Coulomb's Law & Periodic Trends (Quantitative)

You’ll be able to

State Coulomb’s law and connect Coulombic attraction to effective nuclear charge (Zₑff)
Use charge and distance to rank ionization energies and atomic radii quantitatively
Interpret successive ionization energies, reading large jumps as shell boundaries

The one force behind every trend

Every periodic trend traces back to a single electrostatic idea: the nucleus attracts electrons, and the strength of that pull follows Coulomb’s law. The attraction grows with the product of the charges and falls off with the square of the distance. Two levers therefore control everything — how much positive charge an electron feels, and how far it sits from the nucleus.

Coulomb's law (attraction)

F ∝ q₁q₂ / r²

q₁ is the effective nuclear charge (Zₑff), q₂ the electron’s charge, and r the distance between them. Doubling the distance quarters the force — the inverse-square term dominates.

Effective nuclear charge is the q₁ that matters

A valence electron does not feel the full nuclear charge, because inner electrons shield it. The net pull is the effective nuclear charge, Zₑff ≈ (protons − core electrons). Across a period, protons are added to the same shell while shielding barely changes, so Zₑff climbs, r shrinks, and F rises — that is why atoms contract and hold their electrons harder toward the right. Down a group, each new shell increases r sharply, and the 1/r² term weakens the pull despite more protons.

Successive ionization energies reveal shells

Pull electrons off one at a time and each removal costs more (the atom grows more positive). But when you finish a shell and start digging into the next one in, the cost jumps enormously — that electron is far closer to the nucleus with far less shielding, so Coulombic attraction is huge. The position of the big jump counts the valence electrons: a jump after the 2nd removal means 2 valence electrons (group 2).

Worked example

Use Coulombic reasoning to rank the first ionization energies of Na, Mg, and K, largest first.

1.First ionization energy reflects the attraction on the valence electron: larger F ∝ Zₑff / r² means it is harder to remove.
2.Na and Mg share period 3 (valence electron in n = 3, similar r), but Mg has one more proton, so its Zₑff is larger → stronger pull → higher IE than Na.
3.K sits one period below Na: its valence electron is in n = 4, a markedly larger r. The 1/r² term sharply weakens the pull, so K has the lowest IE of the three.
4.Combine: Mg > Na > K.

Answer: Mg > Na > K

Checkpoint

The successive ionization energies of an element (kJ/mol) are 738, 1451, 7733, 10 540, … The large jump between the 2nd and 3rd values tells you that:

Watch out

The distance term is squared, so it usually wins ties. When comparing an atom to one directly below it, the added shell (bigger r) outweighs the extra protons — which is why ionization energy falls down a group even though nuclear charge rises.

Checkpoint

Using Coulomb’s law, why does atomic radius decrease from Li to Ne across period 2?

On the exam

When a free-response asks you to justify a ranking, cite both Coulomb levers explicitly: state whether Zₑff or r changed and in which direction, then conclude about the force. “Higher Zₑff, same shell → stronger attraction → higher IE / smaller radius” is the sentence that earns the point.

Answer the 2 checkpoints as you read.