Revise atoms, ions, isotopes, mass spectra, electron configurations and ionisation energies with explanations and worked practice.
Created by Dave Miller, Chemistry teacher, tutor and current AQA examiner with 15+ years of teaching experience. Independent resources for AQA and OCR A.
OCR A · Atomic structure · Notes and practice
Learn → practise → check
Atoms & ions
5 practice questions
How evidence changed the atomic model
Learning goalLink experimental observations to changes in the accepted atomic model.
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Scientists do not keep a model just because it is familiar. When new evidence does not fit, the model needs to change. For each model below, connect what scientists observed with what that tells us about the atom.
For example, most alpha particles passing through the foil suggests that most of the atom is empty space. The few large deflections suggest that the positive charge is concentrated in a very small region. Those are two different observations supporting two different conclusions.
Model
Evidence and interpretation
Dalton (solid sphere model)
Atoms pictured as tiny, solid, indivisible spheres.
J. J. Thomson (plum pudding model)
Discovery of the electron showed atoms were divisible; electrons were placed within diffuse positive charge.
Rutherford (nuclear model)
Alpha scattering showed a tiny, dense, positively charged nucleus and an atom that is mostly empty space.
Bohr (planetary model)
Electrons occupy fixed energy levels, helping explain periodic properties.
Chadwick
Discovery of the neutron completed the nuclear model and helped explain isotope masses.
Worked example
Observation → inference
Most alpha particles pass through thin foil; some undergo large deflections and a few rebound.
Explain what these observations tell you about the structure of the atom.
Worked answer
Most pass through: most of the atom is empty space.
Rare large deflections: positive charge and most mass are concentrated in a tiny, dense nucleus.
Connect each observation to the conclusion it supports.
EXAM TRAPA list of scientists is not an explanation. Link the evidence to the change in the model.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Quick check 1
Explain how the discovery of electrons challenged Dalton’s indivisible-atom model. [2]
State one feature of Bohr’s model and explain how it helps account for repeating chemical properties. [2]
Total: 4 marks
Show answer & marking points
Atoms contain smaller negatively charged particles [1], so they are not indivisible [1].
Electrons occupy fixed energy levels [1]; repeating outer-shell arrangements help explain repeating chemical properties [1].
✓ DONE WHENYou can explain how evidence led from a solid-sphere model to a nuclear atom with energy levels.
Fundamental particles
Learning goalRecall the relative charge, mass and location of each fundamental particle.
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An atom has a tiny central nucleus containing protons and neutrons. Electrons occupy regions outside the nucleus. The table gives the relative masses and charges you need to know.
Protons and neutrons are much heavier than electrons, so almost all the atom's mass is in its nucleus. An electron's mass is very small, but it is not zero.
In a neutral atom, each positive proton is balanced by one negative electron. That is why a neutral atom has equal numbers of protons and electrons.
Particle
Relative charge
Relative mass
Location
Proton
+1
1
Nucleus
Neutron
0
1
Nucleus
Electron
−1
Approximately 1/1840
Orbitals around the nucleus
EXAM TRAPElectron mass is very small, not zero.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 2
Complete the table for the three fundamental particles. [9]
Particle
Relative charge
Relative mass
Location
Proton
…
…
…
Neutron
…
…
…
Electron
…
…
…
Total: 9 marks
Show answer & marking points
Particle
Relative charge
Relative mass
Location
Proton
+1
1
Nucleus
Neutron
0
1
Nucleus
Electron
−1
Approximately 1/1840
Region around the nucleus
One mark for each correct entry: 9 marks. The electron has a very small mass, not zero mass.
✓ DONE WHENYou can reproduce the particle table and describe the nuclear atom.
Particle numbers in atoms and ions
Learning goalCalculate protons, neutrons and electrons using A, Z and charge.
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The atomic number tells you which element you have: it is the number of protons. The mass number counts protons and neutrons, so subtract the atomic number to find the neutrons. Both numbers are whole numbers with no units.
Find the neutral atom first. Its proton number and electron number both equal the atomic number.
For a positive ion, remove electrons. A 2+ charge means two electrons have been lost.
For a negative ion, add electrons. A 1- charge means one electron has been gained.
Leave the proton and neutron numbers alone. Making an ion changes the electrons, not the nucleus.
Atomic number, Z
The number of protons in the nucleus. It identifies the element.
Mass number, A
The total number of protons and neutrons in one atom of an isotope.
Protons = Z; neutrons = A − Z
A mass number: total number of protons and neutrons; no unit; Z atomic number: number of protons; no unit.
Worked example
Count an ion’s particles
State the numbers of protons, neutrons and electrons in a ⁵⁹Co²⁺ ion. Cobalt has atomic number 27.
Worked answer
Protons = 27.
Neutrons = 59 − 27 = 32.
Electrons = 27 − 2 = 25.
27 protons, 32 neutrons and 25 electrons.
EXAM TRAPApply the sign of the charge correctly: positive means fewer electrons.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Quick check 3
An ion is represented as ⁵⁶Fe²⁺. The atomic number of iron is 26. State the numbers of protons, neutrons and electrons. [3]
Total: 3 marks
Show answer & marking points
Protons = 26. [1]
Neutrons = 56 − 26 = 30. [1]
Electrons = 26 − 2 = 24. [1]
Quick check 4
An ion contains 17 protons, 18 neutrons and 18 electrons. State its element, mass number and charge. Use atomic numbers S = 16, Cl = 17 and Ar = 18. [3]
Total: 3 marks
Show answer & marking points
Chlorine: Z = 17. [1]
Mass number = 17 + 18 = 35. [1]
Charge = −1: one more electron than proton. [1]
Exam-style question 5
Complete the missing entries. [6]
Species
Atomic number
Mass number
Protons
Neutrons
Electrons
²⁷Al³⁺
13
27
…
…
…
³⁷Cl⁻
17
37
…
…
…
Total: 6 marks
Show answer & marking points
Species
Protons
Neutrons
Electrons
²⁷Al³⁺
13
14
10
³⁷Cl⁻
17
20
18
One mark for each correct particle count: 6 marks.
