# 10. Geometry Hybridization Unhybridized p atomic orbitals

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1 APTER 14 VALET BDIG: RBITALS The Localized Electron Model and ybrid rbitals 9. The valence orbitals of the nonmetals are the s and p orbitals. The lobes of the p orbitals are 90E and 180E apart from each other. If the p orbitals were used to form bonds, then all bond angles shoud be 90E or 180E. This is not the case. In order to explain the observed geometry (bond angles) that molecules exhibit, we need to make up (hybridize) orbitals that point to where the bonded atoms and lone pairs are located. We know the geometry; we hybridize orbitals to explain the geometry. Sigma bonds have shared electrons in the area centered on a line joining the atoms. The orbitals that overlap to form the sigma bonds must overlap head to head or end to end. The hybrid orbitals about a central atom always are directed at the bonded atoms. ybrid orbitals will always overlap head to head to form sigma bonds. 10. Geometry ybridization Unhybridized p atomic orbitals linear sp 2 trigonal planar sp 2 1 tetrahedral sp 3 0 The unhybridized p atomic orbitals are used to form π bonds. Two unhybridized p atomic orbitals each from a different atom overlap side to side, resulting in a shared electron pair occupying the space above and below the line joining the atoms (the internuclear axis). 11. We use d orbitals when we have to, i.e., we use d orbitals when the central atom on a molecule has more than eight electrons around it. The d orbitals are necessary to accommodate the electrons over eight. Row 2 elements never have more than eight electrons around them, so they never hybridize d orbitals. We rationalize this by saying there are no d orbitals close in energy to the valence 2s and 2p orbitals (2d orbitals are forbidden energy levels). owever, for Row 3 and heavier elements, there are 3d, 4d, 5d, etc. orbitals that will be close in energy to the valence s and p orbitals. It is Row 3 and heavier nonmetals that hybridize d orbitals when they have to. or phosphorus, the valence electrons are in 3s and 3p orbitals. Therefore, 3d orbitals are closest in energy and are available for hybridization. Arsenic would hybridize 4d orbitals to go with the valence 4s and 4p orbitals, whereas iodine would hybridize 5d orbitals since the valence electrons are in n =

2 APTER 14 VALET BDIG: RBITALS Rotation occurs in a bond as long as the orbitals that go to form that bond still overlap when the atoms are rotating. Sigma bonds, with the head-to-head overlap, remain unaffected by rotating the atoms in the bonds. Atoms that are bonded together by only a sigma bond (single bond) exhibit this rotation phenomenon. The π bonds, however, cannot be rotated. The p orbitals must be parallel to each other to form the π bond. If we try to rotate the atoms in a π bond, the p orbitals would no longer have the correct alignment necessary to overlap. Because π bonds are present in double and triple bonds (a double bond is composed of 1 σ and 1 π bond, and a triple bond is always 1 σ and 2 π bonds), the atoms in a double or triple bond cannot rotate (unless the bond is broken) has 2(1) + 6 = 8 valence electrons. 2 has a tetrahedral arrangement of the electron pairs about the atom that requires sp 3 hybridization. Two of the four sp 3 hybrid orbitals are used to form bonds to the two hydrogen atoms, and the other two sp 3 hybrid orbitals hold the two lone pairs on oxygen. The two bonds are formed from overlap of the sp 3 hybrid orbitals from oxygen with the 1s atomic orbitals from the hydrogen atoms. Each covalent bond is called a sigma (σ) bond since the shared electron pair in each bond is centered in an area on a line running between the two atoms has 2(1) = 12 valence electrons. The central carbon atom has a trigonal planar arrangement of the electron pairs that requires sp 2 hybridization. The two sigma bonds are formed from overlap of the sp 2 hybrid orbitals from carbon with the hydrogen 1s atomic orbitals. The double bond between carbon and oxygen consists of one σ and one π bond. The oxygen atom, like the carbon atom, also has a trigonal planar arrangement of the electrons that requires sp 2 hybridization. The σ bond in the double bond is formed from overlap of a carbon sp 2 hybrid orbital with an oxygen sp 2 hybrid orbital. The π bond in the double bond is formed from overlap of the unhybridized p atomic orbitals. arbon and oxygen each has one unhybridized p atomic orbital that is parallel with the other. When two parallel p atomic orbitals overlap, a π bond results where the shared electron pair occupies the space above and below a line joining the atoms in the bond. 2 2 has 2(4) + 2(1) = 10 valence electrons.

3 542 APTER 14 VALET BDIG: RBITALS Each carbon atom in 2 2 is sp hybridized since each carbon atom is surrounded by two effective pairs of electrons; i.e., each carbon atom has a linear arrangement of the electrons. Since each carbon atom is sp hybridized, then each carbon atom has two unhybridized p atomic orbitals. The two sigma bonds are formed from overlap of carbon sp hybrid orbitals with hydrogen 1s atomic orbitals. The triple bond is composed of one σ bond and two π bonds. The sigma bond between to the carbon atoms is formed from overlap of sp hybrid orbitals from each carbon atom. The two π bonds of the triple bond are formed from parallel overlap of the two unhybridized p atomic orbitals from each carbon. 15. Ethane, 2 6, has 2(4) + 6(1) = 14 valence electrons. The Lewis structure is: The carbon atoms are sp 3 hybridized. The six sigma bonds are formed from overlap of the sp 3 hybrid orbitals on with the 1s atomic orbitals from the hydrogen atoms. The carboncarbon sigma bond is formed from overlap of an sp 3 hybrid orbital on each atom. Ethanol, 2 6 has 2(4) + 6(1) + 6 = 20 e The two atoms and the atom are sp 3 hybridized. All bonds are formed from overlap with these sp 3 hybrid orbitals. The and sigma bonds are formed from overlap of sp 3 hybrid orbitals with hydrogen 1s atomic orbitals. The and sigma bonds are formed from overlap of the sp 3 hybrid orbitals on each atom. 16., = 10 valence electrons Assuming is hybridized, both and atoms are sp hybridized. The σ bond is formed from overlap of a carbon sp hybrid orbital with a hydrogen 1s atomic orbital. The triple bond is composed of one σ bond and two π bonds. The sigma bond is formed from head-to-head overlap of the sp hybrid orbitals from the and atoms. The two π bonds in the triple bond are formed from overlap of the two unhybridized p atomic orbitals on each and atom. l 2, (7) = 24 valence electrons l l

