Part 1: Introduction to Periodic Trends

Welcome to Periodic Trends!

The periodic table is one of the most powerful tools in all of chemistry. It isn't just a list of elements — it's a map that reveals predictable patterns in how elements behave. These patterns, called periodic trends, let you compare and predict properties like size, energy, and reactivity without memorizing data for every element.

In this seven-part series, you'll master the major periodic trends tested on the AP Chemistry exam.

Organization of the Periodic Table

The modern periodic table arranges elements by increasing atomic number (Z).

Key Regions

• Alkali metals — Group 1 (Li, Na, K, …): 1 valence electron, very reactive
• Alkaline earth metals — Group 2 (Be, Mg, Ca, …): 2 valence electrons
• Halogens — Group 17 (F, Cl, Br, …): 7 valence electrons, very reactive nonmetals
• Noble gases — Group 18 (He, Ne, Ar, …): full valence shell, very low reactivity

Quick Check: Table Organization

Effective Nuclear Charge ($Z_{eff}$)

The single most important concept for understanding periodic trends is effective nuclear charge.

The Basic Idea

An atom's nucleus has a charge of $+Z$ (where $Z$ = atomic number). But the outer (valence) electrons don't feel the full $+Z$ because inner-shell electrons block or shield some of that positive charge.

The net positive charge experienced by a valence electron is called the effective nuclear charge:

$Z_{eff} = Z - S$

where:

• $Z$ = actual nuclear charge (atomic number)
• $S$ = shielding constant (approximately equal to the number of core electrons)

Example: Sodium (Na, $Z = 11$)

Sodium has the electron configuration $1s^2\,2s^2\,2p^6\,3s^1$.

• Core electrons (inner shells): $2 + 2 + 6 = 10$
• $Z_{eff} \approx 11 - 10 = +1$

The single valence electron in sodium feels an effective pull of only about $+1$, even though the nucleus has $+11$ protons.

The Shielding Effect

Shielding (or screening) is the reduction of nuclear attraction experienced by valence electrons due to the repulsion from inner-shell electrons.

Key Points

1. Core electrons are effective shielders. Electrons in inner shells ($n-1$, $n-2$, etc.) are located between the nucleus and the valence electrons, effectively canceling some nuclear charge.

2. Valence electrons are poor shielders of each other. Electrons in the same shell do not effectively shield one another because they are at similar distances from the nucleus.

3. Across a period (left → right):

   • $Z$ increases by 1 with each element
   • Electrons are added to the same shell (poor shielding)
   • Result: $Z_{eff}$ increases across a period

4. Down a group (top → bottom):

   • Each new period adds a new shell of core electrons (good shielding)
   • The increase in $Z$ is largely offset by the increase in $S$
   • Result: $Z_{eff}$ felt by valence electrons stays roughly constant or increases only slightly

$Z_{eff}$ Concept Check

Calculate $Z_{eff}$

Using the approximation $Z_{eff} \approx Z - S$ (where $S$ is the number of core electrons), calculate the effective nuclear charge for a valence electron in each element.

1. Beryllium (Be, $Z = 4$, config: $1s^2\,2s^2$). Core electrons = 2.

2. Oxygen (O, $Z = 8$, config: $1s^2\,2s^2\,2p^4$). Core electrons = 2.

3. Potassium (K, $Z = 19$, config: $[Ar]\,4s^1$). Core electrons = 18.

Part 1 Summary Check

Fill in the blanks to review the key ideas.

Part 2: Atomic Radius

Part 2 of 7 — How Big Are Atoms?

Atomic radius is one of the most intuitive periodic trends. You can often predict which atom is larger just by knowing its position on the periodic table. In this lesson, we'll learn the two rules that govern atomic size and practice ranking atoms by radius.

What Is Atomic Radius?

An atom doesn't have a sharp boundary — the electron cloud fades gradually. So chemists define atomic radius in practical terms:

• Covalent radius: Half the distance between the nuclei of two identical bonded atoms.
• Van der Waals radius: Half the distance between nuclei of adjacent atoms in a solid that are not chemically bonded.

