3.1 Intermolecular and Interparticle Forces

1. AAttraction between permanent dipoles that the two atoms carry at all times
   A student who thinks London dispersion forces come from permanent partial charges picks this. The partial charges shown exist only at one instant, because the electrons are moving; an isolated Ar atom has no permanent dipole.
2. BA covalent bond formed by a pair of electrons that the two atoms share between them
   A student who thinks the attractions between particles are covalent bonds picks this. The diagram shows two separate electron clouds; no electrons are shared, and Ar atoms do not form covalent bonds with each other.
3. CAttraction due to the masses of the atoms, which is greater for heavier atoms
   A student who thinks mass causes attractions between particles picks this. The attraction shown is Coulombic, between the temporary partial charges. Heavier noble gas atoms do attract more strongly, but because they have more electrons and are more polarizable, not because of their mass.
4. DAttraction between temporary dipoles that form as electron clouds shift Correct
   Ar atoms have no permanent dipoles. At the instant shown, each electron cloud is shifted to one side, giving a temporary dipole; the δ− side of one atom faces the δ+ side of the other, and the two attract. A moment later the electron distributions change. This is a London dispersion force.

CED 3.1.A.1 · Read this in Fix

1. Aa CH₃F molecule, by dipole-dipole forces
   A student who thinks like particles attract most strongly picks this. Two CH₃F molecules do attract by dipole-dipole forces, but partial charges interacting with partial charges give a weaker attraction than a full ionic charge interacting with a partial charge.
2. Ban Ar atom, by dispersion forces
   A student who thinks heavier particles attract more strongly picks this. Ar (39.95 g/mol) is the heaviest particle listed, but it has no charge or permanent dipole, so its attraction to CH₃F, by dispersion and dipole-induced dipole forces, is weaker than the ion-dipole attraction.
3. Ca Na⁺ ion, by an ion-dipole force Correct
   CH₃F is polar. The full positive charge of Na⁺ attracts the partially negative F end of the molecule more strongly than any partial charge on a neutral neighbor could, so the ion-dipole attraction is the strongest of the four.
4. Da CH₄ molecule, by hydrogen bonding
   A student who thinks H atoms bonded to carbon can hydrogen bond picks this. The H atoms of CH₄ and of CH₃F are bonded to C, so no hydrogen bonds form; CH₄ is attracted only by dispersion and dipole-induced dipole forces.

CED 3.1.A.2 · Read this in Fix

1. AThe attraction is stronger in orientation II, where the larger I atom faces the ion.
   A student who thinks the larger atom attracts an ion more strongly picks this. I is larger, but it carries the partial positive charge, so it is repelled by Na⁺; size does not decide which end is attracted.
2. BThe attraction is stronger in orientation I, where the Cl atom faces the ion. Correct
   Na⁺ attracts the δ− end of the molecule and repels the δ+ end. In orientation I the δ− Cl atom is nearer the ion, so the attraction outweighs the repulsion; in orientation II the δ+ I atom is nearer, so the interaction is repulsive overall.
3. CThe attraction is equally strong in I and II, since the molecule as a whole is polar.
   A student who thinks orientation does not matter picks this. The interaction depends on which partial charge is nearer the ion: unlike charges facing each other attract, like charges facing each other repel.
4. DThere is no attraction in either orientation, since ICl has no net charge.
   A student who thinks an ion cannot attract a neutral molecule picks this. ICl is neutral overall, but its partial charges are at different distances from the ion, so there is a net attraction in orientation I.

CED 3.1.A.3 · Read this in Fix

1. ALine 1
   A student who thinks the O–H covalent bond is a hydrogen bond picks this. This line is the covalent bond between O and H within molecule A; the hydrogen bond is the attraction between that H atom and the O atom of molecule B.
2. BLine 2
   A student who thinks any H atom can hydrogen bond picks this. This H atom is bonded to C, not to N, O or F, so its attraction to the O atom of molecule B is not a hydrogen bond.
3. CLine 3
   A student who thinks a hydrogen bond joins two H atoms picks this. Both atoms joined by this line are H atoms bonded to C; a hydrogen bond needs an H bonded to N, O or F and a separate N, O or F atom.
4. DLine 4 Correct
   A hydrogen bond is the attraction between an H atom covalently bonded to O (here, the H of the O–H group in molecule A) and an O atom of another molecule (molecule B). This line joins exactly those two atoms.

