Periodic trends — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: effective nuclear charge calculation, atomic radius, ionic radius, first ionization energy (including common exceptions), electron affinity, and electronegativity, with reasoning for periodic comparisons across periods and groups.

You should already know: electron configuration for neutral atoms and ions, Coulomb’s law of electrostatic attraction, the structure of the modern periodic table (periods vs groups).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Periodic trends?

Periodic trends are systematic, predictable patterns in the chemical and physical properties of elements that emerge when elements are arranged by increasing atomic number in the modern periodic table. All periodic trends can be rationalized using two core atomic properties: effective nuclear charge experienced by valence electrons and electron shielding by inner core electrons. This topic makes up approximately 4-6% of the total AP Chemistry exam weight, and is tested on both multiple-choice (MCQ) and free-response (FRQ) sections. On MCQs, you will most often be asked to rank elements by a given property or identify the element with the highest/lowest value of a property. On FRQs, you are almost always required to justify your comparison using periodic trend reasoning, which is a common scoring point that many students miss. The periodic table was originally developed based on observation of these trends before the structure of the atom was fully understood, but modern atomic theory explains why these patterns occur, making this topic a key application of atomic structure principles.

2. Effective Nuclear Charge ($Z_{eff}$)

All periodic trends originate from differences in effective nuclear charge, the net positive attraction a valence electron experiences from the nucleus, after accounting for shielding by inner core electrons. The simplified formula used in AP Chemistry for $Z_{eff}$ is: $$Z_{eff}=Z-S$$ where $Z$ is the atomic number (total number of protons) and $S$ is the shielding constant, approximated by the total number of inner core electrons. Intuitively, inner core electrons sit between the nucleus and valence electrons, so they block most of the positive charge from the nucleus, so valence electrons only experience a fraction of the full nuclear charge.

Trends in $Z_{eff}$ are consistent: moving left to right across a period (row), $Z$ increases by 1 for each element, while the number of core electrons stays the same, so $S$ is constant. This means $Z_{eff}$ increases steadily across a period. Moving down a group (column), $Z$ increases, but the number of core electrons also increases by a full new shell, so $S$ increases almost as much as $Z$, leading to only a very small increase in $Z_{eff}$ down most groups.

Worked Example

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

1. Write the full electron configuration for each element: Sodium = $1s^{2}2s^{2}2p^{6}3s^1$, so core electrons = 10, valence electrons = 1. Rubidium = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^1$, so core electrons = 36, valence electrons = 1.
2. Calculate approximate $Z_{eff}$ for each using the formula $Z_{eff}=Z-S$: $Z_{eff}(\text{Na})=11-10=+1$; $Z_{eff}(\text{Rb})=37-36=+1$.
3. Both are group 1 elements, so the net $Z_{eff}$ for valence electrons is approximately equal. The small difference between them is negligible for AP-level reasoning.
4. If forced to select the lower $Z_{eff}$, rubidium has a very slightly lower net $Z_{eff}$ due to increased shielding from additional core electrons that is not fully offset by the increase in proton number.

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 $Z_{eff}$ and shielding to get full credit.

3. Atomic and Ionic Radius

Atomic radius is defined as half the distance between the nuclei of two adjacent bonded atoms of the same element. The trend in atomic radius directly follows from $Z_{eff}$ and principal quantum number $n$ (the energy level of valence electrons): moving left to right across a period, increasing $Z_{eff}$ pulls valence electrons closer to the nucleus, so atomic radius decreases. Moving down a group, valence electrons occupy higher energy levels with larger average distance from the nucleus, and this effect dominates over the small increase in $Z_{eff}$, so atomic radius increases down a group.

For ionic radius, additional rules apply: Cations are smaller than their parent neutral atom, because they lose an entire valence shell of electrons and have a higher $Z_{eff}$ per electron. Anions are larger than their parent neutral atom, because they add electrons to the valence shell, increasing electron-electron repulsion without changing $Z$. For isoelectronic ions (ions with the same number of electrons), ionic radius decreases with increasing atomic number, because higher $Z$ gives higher $Z_{eff}$ 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 identify the number of electrons for each species: $F^{-}$ = 10, $O^{2-}$ = 10, $M{g}^{2+}$ = 10, $C{l}^{-}$ = 18. All three 10-electron species are isoelectronic, while $C{l}^{-}$ has valence electrons in n=3, a higher energy level.
2. For isoelectronic species, radius decreases with increasing Z. Atomic numbers: $Mg$=12, $F$=9, $O$=8. So order of radius for the 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 (smallest to largest, increasing radius): $M{g}^{2+}<F^{-}<O^{2-}<C{l}^{-}$.

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.

4. First Ionization Energy

First ionization energy ($I{E}_1$) is the minimum energy required to remove the outermost valence electron from a gaseous neutral atom, represented by the reaction: $$X(g)\to X^{+}(g)+e^{-}\Delta H=I{E}_1$$ Ionization energy is always endothermic (positive $\Delta H$), because energy must be added to remove an electron attracted to the nucleus. The general trend follows $Z_{eff}$ and radius: $I{E}_1$ generally increases left to right across a period, because higher $Z_{eff}$ holds electrons tighter, so more energy is needed to remove one. $I{E}_1$ decreases down a group, because 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 = $3s^{2}3p^3$, S = $3s^{2}3p^4$.
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.

