From trends to reasons
Chapter 2.3 described how ionization energy, radius, electronegativity, and electron affinity change across the periodic table. AP questions rarely stop at “which is larger”; they almost always ask “why”, and they accept only one kind of reason: Coulomb’s law, the attraction between the nucleus and the electrons (plus the repulsion between electrons).
This lesson puts all four properties into one framework and points out which explanations earn no credit on the AP exam.
Three factors
The attraction an electron feels depends on:
- Nuclear charge (number of protons). More protons, stronger attraction.
- Distance (which shell the electron is in). Larger n means farther from the nucleus and weaker attraction, falling off with the square of the distance.
- Shielding. Core electrons sit between the nucleus and the valence electrons and cancel part of the nuclear charge. Electrons in the same shell shield each other very little.
Putting them together:
- Across a period: each step adds a proton, while the core stays the same and the valence electrons stay in the same shell. Effective nuclear charge rises at constant distance, so the valence electrons are held more tightly.
- Down a group: protons increase, but so do core electrons, so effective nuclear charge stays about the same, while each step adds a shell and the valence electrons are farther away. The deciding factor is distance, not “less shielding”; there are actually more core electrons.
| Property | Across a period (left to right) | Down a group |
|---|---|---|
| First ionization energy | generally increases | decreases |
| Atomic radius | decreases | increases |
| Electron affinity (energy released) | generally increases | generally decreases |
| Electronegativity | increases | decreases |
All four describe the same thing: how strongly the nucleus attracts a valence electron (or an added or shared electron). Stronger attraction means a smaller radius and larger ionization energy, electron affinity, and electronegativity.
Explain the exceptions with Coulomb’s law too
Lesson 2.3.4 covered the dips in ionization energy, and many textbooks (including the Chinese 人教版) explain them by saying “filled and half-filled arrangements are more stable.” That isn’t false, but AP scoring doesn’t accept it as a reason on its own, because it says nothing about a change in attraction or repulsion. AP wants:
- Be > B and Mg > Al: B loses a 2p electron. 2p is higher in energy than 2s: it’s partly shielded by the 2s electrons and farther from the nucleus on average, so it’s easier to remove than beryllium’s 2s electron.
- N > O and P > S: oxygen’s fourth 2p electron must share an orbital, and the repulsion between the paired electrons makes one of them easier to remove.
- Ar > Cl: both lose an electron from n = 3, and Ar has one more proton, so its effective nuclear charge and attraction are greater. Don’t say “because argon is a filled noble gas.”
- F’s electron affinity (328 kJ/mol) is smaller than Cl’s (349 kJ/mol): fluorine is so small that an added electron is crowded into the n = 2 shell with seven other electrons and is strongly repelled by them.
Radii of the transition metals
From Cr to Co, atomic radius hardly changes. Across this stretch the added electrons go into the inner 3d sublevel (n = 3), while the outermost electrons remain 4s electrons. The 3d electrons shield the 4s electrons fairly well, canceling most of each added proton, so the effective nuclear charge on the 4s electrons rises only slightly and the size barely changes. That’s different from the clear shrinking across a period in the main groups.
Successive ionization energies and formulas
As lesson 2.3.4 showed, the electrons removed before the big jump in successive ionization energies are the valence electrons. The jump happens because the next electron must come from the core, much closer to the nucleus and barely shielded. AP often asks you to go from the jump to a formula:
| Element | I₁ | I₂ | I₃ | I₄ | Valence e⁻ | Ion | Oxide | Fluoride | Phosphide |
|---|---|---|---|---|---|---|---|---|---|
| Na | 496 | 4562 | 6910 | 9543 | 1 | NaF | |||
| Mg | 738 | 1451 | 7733 | 10543 | 2 | MgO | |||
| Al | 578 | 1817 | 2745 | 11577 | 3 | AlP |
(In kJ/mol; the first value after each jump is in bold.) For elements with the valence configuration ns² (group 2), the jump comes between I₂ and I₃.
