Blocks: s, p, d, f based on electron configuration
Key Regions
Metals: Left and center (conduct electricity, malleable, lustrous)
Nonmetals: Right side (poor conductors, brittle as solids)
Metalloids: Staircase line (properties between metals and nonmetals)
Four Major Periodic Trends
1. Atomic Radius
Definition: Distance from nucleus to outermost electron (half the distance between nuclei of bonded atoms)
Trends:
Across a period (left → right): Decreases
Down a group (top → bottom): Increases
Explanation:
Across a period:
Same energy level, but increasing nuclear charge
More protons pull electrons closer
Effective nuclear charge increases
Down a group:
Each period adds a new energy level (shell)
Electrons are farther from nucleus
Shielding from inner electrons
Example: Na > Mg > Al > Si > P > S > Cl (Period 3)
2. Ionization Energy (IE)
Definition: Energy required to remove an electron from a gaseous atom
X(g)+energy→X+(g)+e−
Trends:
Across a period (left → right): Increases
Down a group (top → bottom): Decreases
Explanation:
Across a period:
Smaller atomic radius
Stronger attraction between nucleus and electrons
Harder to remove electron
Down a group:
Larger atomic radius
Electrons farther from nucleus
Easier to remove electron
Exceptions:
Slight decrease from Group 15 to 16 (half-filled stability)
Slight decrease from Group 2 to 13 (new sublevel)
Example: He has highest IE; Cs has lowest IE
3. Electron Affinity (EA)
Definition: Energy change when an electron is added to a gaseous atom
X(g)+e−→X−(g)+energy
Convention: Negative EA means energy is released (favorable)
Trends:
Across a period (left → right): Generally becomes more negative (more favorable)
Down a group (top → bottom): Generally becomes less negative
Explanation:
Across a period:
Smaller atoms can accommodate extra electron more easily
Higher effective nuclear charge attracts electron
Down a group:
Larger atoms have more diffuse electron cloud
Added electron is farther from nucleus
Exceptions:
Noble gases have positive EA (stable configuration)
Group 2 and 15 have less negative EA (filled/half-filled stability)
Example: Cl has most negative EA (excluding noble gases)
4. Electronegativity
Definition: Ability of an atom to attract electrons in a chemical bond
Pauling Scale: 0.7 (Cs) to 4.0 (F)
Trends:
Across a period (left → right): Increases
Down a group (top → bottom): Decreases
Explanation:
Similar to ionization energy:
Smaller atoms attract bonding electrons more strongly
Higher nuclear charge increases pull on electrons
Example: F (4.0) > O (3.5) > N (3.0) > C (2.5)
Note: Noble gases are not assigned electronegativity values (don't form bonds)
Ionic Radius Trends
Cations (Positive Ions)
Always smaller than parent atom
Lost electrons → less electron-electron repulsion
Same nuclear charge pulls fewer electrons
Example: Na (186 pm) > Na⁺ (102 pm)
Anions (Negative Ions)
Always larger than parent atom
Gained electrons → more electron-electron repulsion
Same nuclear charge pulls more electrons
Example: Cl (99 pm) < Cl⁻ (181 pm)
Isoelectronic Series
Atoms/ions with same number of electrons:
Example: O²⁻, F⁻, Ne, Na⁺, Mg²⁺ (all have 10 electrons)
Trend: As nuclear charge increases, radius decreases
O2−>F−>Ne>Na+>Mg2+
More protons → stronger pull → smaller radius
Successive Ionization Energies
First ionization energy (IE₁): Remove first electron
Second ionization energy (IE₂): Remove second electron
Third ionization energy (IE₃): Remove third electron, etc.
Trend: IE₁ < IE₂ < IE₃ < ...
Why: Each successive electron is removed from a more positive ion
Big Jump: Large increase when removing electron from inner shell
Predicting reactivity: Most reactive metals (lower left), most reactive nonmetals (upper right)
Bond polarity: Difference in electronegativity determines bond type
Ion formation: Elements lose/gain electrons to achieve noble gas configuration
Chemical behavior: Trends explain why elements in same group have similar properties
📚 Practice Problems
1Problem 1easy
❓ Question:
Arrange the following in order of increasing atomic radius: O, S, Se, Te
💡 Show Solution
Solution:
Given: O, S, Se, Te (all Group 16 elements)
Find: Order of increasing atomic radius
Step 1: Identify the trend
All elements are in Group 16 (same group, different periods).
