AP Chemistry Quiz: Intramolecular And Interparticle Force

What this quiz covers

This quiz focuses on Intramolecular And Interparticle Force, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. London dispersion forces only because linear molecules cannot be polar
2. Hydrogen bonding between H and N because any molecule with H and N hydrogen-bonds
3. Dipole–dipole attractions between polar HCN molecules (correct answer)
4. Ion–ion attractions because HCN fully dissociates into ions in the pure liquid
5. Covalent bonding between molecules because HCN forms a polymer at room temperature

Explanation: This question tests the recognition of dipole-dipole forces as the strongest intermolecular attraction in a polar molecule without hydrogen bonding. HCN is linear and polar due to the electronegativity difference between H, C, and N, leading to dipole-dipole attractions between molecules. Although it has H and N, the H is bonded to C, not N, so it does not qualify for hydrogen bonding in AP Chem. The polarity distinguishes it from nonpolar CO₂, explaining stronger forces despite similar masses. A tempting distractor is B, hydrogen bonding due to H and N presence, but this is wrong because it ignores the requirement for H directly bonded to N, O, or F. To identify forces in polar molecules, confirm hydrogen bonding criteria before defaulting to dipole-dipole.

Question 2

1. HF: hydrogen bonding; F$_2$: London dispersion (correct answer)
2. HF: ion–ion attraction; F$_2$: dipole–dipole attraction
3. HF: covalent bonding between molecules; F$_2$: covalent bonding between molecules
4. HF: London dispersion only; F$_2$: hydrogen bonding
5. HF: metallic bonding; F$_2$: metallic bonding

Explanation: This question tests the differentiation of intermolecular forces in polar and nonpolar substances. HF is polar with an H–F bond, enabling hydrogen bonding as the strongest force, while F₂ is nonpolar, relying on London dispersion. The hydrogen bonding in HF involves H of one molecule attracting F of another, stronger than dispersion in F₂. This explains property differences like boiling points. A tempting distractor is D, HF London only and F₂ hydrogen bonding, which is wrong because it reverses forces, stemming from misunderstanding that nonpolar molecules can't have dispersion. Always verify polarity and hydrogen bonding potential to assign forces accurately.

Question 3

1. CCl$_4$: London dispersion; SO$_2$: dipole–dipole (plus dispersion) (correct answer)
2. CCl$_4$: hydrogen bonding; SO$_2$: hydrogen bonding
3. CCl$_4$: dipole–dipole; SO$_2$: London dispersion only
4. CCl$_4$: ion–dipole; SO$_2$: ion–ion
5. CCl$_4$: covalent network; SO$_2$: covalent network

Explanation: This question evaluates the identification of dominant intermolecular forces in polar and nonpolar molecular substances. CCl₄ is tetrahedral and nonpolar, so its dominant force is London dispersion, while SO₂ is bent and polar, leading to dipole-dipole attractions plus dispersion. The polarity of SO₂ arises from its geometry and electronegativity differences, enabling stronger dipole-dipole forces. Both have dispersion, but SO₂'s polarity adds an additional force. A tempting distractor is C, CCl₄ dipole-dipole and SO₂ London only, which is wrong because it reverses polarities, stemming from the misconception that symmetry always implies polarity. When comparing substances, use molecular geometry to determine polarity and thus the presence of dipole-dipole forces.

Question 4

1. CO$_2$(s): London dispersion between molecules; SiO$_2$(s): covalent bonds in a network (correct answer)
2. CO$_2$(s): ionic bonds; SiO$_2$(s): ion–dipole attractions
3. CO$_2$(s): hydrogen bonding; SiO$_2$(s): hydrogen bonding
4. CO$_2$(s): covalent bonds between molecules; SiO$_2$(s): London dispersion forces
5. CO$_2$(s): metallic bonding; SiO$_2$(s): metallic bonding

Explanation: This question tests the distinction between forces in molecular and network covalent solids. Dry ice, CO₂(s), is a molecular solid held by London dispersion between nonpolar molecules, while quartz, SiO₂(s), is a covalent network solid with strong covalent bonds throughout the lattice. The network structure in SiO₂ leads to much higher melting points than the weak dispersion in CO₂. This difference arises from atomic bonding versus molecular attractions. A tempting distractor is D, CO₂ covalent between molecules, which is incorrect because it confuses molecular solids with network solids, misunderstanding that CO₂ molecules don't bond covalently to each other. Classify solids by type to differentiate holding forces correctly.