✓ DONE WHENYou can move between nuclide notation and particle counts in both directions.
Isotopes & relative masses
6 practice questions
Isotopes and their properties
Learning goalIdentify isotopes and explain similarities and differences.
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Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. The proton number keeps the element the same; the extra neutrons change its mass.
Neutral atoms of the isotopes have the same electron configuration. Since chemical reactions depend on electrons, the isotopes have the same chemical properties.
Their masses are different, so physical properties that depend on mass can differ. Keep mass number, which describes one isotope, separate from the element's weighted mean relative atomic mass.
Isotopes
Atoms of the same element with the same number of protons but different numbers of neutrons and different masses.
Worked example
Compare ³⁵Cl and ³⁷Cl
For neutral atoms of ³⁵Cl and ³⁷Cl, state the numbers of protons, neutrons and electrons. Explain why they have the same chemical properties but different masses. Chlorine has atomic number 17.
Worked answer
Isotope
Protons
Neutrons
Electrons
³⁵Cl
17
18
17
³⁷Cl
17
20
17
Same electron configuration; different neutron numbers and masses.
EXAM TRAPIon charge does not change which isotope a species belongs to. Use proton and neutron numbers.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 6
Define isotope. [2]
Explain why isotopes of the same element have the same chemical properties. [1]
Explain why they can have different physical properties. [2]
Total: 5 marks
Show answer & marking points
Atoms of the same element with the same number of protons but different numbers of neutrons and different masses. [2]
Their neutral atoms have the same electron configuration. [1]
Different neutron numbers give different masses [1], so mass-dependent physical properties can differ [1].
Exam-style question 7
Species
Protons
Neutrons
Electrons
A
17
18
18
B
17
20
17
C
18
20
18
D
18
22
20
Identify both pairs that are species of different isotopes of the same element. [1]
Justify the pairs using the particle data. [1]
Total: 2 marks
Show answer & marking points
A and B; C and D. Both pairs required. [1]
Each pair has the same number of protons but different numbers of neutrons. Ionic charge does not change isotope identity. [1]
✓ DONE WHENYou can define isotopes precisely and identify every valid isotope pair from particle data.
The carbon-12 relative-mass scale
Learning goalDistinguish mass number, relative isotopic mass and relative atomic mass.
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A relative mass compares a particle's mass with a common reference: one-twelfth of the mass of one carbon-12 atom. It is a ratio of two masses, so it has no unit.
Read the particle named in the definition. Relative isotopic mass refers to one isotope. Relative atomic mass is a weighted mean that allows for an element's isotope abundances.
Relative molecular mass refers to molecules. For a substance without separate molecules, such as an ionic compound, use relative formula mass. Molar mass is a different quantity: the mass of one mole, usually in g mol⁻¹.
Relative isotopic mass
The mass of an atom of one isotope compared with one-twelfth of the mass of a carbon-12 atom.
Relative atomic mass, Aᵣ
The weighted mean mass of an atom of an element compared with one-twelfth of the mass of a carbon-12 atom.
Quantity
Meaning
Mass number
Whole-number count of protons plus neutrons in one atom.
Relative isotopic mass
Measured relative mass of one isotope; close to, but not necessarily exactly equal to, mass number.
Relative atomic mass
Weighted mean across the isotopes present; often not a whole number.
EXAM TRAPInclude “weighted mean”, “atom of an element” and the one-twelfth carbon-12 comparison in Aᵣ.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Quick check 8
Define atomic number. [1]
Define mass number. [1]
Explain why mass number is a whole number whereas relative atomic mass often is not. [2]
Total: 4 marks
Show answer & marking points
The number of protons in the nucleus of an atom. [1]
The total number of protons and neutrons in the nucleus of one atom. [1]
Mass number counts whole particles in one isotope [1]. Relative atomic mass is a weighted mean including isotope abundances [1].
Quick check 9
Define relative isotopic mass. [2]
Define relative atomic mass, Aᵣ. [2]
Total: 4 marks
Show answer & marking points
The mass of an atom of one isotope compared with one-twelfth of the mass of a carbon-12 atom. [2]
The weighted mean mass of an atom of an element compared with one-twelfth of the mass of a carbon-12 atom. [2]
✓ DONE WHENYou can give the required definitions and distinguish a particle count from a relative mass.
Relative molecular and formula masses
Learning goalChoose the correct term and calculate relative masses from formulae.
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The formula tells you how many atoms to include. Add their Aᵣ contributions to find the relative molecular mass or relative formula mass.
Use relative molecular mass for substances made of separate molecules. Use relative formula mass when the formula describes a ratio in a giant structure, such as NaCl. Both relative masses have no units.
Use these terms and calculate the values; formal definitions of these two terms are not required here.
Structure
Term
Examples
Discrete molecules
Relative molecular mass, Mᵣ
CO₂, H₂O, NH₃
Giant ionic or covalent structure
Relative formula mass
NaCl, MgCl₂, SiO₂
Relative mass = Σ(Aᵣ × number of each atom in the formula)
Ar relative atomic mass; no unit; Σ sum of all the indicated contributions.
Worked example
Choose the term and calculate
Use C = 12.0, O = 16.0, Mg = 24.3 and Cl = 35.5 to calculate relative masses for CO₂ and MgCl₂.
Worked answer
CO₂: 12.0 + 2(16.0) = 44.0. It contains molecules.
MgCl₂: 24.3 + 2(35.5) = 95.3. It has a giant ionic lattice.
Mᵣ(CO₂) = 44.0; relative formula mass(MgCl₂) = 95.3.