CED 3.1.A.4 · Read this in Fix

1. ANo intermolecular force acts here, since the two groups are in one molecule.
   A student who thinks hydrogen bonds form only between separate molecules picks this. In a large molecule, an N–H group in one region can hydrogen bond to an O atom in another region of the same molecule.
2. BIt is a covalent bond, and bonds like it hold the folded shape in place.
   A student who thinks a biomolecule's shape is held only by covalent bonds picks this. The dashed line joins atoms in different regions of the chain without electron sharing; it is a noncovalent hydrogen bond.
3. CIt is a hydrogen bond, even though both groups belong to the same molecule. Correct
   The H atom is covalently bonded to N, and it is attracted to the O atom of a C=O group in another region of the chain. This is a hydrogen bond, a noncovalent interaction; in large biomolecules such interactions can form between different regions of the same molecule.
4. DIt is an ionic bond between the fully charged H and O atoms of the chain.
   A student who treats partial charges as full charges picks this, reading the δ+ H atom and δ− O atom as ions. Both atoms carry only partial charges, and the attraction between them is a hydrogen bond.

CED 3.1.A.5 · Read this in Fix

1. ACCl₄ molecules attract one another most strongly, as four polar C–Cl bonds make CCl₄ the most polar.
   A student who thinks any molecule with polar bonds is polar picks this. In tetrahedral CCl₄ the four equal C–Cl bond dipoles cancel, so CCl₄ is the only nonpolar molecule of the three; its strong attractions are dispersion forces.
2. BCCl₄ boils highest because boiling breaks more C–Cl bonds per molecule than in CHCl₃ or CH₂Cl₂.
   A student who thinks boiling breaks covalent bonds picks this. The vapor of each liquid consists of intact molecules; boiling overcomes only the attractions between molecules.
3. CCCl₄ molecules attract one another most strongly, as its large electron cloud is very polarizable. Correct
   CCl₄ has the highest boiling point, so its molecules attract one another most strongly. Its four equal C–Cl bond dipoles cancel in the tetrahedral molecule, so CCl₄ is nonpolar and only dispersion forces act between its molecules, while CHCl₃ and CH₂Cl₂ are polar. With 74 electrons, CCl₄ has the largest, most polarizable electron cloud, and its dispersion forces outweigh the combined dispersion and dipole-dipole forces of the smaller polar molecules.
4. DCCl₄ boils highest because its heavy molecules pull one another together by their large mass.
   A student who thinks mass itself causes the attraction between molecules picks this. The attractions between molecules are Coulombic; CCl₄'s large mass goes with many electrons and strong dispersion forces, but the mass itself does not attract.

CED 3.1.A.1 · Read this in Fix

1. AIt is limited: it ranks types of force but ignores how dispersion forces grow with size. Correct
   The ranking compares typical interactions between small molecules. Decane (82 electrons) is much larger and more polarizable than methanol (18 electrons), so its total dispersion forces exceed methanol's combined hydrogen bonding and dispersion forces. The model compares one interaction of each type and breaks down when the molecules differ greatly in size.
2. BIt fails: molecular mass, not the type of force, decides which of the liquids boils higher.
   A student who thinks mass itself decides boiling point picks this. The type and strength of the attractions decide the boiling point; molar mass is only a rough guide to the number of electrons. Water (18 g/mol) boils far higher than CH₄ (16 g/mol), for example.
3. CIt holds; decane boils higher only because its C–C and C–H bonds are harder to break.
   A student who thinks boiling breaks covalent bonds picks this. When decane boils, its molecules are separated from one another but stay intact, so its covalent bonds play no part in its boiling point.
4. DIt was misapplied: decane molecules also hydrogen bond, through their many C–H hydrogens.
   A student who thinks any H atom can hydrogen bond picks this. Decane's H atoms are bonded to C, not to N, O or F, so decane molecules cannot hydrogen bond.

CED 3.1.A.1 · Read this in Fix

1. APentane's chain is not symmetrical, so pentane molecules are polar and attract by dipole-dipole forces.
   A student who thinks unsymmetrical hydrocarbon chains are polar picks this. C–H bonds are effectively nonpolar, so pentane molecules are essentially nonpolar whatever their shape; the difference comes from contact area.
2. BPentane's elongated molecules have more contact area with neighbors, so their dispersion forces are stronger. Correct
   Both isomers have the same number of electrons (42), and both are nonpolar. Pentane's long, narrow molecules can lie alongside one another and touch over a larger area than the compact, nearly spherical 2,2-dimethylpropane molecules, so the dispersion forces between pentane molecules are stronger and its boiling point is higher.
3. CThe H atoms along pentane's chain are exposed to its neighbors, so pentane molecules form more hydrogen bonds.
   A student who thinks any H atom can hydrogen bond picks this. All of the H atoms in both isomers are bonded to C, so neither substance hydrogen bonds.
4. DPentane's chain of C–C bonds is harder to break when it boils than the compact, branched molecule is.
   A student who thinks boiling breaks covalent bonds picks this. Both liquids boil without any C–C or C–H bonds breaking; only the attractions between molecules are overcome.