4 APTER 14 VALET BDIG: RBITALS 543 Assuming all atoms are hybridized, the carbon and oxygen atoms are sp 2 hybridized, and the two chlorine atoms are sp 3 hybridized. The two l σ bonds are formed from overlap of sp 2 hybrids from with sp 3 hybrid orbitals from l. The double bond between the carbon and oxygen atoms consists of one σ and one π bond. The σ bond in the double bond is formed from head-to-head overlap of an sp 2 orbital from carbon with an sp 2 hybrid orbital from oxygen. The π bond is formed from parallel overlap of the unhybridized p atomic orbitals on each atom of and. 17. See Exercises 13.51, 13.52, and for the Lewis structures. To predict the hybridization, first determine the arrangement of electron pairs about each central atom using the VSEPR model, then utilize the information in igure of the text to deduce the hybridization required for that arrangement of electron pairs a. ; is sp hybridized. b. P 3 ; P is sp 3 hybridized. c. l 3 ; is sp 3 hybridized. d. 4 + ; is sp 3 hybridized. e. 2 ; is sp 2 hybridized. f. Se 2 ; Se is sp 3 hybridized. g. 2 ; is sp hybridized. h. 2 ; Each atom is sp 2 hybridized. i. Br; Br is sp 3 hybridized a. All the central atoms are sp 3 hybridized. b. All the central atoms are sp 3 hybridized. c. All the central atoms are sp 3 hybridized a. In 2 and 3, is sp 2 hybridized. In 2 4, both central atoms are also sp 2 hybridized. b. In and S, the central carbon atoms in each ion are sp hybridized, and in 3, the central atom is also sp hybridized. 18. or the molecules in Exercise 13.79, all have central atoms with dsp 3 hybridization because all are based on the trigonal bipyramid arrangement of electron pairs. See Exercise for the Lewis structures. or the molecules in Exercise 13.80, all have central atoms with d 2 sp 3 hybridization because all are based on the octahedral arrangement of electron pairs. See Exercise for the Lewis structures.

5 544 APTER 14 VALET BDIG: RBITALS 19. a. b. tetrahedral sp 3 trigonal pyramid sp E nonpolar < 109.5E polar The angles in 3 should be slightly less than because the lone pair requires more space than the bonding pairs. c. d. B V-shaped sp 3 trigonal planar sp 2 < polar 120 nonpolar e. f. Be a b Te linear sp see-saw dsp nonpolar a).120, b).90 polar g. h. a b As Kr trigonal bipyramid dsp 3 linear dsp 3 a) 90E, b) 120E nonpolar 180E nonpolar

6 APTER 14 VALET BDIG: RBITALS 545 i. j. Kr 90 o Se square planar d 2 sp 3 octahedral d 2 sp 3 90 nonpolar 90 nonpolar k. l. I I square pyramid d 2 sp 3 T-shaped dsp 3.90E polar.90e polar 20. a. S V-shaped, sp 2, 120E nly one resonance form is shown. Resonance does not change the position of the atoms. We can predict the geometry and hybridization from any one of the resonance structures. b. c. S S S 2- trigonal planar, 120E, sp 2 (plus two other resonance structures) tetrahedral, 109.5E, sp 3

7 546 APTER 14 VALET BDIG: RBITALS d. e. S S 2- S 2- tetrahedral geometry about each S, 109.5E, sp 3 hybrids; V-shaped arrangement about peroxide s,.109.5e, sp 3 hybrids trigonal pyramid, < 109.5E, sp 3 f. g. S 2- S tetrahedral, 109.5E, sp 3 V-shaped, < 109.5E, sp 3 h. i. 90 o 120 o S S see-saw,.90e and.120e, dsp 3 octahedral, 90E, d 2 sp 3 j. S b c a S a).109.5e b).90e c).120e see-saw about S atom with one lone pair (dsp 3 ); bent about S atom with two lone pairs (sp 3 )

8 APTER 14 VALET BDIG: RBITALS or the p orbitals to properly line up to form the π bond, all six atoms are forced into the same plane. If the atoms are not in the same plane, then the π bond could not form since the p orbitals would no longer be parallel to each other. 22. o, the 2 planes are mutually perpendicular to each other. The center atom is sp hybridized and is involved in two π bonds. The p orbitals used to form each π bond must be perpendicular to each other. This forces the two 2 planes to be perpendicular. 23. To complete the Lewis structures, just add lone pairs of electrons to satisfy the octet rule for the atoms with fewer than eight electrons. Biacetyl ( ) has 4(4) + 6(1) + 2(6) = 34 valence electrons. sp 2 sp 3 All angles are 120. The six atoms are not forced to lie in the same plane because of free rotation about the carboncarbon single (sigma) bonds. There are 11 σ and 2 π bonds in biacetyl. Acetoin ( ) has 4(4) + 8(1) + 2(6) = 36 valence electrons. sp 3 sp 2 b a sp 3 The carbon with the doubly bonded is sp 2 hybridized. The other 3 atoms are sp 3 hybridized. Angle a = 120 and angle b = There are 13 σ and 1 π bonds in acetoin. ote: All single bonds are σ bonds, all double bonds are one σ and one π bond, and all triple bonds are one σ and two π bonds.

9 548 APTER 14 VALET BDIG: RBITALS 24. Acrylonitrile: 3 3 has 3(4) + 3(1) + 5 = 20 valence electrons. a) 120 b b) 120 a c c) σ and 3 π bonds sp 2 sp All atoms of acrylonitrile must lie in the same plane. The π bond in the double bond dictates that the and atoms are all in the same plane and the triple bond dictates that is in the same plane with the other atoms. Methyl methacrylate ( ) has 5(4) + 8(1) + 2(6) = 40 valence electrons. d) 120 e) 120 f) * * * sp 2 * * + d sp 3 + * + e + f 14 σ and 2 π bonds The atoms marked with an asterisk must be coplanar with each other, as well as the atoms marked with a plus. The two planes, however, do not have to coincide with each other due to the rotations about sigma (single) bonds. 25. To complete the Lewis structure, just add lone pairs of electrons to satisfy the octet rule for the atoms that have fewer than eight electrons.