EXAM TRAPMolar mass in g mol⁻¹ is different from a unitless relative mass.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Quick check 10
Explain the difference between relative molecular mass and relative formula mass. State when each term is used and whether either quantity has a unit. [4]
Total: 4 marks
Show answer & marking points
Relative molecular mass sums the Aᵣ values in one molecule and is used for discrete molecules. [1]
Relative formula mass sums the Aᵣ values in one formula unit. [1]
Use relative formula mass for ionic compounds or giant structures without discrete molecules. [1]
Neither has units. [1]
Exam-style question 11
Use Aᵣ: H 1.0, C 12.0, O 16.0, S 32.1, Ca 40.1.
Calculate Mᵣ of H₂SO₄. [2]
Calculate the relative formula mass of CaCO₃. [2]
Total: 4 marks
Show answer & marking points
2(1.0) + 32.1 + 4(16.0) [1] = 98.1 [1].
40.1 + 12.0 + 3(16.0) [1] = 100.1 [1].
✓ DONE WHENYou can select molecular or formula mass from structure and calculate it without units.
Mass spectra & calculations
11 practice questions
Reading elemental mass spectra
Learning goalRead relative isotopic masses and abundances from singly charged ions.
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Use the information on the spectrum. You do not need TOF instrument stages or kinetic-energy calculations here.
Peak position: read m/z from the x-axis. For a singly charged monatomic ion, this is numerically equal to its relative isotopic mass.
Peak height: read relative abundance or intensity from the y-axis. The tallest isotope peak represents the most abundant isotope.
Check whether abundances are percentages, ratios or raw intensities before calculating the weighted mean.
EXAM TRAPRelative intensities need not add to 100. Only percentages must total 100%.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Quick check 12
State the quantity on the horizontal axis of a mass spectrum. [1]
State the quantity on the vertical axis. [1]
Explain why m/z is numerically equal to relative isotopic mass in a simple elemental spectrum of singly charged ions. [1]
State what the relative abundance of a peak represents. [1]
Total: 4 marks
Show answer & marking points
Mass-to-charge ratio, m/z. [1]
Relative abundance or relative intensity. [1]
The charge number z = 1, so dividing by z does not alter the numerical mass value. [1]
The abundance of that isotope ion relative to the others detected. [1]
Quick check 13
m/z
Relative abundance
24
79
25
10
26
11
Calculate Aᵣ and identify the element. Reference Aᵣ values: Na = 23.0, Mg = 24.3, Al = 27.0. [4]
✓ DONE WHENYou can extract isotope masses and abundances without describing the instrument.
Calculating relative atomic mass
Learning goalSolve weighted means and unknown masses or abundances.
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Relative atomic mass is an average, but the isotopes do not all contribute equally. A common isotope must count more than a rare one. That is what weighted mean means.
Multiply each isotope's relative mass by its abundance.
Add those contributions.
Divide by the total abundance. Use 100 for percentages, but add the values yourself for a ratio or raw intensities.
Your answer must lie between the lightest and heaviest isotope masses. It should be closer to the more abundant isotope. Relative atomic mass has no unit.
Aᵣ = Σ(relative isotopic mass × abundance) / Σ(abundances)
Ar relative atomic mass; no unit; Σ sum of all the indicated contributions.
Worked example
All masses and abundances known
Masses 28, 29 and 30 have abundances 92%, 5% and 3%.
Calculate the relative atomic mass of the element.
Mass 63 has abundance 70.0%. The other isotope has abundance 30.0%. Aᵣ = 63.60.
Calculate the relative isotopic mass of the second isotope.
Worked answer
63.60 = [63(70) + 30x]/100.
6360 = 4410 + 30x.
x = (6360 − 4410)/30.
The other relative isotopic mass is 65.0.
Worked example
Two unknown abundances
Isotope masses are 50, 52 and 54. Mass 50 has abundance 20.0%, and Aᵣ = 52.20.
Calculate the percentage abundances of isotopes 52 and 54.
Worked answer
Let mass 52 have abundance x%. Mass 54 has (80 − x)%.
52.20 = [50(20) + 52x + 54(80 − x)]/100.
5220 = 5320 − 2x, so x = 50.0.
Mass 52: 50.0%; mass 54: 30.0%.
EXAM TRAPFor three isotopes, the final abundance is 100 − known abundance − x. Keep unrounded intermediate values.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 14
Two isotopes have accurate relative isotopic masses 78.918 and 80.916. The lighter : heavier abundance ratio is 4.35 : 1.00. Calculate Aᵣ to three decimal places. [4]
Total: 4 marks
Show answer & marking points
Use the mass–ratio products: 78.918(4.35) + 80.916(1.00). [1]
An isotope of relative isotopic mass 63 has abundance 69.0%. A second isotope has relative isotopic mass x and abundance 31.0%. Aᵣ = 63.62. Calculate x. [4]
Total: 4 marks
Show answer & marking points
63.62 = [63(69.0) + 31.0x]/100. [1]
6362 = 4347 + 31x. [1]
31x = 2015. [1]
x = 65. [1]
Exam-style question 16
An element has relative isotopic masses 24, 25 and 26. Mass 24 has abundance 79.0%, and Aᵣ = 24.32. Let the abundance of mass 25 be x%. Calculate the abundances of masses 25 and 26. [5]
Total: 5 marks
Show answer & marking points
Mass 26 has abundance (21 − x)%. [1]
24.32 = [24(79) + 25x + 26(21 − x)]/100. [1]
2432 = 2442 − x. [1]
Mass 25: x = 10.0%. [1]
Mass 26: 11.0%. [1]
Exam-style question 17
An element has isotopes of relative isotopic mass 10 and 11. The mass-10 abundance is x%, and Aᵣ = 10.81. Calculate each percentage abundance. [4]
Total: 4 marks
Show answer & marking points
Mass 11 has abundance (100 − x)%. [1]
10.81 = [10x + 11(100 − x)]/100. [1]
1081 = 1100 − x, so x = 19. [1]
Mass 10: 19%; mass 11: 81%. [1]
Exam-style question 18
A student says the following data cannot be used because the intensities do not add to 100.
m/z
Relative intensity
90
74.0
91
15.0
92
24.5
94
25.0
96
4.0
Explain why the student is incorrect. [1]
Use the data to show that Aᵣ = 91.32. [4]
Total: 5 marks
Show answer & marking points
Relative intensities are a comparative scale. They need not total 100; use their actual total. [1]
R has isotopes of relative isotopic mass 64, 66 and 68. Mass 68 has abundance 25.0%; the mass-64 : mass-66 abundance ratio is 3 : 2.
