From C1 Atomic Structure you already know the nuclear atom, isotopes, and electron shells like 2,8,1. A-Level keeps all of that and adds three upgrades: shells split into sub-shells and orbitals (so sodium becomes 1s22s22p63s1), relative atomic mass becomes something you measure with a mass spectrometer rather than read off the periodic table, and ionisation energies provide the experimental evidence for the whole picture.
Fundamental particles
The model of the atom has evolved over time — from Dalton’s indivisible spheres, through Thomson’s discovery of the electron and Rutherford’s nuclear atom, to Bohr’s shells and today’s quantum-mechanical model of orbitals.
Dalton · 1803
solid, indivisible spheres
Thomson · 1897
“plum pudding” — electrons in a positive ball
Rutherford · 1911
nuclear model — tiny dense nucleus
Bohr · 1913
electrons in fixed shells
Today · 1926→
sub-shells & orbitals — this course
The version you use at A-Level: an atom consists of a tiny, dense nucleus containing protons and neutrons, surrounded by electrons. Treat the current model as the best fit to today’s evidence — you should appreciate that the model has changed before, and can change again, as new evidence arrives.
| Particle | Relative mass | Relative charge |
|---|---|---|
| Proton | 1 | +1 |
| Neutron | 1 | 0 |
| Electron | 1/1836 (≈ 0) | −1 |
In a neutral atom protons = electrons, and almost all of the mass is in the nucleus — the electron’s mass is negligible, which is exactly why the mass number in the next section counts only protons and neutrons.
Mass number, atomic number & isotopes
Two numbers pin down any nuclide. The atomic (proton) number, Z, is the number of protons in the nucleus — it defines the element. The mass number, A, is the total number of protons plus neutrons. They are written around the symbol as AZX — so 2311Na has 11 protons, 11 electrons and 23 − 11 = 12 neutrons.
Protons = Z. Neutrons = A − Z. Electrons = Z, adjusted for any charge.
For ions, only the electron count changes: the charge tells you how many electrons were lost or gained. 2412Mg2+ has 12 protons, 12 neutrons and 10 electrons; 168O2− has 8 protons, 8 neutrons and 10 electrons.
Change the number of protons → a different element. Change the number of electrons → an ion of the same element. Change the number of neutrons → an isotope of the same element.
Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons — the same atomic number, different mass numbers.
Isotopes of an element have identical chemical properties, because chemical behaviour is decided by the number and arrangement of electrons, and isotopes have the same electron configuration. Their physical properties (density, rate of diffusion, boiling point) differ slightly because the masses differ.
🧪 Exam-style questions
Which row shows the number of each fundamental particle in one 25Mg2+ ion? Tick (✓) one box.
Which atom has the smallest number of neutrons? Tick (✓) one box.
Which statement about isotopes of an element is not correct? Tick (✓) one box.
Time of flight mass spectrometry
The mass spectrometer gives accurate information about relative isotopic mass and relative abundance — it is how relative atomic masses are actually measured. You need the time of flight (TOF) design: four stages, in order, each with a one-line reason. The principle is simple: give every ion the same kinetic energy, and the lighter ones move faster — so they arrive first, and arrival time measures mass. The output is a mass spectrum: a plot of mass-to-charge ratio, m/z (for a 1+ ion, m/z is simply the ion’s mass), against the abundance of ions arriving at each value.
Interactive — race two isotopes at the same kinetic energy
Both ions are accelerated to the same kinetic energy (1.0 × 10−13 J) into a 1.0 m flight tube — press Play to race them.
The flight times stamp in µs (microseconds) — the prefix µ (micro) means 10−6, so 4.4 µs = 4.4 × 10−6 s. Microseconds are the natural scale for a TOF instrument, and a useful sense-check on your own calculated answers.
Stage 1 — ionisation
Two methods; you need to be able to choose and describe the right one for a given sample.
