Part 1: Periodic patterns and ionisation energy

Explain repeating properties from electron configuration, and use ionisation energies as evidence for shells and sub-shells.

Order by proton number

The periodic table arranges elements in increasing atomic number. A period corresponds to filling a principal shell across the main-group elements; members of a group share a similar outer-shell configuration. The s, p and d blocks identify the sub-shell being filled. Periodicity means repeating patterns across successive periods.

Period 2 runs Li, Be, B, C, N, O, F, Ne; Period 3 runs Na, Mg, Al, Si, P, S, Cl, Ar. The outer configurations progress from ns¹ and ns² to ns²np¹ through ns²np⁶. Across a period, increasing nuclear charge with broadly similar inner-shell shielding pulls the outer electrons closer, reducing atomic radius.

First and successive ionisation energies

First ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions. Use gaseous states even if the element is normally solid. Second ionisation removes an electron from each gaseous 1+ ion to form gaseous 2+ ions.

Successive ionisation energies rise because remaining electrons are attracted to an increasingly positive ion. A particularly large jump signals removal from an inner shell, closer to the nucleus and less shielded. An illustrative sequence 600, 1200, 5000, 6500 kJ mol⁻¹ has its large jump after two removals: two outer-shell electrons, consistent with Group 2.

Explain the trend and both exceptions

Across Periods 2 and 3, first ionisation energy generally increases: nuclear charge rises, electrons occupy the same principal shell and shielding changes relatively little, so outer electrons are more strongly attracted. Down a group, additional shells increase radius and shielding; these outweigh the increased nuclear charge, so first ionisation energy generally falls.

Be → B and Mg → Al are downward steps because the removed electron switches from an s to a higher-energy p sub-shell of the same principal shell; it is easier to remove. N → O and P → S are downward steps because pairing begins in one p orbital, and electron–electron repulsion makes a paired electron easier to remove. Do not use the same explanation for both drops.

Read a position from an electron configuration

For [Ne]3s²3p³, the highest occupied principal shell is n = 3, so the atom belongs in Period 3. Its outer shell contains five electrons and the last sub-shell being filled is p: it is a Group 15 p-block element, phosphorus. A main-group atom with ns²np⁵ is in Group 17; the p superscript alone is not its number of outer electrons.

The repetition begins when the next principal shell starts filling. Sodium begins 3s after neon completes 2p, so sodium resembles lithium more than neon chemically. Similar outer configurations explain similar common ion charges; the period tells you how many occupied shells the neutral atom has.

Worked inference: what does a large jump actually tell you?

Consider these invented successive ionisation energies, in kJ mol⁻¹: 590, 1150, 4920, 6490, 8150. First locate the unusually large relative increase: the second-to-third jump is over fourfold, much larger than the neighbouring increases. Two electrons can be removed before the next electron comes from an inner shell. For a main-group atom this supports Group 2.

This sequence alone does not identify which Group 2 element it is. Combine it with proton number, electron configuration or other data. Saying “the third electron is in a higher shell” reverses the reasoning: after the outer shell is emptied, the next removed electron is in a lower-energy inner shell, with stronger attraction to the nucleus.

An ordinary increase between successive removals does not by itself show a new shell. Each remaining electron is being removed from a more positively charged species, so increases are expected even within one shell.

Build a comparative explanation in the right order

Start with what changes, then explain its effect on attraction, then state the energy consequence. Across Na to Mg: proton number increases; the removed electrons are in the same principal shell with similar shielding; attraction to the nucleus is stronger; more energy is required. Down Mg to Ca: the outer electron occupies a shell farther from the nucleus and experiences more shielding; attraction is weaker despite the greater nuclear charge; less energy is required.

For a downward exception across a period, identify the actual electron removed. Aluminium loses a 3p electron whereas magnesium loses a 3s electron. Sulfur and phosphorus both lose 3p electrons, but sulfur has a paired electron. “More shielding” alone does not explain the pairing exception.

Quick checks

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

Sources