Worked example 1: solution

Step 1 · Identify the pieces

Methane is a molecule, not ions in a solvent. Its four C–H bonds sit in a symmetric tetrahedron, so the molecule is nonpolar. No H is bonded to N, O, or F.

Worked example 1: solution

Dispersion is present in every substance. Nothing here adds dipole–dipole, hydrogen bonding, or ion–dipole.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the chart

Worked example 2: forces present

A common first answer

Methanol is polar, so dipole–dipole is real. It is not the strongest, because the H bonded to oxygen does more.

Worked example 2: forces present

A molecule, not ions. Polar. And one H is bonded directly to O: the trigger for hydrogen bonding.

Worked example 2: forces present

Worked example 2: the strongest force

Step 3 · Pick the strongest

Worked example 2: the route on the chart

Practice 1 answer: C

B stopped one rung early: HF is polar, but the H bonded to fluorine upgrades the strongest force to hydrogen bonding, above ordinary dipole–dipole. A defaulted to the weakest force; dispersion is in every molecule but never the strongest when a stronger one is present. D named the H–F bond inside the molecule, which is intramolecular and does not act between molecules.

Worked example 3: forces present

Step 1 · Identify the pieces

Propane is a nonpolar hydrocarbon. Dimethyl ether is polar, but every H sits on carbon. Ethanol is polar and has an O–H.

Worked example 3: forces present

Worked example 3: the ranking

Step 3 · Pick the strongest

Worked example 3: the route on the chart

Practice 2 answer: D

Argon is a lone nonpolar atom: dispersion only. CH₂O is polar, but both H atoms sit on carbon: dipole–dipole. CH₃NH₂ has H on N: hydrogen bonding. Na⁺ pulls on water's dipoles: ion–dipole.

B put hydrogen bonding at the top; ion–dipole outranks it. C ranked by molar mass, 22.99 < 30.03 < 31.06 < 39.95 g/mol, as if dispersion decided everything. A is the right order read strongest first.

Worked example 1: solution

Step 1 · Identify each liquid's strongest force

Hexane is a nonpolar hydrocarbon: dispersion only. 1-Pentanol carries an O–H, so it hydrogen bonds. Nearly equal size means nearly equal dispersion; the O–H is the difference.

Worked example 1: solution

Viscosity measures resistance to flow. To flow, each molecule must slide past its neighbors, and sliding means briefly pulling away from them.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the strip

Worked example 2: solution

The common first answer, tested

Sealed above any liquid, at any temperature, some molecules escape while others return. Both flasks hold vapor at a steady pressure, so both liquids show a vapor pressure. Boiling is not required.

Worked example 2: solution

Water hydrogen bonds through its O–H. Diethyl ether is polar, but every H sits on carbon: ordinary dipole–dipole, far weaker.

Worked example 2: solution

Vapor pressure measures escape: how much vapor builds before condensation catches up with evaporation.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the strip

Practice 1 answer: A

B describes water: when adhesion wins, the edge is dragged up the wall, giving a concave dip and a climbing column. C invents a repulsion; glass attracts every liquid at least weakly, and the dome comes from the liquid's own inward pull. D cannot happen: every substance has at least dispersion forces.

Practice 2 answer: B

The masses sit close, so force type decides. 1-Propanol's O–H hydrogen bonds. Acetone is polar with no O–H. Pentane is nonpolar. Vapor pressure measures escape, so it rises as the forces weaken.

A ranked the attractions, not the vapor pressure; stronger forces lower it. C ranked by molar mass alone, but pentane's extra mass cannot outpull a dipole or a hydrogen bond. D read acetone as nonpolar; its C=O makes it polar, so it holds on harder than pentane.

Worked example 1: solution

Step 1 · Identify each substance's strongest force

Ethanol has an O–H bond, so it hydrogen bonds. Dimethyl ether is polar but has no O–H, so its strongest force is ordinary dipole–dipole attraction.

Worked example 1: solution

Equal molar mass means dispersion is the same for both, so only the force type differs. Hydrogen bonding is stronger than dipole–dipole, so ethanol boils higher.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Identify each substance's strongest force

Both are nonpolar, so dispersion is the only force acting in each.