10 APTER 14 VALET BDIG: RBITALS 549 a. 6 b. 4 c. The center in == group d. 33 σ e. 5 π f. 180 g h. sp a. Piperine and capsaicin are molecules classified as organic compounds, i.e., compounds based on carbon. The majority of Lewis structures for organic compounds have all atoms with zero formal charge. Therefore, carbon atoms in organic compounds will usually form four bonds, nitrogen atoms will form three bonds and complete the octet with one lone pair of electrons, and oxygen atoms will form two bonds and complete the octet with two lone pairs of electrons. Using these guidelines, the Lewis structures are: a d b c piperine e f 3 g 2 h 2 i ( 2 ) 3 j k capsaicin 3 l 3 ote: The ring structures are all shorthand notation for rings of carbon atoms. In piperine the first ring contains six carbon atoms and the second ring contains five carbon atoms (plus nitrogen). Also notice that 3, 2, and are shorthand for a carbon atoms singly bonded to hydrogen atoms. b. piperine: 0 sp, 11 sp 2, and 6 sp 3 carbons; capsaicin: 0 sp, 9 sp 2, and 9 sp 3 carbons c. The nitrogens are sp 3 hybridized in each molecule. d. a) 120 e) i) 120 b) 120 f) j) c) 120 g) 120 k) 120 d) 120 h) l) 109.5

11 550 APTER 14 VALET BDIG: RBITALS 27., = 10 e ; 2, 4 + 2(6) = 16 e ; 3 2, 3(4) + 2(6) = 24 e There is no molecular structure for the diatomic molecule. The carbon in is sp hybridized. 2 is a linear molecule, and the central carbon atom is sp hybridized. 3 2 is a linear molecule with all the central carbon atoms exhibiting sp hybridization. The Molecular rbital (M) Model 28. See igure for the 2s σ bonding and σ antibonding molecular orbitals, and see igure for the 2p σ bonding, σ antibonding, π bonding, and π antibonding molecular orbitals. 29. Bonding and antibonding molecular orbitals are both solutions to the quantum mechanical treatment of the molecule. Bonding orbitals form when in-phase orbitals combine to give constructive interference. This results in enhanced electron probability located between the two nuclei. The end result is that a bonding M is lower in energy than the atomic orbitals from which it is composed. Antibonding orbitals form when out-of-phase orbitals combine. The mismatched phases produce destructive interference leading to a node of electron probability between the two nuclei. With electron distribution pushed to the outside, the energy of an antibonding orbital is higher than the energy of the atomic orbitals from which it is composed. 30. Bond energy is directly proportional to bond order. Bond length is inversely proportional to bond order. Bond energy and bond length can be measured; bond order is calculated from the molecular orbital energy diagram (bond order is the difference between the number of bonding electrons and the number of antibonding electrons divided by two). Paramagnetic: a kind of induced magnetism, associated with unpaired electrons, that causes a substance to be attracted into an inducing magnetic field. Diamagnetic: a type of induced magnetism, associated with paired electrons, that causes a substance to be repelled from the inducing magnetic field. The key is that paramagnetic substances have unpaired electrons in the molecular orbital diagram, whereas diamagnetic substances have only paired electrons in the M diagram. To determine the type of magnetism, measure the mass of a substance in the presence and absence of a magnetic field. A substance with unpaired electrons will be attracted by the magnetic field, giving an apparent increase in mass in the presence of the field. A greater number of unpaired electrons will give a greater attraction and a greater observed mass increase. A diamagnetic species will not be attracted by a magnetic field and will not show a mass increase (a slight mass decrease is observed for diamagnetic species). 31. a. 2 has two valence electrons to put in the M diagram, whereas e 2 has four valence electrons. 2 : (σ 1s ) 2 Bond order = B.. = (2!0)/2 = 1 e 2 : (σ 1s ) 2 (σ 1s *) 2 B.. = (2!2)/2 = 0

12 APTER 14 VALET BDIG: RBITALS has a nonzero bond order, so M theory predicts it will exist. The 2 molecule is stable with respect to the two free atoms. e 2 has a bond order of zero, so it should not form. The e 2 molecule is not more stable than the two free e atoms. b. See igure for the M energy-level diagrams of B 2, 2, 2, 2, and 2. B 2 and 2 have unpaired electrons in their electron configuration, so they are predicted to be paramagnetic. 2, 2, and 2 have no unpaired electrons in the M diagrams; they are all diamagnetic. c. rom the M energy diagram in igure 14.41, 2 maximizes the number of electrons in the lower-energy bonding orbitals and has no electrons in the antibonding 2p molecular orbitals. 2 has the highest possible bond order of three, so it should be a very strong (stable) bond. d. + has = 10 valence electrons to place in the M diagram, and has = 12 valence electrons. The M diagram for these two ions is assumed to be the same as that used for 2. + : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2 B.. = (8!2)/2 = 3 : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2 (π 2p *) 2 B.. = (8!4)/2 = 2 + has a larger bond order than, so + should be more stable than. 32. The localized electron model does not deal effectively with molecules containing unpaired electrons. We can draw all of the possible structures for with its odd number of valence electrons but still not have a good feel for whether the bond in is weaker or stronger than the bond in. M theory can handle odd electron species without any modifications. rom the M electron configurations, the bond order is 2.5 for and 2 for (see Exercise 14.31d for the electron configuration of ). Therefore, should have the stronger bond (and it does). In addition, hybrid orbital theory does not predict that is paramagnetic. The M theory correctly makes this prediction σ* (out-of-phase; the signs oppose each other) σ (in-phase; the signs match up) These molecular orbitals are sigma Ms because the electron density is cylindrically symmetric about the internuclear axis.