CED 3.1.A.1.i · Read this in Fix

1. APentane and 2,2-dimethylpropane, isomers of C₅H₁₂ with different shapes Correct
   The two isomers have the same formula and so the same number of electrons, and both are nonpolar. They differ in shape and therefore in contact area, so a difference in their boiling points can be attributed to contact area.
2. BHexane and 2,2-dimethylpropane, two alkanes with very different shapes
   A student who accepts a comparison in which several factors change at once picks this. Hexane is long and 2,2-dimethylpropane is compact, but hexane also has more electrons (50 vs 42), so a difference in boiling point could be due to polarizability rather than contact area.
3. C2,2-Dimethylpropane and 1-butanol, two compounds of nearly equal molar mass
   A student who thinks molecules of nearly equal molar mass have the same intermolecular forces picks this, seeing two equal-mass molecules with different shapes. 1-Butanol molecules are polar and hydrogen bond, and 2,2-dimethylpropane molecules do neither, so a difference in boiling point would not isolate contact area.
4. D2,2-Dimethylpropane and methane, two molecules that are both close to spherical
   A student who thinks the factor being tested should be kept the same picks this. These molecules have similar shapes, so the comparison does not vary contact area; they also differ greatly in number of electrons.

CED 3.1.A.1.i · Read this in Fix

1. ACovalent bonds link neighboring molecules to one another, so the chains are held in place.
   A student who thinks the attractions between molecules are covalent bonds picks this. Each wax molecule is a separate molecule; the chains are held to one another by dispersion forces, which is why wax melts at a low temperature.
2. BMolecules this large are polar, so strong dipole-dipole forces hold the neighboring chains together.
   A student who confuses polarizability with polarity picks this. Large alkane molecules are highly polarizable but nonpolar; C–H and C–C bonds are effectively nonpolar, so there are no dipole-dipole forces.
3. CThe molecules are very heavy, and their large masses pull them together into a solid at this temperature.
   A student who thinks mass causes intermolecular attraction picks this. The attraction is Coulombic, between temporary dipoles; the large number of electrons, not the mass, makes the dispersion forces strong.
4. DThe long molecules lie side by side over large areas, so the total dispersion forces are large. Correct
   Each C₂₅H₅₂ molecule has many electrons and a long chain that can lie alongside its neighbors. The large contact area and polarizability give strong total dispersion forces, enough to hold the molecules in place at room temperature even though each molecule is nonpolar.

CED 3.1.A.1.i · Read this in Fix

1. AMolecules with more electrons are more polar, so the dipole-dipole forces between them are stronger.
   A student who confuses polarizability with polarity picks this. Each halogen molecule consists of two identical atoms, so it has no dipole moment; the larger molecules are more polarizable, not more polar.
2. BMolecules with more electrons are heavier, and heavier molecules attract each other more strongly.
   A student who thinks mass causes intermolecular attraction picks this. The mass and the boiling point both rise with the number of electrons, but the attraction comes from temporary dipoles in the electron clouds, not from mass.
3. CLarger electron clouds are more easily distorted, so stronger dispersion forces act between molecules. Correct
   All four halogens are nonpolar, so dispersion forces are the only forces between their molecules. More electrons in a larger electron cloud make a molecule more polarizable, so larger temporary dipoles form and the dispersion forces, and therefore the boiling points, increase from F₂ to I₂.
4. DThe bonds in the larger halogen molecules are stronger, so more energy is needed to break them on boiling.
   A student who thinks boiling breaks covalent bonds picks this. Boiling separates whole halogen molecules without breaking the halogen–halogen bonds; in any case, the I–I bond is weaker than the Cl–Cl bond.

CED 3.1.A.1.ii · Read this in Fix

1. AThe Ar–Ar pair, as its minimum lies at the shorter distance
   A student who thinks a shorter distance at the minimum means a stronger attraction picks this. The position of the minimum shows the distance at which the atoms settle; the depth shows the strength. Larger Xe atoms settle farther apart but are held more strongly.
2. BThe Xe–Xe pair, as its potential-energy minimum is the deeper of the two Correct
   The depth of the minimum is the energy needed to separate the pair: about 2.3 kJ/mol for Xe–Xe and about 1.2 kJ/mol for Ar–Ar. Xe atoms have more electrons and a larger, more polarizable electron cloud, so the dispersion forces between them are stronger.
3. CThe Ar–Ar pair, as its minimum energy is the less negative value
   A student who reads a less negative energy as 'more' energy picks this. Separating the Ar pair from its minimum to large distance takes about 1.2 kJ/mol, while the Xe pair takes about 2.3 kJ/mol, so the deeper Xe–Xe well means the stronger attraction.
4. DNeither; both are held by dispersion forces, which are equally strong
   A student who thinks attractions of the same type are equally strong picks this. Both pairs are held by dispersion forces, but the graph shows a well about twice as deep for Xe–Xe, because Xe is more polarizable.

