ar}] } ]

2.25 First Ionisation Energy

DEFINITION — First ionisation energy
The energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous (1+) ions.

The three details that must appear are:

1. one electron;
2. one mole of gaseous atoms;
3. formation of one mole of gaseous (1+) ions.

Equation:

[ \boxed{ X(g)\rightarrow X^+(g)+e^- } ]

For example:

[ \text{Mg}(g)\rightarrow\text{Mg}^+(g)+e^- ]

2.26 Successive Ionisation Energies

The second ionisation energy removes one electron from each ion in one mole of gaseous (1+) ions:

[ \boxed{ X^+(g)\rightarrow X^{2+}(g)+e^- } ]

The third:

[ \boxed{ X^{2+}(g)\rightarrow X^{3+}(g)+e^- } ]

In general, the (n)th ionisation process removes an electron from the appropriate gaseous ion.

Common mistake

Do not write:

[ X(g)\rightarrow X^{2+}(g)+2e^- ]

as the equation for second ionisation energy.

The second ionisation energy is a separate one-electron removal from (X^+).

2.27 Factors Affecting Ionisation Energy

Three major ideas determine ionisation energy:

1. nuclear charge;
2. distance of the electron from the nucleus;
3. shielding by inner-shell electrons.

A fourth useful factor is the particular subshell/orbital environment.

2.27.1 Nuclear charge

More protons means greater nuclear attraction for an electron, all else equal.

Therefore ionisation energy tends to increase with nuclear charge if shielding and distance do not offset the effect.

2.27.2 Distance from nucleus

An electron further from the nucleus experiences weaker electrostatic attraction.

Therefore it is easier to remove.

2.27.3 Shielding

Inner-shell electrons repel outer-shell electrons and reduce the effective attraction between the nucleus and outer electrons.

DEFINITION — Shielding
Reduction in the electrostatic attraction between the nucleus and an outer electron due to repulsion by electrons in inner shells.

More shielding generally lowers ionisation energy.

2.28 First Ionisation Energies Across Period 3

Period 3:

[ \text{Na, Mg, Al, Si, P, S, Cl, Ar} ]

General trend:

[ \boxed{ \text{first ionisation energy generally increases across Period 3} } ]

Reason:

• proton number increases;
• nuclear charge increases;
• electrons are added to the same principal shell;
• shielding changes relatively little;
• atomic radius generally decreases;
• attraction between nucleus and outer electron increases.

High-mark sentence

“Across Period 3, nuclear charge increases while the electron removed remains in the same principal shell and shielding changes only slightly, so nuclear attraction increases and more energy is required to remove the electron.”

2.29 Period 3 Exception: Mg to Al

Mg:

[ [\text{Ne}],3s^2 ]

Al:

[ [\text{Ne}],3s^2,3p^1 ]

The first ionisation energy of Al is lower than Mg.

Why?

The electron removed from Al is in a:

[ 3p ]

subshell, which is higher in energy than the:

[ 3s ]

electron removed from Mg.

The (3p) electron is also more shielded / less strongly attracted to the nucleus.

Therefore less energy is required to remove it.

Mark-scheme-quality wording

“The electron removed from Al is in the higher-energy 3p subshell, whereas the electron removed from Mg is in 3s. The 3p electron is less strongly attracted to the nucleus, so Al has the lower first ionisation energy.”

Avoid saying:

“Al has more shielding because it has more electrons.”

That alone is too crude because both outer electrons are in the third shell and the key distinction is the (3p) vs (3s) subshell.

2.30 Period 3 Exception: P to S

P:

[ [\text{Ne}],3s^2,3p^3 ]

S:

[ [\text{Ne}],3s^2,3p^4 ]

Phosphorus has three (3p) electrons occupying separate orbitals.

Sulfur has four (3p) electrons, so one (3p) orbital contains a pair.

Paired electrons repel each other.

Therefore it is easier to remove one of the paired electrons from sulfur.

So:

[ \boxed{ IE_1(\text{S})<IE_1(\text{P}) } ]

despite sulfur's greater nuclear charge.

