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9.1 9.2 (a) (b) (c) 9.3 (a) (b) 120° Molecular Geometry and Bonding Theories Visualizing Concepts
Removing an atom from the equatorial plane of trigonal bipyramid in Figure 9.3 creates a seesaw shape. It might appear that you could also obtain a seesaw by removing two atoms from the square plane of the octahedron. However, one of the B–A–B angles in the seesaw is 120°, so it must be derived from a trigonal bipyramid. If the blue balloon expands, the angle between red and green balloons decreases. Nonbonding (lone) electron pairs exert greater repulsive forces than bonding pairs, resulting in compression of adjacent bond angles. Square pyramidal Yes, there is one nonbonding electron domain on A. If there were only five bonding domains, the shape would be trigonal bipyramidal. With five bonding and one nonbonding electron domains, the molecule has octahedral domain geometry. Yes. If the B atoms are halogens, each will have three nonbonding electron pairs; there are five bonding pairs, and A has one nonbonded pair, for a total of [5(3) + 5 + 1] = 21 e – pairs and 42 electrons in the Lewis structure. If the five halogens contribute 35 e – , A must contribute seven valence electrons. A is also a halogen. 4 e – domains The molecule has a nonzero dipole moment, because the C–H and C–F bond dipoles do not cancel each other. The dipole moment vector bisects the F–C–F and H–C–H angles, with the negative end of the vector toward the F atoms. The reference point for zero energy on the diagram corresponds to a state where the two Cl atoms are separate and not interacting. This corresponds to an infinite Cl–Cl distance beyond the right extreme of the horizontal axis. The point near the left side of the plot where the curve intersects the xaxis at E = 0 has no special meaning. According to the valencebond model, as atoms approach, their valence atomic orbitals overlap, allowing two electrons of opposite spin to mutually occupy space between the two nuclei. Energy decreases as atom separation decreases because the valence electrons of one atom come close enough to the other atom to be stabilized by both nuclei instead of just one nucleus. (c) 9.4 (a) (b) (c) 9.5 (a) (b) 226 9 Molecular Geometry
(c) (d) (e) Solutions to Exercises The Cl–Cl distance at the energy minimum on the plot is the Cl–Cl bond length. At interatomic separations shorter than the bond distance, the two nuclei begin to repel each other, increasing the overall energy of the system. Energy is shown on the yaxis of the plot. The minimum energy for the two atoms represents the stabilization obtained by bringing two Cl atoms together at the optimum (bond) distance. The ycoordinate of the minimum point on the plot corresponds to the Cl–Cl bond energy, or bond strength. 90° angles are characteristic of hybrids with a d atomic orbital contribution. This pair of orbitals could be sp 3 d or sp 3 d 2 . Angles of 109.5° are characteristic of sp 3 hybrid orbitals only. Angles of 120° can be formed by sp 2 hybrids or sp 3 d hybrids. 9.6 (a) (b) (c) 9.7 9.8 The diagram illustrates mixing of a single s and three p atomic orbitals to form sp 3 hybrids. (a) Recall that π bonds require p atomic orbitals, so the maximum hybridization of a C atom involved in a double bond is sp 2 and in a triple bond is sp. There are 6 C atoms in the molecule. Starting on the left, the hybridizations are: sp 2 , sp 2 , sp 3 , sp, sp, sp 3 . All single bonds are σ bonds. Double and triple bonds each contain 1 σ bond. This molecule has 8 C–H σ bonds and 5 C–C σ bonds, for a total of 13 σ bonds. Double bonds have 1 π bond and triple bonds have 2 π bonds. This molecule has a total of 3 π bonds. (b) (c) 9.9 Analyze/Plan. σ molecular orbitals (MOs) are symmetric about the internuclear axis, π MOs are not. Bonding MOs have most of their electron density in the area between the nuclei, antibonding MOs have a node between the nuclei. (a) (i) (ii) The shape of the molecular orbital (MO) indicates that it is formed by two s atomic orbitals (electron density at each nucleus). σtype MO (symmetric about the internuclear axis, s orbitals can produce only σ overlap). (iii) antibonding MO (node between nuclei) (b) (i) (ii) Shape indicates this MO is formed by two p atomic orbitals overlapping endtoend (node near each nucleus). σtype MO (symmetric about internuclear axis) (iii) bonding MO (concentration of electron density between nuclei) (c) (i) (ii) Shape indicates this MO is formed by two p atomic orbitals overlapping sidetoside (node near each nucleus). πtype MO (not symmetric about internuclear axis, sidetoside overlap) (iii) antibonding MO (node between nuclei) 227 9 Molecular Geometry
9.10 (a) (b) (c) Solutions to Exercises The diagram has five electrons in MOs formed by 2p atomic orbitals. C has two 2p electrons, so X must have three 2p electrons. X is N. The molecule has an unpaired electron, so it is paramagnetic. Atom X is N, which is more electronegative than C. The atomic orbitals of the more electronegative N are slightly lower in energy than those of C. The lower energy π 2 p bonding molecular orbitals will have a greater contribution from the * lower energy N atomic orbitals. (Higher energy π 2p MOs will have a greater contribution from higher energy C atomic orbitals.) Molecular Shapes; the VSEPR Model
9.11 (a) (b) 9.12 (a) Yes. The stated shape, linear, defines the bond angle (180°) and the A–B bond length tells the size. No. Atom A could have 0, 2, 3 or 4 nonbonding electron pairs, depending on the total number of electron domains around atom A. In a symmetrical tetrahedron, the four bond angles are equal to each other, with values of 109.5°. The H–C–H angles in CH 4 and the O–Cl–O angles in ClO 4 – will have values close to 109.5°. ‘Planar’ molecules are flat, so trigonal planar BF3 is flat. In the trigonal pyramidal NH3 molecule, the central N atom sits out of the plane of the three H atoms; this molecule is not flat. An electron domain is a region in a molecule where electrons are most likely to be found. Each balloon in Figure 9.5 occupies a volume of space. The best arrangement is one where each balloon has its “own” space, where they are as far apart as possible and repulsions are minimized. Electron domains are negatively charged regions, so they also adopt an arrangement where repulsions are minimized. The number of electron domains in a molecule or ion is the number of bonds (double and triple bonds count as one domain) plus the number of nonbonding (lone) electron pairs. A bonding electron domain is a region between two bonded atoms that contains one or more pairs of bonding electrons. A nonbonding electron domain is localized on a single atom and contains one pair of nonbonding electrons (a lone pair). (b) 9.13 (a) (b) 9.14 (a) (b) 9.15 Analyze/Plan. Draw the Lewis structure of each molecule and count the number of nonbonding (lone) electron pairs. Note that the question asks ‘in the molecule’ rather than just around the central atom. Solve. (a) (CH3)2S, 20 valence e−, 10 e− pr, 2 nonbonding pairs
H H C H S H C H H 228 9 Molecular Geometry
(b) HCN, 10 valence e−, 5 e− pr, 1 nonbonding pair
H C N Solutions to Exercises (c) H2C2, 10 valence e−, 5 e− pr, 0 nonbonding pairs
H C C H (d) CH3F, 14 valence e−, 7 e− pr, 3 nonbonding pairs
H H C H F 9.16 Analyze/Plan. See Table 9.1. Solve. (a) (c) trigonal planar trigonal bipyramidal (b) (d) tetrahedral octahedral 9.17 The electrondomain geometry indicated by VSEPR describes the arrangement of all bonding and nonbonding electron domains. The molecular geometry describes just the atomic positions. H 2 O has the Lewis structure given below; there are four electron domains around oxygen so the electrondomain geometry is tetrahedral, but the molecular geometry of the three atoms is bent. 9.18 If the electrondomain geometry is trigonal bipyramidal, there are five total electron domains around the central atom. An AB 3 molecule has three bonding domains, so there must be two nonbonding domains on A. Analyze/Plan. See Tables 9.2 and 9.3. Solve. 9.19 229 9 Molecular Geometry Solutions to Exercises (d) X X X
square planar oc tahedral 9.20 (c) X X X tetrahedral bent 9.21 Analyze/Plan. Follow the logic in Sample Exercise 9.1. Solve. bent (b), linear (l), octahedral (oh), seesaw (ss), square pyramidal (sp), square planar (spl), tetrahedral (td), trigonal bipyramidal (tbp), trigonal planar (tr), trigonal pyramidal (tp), Tshaped (T) – 230 9 Molecular Geometry
+ Solutions to Exercises :
*More than one resonance structure is possible. All equivalent resonance structures predict the same molecular geometry. 9.22 bent (b), linear (l), octahedral (oh), seesaw (ss) square pyramidal (sp), square planar (spl), tetrahedral (td), trigonal bipyramidal (tbp), trigonal planar (tr), trigonal pyramidal (tp), Tshaped (T) *More than one resonance structure is possible. All equivalent resonance structures predict the same molecular geometry. 9.23 Analyze/Plan Work backwards from molecular geometry, using Tables 9.2 and 9.3. Solve. (a) (b) Electrondomain geometries: i, trigonal planar; ii, tetrahedral; iii, trigonal bipyramidal nonbonding electron domains: i, 0; ii, 1; iii, 2 231 9 Molecular Geometry
