2.7 VSEPR and Hybridization

1. ARepulsion between the two O atoms, which are farthest apart on opposite sides of C
   A student who thinks shape is set by repulsion between the outer atoms picks this. VSEPR theory treats the electron domains around the central atom as repelling one another; the positions of the O atoms follow from where the domains point.
2. BThe polarity of the two C=O bonds, which holds the two bonds on opposite sides of C
   A student who thinks bond polarity decides shape picks this. VSEPR predicts the same 180° arrangement for any central atom with two domains and no lone pairs, polar bonds or not; the polarity of the bonds plays no part in the prediction.
3. CRepulsion between the electron domains on C, which get as far apart as possible Correct
   VSEPR theory rests on Coulombic repulsion between electrons. The C atom in CO₂ has two electron domains (two double bonds) and no lone pairs; two domains are farthest apart on opposite sides of the atom, so the O atoms at the ends of those domains are 180° apart.
4. DThe Lewis diagram of CO₂, which is written with all three of its atoms in a straight line
   A student who reads shape from the layout of a Lewis diagram picks this. A Lewis diagram can be drawn at any angle on paper; the 180° angle comes from applying VSEPR to the two electron domains around C.

CED 2.7.A.1 · Read this in Fix

1. AThe C≡C bond is stronger, as more electron pairs are shared between the two C nuclei Correct
   Bond order is 3 in ethyne and 1 in ethane. Three shared pairs (one sigma and two pi bonds) hold the two C nuclei together more strongly than one, so the C≡C bond has the greater bond energy.
2. BThe C≡C bond is weaker, as each of its two pi bonds is weaker than a sigma bond
   A student who thinks adding pi bonds makes a bond weaker picks this. Each pi bond is weaker than the sigma bond, but the pi bonds add to the sigma bond, so the triple bond is the stronger bond (839 vs 347 kJ/mol).
3. CThe two bonds are equally strong, as each joins the same two kinds of atom
   A student who thinks bond energy depends only on the two elements picks this. Bonds between the same two atoms differ greatly with bond order: C≡C needs much more energy to break than C–C.
4. DThe C≡C bond is three times as strong, as it consists of three identical C–C bonds
   A student who thinks the bonds within a multiple bond are identical picks this. A triple bond is one sigma and two weaker pi bonds, so its bond energy (839 kJ/mol) is much less than three times that of C–C (347 kJ/mol).

CED 2.7.A.2.iii · Read this in Fix

1. AAtom 1 is sp and atom 2 is sp³, so the ideal angle at atom 2 is 109.5°
   A student who matches the number in the label to the bond order picks this, calling the singly bonded N atom sp and the triply bonded C atom sp³. Atom 1 has four electron domains (sp³) and atom 2 has two (sp, 180°).
2. BAtom 1 is sp³ and atom 2 is sp², so the ideal angle at atom 2 is 120°
   A student who thinks any atom with a multiple bond is sp² picks this. Atom 2 has only two domains, the single bond and the triple bond, so it is sp, with a 180° angle.
3. CAtom 1 is sp³ and atom 2 is sp³, so the ideal angle at atom 2 is 109.5°
   A student who thinks every carbon atom is sp³ because carbon forms four bonds picks this. Atom 2 forms four bonds but has only two electron domains, so it is sp.
4. DAtom 1 is sp³ and atom 2 is sp, so the ideal angle at atom 2 is 180° Correct
   Atom 1 (N) has three bonding domains and one lone pair: four domains, sp³, ideal angles 109.5°. Atom 2 (C) has two domains, the C–C single bond and the C≡N triple bond: sp, ideal angle 180°.

CED 2.7.A.3 · Read this in Fix

1. A4 sigma and 5 pi
   A student who thinks every bond in a multiple bond is a pi bond picks this, counting only the four single bonds as sigma and all five bonds of C=C and C≡N as pi. Each multiple bond contains one sigma bond.
2. B7 sigma and 2 pi
   A student who thinks a multiple bond contains only one pi bond picks this, treating C≡N as two sigma bonds and one pi bond. Only one sigma bond can join two atoms; a triple bond is one sigma and two pi.
3. C9 sigma and 0 pi
   A student who counts every line in the Lewis diagram as a sigma bond picks this. Only one bond between each pair of atoms is a sigma bond; the second bond of C=C and the second and third bonds of C≡N are pi bonds.
4. D6 sigma and 3 pi Correct
   Every pair of bonded atoms is joined by exactly one sigma bond: three C–H, C=C, C–C and C≡N give 6 sigma bonds. The extra bonds are pi bonds: one in C=C and two in C≡N, 3 pi bonds.