For periodic trend discussions, we usually refer to the covalent (bonding) atomic radius.

Approximate Atomic Radii (in pm)

Trend Across a Period (Left → Right)

Atomic radius decreases across a period.

Why?

As you move from left to right across a period:

1. Each element has one more proton in the nucleus
2. Each element has one more electron, but in the same shell
3. Electrons in the same shell shield each other poorly
4. $Z_{eff}$ increases → the nucleus pulls valence electrons closer
5. The atom shrinks

Example: Period 2

$\text{Li} > \text{Be} > \text{B} > \text{C} > \text{N} > \text{O} > \text{F}$ $152 > 112 > 87 > 77 > 75 > 73 > 72 \text{ pm}$

The radius drops by more than half from lithium to fluorine — a dramatic shrinkage driven entirely by increasing $Z_{eff}$.

Trend Down a Group (Top → Bottom)

Atomic radius increases down a group.

Why?

As you move down a group:

1. Each period adds a new principal energy level (shell)
2. Valence electrons are farther from the nucleus
3. Although $Z$ increases, so does shielding (new core electrons added)
4. $Z_{eff}$ stays roughly constant but the distance from nucleus increases
5. The atom expands

Example: Group 1 (Alkali Metals)

$\text{Li} < \text{Na} < \text{K} < \text{Rb} < \text{Cs}$ $152 < 186 < 227 < 248 < 265 \text{ pm}$

Each step down adds an entire new electron shell, making the atom significantly larger.

Visual Summary

$\boxed{\text{Atomic Radius Trends}}$

Quick Rule of Thumb

The largest atoms are in the bottom-left of the periodic table (Cs, Fr).

The smallest atoms are in the top-right (excluding noble gases, which aren't typically measured by covalent radius).

Think: "Down and to the left = bigger."

Ranking Atomic Radii

Comparing Radii

For each pair, type the chemical symbol of the atom with the larger atomic radius.

1. K vs. Ca (both in Period 4)

2. Br vs. I (both in Group 17)

3. Al vs. N (Al is in Period 3 Group 13; N is in Period 2 Group 15)

Exit Check: Atomic Radius

Part 3: Ionization Energy

Part 3 of 7 — How Tightly Do Atoms Hold Their Electrons?

Ionization energy measures how difficult it is to remove an electron from an atom. This property is central to understanding reactivity: metals lose electrons easily (low ionization energy), while nonmetals hold onto them tightly (high ionization energy).

What Is Ionization Energy?

Ionization energy (IE) is the minimum energy required to remove the most loosely bound electron from a gaseous atom or ion.

$X(g) \rightarrow X^+(g) + e^- \qquad \Delta E = IE_1$

• $IE_1$ = first ionization energy (removing the first electron)
• $IE_2$ = second ionization energy (removing a second electron from $X^+$)
• And so on for $IE_3$, $IE_4$, etc.

Important Details

• Ionization energy is always positive (energy must be added to remove an electron)
• Measured in kJ/mol or eV
• We usually focus on the first ionization energy ($IE_1$) when discussing periodic trends

Trend Across a Period (Left → Right)

First ionization energy generally increases across a period.

Why?

Moving left to right across a period:

1. $Z_{eff}$ increases (more protons, same shielding)
2. Valence electrons are held more tightly
3. It takes more energy to remove an electron

First Ionization Energies in Period 2 (kJ/mol)

Notice the Exceptions!

The trend is generally upward, but there are two dips:

1. B < Be: Boron's outermost electron is in a $2p$ orbital (higher energy, easier to remove) versus beryllium's $2s$ orbital.

2. O < N: Nitrogen has a half-filled $2p^3$ configuration (extra stability). Oxygen's fourth $2p$ electron is paired, experiencing electron-electron repulsion that makes it easier to remove.

These exceptions are frequently tested on the AP exam!

Trend Down a Group (Top → Bottom)

First ionization energy decreases down a group.

Why?

Moving down a group:

1. Valence electrons are in higher energy levels, farther from the nucleus
2. More shielding from additional core electron shells
3. The outermost electron is easier to remove → lower $IE_1$

Example: Group 1 (Alkali Metals) $IE_1$ (kJ/mol)

This explains why cesium is more reactive than lithium — its valence electron is much easier to remove.

