1. Atomic structure

Exam analysis

Most tested topics

1.1 Particles in the atom and atomic radius

An atom has a very small, dense nucleus containing protons and neutrons. Electrons occupy shells in the much larger region around the nucleus, so most of the atom’s volume is empty space.

The nucleus is tiny compared with the whole atom but contains almost all its mass. The electron shells describe allowed energy regions around the nucleus; they do not fill the space with solid matter.

“Mostly empty” does not mean the atom has no structure. The positive nucleus and negative electrons are held together by electrostatic attraction, while the shell model records where electrons may be found rather than fixed planetary paths.

Protons, neutrons and electrons are distinguished by relative charge and relative mass. Relative values compare the particles on a convenient scale; they are not their masses in grams or charges in coulombs.

Particle	Relative charge	Relative mass
proton	+1	1
neutron	0	1
electron	−1	$\frac{1}{1836}$

A proton and an electron have equal-magnitude opposite charges. A neutron is uncharged. Because an electron has far less mass than either nuclear particle, electron mass is usually negligible when accounting for an atom’s mass.

Atomic number and proton number are two names for the number of protons in an atom’s nucleus. Mass number and nucleon number are two names for the total number of protons plus neutrons in that nucleus.

Quantity	Symbol	What it counts
atomic (proton) number	$Z$	protons
mass (nucleon) number	$A$	protons + neutrons

The atomic number identifies the element. The mass number describes one particular nuclide and is always a whole number; it is not the relative atomic mass shown in many Periodic Tables, which can be a weighted mean.

Nearly all the mass of an atom and all its positive charge are concentrated in the nucleus. The surrounding electron region contributes negative charge but very little mass.

Protons and neutrons each have relative mass about 1, whereas an electron has relative mass $\frac{1}{1836}$ . Neutrons add mass but no charge; protons add mass and positive charge; electrons add negative charge outside the nucleus.

A neutral atom has equal numbers of protons and electrons, so its total charge is zero even though positive and negative charge occupy different regions. Zero overall charge does not mean that charge is absent.

An electric field exerts a force on charged particles. For proton, neutron and electron beams moving at the same velocity, the direction of bending reveals charge and the amount of bending reflects mass as well as charge.

Beam	Charge	Path between charged plates	Relative deflection
proton	+1	toward the negative plate	small
neutron	0	straight, with no deflection	none
electron	−1	toward the positive plate	large

The proton and electron experience forces in opposite directions because their charges have opposite signs. Their charge magnitudes are equal, but the electron’s mass is much smaller, so it accelerates and bends much more strongly. The comparison depends on the stated equal velocity and electric-field arrangement.

Use $A$ for mass number, $Z$ for atomic number and $q$ for the signed ionic charge. Read the nuclear counts first; only then use the charge to adjust the electron count.

\text{protons}=Z,\qquad \text{neutrons}=A-Z,\qquad \text{electrons}=Z-q

For $^{40}_{20}\ce{Ca}^{2+}$ , $A=40$ , $Z=20$ and $q=+2$ . It therefore has 20 protons, $40-20=20$ neutrons and $20-(+2)=18$ electrons. A positive ion has lost electrons; a negative value of $q$ makes $Z-q$ larger because electrons were gained.

Charge changes only the electron count. If a calculation changes the proton number, it has changed the element rather than formed an ion.

Radius reflects the balance between nuclear attraction, the distance of the outer occupied shell and shielding by inner electrons. Use these causes—not a memorised arrow alone—to explain each trend.

Direction	Atomic radius	Why
across a period	generally decreases	proton number rises while electrons enter the same principal shell, so nuclear attraction strengthens
down a group	increases	an extra occupied shell increases distance and shielding

A cation is smaller than its atom because losing electrons reduces electron–electron repulsion and may remove an outer shell. An anion is larger because added electrons increase repulsion within the electron region. Down a group, ions of comparable charge become larger as shells are added.

Across one period, radii generally decrease within the cation series and within the anion series as nuclear charge increases. There is a large jump between the last cation and first anion because the anions occupy an additional outer shell. Compare like species before applying a trend.

1.2 Isotopes

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. The shared proton number keeps the element identity the same; the different neutron number gives a different mass number.

Carbon-12 and carbon-14 are both carbon because each atom has 6 protons. Carbon-12 has 6 neutrons, while carbon-14 has 8, so their mass numbers are 12 and 14.

