🎯⭐ INTERACTIVE LESSON

Periodic Trends

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Periodic Trends - Complete Interactive Lesson

Part 1: Atomic Radius

Part 1: Introduction to Periodic Trends

Part 1 of 7 — Atomic Radius

Topics in This Part

Section
Key Regions
The Basic Idea
Example: Sodium (Na, $Z = 11$ )
Key Points

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 1

Understanding the core concepts covered in Part 1
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

📊 Organization of the Periodic Table

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

Term	Definition
Period	A horizontal row (1–7). Elements in the same period have the same number of electron shells.
Group	A vertical column (1–18). Elements in the same group have the same number of valence electrons and similar chemical properties.
Main-group elements	Groups 1–2 and 13–18 (s- and p-block)
Transition metals	Groups 3–12 (d-block)

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 or some of that positive charge.

�️ 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

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.
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.
Across a period (left → right):
- $Z$ increases by 1 with each element

$Z_{eff}$ Concept Check

Calculate $Z_{eff}$

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

Part 1 Summary Check

Fill in the blanks to review the key ideas.

Part 2: Ionization Energy

Part 2: Atomic Radius

Part 2 of 7 — How Big Are Atoms?

Topics in This Part

Section
Approximate Atomic Radii (in pm)
Why?
Example: Period 2
Why?
Example: Group 1 (Alkali Metals)

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 2

Understanding the core concepts covered in Part 2
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

📖 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)

Element	Radius (pm)	Element

Part 3: Electron Affinity

Part 3: Ionization Energy

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

Topics in This Part

Section
Important Details
Why?
First Ionization Energies in Period 2 (kJ/mol)
Notice the Exceptions!
Why?

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 3

Understanding the core concepts covered in Part 3
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

📖 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.

$X (g) \to X^{+} (g) + e^{-} Δ E = I E_{}$

Part 4: Electronegativity

Part 4: Electron Affinity

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

Topics in This Part

Section
Sign Convention
Example Values (kJ/mol)
Across a Period (Left → Right)
Down a Group (Top → Bottom)
The Fluorine Anomaly

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 4

Understanding the core concepts covered in Part 4
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

📖 What Is Electron Affinity?

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

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

Part 5: Ionic Radius

Part 5: Electronegativity

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

Topics in This Part

Section
The Pauling Scale
Key Facts
Across a Period (Left → Right)
Down a Group (Top → Bottom)
Summary

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 5

Understanding the core concepts covered in Part 5
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

📖 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:

Element	EN	Element	EN
F	4.0	C	2.5

Part 6: Problem-Solving Workshop

Part 6: Ionic Radius

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

Practice Makes Perfect

This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — structured practice is the best preparation.

What You'll Master in Part 6

Working through complete multi-step problems from start to finish
Building problem-solving strategies you can apply on the AP exam
Identifying which concepts to apply and in what order

📌 Cations Are Smaller Than Their Parent Atoms

When an atom loses electrons to form a cation:

💡 Tip: Cations lose electrons → fewer electrons, same protons → electrons pulled in tighter → smaller ion.

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

Example: Na → Na⁺

	Na	Na⁺
Protons	11

Part 7: Synthesis & AP Review

Part 7 of 7 — Putting It All Together

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — success requires connecting ideas across topics.

What You'll Master in Part 7

Solving AP-style questions that integrate multiple concepts from this unit
Writing clear, concise explanations using proper chemistry terminology
Identifying and avoiding common AP exam traps and mistakes

📋 Master Summary of All Trends

Property	Across Period (→)	Down Group (↓)	Driven By
Atomic radius	Decreases	Increases	$Z_{eff}$ and shell count

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

$\boxed{Z_{eff} = Z - S}$

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

🔑 Key Concept: $Z_{eff}$ is the single most important idea for explaining periodic trends. Almost every trend can be traced back to how $Z_{eff}$ changes across a period or down a group.

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

Problem: Calculate $Z_{eff}$ for the valence electron of sodium ( $Z = 11$ , config: $1s^2\,2s^2\,2p^6\,3s^1$ ).

Solution: Core electrons (inner shells) = $2 + 2 + 6 = 10$ . Therefore $Z_{eff} \approx 11 - 10 = +1$ . The single valence electron feels an effective pull of only about $+1$ , even though the nucleus has $+11$ protons.