Electron impact (for smaller molecules and elements): the vaporised sample is bombarded with high-energy electrons from an electron gun, knocking one electron off each particle to form a 1+ ion:
X(g) → X+(g) + e−
Electrospray (for larger molecules such as proteins): the sample is dissolved in a volatile, polar solvent such as water or methanol, then injected through a fine hypodermic needle whose tip is attached to the positive terminal of a high-voltage supply. The needle turns the solution into a fine mist (an aerosol), and each particle gains a proton from the solvent as it leaves the needle; the solvent evaporates away while the ions are drawn towards a negative plate:
X(g) + H+ → XH+(g)
A polar substance has a small separation of charge across its molecules — water and methanol both qualify. Polarity is taught properly in Bonding (3.1.3); for now, just learn volatile, polar solvent as part of the electrospray description.
Why two methods? Electron impact dumps a lot of energy into the particle — a larger molecule’s ion often breaks apart (fragments), giving extra peaks at lower m/z values, and in the worst case no intact molecular ion survives to be detected. Electrospray is a soft technique: the particle just gains a proton, so the intact XH+ ion reaches the detector — which is exactly why it is the choice for proteins. (Interpreting fragment patterns is not examined here — it returns in organic analysis.)
- With electrospray the detected ion is XH+, so its m/z is Mr + 1 — subtract 1 to get the relative molecular mass, because the particle carries an extra proton.
- With electron impact, the electron sits on the product side: X(g) → X+(g) + e− (X + e− → X+ + 2e− says the same thing).
- When describing electrospray, name the solvent type in full: volatile and polar.
Stage 2 — acceleration
The positive ions are attracted towards a negatively charged plate, and the electric field accelerates them so that all ions of the same charge gain the same kinetic energy; they pass through a hole in the plate into the flight tube. Because KE = ½mv2 is fixed, velocity depends only on mass: the lighter the ion, the faster it travels.
Interactive — same kinetic energy, so mass sets the speed
KE = ½mv2 = 1.0 × 10−13 J — the same for every ion
Ion A keeps a relative mass of 20. Change the mass of ion B, and watch its speed adjust so that ½mv2 stays exactly the same: more m forces less v.
Ion B is 2× the mass of ion A, so it moves √2 ≈ 1.4× slower — the two kinetic energies are identical.
Stage 3 — ion drift
The ions enter a field-free flight tube (the drift region). With no force acting, each ion keeps a constant speed, so the ions separate: light ions pull ahead, heavy ions lag. Each ion crosses the tube (length d) at its own steady speed v, so its flight time is:
t = d / v
Stage 4 — detection
The detector is a negatively charged plate. Each positive ion arriving is discharged by gaining an electron from the plate — so electrons flow from the detector to the ion. That electron flow is a current, and the size of the current is proportional to the abundance of ions arriving at that flight time. The spectrometer converts arrival times to m/z and plots the spectrum.
Detector answers are marked tightly — examiner reports repeatedly flag answers that mention a current without linking it to electron gain and ion abundance. Give all three ideas:
- each positive ion gains an electron at the detector plate…
- …so electrons flow from the detector to the ion — that flow is a current…
- …and the size of the current is proportional to the abundance of ions arriving.
Saying only “the ions reach the detector” describes the geography, not the chemistry.
From KE = ½mv2 and v = d/t, every TOF calculation is the same four moves:
- Find the mass of one ion, in kg — divide the relative mass by the Avogadro constant and by 1000:
m = relative mass6.022 × 1023 × 1000 kg
- Rearrange for the quantity you need. Two rearrangements cover everything:
v = √2KEm t = d √m2KE
- Substitute the numbers — and do not lose the square root on the way.
- Round to sensible significant figures, with the unit.
