Intermolecular Forces (IMF) practice questions

Question 1

A student dissolves a sample of solid magnesium chloride, $\text{MgCl}_2\text{(s)}$, in distilled water. Which of the following best predicts the orientation of the surrounding $\text{H}_2\text{O}$ molecules in the hydration shells around the dissolved $\text{Mg}^{2+}\text{(aq)}$ and $\text{Cl}^-\text{(aq)}$ ions?

Accepted Answer

The oxygen atoms of $\text{H}_2\text{O}$ point toward $\text{Mg}^{2+}$, and the hydrogen atoms point toward $\text{Cl}^-$.

Answer

The hydrogen atoms of $\text{H}_2\text{O}$ point toward $\text{Mg}^{2+}$, and the oxygen atoms point toward $\text{Cl}^-$.

Answer

The oxygen atoms of $\text{H}_2\text{O}$ point toward both $\text{Mg}^{2+}$ and $\text{Cl}^-$.

Answer

The hydrogen atoms of $\text{H}_2\text{O}$ point toward both $\text{Mg}^{2+}$ and $\text{Cl}^-$.

Question 2

A student compares the physical properties of ethanol, $\text{CH}_3\text{CH}_2\text{OH}$, and dimethyl ether, $\text{CH}_3\text{OCH}_3$, which are structural isomers with the same molar mass. Which of the following correctly identifies the substance with the higher boiling point and provides the correct justification?

Accepted Answer

Ethanol has a higher boiling point because ethanol molecules form intermolecular hydrogen bonds, which require more energy to overcome than the dipole-dipole interactions between dimethyl ether molecules.

Answer

Dimethyl ether has a higher boiling point because its symmetrical structure allows for stronger London dispersion forces than ethanol.

Answer

Ethanol has a higher boiling point because the covalent bonds within the $\text{CH}_3\text{CH}_2\text{OH}$ molecule are stronger than the covalent bonds within $\text{CH}_3\text{OCH}_3$.

Answer

Dimethyl ether has a higher boiling point because the central oxygen atom creates a larger permanent dipole moment, resulting in stronger dipole-dipole forces than in ethanol.

Question 3

A student performs paper chromatography on a mixture of two dyes, Dye 1 and Dye 2, using cellulose paper as the stationary phase and ethanol, $\text{C}_2\text{H}_5\text{OH}$ (a polar solvent), as the mobile phase. The experimental results are shown in the table below.

| Dye | Solvent Front Distance (cm) | Dye Distance (cm) |
|---|---|---|
| Dye 1 | $10.0$ | $8.0$ |
| Dye 2 | $10.0$ | $3.0$ |

If the student repeats the experiment using hexane, $\text{C}_6\text{H}_{14}$ (a nonpolar solvent), as the mobile phase, which of the following predictions regarding Dye 1 is correct, and why?

Accepted Answer

Dye 1 will travel a shorter distance because it forms weaker intermolecular attractions with nonpolar hexane than with polar ethanol.

Answer

Dye 1 will travel a greater distance because nonpolar hexane moves faster up the paper, carrying all solutes farther.

Answer

Dye 1 will travel a greater distance because it forms stronger London dispersion forces with hexane than the dipole-dipole forces it formed with ethanol.

Answer

Dye 1 will travel the same distance because the retention factor ($R_f$) depends only on the stationary phase and is independent of the mobile phase.

Question 4

The normal boiling point of 
$\text{HF}$ is $293 \text{ K}$, whereas the normal boiling point of $\text{HCl}$ is $188 \text{ K}$. Which of the following best explains why $\text{HF}$ has a significantly higher boiling point than $\text{HCl}$?

Accepted Answer

$\text{HF}$ has a higher boiling point because $\text{HF}$ molecules form intermolecular hydrogen bonds, which are stronger than the dipole-dipole forces between $\text{HCl}$ molecules.

Answer

$\text{HF}$ has a higher boiling point because the covalent $\text{H-F}$ bond is stronger than the covalent $\text{H-Cl}$ bond, requiring more energy to break during boiling.

Answer

$\text{HCl}$ has a higher boiling point because $\text{HCl}$ has a larger electron cloud, resulting in greater polarizability and stronger London dispersion forces than in $\text{HF}$.

Answer

$\text{HF}$ has a higher boiling point because $\text{HCl}$ molecules form hydrogen bonds that are weaker due to the lower electronegativity of chlorine.

