GENERAL CHEMISTRY · LIQUIDS AND SOLIDS

Name the particles.
Predict the behavior.

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

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6guided lessons

24practice questions

18study objectives

5choices per item

A repeatable condensed-phase routine

Particles. Attractions. Conditions. Evidence.

01Name the particles
Identify atoms, ions, discrete molecules, or an extended network before naming forces.
02Inventory attractions
Include every interaction supported by charge, polarity, donor structure, size, and shape.
03Mark conditions
Track temperature, pressure, phase, and whether the system is open or closed.
04Bound the claim
Predict only what the controlled evidence supports; do not turn a trend into an absolute rule.

Six prerequisite-aware lessons

Connect structure to phase behavior.

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

LESSON 1 · 20 MIN

Study + retrieve

Intermolecular forces from structure to attraction

Separate phase-changing attractions from identity-defining bonds, inventory every supported molecular attraction, and compare dispersion only when size and shape evidence is controlled.

ESSENTIAL QUESTIONWhich particles are attracting, and which structural evidence permits each attraction?

Attraction inventory ledger

Structural evidence required for common molecular attractions
Evidence	Supported attraction	Boundary
Any atom or molecule	London dispersion	Never omit it
Permanent molecular dipole	Dipole–dipole	Polar bonds may cancel
H bonded to N, O, or F + partner	Conventional hydrogen bond	Polarity alone is insufficient
Ion + polar molecule	Ion–dipole	Requires an actual ion

Controlled comparison checklist
Hold or inspect	Useful evidence	Unsafe shortcut
Particle identity	Discrete molecule, ion, or network	Treating phase change as decomposition
Electron cloud	Size and polarizability	Ranking only by force name
Molecular shape	Available surface contact	Ignoring branching or compactness

STUDY TABLES · CAPTIONS INCLUDED

Separate between from within

Boiling or melting a molecular substance separates intact molecules by overcoming attractions between them. Covalent bonds within each molecule remain unless a chemical reaction occurs; ionic, metallic, and network solids require their own extended bonding models.

Phase change ≠ decomposition
Name the particles before the force

Inventory; do not choose only one

Every atom and molecule has London dispersion attraction. Polar molecules add permanent dipole–dipole attraction. A conventional hydrogen-bond donor has H bonded to N, O, or F, and an ion with a polar molecule can add ion–dipole attraction.

Dispersion is universal
Hydrogen bonding needs structure, not polarity alone

Control the comparison

Polarizability generally rises as an electron cloud becomes easier to distort, and greater molecular contact can strengthen dispersion. Force labels alone do not create a universal property ranking when size and shape also change.

Compare all relevant attractions
Branching can reduce surface contact

Worked example

Why does straight-chain n-pentane boil above its compact isomer neopentane even though both have formula C₅H₁₂?

1
Both substances are nonpolar hydrocarbons, so their molecular attraction is dominated by dispersion.
2
They have the same formula and electron count, so the key controlled difference is shape.
3
The elongated n-pentane molecules make greater surface contact than compact neopentane molecules.

ConclusionGreater contact strengthens dispersion in n-pentane, supporting its higher boiling point without inventing a polarity difference.

Close the notes first

Retrieve the model.

01Does boiling molecular water normally break O–H covalent bonds?

No; it overcomes attractions between intact H₂O molecules.

A physical phase change preserves chemical identity.

02What attraction is present in every molecular substance?

London dispersion attraction.

Temporary electron-cloud fluctuations occur even in nonpolar molecules.

03Is molecular polarity alone enough to rank two very different boiling points?

Not always.

Polarizability, size, shape, and multiple attractions can compete.

LESSON 2 · 21 MIN

Study + retrieve

Phase changes, heating curves, and phase diagrams

Track phase direction and system heat, distinguish temperature-changing from phase-changing energy, and read phase regions and special points from pressure–temperature evidence.

ESSENTIAL QUESTIONIs energy changing particle motion, particle separation, or the stable phase region?

