Atomic structure: AQA A-level Chemistry

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.

AQA · Atomic structure · Notes and practice

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Atoms, ions & isotopes

6 practice questions

How the atomic model changed

Learning goalExplain how new evidence changed scientific models of the atom.

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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.

Dalton (solid sphere model)
Atoms pictured as tiny, solid, indivisible spheres.
J. J. Thomson (plum pudding model)
The discovery of electrons showed atoms were divisible. Electrons were placed within diffuse positive charge.
Rutherford (nuclear model)
Alpha scattering supported a tiny, dense, positive nucleus, with most of the atom empty space.
Bohr (planetary model)
Electrons placed in fixed energy levels.
Chadwick
The neutron completed the nuclear model and helped explain isotope masses.
Worked example

Alpha scattering: observation → inference

Most alpha particles passed through; some were deflected through large angles; very few rebounded.

Explain what each observation tells you about the structure of the atom.

Worked answer

  1. Most pass through → most of the atom is empty space.
  2. Rare large deflections or rebounds → positive charge and most mass are concentrated in a tiny, dense nucleus.
Use the evidence in your explanation.
EXAM TRAPDo not merely list the models. State the evidence, then the conclusion.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 1

Positively charged particles are directed at a very thin metal foil. Most pass straight through. A very small proportion are deflected through large angles. Use these observations to explain two conclusions about atomic structure and why the results do not support a model in which positive charge is spread uniformly through the atom.

Total: 4 marks

Show answer & marking points
  • Most particles pass through, so most of the atom is empty space. [1]
  • Large deflections require positive charge concentrated in a very small central nucleus. [1]
  • Most mass is also concentrated in the nucleus. [1]
  • Diffuse positive charge would cause only small deflections, so cannot explain the rare large deflections. [1]
✓ DONE WHENYou can explain how observations led to changes in the atomic model.

Fundamental particles and the atom

Learning goalProtons, neutrons and electrons: relative charge and relative mass. An atom consists of a nucleus containing protons and neutrons surrounded by electrons.

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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 mass
Proton+11
Neutron01
Electron−1≈ 1/1840
MEMORY HOOKProtons are positive. Neutrons are neutral. Electrons are negative and have negligible relative mass.

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Exam-style question 2

State the relative charge and relative mass of a proton, a neutron and an electron.

Total: 6 marks

Show answer & marking points
  • Proton: +1; relative mass 1. [2]
  • Neutron: 0; relative mass 1. [2]
  • Electron: −1; relative mass approximately 1/1840. [2]

The electron has a very small mass, not zero mass.

Quick check 3

State where protons, neutrons and electrons are found in an atom. State where almost all of the mass of an atom is concentrated.

Total: 4 marks

Show answer & marking points
  • Protons: nucleus. [1]
  • Neutrons: nucleus. [1]
  • Electrons: regions around the nucleus. [1]
  • Almost all mass is in the nucleus. [1]
✓ DONE WHENYou can reproduce the particle table and describe the structure of an atom.

Mass number, atomic number and ions

Learning goalMass number (A) and atomic (proton) number (Z). Determine the number of fundamental particles in atoms and ions using mass number, atomic number 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 a particular isotope. It is a whole number.

A = Z + number of neutronsA mass number: total number of protons and neutrons; no unit; Z atomic number: number of protons; no unit.

Worked example

Finding particle numbers in an ion

State the number of protons, neutrons and electrons in 56Fe2+. The atomic number of iron is 26.

Worked answer

  1. Iron has Z = 26, so it contains 26 protons.
  2. Neutrons = 56 − 26 = 30.
  3. Start with 26 electrons, then subtract 2 because the ion is 2+.
26 protons, 30 neutrons and 24 electrons.
EXAM TRAPWork out the neutral atom first. Then subtract electrons for a positive ion or add electrons for a negative ion. This is easy to reverse under pressure.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Quick check 4

An ion contains 17 protons, 18 neutrons and 18 electrons. Give its element, mass number and charge.

Total: 3 marks

Show answer & marking points
  • Chlorine, because Z = 17. [1]
  • A = number of protons + number of neutrons
    A = 17 + 18 = 35. [1]
  • One extra electron gives a 1− charge: ³⁵Cl⁻. [1]
Quick check 5

For the ion 3416Q2−, state the number of protons, neutrons and electrons.

Total: 3 marks

Show answer & marking points
  • Protons = 16. [1]
  • Neutrons = 34 − 16 = 18. [1]
  • Electrons = 16 + 2 = 18. [1]
✓ DONE WHENYou can move between ion notation and particle numbers in both directions.

Isotopes

Learning goalExplain the existence of isotopes.

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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.

Worked example

Comparing isotopes

Compare neutral atoms of 35Cl and 37Cl. State their numbers of protons, neutrons and electrons, then explain why they have the same chemical properties but different masses. The atomic number of chlorine is 17.

Worked answer

  1. 35Cl and 37Cl both contain 17 protons.
  2. 35Cl contains 18 neutrons; 37Cl contains 20 neutrons.
  3. Both neutral atoms contain 17 electrons and have the same electron configuration, so they react in the same way.
Their identical electron configurations make them react in the same way, while their different neutron numbers give them different masses.
DO NOT MIX THESE UPMass number describes one isotope and is a whole number. Relative atomic mass is a weighted mean for the element and is often not a whole number.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Quick check 6
  1. Define the term isotope. [2]
  2. Explain why isotopes of the same element have the same chemical properties. [1]

Total: 3 marks

Show answer & marking points
  1. Atoms with the same number of protons but different numbers of neutrons. [2]
  2. They have the same electron configuration. [1]
✓ DONE WHENYou can define isotopes and explain their similarities and differences.

How mass spectrometry works

5 practice questions

The simple TOF mass spectrometer

Learning goalDescribe ionisation, acceleration, ion drift, detection and data analysis.

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A time-of-flight mass spectrometer identifies ions by timing their journey to a detector. Follow what happens to the particles at each stage, rather than learning five names without their meanings.

  1. Ionise: make positive ions using electrospray ionisation or electron impact, as described below.
  2. Accelerate: an electric field accelerates the positive ions. Ions with the same charge gain the same kinetic energy when accelerated through the same potential difference.
  3. Drift: the ions travel through an evacuated, field-free tube, keeping the same kinetic energy they gained during acceleration. For ions of the same charge, this means equal kinetic energies but different speeds. The vacuum prevents collisions with gas particles.
  4. Detect: positive ions gain electrons at the detector. This electron movement produces a current. More arriving ions produce a larger current.
  5. Analyse: flight time is used to find m/z, while current gives relative abundance.
EXAM TRAPAcceleration uses an electric field. The detector current tells you abundance; it does not directly measure mass.

Ionisation context

Electrospray ionisation
The sample is dissolved in a volatile solvent and passed through a fine needle at high voltage to form charged droplets. The solvent evaporates and molecules gain a proton, forming [MH]⁺ ions. Useful for large or thermally unstable molecules; generally causes little fragmentation.
M(g) + H⁺(g) → [MH]⁺(g)

M the original neutral molecule; [MH]+ the protonated molecule: M with an extra H⁺ attached; H+ a proton; (g) gas.

