Atomic Structure

Physical Chemistry Weightage: AS · Papers 1 and 2 (assumed in Paper 4) Topic 1

“Welcome! Cambridge does not ask you to recite this topic; it asks you to use it. Atomic structure is where the exam first checks whether you write the way the mark scheme wants: an exact definition, a state-symbol equation, an explanation that names the cause and then links it to the effect. The same ideas (shielding, nuclear charge, distance) come back in periodicity, Group 2, Group 17, transition elements and bonding, so a strong Topic 1 quietly lifts marks all the way through Paper 2 and Paper 4. Read the notes below, then attempt the ten exam-style questions and check yourself against the mark schemes.”

— SCORECHEM ACADEMIC TEAM

0. What Examiners Want

How marks are earned in this topic

Define questions are marked on key phrases. For first ionisation energy the marking points are one mole (or each atom), gaseous atoms and 1+ ions / one electron removed. Missing “gaseous” is the most common lost mark.
Explain questions need a chain: a factor (e.g. more protons) → its effect (stronger attraction) → the result (higher IE). A bare “because of shielding” earns nothing.
Deduce questions expect you to use the data given: quote the numbers or the jump.
Units and symbols matter: kJ mol⁻¹, state symbols, and the electron (e⁻) in IE equations.
Marks are whole numbers and each marking point is worth exactly one mark, so a 3-mark question wants three distinct points, not one point said three ways.

1. Particles and Beams

Particle	Relative charge	Relative mass	Where
Proton	+1	1	Nucleus
Neutron	0	1	Nucleus
Electron	−1	1/1836 (negligible)	Shells around the nucleus

The atom is mostly empty space. Almost all the mass sits in a tiny, dense, positively charged nucleus; the electrons occupy the space around it.
Proton (atomic) number = number of protons. Nucleon (mass) number = protons + neutrons. In an atom, electrons = protons; in an ion, electrons = protons − charge.

Beams of electrons, protons and neutrons moving at the same speed through an electric field. Charge sets the direction; charge-to-mass ratio sets the size of the deflection. The electron beam bends far more because an electron has about 1/1836 of a proton's mass.

Direction depends on charge: protons bend towards the negative plate, electrons towards the positive plate, neutrons go straight.
Size depends on charge-to-mass ratio. At the same speed and the same field, an electron is deflected much more than a proton, because it is about 1836 times lighter.

⚠️ Examiner Trap: Direction is not the same as amount. Two separate marking points are usually available: “protons deflected towards the negative plate” (charge) and “electrons deflected more than protons” (mass). Say both, and never say that neutrons are “slightly deflected”: they have no charge, so there is no deflection at all.

Worked example: counting particles in an ion. How many protons, neutrons and electrons are in ²⁷Al³⁺?

Protons = proton number of Al = 13.
Neutrons = nucleon number − protons = 27 − 13 = 14.
Electrons = 13 − 3 = 10 (a 3+ ion has lost three electrons).

2. Isotopes

Definition to learn word for word: Isotopes are atoms of the same element with different numbers of neutrons (so the same proton number but different nucleon numbers).

Property type	Same or different?	Reason to write
Chemical (reactions, bonding)	Same	Same number of electrons, so the same electronic configuration; chemistry is decided by the outer electrons
Physical (mass, density, rate of diffusion)	Different	Different number of neutrons, so different mass

Notation: ^A_ZX, with A the nucleon number and Z the proton number. ³⁵Cl and ³⁷Cl both have 17 protons; they have 18 and 20 neutrons.
Cambridge limits the “different physical properties” part to mass-related ones, so use density or rate of diffusion as your examples.

⚠️ Examiner Trap: “Same chemical properties” needs a reason. The mark is for “same number of electrons / same electronic configuration”. Writing “same number of protons” alone does not explain why chemistry is the same, and “different mass so different melting point” is only half a comparison.

3. Shells, Sub-shells and Orbitals

Electrons sit in shells (principal quantum number n = 1, 2, 3, 4). Each shell splits into sub-shells (s, p, d, f), and each sub-shell is made of orbitals. An orbital is a region of space that holds a maximum of two electrons, and the two electrons must have opposite spins.

Sub-shell	Orbitals	Max electrons	First appears in shell
s	1	2	n = 1
p	3	6	n = 2
d	5	10	n = 3
f	7	14	n = 4

Maximum electrons in shell n is 2n²: 2, 8, 18, 32.

An s orbital is a sphere centred on the nucleus. A p orbital is a dumb-bell with two lobes; there are three p orbitals in each shell from n = 2, at right angles to each other.

Sub-shell energy ladder up to 4p. Each box is one orbital holding a maximum of two electrons. The 4s sub-shell sits lower than 3d, so 4s fills first — and 4s electrons are also the first to leave when a transition-metal ion forms.

Filling order to remember: 1s 2s 2p 3s 3p 4s 3d 4p.
Fill the orbitals of a sub-shell one electron each (same spin) before pairing (Hund’s rule), because electrons in the same orbital repel each other.