13 552 APTER 14 VALET BDIG: RBITALS 34. If we calculate a nonzero bond order for a molecule, then we predict that it can exist (is stable). a. + 2 : (σ 1s ) 1 B.. = (1 0)/2 = 1/2, stable 2 : (σ 1s ) 2 B.. = (2 0)/2 = 1, stable 2 : (σ 1s ) 2 (σ 1s *) 1 B.. = (2 1)/2 = 1/2, stable 2 2 : (σ 1s ) 2 (σ 1s *) 2 B.. = (2 2)/2 = 0, not stable b. 2 2 : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2 (π 2p *) 2 B.. = (8 4)/2 = 2, stable 2 2 : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) 4 B.. = (8 6)/2 = 1, stable 2 2 : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) 4 (σ 2p *) 2 B.. = (8 8)/2 = 0, not stable c. Be 2 : (σ 2s ) 2 (σ 2s *) 2 B.. = (2 2)/2 = 0, not stable B 2 : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 2 B.. = (4 2)/2 = 1, stable e 2 : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) 4 (σ 2p *) 2 B.. = (8 8)/2 = 0, not stable : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2 B.. = (8 2)/2 = 3 We need to decrease the bond order from 3 to 2.5. There are two ways to do this. ne is to add an electron to form * 2. This added electron goes into one of the π 2p orbitals, giving a bond order of (8 3)/2 = 2.5. We could also remove a bonding electron to form + 2. The + bond order for 2 is also 2.5 [= (7 2)/2] : (σ 1s ) 2 B 2 : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2 : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) 3 The bond strength will weaken if the electron removed comes from a bonding orbital. f the molecules listed, 2, B 2, and 2 2 would be expected to have their bond strength weaken as an electron is removed. has the electron removed from an antibonding orbital, so its bond strength increases. 37. : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 1 : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2 (π 2p *) : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 If the added electron goes into a bonding orbital, the bond order would increase, making the species more stable and more likely to form. Between and, would most likely form since the bond order increases (unlike, where the added electron goes into an antibonding orbital). Between 2 2+ and 2 2+, 2 + would most likely form since the bond order increases (unlike 2 2+ going to 2 + ).

14 APTER 14 VALET BDIG: RBITALS The electron configurations are (assuming the same orbital order as that for 2 ): : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2 B.. = (8-2)/2 = 3, 0 unpaired e + : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 1 B.. = (7-2)/2 = 2.5, 1 unpaired e 2+ : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 B.. = (6-2)/2 = 2, 0 unpaired e Because bond order is directly proportional to bond energy and, in turn, inversely proportional to bond length, the correct bond length order should be: Shortest longest bond length: < + < The π bonds between S atoms and between and S atoms are not as strong. The atomic orbitals do not overlap with each other as well as the smaller atomic orbitals of and overlap. 40. There are 14 valence electrons in the M electron configuration. Also, the valence shell is n = 3. Some possibilities from Row 3 having 14 valence electrons are l 2, Sl, S 2 2, and Ar Side-to-side overlap of these d orbitals would produce a π molecular orbital. There would be no probability of finding an electron on the axis joining the two nuclei, which is characteristic of π Ms : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) 2 ; 2 should have a lower ionization energy than. The electron removed from 2 is in a π 2p * antibonding molecular orbital, which is higher in energy than the 2p atomic orbitals from which the electron in atomic oxygen is removed. Because the electron removed from 2 is higher in energy than the electron removed from, it should be easier to remove an electron from 2 than from. 43. a. The electron density would be closer to on average. The atom is more electronegative than the atom, and the 2p orbital of is lower in energy than the 1s orbital of. b. The bonding M would have more fluorine 2p character since it is closer in energy to the fluorine 2p atomic orbital. c. The antibonding M would place more electron density closer to and would have a greater contribution from the higher-energy hydrogen 1s atomic orbital. 44. a. See the illustrations in the solution to Exercise for the bonding and antibonding Ms in. b. The antibonding M will have more hydrogen 1s character since the hydrogen 1s atomic orbital is closer in energy to the antibonding M. c. o, the overall overlap is zero. The p x orbital does not have proper symmetry to overlap with a 1s orbital. The 2p x and 2p y orbitals are called nonbonding orbitals.

15 554 APTER 14 VALET BDIG: RBITALS d. σ* 1s 2p x 2p y 2p z 2p x 2p y σ 2s 2s e. Bond order = order. 2 0 = 1. ote: The 2s, 2px, and 2p y electrons have no effect on the bond 2 f. To form +, a nonbonding electron is removed from. Because the number of bonding electrons and antibonding electrons is unchanged, the bond order is still equal to one. 45. Molecules that exhibit resonance have delocalized π bonding. This is a fancy way of saying that the π electrons are not permanently stationed between two specific atoms but instead can roam about over the surface of a molecule. We use the concept of delocalized π electrons to explain why molecules that exhibit resonance have equal bonds in terms of strength. Because the π electrons can roam about over the entire surface of the molecule, the π electrons are shared by all the atoms in the molecule, giving rise to equal bond strengths. The classic example of delocalized π electrons is benzene ( 6 6 ). igure shows the π molecular orbital system for benzene. Each carbon in benzene is sp 2 hybridized, leaving one unhybridized p atomic orbital. All six of the carbon atoms in benzene have an unhybridized p orbital pointing above and below the planar surface of the molecule. Instead of just two unhybridized p orbitals overlapping, we say all six of the unhybridized p orbitals overlap, resulting in delocalized π electrons roaming about above and below the entire surface of the benzene molecule. S 2, 6 + 2(6) = 18 e S S

16 APTER 14 VALET BDIG: RBITALS 555 In S 2 the central sulfur atom is sp 2 hybridized. The unhybridized p atomic orbital on the central sulfur atom will overlap with parallel p orbitals on each adjacent atom. All three of these p orbitals overlap together, resulting in the π electrons moving about above and below the surface of the S 2 molecule. With the delocalized π electrons, the S bond lengths in S 2 are equal (and not different, as each individual Lewis structure indicates) and 2 are isoelectronic, so we only need consider one of them since the same bonding ideas apply to both. The Lewis structures for 3 are: or each of the two resonance forms, the central atom is sp 2 hybridized with one unhybridized p atomic orbital. The sp 2 hybrid orbitals are used to form the two sigma bonds to the central atom. The localized electron view of the π bond utilizes unhybridized p atomic orbitals. The π bond resonates between the two positions in the Lewis structures: In the M picture of the π bond, all three unhybridized p orbitals overlap at the same time, resulting in π electrons that are delocalized over the entire surface of the molecule. This is represented as: or