Ionization Energy Trends

Successive Ionization Energies

Each successive ionization energy is larger than the previous one because:

• You're removing an electron from a more positive ion
• The remaining electrons are held more tightly

The Big Jump

The most important pattern: there is a huge jump in ionization energy when you start removing core electrons.

Example: Magnesium ($1s^2\,2s^2\,2p^6\,3s^2$)

The jump from $IE_2$ to $IE_3$ is enormous — more than 5× larger! This is because $IE_1$ and $IE_2$ remove the two $3s$ valence electrons, but $IE_3$ must remove a core electron from the tightly held $2p$ subshell.

AP Insight

If you see successive IE data, the location of the big jump tells you the number of valence electrons:

• Jump after $IE_2$ → 2 valence electrons (Group 2)
• Jump after $IE_3$ → 3 valence electrons (Group 13)
• Jump after $IE_1$ → 1 valence electron (Group 1)

Successive IE Analysis

Practice Problems

1. Rank these elements from lowest to highest $IE_1$: Na, Mg, K. Type your answer as three symbols separated by commas with the lowest first (e.g., "K, Na, Mg").

2. An element has successive IEs (kJ/mol): 786, 1,577, 3,232, 4,356, 16,091. How many valence electrons does this element have? (Enter a number.)

Part 4: Electron Affinity

Part 4 of 7 — How Much Do Atoms Want More Electrons?

While ionization energy measures how hard it is to remove an electron, electron affinity measures the energy change when an atom gains an electron. Together, these properties explain why some elements form cations and others form anions.

What Is Electron Affinity?

Electron affinity (EA) is the energy change that occurs when a gaseous atom gains an electron:

$X(g) + e^- \rightarrow X^-(g) \qquad \Delta E = EA$

Sign Convention

• A negative EA means energy is released — the atom wants the electron (exothermic).
• A more negative EA means a greater tendency to gain an electron.
• A positive EA means energy must be added — the atom resists gaining an electron.

Example Values (kJ/mol)

Notice that chlorine has the most negative EA of any element — even more than fluorine!

General Trends

Across a Period (Left → Right)

Electron affinity generally becomes more negative (more favorable) across a period.

Why?

• $Z_{eff}$ increases across a period
• The atom has a stronger pull on an incoming electron
• Nonmetals (right side) have the most negative EA values

Down a Group (Top → Bottom)

Electron affinity generally becomes less negative (less favorable) down a group.

Why?

• Valence shell is farther from the nucleus
• The incoming electron feels less attraction
• Exception: fluorine vs. chlorine (see below)

The Fluorine Anomaly

Fluorine's EA (−328 kJ/mol) is less negative than chlorine's (−349 kJ/mol). This is because fluorine is so small that adding an electron to its compact $2p$ orbitals creates significant electron-electron repulsion. Chlorine's larger $3p$ orbitals accommodate the extra electron more easily.

Important Exceptions

Several elements have near-zero or positive electron affinities:

1. Noble Gases (Group 18)

Full valence shells — no room for an additional electron without going to a higher energy level. EA is positive (unfavorable).

2. Nitrogen ($2p^3$)

Has a half-filled $2p$ subshell. Adding an electron would break this stable half-filled configuration. EA ≈ 0.

3. Group 2 Elements (Be, Mg, Ca)

Full $s$ subshell ($ns^2$). An added electron would go into a higher-energy $p$ orbital. EA is near zero or slightly positive.

AP Exam Tip

The AP exam often asks you to explain why certain elements have unexpectedly low (or positive) electron affinities. Always connect your answer to electron configuration stability (filled or half-filled subshells).

Electron Affinity Concepts

Comparing Electron Affinities

Practice

1. Of the following — Na, S, Cl, Ar — which element has the most negative electron affinity? (Type the symbol.)

2. An element in Period 2 has a near-zero electron affinity and a half-filled $p$ subshell. What is this element? (Type the symbol.)

Exit Quiz

Part 5: Electronegativity

Part 5 of 7 — Who Pulls Harder on Shared Electrons?

