Periodic Trends

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

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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)+energyX+(g)+eX(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)+eX(g)+energyX(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 O2>F>Ne>Na+>Mg2+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

  1. Predicting reactivity: Most reactive metals (lower left), most reactive nonmetals (upper right)
  2. Bond polarity: Difference in electronegativity determines bond type
  3. Ion formation: Elements lose/gain electrons to achieve noble gas configuration
  4. 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:

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: 1s22s22p63s21s^2 2s^2 2p^6 3s^2 (12 electrons)

  • Valence electrons: 2 in 3s orbital
  • 3 electron shells

Mg²⁺: 1s22s22p61s^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:

  1. Entire outermost shell is lost (3s)
  2. Same number of protons pulling fewer electrons
  3. Less electron-electron repulsion
  4. 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.

3Problem 3hard

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

IE1=580 kJ/molIE_1 = 580 \text{ kJ/mol} IE2=1815 kJ/molIE_2 = 1815 \text{ kJ/mol} (×3.1 increase) IE3=2740 kJ/molIE_3 = 2740 \text{ kJ/mol} (×1.5 increase) IE4=11,600 kJ/molIE_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: ns2np1ns^2 np^1

Examples: B, Al, Ga, In, Tl

Step 4: Verify with electron configuration

For aluminum (Al) as example:

  • Configuration: [Ne]3s23p1[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.