AP Chemistry Periodic Trends

Atomic radius, ionization energy, and ion size, explained with Coulomb's law in four worked examples.

Memorizing that radius shrinks across a period earns little on an AP question that says “explain.” The credit is in the reasoning: which charge, which distance, and how much shielding. The periodic table guide and the Unit 1 review give you the trends themselves; this page is for practicing the explanation, including the two places where ionization energy does not follow the simple pattern.

The Three-Part Coulomb's Law Explanation Behind Every Trend

Every periodic trend answer rests on Coulomb's law: the attraction between the nucleus and an outer electron grows with the charge the electron actually feels and shrinks with the distance between them. A complete answer addresses three things, in this order.

1. Nuclear charge. How many protons does each atom have?
2. Shielding. How many inner (core) electrons screen the outer electrons from that charge? Together with the proton count this gives the effective nuclear charge, which you can approximate as the number of protons minus the number of core electrons.
3. Distance. In which shell are the outer electrons? A higher shell means a larger average distance.

Across a period, the shell stays the same and shielding barely changes while the proton count rises, so the effective nuclear charge increases. Down a group, the shell number increases while the effective nuclear charge stays about the same, so distance and shielding win. Everything below is those two sentences applied to specific pairs. The approximation of the effective nuclear charge as protons minus core electrons is a simplification, since real values are smaller, but it ranks elements correctly.

Worked Example 1: Comparing Sodium and Magnesium Across Period 3

Which has the larger atomic radius, Na or Mg, and which has the larger first ionization energy? Justify both. The first ionization energies are 496 kJ/mol (Na) and 738 kJ/mol (Mg).

Nuclear charge and shielding: Na has 11 protons and Mg has 12, and both have the same 10 core electrons (1s²2s²2p⁶). The approximate effective nuclear charge is +1 for Na and +2 for Mg.
Distance: the valence electrons of both are in the third shell, so the starting distance is similar.
Radius: Mg's valence electrons feel a stronger pull at about the same distance, so they are drawn in closer. Na has the larger atomic radius.
Ionization energy: the same stronger attraction means more energy is needed to remove a Mg valence electron. Mg has the larger first ionization energy, consistent with 738 > 496.

Worked Example 2: Comparing Lithium and Potassium Down Group 1

Compare the atomic radius and first ionization energy of Li and K. The first ionization energies are 520 kJ/mol (Li) and 419 kJ/mol (K).

Nuclear charge and shielding: K has 19 protons against 3 for Li, but it also has 18 core electrons against 2, so the approximate effective nuclear charge is +1 for both.
Distance: Li's valence electron is in the second shell and K's is in the fourth, much farther from the nucleus.
Radius: the effective charge is equal, so distance decides it. K has the larger atomic radius.
Ionization energy: at equal effective charge, the more distant electron is held more weakly. Li has the larger first ionization energy, matching 520 > 419.

Notice that extra protons did not make potassium harder to ionize. The added shells shield them almost completely, which is the exam-ready way to say why the group trend runs the opposite way to the period trend.

Worked Example 3: Why Aluminum and Oxygen Dip Below the Ionization Energy Trend

The general rule says first ionization energy rises across a period, yet Al (578 kJ/mol) is lower than Mg (738 kJ/mol), and O (1314 kJ/mol) is lower than N (1402 kJ/mol). Explain each dip.

Mg to Al: Mg's outermost electrons are in a filled 3s subshell. Al's outermost electron is in a 3p subshell, which is slightly higher in energy and, on average, a little farther from the nucleus. The filled 3s subshell also shields that 3p electron. These effects outweigh Al's extra proton, so the 3p electron is easier to remove.

N to O: N's valence configuration is 2s²2p³, with one electron in each 2p orbital. O is 2s²2p⁴, so one 2p orbital holds a pair of electrons. Two electrons in the same orbital repel each other, which makes it easier to remove one of them than to remove an unpaired electron from N, even though O has the greater nuclear charge.

A scope note: the course outline does not ask you to know the unusual ground-state configurations of elements such as chromium and copper (the Unit 1 review covers that), and these dips are a separate matter, since they follow from ordinary configurations. They are, however, among the most common ways ionization energy data departs from the simple trend, so it is worth being able to explain them. Check with your teacher on how much weight to give them.

Worked Example 4: Ranking the Radii of Isoelectronic Ions and Comparing Ions to Atoms

(a) Rank O²⁻, F⁻, Na⁺, and Mg²⁺ by radius, largest to smallest. (b) Explain why Na⁺ is smaller than a Na atom and Cl⁻ is larger than a Cl atom.

(a) All four ions have 10 electrons (1s²2s²2p⁶), so they are isoelectronic: same shell, same shielding. They differ only in proton count: 8, 9, 11, and 12. More protons pull the same 10 electrons in harder, so the radius falls as the proton count rises:
O²⁻ > F⁻ > Na⁺ > Mg²⁺.

(b) Na loses its one 3s electron to form Na⁺, which removes the whole third shell, and the 11 protons now hold only 10 electrons, so the remaining electrons are pulled in closer. Cl gains an electron to form Cl⁻ with the same 17 protons, so there are 18 electrons repelling one another against an unchanged nuclear pull, and the cloud expands. That is why metals form cations that are smaller than their atoms and nonmetals form anions that are larger, the size relationship that matters for the ionic bonding reasoning on later units.

Common Periodic Trend Mistakes

• Stopping at “more protons.” The proton count is only one part of the argument; without shielding and distance the explanation is incomplete.
• Saying electrons are “farther away so they are lost easily.” Name the force: the attraction between the nucleus and the electron is weaker at a greater distance.
• Ignoring shielding down a group. Potassium has far more protons than lithium, yet its first ionization energy is lower because of the extra inner shells.
• Comparing ions with different electron counts as if they were isoelectronic. Use the proton-count argument only when the electron configurations match.
• Treating the Mg/Al and N/O dips as errors in the data. They are real, and they have a short configuration-based explanation.

These trends also underlie the polarizability argument for London dispersion forces, and they connect back to the ground-state patterns in electron configuration. For the full set of free tools and guides in one place, the AP Chem Score Calculator is the site's main page.

Related Resources

• AP Chem Score Calculator
• Unit 1 Review: Atomic Structure and Properties
• AP Chemistry Periodic Table
• AP Chemistry Electron Configuration
• AP Chemistry Ionic vs Covalent Bonds
• AP Chemistry London Dispersion Forces
• AP Chemistry Study Guide

Frequently Asked Questions

This page is not affiliated with or endorsed by College Board. AP® is a trademark registered by the College Board.