(c) Solutions to Exercises N and P. Shape ii has three bonding and one nonbonding electron domains. Li and Al would form ionic compounds with F, so there would be no nonbonding electron domains. Assuming that F always has three nonbonding domains, BF 3 and ClF 3 would have the wrong number of nonbonding domains to produce shape ii. Cl (also Br and I, since they have seven valence electrons). This Tshaped molecular geometry arises from a trigonal bipyramidal electrondomain geometry with two nonbonding domains (Table 9.3). Assuming each F atom has three nonbonding domains and forms only single bonds with A, A must have seven valence electrons to produce these electrondomain and molecular geometries. It must be in or below the third row of the periodic table, so that it can accommodate more than four electron domains. Electrondomain geometries: i, octahedral; ii, tedrahedral; iii, trigonal bipyramial nonbonding electron domains: i, 2; ii, 0; iii, 1 S or Se. Shape iii has five electron domains, so A must be in or below the third row of the periodic table. This eliminates Be and C. Assuming each F atom has three nonbonding electron domains and forms only single bonds with A, A must have six valence electrons to produce these electrondomain and molecular geometries. Xe. (See Table 9.3) Assuming F behaves typically, A must be in or below the third row and have eight valence electrons. Only Xe fits this description. (Noble gas elements above Xe have not been shown to form molecules of the type AF 4 . See Section 7.8.) Solve. (d) 9.24 (a) (b) (c) (d) 9.25 Analyze/Plan. Follow the logic in Sample Exercise 9.3. (a) (c) 1 – 109°, 2 – 109° 5 – 180° 1 – 109°, 2 – 120° 5 – 109°, 6 – 109° (b) 3 – 109°, 4 – 109° (d) 6 – 120°, 7 – 109°, 8 – 109° (b) 3 – 109°, 4 – 120° (d) 7 – 180°, 8 – 109° 9.26 (a) (c) 9.27 Analyze/Plan. Given the formula of each molecule or ion, draw the correct Lewis structure and use principles of VSEPR to answer the question. Solve. The three nonbonded electron pairs on each F atom have been omitted for clarity. The F (axial) –A–F (equatorial) angle is largest in PF 5 and smallest in ClF 3 . As the number of nonbonding domains in the equatorial plane increases, they push back the axial A–F bonds, decreasing the F (axial) –A–F (equatorial) bond angles. 232 9 Molecular Geometry
9.28 Solutions to Exercises Each species has four electron domains around the N atom, but the number of nonbonding domains decreases from two to zero, going from NH 2 – to NH 4 + . Since nonbonding domains exert greater repulsive forces on adjacent domains, the bond angles expand as the number of nonbonding domains decreases. 9.29 Analyze. Given: molecular formulas. Find: explain features of molecular geometries. Plan. Draw the correct Lewis structures for the molecules and use VSEPR to predict and explain observed molecular geometry. Solve. (a) BrF 4 – 36 e – , 18 e – pr BF 4 – 32 e – , 16 e – pr 6 e – pairs around Br octahedral e – domain geometry square planar molecular geometry 4 e – pairs around B, tetrahedral e – domain geometry tetrahedral molecular geometry The fundamental feature that determines molecular geometry is the number of electron domains around the central atom, and the number of these that are bonding domains. Although BrF 4 – and BF 4 – are both of the form AX 4 – , the central atoms and thus the number of valence electrons in the two ions are different. This leads to different numbers of e – domains about the two central atoms. Even though both ions have four bonding electron domains, the six total domains around Br require octahedral domain geometry and square planar molecular geometry, while the four total domains about B lead to tetrahedral domain and molecular geometry. (b) H2O, 8 e – , 4 e – pr (H2O)4+, 4 e – , 2 e – p r O H O H 4+ 4 e– pr around O tetrahedral e– domain geometry bent molecular geometry The shape of the (H2O)4+ molecular ion is linear. 2 e– pr around O linear e– domain and molecular geometry 233 9 Molecular Geometry
9.30 (a) ClO 2 – 20 e – , 10 e – pr Solutions to Exercises 4 e – domains around Cl, tetrahedral e – domain geometry, < bent molecular geometry bond angle ~ 109.5° NO 2 – 18 e – , 9 e – pr 3 e – domains about N (both resonance structures), trigonal planar e – domain < geometry bent molecular geometry bond angle ~ 120° Both molecular geometries are described as “bent” because both molecules have two nonlinear bonding electron domains. The bond angles (the angle between the two bonding domains) in the two ions are different because the total number of electron domains, and thus the electron domain geometries are different. (b) XeF 2 22 e – , 11 e – pr 5 e – domains around Xe, trigonal bipyramidal e – domain geometry, linear molecular geometry The question here really is: why do the three nonbonding domains all occupy the equatorial plane of the trigonal bipyramid? In a tbp, there are several different kinds of repulsions, bonding domainbonding domain (bdbd), bonding domainnonbonding domain (bdnd), and nonbonding domainnonbonding domain (ndnd). Each of these can have 90°, 120°, or 180° geometry. Since nonbonding domains occupy more space, 90° ndnd repulsions are most significant and least desirable. The various electron domains arrange themselves to minimize these 90° ndnd interactions. The arrangement shown above has no 90° ndnd repulsions. An arrangement with one or two nonbonding domains in axial positions would lead to at least two 90° ndnd repulsions, a less stable situation. (To convince yourself, tabulate the number and kinds of repulsions for each possible tbp arrangement of 2bd’s and 3nd’s.) Polarity of Polyatomic Molecules
9.31 Analyze/plan. Follow the logic in Sample Exercise 9.4. Solve. (a) SCl2, 20 e – , 1 0 e − p r
Cl S Cl tetrahedral e– domain geometry bent molecular geometry S and Cl have different electronegativities; the S−Cl bonds are polar. The bond dipoles are not opposite each other, so the molecule is polar. The dipole moment 234 9 Molecular Geometry Solutions to Exercises vector bisects the Cl−S−Cl bond angle. (A more difficult question is which end of the dipole moment vector is negative. The resultant of the two bond dipoles has its negative end away from S. However, the partial negative charge due to the lone pairs points opposite to the negative end of the resultant. A reasonable guess is that the negative end of the dipole moment vector is in the direction of the lone pairs.) (b) BeCl2, 16 e – , 8 e – p r
Cl Be Cl linear electron domain and molecular geometry (Resonance structures with Be=Cl can be drawn, but electronegativity arguments predict that most electron density will reside on Cl and that the structure above is the main resonance contributor.) Be and Cl have very different electronegativities, so the Be−Cl bonds are polar. The individual bond dipoles are equal and opposite, so the net molecular dipole moment is zero. 9.32 (a) If PH 3 were planar, the P−H bond dipoles would cancel, and the molecule would be nonpolar. Since PH 3 is polar, the 3 P–H bond dipoles do not cancel, and the molecule can’t be planar. O3, 18 e – , 9 e – p r;
O O O O O O (b) trigonal planar e− domain geometry bent molecular geometry Since all atoms are the same, the individual bond dipoles are zero. However, the central O atom has a lone pair of electrons which cause an unequal electron (and charge) distribution in the molecule. The unopposed lone pair is the source of the dipole moment in O3. 9.33 (a) In Exercise 9.23, molecules ii and iii will have nonzero dipole moments. Molecule i has no nonbonding electron pairs on A, and the 3 A–F dipoles are oriented so that the sum of their vectors is zero (the bond dipoles cancel). Molecules ii and iii have nonbonding electron pairs on A and their bond dipoles do not cancel. A nonbonding electron pair (or pairs) on a central atom guarantees at least a small molecular dipole moment, because no bond dipole exactly cancels a nonbonding pair. AF 4 molecules will have a zero dipole moment if there are no nonbonding electron pairs on the central atom and the 4 A–F bond dipoles are arranged (symmetrically) so that they cancel. Therefore, in Exercise 9.24, molecules i and ii have zero dipole moments and are nonpolar. For a molecule with polar bonds to be nonpolar, the polar bonds must be (symmetrically) arranged so that the bond dipoles cancel. In most cases, nonbonding e – domains must be absent from the central atom. Square planar structures may not meet the second condition. (b) 9.34 (a) 235 9 Molecular Geometry