Question 5

1. $\mathrm{CO_2}$: hydrogen bonding; $\mathrm{SO_2}$: dipoledipole; $\mathrm{NH_3}$: dispersion
2. $\mathrm{CO_2}$: dispersion; $\mathrm{SO_2}$: hydrogen bonding; $\mathrm{NH_3}$: dipoledipole
3. $\mathrm{CO_2}$: ionic bonding; $\mathrm{SO_2}$: covalent bonding; $\mathrm{NH_3}$: metallic bonding
4. $\mathrm{CO_2}$: dispersion; $\mathrm{SO_2}$: dipoledipole; $\mathrm{NH_3}$: hydrogen bonding (correct answer)
5. $\mathrm{CO_2}$: dipoledipole; $\mathrm{SO_2}$: dispersion; $\mathrm{NH_3}$: iondipole

Explanation: This question tests the ability to identify dominant intermolecular forces based on molecular polarity and hydrogen bonding capability. CO₂ is linear with symmetrical C=O bonds, making it nonpolar despite having polar bonds, so its dominant intermolecular force is London dispersion. SO₂ is bent due to a lone pair on sulfur, creating a net dipole moment, so dipole-dipole attractions are its dominant force. NH₃ is polar with a pyramidal shape and, crucially, has N-H bonds that can donate hydrogen bonds and a lone pair on nitrogen that can accept them, making hydrogen bonding its dominant intermolecular force. All molecules also have dispersion forces, but the question asks for the dominant force in each case. A common misconception is that CO₂ has dipole-dipole forces (choice A), but its linear symmetry cancels out the bond dipoles. To determine dominant intermolecular forces, check molecular geometry for polarity, then look for H bonded to N, O, or F for hydrogen bonding capability.

Question 6

1. Propane, because its electrons can form temporary dipoles
2. Acetone, because its $\mathrm{C=O}$ bond forms strong hydrogen bonds
3. Ethanol, because molecules can form $\mathrm{O\!- H\cdots O}$ attractions (correct answer)
4. Propane, because its $\mathrm{C\!- H}$ bonds are highly polar
5. Acetone, because it has the greatest molar mass of the three

Explanation: This question tests the ability to identify which molecules can participate in hydrogen bonding as their dominant intermolecular force. Hydrogen bonding occurs when a hydrogen atom covalently bonded to N, O, or F (donor) interacts with a lone pair on N, O, or F (acceptor) in another molecule. Ethanol (CH₃CH₂OH) has an -OH group that can both donate (through the H) and accept (through the O) hydrogen bonds, making O-H···O attractions its dominant intermolecular force. Propane is nonpolar with only dispersion forces, while acetone can only accept hydrogen bonds through its C=O oxygen but cannot donate them since its hydrogens are bonded to carbon, not to N, O, or F. A common misconception is that acetone forms strong hydrogen bonds (choice B), but without H bonded to N, O, or F, it cannot donate hydrogen bonds. When identifying hydrogen bonding capability, always check if the molecule has H directly bonded to N, O, or F (donor) and/or lone pairs on N, O, or F (acceptor).

