Atomic structure: OCR A A-level Chemistry

Revise atoms, ions, isotopes, mass spectra, electron configurations and ionisation energies with explanations and worked practice.

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.

ModelEvidence 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.
ChadwickDiscovery 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
  1. Explain how the discovery of electrons challenged Dalton’s indivisible-atom model. [2]
  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
  1. Atoms contain smaller negatively charged particles [1], so they are not indivisible [1].
  2. 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.

ParticleRelative chargeRelative massLocation
Proton+11Nucleus
Neutron01Nucleus
Electron−1Approximately 1/1840Orbitals 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]

ParticleRelative chargeRelative massLocation
Proton………
Neutron………
Electron………

Total: 9 marks

Show answer & marking points
ParticleRelative chargeRelative massLocation
Proton+11Nucleus
Neutron01Nucleus
Electron−1Approximately 1/1840Region 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.

  1. Find the neutral atom first. Its proton number and electron number both equal the atomic number.
  2. For a positive ion, remove electrons. A 2+ charge means two electrons have been lost.
  3. For a negative ion, add electrons. A 1- charge means one electron has been gained.
  4. 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

  1. Protons = 27.
  2. Neutrons = 59 − 27 = 32.
  3. 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]

SpeciesAtomic numberMass numberProtonsNeutronsElectrons
²⁷Al³⁺1327………
³⁷Cl⁻1737………

Total: 6 marks

Show answer & marking points
SpeciesProtonsNeutronsElectrons
²⁷Al³⁺131410
³⁷Cl⁻172018

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

IsotopeProtonsNeutronsElectrons
³⁵Cl171817
³⁷Cl172017
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
  1. Define isotope. [2]
  2. Explain why isotopes of the same element have the same chemical properties. [1]
  3. Explain why they can have different physical properties. [2]

Total: 5 marks

Show answer & marking points
  1. Atoms of the same element with the same number of protons but different numbers of neutrons and different masses. [2]
  2. Their neutral atoms have the same electron configuration. [1]
  3. Different neutron numbers give different masses [1], so mass-dependent physical properties can differ [1].
Exam-style question 7
SpeciesProtonsNeutronsElectrons
A171818
B172017
C182018
D182220
  1. Identify both pairs that are species of different isotopes of the same element. [1]
  2. Justify the pairs using the particle data. [1]

Total: 2 marks

Show answer & marking points
  1. A and B; C and D. Both pairs required. [1]
  2. 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.
QuantityMeaning
Mass numberWhole-number count of protons plus neutrons in one atom.
Relative isotopic massMeasured relative mass of one isotope; close to, but not necessarily exactly equal to, mass number.
Relative atomic massWeighted 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
  1. Define atomic number. [1]
  2. Define mass number. [1]
  3. Explain why mass number is a whole number whereas relative atomic mass often is not. [2]

Total: 4 marks

Show answer & marking points
  1. The number of protons in the nucleus of an atom. [1]
  2. The total number of protons and neutrons in the nucleus of one atom. [1]
  3. Mass number counts whole particles in one isotope [1]. Relative atomic mass is a weighted mean including isotope abundances [1].
Quick check 9
  1. Define relative isotopic mass. [2]
  2. Define relative atomic mass, Aᵣ. [2]

Total: 4 marks

Show answer & marking points
  1. The mass of an atom of one isotope compared with one-twelfth of the mass of a carbon-12 atom. [2]
  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.

