AP Chemistry Ionization Energy

The trend, the successive-values pattern, and how to read a data table to find an element's group.

Ionization energy questions come in two shapes: ranking elements by their first ionization energy, and reading a row of successive values to work out which element it is. The first is covered by the Coulomb's law reasoning in periodic trends. This page concentrates on the second, which the trend pages leave out, and gives the one-sentence explanation that earns the points in both.

What Ionization Energy Is and How First Ionization Energy Trends

Ionization energy is the energy needed to remove an electron from a gaseous atom or ion, in kJ/mol. The first ionization energy (IE₁) removes the outermost electron from the neutral atom: X(g) → X⁺(g) + e⁻. It is always positive, since energy is put in.

• Across a period, IE₁ generally rises: more protons, same shell, similar shielding, so a stronger pull on the valence electrons.
• Down a group, IE₁ falls: the valence electrons are in a higher shell, farther away and shielded by more inner electrons.

The two small dips in the period trend, at aluminum and oxygen, are explained in periodic trends. The outermost peak of a photoelectron spectrum has a binding energy equal to IE₁, so the two topics describe the same data.

How Successive Ionization Energies Reveal Valence Electrons

The second ionization energy (IE₂) removes an electron from the +1 ion, IE₃ from the +2 ion, and so on. Each is larger than the last, because the electron is leaving an increasingly positive ion. The key feature is a large jump: when the valence electrons are gone, the next electron comes from a core shell that is much closer to the nucleus and not shielded by the same electrons, so it is held far more strongly.

The rule: the number of electrons removed before the big jump equals the number of valence electrons, which gives the group.

Worked Example 1: Identifying an Element's Group From Ionization Energy Data

An element has these ionization energies in kJ/mol: IE₁ = 578, IE₂ = 1817, IE₃ = 2745, IE₄ = 11,578, IE₅ = 14,831. Which group is it in?

The values rise gradually from 578 to 2745, then jump from 2745 to 11,578, about 4.2 times larger.
Three electrons were removed before the jump, so the element has 3 valence electrons.
An element with 3 valence electrons is in group 13. These are the values for aluminum, shown in the chart above.

Worked Example 2: Explaining the Jump in Magnesium's Ionization Energies

Magnesium's first three ionization energies are 738, 1451, and 7733 kJ/mol. Explain why IE₃ is so much larger than IE₂.

Magnesium's configuration is 1s² 2s² 2p⁶ 3s². IE₁ and IE₂ remove the two 3s valence electrons. The third electron must come from the 2p subshell, in the second shell.
By Coulomb's law, a 2p electron is much closer to the nucleus than a 3s electron, and it is not shielded by the electrons of the third shell, so the attraction is much greater. Removing it takes about 5.3 times as much energy.

Worked Example 3: Comparing the Second Ionization Energies of Sodium and Magnesium

Sodium's IE₂ is 4562 kJ/mol and magnesium's IE₂ is 1451 kJ/mol. Why is sodium's so much larger, even though sodium has the smaller nuclear charge?

Compare what each second electron is removed from.
Na⁺ has the configuration 1s² 2s² 2p⁶, a stable noble-gas core. Its IE₂ removes an electron from the second shell.
Mg⁺ has the configuration 1s² 2s² 2p⁶ 3s¹. Its IE₂ removes the remaining 3s valence electron, from the third shell.
The sodium electron is closer to the nucleus and unshielded by a third shell, so it is held more tightly. The comparison depends on which shell the electron comes from, not only on the proton count.

Worked Example 4: Predicting an Ion's Charge From Ionization Energy Data

Sodium's IE₁ is 496 kJ/mol and its IE₂ is 4562 kJ/mol. Predict whether sodium forms Na⁺ or Na²⁺ in compounds.

IE₂ is about 9.2 times IE₁. Removing a second electron means breaking into the stable core, which costs far more energy than the bonding in a compound can repay. So sodium stops after one electron and forms Na⁺. The same argument applied to magnesium, whose big jump comes after the second electron, predicts Mg²⁺. This is the energy reason metals form the ions that give them a noble-gas configuration, the idea behind ionic compounds.

Five Ionization Energy Practice Questions With Answers

1. An element has IE values (kJ/mol) of 786, 1577, 3232, 4356, and 16,091. Which group is it in?
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   Group 14. The big jump comes after the fourth value (4356 to 16,091), so 4 valence electrons. These are the values for silicon.
2. Which has the higher first ionization energy, lithium or cesium? Explain.
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   Lithium. Its valence electron is in the second shell, close to the nucleus with little shielding, while cesium's is in the sixth shell, far away and heavily shielded.
3. Which has the larger second ionization energy, potassium or calcium?
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   Potassium. After losing one electron, K⁺ has a noble-gas core, so its second electron comes from an inner shell. Ca⁺ still has a 4s valence electron to lose.
4. Why are all ionization energies positive?
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   Removing an electron from an atom or ion requires energy to overcome the attraction to the nucleus, so the process is endothermic.
5. Why does each successive ionization energy of an element exceed the previous one?
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   Each removal leaves a more highly charged positive ion, with fewer electrons around the same nucleus, so the remaining electrons are held more tightly.

Common Ionization Energy Mistakes

• Looking for the biggest number instead of the biggest jump. The values always increase; the group comes from where the increase is abrupt.
• Counting the electrons after the jump. Count those removed before the jump.
• Explaining the jump with “full shell” alone. Add distance and shielding: the core electron is closer and less shielded.
• Comparing second ionization energies by proton count only. Check which shell the electron is leaving.
• Treating ionization energy as exothermic. It is always positive.

The related quantities, electronegativity and electron affinity, are separated from ionization energy in electronegativity. To see how a unit like this one feeds into an overall estimate, try the AP Chem Score Calculator with your practice results.

Related Resources

• AP Chem Score Calculator
• Unit 1 Review: Atomic Structure and Properties
• AP Chemistry Periodic Trends
• AP Chemistry Photoelectron Spectroscopy
• AP Chemistry Electronegativity
• AP Chemistry Electron Configuration
• AP Chemistry Study Guide

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