(b) AB 3 : trigonal planar edg, trigonal planar mg Solutions to Exercises AB 2 : linear e – domain geometry (edg), linear molecular geometry (mg), trigonal bipyramidal edg, linear mg AB 4 : tetrahedral edg, tetrahedral mg; octahedral edg, square planar mg 9.35 Analyze/Plan. Given molecular formulas, draw correct Lewis structures, determine molecular structure and polarity. Solve. (a) Polar, ΔEN > 0 I–F (b) Nonpolar, the molecule is linear and the bond dipoles cancel. (c) Nonpolar, in a symmetrical trigonal planar structure, the bond dipoles cancel. (d) Polar, although the bond dipoles are essentially zero, there is an unequal charge distribution due to the nonbonded electron pair on P. (e) Nonpolar, symmetrical octahedron (f) Polar, square pyramidal molecular geometry, bond dipoles do not cancel. 9.36 (a) Nonpolar, in a symmetrical tetrahedral structure (Figure 9.1) the bond dipoles cancel. (b) Polar, there is an unequal charge distribution due to the nonbonded electron pair on N. 236 9 Molecular Geometry
(c) Solutions to Exercises Polar, there is an unequal charge distribution due to the nonbonded electron pair on S. (d) Nonpolar, the bond dipoles and the nonbonded electron pairs cancel. (e) Polar, the C–H and C–Br bond dipoles are not equal and do not cancel. (f) Nonpolar, in a symmetrical trigonal planar structure, the bond dipoles cancel. 9.37 Analyze/Plan. Follow the logic in Sample Exercise 9.4. Solve. (a) C2H2Cl2, each isomer has 24 e – , 1 2 e – p r. Lewis structures:
H C Cl H C Cl H C Cl Cl C H H C H Cl C Cl Molecular geometries: (b) All three isomers are planar. The molecules on the left and right are polar because the C–Cl bond dipoles do not point in opposite directions. In the middle isomer, the C–Cl bonds and dipoles are pointing in opposite directions (as are the C–H bonds), the molecule is nonpolar and has a measured dipole moment of zero. C2H3Cl (lone pairs on Cl omitted for clarity) There are four possible placements for Cl:
H C H C H Cl H H C H C Cl Cl H C H C H H H C Cl C H (c) = = = 237 9 Molecular Geometry
9.38 Solutions to Exercises By rotating each of these structures in various directions, it becomes clear that the four structures are equivalent; C2H3Cl has only one isomer. Because C2H3Cl has only one C−Cl bond, the bond dipoles do not cancel, and the molecule has a dipole moment. Each C–Cl bond is polar. The question is whether the vector sum of the C–Cl bond dipoles in each molecule will be nonzero. In the ortho and meta isomers, the C–Cl vectors are at 60° and 120° angles, respectively, and their resultant dipole moments are nonzero. In the para isomer, the C–Cl vectors are opposite, at an angle of 180°, with a resultant dipole moment of zero. The ortho and meta isomers are polar, the para isomer is nonpolar. Orbital Overlap; Hybrid Orbitals
9.39 (a) (b) Orbital overlap occurs when a valence atomic orbital on one atom shares the same region of space with a valence atomic orbital on an adjacent atom. A chemical bond is a concentration of electron density between the nuclei of two atoms. This concentration can take place because orbitals on the two atoms overlap. 9.40 (a) (b) (c) 9.41 (a) 4 valence e – , 2 e – pairs H—Mg—H 2 bonding e – domains, linear e – domain and molecular geometry (b) (c) The linear electron domain geometry in MgH 2 requires sp hybridization. Mg 9.42 By analogy to the H2 molecule shown in Figure 9.15, as the distance between the atoms decreases, the overlap between their valence orbitals increases. According to Figure 7.7, the bonding atomic radius for the halogens is on the order F < Cl < Br < I. The order of bond lengths in the molecules is I−F < I−Cl < I−Br < I−I. If the extent of orbital overlap increases as the distance between atoms decreases, I−F has the greatest overlap and I2 the least. The order for extent of orbital overlap is I−I < I−Br < I−Cl < I−F. 238 9 Molecular Geometry
9.43 Solutions to Exercises Analyze/Plan. For entries where the molecule is listed, follow the logic in Sample Exercises 9.4 and 9.5. For entries where no molecule is listed, decide electron domain geometry from hybridization (or viceversa). If the molecule is nonpolar, the terminal atoms will be identical. If the molecule is polar, the terminal atoms will be different, or the central atom will have one or more lone pairs, or both. Solve. Molecule CO2 Molecular Structures
O C O Electron Domain Geometry linear Hybrdization of CentralAtom sp Dipole Moment Yes or No no NH3 N H H
H H tetrahedral sp3 yes CH4
H C H H
H tetrahedral sp3 no BH3 B H H trigonal planar sp2 no F S F4 F S F F F F F trigonal bipyramidal sp3d yes SF6 F F S F
O octahedral sp3d2 no H2CO
H C H trigonal planar sp2 yes F PF5 F P F F F trigonal bipyramidal sp3d no XeF2 F Xe F trigonal bipyramidal sp3d no 239 9 Molecular Geometry
9.44 Solutions to Exercises In order for atomic orbitals to mix or hybridize, they must have the same principal quantum number. In each principal quantum level, there are a maximum of three p orbitals. Any hybrid orbital can have contribution from a maximum of three p orbitals. Hybrid orbitals designated sp4 or sp5 would require contribution from four or five p orbitals, which is not possible. (a) B: [He]2s 2 2p 1 9.45 (b) (c) The hybrid orbitals are called sp 2 . (d) A single 2p orbital is unhybridized. It lies perpendicular to the trigonal plane of the sp 2 hybrid orbitals. S: [Ne]3s 2 3p 4 9.46 (a) (b) (c) The hybrid orbitals are called sp 3 . (d) The hybrid orbitals formed in (a) would not be appropriate for SF 4 . There are five electron domains in SF 4 , four bonding and one nonbonding, so five hybrid orbitals are required. A set of four sp 3 hybrid orbitals could not accommodate all the electron pairs around S. 9.47 Analyze/Plan. Given the molecular (or ionic) formula, draw the correct Lewis structure and determine the electron domain geometry, which determines hybridization. Solve. (a) 24 e – , 12 e – pairs 3 e – pairs around B, trigonal planar e – domain geometry, sp 2 hybridization 240 9 Molecular Geometry
(b) 32 e – , 16 e – pairs Solutions to Exercises 4 e – domains around Al, tetrahedral e – domain geometry, sp 3 hybridization (c) 16 e – , 8 e – pairs 2 e – domains around C, linear e – domain geometry, sp hybridization (d) 22 e – , 11 e – pairs 5 e– pairs around Kr, trigonal bipyramidal e– domain geometry, sp3d hybridization (e) 48 e – , 24 e – pairs 6 e – pairs around P, octahedral e – domain geometry, sp 3 d 2 hybridization 9.48 (a) 32 e – , 16 e – pairs 4 e – pairs around Si, tetrahedral e – domain geometry, sp 3 hybridization (b) 10 e – , 5 e – pairs 2 e – domains around C, linear e – domain geometry, sp hybridization (c) 24 e – , 12 e – pairs (other resonance structures are possible) 3 e – domains around S, trigonal planar e – domain geometry, sp 2 hybridization 241 9 Molecular Geometry
(d) 22 e – , 11 e – pairs Solutions to Exercises 5 e – domains around I, trigonal bipyramidal e – domain geometry, sp 3 d hybridization (In a trigonal bipyramid, placing nonbonding e – pairs in the equatorial position minimizes repulsion.) (e) 36 e – , 18 e – pairs 6 e – domains around Br, octahedral e – domain geometry, sp 3 d 2 hybridization Multiple Bonds
9.49 (a) (b) (c) (d) A σ bond is generally stronger than a π bond, because there is more extensive orbital overlap. Two s orbitals cannot form a π bond. A π bond has no electron density along the internuclear axis. Overlap of s orbitals results in electron density along the internuclear axis. (Another way to say this is that s orbitals have the wrong symmetry to form a π bond.) Two unhybridized p orbitals remain, and the atom can form two pi bonds. It would be much easier to twist or rotate around a single sigma bond. Sigma bonds are formed by endtoend overlap of orbitals and the bonding electron density is symmetric about the internuclear axis. Rotating (twisting) around a sigma bond can be done without disrupting either the orbital overlap or bonding electron density, without breaking the bond. The π part of a double bond is formed by sidetoside overlap of p atomic orbitals perpendicular to the internuclear axis. This π overlap locks the atoms into position and makes twisting difficult. Also, only a small twist (rotation) destroys overlap of the p orbitals and breaks the π bond. 9.50 (a) (b) 242 9 Molecular Geometry
9.51 Solutions to Exercises Analyze/Plan. Draw the correct Lewis structures, count electron domains and decide hybridization. Molecules with π bonds that require all bonded atoms to be in the same plane are planar. For bondtype counting, single bonds are σ bonds, double bonds consist of one σ and one π bond, triple bonds consist of one or two π bonds. Solve. (a)
