Periodic Trends

Understand and predict trends in atomic radius, ionization energy, electron affinity, and electronegativity across the periodic table.

Periodic Trends

Organization of the Periodic Table

The periodic table is organized by:

Periods (rows): Horizontal rows numbered 1-7
Groups (columns): Vertical columns numbered 1-18
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

$X(g) + \text{energy} \rightarrow X^+(g) + e^-$

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

$X(g) + e^- \rightarrow X^-(g) + \text{energy}$

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 $O^{2-} > F^- > Ne > Na^+ > Mg^{2+}$

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

Example: Magnesium (Mg)

IE₁ = 738 kJ/mol (remove 3s¹)
IE₂ = 1451 kJ/mol (remove 3s¹)
IE₃ = 7733 kJ/mol (HUGE jump - removing from filled 2p⁶)

This confirms Mg forms Mg²⁺, not Mg³⁺

Summary of Trends

| Property | Across Period (→) | Down Group (↓) | |----------|-------------------|----------------| | Atomic Radius | Decreases | Increases | | Ionization Energy | Increases | Decreases | | Electron Affinity | More negative | Less negative | | Electronegativity | Increases | Decreases | | Metallic Character | Decreases | Increases |

Applications

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 1medium

❓ 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.

2Problem 2easy

❓ 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 ✓

3Problem 3medium

❓ 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.

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)

Na⁺, Mg²⁺, and Al³⁺ are all isoelectronic (10 electrons each, configuration: 1s² 2s² 2p⁶)

Nuclear charge comparison:

Na⁺: 11 protons pulling on 10 electrons → radius ≈ 102 pm
Mg²⁺: 12 protons pulling on 10 electrons → radius ≈ 72 pm
Al³⁺: 13 protons pulling on 10 electrons → radius ≈ 54 pm

Conclusion: 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: Na⁺ > Mg²⁺ > Al³⁺

5Problem 5hard

❓ Question:

💡 Show Solution

Solution:

(a) Largest radius: Na (neutral sodium atom)

(b) Smallest radius: Al³⁺ (aluminum cation)

Na⁺, Mg²⁺, and Al³⁺ are all isoelectronic (10 electrons each, configuration: 1s² 2s² 2p⁶)

Nuclear charge comparison:

Na⁺: 11 protons pulling on 10 electrons → radius ≈ 102 pm
Mg²⁺: 12 protons pulling on 10 electrons → radius ≈ 72 pm
Al³⁺: 13 protons pulling on 10 electrons → radius ≈ 54 pm

6Problem 6medium

❓ 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: $1s^2 2s^2 2p^6 3s^2$ (12 electrons)

Valence electrons: 2 in 3s orbital
3 electron shells

Mg²⁺: $1s^2 2s^2 2p^6$ (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.

7Problem 7medium

❓ Question:

(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:

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.

8Problem 8hard

❓ 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

Step 1: Analyze the pattern

$IE_1 = 580 \text{ kJ/mol}$ $IE_2 = 1815 \text{ kJ/mol}$ (×3.1 increase) $IE_3 = 2740 \text{ kJ/mol}$ (×1.5 increase) $IE_4 = 11,600 \text{ kJ/mol}$ (×4.2 increase - 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: $ns^2 np^1$

Examples: B, Al, Ga, In, Tl

Step 4: Verify with electron configuration

For aluminum (Al) as example:

Configuration: $[Ne] 3s^2 3p^1$
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.

9Problem 9medium

❓ Question:

💡 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)

N: IE₁ = 1402 kJ/mol
O: IE₁ = 1314 kJ/mol (slightly less)

(b) Na, Mg, Al comparison:

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

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