Working Read the depth of each potential-energy minimum below zero: Ar–Ar about −1.2 kJ/mol at about 0.38 nm; Xe–Xe about −2.3 kJ/mol at about 0.44 nm. The energy needed to separate a pair from its minimum to large distance is about 1.2 kJ/mol for Ar–Ar and about 2.3 kJ/mol for Xe–Xe, so the Xe–Xe attraction is the stronger (about twice as strong). The position of the minimum (0.38 nm vs 0.44 nm) gives the equilibrium distance, not the strength.

CED 3.1.A.1.ii · Read this in Fix

1. AY molecules attract one another more strongly, as pi bonds hold X's electrons more tightly.
   A student who thinks multiple bonds make a molecule less polarizable picks this. A double or triple bond is stronger overall, but its pi electrons are more easily distorted than sigma electrons, so they increase polarizability.
2. BX and Y molecules attract one another equally, as both have the same number of electrons.
   A student who thinks the number of electrons alone sets the strength of dispersion forces picks this. Polarizability also depends on how the electrons are held; pi bonding makes X more polarizable than Y.
3. CNeither X nor Y molecules attract one another, as both compounds are nonpolar.
   A student who thinks nonpolar molecules do not attract one another picks this. London dispersion forces act between all molecules, including nonpolar ones.
4. DX molecules attract one another more strongly, as their pi electrons are more polarizable. Correct
   With the number of electrons, size and shape the same, the difference is the bonding. Pi electrons are held above and below the bond axis and are more easily displaced, so pi bonding enhances polarizability and the dispersion forces between X molecules are stronger.

CED 3.1.A.1.ii · Read this in Fix

1. AVan der Waals forces include dipole-dipole forces and dispersion forces. Correct
   'Van der Waals forces' is a collective term for the attractions between neutral molecules, including dipole-dipole, dipole-induced dipole and London dispersion forces. HCl is polar, so dipole-dipole forces act between its molecules in addition to dispersion forces; saying there are 'only van der Waals forces' does not rule them out.
2. BThe reasoning is sound, since van der Waals forces are the London dispersion forces.
   A student who treats 'van der Waals forces' and 'London dispersion forces' as the same thing picks this. The terms are not synonyms; London dispersion forces are only one kind of van der Waals force.
3. CHCl molecules hydrogen bond to one another rather than being held by van der Waals forces.
   A student who thinks H bonded to Cl can hydrogen bond picks this. Hydrogen bonding needs H bonded to N, O or F; HCl molecules attract one another by dipole-dipole and dispersion forces.
4. DThe reasoning is sound, since HCl is a linear molecule and so has no dipole.
   A student who thinks every linear molecule is nonpolar picks this. HCl has a single polar bond, with δ− on the more electronegative Cl atom, so the molecule is polar.

CED 3.1.A.1.iii · Read this in Fix

1. Aan H₂S molecule and an Ar atom
   A student who thinks a dipole has no effect on a nonpolar particle picks this. H₂S is polar, and its dipole induces a dipole in the Ar atom, so a dipole-induced dipole interaction acts in addition to dispersion forces.
2. Ba CO₂ molecule and an Ar atom Correct
   CO₂ is linear with equal, opposite bond dipoles, so it has no dipole moment, and an Ar atom is nonpolar. With no permanent dipole on either particle, the only attraction is London dispersion forces.
3. Ctwo adjacent HCl molecules
   A student who thinks all linear molecules are nonpolar picks this. HCl is polar, so dipole-dipole forces act between HCl molecules in addition to dispersion forces.
4. Dtwo neighboring CH₂Cl₂ molecules
   A student who judges polarity from a flat Lewis diagram picks this. CH₂Cl₂ can be drawn with its Cl atoms opposite each other, but the molecule is tetrahedral and its bond dipoles do not cancel, so it is polar and dipole-dipole forces act between its molecules.