H H C H H C H H H H C C H H H C C H (b) (c) (d) (e) sp3 nonplanar 7 σ, 0 π sp2 planar 5 σ, 1 π sp planar 3 σ, 2 π Since Si has the same valence electron configuration as C, the structures of the Si analogs would be the same as the Cbased molecules. The hybridization at Si is then the same as the hybridization at C. This argument assumes that the Si analogs exist. The fact that Si lies in the row below C means that Si has a larger bonding atomic radius and larger atomic orbitals than C. In the analogous compounds, Si−Si distances will be greater than C−C distances. It also means that the close approach of Si atoms required to form strong, stable π bonds in Si2H4 and Si2H2 is unlikely; Si−Si multiple bonds are unusual, possibly unknown. It is probably not true that Si2H4 and Si2H2 exist at “normal” (nonextreme) conditions. 9.52 (a) (b) The N atoms in N 2 H 4 are sp 3 hybridized; there are no unhybridized p orbitals available for π bonding. In N 2 , the N atoms are sp hybridized, with two unhybridized p orbitals on each N atom available to form the two π bonds in the N≡N triple bond. The N−N triple bond is N2 is significantly stronger than the N−N single bond in N2H4, because it consists of one σ and two π bonds, rather than a ‘plain’ sigma bond. Generally, bond strength increases as the extent of orbital overlap increases. The additional overlap from the two π bonds adds to the strength of the N−N bond in N2. (c) 9.53 Analyze/Plan. Single bonds are σ bonds, double bonds consist of 1 σ and 1 π bond. Each bond is formed by a pair of valence electrons. Solve. (a) (b) (c) (d) (e) C 3 H 6 has 3(4) + 6(1) = 18 valence electrons 8 pairs or 16 total valence electrons form σ bonds 1 pair or 2 total valence electrons form π bonds no valence electrons are nonbonding The left and central C atoms are sp 2 hybridized; the right C atom is sp 3 hybridized. 243 9 Molecular Geometry
9.54 (a) (b) (c) (d) (e) 9.55 Solutions to Exercises The C bound to O has three electron domains and is sp 2 hybridized; the other three C atoms are sp 3 hybridized. C 4 H 8 O 2 has 4(4) + 8(1) + 2(6) = 36 valence electrons. 13 pairs or 26 total valence electrons form σ bonds 1 pair or 2 total valence electrons form π bonds 4 pairs or 8 total valence electrons are nonbonding Analyze/Plan. Given the correct Lewis structure, analyze the electron domain geometry at each central atom. This determines the hybridization and bond angles at that atom. Solve. (a) (b) (c) ~109° bond angles about the left most C, sp 3 ; ~120° bond angles about the righthand C, sp 2 The doubly bonded O can be viewed as sp 2 , the other as sp 3 ; the nitrogen is sp 3 with approximately 109° bond angles. 9 σ bonds, 1 π bond 1, 120°; 2, 120°; 3, 109° 1, sp 2 ; 2, sp 2 ; 3, sp 3 21 σ bonds In a localized π bond, the electron density is concentrated strictly between the two atoms forming the bond. In a delocalized π bond, parallel p orbitals on more than two adjacent atoms overlap and the electron density is spread over all the atoms that contribute p orbitals to the network. There are still two regions of overlap, above and below the σ framework of the molecule. The existence of more than one resonance form is a good indication that a molecule will have delocalized π bonding. 9.56 (a) (b) (c) 9.57 (a) (b) (c) The existence of more than one resonance form for NO 2 indicates that the π bond is delocalized. From an orbital perspective, the electrondomain geometry around N is trigonal planar, so the hybridization at N is sp 2 . This leaves a p orbital on N and one on each O atom perpendicular to the trigonal plane of the molecule, in the correct orientation for delocalized π overlap. Physically, the two N–O bond lengths are equal, indicating that the two N–O bonds are equivalent, rather than one longer single bond and one shorter double bond. 9.58 (a) 24 e – , 12 e – pairs (b) 3 electron domains around S, trigonal planar electrondomain geometry, sp 2 hybrid orbitals 244 9 Molecular Geometry
(c) Solutions to Exercises The multiple resonance structures indicate delocalized π bonding. All four atoms lie in the trigonal plane of the sp 2 hybrid orbitals. On each atom there is a p atomic orbital perpendicular to this plane in the correct orientation for π overlap. The resulting delocalized π electron cloud is Yshaped (the shape of the molecule) and has electron density above and below the plane of the molecule. Molecular Orbitals
9.59 (a) Hybrid orbitals are mixtures (linear combinations) of atomic orbitals from a single atom; the hybrid orbitals remain localized on that atom. Molecular orbitals are combinations of atomic orbitals from two or more atoms. They are associated with the entire molecule, not a single atom. Each MO, like each AO or hybrid, can hold a maximum of two electrons. Antibonding MOs can have electrons in them. An MO, since the AOs come from two different atoms. A hybrid orbital, since the AOs are on the same atom. Yes. The Pauli principle, that no two electrons can have the same four quantum numbers, means that an orbital can hold at most two electrons. (Since n, l, and m 1 are the same for a particular orbital and m s has only two possible values, an orbital can hold at most two electrons). This is true for atomic and molecular orbitals. (b) (c) 9.60 (a) (b) (c) 9.61 (a) (b) There is one electron in H 2 + .
* σ1s (c) σ1s
(d) (e) Bond order = 1/2 (10) = 1/2 Fall apart. The stability of H 2 + is due to the lower energy state of the σ bonding molecular orbital relative to the energy of a H 1s atomic orbital. If the single electron in H 2 + is excited to the σ* 1 s orbital, its energy is higher than the energy of an H 1s atomic orbital and H 2 + will decompose into a hydrogen atom and a hydrogen ion.
H2+ → H + H+ .
hν 245 9 Molecular Geometry
9.62 (a) Solutions to Exercises (b) (c) (d) Bond order = 1/2 (21) = ½ If one electron moves from σ 1 s to σ* 1 s, the bond order becomes –1/2. There is a net increase in energy relative to isolated H atoms, so the ion will decompose.
H2− → H + H− .
z z
hν 9.63
y x x y (a) One. With three mutually perpendicular p orbitals on each atom, only one set can be oriented for endtoend, sigma overlap. Two. The 2 p orbitals on each atom not involved in σ bonding can be aligned for sidetoside π overlap. Three, 1 σ* and 2 π*. There are a total of 6 p orbitals on the two atoms. When combining AOs to form MOs, total number of orbitals is conserved. If 3 of the 6 MOs are bonding MOs, as described in (a) and (b), then the remaining 3 MOs must be antibonding. They will have the same symmetry as the bonding MOs, 1 σ* and 2 π*. Zero The two π 2 p molecular orbitals are degenerate; they have the same energy, but they have different spatial orientations 90° apart. In the bonding MO the electrons are stabilized by both nuclei. In an antibonding MO, the electrons are directed away from the nuclei, so π 2 p is lower in energy than π* 2 p. (b) (c) 9.64 (a) (b) (c) 246 9 Molecular Geometry
9.65 (a) Solutions to Exercises When comparing the same two bonded atoms, the greater the bond order, the shorter the bond length and the greater the bond energy. That is, bond order and bond energy are directly related, while bond order and bond length are inversely related. When comparing different bonded nuclei, there are no simple relationships (see Solution 8.95). Be 2 , 4 e – Be 2 + , 3 e – (b) BO = 1/2(22) = 0 BO = 1/2(21) = 0.5 Be 2 has a bond order of zero and is not energetically favored over isolated Be atoms; it is not expected to exist. Be 2 + has a bond order of 0.5 and is slightly lower in energy than isolated Be atoms. It will probably exist under special experimental conditions, but be unstable. 9.66 (a) O 2 2 – has a bond order of 1.0, while O 2 – has a bond order of 1.5. For the same bonded atoms, the greater the bond order the shorter the bond, so O 2 – has the shorter bond. The two possible orbital energy level diagrams are: (b) The magnetic properties of a molecule reveal whether it has unpaired electrons. If the σ2p MOs are lower in energy, B2 has no unpaired electrons. If the π2p MOs are lower in energy than the σ2p MO, there are two unpaired electrons. The magnetic properties of B2 must indicate that it has unpaired electrons. (c) According to Figure 9.46, the two highestenergy electrons of O2 are in antibonding π * p MOs and O2 has a bond order of 2.0 Removing these two 2 electrons to form O22+ produces an ion with bond order 3.0. O22+ has a stronger O−O bond than O2, because O22+ has a greater bond order. 9.67 (a), (b) Substances with no unpaired electrons are weakly repelled by a magnetic field. This property is called diamagnetism. (c) O 2 2 –, Be 2 2 + [see Figure 9.46 and Solution 9.65 (b)] 247 9 Molecular Geometry
9.68 (a) (b) Solutions to Exercises Substances with unpaired electrons are attracted into a magnetic field. This property is called paramagnetism. Weigh the substance normally and in a magnetic field, as shown in Figure 9.47. Paramagnetic substances appear to have a larger mass when weighed in a magnetic field. See Figures 9.37 and 9.46. O 2 + , one unpaired electron; N 2 2 –, two unpaired electrons; Li 2 + , one unpaired electron (c) 9.69 (a) (b) B2+ increase Li 2 + (c) increase (d) N2+ increase Ne 2 2 + decrease Addition of an electron increases bond order if it occupies a bonding MO and decreases stability if it occupies an antibonding MO. 9.70 Determine the number of “valence” (noncore) electrons in each molecule or ion. Use the homonuclear diatomic MO diagram from Figure 9.43 (shown below) to calculate bond order and magnetic properties of each species. The electronegativity difference between heteroatomics increases the energy difference between the 2s AO on one atom and the 2p AO on the other, rendering the “no interaction” MO diagram in Figure 9.43 appropriate.