StructureTermExamples
Discrete moleculesRelative molecular mass, MᵣCO₂, H₂O, NH₃
Giant ionic or covalent structureRelative formula massNaCl, 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

  1. CO₂: 12.0 + 2(16.0) = 44.0. It contains molecules.
  2. 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.

  1. Calculate Mᵣ of H₂SO₄. [2]
  2. Calculate the relative formula mass of CaCO₃. [2]

Total: 4 marks

Show answer & marking points
  1. 2(1.0) + 32.1 + 4(16.0) [1] = 98.1 [1].
  2. 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
  1. State the quantity on the horizontal axis of a mass spectrum. [1]
  2. State the quantity on the vertical axis. [1]
  3. Explain why m/z is numerically equal to relative isotopic mass in a simple elemental spectrum of singly charged ions. [1]
  4. State what the relative abundance of a peak represents. [1]

Total: 4 marks

Show answer & marking points
  1. Mass-to-charge ratio, m/z. [1]
  2. Relative abundance or relative intensity. [1]
  3. The charge number z = 1, so dividing by z does not alter the numerical mass value. [1]
  4. The abundance of that isotope ion relative to the others detected. [1]
Quick check 13
m/zRelative abundance
2479
2510
2611

Calculate Aᵣ and identify the element. Reference Aᵣ values: Na = 23.0, Mg = 24.3, Al = 27.0. [4]

Total: 4 marks

Show answer & marking points
  1. Weighted numerator = 24(79) + 25(10) + 26(11) = 2432. [1]
  2. Total abundance = 79 + 10 + 11 = 100. [1]
  3. Aᵣ = Σ(isotopic mass × abundance)/Σ(abundance)
    Aᵣ = 2432/100 = 24.32. [1]
  4. Magnesium. [1]
✓ 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.

  1. Multiply each isotope's relative mass by its abundance.
  2. Add those contributions.
  3. 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.

Worked answer

Aᵣ = Σ(isotopic mass × abundance)/Σ(abundance)
Aᵣ = [28(92) + 29(5) + 30(3)]/100 = 2811/100.

Aᵣ = 28.11.
Worked example

An unknown isotope mass

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

  1. 63.60 = [63(70) + 30x]/100.
  2. 6360 = 4410 + 30x.
  3. 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

  1. Let mass 52 have abundance x%. Mass 54 has (80 − x)%.
  2. 52.20 = [50(20) + 52x + 54(80 − x)]/100.
  3. 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
  1. Use the mass–ratio products: 78.918(4.35) + 80.916(1.00). [1]
  2. Divide by 4.35 + 1.00 = 5.35. [1]
  3. Aᵣ = Σ(isotopic mass × abundance)/Σ(abundance)
    Aᵣ = 424.2093/5.35 = 79.2914579… [1]
  4. Aᵣ = 79.291. [1]
Exam-style question 15

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
  1. 63.62 = [63(69.0) + 31.0x]/100. [1]
  2. 6362 = 4347 + 31x. [1]
  3. 31x = 2015. [1]
  4. 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
  1. Mass 26 has abundance (21 − x)%. [1]
  2. 24.32 = [24(79) + 25x + 26(21 − x)]/100. [1]
  3. 2432 = 2442 − x. [1]
  4. Mass 25: x = 10.0%. [1]
  5. 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
  1. Mass 11 has abundance (100 − x)%. [1]
  2. 10.81 = [10x + 11(100 − x)]/100. [1]
  3. 1081 = 1100 − x, so x = 19. [1]
  4. 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/zRelative intensity
9074.0
9115.0
9224.5
9425.0
964.0
  1. Explain why the student is incorrect. [1]
  2. Use the data to show that Aᵣ = 91.32. [4]

Total: 5 marks

Show answer & marking points
  1. Relative intensities are a comparative scale. They need not total 100; use their actual total. [1]
  2. Total intensity 142.5 [1]; weighted numerator 13013.0 [1]; divide 13013.0/142.5 = 91.319298… [1]; Aᵣ = 91.32 [1].
Exam-style question 19

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.

  1. Calculate each percentage abundance. [3]
  2. Calculate Aᵣ. [3]

Total: 6 marks

Show answer & marking points
  1. 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].
  2. 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 mass58606162
Abundance / %68.0x1.13.6
  1. Calculate x. [1]
  2. Calculate Aᵣ of X to two decimal places. [4]
  3. X forms ionic XCl₂. Calculate its relative formula mass using Aᵣ(Cl) = 35.45. [2]
  4. Explain why relative formula mass is used for XCl₂. [1]

Total: 8 marks

Show answer & marking points
  1. x = 100 − 68.0 − 1.1 − 3.6 = 27.3%. [1]
  2. 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].
  3. Use the unrounded Aᵣ: 58.723 + 2(35.45) [1] = 129.623 = 129.62 [1].
  4. 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.