CED 3.1.A.2 · Read this in Fix

1. AIt is incorrect: Ar's electrons shift toward the H atom, so the near side of Ar becomes δ−. Correct
   The partially positive H end of HCl attracts the electrons of the Ar atom, so the electron cloud shifts toward HCl: the near side of Ar becomes δ− and the far side δ+. The induced dipole is oriented so that the HCl molecule and the Ar atom attract; the student's model has it reversed.
2. BIt is correct: the side of Ar facing the δ+ H atom takes on the same δ+ charge as H does.
   A student who thinks a partial charge is copied onto the side it faces picks this. Charge is not passed on; the δ+ H atom attracts Ar's electrons, so the near side becomes δ−.
3. CIt is incorrect: Ar's filled outer shell keeps HCl from distorting its electron cloud at all.
   A student who thinks filled-shell atoms cannot be polarized picks this. A filled shell resists gaining or losing electrons, but a nearby dipole can still shift Ar's electron cloud, inducing a dipole.
4. DIt is correct: a polar and a nonpolar particle repel, so like partial charges face each other.
   A student who thinks polar and nonpolar particles repel picks this. A dipole-induced dipole interaction is always attractive, because the induced dipole always forms with its unlike charge facing the polar molecule.

CED 3.1.A.2.i · Read this in Fix

1. AReplace the Ar atom with a Ne atom
   A student who thinks smaller particles are always attracted more strongly picks this. Ne can approach slightly closer, but with only 10 electrons it is much less polarizable than Ar, so the dipole induced in it, and the attraction, are weaker.
2. BReplace the H₂O molecule with H₂Se
   A student who thinks larger molecules are more polar picks this. H₂Se is larger than H₂O but much less polar, because Se is far less electronegative than O; its smaller dipole induces a weaker dipole in Ar.
3. CReplace the H₂O molecule with CCl₄
   A student who thinks any molecule with polar bonds is polar picks this. The four C–Cl bond dipoles of tetrahedral CCl₄ cancel, so CCl₄ has no dipole moment and cannot induce a dipole in Ar; only dispersion forces would act.
4. DReplace the Ar atom with a Xe atom Correct
   The strength of a dipole-induced dipole attraction increases with the size of the dipole of the polar molecule and with the polarizability of the nonpolar particle. Xe (54 electrons) has a larger, more easily distorted electron cloud than Ar (18 electrons), so the water molecule induces a larger dipole in it. Each of the other changes weakens or removes the attraction.

CED 3.1.A.2.i · Read this in Fix

1. AMeasure how much Ar dissolves in water at several temperatures, keeping the same pressure
   A student who thinks changing any condition tests the claim picks this. Only the temperature changes; the polarizability of the nonpolar particle (Ar) is the same in every trial, so the procedure does not test the claim.
2. BMeasure how much of each noble gas, He to Xe, dissolves in water at one temperature and pressure Correct
   Polarizability increases from He to Xe, and the polar molecule (water), the temperature and the pressure are kept the same. The only factor that differs is the polarizability of the nonpolar particle, which is the variable named in the claim.
3. CMeasure how much Ar dissolves in water and in hexane at one temperature and pressure
   A student who treats polarity and polarizability as the same property picks this, changing the polarity of the solvent. The claim concerns the polarizability of the nonpolar particle, which is Ar in both trials, so the procedure does not test it.
4. DCompare the boiling points of the noble gases He to Xe with their numbers of electrons
   A student who thinks data about attractions between identical particles show attraction to water picks this. Boiling points of the noble gases reflect dispersion forces between the gas atoms themselves; water is not involved, so the procedure cannot test a claim about attraction to water.

CED 3.1.A.2.i · Read this in Fix

1. ACH₃CN boils highest because its N atom and its H atoms let its molecules hydrogen bond.
   A student who thinks H atoms on carbon can hydrogen bond picks this. In CH₃CN all three H atoms are bonded to C, so there is no H bonded to N, O or F and no hydrogen bonding; its high boiling point comes from its large dipole moment.
2. BC₃H₈ boils lowest because no attractive forces at all act between its nonpolar molecules.
   A student who thinks nonpolar molecules do not attract one another picks this. Propane can be liquefied and has a boiling point, so its molecules attract one another, through London dispersion forces.
3. CBoiling point rises with dipole moment, as dipole-dipole forces add to dispersion forces of similar size. Correct
   The molar masses (41–50 g/mol) are similar, so the dispersion forces are of similar size. Boiling point rises in the same order as dipole moment, from C₃H₈ (0.1 debye, −42 °C) to CH₃CN (3.9 debye, 82 °C), consistent with dipole-dipole forces that act in addition to dispersion forces and grow with the size of the dipole.
4. DBoiling breaks covalent bonds within each molecule, which are weakest in C₃H₈ molecules.
   A student who thinks boiling breaks covalent bonds picks this. The vapor of each substance consists of intact molecules; boiling overcomes only the attractions between molecules.