Electron impact
The sample is vaporised and bombarded with high-energy electrons. A collision knocks an electron from a gaseous atom or molecule, forming a positive ion. Commonly used for small volatile molecules; molecular ions may fragment.
M(g) + e⁻ → M⁺(g) + 2e⁻

M a gaseous atom or molecule of the sample; M+ the singly charged positive ion formed by electron loss; e− an electron; (g) gas.

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Exam-style question 7

The stages below are not in the correct order: acceleration, data analysis, detection, drift, ionisation.

  1. Put the stages in the correct order. [2]
  2. State how the ions are accelerated and the energy condition produced. [2]
  3. Explain why the drift region is evacuated. [1]

Total: 5 marks

Show answer & marking points
  1. Ionisation → acceleration → drift → detection → data analysis. [2]
  2. An electric field accelerates the ions [1], giving ions of the same charge the same kinetic energy. [1]
  3. To prevent ions colliding with gas particles. [1]
Quick check 8

Describe what happens when a positive ion reaches the detector and explain how the detector signal is related to abundance.

Total: 2 marks

Show answer & marking points
  • The positive ion gains an electron, producing a current. [1]
  • The size of the current is proportional to the abundance of that ion. [1]
Exam-style question 9
  1. Write an equation, including state symbols, for the formation of a singly charged positive ion from a gaseous atom Y by electron impact. [1]
  2. Explain how the ion produces a detector signal and how it gives relative abundance. [3]

Total: 4 marks

Show answer & marking points
  1. Y(g) + e⁻ → Y⁺(g) + 2e⁻. The net ionisation equation Y(g) → Y⁺(g) + e⁻ also shows the loss of one electron. [1]
  2. The ion gains an electron at the detector [1]. Electron movement produces a current [1]. A larger current indicates greater abundance [1].
Exam-style question 10

A large organic molecule M is analysed using electrospray ionisation.

  1. Describe how [MH]⁺ ions are produced. [4]
  2. Write the equation for forming [MH]⁺. [1]
  3. A singly charged [MH]⁺ peak is at m/z 587.4. Determine Mᵣ of M. [1]

Total: 6 marks

Show answer & marking points
  1. Dissolve M in a volatile solvent [1]; inject through a fine needle at high pressure [1]; apply high voltage to form charged droplets [1]; M gains a proton as the solvent evaporates, forming [MH]⁺ [1].
  2. M + H⁺ → [MH]⁺. [1]
  3. Mᵣ = 587.4 − 1 = 586.4. [1]
✓ DONE WHENYou can explain all five stages and connect flight time and current to the spectrum.

Why lower-m/z ions arrive first

Learning goalExplain arrival order using kinetic energy, mass, speed and flight time.

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For ions with the same charge, acceleration through the same potential difference gives equal kinetic energy. Equal kinetic energy does not mean equal speed: a heavier ion moves more slowly than a lighter ion.

  1. Using the kinetic-energy equation, KE = ½mv²: if mass is larger while KE stays the same, speed must be smaller.
  2. Both ions travel the same distance. The slower, heavier ion therefore reaches the detector later.

For ions with the same charge, lower mass means lower m/z, so lower-m/z ions arrive first.

KE = ½mv²

KE kinetic energy of one ion, in J; m mass of one ion, in kg; v speed of the ion, in m s⁻¹.

EXAM TRAPState the equal-charge or equal-KE condition before using the mass comparison.

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Quick check 11

Two singly charged ions are given the same kinetic energy. They travel the same flight distance. One ion is heavier. Explain why the heavier ion reaches the detector later.

Total: 3 marks

Show answer & marking points
  • Both ions have the same kinetic energy. [1]
  • From KE = ½mv², the heavier ion must have a lower speed. [1]
  • It takes longer to travel the same distance. [1]
✓ DONE WHENYou can explain the arrival order, including the assumptions behind the comparison.

TOF calculations

4 practice questions
Learning goalChoose equations, convert units and solve forward, reverse and two-ion calculations.

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A TOF calculation is a chain of linked equations. You do not have to guess the whole chain at once. Start with the quantity the question asks for and find the next missing piece.

  1. Write an equation containing your target, then rearrange it. For time, start with t = d/v.
  2. Put in the quantities you know. If speed is still unknown, use the kinetic-energy equation to find it.
  3. Keep going until every quantity needed for the calculation is known.
  4. Convert to kg, m and s before substituting: mass of one ion in kg, distance in m and time in s. In particular, convert microseconds to seconds.
  5. Keep the calculator's full values during the working. Round only your final answer.

The equations below are your toolkit. Relative isotopic mass has no units; it is not the mass of one ion in kg. For a singly charged mononuclear ion, its numerical value equals m/z.

v = d/t  t = d/v  d = vt

d distance travelled by the ion, in m; t flight time of the ion, in s; v speed of the ion, in m s⁻¹.

KE = ½mv²  m = 2KE/v²  v = √(2KE/m)

KE kinetic energy of one ion, in J; m mass of one ion, in kg; v speed of the ion, in m s⁻¹.

Mass of one ion (kg) = relative mass × 10⁻³ / L

L Avogadro constant, 6.022 × 10²³ mol⁻¹.

Relative mass = m × L × 1000

L Avogadro constant, 6.022 × 10²³ mol⁻¹; m mass of one ion, in kg.

Worked example

Forward: relative mass → flight time

A singly charged ⁴⁰Ar⁺ ion has relative mass 40.0 and KE 6.00 × 10⁻¹⁶ J. It travels 1.50 m. Use L = 6.022 × 10²³ mol⁻¹.

Calculate the time of flight of the ion, giving your answer in seconds and microseconds.

Worked answer

  1. Target t: t = d/v. Speed is unknown, so use v = √(2KE/m).
  2. Mass is still unknown: m = (relative isotopic mass × 10⁻³)/L.
    m = (40.0 × 10⁻³)/(6.022 × 10²³) = 6.642 × 10⁻²⁶ kg.
  3. v = √(2KE/m)
    v = √[(2 × 6.00 × 10⁻¹⁶)/(6.642 × 10⁻²⁶)] = 1.34 × 10⁵ m s⁻¹.
  4. t = d/v.
    t = 1.50/(1.344098… × 10⁵) = 1.12 × 10⁻⁵ s.
t = 11.2 µs
Worked example

Reverse: flight time → m/z

A singly charged ion has KE 5.00 × 10⁻¹⁶ J and travels 1.60 m in 13.0 µs.

Calculate the mass of one ion in kg and its mass-to-charge ratio, m/z. Use L = 6.022 × 10²³ mol⁻¹.

Worked answer

  1. Target relative mass: relative mass = m × L × 1000.
  2. Find m from m = 2KE/v². Speed is unknown, so use v = d/t.
  3. t = 13.0 × 10⁻⁶ s.
    v = d/t.
    v = 1.60/(13.0 × 10⁻⁶) = 1.230769… × 10⁵ m s⁻¹.
  4. m = 2KE/v².
    m = (2 × 5.00 × 10⁻¹⁶)/(1.230769… × 10⁵)² = 6.6015625 × 10⁻²⁶ kg = 6.60 × 10⁻²⁶ kg (3 s.f.).
  5. Relative mass = m × L × 1000.
    Relative mass = (6.6015625 × 10⁻²⁶) × (6.022 × 10²³) × 1000 = 39.754609… = 39.8.
m = 6.60 × 10⁻²⁶ kg; m/z = 39.8
Worked example

Two ions: what cancels?