Question 5

A student recovers ethanol used in a fragrance extraction by heating the liquid in a simple distillation apparatus. The boiling process is represented by the following equation.

$$ \text{CH}_3\text{CH}_2\text{OH(l)} \rightarrow \text{CH}_3\text{CH}_2\text{OH(g)} $$

Which statement correctly classifies and explains the boiling of ethanol?

Accepted Answer

Boiling is a physical change because intermolecular attractions are overcome while the covalent bonds within each ethanol molecule remain intact.

Answer

Boiling is a physical change because the covalent $\text{C-C}$, $\text{C-O}$, and $\text{O-H}$ bonds temporarily break and then re-form in the vapor.

Answer

Boiling is a chemical change because intermolecular attractions between intact ethanol molecules are overcome as the molecules move farther apart.

Answer

Boiling is a chemical change because the change from $\text{(l)}$ to $\text{(g)}$ indicates that gaseous ethanol has a different molecular structure.

Question 6

A student compares the normal boiling points of two structural isomers with the molecular formula $\text{C}_2\text{H}_6\text{O}$: ethanol ($\text{CH}_3\text{CH}_2\text{OH}$, boiling point $78^\circ\text{C}$) and dimethyl ether ($\text{CH}_3\text{OCH}_3$, boiling point $-24^\circ\text{C}$). Which of the following best identifies the isomer with the stronger intermolecular forces and provides the correct justification?

Accepted Answer

Ethanol has stronger intermolecular forces because its molecules form hydrogen bonds with one another, whereas dimethyl ether molecules interact primarily through weaker dipole-dipole forces and London dispersion forces.

Answer

Dimethyl ether has stronger intermolecular forces because its oxygen atom forms a more symmetric dipole moment that enhances London dispersion forces.

Answer

Ethanol has stronger intermolecular forces because the $\text{O-H}$ covalent bond in ethanol is stronger and requires more thermal energy to break than the $\text{C-O}$ covalent bonds in dimethyl ether.

Answer

Ethanol has stronger intermolecular forces because ethanol has a larger molar mass and more total electrons than dimethyl ether, resulting in a more polarizable electron cloud.

Question 7

A student compares the physical properties of silicon dioxide, $\text{SiO}_2\text{(s)}$, and carbon dioxide, $\text{CO}_2\text{(s)}$. The student notes that $\text{SiO}_2\text{(s)}$ has a melting point of $1710^\circ\text{C}$, whereas $\text{CO}_2\text{(s)}$ sublimes at $-78^\circ\text{C}$ at atmospheric pressure. Which of the following statements best accounts for the large difference in phase-change temperatures between the two substances?

Accepted Answer

$\text{SiO}_2\text{(s)}$ has a much higher melting point because phase change requires breaking strong covalent bonds within a continuous network lattice, whereas phase change of $\text{CO}_2\text{(s)}$ requires overcoming much weaker intermolecular forces between discrete molecules.

Answer

$\text{SiO}_2\text{(s)}$ has a much higher melting point because melting $\text{SiO}_2\text{(s)}$ breaks the strong $\text{Si}-\text{O}$ covalent bonds within discrete $\text{SiO}_2$ molecules, whereas melting $\text{CO}_2\text{(s)}$ breaks the weaker $\text{C}=\text{O}$ double bonds within discrete $\text{CO}_2$ molecules.

Answer

$\text{CO}_2\text{(s)}$ has a much lower sublimation temperature because the polar $\text{C}=\text{O}$ bonds create repulsive forces between adjacent $\text{CO}_2$ molecules, whereas nonpolar $\text{Si}-\text{O}$ bonds in $\text{SiO}_2$ attract each other.

Answer

$\text{CO}_2\text{(s)}$ has a much lower sublimation temperature because phase change requires breaking covalent bonds in $\text{CO}_2\text{(s)}$, whereas phase change of $\text{SiO}_2\text{(s)}$ requires overcoming only London dispersion forces.

Question 8

A student conducts an experiment to compare the thermal energy required for phase changes of water, $\text{H}_2\text{O}$. The student measures the thermal energy required to completely melt a $18.0 \text{ g}$ sample of ice at $0^\circ\text{C}$ ($q_{\text{fus}}$) and the thermal energy required to completely vaporize a $18.0 \text{ g}$ sample of liquid water at $100^\circ\text{C}$ ($q_{\text{vap}}$). Which of the following correctly compares $q_{\text{vap}}$ to $q_{\text{fus}}$ and provides the correct molecular explanation?

