AQA A Level Chemistry: Atomic Structure — Revision Notes

Condensed for the final weeks. For the full explanation, use the Atomic Structure study guide.

Fundamental particles

Particle	Relative mass	Relative charge
Proton	1	+1
Neutron	1	0
Electron	1/1836	−1

Atomic number = protons. Mass number = protons + neutrons. Isotopes differ only in neutron number and therefore have identical chemical properties, since chemistry depends on electron arrangement.

Time-of-flight mass spectrometry

Five stages, and the marks are for naming all five in order with what happens at each:

1. Ionisation — either electron impact (a high-energy electron knocks one out, forming X⁺; used for elements and low-Mr compounds) or electrospray ionisation (the sample gains a proton, forming XH⁺; used for large biological molecules that would fragment).
2. Acceleration — ions are accelerated by an electric field so that all have the same kinetic energy.
3. Ion drift — ions travel through a field-free region; lighter ions travel faster and arrive first.
4. Detection — ions gain electrons at the detector, generating a current proportional to abundance.
5. Data analysis — the spectrum gives m/z values and relative abundances.

KE = ½mv²    so    t = d √(m / 2KE)

Worked example. An ion of mass 1.4 × 10⁻²⁵ kg has kinetic energy 4.8 × 10⁻¹⁶ J. Rearranging KE = ½mv²: v = √(2 × KE / m) = √(2 × 4.8 × 10⁻¹⁶ / 1.4 × 10⁻²⁵) = √(6.86 × 10⁹) ≈ 8.3 × 10⁴ m/s.

The critical distinction: with electrospray, the ion is XH⁺, so the m/z value is one greater than the Mr and you must subtract 1. Forgetting this is the standard error. Worked example: a sample analysed by electrospray gives a peak at m/z 181, so Mᵣ = 181 − 1 = 180; with electron impact ionisation no such correction is needed, since the ion formed is X⁺ itself, with the same mass as the original molecule.

relative atomic mass = Σ(isotope mass × abundance) / total abundance

Worked example. Magnesium has isotopes of mass 24, 25 and 26 with abundances 79.0%, 10.0% and 11.0%. Aᵣ = [(24 × 79.0) + (25 × 10.0) + (26 × 11.0)] ÷ 100 = (1896 + 250 + 286) ÷ 100 = 2432 ÷ 100 = 24.3 (to one decimal place).

Electron configuration

Sub-shells hold: s = 2, p = 6, d = 10, f = 14.

Filling order: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p.

4s fills before 3d but empties first on ionisation. Fe = [Ar]3d⁶4s²; Fe²⁺ = [Ar]3d⁶, not 3d⁴4s².

Two anomalies, caused by the extra stability of half-filled and filled d sub-shells:

Cr = [Ar] 3d5 4s1
Cu = [Ar] 3d10 4s1

Ionisation energies

First ionisation energy — the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.

Every underlined element is a mark: one mole, gaseous, and the specified charge.

Three factors govern it: nuclear charge, atomic radius, and shielding by inner shells.

The trends and what they prove

Down a group, ionisation energy decreases: more shells, greater atomic radius, more shielding — so the outer electron is held less tightly, and this outweighs the increased nuclear charge.

Across a period, ionisation energy generally increases: nuclear charge rises while shielding stays roughly constant and radius decreases.

The two dips across period 3 are the highest-value part of the topic, because each one is direct evidence for sub-shell structure:

• Al is lower than Mg — aluminium’s outer electron is in a 3p sub-shell, which is higher in energy and slightly shielded by the 3s electrons, so it is easier to remove.
• S is lower than P — sulfur has two electrons paired in one 3p orbital, and the repulsion between them makes one easier to remove.

Successive ionisation energies show the group of an element. A large jump occurs when an electron is removed from a shell closer to the nucleus. Count the electrons removed before the jump: that is the number in the outer shell, and hence the group.

Exam traps

• Omitting “one mole” or “gaseous” from the definition.
• Forgetting to subtract 1 from the m/z value in electrospray ionisation.
• Writing Fe²⁺ as 3d⁴4s².
• Explaining the Al dip by nuclear charge rather than by sub-shell.
• Explaining the S dip by anything other than paired-electron repulsion.
• Giving the mass spectrometry stages out of order or omitting one.

Self-test

1. Name the five stages of TOF mass spectrometry in order.
2. Why must you subtract 1 from the m/z in electrospray ionisation?
3. Define first ionisation energy precisely.
4. Explain why aluminium has a lower first ionisation energy than magnesium.
5. Explain why sulfur has a lower first ionisation energy than phosphorus.

Answers: 1. Ionisation, acceleration, ion drift, detection, data analysis. 2. The sample gains a proton, so the ion detected is XH⁺ and its mass is one greater than the molecule’s Mr. 3. The energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions. 4. Aluminium’s outer electron is in a 3p sub-shell, which is higher in energy and slightly shielded by the 3s electrons, so less energy is needed to remove it. 5. In sulfur, two electrons occupy the same 3p orbital and repel each other, making one easier to remove than an unpaired 3p electron in phosphorus.