Phase and energy ledger

Forward and reverse phase transitions for the substance as system
Transition	Name	Heat direction
Solid → liquid	Melting	Endothermic
Liquid → gas	Vaporization	Endothermic
Solid → gas	Sublimation	Endothermic
Liquid → solid	Freezing	Exothermic
Gas → liquid	Condensation	Exothermic
Gas → solid	Deposition	Exothermic

Heating-curve and phase-diagram evidence
Representation	Evidence	Interpretation
Sloped heating segment	One phase; changing temperature	Use supplied heat capacity
Horizontal transition segment	Two phases; constant temperature	Use supplied phase-change enthalpy
Diagram region / boundary	Area / line	One phase / two-phase equilibrium
Three-line intersection	Three phases coexist	Triple point

STUDY TABLES · CAPTIONS INCLUDED

Pair every transition with its reverse

Melting, vaporization, and sublimation are endothermic for the substance. Freezing, condensation, and deposition are exothermic reverses. The surroundings experience the opposite heat direction.

Solid → liquid → gas absorbs
Reverse arrows release

Read the heating curve by segment

Within one phase, heat changes temperature and q = mcΔT can apply. During an idealized pure-substance transition at fixed pressure, temperature remains constant while phase proportions change and supplied phase-change enthalpy determines q.

Slope → temperature change
Plateau → two phases

Read regions before labels

A phase-diagram region represents one stable phase, a boundary represents two-phase equilibrium, the triple point joins three boundaries, and the critical point ends the liquid–gas boundary.

Area = one phase
Line = coexistence

Worked example

How much heat is absorbed when 2.00 mol of a solid melts if ΔHfus = 6.01 kJ/mol?

1
Melting is endothermic, so heat for the substance is positive.
2
Use the supplied molar phase-change enthalpy rather than mcΔT.
3
q = 2.00 mol × 6.01 kJ/mol = 12.0 kJ to three significant figures.

ConclusionThe substance absorbs 12.0 kJ while solid and liquid coexist at the melting temperature under the stated constant-pressure model.

Close the notes first

Retrieve the model.

01What is the gas-to-solid transition, and what is its heat direction for the substance?

Deposition; exothermic.

It is the reverse of endothermic sublimation.

02What does a horizontal segment on an idealized heating curve mean?

Two phases coexist while added heat changes phase proportion.

Temperature remains constant during that transition under the stated model.

03Where do three phase boundaries meet?

At the triple point.

All three phases are in equilibrium at that temperature and pressure.

LESSON 3 · 18 MIN

Study + retrieve

Vapor pressure from exchange to boiling

Explain dynamic vapor equilibrium, predict temperature and attraction trends, and identify boiling when vapor pressure matches external pressure.

ESSENTIAL QUESTIONWhat controls escape from the liquid, return from the vapor, and the pressure at which bubbles persist?

Vapor-equilibrium ledger

Microscopic exchange in a closed liquid–vapor system
Rate comparison	State	Macroscopic result
Evaporation > condensation	Net evaporation	Vapor amount rises
Evaporation = condensation > 0	Dynamic equilibrium	Amounts remain constant
Condensation > evaporation	Net condensation	Liquid amount rises

Vapor-pressure and boiling trends
Controlled change	Vapor pressure	Boiling temperature
Higher temperature; same liquid	Increases	Not the variable being compared
Stronger attraction; same T	Decreases	Generally increases
Lower external pressure	Same liquid curve	Decreases
Boiling condition	P_vap = P_external

STUDY TABLES · CAPTIONS INCLUDED

Keep both rates moving

In a closed container with liquid and vapor present at fixed temperature, equilibrium vapor pressure is established when evaporation and condensation continue at equal nonzero rates.

Dynamic ≠ stopped
Open containers may not equilibrate

Separate the two trend drivers

For one liquid, vapor pressure rises with temperature. At the same temperature, a defensibly stronger attraction generally lowers vapor pressure and volatility because fewer particles escape the liquid.

Higher T → higher vapor pressure
Stronger attraction → lower vapor pressure

Match vapor pressure to surroundings

Boiling occurs when vapor pressure equals external pressure. Normal boiling point uses 1 atm; lowering external pressure lowers the boiling temperature, while raising external pressure raises it.

Pvap = Pexternal
Boiling point is pressure-dependent

Worked example

A liquid’s vapor pressure is 0.18 atm at 40 °C, 0.46 atm at 60 °C, 0.80 atm at 80 °C, and 1.00 atm at 100 °C. At what listed temperature does it boil under 0.80 atm external pressure?

1
Boiling begins when the liquid’s vapor pressure equals the external pressure.
2
Find 0.80 atm in the vapor-pressure data.
3
The matching temperature is 80 °C.