Working Sigma bonds: one per pair of bonded atoms = 3 (C–H) + 1 (C=C) + 1 (C–C) + 1 (C≡N) = 6. Pi bonds: 1 (in C=C) + 2 (in C≡N) = 3. All-pi count: sigma = 3 + 1 = 4 single bonds, pi = 2 + 3 = 5. One-pi-per-multiple-bond count: sigma = 3 + 1 + 1 + 2 = 7, pi = 1 + 1 = 2. Every line sigma: 3 + 2 + 1 + 3 = 9 sigma, 0 pi.

CED 2.7.A.4 · Read this in Fix

1. AIt increases, as fewer bonded atoms around the central atom can spread farther apart
   A student who ignores lone pairs when predicting shape picks this. N and O carry one and two lone pairs, which are electron domains; they take up positions around the central atom and push the bonds together, so the angle falls.
2. BIt increases, as bonding pairs repel each other more strongly than lone pairs repel them
   A student who thinks bonding pairs repel more strongly than lone pairs picks this. A lone pair, held by one nucleus, spreads out close to the central atom and repels bonding pairs more strongly, so replacing bonding pairs with lone pairs makes the angle smaller.
3. CIt stays at 109.5°, as four electron domains give a tetrahedral molecular shape
   A student who names the molecular geometry from the arrangement of all domains picks this. Four domains give a tetrahedral arrangement of domains, but NH₃ is trigonal pyramidal and H₂O is bent, and the stronger repulsion from lone pairs makes their angles smaller than 109.5°.
4. DIt decreases, as lone pairs repel bonding pairs more strongly than bonding pairs repel each other Correct
   All three central atoms have four electron domains in a tetrahedral arrangement. CH₄ has no lone pairs (109.5°), NH₃ has one and H₂O has two. Lone pairs repel the bonding pairs more strongly than bonding pairs repel one another, so the angle shrinks: about 107° in NH₃ and 104.5° in H₂O.

CED 2.7.A.1 · Read this in Fix

1. AValid: a Lewis diagram shows true bond angles, so NH₃ is flat, with H–N–H angles of 90° and 180°
   A student who reads shape from the layout of a Lewis diagram picks this. The drawing shows which atoms are bonded and that N has a lone pair; its angles on paper are arbitrary. NH₃ is not flat.
2. BNot valid: N has four electron domains, so NH₃ is trigonal pyramidal, with angles near 107° Correct
   The Lewis diagram is needed to find the domains (three bonding pairs and one lone pair on N), but VSEPR theory must then be applied: four domains are arranged tetrahedrally in three dimensions, and with one of them a lone pair the atoms form a trigonal pyramid. The lone pair squeezes the H–N–H angles to about 107°.
3. CNot valid: the three H atoms spread out evenly around N, so NH₃ is trigonal planar with 120° angles
   A student who ignores the lone pair picks this. The lone pair on N is a fourth electron domain, which pushes the three N–H bonds out of a plane into a pyramid.
4. DNot valid: the four electron domains around N make NH₃ a tetrahedral molecule, with 109.5° angles
   A student who names the molecular geometry from the arrangement of all domains picks this. The domains are arranged tetrahedrally, but the molecule (the atoms only) is trigonal pyramidal, and its H–N–H angles are smaller than 109.5° because the lone pair repels more strongly.

