20ac14k - CHAPTER 14 COVALENT BONDING: ORBITALS 457 11. 12....

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Unformatted text preview: CHAPTER 14 COVALENT BONDING: ORBITALS 457 11. 12. CzH2 has 2(4) + 2(1) = 10 valence electrons. H —CEC— H Each carbon atom in CZH2 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 C — H sigma bonds are formed from overlap of carbon sp hybrid orbitals With hydrogen ls atomic orbitals. The triple bond is composed of one a bond and two 1: bonds. The sigma bond between to the carbon atoms is formed from overlap of sp hybrid orbitals from each carbon atom. The two 1: bonds of the triple bond are formed fi'om parallel overlap of the two unhybridized p atomic orbitals from each carbon. See Exercises 13.47. 13.48 and 13.50 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 Figure 14.24 of the text to deduce the hybridization required for that arrangement of electron pairs. 13.47 a. HCN; C is sp hybridized. b. PH3; P is sp3 hybridized. c. CHC13; C is sp3 hybridized. d. NI-If; N is sp3 hybridized. e. HZCO; C is sp2 hybridized. f. Ser; Se is sp3 hybridized. g. C02; C is sp hybridized. h. 02; Each 0 atom is sp2 hybridized. i. HBr; Br is sp3 hybridized. 13.48 a. All the central atoms are sp3 hybridized. b. All the central atoms are sp3 hybridized. c. All the central atoms are sp3 hybridized. 13 .50 a. In NOZ', N is sp2 hybridized in NO3’, N is sp2 hybridized, and in N204, both central N atoms are also sp2 hybridized. b. In OCN' and SCN‘, the central carbon atoms in each ion are sp hybridized and in N3; the central N atom is also sp hybridized. For the molecules/ion in Exercise 13 .75, all have central atoms with dsp3 hybridization since all are based on the trigonal bipyramid arrangement of electron pairs. See Exercise 13 .75 for the Lewis structures. For the molecules/ ion in Exercise 13 .76, all have central atoms with dzsp3 hybridization since all are based on the octahedral arrangement of electron pairs. See Exercise 13 .76 for the Lewis structures. Wham CHAPTER 14 COVALENT BONDING: ORBITALS 463 19. 20. 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. :O: T l C_H C H—l'N/ \lC/Ji H 'C\ /4 14—9—0 .- | H\C——-H H H! N: H II N H ZN: a. 6 b. 4 c. ThecenterNin—N:N,:Ngroup d. 336 e. 51: f. 180° g. z109.5° h. sp3 a. NCNZ' has 5 + 4 + 5 + 2 = 16 valence electrons. to no 2' ' 2.. o . 2_ [ NZCZN ] <—-—-—> [:Nac—NI] <—————> [IN—CENi] HZNCN has 2(1) + 5 + 4 + 5 = 16 valence electrons. H H\+1 o —1_ I0 0 0 /N:C:N_.' <——-> H—N—CENZ H favored by formal charge NCNC(NH2)2 has 5 + 4 + 5 + 4 + 2(5) + 4(1) = 32 valence electrons. oo/H OOI/H -1 0 +1 ‘N o o o N _- 0/ \ -- 0/ \ .'N =C=N =C\ H <——> : N _=_C—N :C\ H .N—H NL—H - o lo H H favored by formal charge 464 CHAPTER 14 COVALENT BONDING: ORBITALS W Melamine (CgNfiHé) has 3(4) + 6(5) + 6(1) = 48 valence electrons. T T _ .- ou— — N’ H N\C/N\C/T H H N\C/ \C/l I I“ H l. u: ._N \C/N, .. \C/ .. H—l\|lZ H—hlll: J, H b. NCNZ': C is sp hybridized. Depending on the resonance form, N can be sp, spz, or sp3 hybridized. For the remaining compounds, we will give hybrids for the favored resonance structures as predicted from formal charge considerations. H " ;NH2 /N—CENZ :NEC—N\=/C\ > SP3 H I \ / \ I ..NH2 SP Sp2 . sp3 Sp Melamine: N in NH2 groups are all sp3 hybridized. Atoms in ring are all sp2 hybridized. c. NCNZ': 2 o and 2 7r bonds; HZNCN: 4 c and 2 1t bonds; dicyandiamide: 9 6 and 3 1: bonds; melamine: 15 o and 3 1: bonds d. The n—system forces the ring to be planar just as the benzene ring is planar. e. The structure: : N EC—N.. is the most important since it has three different CN bonds (a single, double, and triple bond). This structure is also favored on the basis of formal charge. 21. 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: “war... mm .m .mu,.....m.w m.» M mwwwrmmm 22. 