Nitrogen has three unpaired 2p electrons. Oxygen has one 2p orbital containing a spin pair, so the paired electrons repel each other and oxygen's first ionisation energy is a little lower than nitrogen's.

4. Electronic Configurations

You must write configurations for atoms and ions up to Z = 36 in s, p, d notation.

Species	Configuration
O	1s² 2s² 2p⁴
Cl⁻	1s² 2s² 2p⁶ 3s² 3p⁶
Fe	[Ar] 3d⁶ 4s²
Fe²⁺	[Ar] 3d⁶
Cr	[Ar] 3d⁵ 4s¹
Cu	[Ar] 3d¹⁰ 4s¹

⚠️ Examiner Trap: Which electrons leave first? When a transition metal forms a positive ion, the 4s electrons are removed before the 3d electrons, even though 4s is filled first. So Fe²⁺ is 3d⁶, not 3d⁴ 4s². Cr and Cu each promote one 4s electron into 3d to give the half-filled or full 3d sub-shell.

5. Ionisation Energy

First ionisation energy (IE1): the energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.

Equation with state symbols: X(g) → X⁺(g) + e⁻. Second ionisation energy: X⁺(g) → X²⁺(g) + e⁻. Unit: kJ mol⁻¹.

Four factors decide how large an IE is:

Nuclear charge: more protons, stronger attraction.
Distance: a larger atomic radius means weaker attraction.
Shielding: inner electrons repel the outer electron and reduce the pull it feels.
Spin-pair repulsion: two electrons paired in one orbital repel each other, so one is easier to remove.

6. Trends: Across and Down

Atomic radius: decreases across a period (nuclear charge rises while the electrons go into the same shell, so shielding hardly changes) and increases down a group (extra shell each time).

First ionisation energies across Period 3 (Data Booklet values). The overall rise is nuclear charge with similar shielding; the two dips at Al and S each have their own explanation.

Across Period 3 (general rise): the nuclear charge increases, the outer electrons are in the same shell with similar shielding, so the attraction to the outer electrons is stronger and the radius is smaller.

The two dips, with the wording examiners reward:

Dip	Explanation
Mg → Al (736 → 577)	The electron removed from Al is a 3p electron. It is in a higher-energy sub-shell, slightly further from the nucleus and shielded by the 3s electrons, so it is easier to remove
P → S (1060 → 1000)	In S the electron removed comes from a 3p orbital that holds a pair of electrons. Repulsion between the paired electrons makes it easier to remove

The same pattern appears in Period 2: Be → B (2p replaces 2s) and N → O (paired 2p electrons).

Down a group: IE1 decreases. The outer electron is in a shell further from the nucleus and there is more shielding from extra inner shells; these outweigh the increase in nuclear charge, so the attraction is weaker.

⚠️ Examiner Trap: The two dips are different. Using “spin-pair repulsion” for Al, or “3p is higher energy” for S, earns nothing. Identify what happens to the electron removed: different sub-shell (Al) or paired electron (S).

7. Successive IE and Groups

Successive ionisation energies of magnesium. Two small steps (the two 3s outer electrons) then a very large jump when an electron must be removed from a closer, less-shielded inner shell. The size of the first big jump tells you the group number.

Removing each further electron is harder, because the electron is being taken from an ion with a growing positive charge (more protons than electrons).
A very large jump means the next electron comes from an inner shell, which is closer to the nucleus and less shielded. The number of electrons removed before the first big jump equals the number of outer-shell electrons, which is the group number for Groups 1 and 2 and the last digit of the group number for Groups 13 to 18 (three outer electrons means Group 13).
Magnesium: two small steps, then a big jump after IE2, so it has 2 outer electrons and is in Group 2.

Worked example. An element Z has IE1 to IE5 of 577, 1820, 2740, 11600 and 14800 kJ mol⁻¹.

The big jump is between IE3 and IE4 (2740 → 11600), so three electrons are in the outer shell.
Z is in Group 13; with IE1 = 577 it is aluminium (Data Booklet).
In the exam, quote the jump: “large increase between the 3rd and 4th IE, so the fourth electron is removed from an inner shell”.

8. Quick Sheet and Checklist

Relative charge and mass of proton, neutron, electron; where each is found.
Isotope definition and the two-part reason for same chemistry / different physical.
Sub-shell capacities 2, 6, 10, 14; orbitals 1, 3, 5, 7; 2n² per shell.
Filling order to 4p; Cr and Cu exceptions; 4s leaves first in ions.
IE1 definition, equation with state symbols, unit kJ mol⁻¹.
Four factors; explain across a period, down a group, the two dips.
Reading successive IE data: find the big jump, state the group.

Before you leave this topic, can you:

Write the IE1 definition and the equation for the second IE of calcium from memory?
Explain in one sentence each why Al < Mg and S < P?
Write the full configuration of Fe²⁺ and Cu?
Identify an element from a list of successive IE values?

If yes to all four, attempt the ten exam-style questions with the mark schemes covered, then open the flashcards.