  1. State Mᵣ. [1]
  2. State what is meant by the base peak. [1]
  3. Explain why the base peak is not necessarily used to determine Mᵣ. [1]

Total: 3 marks

Show answer & marking points
  1. Mᵣ = 86. [1]
  2. The most intense peak, conventionally assigned relative intensity 100. [1]
  3. 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.

  1. Determine its empirical formula. [4]
  2. State Mᵣ. [1]
  3. Determine its molecular formula. [2]
  4. Explain why the m/z 43 peak is not used to determine Mᵣ. [1]

Total: 8 marks

Show answer & marking points
  1. 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].
  2. Mᵣ = 88. [1]
  3. Empirical-formula mass = 44, so multiplier = 88/44 = 2 [1]. Molecular formula C₄H₈O₂ [1].
  4. 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
  1. State the maximum number of electrons in each of the first four shells. [4]
  2. Use 2n² to calculate the fourth-shell capacity. [1]

Total: 5 marks

Show answer & marking points
  1. Shells 1–4: 2, 8, 18, 32 respectively. [4]
  2. 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
  1. Define atomic orbital. [2]
  2. Identify the type of orbital shown by A and by B–D. [2]
  3. Describe the shape and relative arrangement of the three p orbitals. [2]

Total: 6 marks

Show answer & marking points
  1. A region around the nucleus [1] that can hold up to two electrons with opposite spins [1].
  2. A is an s orbital; B–D are p orbitals. [2]
  3. 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.

🔓 CRACK IT

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.

SubshellOrbitalsElectrons per orbitalMaximum electrons
s122
p326
d5210
Worked example

Why does d hold ten?

Explain the capacity of a d subshell.

Worked answer

  1. There are five d orbitals.
  2. Each holds at most two electrons with opposite spins.
  3. 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]

SubshellNumber of orbitalsMaximum electrons
s……
p……
d……

Total: 6 marks

Show answer & marking points
SubshellOrbitalsMaximum electrons
s12
p36
d510

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.

🔓 CRACK IT

In an electrons-in-boxes diagram, each box is one orbital and each arrow is one electron. The arrow direction represents spin.

  1. For equal-energy orbitals, place one electron in each box before adding a second electron to any box.
  2. Draw the singly occupied orbitals with parallel spins in these ground-state arrangements.
  3. 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.

  1. Carbon: 2p². [2]
  2. Fluorine: 2p⁵. [2]

Total: 4 marks

Show answer & marking points
  1. Carbon: separate orbitals [1] with parallel spins [1].
    SubshellOrbital 1Orbital 2Orbital 3
    2p²↑↑Empty
  2. Fluorine: two pairs and one single electron [1], with opposite spins within each pair [1].
    SubshellOrbital 1Orbital 2Orbital 3
    2p⁵↑↓↑↓↑
Exam-style question 27

Nitrogen and oxygen have 2p³ and 2p⁴ arrangements.

  1. Draw their 2p electrons-in-box diagrams. [2]
  2. Use the diagrams to explain the singly-before-pairing rule. [2]
  3. State the spin requirement for two electrons in the same orbital. [1]

Total: 5 marks

Show answer & marking points
  1. SubshellOrbital 1Orbital 2Orbital 3
    N, 2p³↑↑↑
    SubshellOrbital 1Orbital 2Orbital 3
    O, 2p⁴↑↓↑↑
    One mark for each diagram. [2]
  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].
  3. Opposite spins. [1]
Exam-style question 28

Three proposed 2p⁴ arrangements are shown. Two arrows in the same cell occupy one orbital.