CED 2.7.A.2 · Read this in Fix

1. ANCl₃ and BCl₃ Correct
   B in BCl₃ has three bonding domains and no lone pairs, so BCl₃ is trigonal planar. N in NCl₃ has three bonding domains and one lone pair; the four domains are arranged tetrahedrally and the molecule is trigonal pyramidal. Same formula type, different shapes.
2. BCO₂ and XeF₂
   A student who thinks lone pairs on a central atom always bend a molecule picks this, expecting XeF₂ to be bent. Xe has two bonding domains and three lone pairs; the lone pairs take the three equatorial positions of a trigonal bipyramid, leaving the F atoms on opposite sides, so XeF₂ is linear, like CO₂.
3. CSO₂ and H₂O
   A student who names molecular geometry from the arrangement of all domains picks this, since S in SO₂ has three domains (trigonal planar arrangement) and O in H₂O four (tetrahedral). Both molecules have two bonded atoms and lone pairs on the central atom, so both are bent.
4. DBCl₃ and CO₃²⁻
   A student who counts each bond of a double bond as a separate domain picks this, giving C in CO₃²⁻ four domains. A double bond is one domain, so C has three domains and no lone pairs, and CO₃²⁻ is trigonal planar, like BCl₃.

CED 2.7.A.2.i · Read this in Fix

1. Ax ≈ 120° and y ≈ 180°
   A student who ignores lone pairs picks this, treating the O atom, with only two bonded atoms, as linear. The two lone pairs on O are electron domains, so O has four domains and the C–O–H angle is near 109.5°.
2. Bx ≈ 109.5° and y ≈ 109.5°
   A student who counts each bond of a double bond as a separate domain picks this, giving C four domains. The C=O double bond is one domain, so C has three domains and angle x is near 120°.
3. Cx ≈ 120° and y ≈ 109.5° Correct
   The C atom has three electron domains (C–H, C=O counted as one domain, and C–O) and no lone pairs, so the angles around it are close to 120°. The O atom bonded to H has two bonding domains and two lone pairs, four domains in all, so the C–O–H angle is close to the tetrahedral 109.5° (a little less, because of the lone pairs).
4. Dx ≈ 90° and y ≈ 180°
   A student who reads bond angles from the layout of a Lewis diagram picks this. The diagram is drawn with right angles and straight lines for neatness; the real angles come from the electron domains around each atom.

CED 2.7.A.2.ii · Read this in Fix

1. ASO₂ has an S–O single bond and an S=O double bond, which pull the two O atoms over toward one side
   A student who thinks a single Lewis diagram of a species with resonance shows its real bonds picks this. SO₂ has two identical S–O bonds (resonance), and a difference in bonds would not by itself bend a molecule; the bend comes from the lone pair on S.
2. BS carries a lone pair as well as two bonded O atoms, and these three domains bend the molecule Correct
   In every Lewis diagram of SO₂, S has a lone pair in addition to its two bonding domains, so S has three electron domains in a trigonal planar arrangement and the molecule is bent, with an angle close to 120°. C in CO₂ has two domains and no lone pair, so CO₂ is linear.
3. CThe two O atoms repel each other less across the larger S atom, so they need not lie opposite
   A student who thinks shape is set by repulsion between the outer atoms picks this. VSEPR theory explains shape by repulsion between electron domains; the size of S does not explain why SO₂ is bent and CO₂ is not.
4. DS has three electron domains around it, so SO₂ has a trigonal planar molecular geometry
   A student who names the molecular geometry from the arrangement of all domains picks this. Three domains are arranged trigonal planar, but only two of them hold atoms, so the molecule SO₂ is bent; a three-atom molecule cannot be trigonal planar.

CED 2.7.A.2.ii · Read this in Fix

1. ACurve 2, as its shallower minimum shows that more energy is needed to break the H–Br bond
   A student who reads a less negative minimum as 'more energy' picks this. Breaking the bond means climbing from the minimum to zero: about 366 kJ/mol for curve 2 and about 431 kJ/mol for curve 1, so curve 2 is the weaker bond. Curve 2 is HBr, but not for this reason.
2. BCurve 2, as the larger Br atom holds the shared pair farther away and less tightly Correct
   Br is in the period below Cl, so its valence electrons are in a higher shell and the atom is larger. The H–Br bond is therefore longer (minimum at about 141 pm rather than 127 pm) and weaker (shallower minimum, about −366 rather than −431 kJ/mol): curve 2.
3. CCurve 1, as Br has more electrons than Cl and so forms the stronger bond, with the deeper well
   A student who thinks atoms with more electrons form stronger bonds picks this. H–X bonds get weaker from HCl to HBr because Br is larger and holds the shared pair farther from its nucleus.
4. DCurve 1, as the Br nucleus has more protons and so pulls the H atom closer to it
   A student who applies Coulomb's law to nuclear charge alone picks this. The extra protons of Br are shielded by extra inner electrons, and its valence shell is farther out, so Br is larger than Cl and the H–Br bond is longer, not shorter.

