Test periodic explanations with ionization evidence

Let a reference observation challenge the rule

A first-ionization plot often rises across a period, but “more protons always means a larger first ionization energy” is too strong. Magnesium’s evaluated value is 7.646236 eV, while aluminum’s is 5.985769 eV. Their difference, aluminum minus magnesium, is −1.660467 eV. An explanation that cannot accommodate this pair needs revision.

Select the Na/Mg/Al packet and compare its first pair, then its second pair. Keep the starting configurations beside the energy values. Na to Mg follows the broad rise. Mg to Al reverses it: aluminum’s removed electron is in 3p rather than 3s. Subshell energy, penetration and shielding matter alongside nuclear charge. The numerical values come from an evaluated reference; the qualitative explanation does not calculate those values.

The packets are selected rows from NIST Atomic Spectra Database version 5.12, inspected 28 October 2026. They are not invented detector readings, a complete photoelectron spectrum or new measurements performed in this browser. Each row preserves its source bibliography identifiers and reported uncertainty. Source links are in the live table.

Define the comparison before calculating

An electron volt, eV, is an energy unit for an individual particle-scale event. The listed ionization energy is the minimum energy to remove one electron from an isolated gaseous ground-state atom or ion. It is not the heat released when a solid reacts, a reaction rate, or the energy needed to remove every electron.

For two rows with energies E₁ and E₂, calculate ΔE = E₂ − E₁ and the ratio E₂/E₁. The signed difference answers which is larger and by how much; the ratio answers how many times as large. The table retains the source values; displayed calculations are rounded. A ratio and a difference are distinct summaries, not independent new observations.

In a first-removal packet, each row begins with a different neutral atom. Adding Li’s removal energy to Na’s does not give a successive ionization energy of either species. In a successive-removal packet, each row begins with the progressively charged ion named before the arrow. Its ground state changes, while proton count remains fixed.

Pairing is another reason to inspect structure

Compare carbon, nitrogen and oxygen. Nitrogen has three singly occupied 2p orbitals in the reference description. Oxygen has a paired 2p electron and a smaller first removal energy: 13.618055 eV versus nitrogen’s 14.53413 eV, a difference of −0.916075 eV. Electron repulsion and the energies of the initial and resulting states help explain the departure from a smooth trend.

Avoid the shortcut “a half-filled shell is stable, so no explanation is needed.” State what changes in the electron arrangement, and acknowledge that a qualitative account is not a quantitative many-electron solution. The source uncertainty is not a freely chosen error bar or a claim that the true value must lie inside a hard interval. This exercise does not perform a statistical significance test.

Read a shell transition in successive removals

Neutral Mg starts with [Ne] 3s². The first removal costs 7.646236 eV and leaves Mg+ with one 3s electron. The second costs 15.035271 eV and leaves Mg2+ with the ten-electron closed-shell arrangement. The third costs 80.1436 eV and removes an electron from what was the neutral atom’s n=2 core.

The much larger third cost is consistent with two outer 3s electrons before reaching that core. Reaching charge +2 requires both earlier steps: 7.646236 + 15.035271 = 22.681507 eV. The second-step energy alone is not that sum. The cumulative column adds central values; it does not invent a combined uncertainty from potentially related source errors.

Compare sodium. It begins with only one 3s electron, so the large jump occurs at the second removal. The third sodium entry appears in brackets because NIST identifies it as semiempirical. The NIST quantity documentation explains that distinction. Keep the bracket visible; do not present every number as a directly measured experimental value.

Use a trend within its conditions

Li, Na and K each begin with one outer s electron, but in different shells. Their first removal energies decrease down this group even though nuclear charge rises. Additional shell extent and shielding help explain that comparison. It does not establish how quickly bulk samples react under unspecified conditions.

For practice, consider an original, deliberately synthetic main-group packet: successive removal energies 6, 12, 25 and 120 arbitrary energy units. Predict the likely number of outer electrons before the pronounced jump, and calculate the cumulative energy of the first three steps. Do not identify an element from this information alone.

Check the changed packet

The pronounced fourth-step jump supports three outer electrons under the stated main-group shell interpretation. The first three steps sum to 43 units. That does not uniquely identify an element or prove a particular compound will form. The synthetic values are not additional NIST rows.

Return to configuration checks if the initial species is unclear. Next use controlled charge and distance comparisons to distinguish an explanatory model from a measured atomic property.