Full chapter
Atomic Structure
Use particles, orbitals and ionisation data to explain the behaviour of atoms and ions.
A-Level 8873, revised syllabus (2026-2027)
Count particles before counting electrons
Atomic number identifies the element; electron loss changes charge, not the nucleus.
An atom has a very small, positively charged nucleus containing protons and neutrons. Almost all its mass is concentrated there. Electrons occupy the surrounding space, which accounts for almost all the volume. A neutral atom has equal numbers of protons and electrons; an ion does not.
| Particle | Relative charge | Relative mass | Where it is found |
|---|---|---|---|
| Proton | +1 | 1 | Nucleus |
| Neutron | 0 | 1 | Nucleus |
| Electron | -1 | About 1/1836 | Outside the nucleus |
The proton number, Z, is the number of protons and defines the element. The nucleon number, A, counts protons plus neutrons. Therefore neutrons = A - Z. For an ion with signed charge q, electrons = Z - q: subtract a positive charge, but add the magnitude of a negative charge. Ion formation transfers electrons; it does not change A or Z.
Worked exampleRead an isotope and its ion
An aluminium-27 ion has Z = 13 and charge 3+. How many of each particle are present?
Try it first. Then show the worked stepsHide the worked steps
- Protons = 13, because the element is aluminium.
- Neutrons = 27 - 13 = 14.
- A neutral atom has 13 electrons. Losing three gives 10 electrons.
- Charge check: 13 positive charges and 10 negative charges give +3.
The ion contains 13 protons, 14 neutrons and 10 electrons. It is still aluminium-27.
Isotopes have the same proton number but different neutron numbers. Chlorine-35 and chlorine-37 both have 17 protons, but have 18 and 20 neutrons respectively. Neutral atoms of both have 17 electrons, so their electron arrangements and chemical behaviour are very similar. Their masses differ; isotopes are not different ions.
Check your understanding
Compare 23Na and 23Na+ (Z = 11). Are they isotopes of each other?
Show the answer
No. Both have 11 protons and 12 neutrons; only the electron count differs.
Isotopes have the same proton number but different neutron numbers. These have the same nucleus, so they are the same isotope. The atom has 11 electrons and the ion has 10.
Use a field to identify the charge
Direction identifies the sign; the amount of bending also depends on the beam conditions.
Make a guess. You are not marked.
A beam of neutrons passes between two charged plates. What happens to its path?
Show the answer
It goes straight through, as neutrons have no charge
Neutrons have no charge, so the electric field exerts no force on them.
Deflection of particle beams
Electrons bend towards the positive upper plate, protons towards the negative lower plate, and neutrons continue straight. At the same initial speed an electron bends much more because its mass is much smaller.
Opposite charges attract. A proton beam deflects towards the negative plate; an electron beam deflects towards the positive plate. Neutrons have no charge and are not deflected by the electric field. In the same field, force depends on charge and acceleration depends on force divided by mass.
Check your understanding
A beam of protons, electrons and neutrons passes between charged plates. What happens if the plate polarities are reversed?
Show the answer
Protons and electrons bend the other way; neutrons still go straight.
Reversing the field reverses the force on each charged particle. Neutrons have no charge, so the field has no effect on them.
Replace circular paths with orbitals
An orbital describes where an electron may be found, not a tiny planetary orbit.
A shell is labelled by principal quantum number n. Within a shell, subshells contain orbitals: one s orbital, three p orbitals or five d orbitals. Each orbital holds at most two electrons with opposite spins. Orbitals in the same subshell have the same energy in an isolated atom unless an external influence splits them.
| Shell/subshell | Number of orbitals | Maximum electrons |
|---|---|---|
| n = 1: 1s | 1 | 2 |
| n = 2: 2s; 2p | 1; 3 | 2; 6 |
| n = 3: 3s; 3p; 3d | 1; 3; 5 | 2; 6; 10 |
| 4s; 4p | 1; 3 | 2; 6 |
s and p orbital shapes
A sphere represents an s orbital. Two lobes on opposite sides of the nucleus represent one p orbital. The two colours distinguish lobes, not electric charges.
Within a multi-electron shell, s is lower in energy than p, which is lower than d where present. The useful filling sequence for the atoms studied is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p. Energies also depend on occupancy: this filling order is not a claim that 4s is always lower than 3d in every ion.
Check your understanding
How many orbitals are in the third shell (n = 3), and how many electrons can the shell hold?
Show the answer
9 orbitals, 18 electrons
The third shell has one 3s, three 3p and five 3d orbitals: 1 + 3 + 5 = 9. Each holds two electrons of opposite spin, so 18.
Check your understanding
In what order do the 3d, 4s, 3p and 4p subshells fill, lowest energy first?
Show the answer
3p, 4s, 3d, 4p
4s fills before 3d in the atoms of this course, then 4p follows 3d. An s orbital is spherical; a p orbital has two lobes, and the three p orbitals lie along axes at right angles.