Calculate each percentage abundance. [3]
Calculate Aᵣ. [3]
Total: 6 marks
Show answer & marking points
75.0% remains [1]. Divide this in the ratio 3 : 2, so one part is 15.0% [1]. Abundances are 45.0%, 30.0%, 25.0% for masses 64, 66, 68 respectively [1].
Weighted numerator = 64(45.0) + 66(30.0) + 68(25.0) [1]. Divide by 100 [1] to obtain 65.60 [1].
Exam-style question 20
The table gives the isotope data for element X.
Relative isotopic mass
58
60
61
62
Abundance / %
68.0
x
1.1
3.6
Calculate x. [1]
Calculate Aᵣ of X to two decimal places. [4]
X forms ionic XCl₂. Calculate its relative formula mass using Aᵣ(Cl) = 35.45. [2]
Explain why relative formula mass is used for XCl₂. [1]
Total: 8 marks
Show answer & marking points
x = 100 − 68.0 − 1.1 − 3.6 = 27.3%. [1]
Use all four weighted mass–abundance products [1]: 58(68.0) + 60(27.3) + 61(1.1) + 62(3.6) = 5872.3 [1]. Divide by 100 [1]: Aᵣ = 58.723 = 58.72 [1].
Use the unrounded Aᵣ: 58.723 + 2(35.45) [1] = 129.623 = 129.62 [1].
XCl₂ has a repeating ionic lattice, not discrete molecules. [1]
✓ DONE WHENYou can handle percentages, ratios, unnormalised intensities and missing information.
Molecular ions: analytical crossover
Learning goalConnect an identified molecular ion to Mᵣ and molecular formula.
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This is an analytical-chemistry crossover, included to support the molecular-ion questions in your practice pack. It connects this topic to the later organic-analysis content.
A molecular ion is the intact molecule after losing an electron. For a singly charged molecular ion, its m/z gives the molecule's relative molecular mass.
The molecular-ion peak is usually the peak furthest to the right. A small M+1 isotope peak can appear just beyond it.
A molecule can also break into smaller ions called fragments. The tallest peak is the base peak; it need not be the molecular-ion peak. Identify the peak the question tells you to use.
Molecular ion, M⁺
The intact molecule after removal of one electron. For the identified singly charged molecular ion, m/z gives its relative mass.
Base peak
The most intense peak, conventionally scaled to 100. It may represent a fragment ion.
Mᵣ = m/z of the identified singly charged molecular ion
Mr relative molecular or formula mass; no unit; m/z mass-to-charge ratio: m represents the ion’s relative mass and z its charge number.
Molecular-formula multiplier = Mᵣ / empirical-formula mass
Mr relative molecular or formula mass; no unit.
Worked example
Use the identified ion
A spectrum has an identified singly charged molecular ion at m/z 72 and a base peak at m/z 43.
Determine the relative molecular mass of the compound and explain which peak you use.
Worked answer
Use the intact molecular ion, not the most intense fragment.
Mᵣ = 72.
EXAM TRAPDo not automatically select the tallest or highest-m/z peak. Isotopic peaks and fragmentation require interpretation.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 21
A compound’s spectrum has an identified singly charged molecular-ion peak at m/z 86 and a base peak at m/z 43.
State Mᵣ. [1]
State what is meant by the base peak. [1]
Explain why the base peak is not necessarily used to determine Mᵣ. [1]
Total: 3 marks
Show answer & marking points
Mᵣ = 86. [1]
The most intense peak, conventionally assigned relative intensity 100. [1]
It may be a fragment ion and therefore may not represent the intact molecule. [1]
Exam-style question 22
Stretch / crossover: this combines mass spectra with empirical and molecular formulae. A compound contains 54.5% C, 9.1% H and 36.4% O by mass. Its identified singly charged molecular-ion peak is at m/z 88; its base peak is at m/z 43. Use Aᵣ: C 12.0, H 1.0, O 16.0.
Determine its empirical formula. [4]
State Mᵣ. [1]
Determine its molecular formula. [2]
Explain why the m/z 43 peak is not used to determine Mᵣ. [1]
Total: 8 marks
Show answer & marking points
Divide by Aᵣ: C 54.5/12 = 4.5417…, H 9.1/1 = 9.10, O 36.4/16 = 2.275 [1]. Divide all by 2.275 [1] to obtain approximately 2 : 4 : 1 [1]. Empirical formula C₂H₄O [1].
Mᵣ = 88. [1]
Empirical-formula mass = 44, so multiplier = 88/44 = 2 [1]. Molecular formula C₄H₈O₂ [1].
The m/z 43 base peak represents a fragment rather than the intact molecular ion. [1]
✓ DONE WHENYou can use an identified molecular ion and distinguish molecular mass from fragment mass.
Shells & orbitals
6 practice questions
The first four shells
Learning goalDistinguish shell capacity from filling order.
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A shell is a principal energy level. We number the shells n = 1, 2, 3 and 4, starting nearest the nucleus. Each shell contains subshells, which contain orbitals.
The expression 2n² gives the maximum capacity of a shell: 2, 8, 18 and 32 electrons for the first four shells. It does not tell you the order in which all the subshells fill.