Question 7

1. Liquid water is held together primarily by hydrogen bonding, not by breaking/forming $\mathrm{O\!- H}$ covalent bonds (correct answer)
2. Liquid water is held together primarily by ionic bonds between $\mathrm{H^+}$ and $\mathrm{OH^-}$
3. Liquid water is held together primarily by metallic bonding among oxygen atoms
4. Liquid water is held together primarily by covalent bonds between different water molecules
5. Liquid water is held together primarily by dispersion forces because it is nonpolar

Explanation: This question tests the distinction between intramolecular bonds (within molecules) and intermolecular forces (between molecules). Water molecules are held together internally by covalent O-H bonds, but separate water molecules attract each other primarily through hydrogen bonding, where the hydrogen of one molecule (bonded to oxygen) is attracted to the lone pairs on the oxygen of another molecule. These hydrogen bonds between molecules are much weaker than the covalent bonds within molecules and are constantly breaking and reforming in liquid water. The O-H covalent bonds remain intact in liquid water; they are not broken during normal phase changes or molecular interactions. A common misconception is that water molecules bond covalently to each other (choice D), but covalent bonds only exist within each H₂O molecule. To distinguish intermolecular from intramolecular forces, remember that intermolecular forces are attractions between separate molecules, while intramolecular bonds hold atoms together within a single molecule.

Question 8

1. Hydrogen bonding between Na$^+$ and the H atoms of H$_2$O
2. Covalent bonding between Na$^+$ and O in H$_2$O to form a new compound
3. London dispersion forces only, because ions do not participate in electrostatic forces
4. Ion–dipole attraction between Na$^+$ and the partially negative O end of H$_2$O (correct answer)
5. Dipole–dipole attraction between Na$^+$ and H$_2$O because both are polar molecules

Explanation: This question examines the understanding of ion-dipole attractions in solutions involving ionic compounds and polar solvents. When NaCl dissolves in water, it dissociates into Na⁺ and Cl⁻ ions, and water's polar nature allows its partially negative oxygen to attract the positive Na⁺ ion. This ion-dipole interaction is the dominant force between Na⁺ and water molecules, facilitating solvation. The attraction is electrostatic, stronger than dipole-dipole due to the full charge on the ion. A tempting distractor is B, hydrogen bonding between Na⁺ and H atoms, but this is incorrect because it confuses ion-dipole with hydrogen bonding, which requires a hydrogen covalently bonded to N, O, or F. In analyzing solutions, distinguish ion-dipole forces by identifying ionic species and polar solvent molecules.

Question 9

1. Hexane has stronger London dispersion forces due to its larger, more polarizable electron cloud (correct answer)
2. Hexane has hydrogen bonding because C–H bonds can hydrogen-bond with oxygen
3. Hexane has ion–dipole forces because nonpolar molecules become ions in liquids
4. Hexane forms covalent bonds between molecules, increasing the effective molar mass
5. Acetone has only dispersion forces because the carbonyl group is nonpolar overall

Explanation: This question evaluates understanding of how London dispersion forces can compete with dipole-dipole in nonpolar molecules with large electron clouds. Hexane, C₆H₁₄, is nonpolar but larger with a more polarizable electron cloud, leading to stronger dispersion forces than in smaller, polar acetone. Despite acetone's dipole-dipole attractions, hexane's size allows comparable or higher boiling points through dispersion. This illustrates that dispersion strength increases with molecular size and surface area. A tempting distractor is E, acetone has only dispersion because carbonyl is nonpolar, which is wrong due to the misconception that functional groups don't contribute to overall polarity. Compare molecular size and polarizability in nonpolar compounds to assess dispersion force impact on properties.

Question 10

1. Hydrogen bonding between an H of H$_2$O and the O atom of CH$_2$O (correct answer)
2. Ion–ion attraction because both molecules partially ionize in the mixture
3. Covalent bonding between the two molecules to form a single larger molecule
4. London dispersion only because hydrogen bonding requires an O–H bond on both molecules
5. Dipole–dipole only because water cannot hydrogen-bond to a molecule without O–H

Explanation: This question assesses the identification of intermolecular forces between different molecules in a mixture, specifically hydrogen bonding. Water, H₂O, can donate hydrogen bonds via its O–H, and formaldehyde, CH₂O, has a carbonyl oxygen that accepts hydrogen bonds. Thus, the strongest attraction is hydrogen bonding between water's H and formaldehyde's O. This interaction is stronger than dipole-dipole or dispersion in the mixture. A tempting distractor is D, London dispersion only, which is wrong because it assumes both molecules need O–H for hydrogen bonding, ignoring acceptor capability. In mixtures, check if one molecule can donate and the other accept hydrogen bonds to predict strongest interactions.

