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Understand and predict trends in atomic radius, ionization energy, electron affinity, and electronegativity across the periodic table.
Learn step-by-step with practice exercises built right in.
The periodic table is organized by:
Definition: Distance from nucleus to outermost electron (half the distance between nuclei of bonded atoms)
Trends:
Arrange the following in order of increasing atomic radius: O, S, Se, Te
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
| 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% | ✅ |
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Across a period:
Down a group:
Example: Na > Mg > Al > Si > P > S > Cl (Period 3)
Definition: Energy required to remove an electron from a gaseous atom
Trends:
Explanation:
Across a period:
Down a group:
Exceptions:
Example: He has highest IE; Cs has lowest IE
Definition: Energy change when an electron is added to a gaseous atom
Convention: Negative EA means energy is released (favorable)
Trends:
Explanation:
Across a period:
Down a group:
Exceptions:
Example: Cl has most negative EA (excluding noble gases)
Definition: Ability of an atom to attract electrons in a chemical bond
Pauling Scale: 0.7 (Cs) to 4.0 (F)
Trends:
Explanation:
Similar to ionization energy:
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)
Example: Na (186 pm) > Na⁺ (102 pm)
Example: Cl (99 pm) < Cl⁻ (181 pm)
Atoms/ions with same number of electrons:
Example: O²⁻, F⁻, Ne, Na⁺, Mg²⁺ (all have 10 electrons)
Trend: As nuclear charge increases, radius decreases
More protons → stronger pull → smaller radius
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)
This confirms Mg forms Mg²⁺, not Mg³⁺
| 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 |
Step 3: Apply the trend
Going down the group (increasing period):
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:
Which has the larger radius: Mg or Mg²⁺? Explain your reasoning.
Solution:
Given: Mg (neutral magnesium) and Mg²⁺ (magnesium ion) Find: Which has larger radius and why
Step 1: Write electron configurations
Mg: (12 electrons)
Mg²⁺: (10 electrons)
Step 2: Compare structures
Mg:
Mg²⁺:
Step 3: Apply cation trend
When forming a cation:
Answer: Mg > Mg²⁺ (neutral atom is larger)
Explanation:
Mg²⁺ is smaller because:
Typical values:
General rule: Cations are always smaller than their parent atoms.
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?
Solution:
Given: Successive ionization energies with large jump after IE₃ Find: Group number of element X
Step 1: Analyze the pattern
(×3.1 increase) (×1.5 increase) (×4.2 increase - HUGE jump!)
Step 2: Interpret the large jump
The enormous increase from IE₃ to IE₄ indicates:
Step 3: Determine group
Elements with 3 valence electrons are in Group 13 (IIIA).
Electron configuration pattern:
Examples: B, Al, Ga, In, Tl
Step 4: Verify with electron configuration
For aluminum (Al) as example:
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.