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Coulomb's Law in AP Chemistry

Coulomb's law says the force between two charged particles is proportional to the product of their charges and inversely proportional to the square of the distance between them, F ∝ q₁q₂/r². Bigger charges make the force stronger, and more distance makes it weaker fast, because r is squared. AP Chemistry uses Coulomb's law mainly to explain trends and ion interactions qualitatively. Given charges and distances, you can also use it to compare or calculate forces.

This definition is part of Fiveable's AP Chemistry review, which also has guided practice questions.

Verified for the 2027 AP Chemistry exam•Last updated November 2026•Fiveable Content Team

Reading q₁q₂/r² for an electron and its nucleus

The course introduces Coulomb's law in Unit 1, Topic 1.5 (Atomic Structure and Electron Configuration). It carries it through Topic 1.7 (Periodic Trends) and Unit 2, Topics 2.1-2.3, to answer one question: how does the attraction between opposite charges control the way atoms hold electrons and the way ions hold each other.

For an atom, q₁ is the positive charge of the nucleus and q₂ is an electron. The electron feels less than the full nuclear charge when core electrons sit between it and the nucleus. What's left is the effective nuclear charge. So the Coulomb argument for an electron always weighs two things. One is how much positive charge it feels, and the other is how far it is from the nucleus.

Photoelectron spectroscopy (PES) makes this visible. Carbon's 1s peak sits at a higher binding energy than boron's because carbon has 6 protons instead of 5, and the 1s electrons are in the same shell at about the same distance. In calcium, the 1s electrons have an enormous binding energy and the 4s electrons a tiny one. The 1s electrons are close to a nucleus with 20 protons and almost nothing shielding them. The 4s electrons are far out and shielded by 18 core electrons.

The same logic explains the periodic trends. Across a period, electrons enter the same shell while protons increase, so effective nuclear charge rises. Atoms get smaller, and ionization energy and electronegativity go up. Down a group, each new shell puts valence electrons much farther out. The added distance outweighs the added protons, so ionization energy drops and radius grows.

Cation-anion strength: why MgO outlasts NaCl

In Topic 2.2 (Intramolecular Force and Potential Energy), the course applies Coulomb's law to cations and anions. The attraction is proportional to each ion's charge, and it gets stronger as the distance between ion centers shrinks. That distance depends on the sizes of both ions, so smaller ions mean a stronger attraction.

Compare MgO and NaCl. Mg²⁺ and O²⁻ give a charge product of 2 × 2 = 4, while Na⁺ and Cl⁻ give 1 × 1 = 1. Mg²⁺ and O²⁻ are also smaller than Na⁺ and Cl⁻, so both factors point the same way. MgO has much stronger ionic attractions and a much higher melting point.

When charge and distance point in opposite directions, consider both together in |q₁q₂|/r². Neither factor wins automatically. A larger charge product pushes the attraction up, but a large enough increase in the distance between ion centers can outweigh it, so a justification has to account for both.

When the charge products match, distance decides. In NaF versus NaCl, both are 1+/1−, and the smaller F⁻ brings the ion centers closer together. That gives NaF the stronger attraction.

Topic 2.3 (Structure of Ionic Solids) extends the same idea to a whole crystal. Each ion sits surrounded by ions of the opposite charge, with like charges held farther apart. A particle diagram consistent with Coulomb's law shows that alternating pattern and draws ions with sensible relative sizes. You aren't asked to name specific crystal structures.

Where a bare Coulomb's law argument falls short

Coulomb's law predicts that a smaller atom should attract an added electron more strongly. By that logic, fluorine should have a more negative electron affinity than chlorine. Measured data shows chlorine's is more negative.

The simple model only counts the nucleus pulling on the incoming electron. Fluorine's 2p subshell is small and crowded, so the new electron is also repelled strongly by the electrons already there. The Coulomb model is still correct here. It just has to include electron-electron repulsion, and a quick nucleus-only version leaves that out. Expect questions that ask you to evaluate a model like this, and say what the model accounts for and what it misses.

Turning charge and distance into a scoring justification

In Section I (Multiple-Choice), Coulomb's law sits behind comparison items. These ask you to read a PES spectrum, pick the ion pair with the strongest attraction, rank ionization energies, or judge a student's reasoning. In Section II, the Short questions and Long questions ask you to justify a trend or compare two substances, and the point goes to the reasoning, not the bare claim.

A strong justification includes:

  • the specific charges involved (protons or effective nuclear charge for an atom, ion charges for an ionic compound)
  • the specific distance (which shell the electron is in, or relative ion sizes)
  • the link to the property, such as "stronger attraction, so more energy is needed to remove the electron"

A sentence like "Mg²⁺ and O²⁻ have larger charges and smaller radii than Na⁺ and Cl⁻, so the Coulombic attraction is stronger and MgO melts at a higher temperature" does the job. "MgO has stronger bonds" restates the answer without explaining it.

The most common wrong move is citing only proton count. Saying potassium holds its valence electron more tightly than sodium because it has more protons ignores the extra shell, and distance is the factor that decides down a group. A second slip is treating ionic radius as the radius of one ion. The r in the equation is the distance between the two ion centers, so both ion sizes matter.

Where this sits in AP® Chemistry

Keep studying AP® Chemistry Unit 2