High-mark wording

“In sulfur, one 3p orbital contains a pair of electrons. Repulsion between paired electrons makes one easier to remove than an unpaired 3p electron from phosphorus.”

2.31 First Ionisation Energy Down Group 2

Group 2:

[ \text{Be, Mg, Ca, Sr, Ba} ]

General trend:

[ \boxed{ \text{first ionisation energy decreases down Group 2} } ]

Down the group:

• nuclear charge increases;
• number of occupied shells increases;
• outer electron is further from nucleus;
• shielding increases substantially;
• increased distance and shielding outweigh increased nuclear charge.

Therefore attraction for the outer electron decreases.

Examiner-safe wording

“Down Group 2, the outer electron is in a higher principal energy level and experiences greater shielding. These effects outweigh the increased nuclear charge, so the attraction to the nucleus decreases and first ionisation energy falls.”

2.32 Successive Ionisation Energies and Shell Structure

Successive ionisation energies increase because after each electron is removed:

• the ion becomes more positively charged;
• remaining electrons experience greater attraction per electron.

However, the most important evidence comes from large jumps.

A very large jump indicates that the next electron is being removed from:

• an inner shell;
• closer to the nucleus;
• with less shielding;
• much more strongly attracted.

2.32.1 Example: magnesium

Mg has configuration:

[ 1s^2,2s^2,2p^6,3s^2 ]

The first two ionisation energies remove the two (3s) electrons.

The third electron would come from the (n=2) shell.

Therefore:

[ IE_3 ]

is dramatically larger than:

[ IE_2 ]

This large jump shows there were two electrons in the outer shell.

2.32.2 Inferring group from successive ionisation energies

If the large jump occurs between:

[ IE_2\text{ and }IE_3 ]

then the atom has:

[ 2 ]

outer-shell electrons.

This is characteristic of Group 2.

If the jump occurs between:

[ IE_3\text{ and }IE_4 ]

there are three outer-shell electrons.

2.33 Ionisation Energy as Evidence for Subshells

Small irregularities in first ionisation energies across a period provide evidence that electrons occupy subshells of different energies.

The Mg → Al decrease indicates:

[ 3p ]

is higher in energy than:

[ 3s ]

The P → S decrease provides evidence for occupancy and pairing within individual (p) orbitals.

Thus experimental ionisation-energy data support the electron-configuration model rather than electron shells being completely uniform energy levels.

3. Exact AQA Exam Language

These are original examiner-safe structures aligned with AQA command-word expectations.

3.1 Define

Give the precise meaning.

First ionisation energy

“The energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous (1+) ions.”

Do not omit:

• gaseous;
• one mole;
• one electron.

3.2 State

Give a direct fact.

“A proton has relative charge (+1) and relative mass 1.”

3.3 Calculate

Use:

1. equation;
2. substitution;
3. processing;
4. appropriate significant figures;
5. unit if the quantity has one.

For (A_r), remember there is no unit.

3.4 Explain

Chemistry explanations should use a causal chain.

Ionisation energy:

“The outer electron is further from the nucleus and more shielded, so electrostatic attraction to the nucleus is weaker and less energy is required to remove it.”

3.5 Compare

Address both species.

“Both isotopes contain 17 protons, but chlorine-37 contains two more neutrons than chlorine-35.”

3.6 Deduce

Use data to reach a conclusion.

Example:

“The large jump between the second and third ionisation energies shows that two electrons occupy the outer shell, so the element is in Group 2.”

3.7 Explain evidence

Use:

observation → atomic interpretation → conclusion

Example:

“The decrease from Mg to Al occurs because the electron removed from Al is in the higher-energy 3p subshell. This provides evidence that the 3s and 3p subshells have different energies.”

4. Common Misconceptions & Lost Marks

4.1 Isotopes have different protons

❌ Incorrect.

Isotopes have the same number of protons but different numbers of neutrons.

4.2 Mass number = relative atomic mass

❌ Incorrect.

Mass number is the integer number of protons + neutrons in one nucleus.

Relative atomic mass is a weighted mean across isotopes.