* σ2p * π2p (a) (b) (c) (d) CO + : 9 e – , B.O. = (7 – 2) / 2 = 2.5, paramagnetic NO – : 12 e – , B.O. = (8 – 4) / 2 = 2.0, paramagnetic OF + : 12 e – , B.O. = (8 – 4) / 2 = 2.0, paramagnetic NeF + : 14 e – , B.O. = (8 – 6) / 2 = 1.0, diamagnetic π2p σ2p σ* 2s σ2s 9.71 Analyze/Plan. Determine the number of “valence” (noncore) electrons in each molecule or ion. Use the homonuclear diatomic MO diagram from Figure 9.43 (shown below) to calculate bond order and magnetic properties of each species. The electronegativity 248 9 Molecular Geometry Solutions to Exercises difference between heteroatomics increases the energy difference between the 2s AO on one atom and the 2p AO on the other, rendering the “no interaction” MO diagram in Figure 9.43 appropriate. Solve. CN: CN + : 9 e – , B.O. = (7 – 2) / 2 = 2.5 8 e – , B.O. = (6 – 2) / 2 = 2.0 CN – : 10 e – , B.O. = (8 – 2) / 2 = 3.0 (a) (b) CN− has the highest bond order and therefore the strongest C−N bond. CN and CN+. CN has an odd number of valence electrons, so it must have an unpaired electron. The electron configuration for CN is shown in the diagram. Removing one electron from the π2p MOs to form CN+ produces an ion with two unpaired electrons. Adding one electron to the π2p MOs of CN to form CN− produces an ion with all electrons paired. The bond order of NO is [1/2 (8 – 3)] = 2.5. The electron that is lost is in an antibonding molecular orbital, so the bond order in NO + is 3.0. The increase in bond order is the driving force for the formation of NO + . To form NO – , an electron is added to an antibonding orbital, and the new bond order is [1/2 (8 – 4)] = 2. The order of increasing bond order and bond strength is: NO – < NO < NO + . NO + is isoelectronic with N 2 , and NO – is isoelectronic with O 2 . 3s, 3p x , 3p y , 3p z (b) π 3 p (c) Two 9.72 (a) (b) (c) 9.73 (a) (d) If the MO diagram for P 2 is similar to that of N 2 , P 2 will have no unpaired electrons and be diamagnetic. I: 5s, 5p x , 5p y , 5p z ; Br: 4s, 4p x , 4p y , 4p z By analogy to F 2 , the BO of IBr will be 1. I and Br have valence atomic orbitals with different principal quantum numbers. This means that the radial extensions (sizes) of the valence atomic orbital that contribute to the MO are different. The n = 5 valence AOs on I are larger than the n = 4 valence AOs on Br.
* σ np 9.74 (a) (b) (c) (d) (e) None 249 9 Molecular Geometry
Additional Exercises
9.75 (a) Solutions to Exercises The physical basis of VSEPR is the electrostatic repulsion of likecharged particles, in this case groups or domains of electrons. That is, owing to electrostatic repulsion, electron domains will arrange themselves to be as far apart as possible. The σbond electrons are localized in the region along the internuclear axes. The positions of the atoms and geometry of the molecule are thus closely tied to the locations of these electron pairs. Because the πbond electrons are distributed above and below the plane that contains the σ bonds, these electron pairs do not, in effect, influence the geometry of the molecule. Thus, all σ and πbond electrons localized between two atoms are located in the same electron domain. (b) 9.76 e– domain geometry td molecular shape td tbp seesaw (ss) octahedral (oh) square planar (s) Although there are four bonding electron domains in each molecule, the number of nonbonding domains is different in each case. The bond angles and thus the molecular shape are influenced by the total number of electron domains. 9.77 For any triangle, the law of cosines gives the length of side c as c 2 = a 2 + b 2 – 2ab cosθ. Let the edge length of the cube (uy = vy = vz) = X The length of the face diagonal (uv) is (uv) 2 = (uy) 2 + (vy) 2 – 2(uy)(vy) cos 90 (uv) 2 = X 2 + X 2 – 2(X)(X) cos 90
(uv) 2 = 2X 2 ; uv = 2 X The length of the body diagonal (uz) is (uz) 2 = (vz 2 ) + (uv) 2 – 2(vz)(uv) cos 90
(uz) 2 = X 2 + ( 2 X)2 − 2(X) ( 2 X) cos 90 (uz) 2 = 3X 2 ; uz = 3X For calculating the characteristic tetrahedral angle, the appropriate triangle has vertices u, v, and w. Theta, θ, is the angle formed by sides wu and wv and the hypotenuse is side uv. wu = wv = uz/2 = 3 / 2 X; uv = 2 X
( 2 X) 2 = ( 3 / 2 X) 2 + ( 3 / 2 X) 2 − 2( 3 / 2 X) ( 3 / 2) cos θ 2X 2 = 3/4 X 2 + 3/4 X 2 – 3/2 X 2 cos θ 250 9 Molecular Geometry
2X 2 = 3/2 X 2 – 3/2 X 2 cos θ 1/2 X 2 = –3/2 X 2 cos θ cos θ = –(1/2 X 2 ) / (3/2 X 2 ) = –1/3 = –0.3333 Solutions to Exercises θ = 109.47°
9.78 (a) 40 e – , 20 e – pairs 5 e – domains trigonal pyramidal electron domain geometry (b) The greater the electronegativity of the terminal atom, the larger the negative charge centered on the atom, the greater the effective size of the P–X electron domain. A P–F bond will produce a larger electron domain than a P–Cl bond. The molecular geometry (shape) is also trigonal bipyramidal, because all five electron domains are bonding domains. Because we predicted the P–F electron domain to be larger, the maximum number of three P–F bonds will occupy the equatorial plane of the molecule, minimizing the number of 90° P–F to P–F repulsions. This is the same argument that places a “larger” nonbonding domain in the equatorial position of a molecule like SF 4 . The P–Cl bond is then axial, as shown in the Lewis structure. The molecular geometry is distorted from a perfect trigonal bipyramid because not all electron domains are alike. The 90° P–F to P–F repulsions will be greater than the 90° P–F to P–Cl repulsions, so the F(axial)–P–F angles will be slightly greater than 90°, and the Cl(axial)–P–F angles will be slightly less than 90°. The equatorial F–P–F angles of 120° are distorted little, if at all. CO 2 , 16 valence e – (b) NCS – , 16 valence e – (c) (d) 9.79 (a) (c) H 2 CO, 12 valence e – (d) HCOOH, 18 valence e – 251 9 Molecular Geometry
9.80 (a) 3(4) + 3(6) + 6(1) = 36 e – , 18 e – pr (b) (c) There are 11 σ and 1 π bonds. Solutions to Exercises The C=O on the righthand C atom is shortest. For the same bonded atoms, in this case C and O, the greater the bond order, the shorter the bond. (d, e) The rightmost C has three e – domains, so the hybridization is sp 2 ; bond angles about this C atom are approximately 120°. The middle and lefthand C atoms both have four e – domains, are sp 3 hybridized, and have bond angles of approximately 109°. 9.81 BF 3 is a trigonal planar molecule with the central B atom symmetrically surrounded by the three F atoms (Figure 9.13). The individual B–F bond dipoles cancel, and the molecule has a net dipole moment of zero. PF 3 has tetrahedral electrondomain geometry with one of the positions in the tetrahedron occupied by a nonbonding electron pair. The individual P–F bond dipoles do not cancel and the presence of a nonbonding electron pair ensures an asymmetrical electron distribution; the molecule is polar. 9.82 (a) The compound on the right has a dipole moment. In the square planar trans structure on the left, all equivalent bond dipoles can be oriented opposite each other, for a net dipole moment of zero.