The combined equation is a two-line “show that”:
Line 1 — rearrange the kinetic energy equation for v:
KE = ½mv2 → v = √(2KE/m)
Line 2 — substitute into t = d/v:
t = d ÷ √(2KE/m) = d √(m/2KE)
Because d and KE are fixed, t ∝ √m: for two ions in the same tube, t2 = t1 × √(m2/m1), and the heavier ion always lands later — a built-in sense check. The KE = ½mv2 equation is given where it is needed: the marks are for the algebra and the kg conversion, not recall.
In a TOF spectrometer, each 79Br+ ion has kinetic energy 1.20 × 10−13 J and the flight tube is 1.50 m long. Calculate the time of flight.
Step 1 — mass of one ion, in kg:
m = 79 ÷ (6.022 × 1023 × 1000) = 1.31 × 10−25 kg
Step 2 — speed, from the kinetic energy:
v = √(2KE/m) = √(2 × 1.20 × 10−13 ÷ 1.31 × 10−25) = 1.35 × 106 m s−1
Step 3 — time over the flight tube:
t = d/v = 1.50 ÷ (1.35 × 106) = 1.11 × 10−6 s (3 s.f.)
Sense-check: microseconds is the right order of magnitude for a TOF instrument.
Interactive — your turn, one step at a time
In a TOF spectrometer, each 56Fe+ ion has kinetic energy 1.20 × 10−13 J and the flight tube is 0.960 m long. Calculate the time of flight. (KE = ½mv2; L = 6.022 × 1023 mol−1.) Give each answer as a number to 3 significant figures times a power of ten — get a step right to unlock the next.
-
Step 1 — mass of one ion, in kg:
m = 56 ÷ (6.022 × 1023 × 1000) = kg -
Step 2 — speed, from the kinetic energy:
v = √(2KE/m) = m s−1 -
Step 3 — time over the flight tube:
t = d/v = s
Type the number that goes in front of the power of ten — the routine is the same every time.
🧪 Exam-style questions
In a time of flight mass spectrometer, molecule X is ionised using electrospray ionisation. What is the equation for this ionisation? Tick (✓) one box.
Samples of four different substances are analysed by TOF mass spectrometry. In each case the samples are ionised to form 1+ ions, and the ions are accelerated to the same kinetic energy. Which sample gives ions with the shortest time of flight? Tick (✓) one box.
A sample of antimony analysed in a TOF mass spectrometer contains two isotopes, 121Sb and 123Sb. After electron impact ionisation, all ions are accelerated to the same kinetic energy and travel through a flight tube 1.05 m long. A 121Sb+ ion takes 5.93 × 10−4 s to travel through the flight tube. (KE = ½mv2; L = 6.022 × 1023 mol−1.)
Calculate the mass, in kg, of one 121Sb+ ion, and the time taken for a 123Sb+ ion to travel through the same flight tube.
Show answer
m(121Sb+) = 121 ÷ (1000 × 6.022 × 1023) = 2.009 × 10−25 kg
Mass of one ion in kg 1 mark
v = d/t = 1.05 ÷ (5.93 × 10−4) = 1770.7 m s−1
Speed of 121Sb+ 1 mark
KE = ½mv2 = ½ × 2.009 × 10−25 × (1770.7)2 = 3.149 × 10−19 J
Kinetic energy (the same for both ions) 1 mark
v(123Sb+) = √(2KE/m) = √(2 × 3.149 × 10−19 ÷ 2.0425 × 10−25) = 1756.1 m s−1
Speed of 123Sb+ 1 mark
t = d/v = 1.05 ÷ 1756.1 = 5.98 × 10−4 s
Final answer 1 mark
Alternative route: since KE is equal, m/t2 is constant, so t1232 = (123/121) × t1212 — same answer. Sense-check: the heavier isotope must arrive slightly later.
The two products of a reaction are analysed in a TOF mass spectrometer. Two peaks are observed, at m/z = 104 and m/z = 118. Outline how the TOF mass spectrometer is able to separate these two species to give two peaks.