CED 3.1.A.2.ii · Read this in Fix

1. APolar ICl molecules attract by dipole-dipole forces, while dispersion forces act only between Br₂ molecules.
   A student who thinks polar molecules have dipole forces instead of dispersion forces picks this. Dispersion forces act between all molecules, including ICl molecules; the dipole-dipole forces are added to them.
2. BICl molecules are heavier than Br₂ molecules, and heavier molecules attract one another more strongly.
   A student who thinks mass causes intermolecular attraction picks this. ICl (162 g/mol) is only slightly heavier than Br₂ (160 g/mol); attractions between molecules are Coulombic, and the difference comes from the dipole-dipole forces between polar ICl molecules.
3. CThe I–Cl bond is stronger than the Br–Br bond, so more energy is needed to boil liquid ICl.
   A student who thinks boiling breaks covalent bonds picks this. The I–Cl bond is indeed somewhat stronger than the Br–Br bond, but neither bond breaks when the liquid boils; only the attractions between molecules are overcome.
4. DICl molecules are polar, so dipole-dipole forces act between them in addition to dispersion forces. Correct
   With the same number of electrons and similar sizes, Br₂ and ICl have dispersion forces of similar strength. ICl is polar (Cl is more electronegative than I), so dipole-dipole forces act between ICl molecules as well, and more energy is needed to separate them. Br₂ has only dispersion forces.

CED 3.1.A.2.ii · Read this in Fix

1. ACH₄ and H₂O, both with 10 electrons
   A student who thinks a hydrogen bond is the O–H bond inside a molecule picks this, believing that water has only dipole-dipole forces between its molecules. H₂O molecules hydrogen bond to one another, so this pair does not isolate dipole-dipole forces.
2. BBr₂ and ICl, each having 70 electrons Correct
   Br₂ and ICl molecules have the same number of electrons (70) and are diatomic molecules of similar size, so their dispersion forces are about the same. Br₂ is nonpolar; ICl is polar because Cl is more electronegative than I. Neither can hydrogen bond, so the difference in their boiling points (59 °C and 97 °C) shows the effect of dipole-dipole forces.
3. CF₂ and Ar, each with 18 electrons
   A student who thinks molecules made of highly electronegative atoms are polar picks this, treating F₂ as polar and Ar as nonpolar. F₂ consists of two identical atoms, so it is nonpolar; neither substance is polar, so the pair cannot show an effect of dipoles.
4. DC₂H₆ and Ar, which both have 18 electrons
   A student who thinks hydrocarbons are polar picks this. C–H bonds are effectively nonpolar, so C₂H₆ is nonpolar like Ar; the pair has no polar member.

Working Quantities needed: the number of electrons and the size of the molecules (to keep dispersion forces about the same), whether each molecule is polar, and whether it can hydrogen bond. Br₂ (70 e⁻, nonpolar, 59 °C) and ICl (70 e⁻, polar, no H atoms, 97 °C) are diatomic molecules of similar size that differ in polarity, so the 38 °C difference shows the effect of dipole-dipole forces. CH₄/H₂O includes hydrogen bonding; F₂/Ar and C₂H₆/Ar contain no polar molecule.

CED 3.1.A.2.ii · Read this in Fix

1. A0.50 Correct
   F ∝ q₁q₂/r². Going from arrangement 1 to arrangement 2, the cation's charge doubles (×2) and the distance doubles, which divides the force by 2² = 4. The force changes by a factor of 2/4 = 0.50.
2. B2.00
   A student who thinks the force depends only on the charges picks this, multiplying by 2 for the doubled charge and ignoring the doubled distance.
3. C1.00
   A student who takes the force to be inversely proportional to the distance, not its square, picks this: 2/2 = 1.00. Doubling r divides the force by 4, not by 2.
4. D0.25
   A student who thinks the force depends only on the distance picks this: doubling r gives 1/4 = 0.25, but the doubled charge of the cation also doubles the force.

Working F ∝ q₁q₂/r². Arrangement 1: q = 1+, distance r. Arrangement 2: q = 2+, distance 2r; the partial charge δ− is the same. F₂/F₁ = (2/1) × (r/2r)² = 2 × 1/4 = 0.50. Distractors: charge only, 2/1 = 2.00; inverse (not inverse-square) distance, 2/2 = 1.00; distance only, (1/2)² = 0.25.