That distinction matters when you reach 4s and 3d. Use shell capacities for maximum numbers, and the filling-order diagram for actual configurations.
Maximum shell capacity = 2n²
n the principal energy level (electron shell).
Worked example
Third-shell capacity
Calculate the maximum number of electrons that can occupy the third shell, using 2n².
Worked answer
n = 3, so 2 × 3² = 18.
The third shell can hold 18 electrons.
EXAM TRAPCapacity is not filling order. Potassium starts 4s before its third shell reaches maximum capacity.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 23
State the maximum number of electrons in each of the first four shells. [4]
Use 2n² to calculate the fourth-shell capacity. [1]
Total: 5 marks
Show answer & marking points
Shells 1–4: 2, 8, 18, 32 respectively. [4]
2 × 4² = 32 electrons. [1]
✓ DONE WHENYou can recall 2, 8, 18, 32 and use 2n² without confusing capacity and filling.
Atomic orbitals and shapes
Learning goalDefine an orbital and recognise s and p shapes.
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An orbital describes a region with a high probability of finding an electron. It is not a little circular track followed by the electron.
An s orbital is spherical. A p orbital has two lobes, often called a dumb-bell shape. The nucleus is at the centre.
A p subshell has three orbitals of the same energy. They point in three mutually perpendicular directions. Each orbital can hold up to two electrons, with opposite spins.
When showing where electrons go in a diagram, orbitals are usually drawn as boxes. One box represents one orbital; one arrow represents one electron.
Atomic orbital
A region around the nucleus that can hold up to two electrons with opposite spins.
EXAM TRAPA p orbital is a three-dimensional region, not a figure-of-eight path.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 24
Define atomic orbital. [2]
Identify the type of orbital shown by A and by B–D. [2]
Describe the shape and relative arrangement of the three p orbitals. [2]
Total: 6 marks
Show answer & marking points
A region around the nucleus [1] that can hold up to two electrons with opposite spins [1].
A is an s orbital; B–D are p orbitals. [2]
Each p orbital is dumb-bell shaped [1]; the three have the same shape and energy and point in directions at right angles [1].
✓ DONE WHENYou can give the definition and draw or recognise the required orbital shapes.
Subshells, orbitals and capacities
Learning goalConnect orbital counts with subshell capacities.
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Count the orbitals first, then double the number to find the maximum electrons. Each orbital holds up to two electrons.
An s subshell has one orbital, so it holds up to 2 electrons.
A p subshell has three orbitals, so it holds up to 6 electrons.
A d subshell has five orbitals, so it holds up to 10 electrons.
An orbital and a subshell are different levels of organisation. Only the s subshell happens to contain a single orbital.
Subshell
Orbitals
Electrons per orbital
Maximum electrons
s
1
2
2
p
3
2
6
d
5
2
10
Worked example
Why does d hold ten?
Explain the capacity of a d subshell.
Worked answer
There are five d orbitals.
Each holds at most two electrons with opposite spins.
5 × 2 = 10.
A d subshell holds up to ten electrons.
EXAM TRAPOne p orbital holds two electrons; the whole p subshell holds six.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 25
Complete the table. [6]
Subshell
Number of orbitals
Maximum electrons
s
…
…
p
…
…
d
…
…
Total: 6 marks
Show answer & marking points
Subshell
Orbitals
Maximum electrons
s
1
2
p
3
6
d
5
10
One mark for each correct entry: 6 marks.
✓ DONE WHENYou can move between subshell, orbital count and electron capacity.
Electrons-in-box diagrams
Learning goalFill equal-energy orbitals singly before pairing and show spins.
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In an electrons-in-boxes diagram, each box is one orbital and each arrow is one electron. The arrow direction represents spin.
For equal-energy orbitals, place one electron in each box before adding a second electron to any box.
Draw the singly occupied orbitals with parallel spins in these ground-state arrangements.
When two electrons share a box, draw opposite arrows. One orbital holds no more than two electrons.
A p subshell therefore uses three boxes; a d subshell uses five. Check both the total electron count and the way they are distributed.
EXAM TRAPFor p², use two different orbitals. Do not pair while an equal-energy orbital is empty.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Quick check 26
Complete an electrons-in-box diagram for each arrangement. Draw three p-orbital boxes on paper and show spins clearly.
Carbon: 2p². [2]
Fluorine: 2p⁵. [2]
Total: 4 marks
Show answer & marking points
Carbon: separate orbitals [1] with parallel spins [1].
Subshell
Orbital 1
Orbital 2
Orbital 3
2p²
↑
↑
Empty
Fluorine: two pairs and one single electron [1], with opposite spins within each pair [1].
Subshell
Orbital 1
Orbital 2
Orbital 3
2p⁵
↑↓
↑↓
↑
Exam-style question 27
Nitrogen and oxygen have 2p³ and 2p⁴ arrangements.
Draw their 2p electrons-in-box diagrams. [2]
Use the diagrams to explain the singly-before-pairing rule. [2]
State the spin requirement for two electrons in the same orbital. [1]
Total: 5 marks
Show answer & marking points
Subshell
Orbital 1
Orbital 2
Orbital 3
N, 2p³
↑
↑
↑
Subshell
Orbital 1
Orbital 2
Orbital 3
O, 2p⁴
↑↓
↑
↑
One mark for each diagram. [2]
Equal-energy orbitals are occupied singly before pairing [1]. Nitrogen has one electron in each; oxygen pairs only after all three are occupied [1].
Opposite spins. [1]
Exam-style question 28
Three proposed 2p⁴ arrangements are shown. Two arrows in the same cell occupy one orbital.
