GENERAL CHEMISTRY · PERIODIC PROPERTIES

Find the driver.
Check the exception.

Learn all four official topics through decision ledgers, worked examples, retrieval prompts, and sixteen original five-choice questions.

Start the lessons ↓Open all problems →

4guided lessons

16practice questions

12study objectives

5choices per item

A repeatable trend routine

Classify. Count. Compare. Bound.

01Classify
Name the block, family, species, and property before applying a trend.
02Count
Track valence electrons, ionic charge, shells, or oxidation numbers.
03Compare
Choose the controlling cause: shell count, effective attraction, or supplied data.
04Bound
Check exceptions and refuse a conclusion stronger than the evidence.

Four prerequisite-aware lessons

Reason from structure to behavior.

Each lesson uses a semantic decision ledger, one fully worked example, and closed-note retrieval before multiple-choice practice.

LESSON 1 · 16 MIN

Study + retrieve

Representative elements from position to formula

Classify representative elements, connect group to valence pattern and common ion charge, and construct neutral binary formulas.

ESSENTIAL QUESTIONWhat does the element’s column justify—and what does it not justify?

Main-group position and charge ledger

Recurring introductory patterns for representative-element groups
Group	Outer pattern	Valence electrons	Common monatomic charge
1	ns¹	1	+1
2	ns²	2	+2
13	ns²np¹	3	+3 pattern
15	ns²np³	5	−3 pattern
16	ns²np⁴	6	−2 pattern
17	ns²np⁵	7	−1
18	ns²np⁶	8	0

BoundaryPatterns guide common ions; they do not ban other oxidation states. Helium is 1s².

STUDY TABLE · CAPTION INCLUDED

Classify before predicting

In the common main-group convention used here, representative elements occupy groups 1, 2, and 13–18. Their recurring chemistry follows related outer-shell patterns; helium is a filled-1s exception to a simple p-block shortcut. If a prompt supplies a different classification convention for group 12, follow the prompt.

Groups 1, 2, 13–18
Hydrogen, helium, and group 12 need context

Translate group into valence

For main-group atoms, groups 1 and 2 have one and two valence electrons; groups 13–18 have three through eight, with helium holding two in its first shell.

Group 17 → ns²np⁵
Same group → related valence pattern

Balance common ions

Common monatomic charges are useful introductory patterns, not universal oxidation-state laws. When ions are specified, use the smallest whole-number ratio that makes total charge zero.

Magnitude and sign both matter
Reduce the formula-unit ratio

Worked example

What formula results from Al³⁺ and O²⁻?

1
Find the least common charge magnitude: 6.
2
Two Al³⁺ ions contribute +6; three O²⁻ ions contribute −6.
3
Use the smallest neutral ratio, 2:3.

ConclusionThe neutral formula is Al₂O₃, not AlO or Al₃O₂.

Close the notes first

Retrieve the model.

01Which groups are the main-group elements?

Groups 1, 2, and 13–18.

These are the representative-element columns in the standard introductory classification.

02What outer pattern identifies a group 17 atom?

ns²np⁵.

It has seven valence electrons, one short of a filled main-group shell.

03What is the smallest neutral ratio of M³⁺ to X²⁻?

2:3.

Two +3 charges and three −2 charges sum to zero.

LESSON 2 · 17 MIN

Study + retrieve

Transition elements with electron and charge ledgers

Form d-block cations correctly, calculate oxidation states, and use electron occupancy to interpret bounded magnetic evidence.

ESSENTIAL QUESTIONWhich electrons remain after ion formation, and what can that occupancy support?

Transition-element evidence ledger

Position, electron configuration, and defensible inference
Evidence	Safe inference	Unsafe shortcut
Element lies in groups 3–12	It is in the d block	Every common ion has an incomplete d subshell
Fe: [Ar] 4s²3d⁶	Fe²⁺: [Ar] 3d⁶	Remove 3d before 4s
Ion is d⁵ in free-ion model	Five unpaired electrons; paramagnetic	Predict an exact compound color

KMnO₄ oxidation-number ledger
Contribution	Value
K	+1
4 O	4(−2) = −8
Mn	+7
Neutral sum	+1 + 7 − 8 = 0

STUDY TABLE · CAPTION INCLUDED

Separate location from definition

The d block occupies groups 3–12, but a transition element is identified by an atom or common ion with an incomplete d subshell. This distinction prevents treating every d-block ion identically.

d block is positional
Incomplete d is electronic

Remove the outer shell first

Although 4s fills before 3d in the usual neutral-atom sequence, transition-metal cations lose the highest-n 4s electrons before 3d electrons.

Fe: remove 4s before 3d
Charge fixes total electron loss

Audit oxidation and magnetism

Oxidation numbers sum to net charge. Within a free-ion d subshell, Hund’s rule gives the number of unpaired electrons; unpaired electrons support paramagnetism.

Σ oxidation numbers = charge
d⁵ free ion → five unpaired

Worked example

Determine Mn oxidation state in neutral KMnO₄.

1
Assign K as +1 and each O as −2.
2
Write the charge ledger: +1 + x + 4(−2) = 0.
3
Solve x − 7 = 0, so x = +7.

ConclusionManganese is +7 in KMnO₄; group position alone would not determine it.

Close the notes first

Retrieve the model.

01Which subshell empties first when Fe forms Fe²⁺?

4s.

Cation formation removes electrons from the highest occupied principal shell first.

02Why can one transition metal have several oxidation states?

Its ns and (n−1)d electrons can be close enough in energy to participate differently in bonding and ion formation.

A single group charge is therefore unsafe.

03What magnetic behavior follows from an unpaired electron?

Paramagnetism.

Unpaired magnetic moments interact with an applied field.