Show answer
Positive ions are accelerated by an electric field 1 mark
to a constant kinetic energy 1 mark
The ions at m/z = 104 have the same kinetic energy as those at m/z = 118 and move faster 1 mark
Therefore the ions at m/z = 104 arrive at the detector first 1 mark
The four-step structure is the answer: accelerated → same KE → lighter is faster → lighter arrives first. Missing the “same kinetic energy” step breaks the logic and usually the marks with it.
Source: AQA AS Chemistry past paper.
Mass spectra & relative atomic mass
A mass spectrum of an element shows one peak per isotope: the position on the x-axis is m/z (mass-to-charge ratio — equal to the isotope’s mass for 1+ ions) and the height is the relative abundance. If electron impact knocks out two electrons, the 2+ ion appears at half the isotope’s mass — 40Ar2+ sits at m/z = 20, exactly where 20Ne+ would — so an unexpected peak at half an isotope’s mass is usually a doubly charged ion. From the positions and heights, Ar is a weighted mean.
Relative atomic mass, Ar, is the average mass of an atom of an element compared with 1/12th the mass of an atom of carbon-12.
This is a two-mark definition marked word by word: average mass (never the “mass of an average atom”), mass not “weight”, and the carbon-12 reference. Defining per mole is equally credited (the average mass of one mole of atoms compared with 1/12th the mass of one mole of 12C) — but never mix atoms and moles in one sentence.
Interactive — drag the peaks, watch Ar follow
Peak heights here are relative intensities — like the numbers a spectrometer reports, they do not have to add up to 100. That is exactly why the calculation divides by their total, never automatically by 100.
Ar = (24 × 79.0) + (25 × 10.0) + (26 × 11.0)79.0 + 10.0 + 11.0 = 2432100.0 =
A mystery sample gives this spectrum. Calculate Ar for the sample (1 decimal place):
Correct. Which element is it?
Ar = (24 × 79.0) + (25 × 10.0) + (26 × 11.0)79.0 + 10.0 + 11.0 = 2432100 = 24.3
Quote to 1 decimal place (or 3 significant figures) here — the answer should not claim more precision than the data (MS 1.1). A sense-check: the answer must sit between the lightest and heaviest isotope, dragged towards the most abundant one.
- Divide by the total abundance, not by the number of isotopes — and not automatically by 100. Reaching for 100 as the denominator when the abundances do not sum to 100 is a recurring error named in examiner reports.
- Show the division by the abundance total explicitly in your working.
- Don’t round Ar to the periodic-table value: the question is about this sample, and the data set the precision.
The same logic runs in reverse: given Ar and all but one abundance, you can solve for the unknown. And spectra identify elements outright — an unknown with peaks at m/z 63 and 65 in a ratio near 7:3 can only be copper.
For molecules, the molecular-ion peak — the significant peak at the highest m/z value — gives Mr directly with electron impact, or Mr + 1 with electrospray. Ignore the very small peak one unit higher still, which comes from molecular ions containing one 13C or 2H atom; with electron impact, the peaks at lower m/z are fragments of the molecular ion. Ar calculations stay with mononuclear (single-atom) ions, but identifying the ions behind a diatomic’s peaks is fair game: chlorine gas shows peaks at m/z 70, 72 and 74 from (35Cl2)+, (35Cl37Cl)+ and (37Cl2)+.
- Name every peak as an ion, with the + charge shown and the mass number matched to its m/z: the peak at 72 in a germanium spectrum is 72Ge+, not “Ge”.
- When asked how a spectrum gives Mr, say “Mr = the highest m/z value” — not the “largest peak” (that means most abundant) or the “right-hand peak” (that describes the picture).
🧪 Exam-style questions
Define the term relative atomic mass.
Show answer
The average (mean) mass of an atom of an element 1 mark
compared to 1/12th the mass of an atom of carbon-12 1 mark
Do not accept “mass of average atom”. “Weight” is penalised. Defining with moles is fine (mass of 1 mole of atoms vs 1/12th the mass of 1 mole of 12C) — but mixing atoms and moles caps the answer at 1 mark.