CED 3.1.A.2.iii · Read this in Fix

1. ANa⁺ forms a covalent bond with the O atom, and this is stronger than any attraction to HCl.
   A student who thinks ions form covalent bonds with water picks this. Na⁺ is attracted to water's partially negative O atom by an ion-dipole force; no electrons are shared.
2. BHCl molecules are nonpolar, so only the Na⁺ ion has a charge that can attract the O atom.
   A student who thinks every linear molecule is nonpolar picks this. HCl is polar, with δ+ on H, so it is attracted to water's O atom, but less strongly than Na⁺ is.
3. CNa⁺ has a full positive charge, larger than the partial positive charge on HCl's H atom. Correct
   With r the same, the force depends on the product of the charges. The full 1+ charge of Na⁺ is larger than the δ+ partial charge on the H atom of HCl, so q₁q₂, and therefore the force, is larger for Na⁺. This is why ion-dipole forces tend to be stronger than dipole-dipole forces.
4. DNa⁺ is a much heavier particle than the H atom, and heavier particles attract more strongly.
   A student who thinks mass causes the attraction picks this. Coulomb's law contains the charges and the distance, not the masses; the larger charge of Na⁺ is what makes the attraction stronger.

Working F ∝ q₁q₂/r² with r the same for both: q₂ (the δ− on O) is the same, so F is proportional to the charge of the particle facing O. Na⁺ carries a full 1+ charge; the H atom of HCl carries only a partial positive charge δ+ (less than 1+). Therefore F(Na⁺) > F(HCl).

CED 3.1.A.2.iii · Read this in Fix

1. AAn ion-dipole attraction between X⁻ and a partially positive H atom of water Correct
   X⁻ is an anion, and water is polar with δ+ H atoms, so each water molecule turns an H atom toward X⁻. The attraction between the ion's negative charge and the partial positive charge on H is an ion-dipole force.
2. BA covalent bond between X⁻ and an H atom, formed as X⁻ reacts with water
   A student who thinks ions form covalent bonds with water picks this. The water molecules keep their own O–H bonds; they are attracted to X⁻, but no electron pair is shared with it.
3. CAn ionic bond between X⁻ and an H⁺ ion that is present in each water molecule
   A student who treats partial charges as full ionic charges picks this. The H atoms of a water molecule carry partial positive charges; they are covalently bonded to O, not present as H⁺ ions.
4. DOnly a dispersion force, as a neutral water molecule cannot attract an ion
   A student who thinks an ion cannot attract a neutral molecule picks this. Water is neutral overall, but its δ+ H atom is closer to X⁻ than its δ− O atom, so there is a net Coulombic attraction.

CED 3.1.A.2.iii · Read this in Fix

1. AIt is correct: water molecules turn their O atoms toward any ion that they surround.
   A student who thinks water always points its O atom at an ion picks this. A δ− O atom is attracted to a cation but repelled by an anion, so water molecules face Cl⁻ with an H atom.
2. BIt is incorrect: around Na⁺, the H atoms of each water molecule should face the ion instead.
   A student who reverses the partial charges of water picks this. O is the δ− end of water, so the O atoms correctly face the Na⁺ cation; the error is around Cl⁻.
3. CIt is incorrect: each ion should be shown covalently bonded to the water molecules nearby.
   A student who thinks ions form covalent bonds with water picks this. The ions attract the water molecules by ion-dipole forces; no electrons are shared.
4. DIt is incorrect: around Cl⁻, a partially positive H atom of each water molecule should face the ion. Correct
   O is more electronegative than H, so in water O is δ− and the H atoms are δ+. The O atoms correctly face Na⁺, but around the anion Cl⁻ the water molecules should be turned so that a δ+ H atom faces the ion.

CED 3.1.A.3 · Read this in Fix

1. AIt leaves out the attraction due to the masses of the ion and the molecule, which is larger than the Coulombic one.
   A student who thinks mass causes attractions between particles picks this. Attractions between ions and molecules are Coulombic; any gravitational attraction between them is negligible.
2. BIt leaves out the covalent bond that forms between Na⁺ and the O atom when the ion dissolves in water.
   A student who thinks ions form covalent bonds with water picks this. Na⁺ is attracted to water by an ion-dipole force; no covalent bond forms.
3. CIt leaves out the repulsion between Na⁺ and the H atoms' partial positive charges, which depends on orientation. Correct
   A water molecule carries δ+ charges on its H atoms as well as the δ− charge on O. Na⁺ repels the δ+ H atoms, so the net attraction depends on which end of the molecule faces the ion. A model with only the O charge cannot show this orientation dependence.
4. DIt treats the charge on O as partial, but the O atom in a water molecule carries a full 2− charge.
   A student who treats partial charges as full ionic charges picks this. Water is a covalent molecule; its O atom carries a partial negative charge, smaller than the charge of an electron.