LESSON 3 · 18 MIN

Study + retrieve

Periodic trends with mechanism and exception checks

Rank radius, ionization energy, electronegativity, and electron-affinity tendency using shell and attraction evidence while respecting exceptions.

ESSENTIAL QUESTIONDoes shell count, effective nuclear attraction, or an exception control this comparison?

Trend decision ledger

General directions and their controlling ideas
Property	Across a period	Down a group	Primary model
Atomic radius	Generally decreases	Increases	Effective attraction · shell count
First ionization energy	Generally increases	Generally decreases	Energy needed to remove an electron
Electronegativity	Generally increases	Generally decreases	Attraction within a bond
Metallic character	Decreases	Increases	Ease of electron loss

Ten-electron isoelectronic series, largest to smallest radius
Species	Protons	Relative radius
N³⁻	7	Largest
O²⁻	8	↓
F⁻	9	↓
Na⁺	11	↓
Mg²⁺	12	Smallest

Exception checkUse supplied values for local IE or EA exceptions; never force every property into one arrow.

STUDY TABLE · CAPTION INCLUDED

Compare shells before arrows

Atomic radius generally increases down a group as the highest occupied shell moves farther out, and decreases across a period as effective nuclear attraction increases.

Down → more shells → larger
Across → stronger effective pull → smaller

Treat ions as new electron counts

Cations are generally smaller than their parent atoms and anions larger. In an isoelectronic series, the species with more protons is smaller because the same electron count feels greater attraction.

Same electrons: compare protons
N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺

Name the property and check exceptions

First ionization energy generally rises across and falls down; electronegativity generally rises toward fluorine. Electron affinity uses its own sign convention and has important group 2, 15, and 18 exceptions.

Do not merge IE, EN, and EA
Use supplied data for local exceptions

Worked example

Rank N³⁻, O²⁻, F⁻, Na⁺, and Mg²⁺ from largest to smallest radius.

1
Each species has 10 electrons, so they are isoelectronic.
2
Compare proton counts: 7, 8, 9, 11, and 12.
3
Fewer protons means weaker attraction for the same electrons and therefore a larger radius.

ConclusionN³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺.

Close the notes first

Retrieve the model.

01Why does radius generally increase down a group?

The highest occupied principal shell increases.

The valence region lies farther from the nucleus despite increased nuclear charge.

02In an isoelectronic series, which species is smallest?

The one with the most protons.

It exerts the strongest nuclear attraction on the same electron count.

03A huge jump occurs between IE₂ and IE₃. How many valence electrons were removed before the jump?

Two.

The third removal reaches a much more tightly bound core shell.

LESSON 4 · 17 MIN

Study + retrieve

Descriptive chemistry from converging evidence

Apply bounded family reactions, oxide acid–base patterns, and multiple physical or chemical clues without overclaiming from one observation.

ESSENTIAL QUESTIONWhich conclusion survives every observation and the stated conditions?

Descriptive chemistry evidence matrix

Bounded family and oxide observations used in this lesson
Observation or species	Supported conclusion	Required boundary
2Na + 2H₂O	2NaOH + H₂	Standard aqueous reaction
F₂ / Cl₂ / Br₂ / I₂	Gas / gas / liquid / solid	Ordinary room conditions
Na₂O	Basic oxide	Introductory acid–base model
Al₂O₃	Amphoteric oxide	Reacts with acid or base
SO₃	Acidic oxide	Common oxide of sulfur(VI)

Identity ruleCombine independent observations; do not identify an element from color alone.

STUDY TABLE · CAPTION INCLUDED

Use family patterns with conditions

Alkali metals commonly form +1 ions and can react with water to produce hydrogen and a basic hydroxide solution. Halogens are diatomic, but their ordinary physical states change down the group.

Check conditions
Conserve atoms and charge

Track oxide character across a period

Common oxides broadly shift from more ionic and basic on the left toward more covalent and acidic on the right, with amphoteric behavior such as Al₂O₃ between the ends.

Na₂O basic
Al₂O₃ amphoteric · SO₃ acidic

Require converging evidence

Conductivity, brittleness, state, formulas, and reaction products can narrow identity. A color or single trend cue rarely identifies an element uniquely without context.

Use at least two independent clues
Reject any candidate that contradicts reliable evidence

Worked example

Classify Na₂O, Al₂O₃, and SO₃ by acid–base character.

1
Na is a strongly metallic element on the left; its oxide is basic in the introductory pattern.
2
Al lies near the metal–metalloid boundary; Al₂O₃ reacts with both acids and bases and is amphoteric.
3
S is a nonmetal on the right; SO₃ is an acidic oxide.

ConclusionNa₂O is basic, Al₂O₃ amphoteric, and SO₃ acidic.

Close the notes first

Retrieve the model.

01What products form when sodium reacts with water in the standard introductory equation?

Sodium hydroxide and hydrogen gas.

The balanced equation is 2Na + 2H₂O → 2NaOH + H₂.

02What is amphoteric behavior?

The ability to react as an acid or as a base, depending on the other reactant.

Al₂O₃ is a standard representative example.

03Why is one observed color usually insufficient for identification?

Color may be shared and can depend on species, oxidation state, ligands, or conditions.

Independent evidence prevents an unsupported unique conclusion.

All sixteen Periodic Properties problems

Name the governing ledger.

Question order and all five answer options shuffle each time. Reports automatically include the exact question, content version, skill, and seed.

16 PRACTICE QUESTIONS

Retrieve before you review.

Question order and all five answer options are shuffled when you begin. The correct answer stays attached to the same underlying choice.

Interpret results carefully

Raw accuracy directs review—not score prediction.

Every question is original, draft, and uncalibrated. Use explanations to repair the specific reasoning step, then use balanced practice to switch among all four topics.

DAT TRAIN does not claim topic quotas because the official manual does not publish them.