23. CHAPTER 14 COVALENT BONDING: ORBITALS 465 capsaicin Note: The ring structures are all shorthand notation for rings of carbon atoms. In piperine, the first ring contains 6 carbon atoms and the second ring contains 5 carbon atoms (plus nitrogen). Also notice that CH3, CH2 and CH are shorthand for a carbon atoms singly bonded to hydrogen atoms. b. piperine: 0 sp, 11 sp2 and 6 sp3 carbons; capsaicin: 0 sp, 9 sp2 and 9 sp3 carbons c. The nitrogens are sp3 hybridized in each molecule. d. a. 120° e. = 109.50 i. 120° b. 120° f. 1095" j. 109.50 0 120° g. 120° k. 120° d 120° b. 109.50 l. 1095" co, 4 + 6 = 10 e“; coz, 4 + 2(6) = 16 e‘; (3302, 3(4) + 2(6) = 24 e" =C There is no molecular structure for the diatomic CO molecule. The carbon in CO is sp hybridized. CO2 is a linear molecule, and the central carbon atom is sp hybridized. C302 is a linear molecule with all of the central carbon atoms exhibiting sp hybridization. To complete the Lewis structure, just add lone pairs of electrons to satisfy the octet rule for the atoms with fewer than eight electrons. 466 CHAPTER 14 COVALENT BONDING: ORBITALS H H H \/ \/H C—C H\ M /H H/C b C\H \C p/ .. "/H —.. N H/IL N\Cl:/\:>\C/ \H d C e H/ \Viq'; H—til: H a. The two nitrogens in the ring with double bonds are sp2 hybridized. The other three nitrogens are sp3 hybridized. ' b. The five carbon atoms in the ring with one nitrogen are all sp3 hybridized. The four carbon atoms in the other ring with double bonds are all sp2 hybridized. c. Angles a and b: :109.5°; angles c, d, and e: =120° d. 31 sigma bonds e. 3 pi bonds (Each double bond consists of one sigma and one pi bond.) The Molecular Orbital (lVIO) Model 24. 25. 26. Bonding molecular orbitals have maximum electron density between the bonded atoms and are lower in energy than the atomic orbitals from which they are formed. Antibonding molecular orbitals have minimal electron density between the bonded atoms and are higher in energy than the atomic orbitals from which they are formed. Bond energy is directly proportional to bond order. Bond length is inversely proportional to bond order. Bond energy and bond length can be measured. If we calculate a non-zero bond order for a molecule, we predict that it can exist (is stable). a. Hf: (6515)1 H2: (0-192 H23 (5132(615’“)1 H227 (5192051332 13.0. = (1-0)/2 = 1/2, stable B0. = (2-0)/2 =1, stable 13.0. = (2-1 )/2 = 1/2, stable B0. = (2-2)/2 = 0, not stable 468 CHAPTER 14 COVALENT BONDING: ORBITALS _____.___—_______________________ The bond order is 2. In order to make the bond order greater (to 2.5) we need to take away a nonbonding electron. Thus, the charges would be 1+. So 02+ would have a bond order of 2.5. (We could also take away 3 electrons, which is much less likely, to have 5 bonding and 0 nonbonding 2p electrons, so Oz3+ would work as well). 31. a. H2: (61,)2 B0. = (2-0)/2 = 1, diamagetic (0 unpaired e') b. B2: (6292(625’”)2(1r,21,)2 B0. = (4-2)/2 = 1, paramagnetic (2 unpaired e‘) c. NO: (62,926»),"‘)2(1%)“(621)2(11521,*)l B0. = (8-3)/2 = 2.5, paramagnetic (1 unpaired e') d. CN“: (625)2(625*)2(1sz)‘ B0. = (6-2)/2 = 2, diamagnetic e. CN: ((523)2(623‘)2(1r21,)“(621,)l B0. = (7-2)/2 = 2.5, paramagnetic (1 unpaired e‘) f. CN': (625)2(623*)2(1D2p)4(621,)2 B0. = 3, diamagnetic g. N2: (629)2(625*)2(152P)4(62P)2 B0. = 3, diamagnetic h. N;: ((52,)2(62;")2('tr,zl,)"(c52p)l B.O. = 2.5, paramagnetic (1 unpaired e‘) i NZ': (625)2(625*)2(152p)4(62P)2(7f/1p*)1 B0. = 2.5, paramagnetic (l unpaired e') 32. H2: (61,)2 B2 : (625)2(625*)2(752p)2 N2: (625)2(625*)2(752p)4(62p)2 0F: (6292(st*)2(62p)2(7r2p)4(7r2p*)3 The bond strength will weaken if the electron removed comes from a bonding orbital. Of the molecules listed, H2, B2, and N2 would be expected to have their bond strength weaken as an electron is removed. OF has the electron removed from an antibonding orbital, so its bond strength increases. 33- CN: (625)2(625"‘)2(7T2p)4(<52p)1 N01 (625)2(023‘)2(1sz)4(7tzp)2(792p*)1 022+: (6292852?)2(<52p)2(7bzp)4 N22+5 (625)2(62s*)2(7t2p)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 CN and NO, CN would most likely form CN‘ since i. the bond order increases (unlike NO‘ where the added electron goes into an antibonding orbital). Between 022+ and N2“, N; would most likely form since the bond order increases (unlike 02+). 