Intramolecular and Interparticle Force
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AP Chemistry › Intramolecular and Interparticle Force
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?
Dipole–dipole attractions between polar HCN molecules
Covalent bonding between molecules because HCN forms a polymer at room temperature
Ion–ion attractions because HCN fully dissociates into ions in the pure liquid
Hydrogen bonding between H and N because any molecule with H and N hydrogen-bonds
London dispersion forces only because linear molecules cannot be polar
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.
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?
HF: hydrogen bonding; F$_2$: London dispersion
HF: London dispersion only; F$_2$: hydrogen bonding
HF: metallic bonding; F$_2$: metallic bonding
HF: covalent bonding between molecules; F$_2$: covalent bonding between molecules
HF: ion–ion attraction; F$_2$: dipole–dipole attraction
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.
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?
CCl$_4$: hydrogen bonding; SO$_2$: hydrogen bonding
CCl$_4$: ion–dipole; SO$_2$: ion–ion
CCl$_4$: London dispersion; SO$_2$: dipole–dipole (plus dispersion)
CCl$_4$: covalent network; SO$_2$: covalent network
CCl$_4$: dipole–dipole; SO$_2$: London dispersion only
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.
Two solids are compared: dry ice, CO$_2$(s) (molecular solid of nonpolar molecules), and quartz, SiO$_2$(s) (covalent network solid). Which statement correctly distinguishes the primary forces holding each solid together?
CO$_2$(s): ionic bonds; SiO$_2$(s): ion–dipole attractions
CO$_2$(s): London dispersion between molecules; SiO$_2$(s): covalent bonds in a network
CO$_2$(s): hydrogen bonding; SiO$_2$(s): hydrogen bonding
CO$_2$(s): metallic bonding; SiO$_2$(s): metallic bonding
CO$_2$(s): covalent bonds between molecules; SiO$_2$(s): London dispersion forces
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.
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?
$\mathrm{CO_2}$: dispersion; $\mathrm{SO_2}$: hydrogen bonding; $\mathrm{NH_3}$: dipoledipole
$\mathrm{CO_2}$: dispersion; $\mathrm{SO_2}$: dipoledipole; $\mathrm{NH_3}$: hydrogen bonding
$\mathrm{CO_2}$: ionic bonding; $\mathrm{SO_2}$: covalent bonding; $\mathrm{NH_3}$: metallic bonding
$\mathrm{CO_2}$: hydrogen bonding; $\mathrm{SO_2}$: dipoledipole; $\mathrm{NH_3}$: dispersion
$\mathrm{CO_2}$: 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.
Three pure liquids are compared at the same temperature: (1) propane, $\mathrm{C_3H_8}$ (nonpolar, no H-bond donors/acceptors); (2) acetone, $\mathrm{(CH_3)_2CO}$ (polar, H-bond acceptor only); and (3) ethanol, $\mathrm{CH_3CH_2OH}$ (polar, H-bond donor and acceptor). Which liquid has hydrogen bonding as its dominant intermolecular force?
Acetone, because its $\mathrm{C=O}$ bond forms strong hydrogen bonds
Ethanol, because molecules can form $\mathrm{O!- H\cdots O}$ attractions
Propane, because its $\mathrm{C!- H}$ bonds are highly polar
Acetone, because it has the greatest molar mass of the three
Propane, because its electrons can form temporary dipoles
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).
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?
Liquid water is held together primarily by covalent bonds between different water molecules
Liquid water is held together primarily by metallic bonding among oxygen atoms
Liquid water is held together primarily by ionic bonds between $\mathrm{H^+}$ and $\mathrm{OH^-}$
Liquid water is held together primarily by hydrogen bonding, not by breaking/forming $\mathrm{O!- H}$ covalent bonds
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.
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?
Hydrogen bonding between Na$^+$ and the H atoms of H$_2$O
Ion–dipole attraction between Na$^+$ and the partially negative O end of H$_2$O
London dispersion forces only, because ions do not participate in electrostatic forces
Covalent bonding between Na$^+$ and O in H$_2$O to form a new compound
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.
Two pure liquids are compared: hexane, C$6$H${14}$ (nonpolar; larger), and acetone, (CH$_3$)$_2$CO (polar; smaller). Which statement best describes why hexane can have a boiling point comparable to or higher than acetone despite being nonpolar?
Hexane has ion–dipole forces because nonpolar molecules become ions in liquids
Hexane has hydrogen bonding because C–H bonds can hydrogen-bond with oxygen
Hexane has stronger London dispersion forces due to its larger, more polarizable electron cloud
Hexane forms covalent bonds between molecules, increasing the effective molar mass
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.
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?
London dispersion only because hydrogen bonding requires an O–H bond on both molecules
Ion–ion attraction because both molecules partially ionize in the mixture
Covalent bonding between the two molecules to form a single larger molecule
Dipole–dipole only because water cannot hydrogen-bond to a molecule without O–H
Hydrogen bonding between an H of H$_2$O and the O atom of CH$_2$O
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.