A sample of krypton is ionised using electron impact. The mass spectrum of this sample has four peaks, at m/z = 82, 83, 84 and 86 with relative intensities 6, 1, 28 and 8 respectively. Calculate the relative atomic mass (Ar) of this sample of krypton. Give your answer to 1 decimal place.
Show answer
Ar = [(82 × 6) + (83 × 1) + (84 × 28) + (86 × 8)] ÷ (6 + 1 + 28 + 8) = 3615 ÷ 43
Working shown as above 1 mark
Ar = 84.1 (1 d.p.) 1 mark
In a TOF mass spectrometer, ions are accelerated to the same kinetic energy (KE = ½mv2). Each 84Kr+ ion is accelerated to a kinetic energy of 4.83 × 10−16 J and the time of flight is 1.72 × 10−5 s. Calculate the length, in metres, of the TOF flight tube. (The Avogadro constant L = 6.022 × 1023 mol−1.)
Show answer
m = 84 ÷ (1000 × 6.022 × 1023) = 1.395 × 10−25 kg
Mass of one ion in kg 1 mark
v2 = 2KE/m = (2 × 4.83 × 10−16) ÷ (1.395 × 10−25)
Rearrangement and substitution 1 mark
v = 8.32 × 104 m s−1
Speed 1 mark
d = v × t = 8.32 × 104 × 1.72 × 10−5 = 1.43 m
Accept 1.43–1.44 m 1 mark
If the mass is not converted to kg, d comes out as 0.045 m — max 3 of the 4 marks.
A sample of chromium containing the isotopes 50Cr, 52Cr and 53Cr has a relative atomic mass of 52.1. The sample contains 86.1% of the 52Cr isotope. Calculate the percentage abundance of each of the other two isotopes.
Show answer
% of 50Cr + 53Cr = 100 − 86.1 = 13.9%
Let % of 53Cr = x, so % of 50Cr = (13.9 − x) 1 mark
52.1 = [(86.1 × 52) + 50(13.9 − x) + 53x] ÷ 100 → 3x = 37.8
Equation set up and simplified 1 mark
x = % of 53Cr = 12.6% 1 mark
% of 50Cr = 13.9 − 12.6 = 1.3% 1 mark
State, in terms of the numbers of fundamental particles, one similarity and one difference between atoms of 50Cr and 53Cr.
Show answer
Similarity: (same) number of protons OR electrons 1 mark
Difference: (different) number of neutrons 1 mark
“Same electron configuration” is not allowed for the similarity mark — the question asks in terms of numbers of fundamental particles.
The sample of chromium is analysed in a TOF mass spectrometer. Give two reasons why it is necessary to ionise the isotopes of chromium before they can be analysed.
Show answer
So the particles can be accelerated (by the electric field) 1 mark
So they create a current when hitting the detector 1 mark
A 53Cr+ ion travels along a flight tube of length 1.25 m with a constant kinetic energy of 1.102 × 10−13 J. Calculate the time, in s, for the ion to reach the detector. (KE = ½mv2; L = 6.022 × 1023 mol−1.)
Show answer
m = 53 ÷ (1000 × 6.022 × 1023) = 8.8 × 10−26 kg
Mass of one ion in kg 1 mark
v = √(2KE/m)
Rearrangement 1 mark
v = √(2 × 1.102 × 10−13 ÷ 8.8 × 10−26) = 1.58 × 106 m s−1
Speed 1 mark
v = d/t so t = d/v 1 mark
t = 1.25 ÷ (1.58 × 106) = 7.90 × 10−7 s (2 s.f. or more)
Final answer 1 mark
The element rubidium exists as the isotopes 85Rb and 87Rb. In a sample, the isotope 85Rb has an abundance 2.5 times greater than that of 87Rb. Calculate the relative atomic mass of rubidium in this sample. Give your answer to 1 decimal place.