Electrons are added to the same shell (poor shielding)

Result:

Z_{eff}

increases across a period

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

🔑 Key Concept: Across a period → $Z_{eff}$ rises (same shielding, more protons). Down a group → $Z_{eff}$ stays roughly constant (shielding increases with $Z$ ).

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.

� Trend Across a Period (Left → Right)

Atomic radius decreases across a period.

🔑 Key Concept: More protons + same shielding = higher $Z_{eff}$ = electrons pulled closer = smaller atom.

Why?

As you move from left to right across a period:

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

Example: Period 2

$\boxed{\text{Li} > \text{Be} > \text{B} > \text{C} > \text{N} > \text{O} > \text{F}}$

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.

🔑 Key Concept: Each new period adds an entire electron shell, pushing valence electrons farther from the nucleus.

Why?

As you move down a group:

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

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

Direction	Trend	Reason
→ Across period	Decreases	$Z_{eff}$ increases; same shell, poor shielding
↓ Down group	Increases	New shell added; greater distance from nucleus

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."

💡 Tip: When comparing atoms in different groups AND different periods, check both directions. If both effects agree (e.g., lower period AND farther left), the answer is clear. If they conflict, you may need data.

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

X (g) \to X^{+} (g) + e^{-} Δ E = I E_{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.

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.

🔑 Key Concept: Higher $Z_{eff}$ across a period means valence electrons are held more tightly, requiring more energy to remove.

Why?

Moving left to right across a period:

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

First Ionization Energies in Period 2 (kJ/mol)

Li	Be	B	C	N	O	F	Ne
520	900	801	1086	1402	1314	1681	2081

Notice the Exceptions!

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

⚠️ Warning: These two IE exceptions are frequently tested on the AP exam.

1. B < Be (Group 13 < Group 2): Boron's outermost electron is in a $2p$ orbital (higher energy, easier to remove) versus beryllium's $2s$ orbital.

2. O < N (Group 16 < Group 15): 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.

📈 Trend Down a Group (Top → Bottom)

First ionization energy decreases down a group.

Why?

Moving down a group:

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

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

Li	Na	K	Rb	Cs
520	496	419	403	376

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

$IE_1$	$IE_2$	$IE_3$

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

AP Insight

🔑 Key Concept: The location of the "big jump" in successive IEs reveals the number of valence electrons.

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.)

X (g) + e^{-} \to X^{-} (g) Δ E = E A

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)

Element	EA (kJ/mol)
F	−328
Cl	−349
Br	−325
O	−141
N	≈ 0
Ne	> 0

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

⚠️ Warning: Fluorine's EA (−328 kJ/mol) is less negative than chlorine's (−349 kJ/mol). This is a common AP exam question!

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

💡 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.)

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

Direction	Trend
→ Across period	EN increases
↓ Down group	EN decreases

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).

🔑 Key Concept: Electronegativity follows the same pattern as ionization energy and is the opposite of atomic radius. Small atoms with high $Z_{eff}$ pull hardest on bonding electrons.

💡 Tip: If you know the atomic radius trend, you know the EN trend — just flip the direction!

🔋 Electronegativity and Bond Polarity

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

$\boxed{\Delta EN = |EN_A - EN_B|}$

$\Delta EN$	Bond Type	Example
$0$	Nonpolar covalent	$\text{H}_2$ ( $\Delta EN = 0$ )

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.)

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

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:

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

Example: Cl → Cl⁻

	Cl	Cl⁻
Protons	17	17
Electrons	17	18
Config	$[Ne]\,3s^2\,3p^5$	$[Ne]\,3s^2\,3p^6$
Radius	99 pm	181 pm

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

Summary Table

Ion Type	Size vs. Parent Atom	Example
Cation ( $+$ )	Smaller	Na⁺ (95 pm) < Na (186 pm)
Anion ( $-$ )	Larger	Cl⁻ (181 pm) > Cl (99 pm)

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):

Species	Protons	Electrons	Radius (pm)
N³⁻	7	10	146
O²⁻	8	10	140
F⁻	9	10	133
Ne	10	10	—
Na⁺	11	10	95
Mg²⁺	12	10	65
Al³⁺	13	10	50

The Rule

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

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

🔑 Key Concept: In an isoelectronic series, rank by proton count — the species with the most protons 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

🔑 Key Concept: Atomic radius goes in the opposite direction from IE, EA, and EN. All four trends are driven by $Z_{eff}$ (across periods) and distance/shells (down groups).