Build a configuration, then make the ion
Fill the available orbitals and count the electrons again after every change.
- Count electrons
Use Z for a neutral atom; adjust for the signed ion charge.
- Fill lower levels first
Follow the filling sequence, respecting the capacity of each subshell.
- Distribute within a subshell
Place electrons singly with parallel spins in equal-energy orbitals before pairing them.
- Check the ion
For transition-metal cations, remove 4s electrons before 3d electrons.
| Species | Electrons | Configuration |
|---|---|---|
| N (Z = 7) | 7 | 1s2 2s2 2p3; one electron in each p orbital |
| O (Z = 8) | 8 | 1s2 2s2 2p4; one p orbital has a pair |
| Cl- (Z = 17) | 18 | 1s2 2s2 2p6 3s2 3p6 |
| Fe (Z = 26) | 26 | [Ar] 3d6 4s2 |
| Fe2+ | 24 | [Ar] 3d6 |
| Fe3+ | 23 | [Ar] 3d5 |
The symbol [Ar] replaces the filled first 18 electrons. For Cr (Z = 24), use [Ar] 3d5 4s1; for Cu (Z = 29), use [Ar] 3d10 4s1. These are exceptions to simply assigning two electrons to 4s before 3d. When forming their positive ions, remove the 4s electron first.
Worked exampleAn ion does not fill like its neutral parent
Write the configuration of Cu2+ (Z = 29).
Try it first. Then show the worked stepsHide the worked steps
- Start with Cu: [Ar] 3d10 4s1.
- Remove the 4s electron, then one 3d electron.
- Count: 18 + 9 = 27 electrons, two fewer than 29 protons.
Cu2+ is [Ar] 3d9, not [Ar] 3d7 4s2.
Check your understanding
What is the electronic configuration of S2- (Z = 16)?
Show the answer
1s2 2s2 2p6 3s2 3p6
S2- has 16 + 2 = 18 electrons. The two gained electrons complete 3p; the nucleus still has 16 protons.
Explain how strongly an electron is held
Balance nuclear attraction, shielding, distance and electron pairing.
Make a guess. You are not marked.
First ionisation energy generally rises across Period 3. How does sulfur compare with phosphorus?
Show the answer
Sulfur is lower, as it loses an electron from a paired 3p orbital
Sulfur loses an electron from a 3p orbital that holds a pair. Repulsion between the paired electrons makes that electron easier to remove.
The first ionisation energy is the energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions. For X: X(g) → X+(g) + e-. Values are positive and usually quoted in kJ mol-1: energy is supplied to overcome attraction to the nucleus.
| Factor | Effect when other factors are similar | How to use it |
|---|---|---|
| Greater nuclear charge | Stronger attraction, higher ionisation energy | Compare proton number, but also check shielding. |
| Greater shielding by inner electrons | Weaker attraction to the outer electron | A new shell often matters more than extra protons. |
| Greater electron-nucleus distance | Weaker attraction, lower ionisation energy | Name the shell from which the electron leaves. |
| Higher-energy subshell or electron pairing | Often makes an electron easier to remove | Identify the specific subshell and paired orbital. |
Across Period 3, first ionisation energy generally increases: nuclear charge rises while added electrons enter the same principal shell, so shielding changes relatively little. Two dips need more precise reasoning. Al loses a higher-energy 3p electron, whereas Mg loses 3s; Al is easier to ionise. S loses an electron from a paired 3p orbital, whereas P has three singly occupied 3p orbitals; pair repulsion makes the S electron easier to remove.
Down a group, the outer electron occupies a higher shell farther from the nucleus and is more shielded. These effects usually outweigh the greater nuclear charge, so first ionisation energy decreases. Say which effects dominate; merely saying "there are more electrons" does not explain the attraction.
Check your understanding
First ionisation energy dips from Mg to Al, even though Al has more protons. Why?
Show the answer
Al loses a 3p electron, which is higher in energy and more shielded than Mg's 3s.
The 3p electron is further from the nucleus on average and shielded by the 3s electrons, so it is easier to remove. This outweighs the extra nuclear charge.
Find the jump that reveals the outer electrons
A large step means removal has reached an inner shell.
Successive ionisation energies refer to removing electrons one at a time from increasingly positive gaseous ions. The second process is X+(g) → X2+(g) + e-. Values rise because the remaining electrons are held more strongly. A particularly large jump indicates that the next electron comes from a lower, inner shell much closer to the nucleus.
| Electron removed | 1st | 2nd | 3rd | 4th |
|---|---|---|---|---|
| Energy / kJ mol-1 | 740 | 1450 | 7730 | 10500 |
Worked exampleLocate a group without overclaiming the element
Use the data to infer the outer-electron count. Then write the configuration if Z = 12.
Try it first. Then show the worked stepsHide the worked steps
- Compare steps: 740 to 1450 is an increase, but 1450 to 7730 is the pronounced jump.