DiagramOrbital 1Orbital 2Orbital 3
A↑↓↑↑
B↑↓↑↓Empty
C↑↑↑↑
  1. Identify the correct diagram. [1]
  2. Explain the error in each other diagram. [4]

Total: 5 marks

Show answer & marking points
  1. A. [1]
  2. 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.

🔓 CRACK IT

Start by counting the electrons. A neutral atom has one electron for every proton, so its total equals its atomic number.

  1. Follow the filling order shown in the diagram.
  2. Use the capacities s², p⁶ and d¹⁰ as you fill each subshell.
  3. Stop when all the atom's electrons have been placed.
  4. 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.

Worked answer

  1. A neutral Ti atom has 22 electrons.
  2. Fill 18 core electrons, then 4s² and 3d².
  3. Check: 18 + 2 + 2 = 22.
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²; shorthand [Ar] 3d² 4s².
EXAM TRAPWriting 3d before 4s is a common convention. It does not mean 3d filled first.

🧪 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 29
  1. Write the subshell filling order from 1s to 4p. [2]
  2. Which subshell fills immediately before 3d, and which immediately after it? [2]
  3. Explain why 4s fills before 3d in the building-up order. [1]

Total: 5 marks

Show answer & marking points
  1. 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p. [2]
  2. Before: 4s; after: 4p. [2]
  3. For building up these neutral-atom configurations, 4s is filled before 3d because it is initially lower in energy. [1]
Quick check 30
  1. Write the full configuration of Ca, Z = 20, and check its electron total. [2]
  2. Write the full configuration of Kr, Z = 36, and check its electron total. [2]

Total: 4 marks

Show answer & marking points
  1. Ca: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² [1]; total 20 [1].
  2. Kr: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ [1]; total 36 [1].
Exam-style question 31

Atomic numbers: K 19, Ca 20, Zn 30, Kr 36.

Each student configuration contains an error.

SpeciesStudent configuration
K[Ar] 3d¹
Ca[Ar] 3d²
Zn[Ar] 3d⁸ 4s⁴
Kr[Ar] 3d¹⁰ 4s² 4p⁵
  1. Write the correct configuration for each. [4]
  2. For each correction, state the rule broken. [4]

Total: 8 marks

Show answer & marking points
SpeciesCorrect configuration [1 each]Rule [1 each]
K[Ar] 4s¹4s fills before 3d.
Ca[Ar] 4s²4s fills before 3d.
Zn[Ar] 3d¹⁰ 4s²An s subshell holds at most two electrons.
Kr[Ar] 3d¹⁰ 4s² 4p⁶Neutral Kr must have 36 electrons.
Exam-style question 32

A student writes selenium as [Ar] 3d¹⁰ 4s² 4p⁴ and says this proves that 3d fills before 4s.

  1. State the filling sequence immediately after 3p. [1]
  2. Write selenium’s full configuration. [2]
  3. Explain why the written order does not prove the filling order. [2]
  4. Check its electron total. [1]

Total: 6 marks

Show answer & marking points
  1. 4s → 3d → 4p. [1]
  2. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴. [2]
  3. 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].
  4. 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.

🔓 CRACK IT

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.

AtomSimple predictionObserved 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].

  1. Cr, Z = 24. [2]
  2. Cu, Z = 29. [2]

Total: 4 marks

Show answer & marking points
  1. Predicted [Ar] 3d⁴ 4s² [1]; observed [Ar] 3d⁵ 4s¹ [1].
  2. Predicted [Ar] 3d⁹ 4s² [1]; observed [Ar] 3d¹⁰ 4s¹ [1].
Exam-style question 34

Neutral atom Y contains 29 electrons.

Atomic numbers: Ni 28, Cu 29, Zn 30.

  1. Write the simple predicted configuration using [Ar]. [2]
  2. Identify Y and write its observed configuration. [2]
  3. Represent the 3d and 4s electrons of the observed atom in boxes. [3]

Total: 7 marks

Show answer & marking points
  1. [Ar] 3d⁹ 4s². [2]
  2. Copper [1]: [Ar] 3d¹⁰ 4s¹ [1].
  3. SubshellOrbital 1Orbital 2Orbital 3Orbital 4Orbital 5
    3d↑↓↑↓↑↓↑↓↑↓
    SubshellOrbital 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.

