Full chapter
Atomic Structure
Count particles, build electron configurations and read the evidence in ionisation energies.
A-Level 9476 (2026-2027)
Almost all the mass is in the nucleus
Proton number identifies the element; electron number determines its charge.
An atom contains a tiny, positively charged nucleus surrounded by electrons. Protons and neutrons supply almost all its mass. Electrons occupy the much larger surrounding region and provide negative charge. A neutral atom has equal numbers of protons and electrons; neutral does not mean that it contains no charged particles.
| Particle | Relative charge | Relative mass | Location |
|---|---|---|---|
| Proton | +1 | 1 | Nucleus |
| Neutron | 0 | 1 | Nucleus |
| Electron | -1 | About 1/1840 | Orbitals outside the nucleus |
In AZX, the proton number Z is the number of protons and the nucleon number A is 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. Gaining or losing electrons does not change A or Z.
Worked example
Read a charged isotope correctly
How many protons, neutrons and electrons are in 5626Fe3+ and 3717Cl-?
- Fe has 26 protons and 56 - 26 = 30 neutrons. Its 3+ charge means it has lost three electrons: 26 - 3 = 23.
- Cl has 17 protons and 37 - 17 = 20 neutrons. Its negative charge means one extra electron: 17 + 1 = 18.
Fe3+: 26 p, 30 n, 23 e. Cl-: 17 p, 20 n, 18 e. The charge changes the electron count, not the proton count.
Isotopes are atoms of the same element with the same proton number but different neutron numbers. Chlorine-35 and chlorine-37 both have 17 protons, but have 18 and 20 neutrons. Their neutral atoms have the same electron configuration, so their chemical properties are very similar; their masses differ. Do not confuse an isotope difference with an ionisation change.
Check your understandingTwo species each have 18 electrons. Must they be isotopes?Think it through, then reveal the answer
A beam bends according to charge and mass
Positive particles bend towards the negative plate; negative particles bend towards the positive plate.
Three beams in the same electric field
A positive upper plate and negative lower plate deflect an electron upwards and a proton downwards. A neutron continues straight. The electron path bends more strongly. These are schematic paths for equal initial speed and the same field length.
The electric force is F = qE. A neutron has no net charge and is not deflected by the uniform electric field. A proton and electron have equal charge magnitudes, so the force magnitudes are equal but opposite. The electron has much less mass, giving much greater acceleration under that force.
For particles entering at the same speed, the time inside the field is the same. Deflection is then proportional to charge-to-mass ratio, |q|/m. If speeds or initial kinetic energies differ, that comparison alone is insufficient. Read the beam conditions before choosing the greatest deflection.
Worked example
Compare ion beams
At equal initial speeds, which bends more: 4He2+ or 20Ne+?
- Use relative |q|/m values: He2+ gives 2/4 = 0.50; Ne+ gives 1/20 = 0.050.
- The helium ion has ten times the ratio, so its deflection is greater under the stated conditions.
- Both are positive, so both bend towards the negative plate.
He2+ bends more strongly; its sign determines direction and its charge-to-mass ratio determines the comparison in these conditions.
A shell contains subshells; a subshell contains orbitals
An orbital holds at most two electrons. A p subshell has three orbitals, not one.
The principal quantum number n labels a shell. Within a shell, subshells are labelled s, p and d. An orbital describes a region where an electron is likely to be found; it is not a circular track. Each orbital can hold at most two electrons with opposite spins.
| Subshells required here | Orbitals in each subshell | Maximum electrons | Relative energies in a many-electron atom |
|---|---|---|---|
| 1s | 1 | 2 | Lowest occupied subshell |
| 2s, 2p | 1 and 3 | 2 and 6 | 2s below 2p |
| 3s, 3p, 3d | 1, 3 and 5 | 2, 6 and 10 | 3s below 3p below 3d within n = 3 |
| 4s, 4p | 1 and 3 | 2 and 6 | 4s below 4p; the position of 3d relative to 4s depends on occupancy |
Orbitals within the same subshell have equal energy in an isolated atom. In many-electron atoms, penetration and shielding split the energies of different subshells. This differs from hydrogen, where orbitals with the same n have the same energy in the simple atomic model.
- Fill the inner subshells
1s, then 2s, 2p, 3s and 3p.
- Start period 4
K and Ca add electrons to 4s.
- Build the transition series
3d occupancy increases, with the Cr and Cu exceptions. Once 3d is occupied, do not assume that 4s electrons are more tightly held.
- Continue to 4p
After the 3d subshell fills, the period continues into 4p.