Cl N Cl Pt NH3 N cisplatin DNA transplatin NH3 Cl NH3 Pt NH3 Cl (b) N Pt N NH3 NH3 N H 3N Pt NH3 Cl N The cis orientation of the Cl atoms in cisplatin means that when they leave, the Pt can bind two adjacent N sites on DNA. This “chelate” orientation (see Chapter 24) tightly binds cisplatin to DNA. Transplatin can bind only one DNA N atom 252 9 Molecular Geometry Solutions to Exercises at a time. Thus, to avoid bumping by transplatin NH3 groups and DNA, the plane of transplatin must rotate away from the DNA backbone. This is a much looser bonding situation than for cisplatin. 9.83 (a) The bond dipoles in H 2 O lie along the O–H bonds with the positive end at H and the negative end at O. The dipole moment vector of the H 2 O molecule is the resultant (vector sum) of the two bond dipoles. This vector bisects the H–O–H angle and has a magnitude of 1.85 D with the negative end pointing toward O. Since the dipole moment vector bisects the H–O–H bond angle, the angle between one H–O bond and the dipole moment vector is 1/2 the H–O–H bond angle, 52.25°. Dropping a perpendicular line from H to the dipole moment vector creates the right triangle pictured. If x = the magnitude of the O–H bond dipole, x cos (52.25) = 0.925 D. x = 1.51 D. The X–H bond dipoles (Table 8.3) and the electronegativity values of X (Figure 8.6) are Electronegativity F O Cl 4.0 3.5 3.0 Bond dipole 1.82 1.51 1.08 (b) (c) Since the electronegativity of O is midway between the values for F and Cl, the O–H bond dipole should be approximately midway between the bond dipoles of HF and HCl. The value of the O–H bond dipole calculated in part (b) is consistent with this prediction. 9.84 (a) XeF 6 50 e – , 25 e – pairs (b) There are seven electron domains around Xe, and the maximum number of e – domains in Table 9.3 is six. 253 9 Molecular Geometry
(c) (d) (e) Solutions to Exercises Tie seven balloons together and see what arrangement they adopt (seriously! see Figure 9.5). Alternatively, go to the chemical literature where VSEPR was first proposed and see if there is a preferred orientation for seven e – domains. Since the hybrid orbitals for five e – domains involve one d orbital and for six pairs, two d orbitals, a reasonable suggestion would be sp 3 d 3 . One of the seven e – domains is a nonbonded domain. The question is whether it occupies an axial or equatorial position. The equatorial plane of a pentagonal bipyramid has F–Xe–F angles of 72°. Placing the nonbonded domain in the equatorial plane would create severe repulsions between it and the adjacent bonded domains. Thus, the nonbonded domain will reside in the axial position. The molecular structure is a pentagonal pyramid. 9.85 (a) (b) (c) The molecule is not planar. The CH 2 planes at each end are twisted 90° from one another. Allene has no dipole moment. The bonding in allene would not be described as delocalized. The π electron clouds of the two adjacent C=C are mutually perpendicular. The mechanism for delocalization of π electrons is mutual overlap of parallel p atomic orbitals on adjacent atoms. If adjacent π electron clouds are mutually perpendicular, there is no overlap and no delocalization of π electrons. 16 e – , 8 e – pairs 9.86 (a) (b) The observed bond length of 1.16 Å is intermediate between the values for N=N, 1.24 Å, and N ≡ N, 1.10 Å. This is consistent with the resonance structures, which indicate contribution from formally double and triple bonds to the true bonding picture in N 3 – . In each resonance structure, the central N has two electron domains, so it must be sp hybridized. It is difficult to predict the hybridization of terminal atoms in molecules where there are resonance structures because there are a different number of electron domains around the terminal atoms in each structure. Since the “true” electronic arrangement is a combination of all resonance structures, we will assume that the terminal N–N bonds have some triple bond character and that the terminal N atoms are sp hybridized. (There is no experimental measure of hybridization at terminal atoms, since there are no bond angles to observe.) In each resonance structure, N–N σ bonds are formed by sp hybrids and π bonds are formed by unhybridized p orbitals. Nonbonding e – pairs can reside in sp hybrids or p atomic orbitals. (c) (d) 254 9 Molecular Geometry
(e) Solutions to Exercises Recall that electrons in 2s orbitals are on the average closer to the nucleus than electrons in 2p orbitals. Since sp hybrids have greater s orbital character, it is reasonable to expect the radial extension of sp orbitals to be smaller than that of sp 2 or sp 3 orbitals and σ bonds formed by sp orbitals to be slightly shorter than those formed by other hybrid orbitals, assuming the same bonded atoms. There are no solitary σ bonds in N 3 – . That is, the two σ bonds in N 3 – are each accompanied by at least one π bond between the bonding pair of atoms. Sigma bonds that are part of a double or triple bond must be shorter so that the p orbitals can overlap enough for the π bond to form. Thus, the observation is not applicable to this molecule. (Comparison of C–H bond lengths in C 2 H 2 , C 2 H 4 , C 2 H 6 and related molecules would confirm or deny the observation.) 9.87 (a) To accommodate the π bonding by all 3 O atoms indicated in the resonance structures above, all O atoms are sp 2 hybridized. (b) For the first resonance structure, both sigma bonds are formed by overlap of sp 2 hybrid orbitals, the π bond is formed by overlap of atomic p orbitals, one of the nonbonded pairs on the right terminal O atom is in a p atomic orbital, and the remaining five nonbonded pairs are in sp 2 hybrid orbitals. Only unhybridized p atomic orbitals can be used to form a delocalized π system. The unhybridized p orbital on each O atom is used to form the delocalized π system, and in both resonance structures one nonbonded electron pair resides in a p atomic orbital. The delocalized π system then contains four electrons, two from the π bond and two from the nonbonded pair in the p orbital. Each C atom is surrounded by three electron domains (two single bonds and one double bond), so bond angles at each C atom will be approximately 120°. (c) (d) 9.88 (a) Since there is free rotation around the central C–C single bond, other conformations are possible. (b) According to Table 8.5, the average C–C length is 1.54 Å, and the average C=C length is 1.34 Å. While the C=C bonds in butadiene appear “normal,” the central C–C is significantly shorter than average. Examination of the bonding in butadiene reveals that each C atom is sp 2 hybridized and the π bonds are formed by the remaining unhybridized 2p orbital on each atom. If the central C–C bond is rotated so that all four C atoms are coplanar, the four 2p orbitals are parallel, and some delocalization of the π electrons occurs. The diagram shows two s atomic orbitals with opposite phases. Because they are spherically symmetric, the interaction of s orbitals can only produce a σ 9.89 (a) 255 9 Molecular Geometry Solutions to Exercises molecular orbital. Because the two orbitals in the diagram have opposite phases, the interaction excludes electron density from the region between the nuclei. The resulting MO has a node between the two nuclei and is labeled σ 2* . The principal s quantum number designation is arbitrary, because it defines only the size of the pertinent AOs and MOs. Shapes and phases of MOs depend only on these same characteristics of the interacting AOs. (b) The diagram shows two p atomic orbitals with oppositely phased lobes pointing at each other. Endtoend overlap produces a σtype MO; opposite phases mean a node between the nuclei and an antibonding MO. The interaction results in a σ 2* p MO. The diagram shows parallel p atomic orbitals with likephased lobes aligned. Sidetoside overlap produces a πtype MO; overlap of likephased lobes concentrates electron density between the nuclei and a bonding MO. The interaction results in a π 2 p MO. C 5 H 5 – , 26 e – , 13 e – pair (c) 9.90 (a) No. According to the single Lewis structure above, the four C atoms involved in double bonds would be sp 2 hybridized, but the C atom with the lone pair would be sp 3 hybridized. Not all C atoms would have the same hybridization. (b) Equivalent sp 2 hydridization at all C atoms is consistent with the planar structure of the ion. The VSEPR model applied to the single Lewis structure above does not predict uniform hybridization or planarity of the ion. The VSEPR/Lewis model is consistent with the known structural features of the ion only if the other resonance structures are considered. If all C atoms are sp 2 hybridized, as required by the planar structure of the ion, the three hybrid orbitals on each C are used to form the σ framework of the molecule. The unshared pair would then reside in an unhybridized p orbital. Yes, there are resonance structures equivalent to the Lewis structure above. It is possible to place the unshared pair on any of the five C atoms, resulting in five equivalent resonance structures (four plus the one above). The five resonance structures indicate that there is delocalization of the π electron density in the molecule. The interior circle conveys uniform delocalization over the entire ring, as in benzene. In a rigid planar structure with each C atom sp 2 hybridized, the unhybridized p orbitals are aligned parallel, facilitating overlap to form a delocalized π network above and below the plane of the molecule. Since the unshared pair occupies an unhybridized p orbital (part c), it is part of the delocalized π network, along with the four electrons involved in π bonds; the delocalized system contains a total of six electrons. (c) (d) (e) (f) 256 9 Molecular Geometry
9.91 (a) (b) Solutions to Exercises
N2 in the first excited state has two unpaired electrons and is paramagnetic. N2 in the ground state has a B.O. of 3; in the first excited state (at left) it has a B.O. of 2. Owing to the reduction in bond order, N2 in the first excited state has a weaker (and longer) N−N bond than N2 in the ground state. 9.92 Paramagnetic materials appear to weigh more when the mass measurement is made in the presence of a magnetic field. Air contains O 2 (g), which is paramagnetic because of its two unpaired electrons in a π 2* molecular orbital (Figure 9.46). In the presence of a p magnetic field, mass measurements made in air would be skewed by the paramagnetism of O 2 , and lead to inaccurate results. 9.93 We will refer to azo benzene (on the left) as A and hydrazobenzene (on the right) as H. (a) (b) (c) (d) A: sp 2 ; H: sp 3 A: Each N and C atom has one unhybridized p orbital. H: Each C atom has one unhybridized p orbital, but the N atoms have no unhybridized p orbitals. A: 120°; H: 109° Since all C and N atoms in A have unhybridized p orbitals, all can participate in delocalized π bonding. The delocalized π system extends over the entire molecule, including both benzene rings and the azo “bridge.” In H, the N atoms have no unhybridized p orbitals, so they cannot participate in delocalized π bonding. Each of the benzene rings in H is delocalized, but the network cannot span the N atoms in the bridge. This is consistent with the answer to (d). In order for the unhybridized p orbitals in A to overlap, they must be parallel. This requires a planar σ bond framework where all atoms in the molecule are coplanar. For a molecule to be useful in a solar energy conversion device, it must absorb visible light. This requires a HOMOLUMO energy gap in the visible region. For organic molecules, the size of the gap is related to the number of conjugated π bonds; the more conjugated π bonds, the smaller the gap and the more likely the molecule is to be colored. Azobenzene has seven conjugated π bonds (π network delocalized over the entire molecule) and appears redorange. Hydrazobenzene has only three conjugated π bonds (π network on benzene rings only) and appears white. Thus, the smaller HOMOLUMO energy gap in A causes it to be both intensely colored and a more useful molecule for solar energy conversion. (e) (f) 257 9 Molecular Geometry
9.94 (a) H: 1s 1 ; F: [He]2s 2 2p 5 Solutions to Exercises When molecular orbitals are formed from atomic orbitals, the total number of orbitals is conserved. Since H and F have a total of five valence AOs (H 1 s + F 2 s + 3F 2 p), the MO diagram for HF has five MOs. (b) (c) H and F have a total of eight valence electrons. Since each MO can hold a maximum of two electrons, four of the five MOs would be occupied. No. The MO diagram for NO in Figure 9.49 has eight MOs, and the diagram for HF has only five. The number, type, and energy spacing of the MOs in HF will be different.
