Cambridge AS Level Chemistry 9701

Atomic structure

Electrons fill sub-shells in order of increasing energy — 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p — and the big jump in a set of successive ionisation energies tells you where a new shell begins. Those two facts answer almost every Cambridge AS Chemistry 9701 question on atomic structure, which is syllabus topic 1 and 58 of the 920 Paper 1 questions Quanta has mapped (6.3%).

The detail that separates marks from near-misses is the order: 4s fills before 3d but empties first too, so a transition-metal ion loses its 4s electrons before any 3d ones — and chromium and copper break the filling pattern outright.

Updated 22 September 2026

Subatomic particles

Particle	Relative mass	Relative charge	Where
Proton	1	+1	Nucleus
Neutron	1	0	Nucleus
Electron	1/1840	−1	Orbitals around the nucleus

Proton number ( $Z$ ) identifies the element; nucleon number ( $A$ ) is protons plus neutrons. Isotopes are atoms of the same element with different numbers of neutrons, so they share the chemistry — chemical behaviour comes from the electrons — but differ in mass, which is why they separate in a mass spectrometer.

Relative atomic mass is the weighted mean of the isotope masses: multiply each isotope mass by its abundance, add, and divide by 100. A behaviour worth remembering for Paper 1: because the electron is so light, an ion’s mass is effectively its atom’s mass, but its deflection in an electric or magnetic field depends on its charge-to-mass ratio.

Shells, sub-shells and orbitals

An orbital is a region that can hold up to two electrons, and they must have opposite spins.
s sub-shells have 1 orbital (2 electrons) and are spherical. p sub-shells have 3 orbitals (6 electrons) and are dumbbell-shaped, at right angles along $x$ , $y$ and $z$ . d sub-shells have 5 orbitals (10 electrons).
Shell $n$ holds up to $2 n^{2}$ electrons: 2, 8, 18, 32.

Three rules govern the filling:

Lowest energy first — 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p. The 4s sub-shell is below 3d in energy, which is the whole reason the transition metals sit where they do.
One electron per orbital first — within a sub-shell, electrons occupy separate orbitals with parallel spins before any orbital takes a second, because they repel.
Paired electrons have opposite spins.

Writing electron configurations

Fill in energy order, counting to the element’s proton number:

O (Z = 8) : 1 s^{2} 2 s^{2} 2 p^{4} C a (Z = 20) : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2}

F e (Z = 26) : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}

Note the convention in that last one: 4s is filled before 3d, but configurations are conventionally written with 3d before 4s because that is the order by shell. Both orderings are accepted in 9701, but the ionisation rule below is not optional.

Ions

For a negative ion, add electrons to the next available space. For a positive ion, remove them — and 4s electrons leave before 3d electrons, despite 4s filling first:

F e^{2 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} F e^{3 +} : \dots 3 d^{5}

This is among the most frequently set Paper 1 questions in the topic, and the distractors are always the versions that took the 3d electrons out first.

Ionisation energy

The first ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms:

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

Every word in that definition earns a mark: one mole, the gaseous state, and one electron removed. The second ionisation energy removes the next electron from the 1+ ion, and so on — each is always larger than the last, because the electron is being pulled from an increasingly positive ion.

Three factors set its size:

Nuclear charge — more protons pull harder, so ionisation energy rises.
Distance from the nucleus — an electron in a shell further out is held less tightly, so ionisation energy falls.
Shielding — inner-shell electrons repel the outer one and reduce the effective nuclear charge it feels.

Worked example

The first six successive ionisation energies of an element, in kJ mol⁻¹, are 578, 1817, 2745, 11 578, 14 831 and 18 378. Deduce the group the element is in and identify it, given that it is in Period 3.

Look for the big jump. Each value rises steadily — 578 → 1817 → 2745 — and then leaps to 11 578, more than four times the previous value. The other steps up are small by comparison.

Read it. The jump comes between the third and fourth electrons removed. So three electrons come from the outer shell relatively easily, and the fourth has to be taken from a new, complete shell much closer to the nucleus and much less shielded.

Conclude. Three outer electrons means Group 13 (Group III). In Period 3 that element is aluminium, configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$ — and removing the 3p¹ and both 3s electrons empties the third shell, after which the fourth electron must come from the full 2p sub-shell.

The method never changes: count how many electrons are removed before the jump, and that is the group number. One before the jump is Group 1, two is Group 2, and so on.

Trends and the two exceptions

Across a period

First ionisation energy generally increases: the nuclear charge rises while electrons are added to the same shell, so shielding barely changes and the outer electrons are held more tightly. There are two dips, and Paper 1 tests both:

Group 2 → Group 13 (e.g. Mg → Al). The next electron enters a p sub-shell, which is higher in energy and slightly further out than the s sub-shell below it, so it is easier to remove.
Group 15 → Group 16 (e.g. P → S). The p sub-shell reaches four electrons, so one orbital now holds a pair. The two paired electrons repel, making one easier to remove.

Down a group

First ionisation energy decreases. Nuclear charge rises, but the outer electron is in a shell further out and is shielded by more inner shells — and those two effects outweigh it.

Chromium and copper

Two elements break the filling order because a half-full or full d sub-shell is more stable:

C r (Z = 24) : \dots 3 d^{5} 4 s^{1} C u (Z = 29) : \dots 3 d^{10} 4 s^{1}

Not $3 d^{4} 4 s^{2}$ or $3 d^{9} 4 s^{2}$ . Both appear as options on Paper 1 more or less every series.

Common mistakes

1.Removing 3d electrons before 4s
4s fills first but empties first. $F e^{2 +}$ is $\dots 3 d^{6}$ , not $\dots 3 d^{4} 4 s^{2}$ .
2.Ionisation energy defined without 'gaseous'
The definition needs one mole of gaseous atoms, one electron each. Missing the state symbol loses the mark even when the rest is right.
3.Counting the jump itself as an electron
The group number is how many electrons are removed before the jump. A jump between the 2nd and 3rd values means two outer electrons — Group 2.
4.Cr and Cu written by the normal rule
Both promote an electron into 4s¹ to gain a half-full or full 3d sub-shell. Learn the two exceptions directly.
5.Explaining the Mg → Al dip by shielding
It is the sub-shell: the electron comes from 3p rather than 3s, higher in energy. The P → S dip is the different one — paired electrons in one p orbital repelling.

Common questions

What is the electron configuration of iron?

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}$ . For $F e^{2 +}$ remove the two 4s electrons to give $\dots 3 d^{6}$ , and for $F e^{3 +}$ one more 3d electron, giving $\dots 3 d^{5}$ .

Why does 4s fill before 3d?

Because the 4s sub-shell is lower in energy than 3d at that point in the Periodic Table, and electrons occupy the lowest energy level available. Once occupied, though, the 4s electrons are the outermost, so they are the first to be removed when an ion forms.

How do successive ionisation energies show the group?

Count the electrons removed before the first large jump. That jump happens when an electron is taken from a new, complete inner shell, so the number before it is the number of outer electrons — and therefore the group.

Why is the first ionisation energy of aluminium lower than magnesium's?

Aluminium’s outer electron is in a 3p orbital, which is higher in energy and slightly further from the nucleus than magnesium’s 3s, so it is more easily removed — despite aluminium having the larger nuclear charge.

How much of 9701 Paper 1 is atomic structure?

58 of the 920 questions Quanta has mapped — 6.3%. The full Paper 1 topic weighting →

Practise atomic structure against real mark schemes

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