5. Electron Affinity and Electronegativity

Electron affinity (EA) is the energy change that occurs when a gaseous neutral atom gains an electron to form an anion: $X(g)+e^{-}\to X^{-}(g)\Delta H=EA$. A more negative EA means the process is more energetically favorable, so the atom has a higher affinity for adding an electron. The general trend is that EA 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 (they do not want to add electrons, as it requires placing the new electron in a new higher energy shell), Group 2 has less negative EA than Group 1, and Group 15 has less negative EA than Group 14.

Electronegativity (EN) is the ability of an atom in a covalent bond to attract shared bonding electrons to itself. It 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. It is a relative scale (no units) unlike the other periodic properties.

Worked Example

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

1. Both are in period 3, so general trend predicts more negative EA for P than Si.
2. Valence configurations: Si = $3s^{2}3p^2$, P = $3s^{2}3p^3$.
3. Adding an electron to Si produces a $3p^3$ 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: electron affinity describes isolated gaseous atoms gaining an electron, while electronegativity describes attraction for bonding electrons in a molecule. Mixing these definitions costs points on FRQs.

6. Common Pitfalls (and how to avoid them)

• 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_eff offsets any added electron repulsion. Correct move: When justifying radius across a period, always state that Z_eff 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_eff 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 IE 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_eff for valence electrons; larger distance from the nucleus dominates over the small Z_eff increase. Correct move: Remember electronegativity always decreases down a group, with fluorine as the most electronegative element.

7. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following correctly ranks the elements Ca, K, Mg, Al in order of decreasing first ionization energy (highest to lowest)? A) Al > Mg > K > Ca B) Mg > Al > Ca > K C) Mg > Ca > Al > K D) Al > Mg > Ca > K

Worked Solution: First, apply the general trend that ionization energy decreases down a group: all period 3 elements (Mg, Al) have higher IE than period 4 elements (Ca, K), so eliminate option A. Next, across period 3, Mg (group 2) has higher IE than Al (group 13) due to the exception for the higher-energy 3p electron in Al, so eliminate option C. For the period 4 elements Ca (group 2) and K (group 1), Ca has higher Z_eff than K, so Ca has higher IE than K. The final order is Mg > Al > Ca > K. The correct answer is B.

Question 2 (Free Response)

Selenium (Se, Z=34), bromine (Br, Z=35), and arsenic (As, Z=33) are consecutive elements in period 4 of the periodic table. (a) Identify the element with the largest atomic radius and justify your answer. (b) Identify the element with the highest first ionization energy, and explain why it has higher IE than the adjacent elements. (c) Explain why the third ionization energy of As is lower than the third ionization energy of Se.

Worked Solution: (a) Arsenic has the largest atomic radius. All three elements are in period 4, so they have the same number of core electrons, so shielding is constant. Arsenic has the lowest atomic number (lowest Z), so it has the lowest Z_eff, which pulls valence electrons less strongly than in Se or Br. Lower Z_eff leads to a larger atomic radius. (b) Arsenic has the highest first ionization energy. Arsenic has valence electron configuration $4s^{2}4p^3$, which is a stable half-filled 4p subshell. Selenium has $4s^{2}4p^4$, so removing an electron from Se removes a paired electron with higher electron-electron repulsion, so Se has lower IE. Bromine has higher Z_eff than As, but it has one paired p electron that is easier to remove than the half-filled p subshell of As. Thus As has the highest IE. (c) After removing two electrons, $A{s}^{2+}$ has valence configuration $4s^{2}4p^1$, while $S{e}^{2+}$ has $4s^{2}4p^3$, a stable half-filled p subshell. Removing the third electron from $S{e}^{2+}$ requires breaking the stable half-filled subshell, which takes more energy than removing the single 4p electron from $A{s}^{2+}$. Thus the third IE of As is lower than that of Se.

Question 3 (Application / Real-World Style)

Water purification systems use ion-exchange resins that trap ions based on their size: smaller ions bind more tightly to the resin, so larger ions are displaced first from the resin when a solution is run through. A solution containing the following ions is passed through an ion-exchange resin: $S{r}^{2+}$, $R{b}^{+}$, $C{a}^{2+}$, $K^{+}$. Which ion will be displaced first (has the largest radius)? Justify your answer with periodic reasoning.

Worked Solution: First, group the ions by period: Ca and K are period 4, Sr and Rb are period 5. Ions in period 5 have valence electrons in n=5, which is farther from the nucleus than n=4 for period 4, so Sr and Rb are larger than Ca and K. For Sr²⁺ and Rb⁺, they are isoelectronic, both have 36 electrons. Sr²⁺ has Z=38, Rb⁺ has Z=37. Higher Z gives higher Z_eff for the same number of electrons, so Sr²⁺ has a smaller radius than Rb⁺. The order of radii from smallest to largest is $C{a}^{2+}<K^{+}<S{r}^{2+}<R{b}^{+}$. Rb⁺ is the largest ion, so it will be displaced first from the ion-exchange resin, consistent with the size-based binding rule.

8. Quick Reference Cheatsheet

9. What's Next

Mastering periodic trends is a critical prerequisite for all subsequent topics involving bonding, intermolecular forces, and chemical reactivity. Next, you will apply the concept of electronegativity differences to classify bonds as ionic, polar covalent, or nonpolar covalent, and to predict molecular polarity in covalent compounds. Without understanding periodic trends in electronegativity and atomic radius, you cannot correctly predict bond polarity, bond strength, or the intermolecular forces that govern physical properties like boiling point and solubility. Periodic trends also explain the reactivity of metals and nonmetals, which is foundational for studying all reaction-based topics later in the course. Follow-on topics that build directly on this chapter: Bond polarity and classification, Molecular structure and polarity, Intermolecular force strength, Main group element reactivity