z H 1s F 2pz y x (d) If H and F lie on the z axis, then the 2pz orbital of F will overlap with the 1s orbital of H. (e) Since F is more electronegative than H, the valence orbitals on F are at lower energy than those on H.
σ*
1s 2p n n σ
n 2s H F The HF MO diagram has 6 nonbonding, 2 bonding and 0 antibonding electrons. The BO = [2 – 0]/2 = 1. (Nonbonding electrons do not “count” toward bond order.) (f)
H F In the Lewis structure for HF, the nonbonding electrons are on the (more electronegative) F atom, as they are in the MO diagram. 9.95 (a) CO, 10 e – , 5 e – pair 258 9 Molecular Geometry
(b) (c) Solutions to Exercises The bond order for CO, as predicted by the MO diagram in Figure 9.49, is 1/2[8 – 2] = 3.0. A bond order of 3.0 agrees with the triple bond in the Lewis structure. Applying the MO diagram in Figure 9.49 to the CO molecule, the highest energy electrons would occupy the π 2 p MOs. That is, π 2 p would be the HOMO, highest occupied molecular orbital. If the true HOMO of CO is a σtype MO, the order of the π 2 p and σ 2 p orbitals must be reversed. Figure 9.45 shows how the interaction of the 2s orbitals on one atom and the 2p orbitals on the other atom can affect the relative energies of the resulting MOs. This 2s–2p interaction in CO is significant enough so that the σ 2 p MO is higher in energy than the π 2 p MOs, and the σ 2 p is the HOMO. We expect the atomic orbitals of the more electronegative element to have lower energy than those of the less electronegative element. When atoms of the two elements combine, the lower energy atomic orbitals make a greater contribution to the bonding MOs and the higher energy atomic orbitals make a larger contribution to the antibonding orbitals. Thus, the π 2 p bonding MOs will have a greater contribution from the more electronegative O atom. (d) Integrative Exercises
9.96 (a) Assume 100 g of compound
2.1 g H × 1 mol H = 2.1 mol H; 2.1/2.1 = 1 1.008 g H 1 mol N = 2.13 mol N; 2.13/2.1 ≈ 1 14.01 g N
1 mol O = 4.26 mol O; 4.26/2.1 ≈ 2 16.00 g O 29.8 g N ×
68.1 g O × The empirical formula is HNO 2 ; formula weight = 47. Since the approximate molar mass is 50, the molecular formula is HNO 2 . (b) Assume N is central, since it is unusual for O to be central, and part (d) indicates as much. HNO 2 : 18 valence e – The second resonance form is a minor contributor due to unfavorable formal charges. (c) The electron domain geometry around N is trigonal planar with an O–N–O angle of approximately 120°. If the resonance structure on the right makes a significant contribution to the molecular structure, all four atoms would lie in a plane. If only the left structure contributes, the H could rotate in and out of the molecular plane. The relative contributions of the two resonance structures could be determined by measuring the O–N–O and N–O–H bond angles. 3 VSEPR e – domains around N, sp 2 hybridization 3 σ, 1 π for both structures (or for H bound to N). (d) (e) 259 9 Molecular Geometry
9.97 (a) (b) 2SF 4 (g) + O 2 (g) → 2OSF 4 (g) 40 e – , 20 e – pairs Solutions to Exercises There must be a double bond drawn between O and S in order for their formal charges to be zero. (c) ΔH = 8D(S–F) + D(O=O) – 8D(S–F) – 2D(S=O) ΔH = D(O=O) – 2D(S=O) = 495 – 2(523) = –551 kJ, exothermic (d) trigonal bipyramidal electrondomain geometry (e) Because F is more electronegative than O, the structure that minimizes S–F repulsions is more likely (see Solution 9.98). That is, the structure with fewer 90° F–S–F angles and more 120° F–S–F angles is favored. The structure on the left, with O in the axial position is more likely. Note that a double bond involving an atom with an expanded octet of electrons, such as the S=O in this molecule, does not have the same geometric implications as a double bond to a first row element. PX 3 , 26 valence e – , 13 – pairs 9.98 (a) 4 electron domains around P, tetrahedral e – domain geometry, bond angles ≤ 109° (b) (c) As electronegativity increases (I < Br < Cl < F), the X–P–X angles decreases. For P–I, ΔEN is (2.5 – 2.1) = 0.4 and the bond dipole is small. For P–F, ΔEN is (4.0 – 2.1) = 1.9 and the bond dipole is large. The greater the ΔEN and bond dipole, the larger the magnitude of negative charge centered on X. The more negative charge centered on X, the greater the repulsion between X and the nonbonding electron pair on P and the smaller the bond angle. (Since electrostatic attractions and repulsions vary as 1/r, the shorter P–F bond may also contribute to this effect.) PBrCl 4 , 40 valence electrons, 20 e – pairs. The molecule will have trigonal bipyramidal electrondomain geometry (similar to PCl 5 in Table 9.3.) Based on the argument in part (c), the P–Br bond will have smaller repulsions with P–Cl bonds than P–Cl bonds have with each other. Therefore, the Br will occupy an axial position in the trigonal bipyramid, so that the more unfavorable P–Cl to P–Cl repulsions can be situated at larger angles in the equatorial plane. (d) 260 9 Molecular Geometry Solutions to Exercises 9.99 (a) (b) Three electron domains around each central C atom, sp 2 hybridization A 180° rotation around the C=C double bond is required to convert the trans isomer into the cis isomer. A 90° rotation around the bond eliminates all overlap of the p orbitals that form the π bond and it is broken. average bond enthalpy C=C C–C 614 kJ/mol 348 kJ/mol (c) The difference in these values, 266 kJ/mol, is the average bond enthalpy of a C–C π bond. This is the amount of energy required to break 1 mol of C–C π bonds. The energy per molecule is
266 kJ/mol × 1000 J 1 mol × = 4.417 × 10 −19 1 kJ 6.022 × 10 23 molecules = 4.42 × 10 −19 J/molecule (d) (e) λ = hc/E = 6.626 × 10 −34 J  s × 2.998 × 10 8 m/s = 4.50 × 10 − 7 m = 450 nm 4.417 × 10 − 19 J Yes, 450 nm light is in the visible portion of the spectrum. A cistrans isomerization in the retinal portion of the large molecule rhodopsin is the first step in a sequence of molecular transformations in the eye that leads to vision. The sequence of events enables the eye to detect visible photons, in other words, to see. C≡C C=C C–C 839 kJ/mol 614 kJ/mol 348 kJ/mol (1 σ, 2 π) (1 σ, 1 π) (1 σ) 9.100 (a) The contribution from 1 π bond would be (614–348) 266 kJ/mol. From a second π bond, (839 – 614), 225 kJ/mol. An average π bond contribution would be (266 + 225)/2 = 246 kJ/mol. This is (b) N≡N N=N N–N 246 kJ/π bond × 100 = 71% of the average enthalpy of a σ bond. 348 kJ/σ bond
941 kJ/mol 418 kJ/mol 163 kJ/mol first π = (418 – 163) = 255 kJ/mol second π = (941 – 418) = 523 kJ/mol 261 9 Molecular Geometry
This is Solutions to Exercises average π bond enthalpy = (255 + 523)/2 = 389 kJ/mol 389 kJ/π bond × 100 = 240% of the average enthalpy of a σ bond. 163 kJ/σ bond N–N σ bonds are weaker than C–C σ bonds, while N–N π bonds are stronger than C–C π bonds. The relative energies of C–C σ and π bonds are similar, while N–N π bonds are much stronger than N–N σ bonds. (c) N 2 H 4 , 14 valence e – , 7 e – pairs 4 electron domains around N, sp 3 hybridization N 2 H 2 , 12 valence e – , 6 e – pairs 3 electron domains around N, sp 2 hybridization N 2 , 10 valence e – , 5 e – pairs 2 electron domains around N, sp hybridization (d) In the three types of N–N bonds, each N atom has a nonbonding or lone pair of electrons. The lone pair to bond pair repulsions are minimized going from 109° to 120° to 180° bond angles, making the π bonds stronger relative to the σ bond. In the three types of C–C bonds, no lonepair to bondpair repulsions exist, and the σ and π bonds have more similar energies. 9.101 ΔH = 6D(C–H) + 3D(C–C) + 3D(C=C) – 0 = 6(413 kJ) + 3(348 kJ) + 3(614 kJ) = 5364 kJ