Diagram
Orbital 1
Orbital 2
Orbital 3
A
↑↓
↑
↑
B
↑↓
↑↓
Empty
C
↑↑
↑
↑
Identify the correct diagram. [1]
Explain the error in each other diagram. [4]
Total: 5 marks
Show answer & marking points
A. [1]
B pairs electrons while one equal-energy orbital is empty [1], breaking singly-before-pairing [1]. C places same-spin electrons in one orbital [1]; a pair must have opposite spins [1].
✓ DONE WHENYou can construct and diagnose box diagrams using occupancy and spin rules.
Electron configurations
14 practice questions
Filling order and electron configurations
Learning goalBuild full and shorthand configurations for atoms up to Z = 36.
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Start by counting the electrons. A neutral atom has one electron for every proton, so its total equals its atomic number.
Follow the filling order shown in the diagram.
Use the capacities s², p⁶ and d¹⁰ as you fill each subshell.
Stop when all the atom's electrons have been placed.
Add the superscripts as a final check: their total must equal Z.
In a configuration, the number names the shell, the letter names the subshell and the superscript counts its electrons. [Ar] is shorthand for the complete 18-electron argon core.
The 4s and 3d energies are close and depend on occupancy. Use the stated filling order without assuming one fixed energy ordering applies to every atom and ion.
1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p
Worked example
Titanium, Z = 22
Write the full electron configuration and the noble-gas shorthand configuration of a neutral titanium atom. Titanium has atomic number 22.
A student writes selenium as [Ar] 3d¹⁰ 4s² 4p⁴ and says this proves that 3d fills before 4s.
State the filling sequence immediately after 3p. [1]
Write selenium’s full configuration. [2]
Explain why the written order does not prove the filling order. [2]
Check its electron total. [1]
Total: 6 marks
Show answer & marking points
4s → 3d → 4p. [1]
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴. [2]
Writing 3d before 4s is a convention [1]. The writing order does not change the building-up sequence, in which 4s fills before 3d [1].
2 + 2 + 6 + 2 + 6 + 2 + 10 + 4 = 34. [1]
✓ DONE WHENYou can write, translate and check any required neutral-atom configuration through krypton.
The chromium and copper exceptions
Learning goalRecall the observed neutral-atom configurations of Cr and Cu.
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EXCEPTIONS · CHROMIUM & COPPER
Chromium and copper are the two exceptions to remember when using the simple filling order. Their measured configurations contain 4s¹, rather than the 4s² you might first predict.
Chromium has a half-filled 3d⁵ subshell; copper has a filled 3d¹⁰ subshell. The 3d and 4s energies are close, and these observed arrangements have lower total energy than the simple predictions.
Use Cr, Z = 24, and Cu, Z = 29, to check the total electrons. Learn their observed configurations without turning the half-filled/full-filled explanation into a rule for every element.
Atom
Simple prediction
Observed configuration
Cr
[Ar] 3d⁴ 4s²
[Ar] 3d⁵ 4s¹
Cu
[Ar] 3d⁹ 4s²
[Ar] 3d¹⁰ 4s¹
EXAM TRAPThese are neutral-atom exceptions. The ion configurations in this topic are limited to s- and p-block elements.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Quick check 33
For each atom, write the simple predicted configuration and the observed configuration using [Ar].
Write the simple predicted configuration using [Ar]. [2]
Identify Y and write its observed configuration. [2]
Represent the 3d and 4s electrons of the observed atom in boxes. [3]
Total: 7 marks
Show answer & marking points
[Ar] 3d⁹ 4s². [2]
Copper [1]: [Ar] 3d¹⁰ 4s¹ [1].
Subshell
Orbital 1
Orbital 2
Orbital 3
Orbital 4
Orbital 5
3d
↑↓
↑↓
↑↓
↑↓
↑↓
Subshell
Orbital 1
4s
↑
Correct 3d occupancy and paired spins [2]; one 4s electron [1].
✓ DONE WHENYou can state both exceptions exactly and check their electron totals.
Electron configurations of s- and p-block ions
Learning goalApply charge and remove or add electrons to the correct subshell.
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Required scope: s- and p-block ions of elements up to Z = 36. Transition-metal ion configurations are outside this requirement.
Write the neutral atom's electron configuration.
Use the charge to work out how many electrons it has lost or gained.
For a positive ion, remove electrons from the highest occupied principal shell, taking p before s within that shell.
For a negative ion, add electrons to the available outer subshell.
Check the final electron total.
For Ga³⁺, remove the 4p electron and both 4s electrons, leaving [Ar] 3d¹⁰. Do not treat the ion as a new neutral atom and blindly refill its electrons.
Ion
Neutral atom configuration
Ion configuration
Electron count
O²⁻ (Z = 8)
O: 1s² 2s² 2p⁴
[Ne]
10
Na⁺ (Z = 11)
Na: [Ne] 3s¹
[Ne]
10
Ca²⁺ (Z = 20)
Ca: [Ar] 4s²
[Ar]
18
Br⁻ (Z = 35)
Br: [Ar] 3d¹⁰ 4s² 4p⁵
[Kr]
36
Worked example
A p-block ion with an occupied d subshell
A neutral gallium atom has electron configuration [Ar] 3d¹⁰ 4s² 4p¹ and atomic number 31. Write the electron configuration of Ga⁺ and explain which electron is removed.
Worked answer
Ga⁺ has 31 − 1 = 30 electrons.
Remove the outer 4p electron.
Keep 3d¹⁰ and 4s²; check 18 + 10 + 2 = 30.
Ga⁺: [Ar] 3d¹⁰ 4s².
EXAM TRAPA positive charge means fewer electrons; a negative charge means more.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 35
Atomic numbers: Al 13, S 16, Br 35.
Calculate the electron count and write the noble-gas configuration for each ion.
Al³⁺. [2]
S²⁻. [2]
Br⁻. [2]
Total: 6 marks
Show answer & marking points
13 − 3 = 10 electrons [1]; [Ne] [1].
16 + 2 = 18 electrons [1]; [Ar] [1].