4.3 Atomic number = protons + neutrons

❌ Incorrect.

Atomic number:

[ Z=\text{protons} ]

Mass number:

[ A=\text{protons}+\text{neutrons} ]

4.4 Positive ion has gained protons

❌ Incorrect.

Ordinary ion formation changes electrons, not protons.

A positive ion has lost electrons.

4.5 Electron mass = 0

AQA typically uses a very small relative mass:

[ \frac{1}{1836} ]

Do not describe an electron as literally massless.

4.6 TOF ions all have the same speed

❌ Incorrect.

They are accelerated to the same kinetic energy, not the same speed.

Lower-mass ions travel faster.

4.7 Heavier ions arrive first

❌ Incorrect.

For the same charge and kinetic energy:

[ v\propto\frac{1}{\sqrt{m}} ]

so lighter ions arrive first.

4.8 Peak height = mass

❌ Incorrect.

Horizontal position gives (m/z).

Peak intensity/height/area represents relative abundance or ion signal.

4.9 Dividing weighted isotope total by 100 every time

Only divide by 100 when abundances are percentages totalling 100.

If abundance data are arbitrary intensities, divide by the sum of intensities.

4.10 Giving units for (A_r)

❌ Incorrect.

Relative atomic mass is a ratio and has no unit.

4.11 Orbital = shell

❌ Incorrect.

A shell contains subshells.

A subshell contains orbitals.

An orbital holds a maximum of two electrons.

4.12 p subshell has two electrons

❌ Incorrect.

A p subshell contains three orbitals and can hold six electrons.

4.13 d subshell holds five electrons

❌ Incorrect.

It has five orbitals and can hold ten electrons.

4.14 Removing 3d before 4s in transition-metal ions

❌ Incorrect.

4s fills before 3d for neutral atoms, but 4s electrons are removed first when forming transition-metal positive ions.

4.15 Writing second ionisation energy from the neutral atom

❌:

[ X(g)\rightarrow X^{2+}(g)+2e^- ]

✅:

[ X^+(g)\rightarrow X^{2+}(g)+e^- ]

4.16 “Ionisation energy increases because there are more protons”

Incomplete.

You must also consider:

• electron distance;
• shielding;
• subshell;
• pairing where relevant.

4.17 Mg → Al anomaly explained only by shielding

The key point is that Al loses a higher-energy (3p) electron while Mg loses a (3s) electron.

4.18 P → S anomaly described as “S has more electrons”

Insufficient.

You need electron pairing:

“S has a pair of electrons in one 3p orbital, increasing electron–electron repulsion.”

4.19 Down Group 2: forgetting shielding and distance

Strong answer must explain that increased shielding and greater distance outweigh the increased nuclear charge.

4.20 Large successive-IE jump means a new subshell

Not necessarily.

A very large jump generally indicates moving to an inner shell.

Smaller irregularities can indicate subshell structure.

5. Worked Exam-Style Questions

Question 1 — Fundamental Particles [4 marks]

For the ion:

[ {}^{63}_{29}\text{Cu}^{2+} ]

determine the number of protons, neutrons and electrons.

Model answer

Protons:

[ \boxed{29} ]

Neutrons:

[ 63-29=34 ]

[ \boxed{34} ]

Electrons:

[ 29-2=27 ]

[ \boxed{27} ]

Mark allocation:

• B1 29 protons;
• M1 neutron calculation;
• A1 34 neutrons;
• B1 27 electrons.

Question 2 — Relative Atomic Mass [5 marks]

A sample of an element contains:

Calculate the relative atomic mass.

Model answer

[ A_r= \frac{ (23.985\times78.99) + (24.986\times10.00) + (25.983\times11.01) }{100} ]

[ A_r

24.305... ]

To appropriate precision:

[ \boxed{ A_r=24.31 } ]

Mark allocation:

• M1 weighted-mean method;
• M1 first isotope contribution;
• M1 remaining isotope contributions;
• A1 correct numerical total;
• A1 suitable final precision.

