HookNeon came off the machine at two different masses
In 1913, at the Cavendish Laboratory in Cambridge, J. J. Thomson bent a beam of ionised neon through electric and magnetic fields onto a photographic plate — and the beam split. One trace landed where a particle of mass 20 should land; a second, fainter trace landed at mass 22. Neon, one element, one box on the periodic table, was arriving at two different masses. After the First World War his assistant Francis Aston built a far sharper instrument, the mass spectrograph, and settled the matter: about nine neon atoms in every ten are neon-20, roughly one in ten is neon-22, and a trace are neon-21. Chemically they are indistinguishable. Physically they weigh different amounts — which is why neon's relative atomic mass is the un-whole number 20.2. Aston went on to catalogue more than 200 isotopes with his machine and collected the 1922 Nobel Prize in Chemistry.
The modern descendant of that instrument — the time-of-flight (TOF) mass spectrometer — is the one piece of large apparatus AQA expects you to know from the inside, stage by stage. And the whole of 3.1.1 runs on the same three moves Aston made: count the particles (protons, neutrons, electrons), weigh the mixture (isotopes, and the weighted average that gives relative atomic mass), then place the electrons (s, p and d sub-shells — with ionisation energies as the evidence that the arrangement is real). Every mark in this section is one of those three moves.
ModelThree particles and the two numbers that define an atom
An atom is a tiny, dense nucleus of protons and neutrons with electrons arranged in energy levels around it. The exam wants the relative numbers cold: a proton has relative mass 1 and relative charge +1; a neutron has relative mass 1 and charge 0; an electron has relative charge −1 and a relative mass of about 1/1840 — so small that it is ignored in mass calculations. Nearly all of an atom's mass therefore sits in the nucleus.
Two numbers pin down any nuclide. The atomic (proton) number \(Z\) is the number of protons — it defines the element and its position in the periodic table. The mass number \(A\) is protons plus neutrons, so the neutron count is \(A - Z\). In a neutral atom, electrons equal protons; ions break that equality. Take \({}^{56}_{26}\text{Fe}^{2+}\): 26 protons, \(56 - 26 = 30\) neutrons, and \(26 - 2 = 24\) electrons, because a 2+ charge means two electrons have been removed. AQA routinely asks this for ions rather than atoms precisely because the electron arithmetic runs opposite to instinct: positive means fewer electrons, not more protons.
MechanismInside the time-of-flight mass spectrometer
The entire instrument is held under vacuum — any air molecules would collide with the ions and stop them reaching the detector. Then four stages. Ionisation happens one of two ways. In electron impact, the vaporised sample is bombarded by high-energy electrons from an electron gun and each particle loses an electron: \(\text{X(g)} + \text{e}^{-} \rightarrow \text{X}^{+}\text{(g)} + 2\text{e}^{-}\). It suits elements and low-mass molecules, but it is violent enough to break larger molecules into fragments. In electrospray, the sample is dissolved in a volatile solvent and forced through a fine hypodermic needle held at high positive voltage; each particle picks up a proton: \(\text{X(g)} + \text{H}^{+} \rightarrow \text{XH}^{+}\text{(g)}\). This is the gentle option used for large biological molecules — but the ion detected is one unit heavier than the molecule, so you subtract 1 from the m/z to get the true mass.
Acceleration: an electric field gives every ion the same kinetic energy, which is the trick that makes the machine work — for equal \(KE\), lighter ions must travel faster. Flight tube: with no field acting, ions drift at constant speed, and since \(KE = \tfrac{1}{2}mv^{2}\) is fixed, flight time \(t = \dfrac{d}{v}\) grows with \(\sqrt{m}\) — light ions arrive first. Detection: each positive ion gains an electron from the negatively charged detector plate; that flow of electrons is a current whose size is proportional to the ion's abundance. The whole spectrum is captured in microseconds.
Every ion in a TOF instrument is accelerated to a kinetic energy of \(1.20\times10^{-13}\) J and the flight tube is 0.800 m long. Find the flight time of a \({}^{24}\text{Mg}^{+}\) ion. First the mass of one ion: \(m = \dfrac{24}{6.022\times10^{23}} = 3.99\times10^{-23}\) g \(= 3.99\times10^{-26}\) kg — dividing the molar mass by the Avogadro constant, then converting to kilograms. Next rearrange \(KE = \tfrac{1}{2}mv^{2}\) for speed: \[ v = \sqrt{\dfrac{2KE}{m}} = \sqrt{\dfrac{2\times1.20\times10^{-13}}{3.99\times10^{-26}}} = 2.45\times10^{6}\ \text{m s}^{-1} \] Finally \(t = \dfrac{d}{v} = \dfrac{0.800}{2.45\times10^{6}} = 3.26\times10^{-7}\) s — about a third of a microsecond. A \({}^{26}\text{Mg}^{+}\) ion given the same kinetic energy arrives at \(t\propto\sqrt{m}\), so \(3.26\times10^{-7}\times\sqrt{26/24} = 3.39\times10^{-7}\) s. That 0.013-microsecond gap is the entire basis on which the machine separates isotopes.
DataReading the spectrum: relative atomic mass to one decimal place
A mass spectrum of an element plots m/z (mass-to-charge ratio) against relative abundance. For 1+ ions — the normal case — m/z simply equals the isotope's mass. The relative atomic mass is then the weighted mean: multiply each isotopic mass by its abundance, add them up, divide by the total abundance. If the abundances are percentages the divisor is 100; if they are arbitrary detector units, divide by their sum — AQA sets both.
