Periodic Trends

AP Chemistry· AP Chemistry CED — Atomic Structure and Properties· 14 min read

1. Core Concept: Effective Nuclear Charge★★☆☆☆⏱ 3 min

All periodic trends originate from differences in effective nuclear charge, the net positive attraction valence electrons experience from the nucleus, after accounting for shielding by inner core electrons.

📘 Definition

Effective Nuclear Charge

$Z_{e f f} = Z - S$

Net positive charge experienced by a valence electron, after shielding by inner core electrons. For AP Chemistry, $S$ (shielding constant) is approximated by the total number of inner core electrons, and $Z$ is the atomic number (total protons).

Z_{eff} = Z - S

Trends in $Z_{e f f}$ are consistent: moving left to right across a period, $Z$ increases by 1 per element while the number of core electrons (and thus $S$ ) stays constant, so $Z_{e f f}$ increases steadily. Moving down a group, $Z$ increases but $S$ increases by a full new shell of electrons, so $Z_{e f f}$ is nearly constant with only a very small increase for most groups.

📐 Worked Example

Compare the effective nuclear charge for valence electrons in sodium (Z=11) and rubidium (Z=37). Which element has lower $Z_{e f f}$ for valence electrons, and why?

1
Write full electron configurations to count core electrons:
2
$\text{Na}: 1s^22s^22p^63s^1 \implies S=10; \quad \text{Rb}: 1s^22s^22p^63s^23p^64s^1 \implies S=36$
3
Calculate approximate $Z_{e f f}$ using the formula:
4
$Z_{eff}(\text{Na}) = 11 - 10 = +1; \quad Z_{eff}(\text{Rb}) = 37 - 36 = +1$
5
Both are group 1 elements, so $Z_{e f f}$ is approximately equal for AP-level reasoning. If asked to select the lower value, rubidium has a very slightly lower net $Z_{e f f}$ because increased shielding from additional core electrons is not fully offset by the higher proton count.

Exam tip:

On AP FRQs, you will not earn a justification point just saying "this follows the periodic trend". Always explicitly connect the property to $Z_{e f f}$ and shielding to get full credit.

2. Atomic and Ionic Radius★★★☆☆⏱ 4 min

📘 Definition

Atomic Radius

Half the distance between the nuclei of two adjacent bonded atoms of the same element, used to measure the size of a neutral atom.

The trend in atomic radius directly follows $Z_{e f f}$ and principal quantum number $n$ : moving left to right across a period, increasing $Z_{e f f}$ pulls valence electrons closer to the nucleus, so atomic radius decreases. Moving down a group, valence electrons occupy higher energy levels with a larger average distance from the nucleus, and this effect dominates over the small $Z_{e f f}$ increase, so atomic radius increases down a group.

For ionic radius, additional rules apply: Cations are smaller than their parent neutral atom (lose an entire valence shell, higher $Z_{e f f}$ per electron). Anions are larger than their parent neutral atom (add electrons to the valence shell, increasing electron-electron repulsion with no change in $Z$ ). For isoelectronic ions (ions with the same number of electrons), ionic radius decreases with increasing atomic number, because higher $Z$ gives higher $Z_{e f f}$ that pulls the same number of electrons closer.

📐 Worked Example

Rank the following species in order of increasing radius: $F^{-}$ , $O^{2 -}$ , $M g^{2 +}$ , $C l^{-}$ . Justify your ranking.

1
First count electrons for each species to identify isoelectronic groups: $F^{-}$ = 10 electrons, $O^{2 -}$ = 10 electrons, $M g^{2 +}$ = 10 electrons, $C l^{-}$ = 18 electrons. All 10-electron ions have valence electrons in $n = 2$ , while $C l^{-}$ has valence electrons in $n = 3$ , a higher energy level.
2
For isoelectronic species, radius decreases with increasing atomic number. Atomic numbers: $M g = 12$ , $F = 9$ , $O = 8$ . So order for 10-electron ions is: $M g^{2 +} < F^{-} < O^{2 -}$ .
3
$C l^{-}$ has valence electrons in $n = 3$ , which is farther from the nucleus than $n = 2$ for the other ions, so $C l^{-}$ is larger than $O^{2 -}$ .
4
Final order (increasing radius, smallest to largest): $M g^{2 +} < F^{-} < O^{2 -} < C l^{-}$

Exam tip:

Always check for isoelectronic ions first when ranking radii. Students often default to the general atomic radius trend and miss that same-electron-count ions follow a different rule.