Question 3 — Time-of-Flight Mass Spectrometry [6 marks]

Two singly charged positive ions have different masses but are accelerated through the same potential difference.

Explain why the lighter ion reaches the detector first.

Full-mark model answer

“The ions have the same charge and are accelerated through the same potential difference, so they gain the same kinetic energy, (qV). Since (\frac12mv^2=qV), the ion with the lower mass must have the greater velocity. In the field-free drift region the ions travel a fixed distance, and (t=d/v). Therefore the lighter ion, having the larger velocity, has the shorter time of flight and reaches the detector first.”

Indicative marking points:

• same charge;
• same accelerating potential difference;
• same kinetic energy;
• (\frac12mv^2=qV);
• lower mass gives greater speed;
• shorter time for fixed drift distance.

Question 4 — Electron Configuration [5 marks]

Write the electron configurations of:

(a) Fe [1 mark]

[ \boxed{ [\text{Ar}],3d^6,4s^2 } ]

(b) Fe(^{2+}) [2 marks]

Remove 4s electrons first:

[ \boxed{ [\text{Ar}],3d^6 } ]

(c) Fe(^{3+}) [2 marks]

Remove one further electron from 3d:

[ \boxed{ [\text{Ar}],3d^5 } ]

Question 5 — Period 3 Ionisation Energy [6 marks]

Explain why the first ionisation energy generally increases from Na to Ar but decreases from Mg to Al.

Full-mark model answer

“Across Period 3 the number of protons increases, so nuclear charge increases. Electrons are added to the same principal shell and shielding changes only slightly. The atomic radius generally decreases, so attraction between the nucleus and the outer electron increases. Therefore more energy is generally required to remove the outer electron. The decrease from Mg to Al occurs because Mg loses a 3s electron whereas Al loses a 3p electron. The 3p subshell is higher in energy and the electron is less strongly attracted to the nucleus, so less energy is required to remove it from Al.”

Indicative marking points:

• increasing proton number/nuclear charge;
• same principal shell;
• similar shielding;
• stronger nuclear attraction;
• Al electron is 3p vs Mg 3s;
• 3p higher energy / easier to remove.

Question 6 — Successive Ionisation Energies [6 marks]

An element has the following successive ionisation energies:

Deduce the group of the element and explain your reasoning.

Model answer

There is a very large increase between:

[ IE_2 ]

and:

[ IE_3 ]

This shows that after two electrons have been removed, the next electron is in an inner shell.

Therefore the atom has:

[ 2 ]

electrons in its outer shell.

So the element is in:

[ \boxed{\text{Group 2}} ]

Indicative marking points:

• identifies major jump between second and third;
• explains third electron is from inner shell;
• inner-shell electron is closer / less shielded / more strongly attracted;
• therefore two outer-shell electrons;
• Group 2 conclusion.

6. Links to Other Topics

6.1 3.1.2 Amount of Substance

Atomic structure links directly to:

• (A_r);
• (M_r);
• isotope calculations;
• mole calculations.

Mass-spectrometry data are often used before stoichiometric calculations.

6.2 3.1.3 Bonding

Electron configurations explain:

• valence electrons;
• ion formation;
• covalent bonding;
• ionic bonding;
• dative bonding.

6.3 3.2.1 Periodicity

First ionisation-energy trends are central to explaining periodicity.

Period 3 anomalies rely on:

• subshell energy;
• shielding;
• electron pairing.

6.4 3.2.2 Group 2

The trend in first ionisation energy down Group 2 helps explain changing reactivity.

As first ionisation energy decreases, electron loss generally becomes easier.

6.5 3.2.5 Transition Metals

Electron configurations involving:

[ 4s ]

and:

[ 3d ]

become essential.

The rule:

[ \boxed{\text{remove 4s electrons before 3d electrons}} ]

is repeatedly examined.

6.6 3.3 Organic Chemistry — Mass Spectrometry

Mass spectrometry later allows students to:

• determine molecular mass;
• identify molecular ions;
• use precise masses;
• determine molecular formulae.

The TOF principles introduced here remain foundational.