Worked example 2: solution

Same force type does not mean same strength. The molecule held by stronger dispersion boils higher.

Worked example 2: solution

Iodine is far larger than chlorine, with many more electrons, so its dispersion is much stronger. Iodine boils higher.

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 answer: A

Both are nonpolar, so dispersion is the only force. Heptane is the larger molecule, so its dispersion is stronger and it boils higher. Answer A.

B confused forces with bonds: boiling overcomes the attractions between molecules, never the covalent bonds inside them. C reversed the trend: weakly held light molecules escape more easily and boil lower, not higher. D forgot that the same force type can differ in strength, and dispersion grows with size.

Worked example 3: solution

Step 1 · Identify each substance's strongest force

Butane is nonpolar: dispersion only. Acetone is polar with no O–H: dipole–dipole. 1-Propanol has an O–H: hydrogen bonding.

Worked example 3: solution

Mass is nearly equal, so dispersion ties and the force type sets the order. Hydrogen bonding beats dipole–dipole beats dispersion, so 1-propanol > acetone > butane.

Worked example 3: solution

Worked example 3: the route on the map

Practice 2 answer: D

Neon and krypton are nonpolar atoms: dispersion only, and the larger krypton attracts more. PH₃ is polar, but its H sits on P: dipole–dipole. NH₃ has H on N: hydrogen bonding.

A ranked by mass alone; krypton is the heaviest, yet it has only dispersion. B let PH₃ hydrogen bond, but an H on P does not qualify. C reversed the size tie; the bigger atom disperses more and boils higher.

Worked example 1: solution

Step 1 · Find the point

The start sits in the solid region, so the sample begins solid.

Worked example 1: solution

Heating at constant pressure is a horizontal path to the right.

Worked example 1: solution

Above the triple point, the path crosses the fusion curve first, then the vaporization curve.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Find the point

On the 1-atm line, 30 °C lies between the fusion crossing (−15 °C) and the vaporization crossing (60 °C). That is the liquid region.

Worked example 2: solution

Step 1 · Find the point
Step 2 · Choose the path

Constant pressure: move right along the 1-atm line, from 30 °C to 100 °C.

Worked example 2: solution

Step 1 · Find the point
Step 2 · Choose the path
Step 3 · Cross the curves

Worked example 2: the route on the map

Practice 1 answer: C

A put the triple point on the path, but it sits at 0.25 atm, 0.75 atm below it. B read the triple-point temperature, 10 °C, as the melting point: 4 °C too low. D took the critical point, 320 °C, as the boiling point: 202 °C too high.

Practice 2 answer: B

The horizontal path runs 16.8 atm above the triple point, so cooling crosses the vaporization curve, then the fusion curve. It ends at −60 °C, 32 °C below the triple-point temperature: solid.

A took the sublimation route, which runs only below 3.2 atm. D stopped at the liquid; the path continues past the fusion curve. C misread the critical point: the gas starts 24 °C above the critical temperature, but cooling carries it below 96 °C, and at 20 atm, 28 atm under the critical pressure, it condenses.

Worked example 1: solution

Step 1 · Name the particle

A metal with a nonmetal gives ions: one Ca²⁺ for every two Cl⁻, so +2 + 2(−1) = 0.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Name the particle

Shiny, bendable, and conducting as a solid: only a metal fits all three. Its lattice holds metal cations in a sea of mobile electrons.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 answer: C

A matched hardness and melting point but skipped the molten test: a melted ionic solid conducts. B ignored the conductivity; a metal conducts as a solid. D read only the conductivity; a molecular solid melts low and is soft, never near 2000 °C.

Worked example 3: solution

Step 1 · Sort the positions

8 corner atoms and 1 body-center atom. Each corner atom sits in 8 cells at once, so counting it whole counts it 8 times over.

Worked example 3: solution

Worked example 3: solution

Practice 2 answer: A

Conductivity is the sharpest test: only a metal conducts as a solid. Soft and 327 °C still fit, because metallic melting points vary widely. Then count shares, not spheres.

B let the softness and the modest melting point outvote the conductivity; a molecular solid never conducts. C counted each face atom whole: 1 + 6 = 7. D counted every drawn sphere whole: 8 + 6 = 14.