Question 11

1. I$_2$: London dispersion between molecules; NaI: ion–ion attraction in a lattice (correct answer)
2. I$_2$: hydrogen bonding; NaI: dipole–dipole attractions
3. I$_2$: ion–dipole attraction; NaI: London dispersion between molecules
4. I$_2$: covalent network bonding; NaI: covalent bonding between Na and I molecules
5. I$_2$: metallic bonding; NaI: metallic bonding

Explanation: This question tests the distinction between intermolecular forces in molecular solids and ionic lattices. Solid I₂ consists of nonpolar molecules held by London dispersion forces, while solid NaI is an ionic compound with ion-ion attractions between Na⁺ and I⁻ in a lattice. The ionic bonds in NaI are much stronger, leading to higher melting points compared to I₂'s weaker dispersion. This difference arises from the charged particles in NaI versus neutral molecules in I₂. A tempting distractor is D, covalent network for I₂, which is incorrect because it confuses molecular solids with network solids like diamond, misunderstanding that diatomic molecules don't form extended networks. Classify substances as molecular, ionic, or network to correctly identify holding forces.

Question 12

1. Propane: hydrogen bonding; acetaldehyde: London dispersion
2. Propane: London dispersion; acetaldehyde: dipole–dipole (plus dispersion) (correct answer)
3. Propane: ion–dipole; acetaldehyde: ion–ion
4. Propane: dipole–dipole; acetaldehyde: hydrogen bonding
5. Propane: covalent bonding; acetaldehyde: covalent bonding

Explanation: This question assesses the ability to classify dominant intermolecular forces based on molecular polarity and structure. Propane, C₃H₈, is nonpolar, so its dominant force is London dispersion, while acetaldehyde, CH₃CHO, is polar with a carbonyl group, leading to dipole-dipole attractions in addition to dispersion. Both have similar molar masses, but the polarity of acetaldehyde enables stronger dipole-dipole forces. This distinction highlights how polarity influences intermolecular attractions beyond dispersion. A tempting distractor is D, propane dipole-dipole and acetaldehyde hydrogen bonding, which is wrong due to the misconception that nonpolar molecules have dipole forces and that carbonyls alone enable donating hydrogen bonds without O–H or N–H. To identify forces, first determine polarity from molecular geometry, then check for special cases like hydrogen bonding.

Question 13

1. $\mathrm{CO}$: dipoledipole; $\mathrm{CO_2}$: London dispersion (correct answer)
2. $\mathrm{CO}$: hydrogen bonding; $\mathrm{CO_2}$: dipoledipole
3. $\mathrm{CO}$: ionic bonding; $\mathrm{CO_2}$: ionic bonding
4. $\mathrm{CO}$: covalent network bonding; $\mathrm{CO_2}$: metallic bonding
5. $\mathrm{CO}$: London dispersion only; $\mathrm{CO_2}$: hydrogen bonding

Explanation: This question tests identification of dominant intermolecular forces based on molecular polarity. Carbon monoxide (CO) has a small dipole moment due to unequal sharing of electrons between C and O, making it polar, so its dominant intermolecular force in the solid state is dipole-dipole attraction. Carbon dioxide (CO₂) is linear with two identical C=O bonds pointing in opposite directions, canceling out any dipole moment and making the molecule nonpolar, so it can only have London dispersion forces. Neither molecule can form hydrogen bonds because they lack hydrogen atoms bonded to N, O, or F. Both are molecular compounds, not ionic or network solids. A common misconception is that CO₂ can form hydrogen bonds (choice E) because it contains oxygen, but hydrogen bonding requires hydrogen atoms bonded to N, O, or F, which CO₂ lacks. To predict intermolecular forces in molecular solids, determine molecular polarity from geometry and bond polarity—polar molecules have dipole-dipole forces while nonpolar molecules have only dispersion forces.