Show answer
numerator = (85 × 2.5) + (87 × 1) = 299.5
Weighted sum set up with the 2.5 : 1 ratio 1 mark
Ar = 299.5 ÷ 3.5
Divide by the abundance total, 2.5 + 1 = 3.5 1 mark
Ar = 85.6 (1 d.p.)
Answer 1 mark
This is the ratio-form version of “divide by the total abundance, not by 100” — here the total is 3.5. Converting the ratio to percentages (71.4% and 28.6%) and dividing by 100 scores equally.
Source: AQA AS Chemistry past paper.
A sample of sulfur consisting of three isotopes has a relative atomic mass of 32.16. Two of the isotopes have these relative abundances: mass number 32 — 91.0%; mass number 33 — 1.8%. Use this information to determine the relative abundance and hence the mass number of the third isotope. Give your answer to the appropriate number of significant figures.
Show answer
abundance of third isotope = 100 − 91.0 − 1.8 = 7.2%
Abundance found 1 mark
32.16 = [(32 × 91.0) + (33 × 1.8) + 7.2y] ÷ 100
Weighted mean set up with the unknown mass number y 1 mark
7.2y = 3216 − 2912 − 59.4 = 244.6
Rearranged 1 mark
y = 244.6 ÷ 7.2 = 33.97 → mass number = 34
Answer as an integer 1 mark
A mass number is a count of protons and neutrons, so 33.97 scores nothing — round to the integer. Contrast: a relative isotopic mass calculated from data is quoted as a decimal, not rounded.
Source: AQA AS Chemistry past paper.
Electron configuration
At GCSE, electrons sat in shells — 2, 8, 8 — and you were never asked to go beyond the first 20 elements. That was no accident: the shell model breaks down at element 21. The third shell is not actually full at 8 electrons (it holds 18), and from scandium onwards the simple 2,8,8 picture cannot explain where the electrons go. The A-Level model fixes this by looking inside each shell: every shell splits into sub-shells (labelled s, p and d), and each sub-shell is built from orbitals — regions of space that each hold a maximum of two electrons. An s sub-shell has 1 orbital (2 electrons), a p sub-shell has 3 orbitals (6 electrons), and a d sub-shell has 5 orbitals (10 electrons) — so the third shell really holds 2 + 6 + 10 = 18.
One new idea before the filling rules: every electron has a property called spin. You do not need the physics behind it — just that spin comes in exactly two states, drawn as an up arrow (↑) and a down arrow (↓), and that two electrons can only share an orbital if their spins are opposite.
Electrons fill the lowest-energy sub-shells first. That lowest-energy arrangement is the ground state — the configuration every question means unless it says otherwise (promote an electron to a higher sub-shell and you have an excited state). The filling order is regular until you reach the fourth shell, where it crosses over: 4s fills before 3d, because the 4s sub-shell is slightly lower in energy than 3d.
Configurations are written as the sub-shells in order with superscript electron counts:
- Na (Z = 11): 1s22s22p63s1
- Fe (Z = 26): 1s22s22p63s23p63d64s2
- Br (Z = 35): 1s22s22p63s23p63d104s24p5
The [noble gas] shorthand replaces the inner electrons with the previous noble gas in brackets: K is 1s22s22p63s23p64s1 in full, or [Ar]4s1 for short. One caution: when a question asks for the full electron configuration, shorthand loses the mark — write everything from 1s.
You need configurations up to Z = 36 (krypton) — and one step beyond can be deduced: a previous exam question asked for the full configuration of rubidium (Z = 37), where a new shell simply starts (…4p65s1).
Within a sub-shell, the orbitals fill in a fixed pattern. Line by line:
- Electrons all carry the same negative charge, so they repel each other.
- To keep that repulsion low, they spread out: every orbital in the sub-shell takes one electron first.
- The single electrons all keep the same spin direction — drawn as parallel arrows (↑ ↑ ↑).
- Only once every orbital holds one electron do they begin to pair up — and each pair must have opposite spins (↑↓).