Trend: Atomic radius increases down a group.
Step 2: Determine period numbers
O (oxygen): Period 2
S (sulfur): Period 3
Se (selenium): Period 4
Te (tellurium): Period 5
Step 3: Apply the trend
Going down the group (increasing period):
O is smallest (Period 2)
S is next (Period 3)
Se is larger (Period 4)
Te is largest (Period 5)
Answer: O < S < Se < Te
Explanation:
Each successive element has one more electron shell, placing the outermost electrons farther from the nucleus despite increasing nuclear charge. The shielding effect from inner electrons outweighs the increased nuclear charge.
Verification:
All in same group ✓
Order follows period numbers ✓
Radius increases down group ✓
2Problem 2medium
❓ Question:
Arrange the following elements in order of increasing atomic radius: O, S, Se, Te. Explain the trend you observe.
💡 Show Solution
Solution:
Order of increasing atomic radius: O < S < Se < Te
Explanation:
All four elements are in Group 16 (oxygen family)
Going down a group, atomic radius increases
Each successive element adds a new electron shell
O (period 2): 2 shells
S (period 3): 3 shells
Se (period 4): 4 shells
Te (period 5): 5 shells
Although nuclear charge increases down the group, the effect of additional electron shells (increased shielding) dominates, resulting in larger atomic radii.
3Problem 3medium
❓ Question:
Which has the larger radius: Mg or Mg²⁺? Explain your reasoning.
💡 Show Solution
Solution:
Given: Mg (neutral magnesium) and Mg²⁺ (magnesium ion)
Find: Which has larger radius and why
Step 1: Write electron configurations
Mg:1s (12 electrons)
4Problem 4hard
❓ Question:
Consider the species: Na, Na⁺, Mg, Mg²⁺, Al, Al³⁺. (a) Which species has the largest radius? (b) Which has the smallest radius? (c) Explain why Na⁺, Mg²⁺, and Al³⁺ are all isoelectronic but have different radii.
💡 Show Solution
Solution:
(a) Largest radius: Na (neutral sodium atom)
(b) Smallest radius: Al³⁺ (aluminum cation)
(c) Isoelectronic species explanation:
Na⁺, Mg²⁺, and Al³⁺ are all isoelectronic (10 electrons each, configuration: 1s² 2s² 2p⁶)
For isoelectronic species, the one with more protons (higher nuclear charge) has a smaller radius because the electrons are pulled more tightly toward the nucleus. Order of decreasing radius:
5Problem 5hard
❓ Question:
The successive ionization energies for element X are: IE₁ = 580 kJ/mol, IE₂ = 1815 kJ/mol, IE₃ = 2740 kJ/mol, IE₄ = 11,600 kJ/mol. In which group of the periodic table is element X likely found?
💡 Show Solution
Solution:
Given: Successive ionization energies with large jump after IE₃
Find: Group number of element X
(a) Explain why the first ionization energy of oxygen is less than that of nitrogen, even though oxygen is to the right of nitrogen in the periodic table. (b) Compare the first ionization energies of Na, Mg, and Al.
💡 Show Solution
Solution:
(a) Oxygen vs. Nitrogen ionization energy:
N: 1s² 2s² 2p³ (half-filled p subshell, all unpaired)
O: 1s² 2s² 2p⁴ (one paired electrons in p subshell)
Explanation: Nitrogen has a half-filled 2p³ configuration which is particularly stable. Oxygen's fourth 2p electron must pair up, creating electron-electron repulsion. This repulsion makes it slightly easier to remove one electron from oxygen than from nitrogen's stable half-filled configuration.