Question 14

1. Hydrogen bonding between $\mathrm{Na^+}$ and the hydrogen atoms of water
2. Iondipole attractions between the ions and polar water molecules (correct answer)
3. Dipoledipole attractions between $\mathrm{Na^+}$ and $\mathrm{Cl^-}$
4. Covalent bonding between ions and water to form neutral molecules
5. London dispersion forces because ions are nonpolar particles

Explanation: This question tests understanding of ion-solvent interactions in aqueous solutions. When NaCl dissolves in water, it dissociates into Na⁺ and Cl⁻ ions, which are then stabilized by ion-dipole attractions with polar water molecules. The positive Na⁺ ions are attracted to the partially negative oxygen atoms of water molecules, while negative Cl⁻ ions are attracted to the partially positive hydrogen atoms of water. These ion-dipole interactions are stronger than typical dipole-dipole forces because ions have full charges rather than partial charges. The ions do not form covalent bonds with water molecules; they remain as discrete charged particles surrounded by oriented water molecules (hydration shells). A common misconception is that Na⁺ forms hydrogen bonds (choice A), but hydrogen bonding specifically involves H bonded to N, O, or F interacting with lone pairs, not interactions with metal cations. When ions dissolve in polar solvents, the primary stabilizing force is ion-dipole attraction between the charged ions and the solvent's dipole.

Question 15

1. Pentane has stronger dipoledipole forces because it is less symmetric
2. Pentane has a larger surface area, leading to stronger dispersion forces (correct answer)
3. Neopentane can hydrogen bond more effectively due to branching
4. Neopentane has fewer covalent bonds, so it vaporizes more easily
5. Pentane is more polar because it has more $\mathrm{C\!- H}$ bonds

Explanation: This question tests understanding of how molecular shape affects London dispersion forces in nonpolar molecules. Both pentane and neopentane are nonpolar hydrocarbons with the same molecular formula (C₅H₁₂) and thus the same number of electrons, but pentane has a linear, extended shape while neopentane is compact and spherical. The extended shape of pentane provides a larger surface area for contact between molecules, allowing for stronger London dispersion forces compared to the compact neopentane. Stronger intermolecular forces require more energy to overcome, resulting in pentane's higher boiling point. Neither molecule is polar or capable of hydrogen bonding since they contain only C-H and C-C bonds. A common misconception is that pentane is more polar (choice E), but all hydrocarbons containing only carbon and hydrogen are nonpolar. When comparing dispersion forces in molecules with the same formula, consider molecular shape and surface area—more extended shapes generally have stronger dispersion forces.

Question 16

1. Metallic bonding, because oxygen atoms delocalize electrons in ethanol
2. Covalent bonding between molecules, because ethanol molecules link into chains
3. Hydrogen bonding, because ethanol has an O–H bond and acetone does not (correct answer)
4. Ion–ion attraction, because ethanol can form ions more easily than acetone
5. London dispersion only, because polarity does not affect intermolecular forces

Explanation: This question assesses understanding of how hydrogen bonding differentiates intermolecular forces in similar polar molecules. Ethanol, CH₃CH₂OH, has an O–H bond, enabling it to form hydrogen bonds as both donor and acceptor, while acetone, (CH₃)₂CO, lacks an O–H or N–H bond and can only accept hydrogen bonds but not donate them in its pure form. The presence of hydrogen bonding in ethanol provides a stronger attractive force compared to the dipole-dipole interactions in acetone. This explains why hydrogen bonding is the dominant additional attraction in ethanol. A tempting distractor is A, ion–ion attraction, which is wrong because it stems from the misconception that polar molecules like ethanol ionize easily, but neither compound forms ions in pure liquid state. To compare intermolecular forces, evaluate the presence of hydrogen bonding based on specific bond types before assessing polarity.