35 + 1 = 36 electrons [1]; [Kr] [1].
Exam-style question 36
Gallium has Z = 31.
Write the configuration of Ga using [Ar]. [2]
Calculate the electron count of Ga³⁺. [1]
Write its configuration using [Ar]. [1]
Identify the three electrons removed. [2]
Total: 6 marks
Show answer & marking points
Ga: [Ar] 3d¹⁰ 4s² 4p¹. [2]
31 − 3 = 28. [1]
Ga³⁺: [Ar] 3d¹⁰. [1]
Remove 4p¹ [1] and then 4s² [1].
Exam-style question 37
A 2− ion of a p-block element has 18 electrons.
Atomic numbers: Si 14, P 15, S 16, Cl 17, Ar 18.
Calculate the neutral atom’s Z and identify it. [2]
Write the ion’s full configuration. [2]
Draw the ion’s 3p box diagram. [2]
Total: 6 marks
Show answer & marking points
Z = 18 − 2 = 16 [1]; sulfur [1].
1s² 2s² 2p⁶ 3s² 3p⁶. [2]
Subshell
Orbital 1
Orbital 2
Orbital 3
3p⁶
↑↓
↑↓
↑↓
Three filled orbitals [1], opposite spins in each pair [1].
Exam-style question 38
A²⁻, B⁻, C and D⁺ each contain 18 electrons. A–D are consecutive elements in increasing atomic number.
Atomic numbers: Si 14, P 15, S 16, Cl 17, Ar 18, K 19, Ca 20.
Identify A, B, C and D. [4]
Write their shared configuration. [1]
Total: 5 marks
Show answer & marking points
A = S; B = Cl; C = Ar; D = K. [4]
[Ar]. [1]
Exam-style question 39
Gallium has Z = 31; germanium has Z = 32.
Write each neutral configuration using [Ar]. [2]
Calculate the electron counts of Ga³⁺ and Ge⁴⁺. [2]
Write each ion’s configuration. [2]
Explain why these ions have the same configuration. [1]
Total: 7 marks
Show answer & marking points
Ga: [Ar] 3d¹⁰ 4s² 4p¹; Ge: [Ar] 3d¹⁰ 4s² 4p². [2]
Ga³⁺: 31 − 3 = 28; Ge⁴⁺: 32 − 4 = 28. [2]
Both are [Ar] 3d¹⁰. [2]
Both lose their occupied outer 4p and 4s electrons and retain the same 28-electron arrangement. [1]
✓ DONE WHENYou can build and check the permitted ion configurations, including p-block cations with filled 3d.
Putting electron structure together
Learning goalTranslate between particle counts, notation, configurations and box diagrams.
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These questions connect several electron-structure skills. Keep returning to the electron count so you do not lose track of the atom or ion.
Use atomic number to count electrons in a neutral atom.
If an ion is given, use its charge to relate its electron count to the neutral atom.
Use the supplied atomic numbers to identify the element.
Write the configuration, then translate it into orbital boxes if asked. Paired electrons need opposite spins.
Show each step. A correct total alone does not prove the electrons occupy the correct subshells.
Identify representation → count electrons → apply rules → check charge and total
EXAM TRAPDo not answer a request for a full configuration with only a noble-gas core.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 40
Atom X has configuration [Ar] 3d¹⁰ 4s² 4p³.
Atomic numbers: Ge 32, As 33, Se 34, Br 35, Kr 36.
Calculate Z and identify X. [2]
Give the electron count and configuration of X³⁻. [2]
Draw X’s 4p³ box diagram. [2]
Total: 6 marks
Show answer & marking points
18 + 10 + 2 + 3 = 33 [1]; arsenic, As [1].
36 electrons [1]; [Kr] [1].
Subshell
Orbital 1
Orbital 2
Orbital 3
4p³
↑
↑
↑
One electron in each orbital [1], parallel spins [1].
Exam-style question 41
An atom has configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴.
Atomic numbers: Ge 32, As 33, Se 34, Br 35, Kr 36.
Deduce Z and identify the element. [1]
Write the configuration using [Ar]. [1]
Give the count and configuration of its 2− ion. [2]
Draw the neutral atom’s 4p box diagram. [2]
Total: 6 marks
Show answer & marking points
Z = 34: selenium, Se. [1]
[Ar] 3d¹⁰ 4s² 4p⁴. [1]
36 electrons [1]; [Kr] [1].
Subshell
Orbital 1
Orbital 2
Orbital 3
4p⁴
↑↓
↑
↑
One pair and two singles [1], correct spins [1].
Exam-style question 42
Stretch: combine particle counting with configurations. X⁻ contains 35 protons, 44 neutrons and 36 electrons.
Atomic numbers: As 33, Se 34, Br 35, Kr 36.
Identify the element and write the full nuclide symbol, with mass number, atomic number and charge. [2]
Write the full ion configuration and check its total. [2]
Write the neutral atom using [Ar]. [2]
Draw the 4p diagrams of the neutral atom and the ion. [4]
Two marks per correctly occupied diagram with correct spins. [4]
✓ DONE WHENYou can choose and combine the required rules without being told the route.
Ionisation energies
3 practice questions
First ionisation energy and attraction
Learning goalDefine first ionisation energy and explain factors controlling it.
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Periodicity connection: ionisation energies are included here to show how electron structure explains experimental evidence.
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. Keep all those details in an exam definition.
Energy is needed to overcome the electron's attraction to the nucleus, so ionisation is endothermic. The unit is kJ mol⁻¹. A second ionisation starts from gaseous 1+ ions; later ionisations start from increasingly positive ions.
First ionisation energy
The energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.
X(g) → X⁺(g) + e⁻
X an atom of the element in the gas state; X+ a gaseous ion of the element with charge +1; e− an electron; (g) gas.
Factor
Consequence
More protons
Greater nuclear charge usually strengthens attraction.