ConclusionThe listed boiling temperature is 80 °C at 0.80 atm, below the 100 °C normal boiling point at 1 atm.

Close the notes first

Retrieve the model.

01At dynamic vapor equilibrium, how do evaporation and condensation rates compare?

They are equal and nonzero.

Molecular exchange continues even while macroscopic amounts remain constant.

02What happens to one liquid’s vapor pressure when temperature rises?

It increases.

A larger fraction of particles can enter the vapor phase.

03Why does boiling temperature fall at lower external pressure?

A lower temperature is sufficient for vapor pressure to match the surroundings.

The equality condition is reached earlier on the rising vapor-pressure curve.

LESSON 4 · 20 MIN

Study + retrieve

Solid structure from order to unit cells

Separate crystalline from amorphous order, classify the four major crystalline-solid models, and count shared particles in basic cubic unit cells.

ESSENTIAL QUESTIONWhat repeats through the solid, and how much of each boundary particle belongs to one cell?

Solid-structure ledger

Classifying solids from their repeating constituents
Solid type	Repeating constituents	Dominant model
Ionic	Cations and anions	Electrostatic lattice
Metallic	Metal atoms	Delocalized electrons
Covalent network	Bonded atomic network	Extended covalent bonds
Molecular	Discrete molecules	Intermolecular attractions

Particle sharing in a conventional cubic unit cell
Position	Contribution	Common cell count
Corner	1/8	Simple cubic = 1 BCC = 2 FCC = 4
Edge	1/4
Face	1/2
Body	1

STUDY TABLES · CAPTIONS INCLUDED

Start with long-range order

Crystalline solids repeat an ordered pattern and have a characteristic melting temperature. Amorphous solids lack long-range periodic order and generally soften over a temperature range.

Rigid does not guarantee crystalline
Order controls melting behavior

Classify the repeating units

Ionic solids contain ions, metallic solids contain metal atoms with delocalized electrons, covalent-network solids extend covalent bonds through the structure, and molecular solids contain discrete molecules.

Particles + attraction → type
Discrete molecules ≠ network

Share boundary particles

A corner contributes 1/8, an edge 1/4, a face 1/2, and a body particle 1 to one cubic cell. Conventional simple cubic, BCC, and FCC cells contain 1, 2, and 4 effective particles.

Count fractions, not visible spheres
8 corners contribute one total

Worked example

How many effective particles are in a face-centered cubic unit cell with eight corner particles and six face-centered particles?

1
Corners contribute 8 × 1/8 = 1 particle.
2
Faces contribute 6 × 1/2 = 3 particles.
3
Add the contributions: 1 + 3 = 4 particles.

ConclusionA conventional FCC unit cell contains four effective particles.

Close the notes first

Retrieve the model.

01How does an amorphous solid typically respond near melting?

It softens across a range.

Its local environments are not all equivalent in one repeating lattice.

02What distinguishes a molecular solid from a covalent-network solid?

A molecular solid has discrete molecules; a network solid has covalent bonds extending through the structure.

Covalent bonds within a molecule do not automatically make the entire solid a network.

03How much does one corner particle contribute to one cubic unit cell?

One eighth.

Eight neighboring cells share a corner position.

LESSON 5 · 18 MIN

Study + retrieve

Polarity as a vector and evidence problem

Combine bond dipoles with molecular geometry, distinguish polarity from hydrogen bonding, and make property comparisons only when size and shape are controlled.

ESSENTIAL QUESTIONDo the bond-dipole vectors cancel, and what else changes in the comparison?

Dipole-vector ledger

From bond polarity and geometry to molecular polarity
Evidence	Vector result	Conclusion
No polar bonds	No bond-dipole vectors	Nonpolar molecule
Polar bonds in canceling geometry	Vector sum = 0	Nonpolar molecule
Polar bonds in noncanceling geometry	Vector sum ≠ 0	Polar molecule

Polarity claims and their required boundaries
Claim	Required evidence	Do not infer
Dipole–dipole attraction	Net molecular dipole	Hydrogen bonding automatically
Self hydrogen bonding	Valid donor and acceptor	From polarity alone
Property comparison	Controlled size and shape	Polarity always dominates

STUDY TABLES · CAPTIONS INCLUDED

Add vectors through geometry

Polar bonds can cancel in a symmetric molecule such as linear CO₂. A nonsymmetric geometry such as trigonal-pyramidal NH₃ can preserve a net molecular dipole.