Question 17

1. London dispersion forces are stronger in HF because it has more electrons than HCl
2. Metallic bonding present in HF but absent in HCl
3. Covalent bonding between HF molecules, forming HF polymers in all phases
4. Ion–ion attraction present in HF but absent in HCl
5. Hydrogen bonding present in HF but not in HCl (correct answer)

Explanation: This question tests the identification of hydrogen bonding as a key intermolecular force distinguishing boiling points in similar polar compounds. HF is polar and has an H–F bond, allowing hydrogen bonding, while HCl is polar but lacks hydrogen bonding to Cl under AP Chem conventions, relying instead on dipole-dipole and dispersion forces. The hydrogen bonding in HF creates stronger attractions between molecules, leading to a higher boiling point compared to HCl despite similar molar masses. This force involves the hydrogen of one HF molecule attracting the fluorine of another. A tempting distractor is E, stronger London dispersion in HF, which is wrong because it misinterprets electron count as the primary factor, ignoring that HCl has more electrons yet weaker overall forces due to no hydrogen bonding. To explain property differences in polar molecules, check for hydrogen bonding eligibility based on F–H, O–H, or N–H bonds.

Question 18

1. Both have dipole–dipole and London dispersion forces as the key attractions (correct answer)
2. Both have hydrogen bonding because fluorine and chlorine always cause H-bonding
3. Both are held together primarily by covalent bonds between molecules
4. Both are held together primarily by ion–dipole forces due to ionic character
5. Both have only London dispersion forces because polar molecules cannot attract each other

Explanation: This question evaluates the recognition of similar intermolecular forces in polar molecules without hydrogen bonding. Both CH₃Cl and CH₃F are polar due to electronegativity differences, leading to dipole-dipole and London dispersion forces as dominant attractions. Neither has H bonded to N, O, or F, so no hydrogen bonding occurs. Their similar structures result in comparable force types. A tempting distractor is B, both have hydrogen bonding because of F and Cl, which is incorrect due to the misconception that any halogen with H enables hydrogen bonding, ignoring the specific requirement for H-F, H-O, or H-N. For polar haloalkanes, remember they exhibit dipole-dipole but not hydrogen bonding unless H is on F.

Question 19

1. London dispersion forces, because benzene is nonpolar and cannot participate in stronger attractions with water (correct answer)
2. Hydrogen bonding, because water can hydrogen bond with any molecule that has electrons
3. Ion–dipole attraction, because benzene forms ions in the presence of water
4. Dipole–dipole attraction, because benzene has a strong permanent dipole moment
5. Covalent bonding, because water reacts with benzene to link the molecules together

Explanation: This question evaluates intermolecular forces between polar and nonpolar molecules in a mixture. Benzene, C₆H₆, is nonpolar, so the strongest attraction with polar water is London dispersion, as water cannot form hydrogen bonds with benzene's non-acceptor structure. Dispersion arises from temporary dipoles in benzene's large electron cloud interacting with water. Stronger forces like hydrogen bonding are absent due to incompatibility. A tempting distractor is B, hydrogen bonding with any molecule having electrons, which is wrong because it overgeneralizes hydrogen bonding, ignoring the need for specific donors and acceptors. In mixtures, identify compatible forces like dispersion when hydrogen bonding isn't possible between dissimilar molecules.

Question 20

1. Br$_2$ has stronger London dispersion forces due to greater polarizability (correct answer)
2. Br$_2$ has hydrogen bonding because bromine atoms attract hydrogen atoms
3. Br$_2$ has dipole–dipole forces because larger molecules are always more polar
4. Br$_2$ forms covalent bonds between molecules more readily than Cl$_2$
5. Br$_2$ experiences ion–dipole forces because it contains ions in the liquid

Explanation: This question evaluates the recognition of London dispersion forces as the dominant intermolecular force in nonpolar molecules and how molecular size affects their strength. Both Br₂ and Cl₂ are nonpolar diatomic molecules, so their intermolecular forces are primarily London dispersion forces, which depend on the size and polarizability of the electron cloud. Br₂ has a larger electron cloud than Cl₂, leading to stronger temporary dipoles and thus stronger dispersion forces, resulting in a higher boiling point for Br₂. This difference arises from the greater number of electrons in bromine atoms compared to chlorine. A tempting distractor is C, dipole–dipole forces, but this is incorrect due to the misconception that larger molecules are inherently more polar, whereas both are nonpolar. When comparing nonpolar molecules, consider electron cloud size and polarizability to predict dispersion force strength and related properties like boiling points.