Comparing second ionization energies within a group also comes down to distance. Every alkali metal’s I₂ removes a core electron, and lithium’s I₂ (7298 kJ/mol) is far larger than sodium’s (4562 kJ/mol), because Li⁺ loses a 1s electron, much closer to the nucleus than the 2p electron Na⁺ loses.
Estimating from neighbors
One use of periodicity is estimating a value you don’t have. Potassium sits between sodium and rubidium in group 1:
- First ionization energy: Na 496 and Rb 403 kJ/mol, so K should fall in between (measured: 419 kJ/mol).
- Atomic radius: Na 166 and Rb 220 pm, so K should fall in between too (measured: 203 pm).
An estimate gives a range rather than an exact value, but that’s enough for a “closest to” multiple-choice question.
Make a claim, then choose the reason. Every rejected reason comes with an explanation; watch for “filled, so stable”, wrong facts, and reasoning that points the wrong way:
Data: ionization energies, CRC; atomic radii are covalent radii (Cordero 2008); ionic radii, Shannon 1976; electronegativity, revised Pauling scale; electron affinity, CRC (all in trends.ts).
Example 1: Why is oxygen’s first ionization energy below both N and F?
The first ionization energies of N, O, and F are 1402, 1314, and 1681 kJ/mol. Explain O < N and O < F.
-
Identify: write the valence configurations. All three lose a 2p electron from n = 2:
- N: 2s²2p³ (three 2p electrons, one per orbital)
- O: 2s²2p⁴ (one 2p orbital holds 2 electrons)
- F: 2s²2p⁵ (two 2p orbitals hold 2 electrons each)
-
Compare O with N. Same shell. Oxygen has one more proton, which should make it harder to ionize, but its fourth 2p electron shares an orbital, and the two electrons in that orbital repel each other, so one of them is easier to remove. That repulsion outweighs oxygen’s extra proton.
-
Compare O with F. Both have paired 2p electrons, so both feel that repulsion; fluorine has one more proton, and its electrons are still in the same shell, so they are attracted more strongly.
-
Connect to ionization energy.
All values are in kJ/mol. O is below N because of electron–electron repulsion in a shared orbital, and below F because it has fewer protons.
-
Check the wording. The reasons use “number of protons”, “same shell”, and “repulsion between electrons in the same orbital”, not “half-filled is stable”: AP scoring expects a Coulombic explanation.
Example 2: From successive ionization energies to formulas
A period 3 element X has successive ionization energies of 787, 1577, 3232, 4356, and 16091 kJ/mol. Write the formulas of its oxide and its fluoride.
-
Identify: compute each successive ratio. A sudden jump means core electrons are being removed:
-
Compare: find the jump. The first ratios are all 2.1 or less; only I₅/I₄ jumps, so the first 4 electrons are valence electrons and the fifth comes from the core (n = 2, closer to the nucleus and less shielded). X has 4 valence electrons.
-
Place it. Period 3 with 4 valence electrons is group 14: Si.
-
Connect to formulas. All 4 valence electrons take part in bonding, so X combines like a +4 element; O takes −2 and F takes −1:
-
Note. Silicon bonds covalently to O and F, not ionically, but the formula pattern set by the valence electrons is the same.
- 496, 4562, 6910
The jump is at I₂: 1 valence electron, so and the oxide (this is Na).
- 738, 1451, 7733
The jump is at I₃: 2 valence electrons, so and the oxide XO (Mg).
- 578, 1817, 2745, 11577
The jump is at I₄: 3 valence electrons, so and the oxide (Al).
- 787, 1577, 3232, 4356, 16091
The jump is at I₅: 4 valence electrons, so the oxide (Si).
- 520, 7298, 11815
I₂ is 14 times I₁: 1 valence electron (Li).
- 900, 1757, 14849
The jump is at I₃: 2 valence electrons (Be), with the configuration 2s².
- 801, 2427, 3660, 25026
The jump is at I₄: 3 valence electrons (B), so the oxide .
- 1086, 2353, 4621, 6223, 37831
The jump is at I₅: 4 valence electrons (C), so the oxide .
Common mistakes
Mistake 1: “Argon’s first ionization energy is higher than chlorine’s because argon is filled and stable.” That only restates the observation, and AP gives it no credit. Say instead: both lose an electron from n = 3, and argon has one more proton, so its effective nuclear charge and attraction are greater.