17 556 APTER 14 VALET BDIG: RBITALS 47. The Lewis structures for 3 2 are (24 e ): In the localized electron view, the central carbon atom is sp 2 hybridized; the sp 2 hybrid orbitals are used to form the three sigma bonds in 3 2. The central atom also has one unhybridized p atomic orbital that overlaps with another p atomic orbital from one of the oxygen atoms to form the π bond in each resonance structure. This localized π bond moves (resonates) from one position to another. In the molecular orbital model for 3 2, all four atoms in 3 2 have a p atomic orbital that is perpendicular to the plane of the ion. All four of these p orbitals overlap at the same time to form a delocalized π bonding system where the π electrons can roam above and below the entire surface of the ion. The π molecular orbital system for 3 2 is analogous to that for 3 which is shown in igure of the text. 48. Benzoic acid ( ) has 7(4) + 6(1) + 2(6) = 46 valence electrons. The Lewis structure for benzoic acid is: The circle in the ring indicates the delocalized π bonding in the benzene ring. The two benzene resonance Lewis structures have three alternating double bonds in the ring (see igure 14.48). The six carbons in the ring and the carbon bonded to the ring are all sp 2 hybridized. The five sigma bonds are formed from overlap of the sp 2 hybridized carbon atoms with hydrogen 1s atomic orbitals. The seven σ bonds are formed from head to head overlap of sp 2 hybrid orbitals from each carbon. The single bond is formed from overlap of an sp 2 hybrid orbital on carbon with an sp 3 hybrid orbital from oxygen. The σ bond in the double bond is formed from overlap of carbon sp 2 hybrid orbital with an oxygen sp 2 orbital. The π bond in the double bond is formed from overlap of parallel p unhybridized atomic orbitals from and. The delocalized π bonding system in the ring is formed from overlap of all six unhybridized p atomic orbitals from the six carbon atoms. See igure for delocalized π bonding system in the benzene ring.

18 APTER 14 VALET BDIG: RBITALS 557 Spectroscopy 49. reduced mass = µ = m1m2 m + m 1 2 = (1.0078)(78.918) amu amu 27 kg = kg ν o = 1 2 π k c = = cm s cm 1 = s 1 μ λ s 1 = 1 2π k = μ 1 2π k kg Solving for k, the force constant: k = kg s 2 = m 1 ote: one newton = 1 = 1 kg m s ν o = 1 2π k, µ = μ (14.003)(15.995) amu amu 27 kg = kg ν o = 1 2π m kg m s kg = s 1 10 c cm s λ = = ν s 1 = cm Wave number = λ 1 = cm = 1877 cm a. ΔE = 2hB(J i + 1) = hν = hc, λ c λ = 2B(J i + 1) c = 2B(0 + 1) = 2B = λ m s m 1 = s 1 B = s = s 1 I = 8π h 2 = B π ( J s s 1 ) = kg m 2 I = µr e 2, µ = m1m2 m + m 1 2 = (12.000)(15.995) kg amu amu = kg 27

19 558 APTER 14 VALET BDIG: RBITALS 46 R 2 I kg m e = = μ kg 2 = m 2, R e = bond length = m Δ E b. ν = = 2B(Ji + 1) = 2B(2 +1) = 6B h rom part a, B = s 1, so: ν = 6( s 1 ) = s 1 = 113 pm 52. a. There are three sets of magnetically equivalent hydrogens (marked a, b, and c in the following structure). We are assuming that all the benzene ring hydrogens (marked c) are equivalent and do not exhibit spin-spin coupling. c c c 2 3 b a c c Because the hydrogens marked by a and the hydrogens marked by b are separated by more than three sigma bonds, they will also not exhibit spin-spin coupling. Thus we will have three singlet peaks (following our assumptions). Predicting the relative locations of the peaks was not discussed in the text, but they should be in a 3:2:5 relative area ratio (a:b:c). A sketch of the idealized MR spectrum is: c b a b. a 3 b a a The different groups of equivalent hydrogen atoms are labeled a and b. Since all atoms are separated by more than three sigma bonds, we should have no spin-spin coupling. Again, you do not have the information to predict where the two peaks should be in the idealized MR spectrum, but they should have relative areas of 9:3 (or 3:1).

20 APTER 14 VALET BDIG: RBITALS 559 a b c. c 2 3 b a c a b a b c The three different groups of equivalent hydrogen atoms are labeled a, b, and c. The atoms marked c will not exhibit spin-spin coupling, but the atoms marked a and b will. Since the atoms marked a in the 3 groups neighbor 2 groups, a triplet line pattern will result with intensities of 1:2:1. The atoms marked b in the 2 groups neighbor 3 groups, so a quartet peak should result with intensities of 1:3:3:1. The relative area ratios of the three different atoms (a:b:c) should be 9:6:3 (or 3:2:1). c b a 53. a. The 2 group neighbors a 3 group, so a quartet of peaks should result (iv). b. The 3 atoms are separated by more than three sigma bonds from other atoms, so no spin-spin coupling should occur. A singlet peak should result (i). c. The 2 atoms neighbor two atoms, so a triplet peak should result (iii). d. The two atoms in the 2 group neighbor one atom, so a doublet peak should result (ii) MR spectrum: Because only one peak is present, all the hydrogen atoms are equivalent in their environment. Knowing this, then the process is trial and error to come up with the correct structure for ollowing the organic rules in Exercise and knowing a double bond is present, the only possibility to explain the MR pattern is:

21 560 APTER 14 VALET BDIG: RBITALS ere, all the atoms have equivalent neighboring atoms and there would be no spin-spin coupling since all atoms are separated by more than three sigma bonds spectrum: We have a quartet peak and a triplet peak in the spectrum. The quartet peak indicates hydrogen atom(s) that neighbor three atoms; the triplet peak indicates hydrogen atom(s) that neighbor two atoms. Again, following the organic rules in Exercise 14.64, the only possibility to explain this MR pattern is: This compound has two different types of hydrogen atoms ( 3 and 2 ). The 3 hydrogen atoms neighbor the 2 group, which would produce the triplet peak, and the 2 hydrogen atoms neighbor the 3 group, that would produce the quartet peak. rom inspection, the relative areas of the two different patterns seems to confirm a 6:4 (or 3:2) ratio as it should be. Additional Exercises 55. a. Xe 3, 8 + 3(6) = 26 e b. Xe 4, 8 + 4(6) = 32 e Xe Xe trigonal pyramid; sp 3 tetrahedral; sp 3 c. Xe 4, (7) = 42 e d. Xe 2, (7) = 28 e Xe or Xe Xe or Xe square pyramid; d 2 sp 3 T-shaped; dsp 3