Two ions of masses m₁ and m₂ have equal kinetic energies and travel equal distances in times t₁ and t₂. Derive an expression for t₂/t₁ in terms of m₁ and m₂.

Worked answer

  1. ½m₁v₁² = ½m₂v₂². Multiply by 2.
  2. Substitute v = d/t: m₁(d/t₁)² = m₂(d/t₂)².
  3. Only the identical d² cancels: m₁/t₁² = m₂/t₂².
  4. Rearrange: t₂²/t₁² = m₂/m₁, then take the positive square root.
t₂/t₁ = √(m₂/m₁)
Worked example

Comparing ²⁴Mg⁺ and ²⁶Mg⁺

²⁴Mg⁺ and ²⁶Mg⁺ ions have the same kinetic energy and travel the same flight distance. State which ion arrives first and calculate the ratio t(²⁶Mg⁺)/t(²⁴Mg⁺).

Worked answer

t₂/t₁ = √(m₂/m₁)
t₂/t₁ = √(26/24) = 1.0408.

²⁴Mg⁺ arrives first. The heavier ion takes about 4.1% longer.
EXAM TRAPUse the ratio shortcut only for equal KE and equal distance. For a direct m/z comparison, charges must also be equal. Do not cancel different masses, speeds or times.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 12

A singly charged ion has relative isotopic mass 72.0 and kinetic energy 8.25 × 10⁻¹⁶ J. It travels through a 1.40 m flight tube. Calculate its time of flight in µs to three significant figures.

Use L = 6.022 × 10²³ mol⁻¹. KE = ½mv²; v = d/t.

Total: 5 marks

Show answer & marking points
  1. Target t: t = d/v. Find v using v = √(2KE/m).
  2. m = (72.0 × 10⁻³) / (6.022 × 10²³) = 1.1956 × 10⁻²⁵ kg. [1]
  3. Rearrange KE: v = √(2KE/m). [1]
  4. v = √(2KE/m)
    v = √[(2 × 8.25 × 10⁻¹⁶)/(1.1956… × 10⁻²⁵)] = 1.1748 × 10⁵ m s⁻¹. [1]
  5. t = d/v.
    t = 1.40/(1.1747517… × 10⁵) = 1.1917 × 10⁻⁵ s. [1]
  6. t = 11.9 µs. [1]
Exam-style question 13

An ion travels 1.12 m in 11.4 µs. Its kinetic energy is 6.75 × 10⁻¹⁶ J. Calculate its mass number.

Use L = 6.022 × 10²³ mol⁻¹. KE = ½mv²; v = d/t.

Total: 5 marks

Show answer & marking points
  1. Target mass: m = 2KE/v². First find the unknown speed.
  2. v = d/t.
    v = 1.12/(11.4 × 10⁻⁶) = 9.8246 × 10⁴ m s⁻¹. [1]
  3. m = 2KE/v². [1]
  4. m = 2KE/v²
    m = (2 × 6.75 × 10⁻¹⁶)/(9.8245614… × 10⁴)² = 1.3986 × 10⁻²⁵ kg. [1]
  5. Molar mass = m × L × 1000.
    Molar mass = (1.3986… × 10⁻²⁵) × (6.022 × 10²³) × 1000 = 84.23 g mol⁻¹. [1]
  6. Mass number = 84, a whole number. [1]
Exam-style question 14

A singly charged ion has relative isotopic mass 96.0, kinetic energy 7.40 × 10⁻¹⁶ J and time of flight 18.5 µs. Calculate the flight-tube length to three significant figures.

Use L = 6.022 × 10²³ mol⁻¹. KE = ½mv²; v = d/t.

Total: 5 marks

Show answer & marking points
  1. Target d: d = vt. Find v using v = √(2KE/m).
  2. m = (96.0 × 10⁻³) / (6.022 × 10²³) = 1.5942 × 10⁻²⁵ kg. [1]
  3. v = √(2KE/m). [1]
  4. v = √(2KE/m)
    v = √[(2 × 7.40 × 10⁻¹⁶)/(1.5941547… × 10⁻²⁵)] = 9.6353 × 10⁴ m s⁻¹. [1]
  5. d = vt = (9.6353 × 10⁴)(18.5 × 10⁻⁶). [1]
  6. d = 1.78 m. [1]
Exam-style question 15

A⁺ and B⁺ have the same kinetic energy and travel the same distance. Relative isotopic mass of A = 64.0; tₐ = 13.6 µs; tᵦ = 15.3 µs.

  1. Why can you not calculate the mass of B directly from its KE equation alone? [1]
  2. Set the kinetic energies equal and calculate the relative isotopic mass of B. Show why distance cancels. [5]
  3. Why can the masses and speeds not be cancelled? [1]

Total: 7 marks

Show answer & marking points
  1. Mass and speed are both unknown. [1]
    1. ½mₐvₐ² = ½mᵦvᵦ². [1]
    2. Multiply by 2: mₐvₐ² = mᵦvᵦ². [1]
    3. Substitute v = d/t: mₐ(d/tₐ)² = mᵦ(d/tᵦ)². [1]
    4. The identical d² factors cancel: mₐ/tₐ² = mᵦ/tᵦ². [1]
    5. mᵦ = mₐ(tᵦ/tₐ)².
      mᵦ = 64.0(15.3/13.6)² = 81.0. [1]
  3. They have different masses and speeds, so these are not identical factors. [1]
✓ DONE WHENYou can build the route from the requested unknown, use SI units and justify what cancels.

Mass spectra & relative masses

9 practice questions

Reading elemental mass spectra

Learning goalInterpret peak positions and intensities, and use spectra to identify elements.

🔓 CRACK IT

Read a mass spectrum using its axes. The x-axis shows m/z; the y-axis shows the relative abundance of each ion.

  • The x-axis is mass-to-charge ratio, m/z. For singly charged mononuclear ions, it is numerically equal to relative isotopic mass.
  • The y-axis is relative abundance or intensity. A taller peak means more ions of that isotope.
  • Use the isotope pattern and calculated weighted mean to identify the element. Check against its Periodic Table value.
EXAM TRAPPeak heights are relative quantities. They need not add to 100 unless they are given as percentages.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 16
m/zRelative abundance / %
19.992490.48
20.99380.27
21.99149.25
  1. State the quantity shown on each axis of an elemental mass spectrum. [2]
  2. Calculate the relative atomic mass to four significant figures. [3]
  3. Identify the element. [1]

Total: 6 marks

Show answer & marking points
  1. x-axis: m/z. y-axis: relative abundance or intensity. [2]
  2. Aᵣ = [(19.9924 × 90.48) + (20.9938 × 0.27) + (21.9914 × 9.25)] / 100 = 20.18001128 = 20.18. Correct weighted sum [1], denominator [1], answer [1].
  3. Neon. [1]
Exam-style question 17

A singly charged ion has KE = 7.50 × 10⁻¹⁶ J and travels 1.25 m in 10.92 µs. Use L = 6.022 × 10²³ mol⁻¹.