Polar bonds ≠ automatically polar molecule
Use the three-dimensional shape

Do not rename every polar force

A polar molecule adds dipole–dipole attraction but does not necessarily self hydrogen-bond. Conventional donation requires H bonded to N, O, or F; dispersion remains present either way.

Polarity and hydrogen bonding differ
Inventory all attractions

Bound the property claim

With closely matched mass and shape, a permanent dipole can help explain lower vapor pressure or higher boiling point. With large structural differences, polarizability and contact may dominate.

Control size and shape
Do not say always

Worked example

CO₂ has polar C=O bonds. Why is the molecule nonpolar?

1
Draw the valid linear geometry O=C=O.
2
Represent the two equal C=O bond dipoles as vectors pointing in opposite directions.
3
Add the vectors: their sum is zero.

ConclusionCO₂ has polar bonds but no net molecular dipole because its linear geometry cancels them.

Close the notes first

Retrieve the model.

01Can a molecule with polar bonds be nonpolar?

Yes, if its bond-dipole vectors cancel.

Molecular polarity is a vector sum, not a bond checklist.

02Does CH₃Cl self hydrogen-bond in the conventional introductory model?

No.

It is polar but has no H covalently bonded to N, O, or F.

03When is polarity a safer explanation for a boiling-point difference?

When molecular size and shape are closely controlled.

That reduces competing dispersion and contact differences.

LESSON 6 · 19 MIN

Study + retrieve

Liquid and solid properties from particle mobility

Connect viscosity, surface behavior, capillary action, and solid conductivity to attraction, temperature, interface evidence, and charge mobility.

ESSENTIAL QUESTIONWhich particles must move, and which attractions help or resist that motion?

Mobility and interface ledger

Liquid-property evidence from motion and attraction
Property	Particle explanation	Controlled trend
Viscosity	Resistance to particles sliding	Same liquid: higher T → lower viscosity
Surface tension	Unbalanced inward cohesion at surface	Stronger cohesion generally raises it
Capillary action	Adhesion competes with cohesion	Adhesion > cohesion → concave rise

Electrical conduction requires mobile charge
Structure and state	Charge carrier	Conducts?
Metallic solid	Delocalized electrons	Yes
Ionic solid	Ions fixed in lattice	No
Molten or aqueous ionic substance	Mobile ions	Yes
Most molecular solids	No mobile charge	No

STUDY TABLES · CAPTIONS INCLUDED

Treat viscosity as resistance to flow

Stronger attractions and shapes that hinder sliding can raise liquid viscosity. For a given liquid, increasing temperature generally lowers viscosity because particles move past one another more readily.

Higher liquid T → lower viscosity
Compare structure before ranking

Compare cohesion with adhesion

Cohesion acts within a liquid and contributes to surface tension. Adhesion acts between liquid and surface; adhesion stronger than cohesion supports a concave meniscus and capillary rise, while the reverse supports a convex meniscus and depression.

Cohesion = like with like
Adhesion = liquid with surface

Require mobile charge

Metals conduct through delocalized electrons. Ionic solids do not conduct while ions are fixed, but molten or dissolved ionic substances can conduct when ions move. Most molecular solids are nonconducting; structural exceptions must be supplied.

Charge present is not enough
Charge must be mobile

Worked example

Why does solid NaCl fail to conduct while molten NaCl conducts?

1
Both states contain charged Na⁺ and Cl⁻ ions.
2
In the crystal, ions are fixed in lattice positions and cannot carry charge through the sample.
3
Melting frees the ions to move under an electric field.

ConclusionThe change in conductivity comes from charge mobility, not from creating ions during melting.

Close the notes first

Retrieve the model.

01What usually happens to one liquid’s viscosity as temperature rises?

It decreases.

Greater molecular motion helps particles move past one another.

02If liquid–glass adhesion exceeds liquid cohesion, what meniscus and capillary motion are expected?

A concave meniscus and capillary rise.

The liquid wets and climbs the surface.

03Why does an ionic melt conduct when its solid does not?

The ions become mobile.

Electrical current requires movable charge carriers.

All twenty-four Liquids and Solids problems

Select the supported inference.

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

24 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.

These original questions have not been calibrated to the official score scale. Use explanations to repair the specific reasoning step, then use balanced practice to switch among all six topics.

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