Greater atomic radius
The removed electron is further from the nucleus, weakening attraction.
More inner-electron shielding
Weaker attraction to the electron being removed.
Higher-energy subshell
Less additional energy may be needed to remove its electron.
Paired electrons in an orbital
Electron repulsion can make one easier to remove.
Structural change → attraction changes → energy required changes
EXAM TRAPFinish your explanation with attraction and the energy needed; do not stop at “bigger atom”.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 43
Define first ionisation energy. [3]
Write the first-ionisation equation for potassium, including state symbols. [1]
State the units and whether the process is endothermic or exothermic. [2]
Total: 6 marks
Show answer & marking points
Energy to remove one electron from each atom [1] in one mole of gaseous atoms [1], forming one mole of gaseous 1+ ions [1].
K(g) → K⁺(g) + e⁻. [1]
kJ mol⁻¹ [1]; endothermic [1].
✓ DONE WHENYou can define first IE precisely and connect structure to removal energy.
Ionisation-energy trends and exceptions
Learning goalExplain trends across Periods 2 and 3 and down a group.
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Periodicity connection: explain each trend through the attraction between the nucleus and the electron being removed.
Across a period: nuclear charge increases while electrons enter the same main shell. Inner-shell shielding changes little, the radius decreases and attraction strengthens. First ionisation energy generally increases.
Down a group: the outer electron is further away and more shielded. These effects outweigh the increased nuclear charge, so attraction weakens and first ionisation energy generally decreases.
For exceptions, check the subshell and electron pairing. Removing a higher-energy p electron, or an electron repelled by its partner in a p orbital, can require less energy than the simple trend suggests.
EXCEPTIONS TO THE TREND
Decrease
Configurations
Explanation
Be → B
Be 1s² 2s²; B 1s² 2s² 2p¹
B loses a higher-energy, more shielded 2p electron rather than 2s.
N → O
N ends 2p³; O ends 2p⁴
O has one paired p orbital. Pair repulsion makes an electron easier to remove.
Mg → Al
Mg [Ne] 3s²; Al [Ne] 3s² 3p¹
Al loses a higher-energy, more shielded 3p electron rather than 3s.
P → S
P ends 3p³; S ends 3p⁴
S has a paired p orbital; pair repulsion lowers the energy needed.
EXAM TRAPThe s → p decrease and the pairing decrease have different explanations. Name the relevant subshell or orbital.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 44
Be is 1s² 2s²; B is 1s² 2s² 2p¹. N ends in 2p³; O ends in 2p⁴.
Explain why B has a lower first ionisation energy than Be. [3]
Explain why O has a lower first ionisation energy than N. [2]
Explain why first ionisation energy generally decreases down a group. [4]
Total: 9 marks
Show answer & marking points
B loses a 2p electron rather than a 2s electron [1]. It is higher in energy / more shielded [1], so less strongly attracted and easier to remove [1].
O has a paired 2p orbital [1]; repulsion within that pair makes an electron easier to remove [1].
The outer electron occupies a higher shell [1], further from the nucleus [1], with greater shielding [1]. These effects outweigh the greater nuclear charge, reducing attraction and energy needed [1].
✓ DONE WHENYou can explain the general trends and both types of small decrease.
Using successive ionisation energies
Learning goalUse jumps to deduce shell populations and an element’s group.
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Periodicity connection: a sequence of ionisation energies can reveal the atom's shell structure.
The energies rise as electrons are removed from an increasingly positive ion. A much larger jump marks the start of removal from an inner shell, where electrons are closer to the nucleus and less shielded.
Find the first large jump.
Count the electrons removed before that jump. This is the outer-shell electron count.
For a main-group element, link that number to the group.
Use any further large jumps to deduce the populations of inner shells.
Outer-shell electrons
Group
1 or 2
Group 1 or 2
3, 4, 5, 6, 7, 8
Groups 3, 4, 5, 6, 7, 0 respectively
Worked example
A large jump after three removals
Al has successive IEs 578, 1817, 2745 and 11 578 kJ mol⁻¹.
Identify where the largest jump occurs and explain what it shows about the number of outer-shell electrons and the group of the element.
Worked answer
The large jump is between IE₃ and IE₄.
Three outer electrons have been removed before the jump.
The fourth electron comes from an inner shell.
Three outer electrons: Group 3.
Worked example
Read more than one shell
All 13 successive ionisations of an atom show large jumps after removals 3 and 11.
Deduce the electron arrangement by shells and identify the element. Explain how the jumps support your answer.
Worked answer
First 3 removals: outer shell contains 3 electrons.
Next 8 removals: second shell contains 8 electrons.
Last 2 removals: innermost shell contains 2 electrons.
Shell arrangement 2, 8, 3: aluminium.
Second IE: X⁺(g) → X²⁺(g) + e⁻
Third IE: X²⁺(g) → X³⁺(g) + e⁻
EXAM TRAPIf the jump is between IEₙ and IEₙ₊₁, the outer shell had n electrons, not n + 1. Use the period or further data when identifying the exact element.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Exam-style question 45
A Period 3 element has successive ionisation energies 738, 1451, 7733 and 10 543 kJ mol⁻¹. Period 3 elements: Na, Mg, Al, Si, P, S, Cl, Ar.
Locate the large jump and state the outer-shell electron count. [2]
Identify the element. [1]
Explain the jump. [3]
Write the equation for its third ionisation energy. [1]
Total: 7 marks
Show answer & marking points
Between IE₂ and IE₃ [1]; two outer-shell electrons [1].
Magnesium. [1]
The third electron comes from an inner shell [1], closer to the nucleus with less shielding [1], so it is much more strongly attracted and harder to remove [1].
Mg²⁺(g) → Mg³⁺(g) + e⁻. [1]
✓ DONE WHENYou can deduce outer-shell and inner-shell counts from jumps and write the corresponding ionisation equations.
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