3. First Ionization Energy★★★☆☆⏱ 4 min

📘 Definition

First Ionization Energy

$I E_{1}$

Minimum energy required to remove the outermost valence electron from a gaseous neutral atom. The process is always endothermic ( $Δ H > 0$ ) because energy must be added to overcome nuclear attraction.

X(g) \rightarrow X^+(g) + e^- \quad \Delta H = IE_1

The general trend follows $Z_{e f f}$ and radius: $I E_{1}$ generally increases left to right across a period (higher $Z_{e f f}$ holds electrons tighter, so more energy is needed to remove one) and decreases down a group (valence electrons are farther from the nucleus and easier to remove).

AP Chemistry regularly tests two common exceptions to the general trend: (1) Group 13 elements have lower $I E_{1}$ than Group 2 elements in the same period, because the electron removed from Group 13 is in a higher-energy p orbital, compared to the lower-energy s orbital valence electrons of Group 2. (2) Group 16 elements have lower $I E_{1}$ than Group 15 elements in the same period, because Group 15 has a stable half-filled p subshell, while Group 16 has one paired p electron that experiences extra electron-electron repulsion, making it easier to remove.

📐 Worked Example

Which has a higher first ionization energy: phosphorus (Z=15) or sulfur (Z=16)? Justify your answer.

1
Both elements are in period 3, so they have the same number of core electrons, and $Z$ increases from P to S, so the general trend would predict higher $I E_{1}$ for S.
2
Write valence electron configurations: P = $3 s^{2} 3 p^{3}$ , S = $3 s^{2} 3 p^{4}$ .
3
Phosphorus has a half-filled 3p subshell, which has extra stability from symmetric electron distribution and minimal electron-electron repulsion. The electron removed from sulfur is the paired electron in the 3p subshell, which experiences greater repulsion, so less energy is required to remove it.
4
This exception overrides the general trend, so phosphorus has a higher first ionization energy than sulfur.

Exam tip:

Always write the valence electron configuration when justifying an ionization energy exception. AP graders explicitly require this connection to electron configuration to award the point.

4. Electron Affinity and Electronegativity★★★★☆⏱ 3 min

📘 Definition

Electron Affinity

EA

The energy change that occurs when a gaseous neutral atom gains an electron to form an anion. A more negative EA means the process is more energetically favorable, so the atom has a higher affinity for adding an electron.

X(g) + e^- \rightarrow X^-(g) \quad \Delta H = EA

The general trend for EA is that it becomes more negative (higher affinity) moving left to right across a period, and less negative moving down a group. Common exceptions mirror ionization energy: noble gases have positive EA (adding an electron requires placing it in a new higher energy shell, so the process is unfavorable), Group 2 has less negative EA than Group 1, and Group 15 has less negative EA than Group 14.

📘 Definition

Electronegativity

EN

The relative ability of an atom in a covalent bond to attract shared bonding electrons to itself. It is a unitless relative scale.

Electronegativity follows the same general trend as $I E_{1}$ : EN increases left to right across a period, and decreases down a group. Fluorine is the most electronegative element on the periodic table, and francium is the least. A critical distinction: electron affinity describes isolated gaseous atoms gaining an electron, while electronegativity describes attraction for bonding electrons in a molecule.

📐 Worked Example

Which element has a more negative electron affinity: silicon (Z=14) or phosphorus (Z=15)? Justify your answer.