Changing neutron number makes a different isotope. Changing electron number makes an ion, and changing proton number makes a different element.

An isotope is written with its mass number at the upper left of the element symbol and its atomic number at the lower left.

^{A}_{Z}X

$A$ is the mass (nucleon) number, $Z$ is the atomic (proton) number and $X$ is the element symbol. Therefore the isotope contains $Z$ protons and $A-Z$ neutrons.

In $^{37}_{17}\ce{Cl}$ , the lower number 17 identifies chlorine and gives 17 protons. The upper number 37 counts all nucleons, so the neutron number is $37-17=20$ . Isotopes of chlorine keep $Z=17$ but have different values of $A$ .

Isotopes of the same element have the same chemical properties because their atoms have the same electron configuration. Chemical reactions involve electrons, especially those in the outer shell, rather than neutrons in the nucleus.

The isotopes have the same proton number. As atoms, they therefore have the same number of electrons, arranged in the same shells and outer-shell pattern. They form the same types of bonds and undergo the same chemical reactions.

Different neutron numbers change nuclear mass, not the electron arrangement that controls chemistry. Do not use the isotope’s different mass number as a reason for different chemical behaviour.

Isotopes of one element have different physical properties because they contain different numbers of neutrons. The syllabus limit here is mass and density.

Property	Effect of more neutrons	Reason
mass	increases	each added neutron contributes nuclear mass
density	can increase	under the same physical conditions, greater mass is present in approximately the same volume

A $^{22}\ce{Ne}$ atom has two more neutrons and greater mass than a $^{20}\ce{Ne}$ atom. A sample enriched in neon-22 therefore has a higher density than a comparable sample enriched in neon-20 when state, temperature and pressure are the same.

The neutron difference explains mass-related physical changes. It does not change the element’s proton number or the electron arrangement responsible for its chemical properties.

1.3 Electrons, energy levels and atomic orbitals

Throughout section 1.3, every atom or ion is in its ground state: its electrons occupy the lowest-energy arrangement allowed. The assessed elements run from hydrogen, H, to krypton, Kr.

Use this boundary when choosing examples and writing configurations. Excited-state arrangements and elements beyond krypton are outside this section; the boundary limits what is assessed but does not redefine shells, sub-shells or orbitals.

A shell is a principal energy level identified by the principal quantum number, n. Each shell contains one or more sub-shells, labelled s, p or d within the assessed range, and each sub-shell contains one or more orbitals.

An orbital is a region of space that can hold a maximum of two electrons. It is not a fixed path followed by an electron. For example, 2p identifies the p sub-shell in the shell with n = 2.

The ground state is the lowest-energy electronic configuration available to the species. Do not use shell, sub-shell and orbital as interchangeable terms: they are nested levels of the model.

Each orbital holds at most two electrons, so a sub-shell's capacity is twice its number of orbitals.

Sub-shell	Number of orbitals	Maximum electrons
s	1	2
p	3	6
d	5	10

Keep the two counts distinct: a p sub-shell contains three orbitals but can contain six electrons. This objective assesses s, p and d; adding f is not needed for the stated learning outcome.

A ground-state configuration places electrons into lower-energy sub-shells before higher-energy sub-shells. Sub-shell energies overlap, so principal shell number alone does not give the filling order.

$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p$

Potassium therefore ends in 4s¹, not 3d¹. The sequence is an energy order for the sub-shells required here; it is not the simple numerical order 1, 2, 3, 4.

In a term such as 2p⁴, 2 is the principal shell, p is the sub-shell, and the superscript 4 is the number of electrons in that sub-shell. The superscripts in a complete configuration must add to the species' total number of electrons.

$\mathrm{O}:\ 1s^2\,2s^2\,2p^4$

Sub-shell notation gives the total within each sub-shell. Electron-in-box notation adds the orbital-level detail by showing how those electrons are distributed among the individual orbitals.

Do not read the superscript as an orbital count or an atomic number. It is an electron count for that sub-shell.

In the ground state, electrons occupy the available sub-shells in increasing energy. Within one sub-shell, orbitals have the same energy, so electrons occupy separate orbitals before pairing.

Separate occupancy keeps electrons farther apart and reduces inter-electron repulsion. Once every orbital in that sub-shell contains one electron, further electrons must pair; the two electrons in one orbital have opposite spins.