Electronegativity is the property that determines how electrons are distributed in chemical bonds. It's the bridge between atomic properties and molecular behavior — and it shows up constantly on the AP exam.

What Is Electronegativity?

Electronegativity is a measure of an atom's ability to attract electrons toward itself in a chemical bond.

The Pauling Scale

Linus Pauling developed the most widely used electronegativity scale:

Key Facts

• Fluorine has the highest electronegativity of any element (4.0)
• Noble gases are generally not assigned electronegativity values (they rarely form bonds)
• Electronegativity is a relative scale, not an absolute energy measurement
• It combines aspects of both ionization energy and electron affinity

Periodic Trends in Electronegativity

Across a Period (Left → Right)

Electronegativity increases across a period.

Why?

• $Z_{eff}$ increases → atoms pull harder on bonding electrons
• Nonmetals (right side) are the most electronegative

Down a Group (Top → Bottom)

Electronegativity decreases down a group.

Why?

• Bonding electrons are farther from the nucleus
• Additional shielding reduces the attraction for bonding electrons

Summary

The most electronegative elements are in the top-right corner (F, O, N, Cl).

The least electronegative elements are in the bottom-left corner (Cs, Fr, Ba).

This trend mirrors ionization energy and is the opposite of atomic radius.

Electronegativity and Bond Polarity

The difference in electronegativity ($\Delta EN$) between two bonded atoms determines the type of bond:

Dipole Direction

In a polar bond, the more electronegative atom carries a partial negative charge ($\delta^-$) and the less electronegative atom carries a partial positive charge ($\delta^+$).

$\overset{\delta^+}{\text{H}} — \overset{\delta^-}{\text{Cl}}$

The dipole arrow points toward the more electronegative atom.

Electronegativity Trends

Bond Polarity Practice

Calculate the electronegativity difference ($\Delta EN$) for each bond and classify it. Use the values: H = 2.1, C = 2.5, O = 3.5, F = 4.0, Na = 0.9, Cl = 3.0.

1. $\text{C-O}$ bond: What is $\Delta EN$? (Give your answer to 3 significant figures.)

2. $\text{Na-Cl}$ bond: What is $\Delta EN$? (Give your answer to 3 significant figures.)

3. $\text{C-H}$ bond: What is $\Delta EN$? (Give your answer to 3 significant figures.)

Ranking Practice

Exit Quiz

Part 6: Ionic Radius

Part 6 of 7 — How Does Gaining or Losing Electrons Change Size?

When atoms become ions, their size changes dramatically. Cations shrink and anions expand. Understanding ionic radius — especially in isoelectronic series — is a classic AP Chemistry skill.

Cations Are Smaller Than Their Parent Atoms

When an atom loses electrons to form a cation:

1. The outermost shell may be completely emptied, exposing a smaller inner shell
2. Even if the shell isn't emptied, fewer electrons means less electron-electron repulsion
3. The same nuclear charge pulls the remaining electrons closer

Example: Na → Na⁺

Sodium loses its single $3s$ electron, eliminating the entire $n = 3$ shell. The radius drops by nearly half.

General Pattern

The more electrons removed, the smaller the cation:

$\text{Fe} (126\text{ pm}) > \text{Fe}^{2+} (76\text{ pm}) > \text{Fe}^{3+} (65\text{ pm})$

Anions Are Larger Than Their Parent Atoms

When an atom gains electrons to form an anion:

1. More electrons means increased electron-electron repulsion
2. The electron cloud expands as electrons push each other apart
3. The same nuclear charge now must hold more electrons

Example: Cl → Cl⁻

Chlorine gains one electron, nearly doubling its radius due to increased electron repulsion with the same nuclear charge.

Summary Table

Cation vs. Anion Size

Isoelectronic Series

An isoelectronic series is a set of atoms and ions that all have the same number of electrons.

Example: The 10-Electron Series

All of the following species have 10 electrons (same as neon):

The Rule

In an isoelectronic series, all species have the same electron count, so the size is determined entirely by nuclear charge:

$\boxed{\text{More protons} = \text{smaller radius}}$

The ion with the most protons pulls its electrons in the tightest and is the smallest.