Two refinements earn the top marks. First, if the question says 'identify the element', finish the calculation and then name the element whose \(A_r\) on the periodic table matches — students lose a mark by stopping at the number. Second, elements that exist as molecules produce molecular ions: chlorine gas gives \(\text{Cl}_2^{+}\) peaks at m/z 70, 72 and 74 (35+35, 35+37, 37+37) in the ratio 9 : 6 : 1, which falls straight out of the 3 : 1 odds of drawing each isotope twice. Spotting a diatomic pattern like that is a classic AQA twist.
A mass spectrum of magnesium shows peaks at m/z = 24 (abundance 79.0%), 25 (10.0%) and 26 (11.0%). \[ A_r = \dfrac{(24\times79.0)+(25\times10.0)+(26\times11.0)}{100} = \dfrac{1896+250+286}{100} = 24.3 \] Quote it to one decimal place unless told otherwise, and check it lies between the lightest and heaviest isotope — a weighted mean always must. Run the logic backwards for chlorine: 75% at mass 35 and 25% at mass 37 gives \(\dfrac{(35\times75)+(37\times25)}{100} = 35.5\). No chlorine atom weighs 35.5; the decimal is the fingerprint of a mixture, exactly what Aston saw in neon.
ModelElectron configuration: s, p, d and the 4s wobble
Electrons sit in orbitals — regions of space that each hold at most two electrons of opposite spin. Orbitals group into sub-shells: an s sub-shell is one orbital (2 electrons), p is three orbitals (6), d is five orbitals (10). Sub-shells group into shells, and the filling follows energy, lowest first: 1s, 2s, 2p, 3s, 3p, then — the wobble — 4s before 3d, because the empty 4s sub-shell sits slightly lower in energy than 3d. Within a p or d sub-shell, electrons occupy orbitals singly before any orbital takes a second electron. So iron (\(Z = 26\)) is 1s\(^{2}\) 2s\(^{2}\) 2p\(^{6}\) 3s\(^{2}\) 3p\(^{6}\) 3d\(^{6}\) 4s\(^{2}\), writable in shorthand as \([\text{Ar}]3\text{d}^{6}4\text{s}^{2}\). AQA expects you to write configurations for atoms and ions up to krypton (\(Z = 36\)).
Two atoms disobey the pattern and both are named in mark schemes: chromium is \([\text{Ar}]3\text{d}^{5}4\text{s}^{1}\) and copper is \([\text{Ar}]3\text{d}^{10}4\text{s}^{1}\) — a half-filled or full d sub-shell is stable enough to poach one 4s electron. And when d-block atoms form positive ions, the 4s electrons leave first: \(\text{Fe}^{2+}\) is \([\text{Ar}]3\text{d}^{6}\) and \(\text{Fe}^{3+}\) is \([\text{Ar}]3\text{d}^{5}\). Filled first, emptied first — it feels wrong, which is why examiners keep asking it.
DataIonisation energy: the evidence that shells and sub-shells are real
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: \(\text{Na(g)} \rightarrow \text{Na}^{+}\text{(g)} + \text{e}^{-}\), which for sodium costs +496 kJ mol\(^{-1}\). Three factors control it: nuclear charge, distance of the outer electron from the nucleus, and shielding by inner electrons. Successive ionisation energies expose the shell structure directly. Sodium's second ionisation energy leaps to 4,562 kJ mol\(^{-1}\) — nine times the first — because the second electron must come from the n = 2 shell, far closer to the nucleus and barely shielded. Aluminium's energies climb gently (578, 1,817, 2,745 kJ mol\(^{-1}\)) then explode to 11,577 at the fourth — three outer electrons, then a new shell. Give a chemist the full list of jumps and they can read off the group number.
Across Period 3 the first ionisation energy generally rises — nuclear charge increases while electrons enter the same shell with near-constant shielding — but with two dips that are the exam's favourite question. Aluminium (578) sits below magnesium (738) because its outer electron occupies the higher-energy 3p sub-shell, further out than 3s and slightly shielded by it: evidence that sub-shells exist. Sulfur (1,000) sits below phosphorus (1,012) because sulfur's fourth 3p electron is the first to pair up, and the repulsion between two electrons in one orbital makes it easier to remove: evidence that orbitals hold two electrons. Down any group, ionisation energy falls — extra shells and shielding beat the rising nuclear charge.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Definitions in this section are marked word by word, so learn them tight: ionisation-energy answers must say one mole of gaseous atoms, and the supporting equation must carry (g) on both species — \(\text{Mg(g)} \rightarrow \text{Mg}^{+}\text{(g)} + \text{e}^{-}\) scores, an equation without state symbols does not. In mass-spectrometry questions, name the ionisation method and write its equation: electron impact gives \(\text{X}^{+}\), electrospray gives \(\text{XH}^{+}\) — and remember the electrospray peak sits at Mr + 1. When asked why ions accelerate to the detector or how detection works, answer in the mark scheme's currency: same kinetic energy for all ions; positive ions gain electrons at the detector; the current is proportional to abundance.
TOF calculations are three method marks: convert the ion's mass to kilograms by dividing molar mass by the Avogadro constant, rearrange \(KE = \tfrac{1}{2}mv^{2}\) for \(v\), then \(t = d/v\) — show each line, because a slip in the last step still banks the first two marks. Quote Ar values to one decimal place and sanity-check that your answer lies between the lightest and heaviest isotope; if the question says 'identify the element', write its name.
For electron configurations, write every sub-shell unless the question allows shorthand, know chromium and copper as the two exceptions, and remove 4s electrons first when forming d-block ions. Trend questions on ionisation energy are AO2, not recall: state the pattern, name the outlier, then explain at the level of sub-shells and orbital pairing. Section 3.1.1 is examined on Paper 1 (and the synoptic Paper 3), and AQA weights application at 40–45% of the qualification — expect your knowledge to arrive dressed in an unfamiliar element or an unlabelled spectrum.