22 APTER 14 VALET BDIG: RBITALS 561 e. Xe 3 2 has 8 + 3(6) + 2(7) = 40 valence electrons. Xe or Xe or Xe trigonal bipyramid; dsp l l 2 2 l 2, 2(7) (6) + 1 = 34 e - l dsp 3 hybridization 3 l + 4 l 4 l, 4(7) = 42 e - l d 2 sp 3 hybridization ote: Similar to Exercise 14.55c, d, and e, 2 l 2 - has two additional Lewis structures that are possible, and 4 l - has one additional Lewis structure that is possible, depending on the relative placement of the and atoms. The predicted hybridization is unaffected. 3 l + 2 l + 3 l l l +, 2(7) = 26 e 2 l 2 +, 2(7) (6) 1 = 32 e l + l + sp 3 hybridization sp 3 hybridization 57. a. o, some atoms are in different places. Thus these are not resonance structures; they are different compounds. b. or the first Lewis structure, all nitrogens are sp 3 hybridized and all carbons are sp 2 hybridized. In the second Lewis structure, all nitrogens and carbons are sp 2 hybridized.

23 562 APTER 14 VALET BDIG: RBITALS c. or the reaction: Bonds broken: Bonds formed: 3 = (745 kj/mol) 3 = (615 kj/mol) 3 (305 kj/mol) 3 (358 kj/mol) 3 (391 kj/mol) 3 (467 kj/mol) Δ = 3(745) + 3(305) + 3(391) - [3(615) + 3(358) + 3(467)] Δ = 4323 kj 4320 kj = 3 kj The bonds are slightly stronger in the first structure with the carbon-oxygen double bonds since Δ for the reaction is positive. owever, the value of Δ is so small that the best conclusion is that the bond strengths are comparable in the two structures. 58. l l l l l l In order to rotate about the double bond, the molecule must go through an intermediate stage where the π bond is broken and the sigma bond remains intact. Bond energies are 347 kj/mol for a bond and 614 kj/mol for a = bond. If we take the single bond as the strength of the σ bond, then the strength of the π bond is ( = ) 267 kj/mol. Thus 267 kj/mol must be supplied to rotate about a carbon-carbon double bond.

24 APTER 14 VALET BDIG: RBITALS d xz + p z x x z z The two orbitals will overlap side-to-side, so when the orbitals are in-phase, a π bonding molecular orbital would form (ground state): (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2, B.. = 3, diamagnetic (0 unpaired e ) 2 (1st excited state): (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 1 (π 2p *) 1 B.. = (7 3)/2 = 2, paramagnetic (2 unpaired e - ) The first excited state of 2 should have a weaker bond and should be paramagnetic. 61. onsidering only the 12 valence electrons in 2, the M models would be: σ 2p* π 2p* π 2p σ 2p σ 2s* σ 2s 2 ground state Arrangement of electrons consistent with the Lewis structure (double bond and no unpaired electrons). It takes energy to pair electrons in the same orbital. Thus the structure with no unpaired electrons is at a higher energy; it is an excited state. 62. a. Yes, both have four sets of electrons about the P. We would predict a tetrahedral structure for both. See part d for the Lewis structures.

25 564 APTER 14 VALET BDIG: RBITALS b. The hybridization is sp 3 for P in each structure since both structures exhibit a tetrahedral arrangement of electron pairs. c. P has to use one of its d orbitals to form the π bond since the p orbitals are all used to form the hybrid orbitals. d. ormal charge = number of valence electrons of an atom - [(number of lone pair electrons) + 1/2 (number of shared electrons)]. The formal charges calculated for the and P atoms are next to the atoms in the following Lewis structures l P l l +1 l P l l 0 In both structures, the formal charges of the l atoms are all zeros. The structure with the P= bond is favored on the basis of formal charge since it has a zero formal charge for all atoms. 63. a. B 3 has 3 + 3(1) = 6 valence electrons. B trigonal planar, nonpolar, 120E, sp 2 b. 2 2 has 2(5) + 2(7) = 24 valence electrons. an also be: V-shaped about both s; 120E about both s; both atoms: sp 2 polar nonpolar These are distinctly different molecules. c. 4 6 has 4(4) + 6(1) = 22 valence electrons.

26 APTER 14 VALET BDIG: RBITALS 565 All atoms are trigonal planar with 120E bond angles and sp 2 hybridization. Because and have about equal electronegativities, the bonds are essentially nonpolar, so the molecule is nonpolar. All neutral compounds composed of only and atoms are nonpolar. d. Il 3 has 7 + 3(7) = 28 valence electrons. l I l 90o l 90o T-shaped, polar.90e, dsp or carbon, nitrogen, and oxygen atoms to have formal charge values of zero, each atom will form four bonds to other atoms and have no lone pairs of electrons, each atom will form three bonds to other atoms and have one lone pair of electrons, and each atom will form two bonds to other atoms and have two lone pairs of electrons. ollowing these bonding requirements gives the following two resonance structures for vitamin B 6 : a b g f d e c a. 21 σ bonds; 4 π bonds (The electrons in the three π bonds in the ring are delocalized.) b. Angles a), c), and g): ; angles b), d), e), and f): 120 c. 6 sp 2 carbons; the five carbon atoms in the ring are sp 2 hybridized, as is the carbon with the double bond to oxygen. d. 4 sp 3 atoms; the two carbons that are not sp 2 hybridized are sp 3 hybridized, and the oxygens marked with angles a and c are sp 3 hybridized. e. Yes, the π electrons in the ring are delocalized. The atoms in the ring are all sp 2 hybridized. This leaves a p orbital perpendicular to the plane of the ring from each atom. verlap of all six of these p orbitals results in a π molecular orbital system where the electrons are delocalized above and below the plane of the ring (similar to benzene in igure of the text).