- Two electrons are removed before reaching an inner shell, so the atom has two outer electrons.
- For a main-group element this suggests Group 2. The four energies alone do not establish the period.
- With Z = 12 supplied, place 12 electrons: 1s2 2s2 2p6 3s2.
Two outer electrons; Group 2. With the extra information Z = 12, the atom is magnesium in Period 3.
| Electron removed | Ionisation energy / kJ mol-1 |
|---|---|
| 1 | 496 |
| 2 | 4,560 |
| 3 | 6,910 |
| 4 | 9,540 |
| 5 | 13,400 |
| 6 | 16,600 |
| 7 | 20,100 |
| 8 | 25,500 |
| 9 | 28,900 |
| 10 | 141,000 |
| 11 | 159,000 |
Worked exampleRecover the whole configuration from complete data
All eleven successive ionisation energies of a neutral atom are shown. Deduce its shell arrangement, configuration, group and period.
Try it first. Then show the worked stepsHide the worked steps
- All electrons have been removed after the eleventh step, so the neutral atom contained eleven electrons. Its proton number is therefore 11; this comes from the completeness of the data, not from memorising a numerical energy.
- The large jumps are from the 1st to the 2nd energy and from the 9th to the 10th. Separate the removals into 1 | 8 | 2 electrons. These groups are removed from the outside in.
- Reverse their order to write the occupied shells from the nucleus outwards: 2, 8, 1. Three occupied principal shells place the atom in Period 3; one outer electron places it in Group 1.
- Fill the subshells: the first two electrons occupy 1s, the next eight occupy 2s and 2p, and the last occupies 3s. The smaller steps within a shell do not represent additional principal shells.
The atom is Na: 1s2 2s2 2p6 3s1, Group 1, Period 3. If only the first few energies had been supplied, that full conclusion would need additional evidence.
Check your understanding
A main-group atom has a large jump between its third and fourth ionisation energies. What can you infer?
Show the answer
It has three outer electrons, so it is in Group 13.
Three electrons come off before the next one must come from an inner shell, closer to the nucleus. The data do not tell you the period on their own.
Quick revision
Revisit the essentials, then return to an explanation when you need it.
| Term | Definition to write |
|---|---|
| Proton (atomic) number | The number of protons in the nucleus of an atom. |
| Nucleon (mass) number | The total number of protons and neutrons in the nucleus of an atom. |
| Isotopes | Atoms of the same element with the same number of protons but different numbers of neutrons. They have the same chemical properties because they have the same number of electrons (the same electronic configuration). |
| Orbital | A region of space around the nucleus where there is a high probability of finding an electron. Each orbital holds at most two electrons, with opposite spins. |
| First ionisation energy | The energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous singly charged positive ions: M(g) → M+(g) + e-. |
| Question | First move |
|---|---|
| How many neutrons/electrons? | Neutrons = A - Z; electrons = Z - signed charge. |
| Which way does a beam bend? | Positive to negative plate; negative to positive; neutral straight. |
| Which orbital fills? | 1s 2s 2p 3s 3p 4s 3d 4p; two per orbital, singly before pairing. |
| Which electron leaves an ion? | Remove outer-shell electrons; 4s before 3d for transition cations. |
| Why a different ionisation energy? | Compare charge, shielding, distance, subshell and pairing. |
| What does the large jump reveal? | Count electrons removed before reaching the inner shell. |
Scope
Learning outcomes
1. Atomic Structure (8873, 2026 revision; examinations 2026 and 2027). Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
1(a) Identify the subatomic particles.
- Relative charges and masses of protons, neutrons and electrons.
1(b) Predict beam deflection.
- Proton, neutron and electron beams in an electric field; conditions for comparing deflection.
1(c) Locate atomic mass and charge.
- Small positive nucleus; almost all mass in the nucleus; electrons outside.
1(d) Count particles in atoms and ions.
- Given proton number, nucleon number and charge.
1(e) Interpret nuclear notation and isotopes.
- (i) Protons/neutrons in Z and A.
- (ii) Isotopes distinguished by neutron number.
1(f) Organise shells, subshells and orbitals.
- Numbers and relative energies of s, p and d orbitals for n = 1, 2, 3, plus 4s and 4p.
1(g) Describe s and p shapes.
- Spherical s and two-lobed p orbitals; no wave functions.
1(h) Write atom and ion configurations.
- Use proton number and charge; filling, pairing, Cr/Cu exceptions and 4s removal.
1(i) Explain ionisation energies.
- Nuclear charge, shielding, distance, subshell and pairing; Periodic Table/Data Booklet application.
1(j) Infer configurations from successive energies.
- Identify shell jumps; recover the shell populations and full configuration when a complete data set is supplied; distinguish limited data.
1(k) Connect successive energies to periodic position.
- Use outer-electron count for main-group position and the number of occupied shells for period when supported by complete data.