Mistake 2: “Ionization energy falls down a group because shielding by the core electrons decreases.” It’s the other way around: there are more core electrons lower in the group. The real reason is that the valence electrons are one shell farther out, so they’re attracted less.
Mistake 3: “Fluorine has the largest electron affinity.” The measured value for Cl (349 kJ/mol) is larger than for F (328 kJ/mol). Fluorine is so small that an added electron is crowded into the n = 2 shell and strongly repelled by fluorine’s seven valence electrons.
Quick check
Which atom has the largest radius?
K and Ca have valence electrons in n = 4, one shell beyond Na and Mg; of those two, K has fewer protons and weaker attraction. Potassium's covalent radius is 203 pm.
The first ionization energies of Na and Rb are 496 and 403 kJ/mol. The first ionization energy of K is closest to
K lies between Na and Rb in the same group, so its first ionization energy should fall between 403 and 496; only 420 does (measured: 419 kJ/mol).
The first ionization energy of P (1012 kJ/mol) is slightly greater than that of S (1000 kJ/mol). Which is the best explanation?
Both lose a 3p electron. In sulfur's 3p⁴, one pair of electrons shares an orbital, and that repulsion outweighs the extra proton. B earns no credit on its own; C and D are wrong facts.
Atomic radius stays nearly constant from Cr to Co. What is the main reason?
The outermost electrons stay in 4s, while the added electrons go into the n = 3 shell's 3d sublevel and cancel most of each added proton. B and C have nothing to do with radius; D is false, since only Cu and Zn have a full 3d.
AP-style free response
A period 3 element X has successive ionization energies of 738, 1451, 7733, and 10543 kJ/mol.
(a) Write the formulas of the compounds X forms with oxygen, fluorine, and phosphorus.
(b) Explain why the ionization energy jumps from I₂ to I₃.
(c) Calcium is in the same group as X. Explain why calcium’s first ionization energy is lower than that of X.
(d) Explain why the atomic radius of X is larger than that of Al.
Model answer and scoring
(a) The jump comes at I₃, so X has 2 valence electrons and forms (X is Mg): XO, , and , that is, MgO, , and .
(b) The first two electrons are 3s valence electrons; the third must come from the 2p core. A 2p electron is in n = 2, much closer to the nucleus and barely shielded, so it is attracted far more strongly and takes much more energy to remove.
(c) Calcium’s valence electrons are in n = 4, one shell farther from the nucleus than magnesium’s (n = 3). Their effective nuclear charges are about the same, so the greater distance means weaker attraction, and calcium’s valence electron is easier to remove.
(d) The valence electrons of Mg and Al are both in n = 3. Aluminum has one more proton and a larger effective nuclear charge, so it pulls its valence electrons in closer; Al is smaller and Mg is larger.
Scoring: (a) all three formulas correct; (b) states that the third electron comes from the core (n = 2) and that it is closer or less shielded; (c) “one more shell, farther away”, while “less shielding” earns no credit; (d) both “same shell” and “more protons (greater nuclear charge)”.
Summary
- Three factors: nuclear charge (protons), distance (shell), and shielding (core electrons). Across a period, effective nuclear charge rises at constant distance; down a group, distance grows while effective nuclear charge stays about the same.
- Ionization energy, electron affinity, and electronegativity grow as the attraction grows; radius shrinks.
- Explain the exceptions with sublevel energy (2p above 2s) and paired-electron repulsion; “filled or half-filled is more stable” earns no credit on its own.
- The jump in successive ionization energies marks the end of the valence electrons and gives the formulas of oxides, fluorides, and phosphides; neighboring elements’ data can estimate unknown values.
AP tip: Explanation questions almost always have two scoring points: say which shell the electrons being compared are in (distance), and say how the nuclear charge (protons) or effective nuclear charge differs. Within a group, say “one more shell, farther away”; within a period, say “same shell, more protons”. “Stable”, “filled”, “half-filled”, and “heavier atom” are not reasons AP accepts.