27 566 APTER 14 VALET BDIG: RBITALS 65. The Lewis structure for caffeine is: The three atoms each bonded to three atoms are sp 3 hybridized (tetrahedral geometry); the other five atoms with trigonal planar geometry are sp 2 hybridized. The one atom with the double bond is sp 2 hybridized, and the other three atoms are sp 3 hybridized. The answers to the questions are: 6 and atoms are sp 2 hybridized 6 and atoms are sp 3 hybridized 0 and atoms are sp hybridized (linear geometry) 25 σ bonds and 4 π bonds 66. We rotated the molecule about some single bonds from the structure given in the question. ollowing the bonding guidelines outlined in Exercise 14.64, a Lewis structure for aspartame is: Another resonance structure could be drawn having the double bonds in the benzene ring moved over one position.

28 APTER 14 VALET BDIG: RBITALS 567 Atoms that have trigonal planar geometry of electron pairs are assumed to have sp 2 hybridization, and atoms with tetrahedral geometry of electron pairs are assumed to have sp 3 hybridization. All the atoms have tetrahedral geometry, so they are all sp 3 hybridized (no sp 2 hybridization). The oxygens double bonded to carbon atoms are sp 2 hybridized; the other two oxygens with two single bonds are sp 3 hybridized. or the carbon atoms, the six carbon atoms in the benzene ring are sp 2 hybridized, and the three carbons double bonded to oxygen are also sp 2 hybridized (tetrahedral geometry). Answering the questions: 9 sp 2 hybridized and atoms (9 from s and 0 from s) 7 sp 3 hybridized and atoms (5 from s and 2 from s) 39 σ bonds and 6 π bonds (this includes the 3 π bonds in the benzene ring that are delocalized) 67. a. The Lewis structures for and are: The structure is correct. rom the Lewis structures we would predict both and to be linear. owever, we would predict to be polar and to be nonpolar. Since experiments show 2 to be polar, then is the correct structure. b. ormal charge = number of valence electrons of atoms - [(number of lone pair electrons) + 1/2 (number of shared electrons)] The formal charges for the atoms in the various resonance structures are below each atom. The central is sp hybridized in all the resonance structures. We can probably ignore the third resonance structure on the basis of the relatively large formal charges as compared to the first two resonance structures. c. The sp hybrid orbitals from the center overlap with atomic orbitals (or appropriate hybrid orbitals) from the other two atoms to form the two sigma bonds. The remaining two unhybridized p orbitals from the center overlap with two p orbitals from the peripheral atom to form the two π bonds. 2p x sp sp z axis 2py

29 568 APTER 14 VALET BDIG: RBITALS 68. = l: The bond order of the bond in l is 2 (a double bond). : rom molecular orbital theory, the bond order of this bond is 2.5 (see igure of the text). Both reactions apparently only involve the breaking of the l bond. owever, in the reaction l + l, some energy is released in forming the stronger bond, lowering the value of Δ. Therefore, the apparent l bond energy is artificially low for this reaction. The first reaction only involves the breaking of the l bond. hallenge Problems 69. a. 2 (g) (g) + (g) Δ = 2 bond energy Using ess s law: 2 (g) 2 (g) + e Δ = 290. kj (IE for 2 ) 2 (g) 2 (g) Δ = 154 kj (BE for 2 from Table 13.6) (g) + e (g) Δ =!327.8 kj (EA for from Table12.8) 2 (g) (g) + (g) Δ = 116 kj; BE for 2 = 116 kj/mol ote that 2 has a smaller bond energy than 2. b. 2 : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) 4 B.. = (8 6)/2 = 1 2 : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) 4 (σ 2p *) 1 B.. = (8 7)/2 = 0.5 M theory predicts that 2 should have a stronger bond than 2 because 2 has the larger bond order. As determined in part a, 2 indeed has a stronger bond because the 2 bond energy (154 kj/mol) is greater than the 2 bond energy (116 kj/mol). 70. The ground state M electron configuration for e 2 is (σ 1s ) 2 (σ 1s *) 2 giving a bond order of 0. Therefore, e 2 molecules are not predicted to be stable (and are not stable) in the lowestenergy ground state. owever, in a high-energy environment, electron(s) from the antibonding orbitals in e 2 can be promoted into higher-energy bonding orbitals, thus giving a nonzero bond order and a reason to form. or example, a possible excited-state M electron configuration for e 2 would be (σ 1s ) 2 (σ 1s *) 1 (σ 2s ) 1, giving a bond order of (3 1)/2 = 1. Thus excited e 2 molecules can form, but they spontaneously break apart as the electron(s) fall back to the ground state, where the bond order equals zero. 71. a. E 34 hc ( J s)( m / s) = = = J 9 λ m 8

30 APTER 14 VALET BDIG: RBITALS J mol 23 1 kj 1000 J = 4800 kj/mol Using Δ values from the various reactions, 25-nm light has sufficient energy to ionize 2 and and to break the triple bond. Thus 2, 2 +,, and + will all be present, assuming excess 2. b. To produce atomic nitrogen but no ions, the range of energies of the light must be from 941 kj/mol to just below 1402 kj/mol. 941 kj mol 1 mol 1000 J kj = J/photon λ 34 hc ( J s)( m / s) = = = m = 127 nm 18 E J kj mol 1 mol 1000 J kj = J/photon λ 34 hc ( J s)( m / s) = = = m = nm 18 E J 8 Light with wavelengths in the range of nm < λ < 127 nm will produce but no ions. c. 2 : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2 ; the electron removed from 2 is in the σ 2p molecular orbital, which is lower in energy than the 2p atomic orbital from which the electron in atomic nitrogen is removed. Because the electron removed from 2 is lower in energy than the electron removed from, the ionization energy of 2 is greater than that for. 72. The electron configurations are: 2 : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2 2 : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2 (π 2p *) 2 2 : (σ 2s ) 2 (σ 2s *) 2 (π 2p ) 4 (σ 2p ) 2 (π 2p *) : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) 1 ote: The ordering of the σ 2p and π 2p orbitals is not important to this question. The species with the smallest ionization energy has the electron that is easiest to remove. rom the M electron configurations, 2, 2 2, + 2, and 2 all contain electrons in the same higher-energy antibonding orbitals ( π * 2 p ), so they should have electrons that are easier to * remove as compared to 2, which has no π2p electrons. To differentiate which has the easiest * π 2p to remove, concentrate on the number of electrons in the orbitals attracted to the number of protons in the nucleus.