For 2p³, place one electron in each of the three 2p orbitals. For 2p⁴, the fourth electron pairs in one orbital. Pairing earlier would give greater repulsion without lowering the sub-shell energy.

First determine the electron count: a neutral atom has Z electrons; subtract the positive charge for a cation or add the magnitude of the negative charge for an anion. Fill sub-shells in the assessed energy order, then check that all superscripts add to this count.

$\mathrm{Fe}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^6\,4s^2 = [Ar]\,3d^6\,4s^2$

$\mathrm{Fe^{2+}}:\ [Ar]\,3d^6$

When a transition-metal ion forms, remove 4s electrons before 3d electrons. Changing charge changes the electron count, not the proton number or the identity of the element.

Each box represents one orbital and each arrow represents one electron. Two arrows in one box must point in opposite directions; among equal-energy orbitals, place one electron in each box before pairing.

2p orbital	first	second	third
2p⁴ occupancy	↑↓	↑	↑

This 2p⁴ diagram contains four electrons: one pair and two unpaired electrons. A box is an orbital, not a shell, and arrow direction represents spin rather than electron motion along a path.

An s orbital is spherical around the nucleus. A larger principal quantum number gives a larger s orbital, while the assessed overall shape remains spherical.

Each p sub-shell contains three dumbbell-shaped orbitals, labelled pₓ, pᵧ and p_z. Their lobes point along three mutually perpendicular axes, and the nucleus lies at the centre between the two lobes of each orbital.

In a sketch, show one sphere for an s orbital or two equal lobes on a straight axis for a p orbital, centred on the nucleus. The boundary is a probability-region model, not a hard surface or an electron track.

A free radical is a species with one or more unpaired electrons. Radical status is determined by electron pairing, not by whether the species is neutral or charged.

$\mathrm{Cl\boldsymbol{\cdot}}$

The dot represents the unpaired electron. In electron-in-box notation, the same evidence is a singly occupied orbital. A lone pair contains two paired electrons, so a lone pair alone does not make a species a radical.

1.4 Ionisation energy

Throughout section 1.4, every atom or ion is in its ground state, and the assessed elements run from hydrogen, H, to krypton, Kr. Use ground-state configurations when explaining or interpreting ionisation-energy data.

This is an assessment boundary, not a separate ionisation rule. Excited-state configurations and elements beyond krypton are outside the cases required here; definitions, equations and trends are taught in the following objectives.

The first ionisation energy, IE₁, is the energy required to remove one mole of electrons from one mole of gaseous atoms, forming one mole of gaseous 1+ ions. Its unit is kJ mol⁻¹.

$\mathrm{X(g) \rightarrow X^+(g) + e^-}$

All three details matter: the atoms are gaseous, exactly one electron is removed from each atom, and gaseous 1+ ions form. Removing an electron requires energy because attraction between the nucleus and electron must be overcome.

Construct successive ionisation equations one step at a time. The reactant for each step is the gaseous ion produced by the preceding step, and its charge increases by one.

$\mathrm{X(g) \rightarrow X^+(g) + e^-}$

$\mathrm{X^+(g) \rightarrow X^{2+}(g) + e^-}$

$\mathrm{X^{(n-1)+}(g) \rightarrow X^{n+}(g) + e^-}$

Do not combine several removals into one equation. For example, the second ionisation must start with X⁺(g), not X(g), and every atom or ion in the equation must carry the gaseous state symbol.

Direction	General IE₁ trend	Main explanation
Across a period	increases	nuclear charge increases while shielding is similar, so radius decreases and attraction to the outer electron strengthens
Down a group	decreases	the outer electron is in a higher shell, farther from the nucleus and more shielded, so attraction weakens

The across-period rise is not perfectly smooth. Moving from an s to a higher-energy p sub-shell can lower IE₁, and pairing two electrons in one p orbital adds repulsion, making one easier to remove.

Explain the trend by following the electron being removed. Nuclear charge alone is insufficient when distance, shielding, sub-shell energy or spin-pair repulsion changes.

Successive ionisation energies increase because each electron is removed from an increasingly positive ion. The remaining electrons experience stronger attraction to the unchanged nuclear charge.

Calcium ionisation	IE₁	IE₂	IE₃	IE₄
IE / kJ mol⁻¹	590	1150	4940	6480

The large jump between IE₂ and IE₃ shows that two outer-shell electrons have been removed and the third electron comes from an inner shell. Inner-shell electrons are closer to the nucleus and less shielded from it, so much more energy is required.