That is exactly what electrons-in-boxes diagrams show, and why nitrogen’s three 2p electrons sit one per orbital (↑ ↑ ↑) rather than pairing up early (↑↓ ↑ with an orbital left empty).
Interactive — build the configuration, electron by electron
Tap a box to add an electron — tap again to pair it, again to clear. Nothing stops a wrong arrangement, so press Check answer when you think it’s right.
Ions: remove (or add) electrons from the configuration of the atom — but for d-block elements the 4s electrons are lost first: Fe is …3d64s2, so Fe2+ is …3d6 and Fe3+ is …3d5.
Exceptions: chromium is [Ar] 3d54s1 and copper is [Ar] 3d104s1 — a half-filled or full d sub-shell with one 4s electron is more stable than the “expected” arrangement.
An element’s block is the sub-shell type of its highest-energy electrons: Na (…3s1) and Mg are s block, Al–Ar are p block, Sc–Zn are d block. Chromium is d block because its highest-energy electrons sit in 3d — and zinc is d block even though its configuration can be written ending …4s2, because the d sub-shell is what was being filled across that row. Block answers are strict one-letter answers (“s”, “p”, “d” — not a specific orbital such as 2s). Blocks formally belong to Periodicity (3.2.1), but they are asked wherever configurations are.
Interactive — the periodic table in blocks
The table below is coloured by block — the sub-shell being filled across each region. Tap any element up to krypton (Z = 36, the ones you can be asked about) to see its full electron configuration.
Tap an element up to Kr (Z = 36) to see its configuration and block.
- Within a written configuration, 3d and 4s can go either way round (…3d64s2 or …4s23d6) — but when a d-block atom forms an ion, the 4s electrons leave first.
- In an electrons-in-boxes diagram, two arrows sharing a box must point opposite ways — same-direction arrows contradict the opposite spins that let them share an orbital.
🧪 Exam-style questions
What is the electron configuration of V2+ in the ground state? Tick (✓) one box.
Which is the electron configuration of an atom with only two unpaired electrons? Tick (✓) one box.
Ionisation energies
The first ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.
Three phrases carry the marks in that definition:
- one mole — it is a molar quantity, measured in kJ mol−1;
- gaseous — which is why state symbols are compulsory in the equation;
- 1+ ions — one electron removed per atom, no more.
Learn the equation alongside the definition — it says the same thing in symbols, and writing both together is the safest way to secure the marks. For sodium:
Na(g) → Na+(g) + e−
Successive ionisation energies remove further electrons from the increasingly positive ion — the second ionisation energy of sodium is:
Na+(g) → Na2+(g) + e−
Three factors control how tightly an electron is held, and every explanation in this section is built from them: the nuclear charge (more protons, stronger attraction), the distance of the outer electron from the nucleus, and the shielding by inner-shell electrons.
Evidence for shells — successive ionisation energies
Sodium’s successive ionisation energies show a huge jump between the 1st and 2nd: the first electron comes from the outer 3s, but the second must come from the 2p sub-shell in the second shell — much closer to the nucleus, with less shielding. The jumps map the shell structure directly, and they identify an element’s group: a big jump between the nth and (n+1)th ionisation energies means n electrons in the outer shell.
Evidence for sub-shells — first ionisation energies across Period 3
The general increase across the period: nuclear charge increases while the electrons enter the same shell with (almost) the same shielding, so attraction strengthens. The two dips carry the marks:
- Mg → Al: aluminium’s outer electron is in the 3p sub-shell, which is higher in energy than magnesium’s 3s (and slightly shielded by it) — so it is easier to remove despite the greater nuclear charge. Evidence that the third shell has s and p sub-shells.
- P → S: in sulfur, two 3p electrons are paired in the same orbital for the first time; their mutual repulsion makes one easier to remove than phosphorus’s unpaired 3p electrons. Evidence for orbitals within the p sub-shell.