7. Summary Notes

7.1 Fundamental particles

Atom:

• nucleus contains protons + neutrons;
• electrons surround nucleus.

7.2 Atomic and mass number

Atomic number:

[ \boxed{ Z=\text{number of protons} } ]

Mass number:

[ \boxed{ A=\text{protons}+\text{neutrons} } ]

Neutrons:

[ \boxed{ A-Z } ]

Neutral atom:

[ e^- = p^+ ]

Positive ion:

[ \text{electrons lost} ]

Negative ion:

[ \text{electrons gained} ]

7.3 Isotopes

Same number of protons, different numbers of neutrons.

Therefore:

• same element;
• same atomic number;
• different mass number;
• very similar chemical behaviour because electron configuration is the same.

7.4 TOF mass spectrometer

Sequence:

[ \boxed{ \text{ionisation} \rightarrow \text{acceleration} \rightarrow \text{drift} \rightarrow \text{detection} \rightarrow \text{analysis} } ]

Ionisation:

[ X(g)\rightarrow X^+(g)+e^- ]

Acceleration:

[ \boxed{ qV=\frac12mv^2 } ]

Same (q) and (V):

[ v\propto\frac{1}{\sqrt m} ]

Drift:

[ t=\frac{d}{v} ]

Therefore:

[ \boxed{ t\propto\sqrt{\frac{m}{q}} } ]

Lighter ions arrive sooner.

Mass spectrum:

• x-axis → (m/z);
• y-axis → relative abundance/intensity.

7.5 Relative isotopic mass

Mass of one atom of an isotope compared with (1/12) of the mass of one carbon-12 atom.

No unit.

7.6 Relative atomic mass

Weighted mean mass of an atom of an element compared with (1/12) of the mass of one carbon-12 atom.

[ \boxed{ A_r= \frac{\sum(\text{isotopic mass}\times\text{abundance})} {\sum\text{abundance}} } ]

No unit.

7.7 Orbitals and subshells

Orbital = region of space that can hold up to two electrons with opposite spins.

[ s:1\text{ orbital}\rightarrow2e^- ]

[ p:3\text{ orbitals}\rightarrow6e^- ]

[ d:5\text{ orbitals}\rightarrow10e^- ]

Filling order to (Z=36):

[ \boxed{ 1s,2s,2p,3s,3p,4s,3d,4p } ]

7.8 Transition-metal configuration rule

4s fills before 3d in neutral atoms.

When positive transition-metal ions form:

[ \boxed{ 4s\text{ electrons are removed first} } ]

Examples:

[ \text{Fe}=[\text{Ar}]3d^64s^2 ]

[ \text{Fe}^{2+}=[\text{Ar}]3d^6 ]

[ \text{Fe}^{3+}=[\text{Ar}]3d^5 ]

7.9 First ionisation energy

Energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous (1+) ions.

Equation:

[ \boxed{ X(g)\rightarrow X^+(g)+e^- } ]

Factors:

• nuclear charge;
• distance;
• shielding;
• subshell;
• electron pairing where relevant.

7.10 Successive ionisation energy equations

Second:

[ \boxed{ X^+(g)\rightarrow X^{2+}(g)+e^- } ]

Third:

[ \boxed{ X^{2+}(g)\rightarrow X^{3+}(g)+e^- } ]

Large jump:

[ \boxed{ \text{next electron is being removed from an inner shell} } ]

Number removed before large jump:

[ \boxed{ \text{number of outer-shell electrons} } ]

7.11 Period 3 first ionisation trend

General increase Na → Ar because:

• nuclear charge increases;
• similar shielding;
• same principal outer shell;
• smaller atomic radius;
• stronger attraction.

Mg → Al drop:

[ 3s\rightarrow3p ]

Al loses a higher-energy 3p electron.

P → S drop:

• S has paired 3p electrons;
• electron–electron repulsion makes one easier to remove.

7.12 Group 2 first ionisation trend

Down Be → Ba:

[ \boxed{ IE_1\text{ decreases} } ]

because:

• outer electron is further from nucleus;
• shielding increases;
• these outweigh increased nuclear charge.