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:

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

⚠️ Key Exceptions to Remember

The AP exam frequently tests these exceptions:

📌 Exception Reference Table

Exception	What Happens	Why
IE: B < Be	IE drops going from Be to B	B removes a $2p$ e⁻ (higher energy) vs. Be's $2s$ e⁻
IE: O < N	IE drops going from N to O	O has a paired $2p$ e⁻; N's half-filled $2p^3$ is extra stable
EA: F < Cl (less negative)	F gains e⁻ less readily than Cl	F is so small → e⁻–e⁻ repulsion in tiny $2p$ orbitals
EA ≈ 0: N, Be, Ne	Won't accept an extra electron	Stable configs: half-filled $2p^3$ , full $2s^2$ , full octet

🧪 EA Values to Know

Element	$EA$ (kJ/mol)	Why?
F	−328	Small size → repulsion offsets gain
Cl	−349	Larger, less repulsion → most negative EA
N, Be, Ne	≈ 0	Stable subshell configs resist extra e⁻

💡 AP Writing Tip: When explaining a trend exception, always state: (1) what happens to $Z_{eff}$ or distance, (2) why it affects the property, and (3) connect to electron configuration and subshell occupancy.

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."

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

152 > 112 > 87 > 77 > 75 > 73 > 72 \text{ pm}

Li	152	Na	186
Be	112	Mg	160
B	87	Al	143
C	77	Si	117
N	75	P	110
O	73	S	104
F	72	Cl	99

Periodic Trends

Periodic Trends - Complete Interactive Lesson

Part 1: Atomic Radius

Part 1: Introduction to Periodic Trends

Topics in This Part

What You'll Master in Part 1

📊 Organization of the Periodic Table

Key Regions

Quick Check: Table Organization

⚛️ Effective Nuclear Charge (ZeffZ_{eff}Zeff​)

The Basic Idea

�️ The Shielding Effect

Key Points

ZeffZ_{eff}Zeff​ Concept Check

Calculate ZeffZ_{eff}Zeff​

Part 1 Summary Check

Part 2: Ionization Energy

Part 2: Atomic Radius

Topics in This Part

What You'll Master in Part 2

📖 What Is Atomic Radius?

Approximate Atomic Radii (in pm)

Part 3: Electron Affinity

Part 3: Ionization Energy

Topics in This Part

What You'll Master in Part 3

📖 What Is Ionization Energy?

Part 4: Electronegativity

Part 4: Electron Affinity

Topics in This Part

What You'll Master in Part 4

📖 What Is Electron Affinity?

Part 5: Ionic Radius

Part 5: Electronegativity

Topics in This Part

What You'll Master in Part 5

📖 What Is Electronegativity?

The Pauling Scale

Part 6: Problem-Solving Workshop

Part 6: Ionic Radius

Practice Makes Perfect

What You'll Master in Part 6

📌 Cations Are Smaller Than Their Parent Atoms

Example: Na → Na⁺

Part 7: Synthesis & AP Review

Part 7: Synthesis & AP Review

Bringing It All Together

What You'll Master in Part 7

📋 Master Summary of All Trends

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

� Trend Across a Period (Left → Right)

Why?

Example: Period 2

📈 Trend Down a Group (Top → Bottom)

Why?

Example: Group 1 (Alkali Metals)

📋 Visual Summary

Quick Rule of Thumb

Ranking Atomic Radii

Comparing Radii

Exit Check: Atomic Radius

Important Details

📈 Trend Across a Period (Left → Right)

Why?

First Ionization Energies in Period 2 (kJ/mol)

Notice the Exceptions!

📈 Trend Down a Group (Top → Bottom)

Why?

Example: Group 1 (Alkali Metals) IE1IE_1IE1​ (kJ/mol)

Ionization Energy Trends

⚛️ Successive Ionization Energies

The Big Jump

Example: Magnesium (1s2 2s2 2p6 3s21s^2\,2s^2\,2p^6\,3s^21s22s22p63s2)

AP Insight

Successive IE Analysis

Practice Problems

Sign Convention

Example Values (kJ/mol)

📈 General Trends

Across a Period (Left → Right)

⚛️ Effective Nuclear Charge ( $Z_{eff}$ )

$Z_{eff}$ Concept Check

Calculate $Z_{eff}$

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

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

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

2. Nitrogen ( $2p^3$ )