Read the shape and orientation of an orbital
s is spherical; p has two lobes; the five d orbitals have distinct orientations.
s and p orbitals
A spherical s orbital is represented by a circle centred on a nucleus. A p orbital has two lobes on opposite sides of its nucleus. The lobe boundaries indicate a probability region, not a solid surface or electron path.
| Orbital | Where its electron-density lobes lie |
|---|---|
| dxy | Four lobes between the x and y axes in the xy plane. |
| dxz | Four lobes between the x and z axes in the xz plane. |
| dyz | Four lobes between the y and z axes in the yz plane. |
| dx2-y2 | Four lobes pointing along the x and y axes. |
| dz2 | Two main lobes along z and a ring of electron density around the middle in the xy plane. |
Two important d-orbital patterns
The dxy sketch has four lobes between perpendicular axes. The dz2 sketch has two vertical lobes and an equatorial ring. These distinguish between-axis orientation from the special dz2 shape; the shapes are schematic.
A drawing of a two-lobed p orbital is one orbital, not two orbitals. Similarly, the four lobes of most d orbitals belong to one orbital. An orbital shape is a probability description; it is not the shape of an electron. Mathematical wave functions are outside this syllabus.
Fill orbitals, then remove electrons from the outer shell
Use electron count, Pauli exclusion and Hund's rule; remember Cr, Cu and 4s removal.
Count electrons first. Fill lower-energy orbitals before higher-energy orbitals. Two electrons in one orbital have opposite spins (Pauli exclusion). In a set of equal-energy orbitals, put one electron into each with parallel spins before pairing (Hund's rule). Thus the three 2p electrons of nitrogen occupy three different p orbitals, rather than pairing in one immediately.
| Species | Electron count | Ground-state configuration |
|---|---|---|
| N | 7 | 1s2 2s2 2p3 |
| Cl | 17 | 1s2 2s2 2p6 3s2 3p5 |
| Ca | 20 | [Ar]4s2 |
| Cr | 24 | [Ar]3d5 4s1 |
| Fe | 26 | [Ar]3d6 4s2 |
| Cu | 29 | [Ar]3d10 4s1 |
| Fe2+ | 24 | [Ar]3d6 |
| Fe3+ | 23 | [Ar]3d5 |
| Cu2+ | 27 | [Ar]3d9 |
Superscripts count electrons: 3d6 means six electrons distributed over five d orbitals. [Ar] abbreviates the 18-electron argon core. Chromium and copper have the ground-state exceptions shown above; a rigid filling diagram does not capture every balance of orbital energies and electron repulsions.
Worked example
An ion is not filled like a neutral atom with the same electron count
Write the configuration of Cr3+ (Z = 24).
- Neutral chromium is [Ar]3d5 4s1.
- Remove the 4s electron first, then two 3d electrons.
- Check the total: 18 + 3 = 21 electrons, matching 24 - 3.
Cr3+ is [Ar]3d3. For these transition-metal cations, remove 4s electrons before 3d electrons.
Check your understandingFe2+ and Cr both have 24 electrons. Must their configurations be the same?Think it through, then reveal the answer
Explain the attraction to the electron being removed
Nuclear charge, distance, shielding and pairing compete.
First ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms, forming one mole of gaseous singly positive ions: X(g) → X+(g) + e-. It is positive because energy is needed to separate an electron from the attraction of the nucleus. The gas state prevents lattice or intermolecular effects being included.
| Change | Typical effect on ionisation energy | Reason |
|---|---|---|
| Greater nuclear charge, similar shielding and distance | Increases | The electron is attracted more strongly. |
| An outer electron in a more distant shell | Decreases | Greater distance and extra inner-shell shielding weaken attraction. |
| More shielding between nucleus and electron | Decreases | Inner electrons reduce the nuclear attraction felt by the outer electron. |
| Electron removed from a higher-energy subshell | Often decreases | That electron is less tightly bound; compare B 2p with Be 2s. |
| Pairing in one orbital | Can decrease | Repulsion between paired electrons assists removal; compare O with N. |
Across a period, proton number rises while added electrons enter the same principal shell. Shielding does not increase enough to cancel the increased nuclear charge, so the general trend is increasing ionisation energy. Down a group, extra shells increase distance and shielding, usually outweighing the increased nuclear charge.
Worked example
Explain a dip, not just the general trend
Why is oxygen's first ionisation energy lower than nitrogen's despite oxygen having more protons?
- Both lose a 2p electron. Their inner-shell shielding is similar.
- Nitrogen has 2p3: one electron in each p orbital. Oxygen has 2p4, so one p orbital contains a pair.
- Removing a paired oxygen electron relieves additional repulsion. This effect outweighs the increased nuclear attraction in this comparison.