  1. Calculate its mass number. [5]
  2. The element has exactly two isotopes. The other mass number is two greater. Currents are 3.20 nA for the lighter isotope and 1.10 nA for the heavier. Calculate their percentage abundances. [2]
  3. Calculate Aᵣ to three decimal places. [2]
  4. Explain why the heavier isotope gives the smaller current. [1]

Total: 10 marks

Show answer & marking points
  1. v = d/t.
    v = 1.25/(10.92 × 10⁻⁶) = 1.1447 × 10⁵ m s⁻¹ [1].
    m = 2KE/v² [1].
    m = (2 × 7.50 × 10⁻¹⁶)/(1.1446886… × 10⁵)² = 1.14476544 × 10⁻²⁵ kg [1].
    Relative mass = m × L × 1000.
    Relative mass = (1.14476544 × 10⁻²⁵) × (6.022 × 10²³) × 1000 = 68.94 [1].
    Mass number = 69 [1].
  2. Total current = 4.30 nA. Isotope 69: 3.20/4.30 × 100 = 74.4% [1]. Isotope 71: 25.6% [1].
  3. Aᵣ = Σ(isotopic mass × abundance)/Σ(abundance)
    Aᵣ = [69(3.20) + 71(1.10)]/4.30 [1] = 69.512 [1].
  4. It is less abundant, so fewer ions of that isotope arrive. [1]
Exam-style question 18

Stretch: this molecular-isotope probability calculation goes beyond the usual mononuclear-ion Aᵣ calculation requirement.

Element X has isotopes ²⁴X and ²⁶X with abundances 79.0% and 21.0%. X₂ is analysed by electron impact. Assume molecular ions form without fragmentation.

  1. State the three m/z values for X₂⁺. [2]
  2. Calculate the expected percentage ratio of the peaks. [3]
  3. Explain why the middle peak can form in two ways. [1]

Total: 6 marks

Show answer & marking points
  1. m/z = 48, 50, 52. [2]
  2. P(48) = 0.790² = 62.41% [1]; P(50) = 2(0.790)(0.210) = 33.18% [1]; P(52) = 0.210² = 4.41% [1]. Ratio = 62.41 : 33.18 : 4.41.
  3. Mixed pairs can be ²⁴X–²⁶X or ²⁶X–²⁴X, so double the mixed-isotope probability. [1]
✓ DONE WHENYou can interpret both axes, identify an element and distinguish abundance from flight time.

Calculating relative atomic mass

Learning goalCalculate weighted means and solve missing-mass and missing-abundance problems.

🔓 CRACK IT

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(isotopic mass × abundance) / Σ(abundances)

Ar relative atomic mass; no unit; Σ sum of all the indicated contributions.

Worked example

Type 1: all masses and abundances known

Isotopes 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.

Aᵣ = 28.11
Worked example

Type 2: one mass unknown

An element has Aᵣ = 63.62. Isotope 63 has abundance 69%; the other isotope has abundance 31%.

Calculate the relative isotopic mass of the second isotope.

Worked answer

  1. 63.62 = [63(69) + 31x] / 100.
  2. 6362 = 4347 + 31x.
  3. x = (6362 − 4347) / 31 = 65.
The second mass number is 65.
Worked example

Type 3: two abundances unknown

Isotopes 24, 25 and 26 give Aᵣ = 24.32. Isotope 24 has abundance 79.0%.

Calculate the percentage abundances of isotopes 25 and 26.

Worked answer

  1. Let isotope 25 have abundance x%. Isotope 26 then has abundance (21 − x)%.
  2. 24.32 = [24(79) + 25x + 26(21 − x)] / 100.
  3. 2432 = 2442 − x, so x = 10.
Isotope 25: 10%; isotope 26: 11%.
EXAM TRAPFor three isotopes, the last abundance is 100 − known abundance − x.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 19

An element has isotopes with 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.

Total: 4 marks

Show answer & marking points
  1. Use the ratio as weighting factors. [1]
  2. Aᵣ = [(78.918 × 4.35) + (80.916 × 1.00)] / 5.35. [1]
  3. Aᵣ = 79.2914579… [1]
  4. Aᵣ = 79.291. [1]
Exam-style question 20

An element has Aᵣ = 58.573. One isotope has accurate relative isotopic mass 57.935 and abundance 68.0%. The second isotope has abundance 32.0%.

  1. Calculate the accurate relative isotopic mass of the second isotope to three decimal places. [3]
  2. State its mass number. [1]

Total: 4 marks

Show answer & marking points
  1. 58.573 = 57.935(0.680) + 0.320x [1]. x = [58.573 − 57.935(0.680)] / 0.320 [1] = 59.92875 = 59.929 [1].
  2. Mass number = 60. [1]
Exam-style question 21

Two isotopes have relative isotopic masses 10 and 11. Aᵣ = 10.8. Calculate their percentage abundances.

Total: 4 marks

Show answer & marking points
  1. Let mass 10 have abundance x%; mass 11 has (100 − x)%.
  2. 10.8 = [10x + 11(100 − x)]/100.
  3. 1080 = 1100 − x, so x = 20.
  4. Mass 10: 20%; mass 11: 80%.
✓ DONE WHENYou can solve all three weighted-mean question types and check the result.

Relative molecular mass from mass spectra

Learning goalIdentify molecular and protonated molecular ions and distinguish them from the base peak.

🔓 CRACK IT

First identify which ion the question describes. A molecular ion formed by electron impact has lost one electron; its mass is effectively the mass of the original molecule.

The molecular-ion peak is usually the peak furthest to the right, because the intact molecule is heavier than its fragments. Isotope peaks, such as a small M+1 peak, can appear just beyond it.

For a singly charged molecular ion, use its m/z as Mᵣ. An electrospray [MH]⁺ ion includes an extra proton, so subtract 1 from its m/z to find Mᵣ of M.

The base peak is simply the most intense peak. It may be a fragment, so it does not automatically give the mass of the intact molecule. These shortcuts assume singly charged ions; check the charge and ion label first.

Molecular-ion peak, M⁺
The intact molecule after one electron is removed. For a singly charged molecular ion, its m/z gives the relative mass of that molecular ion.
Base peak
The tallest peak, representing the most abundant ion. It is not automatically the molecular ion.
Electron impact: Mᵣ = m/z of the identified M⁺ ion

M+ the singly charged molecular ion; M a neutral molecule; M⁺ is its positive molecular ion and [MH]⁺ is its protonated 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.

Electrospray: Mᵣ = m/z of [MH]⁺ − 1

[MH]+ the protonated molecule: M with an extra H⁺ attached; M a neutral molecule; M⁺ is its positive molecular ion and [MH]⁺ is its protonated 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.

Worked example

Read the right peak

A [MH]⁺ ion appears at 151. In another spectrum, the base peak is 43 and the identified M⁺ peak is 72.

Determine the relative molecular mass of the compound in each spectrum.

Worked answer

For the protonated ion, subtract 1. For the identified molecular ion, use its m/z.

Mᵣ = 150 and 72 respectively.
EXAM TRAPDo not automatically choose the tallest or the highest-m/z peak. Identify the intact molecular ion and account for charge/protonation.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 22

An electron-impact spectrum has peaks at m/z 41, 55 and 70. The base peak is at 41; the molecular-ion peak is at 70.

  1. State the relative molecular mass. [1]
  2. Explain the term base peak and why it does not necessarily give Mᵣ. [2]

Total: 3 marks

Show answer & marking points
  1. Mᵣ = 70. [1]
  2. The base peak is the most intense peak, assigned relative intensity 100 [1]. It may be a fragment ion, so it need not give Mᵣ [1].
Exam-style question 23

A reference ion of relative isotopic mass 125.0 takes 18.6 µs. A singly charged [MH]⁺ ion takes 23.3 µs in the same instrument. Both have the same kinetic energy and travel the same distance.

