Part 1: Electron structure, atomic radius and ionisation energy

Use electron configurations to explain recurring properties, then build complete answers for Period 3 radius and ionisation-energy comparisons.

Start with the periodic table

Elements are ordered by proton number. A period is a horizontal row; a group is a vertical column. Periodicity means that patterns in properties recur as the outer-electron arrangements repeat. Elements in a main group have similar outer configurations, which helps explain similar chemical reactions.

The blocks identify the subshell being filled across that region of the table: s on the left, p on the right, d in the central region and f in the two rows usually printed below. Helium has 1s², so it is s-block even though its chemistry places it with the noble gases. Block is not simply the last symbol written in a configuration: conventions for writing 3d and 4s differ.

Period 3 runs from Na to Ar. All have a filled neon core, 1s² 2s² 2p⁶, and fill the n = 3 shell. There are no d-block elements in this period.

Why atomic radius decreases

Across Period 3, proton number increases while the extra electrons enter the same principal shell. The inner-shell shielding is similar. The stronger attraction between the nucleus and the outer electrons draws that shell closer, so atomic radius decreases.

Extra electrons do not automatically make the atom bigger: ask whether they enter an additional shell or the existing one. Across a period it is the existing shell; down a group it is a new shell.

Use the radius definition supplied with a data table. Covalent, metallic and non-bonded radii are not identical measurements; a noble-gas non-bonded radius should not be treated as a directly comparable covalent radius.

First ionisation energy: the general trend

First ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions. It is positive because energy is needed to separate an electron from the nucleus. Units are kJ mol⁻¹.

In general it increases from Na to Ar: the outer electron is in the same principal shell, with similar shielding, but the nucleus has more protons. Attraction to the electron is stronger, so more energy is needed to remove it. The trend has two dips, so do not describe it as an uninterrupted increase.

Explain the two dips using the electron removed

For Mg → Al, compare 3s² with 3s² 3p¹. Aluminium loses a 3p electron. The 3p subshell is higher in energy than 3s, so that electron is easier to remove even though aluminium has an extra proton. Both subshells belong to n = 3: this is not removal from a new principal shell.

For P → S, compare 3p³ with 3p⁴. Phosphorus has one electron in each of the three p orbitals. Sulfur has a pair in one p orbital. Repulsion between the paired electrons makes removal of one of them easier, causing the dip.

“Full subshells are stable” or “sulfur has more electrons” does not explain which electron is easier to remove. Locate the electron, identify the relevant energy or pairing difference, then connect it to lower ionisation energy.

Transfer the explanation without changing the science

The analogous Period 2 dips are Be → B and N → O. Use 2s/2p, not 3s/3p. In an unfamiliar comparison, write the configurations first and identify the electron actually being removed.

For successive ionisations, also check the charge of the starting ion. A large jump after the third electron, for example, indicates that the fourth is being removed from an inner shell. This is different from the two small first-ionisation dips across a period.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources