Intramolecular and Interparticle Force - AP Chemistry

Consider CO $_2$ (linear; nonpolar) and HCN (linear; polar; H bonded to C, not N). Both have similar molar masses. Which statement correctly identifies the strongest type of intermolecular force present in pure HCN?
Answer: Dipole–dipole attractions between polar HCN molecules 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.
A mixture contains molecules of benzene, C $_6$ H $_6$ (nonpolar; relatively large), and water, H $_2$ O (polar; hydrogen bonding). Which statement best describes the strongest intermolecular attraction between a benzene molecule and a water molecule?
Answer: London dispersion forces, because benzene is nonpolar and cannot participate in stronger attractions with water 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.
Two hydrocarbons are compared: pentane, $\mathrm{C_5H_{12}}$ , and neopentane, $\mathrm{C_5H_{12}}$ (same formula but more compact/branched shape). Both are nonpolar and have no H-bond donors/acceptors. Which statement best explains why pentane has a higher boiling point than neopentane?
Answer: Pentane has a larger surface area, leading to stronger dispersion forces 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.
A student compares three molecular substances: $\mathrm{CO_2}$ (linear, nonpolar), $\mathrm{SO_2}$ (bent, polar), and $\mathrm{NH_3}$ (trigonal pyramidal, polar; H-bond donor and acceptor). Which ranking lists the dominant intermolecular force in each substance correctly?
Answer: $\mathrm{CO_2}$ : dispersion; $\mathrm{SO_2}$ : dipoledipole; $\mathrm{NH_3}$ : hydrogen bonding 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.
Two compounds are compared: CH $_3$ Cl (polar; no H-bond donors) and CH $_3$ F (polar; no H-bond donors). Which statement best describes the dominant intermolecular forces in both pure liquids?
Answer: Both have dipole–dipole and London dispersion forces as the key attractions 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.
Consider the molecules formaldehyde, CH $_2$ O (polar; O is an H-bond acceptor only), and water, H $_2$ O (polar; H-bond donor and acceptor). In a mixture of these two liquids, what is the strongest intermolecular attraction between a water molecule and a formaldehyde molecule?
Answer: Hydrogen bonding between an H of H $_2$ O and the O atom of CH $_2$ O 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.
Methane, $\mathrm{CH_4}$ , and chloromethane, $\mathrm{CH_3Cl}$ , have similar sizes, but $\mathrm{CH_3Cl}$ has a polar $\mathrm{C\!- Cl}$ bond and a net dipole. Which intermolecular force is present in $\mathrm{CH_3Cl}$ but not in $\mathrm{CH_4}$ ?
Answer: Dipoledipole attractions This question tests understanding of how molecular polarity affects intermolecular forces. Methane (CH₄) is tetrahedral with four identical C-H bonds, making it completely nonpolar and capable only of London dispersion forces. Chloromethane (CH₃Cl) has a polar C-Cl bond that creates a net dipole moment because the molecule lacks the symmetry to cancel it out, enabling dipole-dipole attractions between molecules. Both molecules have dispersion forces, but the question asks specifically for the force present in CH₃Cl but not in CH₄, which is dipole-dipole attraction. The molecules do not form new covalent bonds between each other, nor do they dissociate into ions. A common misconception is that polar molecules form covalent bonds between molecules (choice D), but intermolecular forces are weaker attractions, not chemical bonds. When comparing intermolecular forces, identify whether each molecule has a net dipole moment to determine if dipole-dipole forces are possible.
Two substances have similar molar masses: dimethyl ether, $\mathrm{CH_3OCH_3}$ (polar; H-bond acceptor only), and ethanol, $\mathrm{CH_3CH_2OH}$ (polar; H-bond donor and acceptor). Which statement best explains why ethanol has the higher boiling point?
Answer: Ethanol forms hydrogen bonds, while dimethyl ether cannot donate H-bonds This question tests understanding of how hydrogen bonding affects boiling points in molecules with similar molar masses. Ethanol (CH₃CH₂OH) has a hydroxyl group (-OH) that can both donate hydrogen bonds (H bonded to O) and accept them (lone pairs on O), while dimethyl ether (CH₃OCH₃) can only accept hydrogen bonds through its oxygen's lone pairs but cannot donate because its hydrogens are bonded to carbon. This ability to form extensive hydrogen bonding networks gives ethanol stronger intermolecular forces and thus a higher boiling point than dimethyl ether, despite their similar molar masses. Both molecules are polar and have dipole-dipole interactions plus dispersion forces, but ethanol's hydrogen bonding is the key differentiator. A common misconception is that dimethyl ether is nonpolar (choice C), but the bent C-O-C structure creates a net dipole. When comparing boiling points of similar-sized molecules, check for hydrogen bonding capability by looking for H bonded to N, O, or F.
Consider hydrogen chloride, HCl (polar; no H-bonding to Cl in AP Chem convention), and hydrogen fluoride, HF (polar; H-bond donor and acceptor). Which intermolecular force is responsible for the major difference in their boiling points?
Answer: Hydrogen bonding present in HF but not in HCl 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.
A student is asked to identify the strongest intermolecular force between molecules in pure liquid acetic acid, CH $_3$ COOH (polar; contains an O–H donor and a carbonyl O acceptor). Which force is most significant between two acetic acid molecules?
Answer: Hydrogen bonding between the O–H of one molecule and the O of another This question assesses the identification of hydrogen bonding in carboxylic acids. Acetic acid, CH₃COOH, has an O–H bond and a carbonyl oxygen, allowing hydrogen bonding between the O–H of one molecule and the O of another. This is the strongest intermolecular force, contributing to properties like boiling point. The polar nature enhances this attraction. A tempting distractor is D, dipole-dipole only requiring N–H, which is wrong because it limits hydrogen bonding to N–H, ignoring O–H capability. Evaluate functional groups for O–H or carbonyls to recognize hydrogen bonding in organic acids.
A sample of iodine, $\mathrm{I_2}$ , is a nonpolar molecular solid at room temperature. Which statement correctly identifies the strongest attractive force holding $\mathrm{I_2}$ molecules together in the solid?
Answer: London dispersion forces from instantaneous induced dipoles This question tests understanding of intermolecular forces in nonpolar molecular solids. Iodine (I₂) molecules are nonpolar because the two identical atoms share electrons equally, creating no permanent dipole. In solid iodine, the strongest attractive force between separate I₂ molecules is London dispersion forces, which arise from temporary, instantaneous dipoles created by electron movement. These temporary dipoles induce dipoles in neighboring molecules, creating attractive forces that hold the solid together. The I-I bond within each molecule is covalent, but the question asks about forces between molecules, not within them. A common misconception is that I₂ has ionic bonding (choice A), but iodine atoms in I₂ share electrons covalently and do not form ions. To identify intermolecular forces, first determine if the molecule is polar or nonpolar; nonpolar molecules like I₂ can only have London dispersion forces as their primary intermolecular attraction.
A solution is made by dissolving sodium chloride, $\mathrm{NaCl(s)}$ , in water. Which interaction is primarily responsible for stabilizing $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ ions in the aqueous solution?
Answer: Iondipole attractions between the ions and polar water molecules 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.
Two molecular solids are compared: solid $\mathrm{CO}$ (small, polar; no H-bonding) and solid $\mathrm{CO_2}$ (larger, nonpolar). Which statement best predicts the dominant intermolecular force in each solid?
Answer: $\mathrm{CO}$ : dipoledipole; $\mathrm{CO_2}$ : London dispersion 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.
Two molecular substances are compared: carbon tetrachloride, CCl $_4$ (tetrahedral; nonpolar; large electron cloud), and sulfur dioxide, SO $_2$ (bent; polar). Which statement correctly identifies the dominant intermolecular force in each pure substance?
Answer: CCl $_4$ : London dispersion; SO $_2$ : dipole–dipole (plus dispersion) 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.
A student claims that the strong attraction between water molecules is the covalent $\mathrm{O\!- H}$ bond. Which statement correctly distinguishes the key intermolecular force in liquid water from the intramolecular bonds within a water molecule?
Answer: Liquid water is held together primarily by hydrogen bonding, not by breaking/forming $\mathrm{O\!- H}$ covalent bonds 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.
A student analyzes the forces in solid iodine, I $_2$ (nonpolar), and solid sodium iodide, NaI (ionic lattice of Na $^+$ and I $^-$ ). Which statement correctly identifies the primary interaction holding each solid together?
Answer: I $_2$ : London dispersion between molecules; NaI: ion–ion attraction in a lattice 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.
A student compares two isomers with the same formula C $_2$ H $_6$ O: ethanol, CH $_3$ CH $_2$ OH (polar; H-bond donor/acceptor), and dimethyl ether, CH $_3$ OCH $_3$ (polar; H-bond acceptor only; no O–H). Which statement best explains why ethanol has a higher boiling point?
Answer: Ethanol can hydrogen bond with itself, while dimethyl ether cannot donate H-bonds This question assesses why hydrogen bonding affects boiling points in isomers with similar molar masses but different functional groups. Ethanol, CH₃CH₂OH, has an O–H bond allowing it to donate and accept hydrogen bonds, while dimethyl ether, CH₃OCH₃, is polar but can only accept hydrogen bonds without an O–H. This capability in ethanol leads to stronger intermolecular attractions, raising its boiling point. Both are polar, but hydrogen bonding provides the key difference. A tempting distractor is C, more electrons in ethanol strengthening dispersion, which is incorrect because both have the same formula and electron count, misunderstanding that dispersion alone doesn't explain the difference. Compare functional groups in isomers to predict how hydrogen bonding influences physical properties like boiling points.
A student compares the attractions in liquid HF and liquid F $_2$ . HF is polar and has an H–F bond; F $_2$ is nonpolar. Which statement correctly identifies the strongest intermolecular force in each liquid?
Answer: HF: hydrogen bonding; F $_2$ : London dispersion 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.
Consider two pure substances at the same temperature: $\mathrm{HCl}$ (polar molecule; no H-bonding) and $\mathrm{Cl_2}$ (nonpolar molecule). Both have similar molar masses. Which statement best compares their dominant intermolecular forces?
Answer: $\mathrm{HCl}$ : dipoledipole; $\mathrm{Cl_2}$ : London dispersion This question tests identification of intermolecular forces based on molecular polarity and composition. HCl is a polar molecule due to the electronegativity difference between H and Cl, creating a permanent dipole, so its dominant intermolecular force is dipole-dipole attraction. Although HCl contains hydrogen, it cannot form hydrogen bonds because the hydrogen is not bonded to N, O, or F. Cl₂ consists of two identical atoms sharing electrons equally, making it nonpolar, so it can only have London dispersion forces between molecules. Neither substance exhibits ionic bonding in its pure form; they are both molecular compounds with covalent bonds within the molecules. A common misconception is that HCl exhibits hydrogen bonding (choice A) simply because it contains hydrogen, but hydrogen bonding requires H bonded specifically to N, O, or F. When analyzing intermolecular forces, first determine molecular polarity, then check if hydrogen bonding criteria are met (H bonded to N, O, or F).
A solution is made by dissolving solid NaCl in water. Water is polar and can hydrogen bond; NaCl dissociates into Na $^+$ and Cl $^-$ . What is the dominant intermolecular attraction between Na $^+$ ions and water molecules in the solution?
Answer: Ion–dipole attraction between Na $^+$ and the partially negative O end of H $_2$ O 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.

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