Accepted Answer

$q_{\text{vap}} > q_{\text{fus}}$, because vaporization requires completely overcoming all intermolecular hydrogen bonds, whereas fusion requires only partially disrupting the hydrogen bonding network.

Answer

$q_{\text{vap}} < q_{\text{fus}}$, because ice has a lower density than liquid water, requiring more energy to expand the crystal structure during melting.

Answer

$q_{\text{vap}} < q_{\text{fus}}$, because melting requires breaking covalent $\text{O-H}$ bonds within $\text{H}_2\text{O}$ molecules, whereas boiling only disrupts intermolecular hydrogen bonds.

Answer

$q_{\text{vap}} > q_{\text{fus}}$, because vaporization requires breaking intramolecular covalent $\text{O-H}$ bonds, whereas fusion requires only overcoming intermolecular hydrogen bonds.

Question 9

A student evaluates two nonpolar hydrocarbon solvents for a fast-drying coating. Equal volumes of the liquids are placed in separate evacuated flasks of equal volume at $25^\circ\text{C}$. After equilibrium is established, some liquid remains in each flask.

| Solvent | Molecular formula | Condensed structural formula |
|---------|-------------------|------------------------------|
| Straight-chain hexane | $\text{C}_6\text{H}_{14}$ | $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$ |
| 2,2-Dimethylbutane | $\text{C}_6\text{H}_{14}$ | $\text{CH}_3\text{C}(\text{CH}_3)_2\text{CH}_2\text{CH}_3$ |

Which statement best predicts the relative equilibrium vapor pressures of the two liquids and explains the prediction?

Accepted Answer

The branched isomer has the higher vapor pressure because, although the isomers have comparably polarizable electron clouds, its compact shape provides less intermolecular surface contact and therefore weaker London dispersion attractions.

Answer

The branched isomer has the higher vapor pressure because branching reduces the number of electrons in each molecule, making its electron cloud less polarizable.

Answer

The two isomers have the same vapor pressure because they have the same number of electrons and therefore experience London dispersion attractions of identical strength.

Answer

The straight-chain isomer has the higher vapor pressure because its extended shape provides greater intermolecular surface contact, strengthening London dispersion attractions and causing molecules to enter the vapor more readily.

Question 10

A student compares the scratch hardness and cleavage behavior of three crystalline solids. Cleavage is observed when a sharp blow causes a solid to split along flat, repeating surfaces.

| Solid | Approximate Mohs hardness | Observation after a sharp blow | Structural description |
|---|---:|---|---|
| Diamond, $\text{C(s)}$ | 10 | Flat surfaces form along certain orientations | Each $\text{C}$ atom is covalently bonded to four other $\text{C}$ atoms in a three-dimensional network. |
| Quartz, $\text{SiO}_2\text{(s)}$ | 7 | Irregular surfaces form without cleavage | Covalent $\text{Si}-\text{O}$ bonds extend throughout a three-dimensional framework without distinct layers. |
| Graphite, $\text{C(s)}$ | 1–2 | Thin, flat sheets separate readily | Strong covalent bonds connect $\text{C}$ atoms within each sheet, and London dispersion forces act between sheets. |

Which of the following best explains the hardness and cleavage results?

Accepted Answer

Hardness and cleavage test different responses: diamond's strong three-dimensional network resists scratching but can split along planes where fewer bonds oppose separation; quartz lacks layered weak regions, and graphite separates between sheets held by dispersion forces.

Answer

Diamond's high hardness means that its cleavage cannot involve breaking covalent bonds; therefore, diamond must contain unrepresented molecular layers held together by weak intermolecular forces.

Answer

Quartz's lack of cleavage means that quartz should be harder than diamond, because a solid that fractures irregularly always requires more energy to scratch than one that cleaves.

Answer

Graphite is soft and cleaves readily because the $\text{C}-\text{C}$ covalent bonds within each sheet are much weaker than those in diamond, so both tests separate atoms within a sheet.

Question 11

A student plans to separate an equimolar liquid mixture of ethanol, $\text{C}_2\text{H}_5\text{OH}$, and water, $\text{H}_2\text{O}$, by fractional distillation at $1\text{ atm}$.

| Component | Normal boiling point |
|-----------|----------------------|
| $\text{C}_2\text{H}_5\text{OH}$ | $78^\circ\text{C}$ |
| $\text{H}_2\text{O}$ | $100^\circ\text{C}$ |

The apparatus includes an electrically heated distillation flask, a packed fractionating column, a thermometer, a water-cooled condenser, and a receiving flask. Which proposed setup, operating procedure, and prediction will best produce an ethanol-enriched initial distillate while correctly monitoring the vapor temperature?