CED 3.1.A.3 · Read this in Fix

1. ACH₃CH₂F
   A student who thinks any H atom can hydrogen bond picks this. In fluoroethane, CH₃CH₂F, all the H atoms are bonded to C; although each molecule contains F, no H is bonded to N, O or F, so there is no hydrogen bonding.
2. BHCl
   A student who thinks H bonded to Cl can hydrogen bond picks this. Hydrogen bonding needs H bonded to N, O or F; HCl molecules attract by dipole-dipole and dispersion forces.
3. CCH₃CH₂NH₂ Correct
   In ethylamine, CH₃CH₂NH₂, two H atoms are covalently bonded to N, and each N atom has a lone pair, so an N–H hydrogen of one molecule is attracted to the N atom of another: hydrogen bonding.
4. DC₂H₆
   A student who thinks a hydrogen bond is an attraction between H atoms picks this. Ethane, C₂H₆, contains no N, O or F, so its molecules attract only by dispersion forces.

CED 3.1.A.4 · Read this in Fix

1. AHydrogen bonds form between H₂O molecules but not between those of the other group 16 hydrides. Correct
   In the group 14 hydrides and in H₂S, H₂Se and H₂Te, boiling point rises with the number of electrons, as dispersion forces grow. H₂O, the smallest group 16 hydride, breaks the trend because its H atoms are bonded to O, so H₂O molecules hydrogen bond to one another; S, Se and Te are not among N, O and F.
2. BBoiling H₂O breaks its O–H bonds, which are stronger than the bonds in the other group 16 hydrides.
   A student who thinks boiling breaks covalent bonds picks this. The O–H bond is stronger than the S–H bond, but water vapor consists of intact H₂O molecules; boiling overcomes only the attractions between molecules.
3. CEach H₂O molecule is held together by two hydrogen bonds, its O–H bonds, and these are hard to break.
   A student who thinks the O–H bond is a hydrogen bond picks this. The O–H bonds within a water molecule are covalent bonds; hydrogen bonds are the attractions between an H atom of one molecule and the O atom of another.
4. DOnly H₂O molecules are polar, so only H₂O molecules attract one another by dipole forces.
   A student who thinks molecules with only slightly polar bonds are nonpolar picks this. H₂S, H₂Se and H₂Te are bent and polar too, though less polar than H₂O; dipole-dipole forces alone cannot explain how far H₂O lies above the trend.

CED 3.1.A.4 · Read this in Fix

1. ABoth will boil at nearly the same temperature, as they have identical molar masses.
   A student who thinks equal molar mass means equal intermolecular forces picks this. Equal molar mass makes the dispersion forces similar, but only ethanol can hydrogen bond.
2. BEthanol will boil much higher, as only its molecules have H bonded to O. Correct
   Ethanol has an H atom bonded to O, so its molecules hydrogen bond to one another; in dimethyl ether every H atom is bonded to C, so it cannot. The two have the same number of electrons, so if hydrogen bonding has a large effect, ethanol should boil at a much higher temperature.
3. CEthanol will boil much higher, as dimethyl ether's symmetric molecules are nonpolar.
   A student who judges polarity from a flat drawing picks this. Dimethyl ether looks symmetrical when drawn flat, but it is bent at O and is polar; the difference the hypothesis predicts comes from hydrogen bonding in ethanol.
4. DEthanol will boil much higher, as boiling it must break its strong O–H bonds.
   A student who thinks boiling breaks covalent bonds picks this. Ethanol vapor consists of intact molecules; the O–H bond matters because it allows hydrogen bonds between molecules, not because it breaks.

CED 3.1.A.4 · Read this in Fix

1. ACovalent bonds along each strand break, so the double helix falls apart into many small fragments.
   A student who thinks separating the strands means breaking covalent bonds picks this. Each strand stays intact; only the noncovalent attractions between the two strands are overcome.
2. BHydrogen bonds between bases on opposite strands are overcome; the covalent bonds of each strand remain. Correct
   The two strands are separate molecules held together by noncovalent attractions, including hydrogen bonds between bases on opposite strands. Heating gives the strands enough energy to overcome these attractions, while the covalent bonds within each strand are unaffected.
3. CThe N–H and O–H covalent bonds in the bases break, and these are the hydrogen bonds between strands.
   A student who thinks hydrogen bonds are the covalent N–H and O–H bonds picks this. Hydrogen bonds are attractions between such an H atom and an N or O atom on the other strand; the covalent N–H bonds of the bases remain (the bases contain no O–H bonds).
4. DWater molecules form covalent bonds with the bases, which pushes the two strands apart.
   A student who thinks molecules attracted to water form covalent bonds with it picks this. Water molecules interact with the bases through noncovalent attractions; no covalent bonds form between water and DNA.

CED 3.1.A.5 · Read this in Fix