  1. Calculate the relative mass of [MH]⁺ on the carbon-12 scale. [4]
  2. Determine Mᵣ of M. [1]
  3. Why is electrospray often used for a large molecule? [1]
  4. An electron-impact spectrum of M has a base peak at 43 and molecular-ion peak at 195. Which confirms Mᵣ, and why? [2]

Total: 8 marks

Show answer & marking points
  1. Equal KE and distance give m/t² = constant [1]. m₂/m₁ = (t₂/t₁)² [1]. m₂ = m₁(t₂/t₁)²
    m₂ = 125.0(23.3/18.6)² [1] = 196.153… ≈ 196 to 3 s.f. [1].
  2. Mᵣ ≈ 196.153… − 1 = 195.153… ≈ 195. [1]
  3. Less fragmentation, so the large molecule is more likely to remain intact. [1]
  4. Use m/z 195 because it is the intact molecular ion [1]. The intense fragment at 43 does not give Mᵣ [1].
✓ DONE WHENYou can extract Mᵣ from the identified M⁺ or [MH]⁺ peak.

Relative masses: definitions and language

Learning goalDefine relative atomic and molecular mass using carbon-12 and use relative formula mass for ionic compounds.

🔓 CRACK IT

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.
Relative molecular mass, Mᵣ
The mean mass of a molecule compared with one-twelfth of the mass of a carbon-12 atom.
Relative formula mass
The sum of Aᵣ values in one formula unit. Use this term for ionic compounds.
Mᵣ or relative formula mass = Σ(Aᵣ × number of each atom)

Ar relative atomic mass; no unit; Σ sum of all the indicated contributions.

Worked example

Molecules or ionic compounds?

Calculate the relative molecular mass of CO₂ and the relative formula mass of NaCl. Use Aᵣ values: C = 12.0, O = 16.0, Na = 23.0 and Cl = 35.5.

Worked answer

  1. CO₂ is molecular: Mᵣ = sum of the Aᵣ values of all atoms in the formula
    Mᵣ = 12.0 + 2(16.0) = 44.0.
  2. NaCl is ionic: relative formula mass = 23.0 + 35.5 = 58.5.
44.0 and 58.5; neither has units.
EXAM TRAPInclude “weighted mean” in Aᵣ and the one-twelfth carbon-12 reference in both definitions.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 24
  1. Define relative atomic mass. [2]
  2. Define relative molecular mass. [2]
  3. State the term used instead of relative molecular mass for an ionic substance. [1]

Total: 5 marks

Show answer & marking points
  1. The weighted mean mass of an atom of an element compared with one-twelfth of the mass of a carbon-12 atom. [2]
  2. The mean mass of a molecule compared with one-twelfth of the mass of a carbon-12 atom. [2]
  3. Relative formula mass. [1]
✓ DONE WHENYou can give precise definitions and choose the correct term without adding units.

Shells, sub-shells & orbitals

1 practice questions
Learning goalConnect principal energy levels, sub-shells, orbitals and electron capacities.

🔓 CRACK IT

Think of electron structure as three levels of organisation: shells contain subshells, and subshells contain orbitals. These names describe different things, so do not use them interchangeably.

An orbital can hold up to two electrons. An s subshell contains one orbital, a p subshell three and a d subshell five. That gives capacities of 2, 6 and 10 electrons respectively. The definitions and diagram below show how the levels fit together.

When drawing where electrons go, orbitals are usually represented by boxes. Each box is one orbital and each arrow is one electron; a pair of electrons in one box has opposite spins.

Principal energy level (shell)
The number n. In 3p, the 3 identifies the principal energy level.
Sub-shell
The letter, such as s, p or d. In 3p, p is a sub-shell within level 3.
Orbital
A region of space occupied by up to two electrons with opposite spins.
Sub-shellOrbitalsMaximum electrons
s12
p36
d510
EXAM TRAPA p sub-shell contains three orbitals and up to six electrons. One p orbital holds only two.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 25
  1. State the maximum number of electrons in s, p and d sub-shells. [3]
  2. Explain why a p sub-shell holds up to six electrons. [2]

Total: 5 marks

Show answer & marking points
  1. s: 2; p: 6; d: 10. [3]
  2. Three orbitals [1], each holding a maximum of two electrons [1].
✓ DONE WHENYou can connect the three levels and recall the capacities 2, 6 and 10.

Electron configurations

7 practice questions

Filling sub-shells and writing configurations

Learning goalWrite full and shorthand configurations of atoms up to Z = 36.

🔓 CRACK IT

An electron configuration is a record of where an atom's electrons go. Put electrons into the available subshells in increasing energy, following the arrows in the diagram.

  • In 2p⁴, the 2 tells you the shell, p tells you the subshell and 4 tells you how many electrons are in it.
  • Use the s, p and d capacities of 2, 6 and 10. Check that the superscripts add to the atom's total electron number.
  • Within a set of equal-energy orbitals, put one electron in each orbital before pairing. Two electrons in one orbital must have opposite spins.
1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p
Worked example

Bromine, Z = 35

Write the full electron configuration and the noble-gas shorthand configuration of a neutral bromine atom. Bromine has atomic number 35.

Worked answer

  1. A neutral atom has 35 electrons.
  2. Fill in order, respecting s², p⁶ and d¹⁰ capacities.
  3. The first 18 electrons are the [Ar] core.
  4. Check: 18 + 10 + 2 + 5 = 35.
Full: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵.
Shorthand: [Ar] 3d¹⁰ 4s² 4p⁵.
EXAM TRAPAdd the superscripts. For an atom they must equal Z; for an ion apply the charge first.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Quick check 26
  1. Put 3p, 4s, 3d and 4p in filling order. [1]
  2. State the maximum number of electrons in 3d. [1]
  3. State the principal energy level number of an electron in 4p. [1]

Total: 3 marks

Show answer & marking points
  1. 3p → 4s → 3d → 4p. [1]
  2. 10. [1]
  3. n = 4. [1]
Quick check 27

Atomic numbers: P = 15, Ca = 20, Se = 34.

  1. Write the full electron configuration of phosphorus. [1]
  2. Write the full electron configuration of calcium. [1]
  3. Write the full electron configuration of selenium. [1]
  4. Write selenium using [Ar] shorthand. [1]

Total: 4 marks

Show answer & marking points
  1. P: 1s² 2s² 2p⁶ 3s² 3p³. [1]
  2. Ca: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s². [1]
  3. Se: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴. [1]
  4. Se: [Ar] 3d¹⁰ 4s² 4p⁴. [1]
✓ DONE WHENYou can build, check and abbreviate configurations up to krypton.

The chromium and copper exceptions

Learning goalRecall the observed configurations of chromium and copper.

🔓 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.

Cr: [Ar] 3d⁵ 4s¹
Cu: [Ar] 3d¹⁰ 4s¹
EXAM TRAPDo not write Cr as [Ar] 3d⁴ 4s² or Cu as [Ar] 3d⁹ 4s².

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Quick check 28

Atomic numbers: Cr = 24, Cu = 29.

