AP Unit 2 Solutions

1. C–H has a small electronegativity difference and is treated as effectively nonpolar in the AP model.
2. O–H is polar covalent with Oδ− and Hδ+.
3. The second sample supports an ionic model: ions held in a lattice are immobile as a solid but mobile when molten.
4. Across Mg to S, nuclear charge rises while valence electrons enter the same principal shell with broadly similar core shielding; increased effective attraction generally raises electronegativity.
5. From O to S, the valence shell is farther from the nucleus and more core-shielded; this generally reduces attraction to shared electron density despite more protons.
6. H–Cl has ΔEN = 1.0 and H–Br has ΔEN = 0.8. Both have Hδ+ and the halogen δ−. The model predicts greater charge separation, ionic character and bond dipole for H–Cl; this is a qualitative comparison, not an exact dipole calculation.
7. Unequal electron sharing produces partial charge separation, so polar covalent bonds include ionic character. The distribution changes continuously rather than switching at one arbitrary ΔEN cutoff. Bond length and the actual electron distribution are needed to determine an exact dipole; partial charges are not full monatomic-ion charges.

1. A is consistent with a metallic model: delocalized electrons are mobile in the solid and the cohesive metal arrangement tolerates sliding.
2. B is consistent with an ionic model: solid ions are fixed, and displaced layers can align like charges and fracture; melted ions carry charge.
3. Both can have mobile charge carriers when molten, so that single observation is insufficient.

1. Create a graph with internuclear distance in pm horizontally and potential energy in kJ mol⁻¹ vertically. Draw the separated-atom zero reference, steeply rising short-distance regions, the labeled minima X and Y, and curves approaching zero from below at large distance.
2. The shorter, deeper curve Y is consistent with greater bond order C=C; X with C–C.
3. Dissociation takes Y from −610 to zero: ΔE = +610 kJ mol⁻¹.

1. The charge-product ratio is 4.
2. The inverse-square distance factor is 1/(0.80)², so F_B/F_A = 4/0.64 = 6.25.
3. This pairwise force comparison supports stronger interactions in B. Lattice energy is an energy involving all interactions and structure, so it is not the force ratio multiplied by an arbitrary energy.

1. Six Al³⁺ ions contribute +18 charge units; nine oxide ions contribute −18. The sketch must label six Al³⁺ and nine O²⁻ ions in an extended arrangement, favor unlike-charge contacts, and state that the picture is a slice of a three-dimensional structure. Do not draw separate Al₂O₃ molecules.
2. A represents separate Ar atoms. B represents discrete CO₂ molecules: each separated group contains one C and two O atoms joined within that group.
3. C represents an extended ionic sample with a 1:1 Na⁺:Cl⁻ ratio. Its formula is NaCl; the formula unit records that simplest ratio, not a separable two-ion molecule. Each ion interacts with more than one neighbor, and the repeated structure continues across any arbitrarily drawn pair boundary.
4. Two macroscopic pieces of pure C have the same 1:1 elemental number ratio and the same elemental mass percentages. Changing sample size changes total mass and ion counts, but not the proportions of a pure compound. Equal Na/Cl counts do not imply equal Na/Cl masses because their atomic masses differ.

1. Sliding layers can place like charges close together, causing repulsion and fracture.
2. Melting releases ions from fixed lattice sites, allowing charge transport.
3. A two-dimensional picture omits many three-dimensional neighbors and cannot establish an exact crystal structure.

1. In the first created sketch, represent A replacing some host sites: substitutional alloy.
2. In the second created sketch, represent B in gaps between host cores: interstitial alloy.
3. Include delocalized valence electrons throughout both solids; those mobile electrons carry current.

1. Small carbon atoms in spaces between iron cores impede rearrangement of layers and increase rigidity.
2. Delocalized electrons remain available for charge transport.
3. Exact conductivity also depends on composition, electron scattering, temperature and material structure; the schematic model predicts behavior qualitatively.

1. Count 1 + 4 + 5 = 10 valence electrons.
2. Use H–C–N connectivity. A C≡N triple bond plus H–C single bond spends eight electrons; place the remaining lone pair on N.
3. H has two local electrons, C and N each have an octet. H cannot connect two atoms while retaining its ordinary duet.

1. Available electrons = 5 + 4 − 1 = 8. Four bonds already use all eight.
2. The additional lone pair would give ten total electrons and ten around second-period N, violating both the budget and N’s ordinary octet.
3. Remove the lone pair; bracket four single N–H bonds with overall + charge.