CED 2.7.A.2.iv · Read this in Fix

1. AThe carbon–carbon bonds in the molecules C₂H₆, C₂H₄ and C₂H₂ Correct
   These are a single, a double and a triple bond between the same two elements, carbon and carbon. Only the bond order changes, so a trend in their lengths (about 154, 134 and 120 pm) tests the hypothesis directly.
2. BThe C–C bond in C₂H₆, the C=O bond in H₂CO and the N≡N bond in N₂
   A student who accepts a comparison in which several factors change at once picks this. Both the bond order and the bonded elements change, so a trend in bond length could be due to atomic radius rather than bond order.
3. CThe carbon–carbon bonds in the molecules C₂H₆, C₃H₈ and C₄H₁₀
   A student who thinks the factor named in the hypothesis should be held constant picks this. Every carbon–carbon bond in these alkanes is a single bond, so bond order does not change and the data cannot test its effect.
4. DThe three carbon–oxygen bonds within a single carbonate ion, CO₃²⁻
   A student who takes one resonance structure as the real bonding picks this, expecting one C=O and two C–O bonds of different lengths. The three bonds in CO₃²⁻ are identical, each with a bond order of 4/3, so they do not vary in bond order.

CED 2.7.A.2.iv · Read this in Fix

1. ASeF₄
   A student who thinks identical outer atoms always give a nonpolar molecule picks this. Se has four bonding domains and one lone pair; the molecule is seesaw-shaped, and its Se–F bond dipoles do not cancel, so SeF₄ is polar.
2. BPCl₃
   A student who ignores the lone pair on P picks this, picturing a flat, symmetrical molecule. The lone pair makes PCl₃ trigonal pyramidal, so the three P–Cl bond dipoles do not cancel and the molecule is polar.
3. CXeF₄ Correct
   Xe has four bonding domains and two lone pairs; the lone pairs lie opposite each other, and the four F atoms form a square plane around Xe. The four equal Xe–F bond dipoles cancel in pairs, so XeF₄ has no dipole moment.
4. DSCl₂
   A student who reads shape from a Lewis diagram drawn in a straight line, Cl–S–Cl, picks this. S has two lone pairs as well as two bonds, so SCl₂ is bent and its S–Cl bond dipoles do not cancel.

CED 2.7.A.2.v · Read this in Fix

1. AOnly Y, as its two Cl atoms are farther apart, so more charge is separated
   A student who thinks the isomer with its polar atoms farthest apart has the larger dipole moment picks this. In Y the two C–Cl bond dipoles point in opposite directions and cancel (as do the C–H dipoles), so Y has no dipole moment at all.
2. BBoth X and Y, as each of the two molecules contains two polar C–Cl bonds
   A student who thinks any molecule with polar bonds is polar picks this. Polar bonds give a dipole moment only if they do not cancel; in Y they are arranged symmetrically and cancel.
3. CNeither, as both molecules are flat, with their charges spread evenly in one plane
   A student who thinks planar molecules are nonpolar picks this. Both isomers are planar, but in X the C–Cl bond dipoles lie on one side of the molecule and do not cancel, so X is polar.
4. DOnly X, as its two C–Cl bond dipoles point to one side and add to a net dipole Correct
   In X both Cl atoms are on the same side of the rigid C=C bond, so the two C–Cl bond dipoles have components in the same direction that add; X has a dipole moment. In Y the C–Cl bond dipoles point in opposite directions and cancel.

CED 2.7.A.2.v · Read this in Fix

1. AH₂O
   A student who assigns hybridization from the shape name picks this, matching bent H₂O to bent SO₂. O in H₂O has four electron domains (two bonds, two lone pairs), so it is sp³.
2. BSO₂ Correct
   S in SO₂ has two bonding domains (to the two O atoms) and one lone pair: three electron domains, so S is sp² hybridized, with ideal angles of 120°.
3. CNH₃
   A student who assigns hybridization from the number of bonded atoms picks this, since N is bonded to three atoms. N also has a lone pair, so it has four domains and is sp³.
4. DCO₂
   A student who thinks any atom with a multiple bond is sp² picks this. C in CO₂ has two double bonds and no lone pairs: two electron domains, so it is sp hybridized, with a 180° angle.