Accepted Answer

Use the packed column, center the thermometer bulb at the entrance to the condenser, heat slowly, and collect the initial fraction separately; predict \(0.50

Answer

Use the packed column, submerge the thermometer bulb in the boiling mixture, heat slowly, and collect the initial fraction separately; predict \(0.50

Answer

Use the packed column, center the thermometer bulb at the entrance to the condenser, heat rapidly, and collect the initial fraction separately; predict \(0.50

Answer

Use the packed column, center the thermometer bulb at the entrance to the condenser, heat slowly, and collect the initial fraction separately; predict $x_{\text{ethanol}}=1.00$.

Question 12

In a gas-recovery study, a student places separate samples containing identical numbers of gas molecules into identical rigid vessels at the same temperature. The vessels are maintained at high pressure, and both samples remain single phases. The student calculates the compressibility factor, defined as $Z=\dfrac{PV}{nRT}$. For an ideal gas, $Z=1$.

| Gas | Molecular description | Measured $Z$ |
|-----|-----------------------|----------------|
| $\text{CH}_4$ | Tetrahedral and nonpolar | $0.96$ |
| $\text{SO}_2$ | Bent and polar | $0.72$ |

Which choice ranks the gases from greatest to least magnitude of deviation from ideal behavior and best explains the ranking?

Accepted Answer

$\text{SO}_2 > \text{CH}_4$, because dipole-dipole attractions between $\text{SO}_2$ molecules reduce momentum transfer to the vessel walls more strongly than the dispersion forces between $\text{CH}_4$ molecules do.

Answer

$\text{CH}_4 > \text{SO}_2$, because a value of $Z$ closer to 1 indicates a larger fractional pressure decrease caused by intermolecular attractions.

Answer

$\text{CH}_4 > \text{SO}_2$, because the lower molar mass of $\text{CH}_4$ gives its molecules a greater average speed, which produces a greater departure from ideal behavior.

Answer

$\text{SO}_2 > \text{CH}_4$, because the larger volume occupied by individual $\text{SO}_2$ molecules lowers the measured pressure below the ideal pressure.

Question 13

A student compares the relative solubilities of four alcohols in water, $\text{H}_2\text{O}$, and hexane, $\text{C}_6\text{H}_{14}$, at $25^\circ\text{C}$. A small amount of each alcohol is added to a separate vial containing equal volumes of water and hexane. After each mixture reaches equilibrium, the student determines the ratio $R$.

$$
R=\dfrac{[\text{alcohol}]_{\text{water}}}{[\text{alcohol}]_{\text{hexane}}}
$$

| Alcohol | Condensed structural formula |
|---------|------------------------------|
| Methanol | $\text{CH}_3\text{OH}$ |
| Ethanol | $\text{CH}_3\text{CH}_2\text{OH}$ |
| 1-Butanol | $\text{CH}_3(\text{CH}_2)_3\text{OH}$ |
| 1-Hexanol | $\text{CH}_3(\text{CH}_2)_5\text{OH}$ |

Which choice correctly ranks the expected values of $R$ from greatest to least?

Accepted Answer

$\text{CH}_3\text{OH} > \text{CH}_3\text{CH}_2\text{OH} > \text{CH}_3(\text{CH}_2)_3\text{OH} > \text{CH}_3(\text{CH}_2)_5\text{OH}$, because each molecule has one hydroxyl group, while increasing carbon-chain length increases affinity for nonpolar hexane relative to water.

Answer

$\text{CH}_3(\text{CH}_2)_5\text{OH} > \text{CH}_3(\text{CH}_2)_3\text{OH} > \text{CH}_3\text{CH}_2\text{OH} > \text{CH}_3\text{OH}$, because the larger nonpolar portions strengthen dispersion forces with hexane, increasing $R$.

Answer

$\text{CH}_3(\text{CH}_2)_5\text{OH} > \text{CH}_3(\text{CH}_2)_3\text{OH} > \text{CH}_3\text{CH}_2\text{OH} > \text{CH}_3\text{OH}$, because each additional $\text{C-H}$ bond provides another site for hydrogen bonding with water.

Answer

$\text{CH}_3\text{OH} > \text{CH}_3\text{CH}_2\text{OH} > \text{CH}_3(\text{CH}_2)_3\text{OH} > \text{CH}_3(\text{CH}_2)_5\text{OH}$, because London dispersion forces between an alcohol and water become weaker as the carbon chain length increases.

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