31 570 APTER 14 VALET BDIG: RBITALS 2 2 and 2 both have 14 protons in the two nuclei combined. Because 2 2 has more electrons, one would expect 2 2 to have more electron repulsions, which translates into having an easier electron to remove. Between 2 and 2 +, the electron in 2 should be easier to remove. 2 has one more electron than 2 +, and one would expect the fewer electrons in 2 + to be better attracted to the nuclei (and harder to remove). Between 2 2 and 2, both have 16 electrons; the difference is the number of protons in the nucleus. Because 2 2 has two fewer protons than 2, one would expect the 2 2 to have the easiest electron to remove which translates into the smallest ionization energy. 73. The complete Lewis structure follows. All but two of the carbon atoms are sp 3 hybridized. The two carbon atoms that contain the double bond are sp 2 hybridized (see *) * * 2 o; most of the carbons are not in the same plane since a majority of carbon atoms exhibit a tetrahedral structure. ote:, 2, and 3 are shorthand for carbon atoms singly bonded to hydrogen atoms. 2p 2s The order from lowest IE to highest IE is: 2 < 2 < 2 + <.

32 APTER 14 VALET BDIG: RBITALS 571 The electrons for 2, 2, and + 2 that are highest in energy are in the π Ms. But for 2, these electrons are paired. 2 should have the lowest ionization energy (its paired π * 2p electron is easiest to remove). The species + 2 has an overall positive charge, making it harder to remove an electron from + 2 than from 2. The highest energy electrons for (in the 2p atomic orbitals) are lower in energy than the π * 2p electrons for the other species; will have the highest ionization energy because it requires a larger quantity of energy to remove an electron from as compared to the other species. 75. a. The bond is polar with the negative end around the more electronegative oxygen atom. We would expect metal cations to be attracted to and to bond to the oxygen end of on the basis of electronegativity. * 2p b. (carbon) = 4 2 1/2(6) = 1 (oxygen) = 6 2 1/2(6) = +1 rom formal charge, we would expect metal cations to bond to the carbon (with the negative formal charge). c. In molecular orbital theory, only orbitals with proper symmetry overlap to form bonding orbitals. The metals that form bonds to are usually transition metals, all of which have outer electrons in the d orbitals. The only molecular orbitals of that have proper symmetry to overlap with d orbitals are the π 2p * orbitals, whose shape is similar to the d orbitals (see igure 14.36). Since the antibonding molecular orbitals have more carbon character, one would expect the bond to form through carbon. 76. The molecular orbitals for Be 2 are formed from the two hydrogen 1s orbitals and the 2s and one of the 2p orbitals from beryllium. ne of the sigma bonding orbitals forms from overlap of the hydrogen 1s orbitals with a 2s orbital from beryllium. Assuming the z axis is the internuclear axis in the linear Be 2 molecule, then the 2p z orbital from beryllium has proper symmetry to overlap with the 1s orbitals from hydrogen; the 2p x and 2p y orbitals are nonbonding orbitals since they don t have proper symmetry necessary to overlap with 1s orbitals. The type of bond formed from the 2p z and 1s orbitals is a sigma bond since the orbitals overlap head to head. The M diagram for Be 2 is:

33 572 APTER 14 VALET BDIG: RBITALS Be 2 σ* s σ* p 2p 2s 2px 2p y 1s 1s σ p σ s Bond order = (4-0)/2 = 2; the M diagram predicts Be 2 to be a stable species and also predicts that Be 2 is diamagnetic. ote: The σ s M is a mixture of the two hydrogen 1s orbitals with the 2s orbital from beryllium, and the σ p M is a mixture of the two hydrogen 1s orbitals with the 2p z orbital from beryllium. The Ms are not localized between any two atoms; instead, they extend over the entire surface of the three atoms. 77. ne of the resonance structures for benzene is: To break 6 6 (g) into (g) and (g) requires the breaking of 6 bonds, 3 = bonds and 3 bonds: 6 6 (g) 6 (g) + 6 (g) Δ = 6D + 3D = + 3D Δ = 6(413 kj) + 3(614 kj) + 3(347 kj) = 5361 kj The question asks for Δ f for 6 6 (g), which is Δ for the reaction: 6 (s) (g) 6 6 (g) Δ = Δ f, 6 6 (g)

34 APTER 14 VALET BDIG: RBITALS 573 To calculate Δ for this reaction, we will use ess s law along with the value and the bond energy value for 2 ( D = 432 kj/mol). 2 Δ f for (g) 6 (g) + 6 (g) 6 6 (g) Δ 1 =!5361 kj 6 (s) 6 (g) Δ 2 = 6(717 kj) 3 2 (g) 6 (g) Δ 3 = 3(432 kj) 6 (s) (g) 6 6 (g) Δ = Δ 1 + Δ 2 + Δ 3 = 237 kj; Δ f, = 237 kj/mol 6 6 (g) The experimental Δ f for 6 6 (g) is more stable (lower in energy) by 154 kj than the Δ f calculated from bond energies (83! 237 =!154 kj). This extra stability is related to benzene s ability to exhibit resonance. Two equivalent Lewis structures can be drawn for benzene. The π bonding system implied by each Lewis structure consists of three localized π bonds. This is not correct because all bonds in benzene are equivalent. We say the π electrons in benzene are delocalized over the entire surface of 6 6 (see Section 14.5 of the text). The large discrepancy between Δ f values is due to the delocalized π electrons, whose effect was not accounted for in the calculated Δ f value. The extra stability associated with benzene can be called resonance stabilization. In general, molecules that exhibit resonance are usually more stable than predicted using bond energies. 78. a. The MR indicates that all hydrogens in this compound are equivalent. A possible structure for 2 3 l 2 that explains the MR data is: l l l ote that for organic compounds to have a zero formal charge for all atoms, carbon will always satisfy the octet rule by having four bonds and no lone pairs, oxygen will always satisfy the octet rule by having two bonds and two lone pairs, and l will always satisfy the octet rule by having one bond and three lone pairs of electrons. b. We have two types of hydrogens. The triplet signal is produced by having two neighboring bonds, whereas the quintet signal is produced by having four neighboring bonds. A possible structure that explains the MR (number of peaks, types of peaks, and the 2:1 relative intensities of the signals) is: l l

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