CED 2.7.A.2.vi · Read this in Fix

1. ACorrect: each C atom forms four bonds in all, so each one has an sp³ arrangement
   A student who thinks every carbon atom is sp³ because carbon forms four bonds picks this. Only the CH₃ carbon has four single bonds; the two carbons of the triple bond each have only two electron domains.
2. BCorrect: the triple bond gives each of its two C atoms four electron domains
   A student who counts each bond of a multiple bond as a separate domain picks this. The triple bond is one domain, so each carbon in it has two domains (the triple bond and one other bond).
3. CIncorrect: each C atom in the triple bond has two electron domains, so it is sp Correct
   The CH₃ carbon has four single bonds (four domains) and is sp³. Each carbon of the triple bond has two domains, the triple bond and one single bond, so it is sp hybridized and the C–C≡C–H chain is linear. The claim is wrong for those two atoms.
4. DIncorrect: the two C atoms in the triple bond are sp², as each forms a multiple bond
   A student who thinks any atom with a multiple bond is sp² picks this. Each carbon of the triple bond has two domains, not three, so it is sp, with 180° bond angles.

CED 2.7.A.2.vi · Read this in Fix

1. AClHC=CHCl exists as two separable compounds Correct
   If the pi bond prevents rotation, the Cl atoms of ClHC=CHCl are fixed on the same side or on opposite sides of the double bond, giving two different compounds (geometric isomers). The C–C single bond of ClH₂C–CH₂Cl has no pi bond, so rotation interconverts its arrangements and there is one compound.
2. BClH₂C–CH₂Cl exists as two separable compounds
   A student who thinks groups drawn on different sides of a single bond are fixed there picks this. Rotation about the C–C single bond is free, so ClH₂C–CH₂Cl is one compound; the hypothesis predicts isomers only where there is a pi bond.
3. CH₂C=CHCl exists as two separable compounds
   A student who thinks every C=C compound has geometric isomers picks this. One carbon of H₂C=CHCl carries two identical H atoms, so exchanging them gives the same molecule; there is one compound.
4. DClC≡CCl exists as two separable compounds
   A student who thinks a triple bond also gives cis and trans isomers picks this. The four atoms Cl–C≡C–Cl lie in a straight line, so there are no 'same side' and 'opposite side' forms; ClC≡CCl is one compound.

CED 2.7.A.4 · Read this in Fix

1. A559 kJ/mol
   A student who thinks the three bonds of a triple bond are identical picks this, giving each 839 ÷ 3 ≈ 280 kJ/mol and the two pi bonds 2 × 280 ≈ 559 kJ/mol. The pi bonds are weaker than the sigma bond, so they cannot be equal thirds.
2. B145 kJ/mol
   A student who thinks a multiple bond has only one pi bond picks this, treating C≡C as two sigma bonds and one pi bond: 839 − 2(347) = 145 kJ/mol. Only one sigma bond joins two atoms.
3. C839 kJ/mol
   A student who thinks all the bonds in a multiple bond are pi bonds picks this, taking the whole triple-bond energy as pi bonding. One of the three bonds is a sigma bond (about 347 kJ/mol), which must be subtracted.
4. D492 kJ/mol Correct
   A triple bond is one sigma bond and two pi bonds. Taking the sigma bond as 347 kJ/mol, the two pi bonds together contribute 839 − 347 = 492 kJ/mol (about 246 kJ/mol each, less than the sigma bond).

Working C≡C = 1 sigma + 2 pi. Sigma ≈ E(C–C) = 347 kJ/mol. Total pi ≈ 839 − 347 = 492 kJ/mol. Identical-bonds belief: 2 × 839/3 = 559 kJ/mol. One-pi belief (2 sigma + 1 pi): 839 − 2 × 347 = 145 kJ/mol. All-pi belief: 839 kJ/mol.