🔓 CRACK IT

Required scope: s- and p-block ions of elements up to Z = 36. Transition-metal ion configurations are outside this requirement.

  1. Write the neutral atom's electron configuration.
  2. Use the charge to work out how many electrons it has lost or gained.
  3. For a positive ion, remove electrons from the highest occupied principal shell, taking p before s within that shell.
  4. For a negative ion, add electrons to the available outer subshell.
  5. 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.

IonNeutral atom configurationIon configurationElectron 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

  1. Ga⁺ has 31 − 1 = 30 electrons.
  2. Remove the outer 4p electron.
  3. 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.

  1. Al³⁺. [2]
  2. S²⁻. [2]
  3. Br⁻. [2]

Total: 6 marks

Show answer & marking points
  1. 13 − 3 = 10 electrons [1]; [Ne] [1].
  2. 16 + 2 = 18 electrons [1]; [Ar] [1].
  3. 35 + 1 = 36 electrons [1]; [Kr] [1].
Exam-style question 36

Gallium has Z = 31.

  1. Write the configuration of Ga using [Ar]. [2]
  2. Calculate the electron count of Ga³⁺. [1]
  3. Write its configuration using [Ar]. [1]
  4. Identify the three electrons removed. [2]

Total: 6 marks

Show answer & marking points
  1. Ga: [Ar] 3d¹⁰ 4s² 4p¹. [2]
  2. 31 − 3 = 28. [1]
  3. Ga³⁺: [Ar] 3d¹⁰. [1]
  4. 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.

  1. Calculate the neutral atom’s Z and identify it. [2]
  2. Write the ion’s full configuration. [2]
  3. Draw the ion’s 3p box diagram. [2]

Total: 6 marks

Show answer & marking points
  1. Z = 18 − 2 = 16 [1]; sulfur [1].
  2. 1s² 2s² 2p⁶ 3s² 3p⁶. [2]
  3. SubshellOrbital 1Orbital 2Orbital 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.

  1. Identify A, B, C and D. [4]
  2. Write their shared configuration. [1]

Total: 5 marks

Show answer & marking points
  1. A = S; B = Cl; C = Ar; D = K. [4]
  2. [Ar]. [1]
Exam-style question 39

Gallium has Z = 31; germanium has Z = 32.

  1. Write each neutral configuration using [Ar]. [2]
  2. Calculate the electron counts of Ga³⁺ and Ge⁴⁺. [2]
  3. Write each ion’s configuration. [2]
  4. Explain why these ions have the same configuration. [1]

Total: 7 marks

Show answer & marking points
  1. Ga: [Ar] 3d¹⁰ 4s² 4p¹; Ge: [Ar] 3d¹⁰ 4s² 4p². [2]
  2. Ga³⁺: 31 − 3 = 28; Ge⁴⁺: 32 − 4 = 28. [2]
  3. Both are [Ar] 3d¹⁰. [2]
  4. 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.

🔓 CRACK IT

These questions connect several electron-structure skills. Keep returning to the electron count so you do not lose track of the atom or ion.

  1. Use atomic number to count electrons in a neutral atom.
  2. If an ion is given, use its charge to relate its electron count to the neutral atom.
  3. Use the supplied atomic numbers to identify the element.
  4. 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.

  1. Calculate Z and identify X. [2]
  2. Give the electron count and configuration of X³⁻. [2]
  3. Draw X’s 4p³ box diagram. [2]

Total: 6 marks

Show answer & marking points
  1. 18 + 10 + 2 + 3 = 33 [1]; arsenic, As [1].
  2. 36 electrons [1]; [Kr] [1].
  3. SubshellOrbital 1Orbital 2Orbital 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.