1
Both are in period 3, so the general trend predicts more negative EA for P than Si.
2
Valence configurations: Si = $3 s^{2} 3 p^{2}$ , P = $3 s^{2} 3 p^{3}$ .
3
Adding an electron to Si produces a $3 p^{3}$ configuration, which is a stable half-filled subshell, so this process is very energetically favorable (large negative EA). Adding an electron to P requires placing the new electron in an already occupied 3p orbital, leading to increased electron-electron repulsion, making the process less favorable (less negative EA).
4
Therefore, silicon has a more negative electron affinity than phosphorus.

Exam tip:

Do not confuse electron affinity with electronegativity: mixing these definitions costs points on FRQs.

5. Common Pitfalls

Wrong move:

Claiming atomic radius increases across a period because the number of electrons increases

Why:

Students confuse adding valence electrons to the same shell with adding new shells; across a period, shielding is constant so higher $Z_{e f f}$ offsets any added electron repulsion

Correct move:

When justifying radius across a period, always state that $Z_{e f f}$ increases with constant shielding, pulling electrons closer to decrease radius

Wrong move:

Ranking isoelectronic ions to give larger radius for higher atomic number

Why:

Students assume more protons mean larger size, forgetting the number of electrons is identical

Correct move:

For isoelectronic ions, always note that more protons = higher $Z_{e f f}$ pulls the same number of electrons closer, giving smaller radius

Wrong move:

Stating that second ionization energy is lower than first ionization energy for an element

Why:

Students think removing a second electron is easier after the first, but the second electron is removed from a positively charged ion that holds electrons tighter

Correct move:

Remember ionization energy always increases for each subsequent electron removed, with a large jump when removing core electrons after all valence electrons are gone

Wrong move:

Claiming oxygen has higher first ionization energy than nitrogen, or aluminum higher than magnesium, following only the general trend

Why:

Students memorize the general trend but forget the common orbital-based exceptions that are frequently tested

Correct move:

Whenever comparing $I E$ of adjacent elements in a period, always check their valence electron configurations to see if an exception applies

Wrong move:

Calling a less negative EA "higher" electron affinity

Why:

Students mix up the sign convention: more negative means more energy is released when adding an electron, so higher affinity

Correct move:

When asked "which has higher electron affinity", select the element with the more negative EA value, corresponding to greater favorable attraction for an added electron

Wrong move:

Claiming electronegativity increases down the halogens group because atomic number increases

Why:

Students confuse increasing $Z$ with $Z_{e f f}$ for valence electrons; larger distance from the nucleus dominates over the small $Z_{e f f}$ increase

Correct move:

Remember electronegativity always decreases down a group, with fluorine as the most electronegative element

6. Quick Reference Cheatsheet

Property	Trend Across Period (Left → Right)	Trend Down Group (Top → Bottom)	Key AP Notes
Effective Nuclear Charge ( $Z_{e f f}$ )	Increases	~Constant (very slight increase)	Calculated as $Z_{e f f} = Z - S$ ; basis for all trends
Atomic Radius	Decreases	Increases	Driven by increasing principal quantum number $n$ down groups
Isoelectronic Ionic Radius	N/A	N/A	Decreases with increasing atomic number (same electron count)
First Ionization Energy ( $I E_{1}$ )	Generally increases	Decreases	Exceptions: Group 13 < Group 2, Group 16 < Group 15
Electron Affinity (EA)	Generally more negative	Generally less negative	Noble gases have positive EA (no affinity for added electrons)
Electronegativity (EN)	Increases	Decreases	Fluorine = most EN; unitless relative scale

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Rank ions by ionic radius
2022 · FRQ
Justify ionization energy trend
2021 · MCQ
Compare atomic radius values

Going deeper

unit overviewAP Chemistry Unit 1 OverviewFoundational concepts for periodic trends

What's Next

Periodic trends are a foundational concept for AP Chemistry that you will apply across all units, from bonding and intermolecular forces to acid-base chemistry and electrochemistry. Mastering the skill of justifying trend comparisons with $Z_{e f f}$ and electron configuration is critical for earning full points on FRQs throughout the entire exam. This topic builds directly on your understanding of atomic structure and electron configuration, and it is the basis for predicting bonding behavior and chemical reactivity in all subsequent units.