The dip is explained by paired-electron repulsion, not by claiming oxygen has fewer protons or a new principal shell.
A large jump reveals an inner shell
Count how many electrons are removed before the jump; use other evidence to determine the period.
Successive ionisation energies remove electrons one at a time from increasingly positive gaseous species. The second process is X+(g) → X2+(g) + e-. Values rise because the remaining electrons are attracted more strongly to the increasingly positive ion. A particularly large jump means the next electron is removed from a more tightly held inner shell.
Worked example
Reconstruct a period-3 atom
A main-group period-3 element has the first four ionisation energies 580, 1820, 2740 and 11600 kJ mol-1 (rounded teaching data). Deduce its outer configuration and group.
- The large jump is between the third and fourth removals: 11600/2740 is about 4.2, much larger than the previous ratio.
- Three electrons lie outside the inner core. For a period-3 main-group atom, they occupy 3s2 3p1.
- The group is 13. Combining period and group identifies aluminium; the complete configuration is [Ne]3s2 3p1.
Three outer electrons, group 13, period 3, [Ne]3s2 3p1. The period came from the question, not from the first jump alone.
For a main-group atom, one outer electron suggests group 1; two suggest group 2; three to eight suggest groups 13 to 18. Helium is the familiar two-electron exception in group 18. Do not apply this simple outer-count rule uncritically to transition elements, where 3d and 4s energies and occupations also matter.
- Find unusually large jumps
Compare the changes or ratios. On a logarithmic plot, read the axis before interpreting spacing.
- Count removals before each jump
This gives groups of electrons with substantially different binding energies.
- Combine with independent clues
Use proton number, total electron count or a stated period to identify the shell numbers.
- Check the proposed configuration
Its electron count and shell boundaries must explain the observed pattern.
Check your understandingThe first large jump is after removal of two electrons. Does this alone prove that the element is magnesium?Think it through, then reveal the answer
Quick revision
Revisit the essentials, then return to an explanation when you need it.
| Task | Reliable starting point |
|---|---|
| Count particles | p = Z; n = A - Z; e = Z - signed charge. |
| Compare deflection | Direction follows charge. At equal speed and field, compare |q|/m. |
| Count orbital capacity | s: 1 orbital/2 e; p: 3/6; d: 5/10. |
| Form transition-metal cations | Write the neutral configuration, then remove 4s before 3d. |
| Explain first IE | Compare nuclear charge, shielding, distance, subshell and pairing. |
| Read successive IE | A large jump indicates inner-shell removal; count electrons before it. |
Keep the distinctions: isotope versus ion; orbital versus subshell; electron count versus proton number; filling order versus removal order; general periodic trend versus a subshell or pairing exception.
Scope and references
Learning outcomes and sources
1. Atomic Structure. Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
1(a) Describe the three subatomic particles.
- Relative proton, neutron and electron charges
- Relative masses
1(b) Deduce particle-beam behaviour in an electric field.
- Proton, neutron and electron beam directions
- Mass and charge comparisons under stated speed conditions
1(c) Locate the mass and charges of an atom.
- Concentrated nuclear mass and positive charge
- Negative electrons outside the nucleus
1(d) Calculate particle counts in atoms and ions.
- Proton and nucleon numbers
- Positive and negative ion charges
1(e) Explain nuclear composition and isotopes.
- (i) Proton number and nucleon number
- (ii) Same protons, different neutron counts in isotopes
1(f) Describe orbital numbers and relative energies.
- s, p and d subshells for n = 1, 2, 3
- 4s and 4p
- One, three and five orbitals in s, p and d
1(g) Recognise s, p and d orbital shapes.
- Spherical s
- Three two-lobed p orientations
- Five d orbitals including dz2
- No wave-function mathematics
1(h) Write configurations of atoms and ions.
- Proton number and charge
- Orbital filling and electron pairing
- Cr and Cu; 4s removal in transition-metal ions
1(i) Explain ionisation-energy differences.
- Nuclear attraction, shielding, distance
- Subshell energy and paired-electron repulsion
- Data Booklet and periodic trends
1(j) Deduce configurations from successive ionisation energies.
- Increasing positive charge
- Large jumps and shell boundaries
- Combining energy evidence with particle count or period
1(k) Link successive ionisation energies to periodic position.
- Main-group outer-electron count
- Group inference
- Limits on identifying a period from a partial succession
- SEAB H2 Chemistry 9476, examination 2026
Topic 1, printed page 13; all 11 lettered outcomes and nested parts inspected. Explanations, sketches and worked questions are original.
- SEAB H2 Chemistry 9476, examination 2027
Current 9476 course identity; local 2027 syllabus text supplied for comparison.