  1. Deduce Z and identify the element. [1]
  2. Write the configuration using [Ar]. [1]
  3. Give the count and configuration of its 2− ion. [2]
  4. Draw the neutral atom’s 4p box diagram. [2]

Total: 6 marks

Show answer & marking points
  1. Z = 34: selenium, Se. [1]
  2. [Ar] 3d¹⁰ 4s² 4p⁴. [1]
  3. 36 electrons [1]; [Kr] [1].
  4. SubshellOrbital 1Orbital 2Orbital 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.

  1. Identify the element and write the full nuclide symbol, with mass number, atomic number and charge. [2]
  2. Write the full ion configuration and check its total. [2]
  3. Write the neutral atom using [Ar]. [2]
  4. Draw the 4p diagrams of the neutral atom and the ion. [4]

Total: 10 marks

Show answer & marking points
  1. Bromine [1]: ⁷⁹₃₅Br⁻, since 35 + 44 = 79 [1].
  2. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ [1]; total 36 [1].
  3. [Ar] 3d¹⁰ 4s² 4p⁵. [2]
  4. SubshellOrbital 1Orbital 2Orbital 3
    Br, 4p⁵↑↓↑↓↑
    SubshellOrbital 1Orbital 2Orbital 3
    Br⁻, 4p⁶↑↓↑↓↑↓
    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.

🔓 CRACK IT

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.

FactorConsequence
More protonsGreater nuclear charge usually strengthens attraction.
Greater atomic radiusThe removed electron is further from the nucleus, weakening attraction.
More inner-electron shieldingWeaker attraction to the electron being removed.
Higher-energy subshellLess additional energy may be needed to remove its electron.
Paired electrons in an orbitalElectron 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
  1. Define first ionisation energy. [3]
  2. Write the first-ionisation equation for potassium, including state symbols. [1]
  3. State the units and whether the process is endothermic or exothermic. [2]

Total: 6 marks

Show answer & marking points
  1. Energy to remove one electron from each atom [1] in one mole of gaseous atoms [1], forming one mole of gaseous 1+ ions [1].
  2. K(g) → K⁺(g) + e⁻. [1]
  3. 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.

🔓 CRACK IT

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
DecreaseConfigurationsExplanation
Be → BBe 1s² 2s²; B 1s² 2s² 2p¹B loses a higher-energy, more shielded 2p electron rather than 2s.
N → ON ends 2p³; O ends 2p⁴O has one paired p orbital. Pair repulsion makes an electron easier to remove.
Mg → AlMg [Ne] 3s²; Al [Ne] 3s² 3p¹Al loses a higher-energy, more shielded 3p electron rather than 3s.
P → SP ends 3p³; S ends 3p⁴S 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.

  1. Explain why B has a lower first ionisation energy than Be. [3]
  2. Explain why O has a lower first ionisation energy than N. [2]
  3. Explain why first ionisation energy generally decreases down a group. [4]

Total: 9 marks

Show answer & marking points
  1. 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].
  2. O has a paired 2p orbital [1]; repulsion within that pair makes an electron easier to remove [1].
  3. 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.

🔓 CRACK IT

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.

  1. Find the first large jump.
  2. Count the electrons removed before that jump. This is the outer-shell electron count.
  3. For a main-group element, link that number to the group.
  4. Use any further large jumps to deduce the populations of inner shells.
Outer-shell electronsGroup
1 or 2Group 1 or 2
3, 4, 5, 6, 7, 8Groups 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

  1. The large jump is between IE₃ and IE₄.
  2. Three outer electrons have been removed before the jump.
  3. 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

  1. First 3 removals: outer shell contains 3 electrons.
  2. Next 8 removals: second shell contains 8 electrons.
  3. 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.

  1. Locate the large jump and state the outer-shell electron count. [2]
  2. Identify the element. [1]
  3. Explain the jump. [3]
  4. Write the equation for its third ionisation energy. [1]

Total: 7 marks

Show answer & marking points
  1. Between IE₂ and IE₃ [1]; two outer-shell electrons [1].
  2. Magnesium. [1]
  3. 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].
  4. 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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