N: IE₁ = 1402 kJ/mol
O: IE₁ = 1314 kJ/mol (slightly less)
(b) Na, Mg, Al comparison:
Explain using:
📋 AP Chemistry — Exam Format Guide
⏱ 3 hours 15 minutes📝 67 questions📊 3 sections
Section
Format
Questions
Time
Weight
Calculator
Multiple Choice
MCQ
60
90 min
50%
✅
Free Response (Long)
FRQ
3
69 min
30%
✅
Free Response (Short)
FRQ
4
36 min
20%
✅
📊 Scoring: 1-5
5
Extremely Qualified
~12%
4
Well Qualified
~16%
3
Qualified
~24%
2
Possibly Qualified
~24%
1
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~24%
💡 Key Test-Day Tips
✓Memorize common polyatomic ions
✓Practice dimensional analysis
✓Know your gas laws
⚠️ Common Mistakes: Periodic Trends
Avoid these 3 frequent errors
🌍 Real-World Applications: Periodic Trends
See how this math is used in the real world
📝 Worked Example: Stoichiometry — Limiting Reagent
Problem:
2 mol of H2 reacts with 1 mol of O2. How many grams of water are produced? Which is the limiting reagent? (2H2+O2→2H2O)
Understand and predict trends in atomic radius, ionization energy, electron affinity, and electronegativity across the periodic table.
How can I study Periodic Trends effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 6 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Periodic Trends study guide free?▾
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What course covers Periodic Trends?▾
Periodic Trends is part of the AP Chemistry course on Study Mondo, specifically in the Atomic Structure and Properties section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Periodic Trends?▾
Yes, this page includes 6 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
2
2
s2
2
p6
3
s2
Valence electrons: 2 in 3s orbital
3 electron shells
Mg²⁺:1s22s22p6 (10 electrons)
Lost both 3s electrons
2 electron shells (same as Ne)
Step 2: Compare structures
Mg:
12 protons
12 electrons
Outermost shell: n = 3
Mg²⁺:
12 protons (unchanged)
10 electrons (lost 2)
Outermost shell: n = 2
Step 3: Apply cation trend
When forming a cation:
Entire outermost shell is lost (3s)
Same number of protons pulling fewer electrons
Less electron-electron repulsion
Electrons pulled closer to nucleus
Answer:Mg > Mg²⁺ (neutral atom is larger)
Explanation:
Mg²⁺ is smaller because:
Lost entire n=3 shell → fewer shells
Same nuclear charge (12+) pulling only 10 electrons instead of 12
Higher charge-to-electron ratio
Tighter electron cloud
Typical values:
Mg radius: ~160 pm
Mg²⁺ radius: ~72 pm (less than half!)
General rule: Cations are always smaller than their parent atoms.
Conclusion:
Na⁺ > Mg²⁺ > Al³⁺
1
=
580 kJ/mol
IE2=1815 kJ/mol
IE3=2740 kJ/mol
IE4=11,600 kJ/mol
HUGE jump!
Step 2: Interpret the large jump
The enormous increase from IE₃ to IE₄ indicates:
First 3 electrons are relatively easy to remove (valence electrons)
Fourth electron is much harder to remove (core electron)
Conclusion: Element X has 3 valence electrons
Step 3: Determine group
Elements with 3 valence electrons are in Group 13 (IIIA).
Electron configuration pattern: ns2np1
Examples: B, Al, Ga, In, Tl
Step 4: Verify with electron configuration
For aluminum (Al) as example:
Configuration: [Ne]3s23p1
IE₁: Remove 3p¹ electron (easiest)
IE₂: Remove 3s¹ electron (harder, same shell)
IE₃: Remove 3s¹ electron (even harder, now +3 charge)
IE₄: Remove 2p⁶ electron (HUGE jump - breaking into filled shell)
Answer:Group 13 (or IIIA)
Reasoning:
The large jump after the third ionization indicates that the element has 3 valence electrons. Removing the fourth electron breaks into a stable, filled inner shell (noble gas configuration), requiring significantly more energy.
General principle: The position of the large jump in successive ionization energies reveals the number of valence electrons.
Na: [Ne] 3s¹ → IE₁ = 496 kJ/mol
Mg: [Ne] 3s² → IE₁ = 738 kJ/mol
Al: [Ne] 3s² 3p¹ → IE₁ = 578 kJ/mol
Trend: Mg > Al > Na
Magnesium has the highest because removing an electron from a filled 3s² is difficult. Aluminum's drop is because its 3p¹ electron is in a higher energy orbital and experiences more shielding from the 3s² electrons.