  1. Write chromium using [Ar] shorthand. [1]
  2. Write copper using [Ar] shorthand. [1]
  3. Describe the 3d occupancy in each. [2]

Total: 4 marks

Show answer & marking points
  1. Cr: [Ar] 3d⁵ 4s¹. [1]
  2. Cu: [Ar] 3d¹⁰ 4s¹. [1]
  3. Chromium: half-filled 3d [1]. Copper: full 3d [1].
✓ DONE WHENYou can recall Cr and Cu correctly and describe their 3d occupancies.

Electron configurations of ions

Learning goalApply charge and remove 4s electrons before 3d when forming transition-metal ions.

🔓 CRACK IT

Start with the neutral atom's configuration. The charge then tells you how many electrons to remove or add. A positive ion has lost electrons; a negative ion has gained them.

When removing electrons, start with the highest occupied principal shell. For transition-metal ions, remove the 4s electrons before the 3d electrons. For example, losing two electrons from 4s² leaves 4s empty.

Check the final electron total against atomic number and charge. Counting is a useful check, but remember that removing electrons from an atom is not the same as refilling a new neutral atom from scratch.

Transition-metal ions: remove 4s before 3d
IonNeutral atom configurationIon configurationElectrons removed or added
O²⁻O: 1s² 2s² 2p⁴1s² 2s² 2p⁶Gain 2 electrons
Ca²⁺Ca: [Ar] 4s²[Ar]Remove 4s²
Fe²⁺Fe: [Ar] 3d⁶ 4s²[Ar] 3d⁶Remove both 4s electrons
Fe³⁺Fe: [Ar] 3d⁶ 4s²[Ar] 3d⁵Remove both 4s electrons and one 3d electron
Worked example

Work backwards from an ion

X²⁺ has configuration [Ar] 3d⁷. Identify X and reconstruct its neutral configuration. Atomic numbers: Fe = 26, Co = 27, Ni = 28.

Worked answer

  1. X²⁺ has 18 + 7 = 25 electrons.
  2. The neutral atom has 27 electrons, so Z = 27: cobalt.
  3. Reverse ion formation by returning two electrons to 4s.
X is Co: [Ar] 3d⁷ 4s².
EXAM TRAP4s fills before 3d in the simple building-up order, but 4s electrons are removed first from these ions.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 29

Atomic numbers: O = 8, Ca = 20, Fe = 26.

  1. Write the electron configuration of O²⁻. [1]
  2. Write the electron configuration of Ca²⁺. [1]
  3. Write the electron configuration of Fe²⁺. [1]
  4. Write the electron configuration of Fe³⁺. [1]
  5. Which sub-shell loses electrons first when iron forms a positive ion? [1]

Total: 5 marks

Show answer & marking points
  1. O²⁻: 1s² 2s² 2p⁶. [1]
  2. Ca²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶, or [Ar]. [1]
  3. Fe²⁺: [Ar] 3d⁶. [1]
  4. Fe³⁺: [Ar] 3d⁵. [1]
  5. 4s is emptied before 3d. [1]
Quick check 30

X³⁺ has configuration [Ar] 3d². Relevant atomic numbers: Sc = 21, Ti = 22, V = 23, Cr = 24.

  1. How many electrons are in X³⁺? [1]
  2. Identify X. [1]
  3. Write the neutral atom’s configuration. [1]
  4. Explain how you reconstructed it. [1]

Total: 4 marks

Show answer & marking points
  1. 18 + 2 = 20 electrons. [1]
  2. Vanadium: 20 + 3 = 23 electrons in the neutral atom. [1]
  3. V: [Ar] 3d³ 4s². [1]
  4. Reverse the 3+ charge by returning one electron to 3d and two to 4s. [1]
Exam-style question 31

Y²⁺ has configuration [Ar] 3d⁵. Y has Z ≤ 36. Relevant atomic numbers: Cr = 24, Mn = 25, Fe = 26, Co = 27.

  1. How many electrons are in Y²⁺? [1]
  2. Determine the proton number and identity of Y. [2]
  3. Write the neutral Y configuration. [1]
  4. Write the Y³⁺ configuration. [1]
  5. Which sub-shell loses electrons first? [1]
  6. A student says Y is iron because Fe³⁺ is also [Ar] 3d⁵. Explain the error. [2]

Total: 8 marks

Show answer & marking points
  1. 23. [1]
  2. Z = 23 + 2 = 25 [1], manganese [1].
  3. Mn: [Ar] 3d⁵ 4s². [1]
  4. Mn³⁺: [Ar] 3d⁴. [1]
  5. 4s before 3d. [1]
  6. Charge must be used as well as electron count [1]. Fe³⁺ is [Ar] 3d⁵, but Fe²⁺ is [Ar] 3d⁶, so Y²⁺ cannot be Fe²⁺ [1].
✓ DONE WHENYou can count electrons, apply charge and reconstruct atoms and ions.

Electron configuration and the Periodic Table

Learning goalIdentify s, p, d and f blocks from electron configurations or table position.

🔓 CRACK IT

The Periodic Table helps you predict electron structure. An element's block tells you the type of subshell being filled across that part of the table.

Look for the differentiating electron: the electron that distinguishes this element's configuration from the preceding element's. Do not simply choose the subshell written last on the page.

In the table below, n means the outer-shell principal energy level. Electron configurations up to Z = 36 use s, p and d subshells. You should also recognise where the f block sits.

BlockPositionTypical ending
sGroups 1 and 2, plus heliumns¹ or ns²
pGroups 3–7 and 0, except heliumnp¹ to np⁶
dCentral block(n − 1)d¹ to (n − 1)d¹⁰
fTwo rows below the main table(n − 2)f¹ to (n − 2)f¹⁴
Worked example

Identify the block

An atom has configuration [Ar] 3d⁵ 4s². State its block.

Worked answer

The differentiating electron enters 3d, even though 4s² may be written last.

d block.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 32

Element E has proton number 34 and forms E²⁻.

ElementGeAsSeBrKr
Atomic number3233343536
  1. Identify E. [1]
  2. Write the full configuration of E. [1]
  3. Write the full configuration of E²⁻. [1]
  4. State the number of electrons in E²⁻. [1]
  5. Final electrons enter 3s, 3p, 3d and 4f in four different atoms. State their blocks. [4]
  6. Explain how a block is named. [1]

Total: 9 marks

Show answer & marking points
  1. Selenium, Se. [1]
  2. Se: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴. [1]
  3. Se²⁻: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶. [1]
  4. 36 electrons. [1]
  5. s, p, d and f blocks respectively. [4]
  6. The type of sub-shell receiving the differentiating electron gives the block name. [1]
✓ DONE WHENYou can connect sub-shell occupancy to block position, including helium as s block.

Ionisation energies

14 practice questions

First and successive ionisation energies

Learning goalDefine first ionisation energy and write equations for successive ionisations.

🔓 CRACK IT

Ionisation energy measures how much energy is needed to remove electrons. First ionisation energy removes one electron from each gaseous atom. The precise definition below includes the amount, gaseous state and ions formed.

Removing a negative electron from the attraction of a positive nucleus needs energy, so ionisation is endothermic. The unit is kJ mol⁻¹.

The second ionisation energy starts with gaseous 1+ ions, not neutral atoms. Each successive ionisation removes another electron from an increasingly positive ion.

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.

