Unit 1 · Atomic Structure and Properties

AP Chemistry Electron Affinity

What electron affinity measures, how its sign convention works, why it rises across a period, and how electron configurations explain the exceptions, in four worked examples.

Electron affinity is the third member of the periodic-trend family, alongside ionization energy and atomic radius. Ionization energy measures how hard it is to remove an electron; electron affinity measures how much energy is released when an atom gains one. The exam asks you to explain it qualitatively, using Coulomb’s law and shielding, and to account for the elements that break the pattern.

What Electron Affinity Measures and Its Two Sign Conventions

Electron affinity is the energy change when an isolated gaseous atom gains one electron to form a gaseous anion:

X(g) + e⁻ → X⁻(g)

For most nonmetals the process releases energy, because the incoming electron is attracted to the nucleus. Two sign conventions are in use, and both describe the same physics:

• As an energy change (ΔE): energy released is negative. Chlorine’s electron affinity is −349 kJ/mol.
• As energy released: many data tables list the released energy as a positive number. Chlorine’s is 349 kJ/mol.

This page uses “energy released” so that a bigger number means a stronger pull on the added electron. Atoms that cannot form a stable gaseous anion, such as the noble gases, release no energy at all.

The Electron Affinity Trend Across the Periodic Table

• Across a period: more energy is generally released from left to right. Effective nuclear charge increases while the added electron enters the same shell, so it is held more strongly.
• Down a group: less energy is generally released, because the added electron sits in a shell farther from the nucleus and is better shielded. This trend is weaker and less regular than the one across a period.
• Exceptions: group 2, group 15, and the noble gases release little or no energy, for reasons that come from their electron configurations (Example 3).

Values are experimental electron affinities as compiled on Wikipedia’s electron affinity data page, rounded to the nearest kJ/mol. Elements marked ≤0 or ≈0 form no stable gaseous anion.

How Electron Affinity Differs From Ionization Energy and Electronegativity

All three grow from left to right across a period for the same reason, stronger attraction to the nucleus, which is why they usually trend together. The difference is the situation each describes: removing an electron, adding one, or sharing one in a bond. Electronegativity is the one used to judge bond polarity.

Worked Example 1: Writing the Electron Affinity Process for Chlorine

Write the process that corresponds to chlorine’s electron affinity, give ΔE, and find the energy released per atom.

Process: Cl(g) + e⁻ → Cl⁻(g).
ΔE: −349 kJ/mol; the negative sign means energy is released.
Per atom: (349 × 10³ J/mol) ÷ (6.022 × 10²³ atoms/mol) = 5.80 × 10⁻¹⁹ J.

Both species must be gaseous, so the value describes isolated atoms, not chlorine in a crystal or a solution. The per-atom conversion uses Avogadro’s number.

Worked Example 2: Explaining Why Chlorine Releases More Energy Than Sodium

Sodium releases 53 kJ/mol when it gains an electron, and chlorine releases 349 kJ/mol. Explain the difference.

Same shell: in both atoms the added electron goes into the third shell (n = 3), so the distance and shielding by inner electrons are similar.
Different nuclear charge: chlorine has 17 protons and sodium has 11, so chlorine’s effective nuclear charge on the outer shell is much larger.
Conclusion: by Coulomb’s law, the added electron is attracted more strongly to chlorine’s nucleus, so more energy is released.

This “same shell, more protons” argument is the same one that explains the rise in ionization energy across a period.

Worked Example 3: Explaining the Dips at Magnesium, Phosphorus, and Argon

Use electron configurations to explain why Mg and Ar release no energy on gaining an electron, and why P (72 kJ/mol) releases less than Si (134) or S (200).

Mg, [Ne]3s²: the 3s subshell is full, so the added electron must go into the higher-energy 3p subshell, where the 3s electrons partly shield it. No stable anion forms.
Ar, [Ne]3s²3p⁶: the third shell is full, so the added electron would start the fourth shell, far from the nucleus and heavily shielded. No stable anion forms.
P, [Ne]3s²3p³: each 3p orbital already holds one electron, so the added electron must pair up in an occupied orbital and is repelled by its partner. Less energy is released than for Si, whose added electron enters an empty 3p orbital, or S.

The same reasoning explains the period 2 dips at Be, N, and Ne in Fig. 1. These are the same configurations that cause the dips in ionization energy, explained in electron configuration terms.

Worked Example 4: Comparing the Halogens Down Group 17

Fluorine releases 328 kJ/mol, chlorine 349, bromine 325, and iodine 295. Describe the trend and explain the one value that does not fit.

Cl, Br, I: energy released falls down the group because the added electron enters a shell farther from the nucleus and better shielded, so the attraction is weaker.
F: releases less than Cl even though it is higher in the group. Fluorine’s 2p orbitals are small, so the added electron is crowded close to the electrons already there, and the extra repulsion outweighs the closer approach to the nucleus.

Chlorine therefore releases the most energy of any element on gaining an electron. On the exam, the general trend and its Coulomb’s-law explanation matter more than this exception, but it is a common multiple-choice trap.

Five Electron Affinity Practice Questions With Answers

1. Write the process for oxygen’s first electron affinity. Oxygen releases 141 kJ/mol; what is ΔE?
   Show answer

   O(g) + e⁻ → O⁻(g), ΔE = −141 kJ/mol.
2. Which releases more energy on gaining an electron, sulfur or phosphorus? Explain.
   Show answer

   Sulfur (200 vs 72 kJ/mol). Phosphorus’s half-filled 3p subshell forces the added electron to pair up, which adds repulsion.
3. Adding a second electron to O⁻ requires 744 kJ/mol. Why does the second step absorb energy?
   Show answer

   The electron is being added to a negative ion, which repels it, so energy must be supplied.
4. Rank C, N, and O by the energy released when each gains an electron.
   Show answer

   O (141) > C (122) > N (about 0). Nitrogen’s half-filled 2p subshell gives the lowest value.
5. Sodium releases 53 kJ/mol and potassium 48 kJ/mol. Explain the difference.
   Show answer

   Potassium’s added electron enters the fourth shell, farther from the nucleus and better shielded than sodium’s third shell, so it is attracted less strongly.

Common Electron Affinity Mistakes

• Mixing up the sign conventions. −349 kJ/mol as ΔE and 349 kJ/mol as energy released describe the same thing.
• Confusing electron affinity with electronegativity. One is an energy change for an isolated atom; the other is a unitless pull on shared electrons.
• Confusing it with ionization energy. Ionization energy removes an electron; electron affinity adds one.
• Assuming fluorine releases the most energy. Chlorine does; fluorine’s small 2p subshell adds repulsion.
• Explaining with “wants a full octet.” Use Coulomb’s law: nuclear charge, distance, and shielding.
• Forgetting the gaseous state. The definition applies to isolated gaseous atoms.

Periodic trend explanations are some of the most reliable free-response points in Unit 1. To see how stronger Unit 1 answers would move your overall estimate, try the AP Chem Score Calculator.

Related Resources

• AP Chem Score Calculator
• Unit 1 Review: Atomic Structure and Properties
• AP Chemistry Periodic Trends
• AP Chemistry Ionization Energy
• AP Chemistry Electronegativity
• How to Write Electron Configurations
• AP Chemistry Photoelectron Spectroscopy
• AP Chemistry Study Guide

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