1. There are 4 + 3(6) + 2 = 24 valence electrons. Each contributor has one C=O and two C–O bonds; the double bond can occupy each of three equivalent positions.
2. C formal charge is 4 − 0 − 4 = 0; double-bonded O is 0; single-bonded O atoms each are −1. Sum is −2.
3. Equivalent contributors give average order 4/3 for each C–O position.

1. O=C=O gives FC(C) = 4 − 4 = 0 and each O = 6 − 4 − 2 = 0.
2. In the single/triple model, C has four bond lines and FC = 0. Single O has three lone pairs: −1; triple O has one lone pair: +1.
3. Both obey ordinary octets and the net zero charge, but the equal-double-bond model avoids unnecessary charge separation.

1. Eleven is an odd electron total, so not all electrons can be arranged in pairs.
2. A simple Lewis diagram must accommodate an unpaired electron, and cannot give both atoms ordinary complete paired-electron octets while preserving this budget.
3. Lewis octet rules are useful but have known odd-electron limitations.

1. H₂O has two bonds and two lone pairs: four domains, tetrahedral electron geometry, bent atom geometry, angle below 109.5°, sp³. O–H dipoles do not cancel.
2. BF₃ has three bonds and no central lone pair: three domains, trigonal-planar electron and atom geometry, 120°, sp². Equal planar B–F bond vectors cancel.
3. NH₃ has 5 + 3 = 8 valence electrons. Three N–H bonds use six, leaving one lone pair on N. Four domains give tetrahedral electron geometry and sp³ nomenclature; considering only the three atom positions gives trigonal-pyramidal molecular geometry. Lone-pair repulsion compresses the H–N–H angles below 109.5° (about 107°).
4. NH₃ has a net dipole because its N–H bond contributions do not cancel in the pyramidal arrangement. Three bonded atoms do not imply three total domains when a lone pair is present.

1. Five-domain cases use a trigonal-bipyramidal arrangement: PCl₅ trigonal bipyramidal; SF₄ seesaw; ClF₃ T-shaped. Equatorial lone pairs have fewer 90° close contacts.
2. Six-domain cases use an octahedral arrangement: SF₆ octahedral; BrF₅ square pyramidal; XeF₄ square planar.
3. For XeF₄, the two lone pairs occupy opposite directions, leaving four coplanar atom positions.

A suitable structural sketch is H–C≡C–H in a straight line, with the C–C connection labeled one σ + two π and each C–H labeled one σ. The shown separation is schematic, not a measured length.

1. Each carbon has two domains, one C–H direction and one C≡C direction: a linear sp model with 180° angles. A multiple bond still occupies one overall direction.
2. The triple bond has one σ and two π; two C–H bonds each add one σ. Total: three σ and two π.
3. For this same C–C pair, higher bond order gives shorter, stronger bonding: length ethyne < ethene < ethane, while total C–C dissociation energy ethyne > ethene > ethane. These trends do not rank arbitrary bonds between different atom pairs.
4. Head-on σ overlap places more overlapping density along the line between nuclei than corresponding side-by-side π overlap. Greater overlap generally gives greater stabilization and a larger individual bond-energy contribution in this model.
5. No. A double bond contains both σ and π contributions, so its larger total energy does not establish that π alone is stronger than σ. The π contribution is ordinarily weaker than the corresponding σ contribution, even while adding π bonding strengthens the total connection. These are qualitative model comparisons, not exact independent bond-energy measurements.

1. A C=C includes side-to-side pi overlap as well as sigma overlap. Free rotation disrupts that pi overlap.
2. Each carbon in 2-butene is attached to H and CH₃, two different groups; distinct same-side and opposite-side arrangements are possible.
3. Each ethene carbon has two identical H substituents, so swapping the apparent sides produces the same arrangement.

1. CH₃Cl has four central domains and an sp³ model, but its four substituents are not equivalent; the C–Cl dipole does not cancel the other bond contributions.
2. SO₂ has three central domains; each multiple bond is one domain. Its electron arrangement is trigonal planar and its molecular shape is bent, with an sp² descriptor in this simple model.
3. The bent S–O bond dipoles give a nonzero vector.
4. Br has its valence shell at a higher principal level and generally a larger covalent radius than Cl. With the same H partner and single-bond order, the larger bonded radius supports a longer H–Br bond than H–Cl. Controlling bond order matters because multiple bonding also shortens bonds; radius alone cannot rank arbitrary bond lengths or bond energies.