StepEquation
FirstMg(g) → Mg⁺(g) + e⁻
SecondMg⁺(g) → Mg²⁺(g) + e⁻
ThirdMg²⁺(g) → Mg³⁺(g) + e⁻
EXAM TRAPFor the nth ionisation, begin with charge (n − 1)+ and finish with n+. Include (g) for the atom or ion on both sides.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 33
  1. Define first ionisation energy. [3]
  2. Write the first-ionisation equation for magnesium, with state symbols. [1]
  3. State the units. [1]

Total: 5 marks

Show answer & marking points
  1. Energy required to remove one electron from each atom [1] in one mole of gaseous atoms [1] to form one mole of gaseous 1+ ions [1].
  2. Mg(g) → Mg⁺(g) + e⁻. [1]
  3. kJ mol⁻¹. [1]
Quick check 34
  1. Write the equation for the second ionisation energy of magnesium. [1]
  2. Write the equation for the third ionisation energy of aluminium. [1]
  3. Is ionisation endothermic or exothermic? [1]

Total: 3 marks

Show answer & marking points
  1. Mg⁺(g) → Mg²⁺(g) + e⁻. [1]
  2. Al²⁺(g) → Al³⁺(g) + e⁻. [1]
  3. Endothermic. [1]
✓ DONE WHENYou can give the precise definition and write each ionisation equation correctly.

What controls ionisation energy?

Learning goalExplain changes using nuclear attraction, distance, shielding, sub-shell energy and pairing.

🔓 CRACK IT

Ask one central question: how strongly is the electron being removed attracted to the nucleus? Stronger attraction means more energy is needed to remove it.

  • More protons give a greater nuclear charge and tend to strengthen attraction.
  • An electron further from the nucleus is less strongly attracted.
  • Inner electrons shield an outer electron from some of the nuclear attraction.
  • Subshell energy and repulsion between paired electrons can explain exceptions to a simple trend.

Compare the relevant factors together. Saying only that an atom has more protons is not enough if the electron is also further away and more shielded.

FactorEffect
More protonsGreater nuclear charge strengthens attraction.
Greater distanceWeaker attraction to the electron.
More inner-shell shieldingWeaker attraction to the electron being removed.
Higher-energy sub-shellLess additional energy needed to remove the electron.
Electron pairingRepulsion within an orbital can make one electron easier to remove.
Structural change → attraction changes → energy changes
EXAM TRAP“The atom is bigger” or “more shielding” is incomplete. Finish the chain with attraction and the energy required.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 35
  1. State the three main factors controlling attraction between the nucleus and the electron being removed. [3]
  2. Why is an electron in a higher-energy sub-shell easier to remove? [2]

Total: 5 marks

Show answer & marking points
  1. Nuclear charge [1], distance from the nucleus [1], shielding by inner electrons [1].
  2. It is already at higher energy / less strongly attracted [1], so less additional energy is required to remove it [1].
Exam-style question 36

Na has IE₁ = 496 and IE₂ = 4562 kJ mol⁻¹. Mg has IE₁ = 738 and IE₂ = 1451 kJ mol⁻¹. Explain why the second IE of sodium is much greater than that of magnesium. Atomic numbers: Na = 11, Mg = 12.

Total: 4 marks

Show answer & marking points
  • Na⁺ is 1s² 2s² 2p⁶: the second electron is removed from n = 2. [1]
  • Mg⁺ is 1s² 2s² 2p⁶ 3s¹: its second electron is removed from 3s. [1]
  • The electron in Na⁺ is closer to the nucleus and less shielded. [1]
  • It is more strongly attracted and requires much more energy to remove. [1]
✓ DONE WHENYou can choose the relevant factor and finish a cause-and-effect explanation.

First ionisation energy across Period 3

Learning goalExplain the general trend and the magnesium–aluminium and phosphorus–sulfur decreases.

🔓 CRACK IT

Across Period 3, first ionisation energy generally rises. The nucleus gains protons, while the added electrons enter the same main shell. Inner-shell shielding stays broadly similar.

The outer electron is attracted more strongly and the atomic radius decreases. More energy is therefore needed to remove that electron.

EXCEPTIONS · THE TWO DIPS

From Mg to Al, the removed electron is in the higher-energy 3p subshell. From P to S, the removed 3p electron is paired, so electron-electron repulsion makes removal easier. Use these specific explanations for the exceptions.

Mg → Al
Mg: [Ne] 3s². Al: [Ne] 3s² 3p¹. The electron removed from Al is in a higher-energy, slightly more shielded 3p sub-shell, so is easier to remove.
P → S
P: [Ne] 3s² 3p³. S: [Ne] 3s² 3p⁴. Sulfur has a pair in one 3p orbital. Repulsion in that pair makes an electron easier to remove.
EXAM TRAPUse sub-shell energy for Mg → Al and paired-electron repulsion for P → S. They are different explanations.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 37
ElementFirst IE / kJ mol⁻¹
Na496
Mg738
Al578
Si787
P1012
S1000
Cl1251
Ar1521
  1. Describe the general trend across Period 3. [1]
  2. Explain the trend. [3]
  3. Identify the two decreases. [1]
  4. What do the decreases show about electron arrangement? [1]

Total: 6 marks

Show answer & marking points
  1. First IE generally increases from Na to Ar. [1]
  2. Nuclear charge increases [1]. Electrons enter the same principal shell, so inner-shell shielding is broadly similar [1]. Stronger attraction means more energy is required [1].
  3. Mg → Al and P → S. Both needed. [1]
  4. Evidence for different sub-shell energies and electron pairing within orbitals. [1]
Exam-style question 38

Mg is [Ne] 3s²; Al is [Ne] 3s² 3p¹.

  1. Which sub-shell loses the first electron in Al? [1]
  2. Explain why Al has a lower first IE than Mg. [3]

Total: 4 marks

Show answer & marking points
  1. 3p. [1]
  2. The 3p electron is higher in energy than 3s [1] and slightly more shielded [1], so it is less strongly attracted / easier to remove [1].
Exam-style question 39
  1. State the final sub-shell occupancy of phosphorus and sulfur. Atomic numbers: P = 15, S = 16. [2]
  2. Explain why sulfur has a slightly lower first ionisation energy than phosphorus. [2]

Total: 4 marks

Show answer & marking points
  1. P: 3p³ [1]. S: 3p⁴ [1].
  2. Sulfur has an electron pair in one 3p orbital [1]. Repulsion within that pair makes an electron easier to remove [1].
Exam-style question 40
SpeciesConfigurationFirst IE / kJ mol⁻¹
A[Ne] 3s²738
B[Ne] 3s² 3p¹578
C[Ne] 3s² 3p³1012
D[Ne] 3s² 3p⁴1000

All four are in Period 3: Na, Mg, Al, Si, P, S, Cl, Ar.

  1. Identify A, B, C and D. [4]
  2. Explain why the first ionisation energy of B is lower than that of A. [3]
  3. Explain why the first ionisation energy of D is lower than that of C. [2]

Total: 9 marks

Show answer & marking points
  1. A = Mg; B = Al; C = P; D = S. [4]
  2. B loses 3p rather than 3s [1]. The 3p electron is higher in energy / slightly more shielded [1], so it is easier to remove [1].
  3. D has a pair in a 3p orbital [1]. Pair repulsion makes one electron easier to remove [1].
✓ DONE WHENYou can explain the overall trend and both anomalies with the correct mechanism.

First ionisation energy down Group 2

Learning goalExplain the Group 2 trend using occupied shells, distance, shielding and attraction.