(The products are isolated atoms; there is no bond making.) According to Hess’ law: ΔH° = 6 ΔH° C(g) + 6 ΔH° H(g) − ΔH° C 6 H 6 (g) f f f
= 6(718.4 kJ) + 6(217.94 kJ) – (82.9 kJ) = 5535 kJ The difference in the two results, 171 kJ/mol C 6 H 6 is due to the resonance stabilization in benzene. That is, because the π electrons are delocalized, the molecule has a lower overall energy than that predicted for the presence of 3 localized C–C and C=C bonds. 262 9 Molecular Geometry
9.102 (a) 1 eV = 96.485 kJ/mol
H 2 : 15.4 eV × N 2 : 15.6 eV × O 2 : 12.1 eV × Solutions to Exercises Thus, the amount of energy actually required to decompose 1 mole of C 6 H 6 (g), represented by the Hess’ law calculation, is greater than the sum of the localized bond enthalpies (not taking resonance into account) from the first calculation above. 96.485 kJ/mol = 1486 = 1.49 × 10 3 kJ/mol 1 eV 96.485 kJ/mol = 1505 = 1.51 × 10 3 kJ/mol 1 eV 96.485 kJ/mol = 1167 = 1.17 × 10 3 kJ/mol 1 eV 96.485 kJ/mol F2 : 15.7 eV × = 1515 = 1.52 × 10 3 kJ/mol 1 eV (b)
I1 (kJ/mol) (c) In general, I 1 for atoms and molecules increases going across a row of the periodic chart. In both cases, there is a discontinuity at oxygen. The details of the trends are different. The deviation at O is larger for the molecules than the atoms, while the increase at F is much greater for the atoms than the molecules. According to Figures 9.35 and 9.46, H 2 , N 2 , and F 2 are diamagnetic and O 2 is paramagnetic. That is, ionization in H 2 , N 2 , and F 2 has to overcome spinpairing energy, while ionization of O 2 removes an already unpaired electron. Thus, the ionization energy of O 2 is much less than I 1 for H 2 , N 2 , and F 2 . Despite differences in bond order, bond length, and the bonding or antibonding nature of the HOMO in H 2 , N 2 , and F 2 , the ionization energies for these molecules are very similar. (d) 9.103 (a) (b) 3d z 2 Ignoring the donut of the d z 2 orbital 263 9 Molecular Geometry
(c) (d) Solutions to Exercises A node is generated in σ 3* because antibonding MOs are formed when AO d lobes with opposite phases interact. Electron density is excluded from the internuclear region and a node is formed in the MO. Sc: [Ar]4s 2 3d 1 Omitting the core electrons, there are six e – in the energy level diagram. (e) 9.104 (a) The bond order in Sc 2 is 1/2 (4 – 2) = 1.0. The molecular and empirical formulas of the four molecules are: benzene: molecular, C 6 H 6 ; empirical, CH napthalene: molecular, C 1 0H 8 ; empirical, C 5 H 4 anthracene: molecular, C 1 4H 1 0; empirical, C 7 H 5 tetracene: molecular, C 1 8H 1 2, empirical, C 3 H 2 (b) Yes. Since the compounds all have different empirical formulas, combustion analysis could in principle be used to distinguish them. In practice, the mass % of C in the four compounds is not very different, so the data would have to be precise to at least 3 decimal places and 4 would be better. C 1 0H 8 (s) + 12O 2 (g) → 10CO 2 (g) + 4H 2 O(g) (c) (d) ΔH c omb = 5D(C=C) + 6D(C–C) + 8D(C–H) + 12D(O=O) – 20D(C=O) – 8D(O–H)
= 5(614) + 6(348) + 8(413) + 12(495) – 20(799) – 8(463) = –5282 kJ/mol C 1 0H 8 (e) Yes. For example, the resonance structures of naphthalene are: 264 9 Molecular Geometry
(f) Solutions to Exercises Colored compounds absorb visible light and appear the color of the visible light that they reflect. Colorless compounds typically absorb shorter wavelength, higher energy light. The energy of light absorbed corresponds to the energy gap between the HOMO and LUMO of the molecule. That tetracene absorbs longer wavelength, lower energy visible light indicates that it has the smallest HOMOLUMO energy gap of the four molecules. Tetracene also has the most conjugated double bonds of the four molecules. We might conclude that the more conjugated double bonds in an organic molecule, the smaller the HOMOLUMO energy gap. More information about the absorption spectra of anthracene, naphthalene and benzene is needed to confirm this conclusion. 9.105 (a) – (e) y − M + + +
C O − +
x −− (c) The two lobes of a p AO have opposite phases. These are shown on the diagram as + and −. An antibonding MO is formed when p AOs with opposite phases interact. + − + − + − − + − + (d) (e) (f) Note that the dxy AO has lobes that lie between, not on, the x and y axes. A π bond forms by overlap of orbitals on M and C. There is electron density above and below, but not along, the M−C axis. According to Exercise 9.95, the HOMO of CO is a σtype MO. So the appropriate MO diagram is shown on the left side of Figure 9.46. A lone CO molecule has 10 valence electrons, the HOMO is σ 2 p and the bond order is 3.0. The LUMO is π * p . 2 When M and CO interact as shown in the π * p diagram, dπ back bonding causes 2 * the π * p to become partially occupied. Electron density in the π 2 p decreases 2 electron density in the bonding molecular orbitals and decreases the BO of the bound CO. The strength of the C−O bond in a metal−CO complex decreases relative to the strength of the C−O bond in an isolated CO molecule. 265 9 Molecular Geometry
9.106 (a) (b) (c) (d)
Cl Cl Solutions to Exercises 4 e− domains on each Cl, sp3 hybridization at each Cl Since both Cl atoms employ sp3 hybrid orbitals, the Cl−Cl bond is formed by endtoend, σ overlap of two sp3 hybrid orbitals on different Cl atoms. In the AO model, two of the lone pairs occupy mutually perpendicular p orbitals on the same Cl atom. The angle between these two lone pair electron domains is 90°. In the hybrid orbital model, all three lone pairs occupy sp3 hybrid orbitals on the same Cl atom. The angle between these lone pair domains is 109°. If we could measure the “positions” of the lone pairs, we could distinguish between models by the angles between electron domains. According to the MO diagram for F2, and by inference Cl2, shown in Figure 9.46, the HOMO for Cl2 is π* p and it contains four electrons. If we ionize Cl2 → Cl2+, 3 the number of antibonding electrons decreases and the BO increases. Cl2+ should have a stronger, shorter Cl–Cl bond than Cl2 has. Conversely, adding an electron to form Cl2– adds an electron to the antibonding σ * p . This reduces the BO in Cl2–, 3 resulting in a weaker, longer Cl–Cl bond. So, generate Cl2+ and Cl2– and measure the Cl–Cl distances. If the order of Cl–Cl bond lengths is Cl2+ < Cl2 < Cl2–, the data supports an MO model. (e) 266 ...
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This note was uploaded on 04/04/2009 for the course CHEM 121 taught by Professor Wyzlouzil during the Spring '07 term at Ohio State.
 Spring '07
 Wyzlouzil
 Atom, Mole

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