$\text{N}^{3-} > \text{O}^{2-} > \text{F}^- > \text{Na}^+ > \text{Mg}^{2+} > \text{Al}^{3+}$

Isoelectronic Series Practice

Ranking Exercises

1. Rank the following isoelectronic species from smallest to largest radius: K⁺, Cl⁻, Ca²⁺, S²⁻. All have 18 electrons. Type your answer as symbols separated by commas (smallest first).

2. Consider the pair Mg²⁺ and O²⁻. Both have 10 electrons. Which has the larger radius? (Type the ion, e.g., "Mg2+" or "O2-".)

Exit Check: Ionic Radius

Part 7: Synthesis & AP Review

Part 7 of 7 — Putting It All Together

On the AP Chemistry exam, periodic trend questions rarely test just one property. You'll need to combine your knowledge of atomic radius, ionization energy, electron affinity, electronegativity, and ionic radius to answer complex comparison and explanation questions. This final lesson pulls everything together.

Master Summary of All Trends

The Big Pattern

Atomic radius goes in the opposite direction from IE, EA, and EN.

• Small atom → tightly held electrons → high IE, high EN, very negative EA
• Large atom → loosely held electrons → low IE, low EN, near-zero EA

All four major trends are ultimately explained by the same two factors:

1. Effective nuclear charge ($Z_{eff}$) — across a period
2. Distance from nucleus / number of shells — down a group

Key Exceptions to Remember

The AP exam frequently tests these exceptions:

1. IE: B < Be and O < N

• B < Be: B removes a $2p$ electron (higher energy) vs. Be's $2s$ electron
• O < N: O has a paired $2p$ electron; N's half-filled $2p^3$ has extra stability

2. EA: F less negative than Cl

• Fluorine is so small that electron-electron repulsion in its tiny $2p$ orbitals offsets the energy gain
• $EA_\text{F} = -328$ kJ/mol vs. $EA_\text{Cl} = -349$ kJ/mol

3. EA ≈ 0 for N, Be, Ne

• N: Half-filled $2p^3$ — stable configuration resists an extra electron
• Be: Full $2s^2$ subshell — added electron would go to higher-energy $2p$
• Ne: Full octet — added electron enters $3s$ (much higher energy)

AP Writing Tip

When asked to "explain" a trend, always state:

1. What happens to $Z_{eff}$ (or distance from nucleus)
2. Why that affects the property
3. Connect back to electron configuration when exceptions arise

Multi-Trend AP Questions

Explaining Trends Using Electron Configuration

On the AP exam, you often must explain a trend, not just state it. Here is a model framework:

Template for Across-a-Period Explanations

"As you move from [element A] to [element B] across Period [N], the atomic number increases from [Z₁] to [Z₂]. Electrons are added to the same principal energy level ($n =$ [N]), and same-shell electrons do not effectively shield each other. Therefore, the effective nuclear charge ($Z_{eff}$) increases, which causes [property] to [increase/decrease]."

Template for Down-a-Group Explanations

"As you move from [element A] to [element B] down Group [G], each element has an additional principal energy level occupied by electrons. The valence electrons are farther from the nucleus and are shielded by more core electrons. Although the nuclear charge increases, the effect of increased distance and shielding dominates, causing [property] to [increase/decrease]."

Template for Exceptions

"[Element B] has a [lower/higher] [property] than expected because its electron configuration [describe specific feature: paired electron, half-filled subshell, etc.], which [increases repulsion / provides extra stability], making the electron [easier/harder] to remove."

AP-Style Matching

Mixed Practice

1. Of Li, C, F, and Na — which element has the highest first ionization energy? (Type the symbol.)

2. Rank the following from smallest to largest ionic radius: Na⁺, Mg²⁺, F⁻. All have 10 electrons. (Type as symbols separated by commas, smallest first, e.g., "Mg2+, Na+, F-".)

3. Which element in Period 3 has the smallest atomic radius (excluding noble gases)? (Type the symbol.)

Final Exit Quiz — AP Exam Readiness