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Down Group 2, first ionisation energy falls even though the nucleus contains more protons. The electron being removed is now in a shell further from the nucleus.

  1. There are more occupied shells.
  2. The outer electron is further away and more shielded by inner electrons.
  3. These effects outweigh the larger nuclear charge, so the attraction is weaker.
  4. Less energy is needed to remove the outer electron.

The repeated drop when a new outer shell begins to fill is evidence for the shell model.

EXAM TRAPDo not stop at “larger atoms”. State greater distance and shielding, then weaker attraction and lower IE.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 41
ElementFirst IE / kJ mol⁻¹
Be900
Mg738
Ca590
Sr550
Ba503
  1. Describe the trend in first ionisation energy down Group 2. [1]
  2. Explain this trend in first ionisation energy. [4]

Total: 5 marks

Show answer & marking points
  1. First IE decreases. [1]
  2. The outer electron occupies a higher principal shell [1], is further from the nucleus [1] and has more shielding [1]. These outweigh increased nuclear charge, so attraction is weaker and less energy is required [1].
✓ DONE WHENYou can explain why first IE decreases despite increasing nuclear charge.

Successive ionisation energies and shell structure

Learning goalUse large jumps to determine outer-electron numbers, group and shell arrangement.

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Each successive ionisation removes an electron from an ion that is already positive, so the energy rises. A particularly large jump tells you that the next electron comes from an inner shell.

Inner-shell electrons are closer to the nucleus and less shielded. They are much more strongly attracted than the outer electrons just removed.

  1. Locate the first large jump.
  2. Count the electrons removed before it. That is the number in the outer shell.
  3. For a main-group element, connect this count to the group: three outer electrons means Group 3; six means Group 6.
  4. If more large jumps are shown, use them to work out further shell populations.
Period 3 elementFirst major jump after
NaIE₁
MgIE₂
AlIE₃
SiIE₄
PIE₅
SIE₆
ClIE₇
ArIE₈
Worked example

Identify a Period 3 element

Successive IEs: 578, 1817, 2745 and 11 578 kJ mol⁻¹.

These are the first four successive ionisation energies of a Period 3 element. Identify the element and explain your reasoning. Write its full electron configuration.

Worked answer

  1. Large jump between IE₃ and IE₄.
  2. Three outer-shell electrons, then an inner-shell electron is removed.
  3. The element is in Group 3. In Period 3, this is aluminium.
Al: 1s² 2s² 2p⁶ 3s² 3p¹.
EXAM TRAPA jump after the third electron means three outer electrons, not four.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 42

A Period 3 element has successive IEs of 786, 1577, 3232, 4356 and 16 091 kJ mol⁻¹. Period 3: Na, Mg, Al, Si, P, S, Cl, Ar.

  1. Where is the large jump? [1]
  2. How many outer-shell electrons? [1]
  3. Identify the element. [1]
  4. Write its full electron configuration. [1]
  5. Explain the jump. [1]

Total: 5 marks

Show answer & marking points
  1. Between IE₄ and IE₅. [1]
  2. Four. [1]
  3. Silicon, Si. [1]
  4. 1s² 2s² 2p⁶ 3s² 3p². [1]
  5. The fifth electron comes from an inner shell and is more strongly attracted to the nucleus. [1]
Exam-style question 43

A Period 3 element has successive IEs of 1000, 2250, 3360, 4560, 7000, 8500, 27 100 and 31 800 kJ mol⁻¹. Period 3: Na, Mg, Al, Si, P, S, Cl, Ar.

  1. Identify the largest jump. [1]
  2. State the number of outer-shell electrons. [1]
  3. Identify the element. [1]
  4. Write its full configuration. [1]
  5. Explain the sharp rise. [3]

Total: 7 marks

Show answer & marking points
  1. Between IE₆ and IE₇. [1]
  2. Six. [1]
  3. Sulfur, S. [1]
  4. 1s² 2s² 2p⁶ 3s² 3p⁴. [1]
  5. The seventh electron comes from n = 2 [1]. It is closer and less shielded [1], so is much more strongly attracted and requires far more energy [1].
Exam-style question 44

Al has successive IEs of 578, 1817, 2745 and 11 578 kJ mol⁻¹.

  1. Write the fourth-ionisation equation, including state symbols. [1]
  2. Explain why IE₄ is much greater than IE₃. [3]
  3. Calculate IE₄ / IE₃ to three significant figures. [1]
  4. What does this show about the number of outer-shell electrons? [1]

Total: 6 marks

Show answer & marking points
  1. Al³⁺(g) → Al⁴⁺(g) + e⁻. [1]
  2. Three outer electrons have already been removed [1]. The fourth comes from an inner shell, closer / less shielded [1], so is much more strongly attracted [1].
  3. 11 578 / 2745 = 4.22. [1]
  4. Al has three outer-shell electrons. [1]
Exam-style question 45

The graph shows all thirteen successive IEs of an unknown Period 3 element. Period 3: Na, Mg, Al, Si, P, S, Cl, Ar.

  1. Identify both large jumps. [2]
  2. State the number of electrons in the outer shell and second shell. [2]
  3. Deduce the shell arrangement and identify the element. [2]
  4. Write its full configuration. [1]
  5. Explain the first large jump. [2]
  6. Why do successive IEs increase even before a new shell is reached? [1]

Total: 10 marks

Show answer & marking points
  1. After IE₃ and IE₁₁. [2]
  2. Outer shell: 3. Second shell: 8. [2]
  3. 2,8,3; aluminium. [2]
  4. 1s² 2s² 2p⁶ 3s² 3p¹. [1]
  5. The next electron comes from n = 2 [1], closer / less shielded and more strongly attracted [1].
  6. Each electron is removed from an increasingly positive ion, so remaining electrons are more strongly attracted. [1]
Exam-style question 46

M has isotopes at m/z 24, 25 and 26 with abundances 78.99%, 10.00% and 11.01%. Relevant atomic numbers: Na = 11, Mg = 12, Al = 13, Ca = 20.

  1. Calculate Aᵣ to two decimal places. [2]
  2. Identify M. [1]
  3. Write the full configuration of M. [1]
  4. Write the configuration of M²⁺. [1]
  5. Write the third-ionisation equation. [1]
  6. The first three IEs are 738, 1451 and 7733 kJ mol⁻¹. Explain what these show about the outer electrons and the next electron removed. [4]
  7. Calcium has first IE 590 kJ mol⁻¹. Why is this lower than M’s? [2]

Total: 12 marks

Show answer & marking points
  1. Aᵣ = Σ(isotopic mass × abundance)/Σ(abundance)
    Aᵣ = [24(78.99) + 25(10.00) + 26(11.01)]/100 [1] = 24.32 [1].
  2. Magnesium, Mg. [1]
  3. 1s² 2s² 2p⁶ 3s². [1]
  4. 1s² 2s² 2p⁶ or [Ne]. [1]
  5. Mg²⁺(g) → Mg³⁺(g) + e⁻. [1]
  6. Large jump between IE₂ and IE₃ [1], hence two outer electrons [1]. The third comes from an inner shell [1], closer / less shielded and much more strongly attracted [1].
  7. Ca has another occupied shell, so the outer electron is further away and more shielded [1]. Attraction is weaker, requiring less energy [1].
✓ DONE WHENYou can use the data to infer outer-electron number, group, element and shell structure.

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