CED 2.7.A.4 · Read this in Fix

1. AF–Cl–F angles of 90° and 120°, in a trigonal bipyramidal molecule
   A student who names the molecular geometry from the arrangement of all domains picks this. The five domains are arranged in a trigonal bipyramid, but two are lone pairs; the three F atoms alone form a T shape.
2. BF–Cl–F angles of about 90° and 180°, in a T-shaped molecule Correct
   Cl has three bonding domains and two lone pairs: five domains in a trigonal bipyramidal arrangement. The lone pairs take two equatorial positions, leaving two axial F atoms and one equatorial F atom: a T shape, with F–Cl–F angles near 90° and 180° (slightly less, because the lone pairs repel strongly).
3. CF–Cl–F angles all equal to 120°, in a trigonal planar molecule
   A student who places lone pairs in the axial positions picks this, leaving the three F atoms in the equatorial plane. Lone pairs go in equatorial positions, where they have fewer 90° neighbors, so the F atoms are two axial and one equatorial.
4. DF–Cl–F angles of about 107°, in a trigonal pyramidal molecule
   A student who treats every molecule with three atoms and lone pairs on the central atom as trigonal pyramidal, like NH₃, picks this. Cl has two lone pairs and five domains in all, not four, so the shape is not a pyramid.

CED 2.7.A.2.i · Read this in Fix

1. A2
   A student who counts only bonded atoms picks this. The three lone pairs on the central I atom are electron domains too.
2. B3
   A student who counts only the lone pairs as electron domains picks this. The two bonds to the outer I atoms are also domains.
3. C5 Correct
   The central I atom has two bonding domains (one to each outer I atom) and three lone pairs, five electron domains in all. They are arranged in a trigonal bipyramid with the lone pairs equatorial, so I₃⁻ is linear.
4. D4
   A student who assumes every central atom has an octet of four electron pairs picks this. The diagram shows five pairs around the central I atom: two bonding pairs and three lone pairs.

Working Domains on the central I = bonded atoms + lone pairs = 2 + 3 = 5. Bonded atoms only: 2. Lone pairs only: 3. Octet assumption: 4.

CED 2.7.A.2.i · Read this in Fix

1. AEach O₃ bond is longer and weaker, since its bond order lies partway between one and two Correct
   O₃ has two equivalent resonance structures, so its two O–O bonds are identical, each with a bond order of 1.5. The bond in O₂ is a double bond (bond order 2), so each O₃ bond is weaker and longer than the O₂ bond.
2. BOne O₃ bond matches the O₂ bond, while the other is a longer, weaker single bond
   A student who takes one resonance structure as the real bonding picks this. The two O–O bonds in O₃ are identical, intermediate between a single and a double bond.
3. CEach O₃ bond is stronger, since it has less of the weak pi bonding found in O₂
   A student who thinks pi bonding makes a bond weaker picks this. Pi bonding adds to the sigma bond, so less pi bonding (bond order 1.5 rather than 2) means a weaker bond, not a stronger one.
4. DEach O₃ bond is equally strong, since both molecules join oxygen atoms to oxygen atoms
   A student who thinks bond energy depends only on the elements bonded picks this. Bonds between two O atoms differ with bond order: O–O, O₃'s bonds and O=O get stronger and shorter in that order.

CED 2.7.A.2.iii · Read this in Fix

1. ABond 2 is stronger, as its orbitals overlap in two regions, above and below the axis
   A student who thinks two regions of overlap mean more overlap picks this. The side-by-side overlap in 2 (a pi bond) is less effective than the head-on overlap in 1 (a sigma bond), so pi bonds have smaller bond energies.
2. BBond 2 is stronger, as its two lobes are two bonds, each with its own shared pair
   A student who reads the two regions of a pi bond as two bonds picks this. The regions above and below the axis together form one pi bond holding one shared pair, and it is weaker than a sigma bond.
3. CBond 1 is stronger, as its orbitals overlap directly on the axis between the nuclei Correct
   In 1 the orbitals overlap head-on, concentrating electron density on the axis between the two nuclei (a sigma bond). In 2 they overlap side by side, above and below the axis (a pi bond), and the overlap is weaker. Stronger overlap gives the sigma bond the greater bond energy.
4. DBonds 1 and 2 are equally strong, as each one is formed by one shared pair of electrons
   A student who thinks the bonds within a multiple bond are identical picks this. Each bond does hold one shared pair, but the head-on overlap of a sigma bond is stronger than the side-by-side overlap of a pi bond, so the two bonds differ in energy.

CED 2.7.A.4 · Read this in Fix