Every successive value may be larger, but only a clear change of scale is evidence for moving to a new shell. Smaller changes can reflect different sub-shells or electron pairing within the same shell.

An outer electron is electrostatically attracted to the positively charged nucleus. Ionisation energy is the energy needed to overcome this attraction and separate the electron from the gaseous atom or ion.

Stronger attraction gives a larger ionisation energy; weaker attraction gives a smaller one. The attraction changes with nuclear charge and with how far the electron is from the nucleus and how strongly inner electrons shield it.

The nucleus does not lose protons during ionisation. The positive charge increases because electrons are removed, so the remaining electrons are generally held more strongly.

Factor	Effect on attraction and ionisation energy
Greater nuclear charge	stronger attraction; IE tends to increase
Greater atomic or ionic radius	electron is farther away; IE tends to decrease
More inner-shell shielding	lower effective attraction; IE tends to decrease
Higher-energy, more shielded sub-shell	electron is easier to remove; IE tends to decrease
Spin-pair repulsion	a paired electron is easier to remove; IE tends to decrease

First identify the shell, sub-shell and pairing of the electron removed. Then compare the factors and state which change dominates; do not merely list them.

This explains common dips across a period: Al loses a higher-energy 3p electron whereas Mg loses a 3s electron; S has a paired 3p electron whereas P has three singly occupied 3p orbitals.

Locate the largest jump in successive ionisation energies. The number of electrons removed before that jump is the number in the outer shell. Combine this evidence with the atomic number or other supplied information to build the complete ground-state configuration.

Ionisation	IE₁	IE₂	IE₃	IE₄
IE / kJ mol⁻¹	577	1820	2740	11600

The jump after three removals shows three outer-shell electrons. With atomic number 13, the configuration is 1s² 2s² 2p⁶ 3s² 3p¹: three electrons occupy the third shell before removal begins from the second shell.

$\mathrm{1s^2\,2s^2\,2p^6\,3s^2\,3p^1}$

The jump gives the outer-shell count, not the full configuration by itself. Use the other supplied evidence to determine how many inner electrons and occupied shells are present.

For an assessed s- or p-block element, the first large jump reveals the number of outer-shell electrons and therefore its group pattern. A jump after two removals supports Group 2; a jump after seven supports Group 17.

Unknown Period 3 element	IE₁	IE₂	IE₃	IE₄
IE / kJ mol⁻¹	736	1450	7740	10500

The large jump between IE₂ and IE₃ shows two outer electrons, so the element follows the Group 2 pattern. The supplied Period 3 information fixes the row; Period 3 and Group 2 identify magnesium.

Successive values alone do not always reveal the period. Use the stated period, atomic number, identity or deduced configuration alongside the jump before claiming a complete Periodic Table position.

1. Atomic structure

Exam analysis

Most tested topics

1.1 Particles in the atom and atomic radius

Most of an atom is empty space

Compare the three subatomic particles

Atomic number and mass number count different particles

Mass and charge are distributed differently in an atom

How particle beams move in an electric field

Count particles in atoms and ions

Explain trends in atomic and ionic radius

1.2 Isotopes

Isotopes differ only in neutron number

Read isotope notation without swapping the numbers

Why isotopes have the same chemical properties

Why isotope mass and density can differ

1.3 Electrons, energy levels and atomic orbitals

The assessed 1.3 cases are ground-state species from H to Kr

Shells, sub-shells and orbitals form a hierarchy

Orbital count determines each sub-shell's electron capacity

Use the assessed sub-shell energy order

A configuration records electrons at shell, sub-shell and orbital levels

Lowest energy and minimum repulsion explain electron placement

Determine configurations by count, fill, then check

Electron-in-box notation shows occupancy and spin

s orbitals are spherical; p orbitals are directional

A free radical has at least one unpaired electron

1.4 Ionisation energy

The assessed ionisation cases are ground-state species from H to Kr

First ionisation energy removes one mole of gaseous atoms' electrons

Each ionisation equation removes one electron from a gaseous species

Ionisation energy rises across a period and falls down a group

Successive ionisation energies rise and reveal shell changes

Ionisation energy measures nucleus–outer-electron attraction

Use the actual electron removed to compare ionisation energies

Use successive ionisation data to deduce an electronic configuration

Use successive ionisation data to locate an element in the Periodic Table