Down Group 2 — Be to Ba
First ionisation energy decreases down the group: each step down adds a shell, so the outer electrons are further from the nucleus and more shielded, and those two factors outweigh the increased nuclear charge.
- Ionisation equations need (g) on the atom and on the ion — the definition requires gaseous species, so the equation must show it. The electron takes no state symbol.
- In explanations, handle all three factors explicitly — nuclear charge, distance, shielding — rather than a vague “further out so easier to remove”.
- Say nuclear charge or “more protons”, not “bigger atomic number” — atomic number is a label, not a force.
- Check the trend’s direction before explaining it — stating it backwards is a chemical error that costs the whole part.
- The second ionisation energy of sodium starts from Na+(g) — never Na(g) losing two electrons.
Capstone — read the jumps
Everything in this section comes together in one skill: given a run of successive ionisation energies, find the big jump and count the electrons removed before it — that count is the number of outer-shell electrons, and therefore the group.
Interactive — mystery element: which group?
The first eight successive ionisation energies of a Period 3 element, in kJ mol−1, on a log scale (the jumps are so big that a normal scale would flatten everything else).
Find the biggest jump between consecutive bars, count the electrons removed before it, then pick the group.
🧪 Exam-style questions
Which equation shows the process that occurs during the second ionisation of magnesium? Tick (✓) one box.
The first six ionisation energies, in kJ mol−1, of an element are: 1090, 2350, 4610, 6220, 37 800, 47 000. What is the element? Tick (✓) one box.
In which pair is the first ionisation energy of atom Y greater than that of atom X? Tick (✓) one box.
The Ne atom and the Mg2+ ion have the same electron configuration (1s22s22p6). Give two reasons why the first ionisation energy of neon is lower than the third ionisation energy of magnesium.
Show answer
The Mg2+ ion is smaller than the Ne atom / the electron removed is closer to the nucleus in Mg2+ 1 mark
Mg2+ has more protons (a higher nuclear charge) than Ne / the electron is removed from a 2+ ion rather than a neutral atom 1 mark
Isoelectronic species have identical configurations, so shell and shielding are the same — the whole comparison rests on nuclear charge and the charge on the species losing the electron. Converse arguments about Ne score equally.
Source: AQA AS Chemistry past paper.
- Relative masses/charges — proton 1, +1; neutron 1, 0; electron 1/1836, −1. Nucleus of protons + neutrons, electrons around it; the model has changed as evidence accumulated.
- Z and A — protons = Z; neutrons = A − Z; electrons = Z adjusted for ion charge.
- Isotopes — same protons, different neutrons; identical chemistry because identical electron configuration.
- TOF spectrometer — ionisation (electron impact X(g) → X+(g) + e−, or electrospray X + H+ → XH+); acceleration to constant KE; field-free drift (lighter = faster); detection by electron gain, current ∝ abundance. Vacuum throughout.
- TOF maths — KE = ½mv2, t = d/v; mass of one ion in kg = relative mass ÷ (L × 1000). Electrospray peaks sit at Mr + 1.
- Ar — weighted mean: Σ(mass × abundance) ÷ total abundance; answer between lightest and heaviest isotope, sensible sig figs.
- Configurations — orbitals hold 2 electrons of opposite spin; s/p/d hold 2/6/10; fill 1s 2s 2p 3s 3p 4s 3d 4p; singly first within a sub-shell; up to Z = 36.
- d-block ions lose 4s first — Fe3+ = [Ar]3d5; exceptions Cr [Ar]3d54s1, Cu [Ar]3d104s1.
- First ionisation energy — one mole of electrons from one mole of gaseous atoms: X(g) → X+(g) + e−; controlled by nuclear charge, distance, shielding.
- Evidence — successive-IE jumps reveal shells (and the group); Period 3 dips at Al (3p above 3s) and S (paired 3p repulsion) reveal sub-shells and orbitals; Group 2 IE falls down the group.