34. The electron configurations are: Ff: ((525)2(c525*)2(c52p)2(1clp)4(7L4,"‘)3 RC. = (8-5)/2 = 1.5; 1 unpaired e‘ F23 (625)2(625’")2(62,,)2(752p)4(7£2,,"‘)4 30. = (8-6)/2 = l; 0 unpaired e‘ Fz‘: ((525)2(5251‘)2((SZI,)2(1I:2P)“(1|:2,,"‘)“(c52,,"‘)I B.O. = (8-7)/2 = 0.5; 1 unpaired 6' Since bond order is directly related to bond energy and, in turn, inversely related to bond length, then the bond length order should be: F2+ < F2 < Fz'. g1? CHAPTER 14 COVALENT BONDING: ORBITALS 469 35. The electron configurations are (assuming the same orbital order as that for N2): CO: ((525)2(625"')2(1I:21,)“(c521,)2 B0. = (8-2)/2 = 3; 0 unpaired e‘ CO+2 (625)2(62;")2(1r,21,)4(c521,)l B0. = (7-2)/2 = 2.5; l unpaired e‘ CO”: (6292(625*)2(1L1p)4 B0. = (6-2)/2 = 2; 0 unpaired e' Since bond order is directly related to bond energy and, in turn, inversely related to bond length, then the bond length order should be: CO < CO+ < CO”. 36. N2: The 1: and 1c* orbitals are symmetrical. CO: The 1!: orbitals would place more electron density nearer the more electronegative oxygen atom and the 1:* orbitals would place more electron density nearer the carbon atom. 37. The 1: bonds between S atoms and between C and S atoms are not as strong. The atomic orbitals do not overlap with each other as well as the smaller atomic orbitals of C and O overlap. 38. Cl 02 Cl @ fickp’“ 39W?" TE313"“ \‘irjflfllfl 39 \‘\‘\ TC I’I” ------- an ------- ,--——-""/’ 535* Ssh]; :13] as C$35 The bond order for C12 is l and since all electrons are paired, we would expect Cl2 to be diamagnetic. 39. Side to side overlap of these d-orbitals would produce a 1: molecular orbital. There would be no probability of finding an electron on the axis joining the two nuclei, which is characteristic of 11: MOS. 0 CHAPTER 14 COVALENT BONDING: ORBITALS 40. Oz: (625)2(62;")2(c521,)2(792194054,*)2 ; 02 should have a lower ionization energy than 0. The electron removed from O2 is in a 1521;“ antibonding molecular orbital which is higher in energy than the 2p atomic orbitals from which the electron in atomic oxygen is removed. Since the electron removed from O2 is higher in energy than the electron removed from 0, then it should be easier to remove an electron from 02 than from O. 41. a. The electron density would be closer to F on the average. The F atom is more electronegative than the H atom and the 2p orbital of F is lower in energy than the ls orbital of H. b. The bonding MO would have more fluorine 2p character since it is closer in energy to the fluorine 2p atomic orbital. c. The antibonding MO would place more electron density closer to H and would have a greater contribution from the higher energy hydrogen 1s atomic orbital. 42. a. See the illustrations in Exercise 14.29 for the bonding and antibonding MOS in OH. b. The antibonding MO will have more hydrogen ls character since the hydrogen ls atomic orbital is closer in energy to the antibonding MO. c. No, the overall overlap is zero. The px orbital does not have proper symmetry to overlap with a ls orbital. The 2px and 2py orbitals are called nonbonding orbitals. d. H HO O * /—---——fi 0 I, \\ I \ T ,” \\ .____‘ \\ \\ I O I u 2.3 Ti 2s 5 6. Bond order = 2&9 = 1; Note: The 25, 2px, and 2py electrons have no effect on the bond order. electrons and antibonding electrons are unchanged, then the bond order is still equal to one. f. To form OH+ a nonbonding electron is removed from OH. Since the number of bonding ‘ s i CHAPTER 14 COVALENT BONDING: ORBITALS 471 Spectroscopy 43. reduced mass =11, = mlmz =W amu x = 1.6524 X 10‘27 kg m1+ m2 1.0078 + 78.918 amu v0 = 31— 5 = % = 2.9979 x 101° cm s-1 x 2650. cm'1= 7.944 x 1013 5" W M 7944 x1013 5-1 =__1_ E = _1_ ______k___ 211: ,u 27‘ 1.6524-X 10—27 kg Solving for k, the force constant: k = 411.7 kg 5'2 = 411.7 N m‘1 Note: 1 newton = IN = 1 kg m 5'2 1 k -27 44_ v0: —— — , =W amu x =1.2393 x10-26kg 27! M 14.003 + 15.995 amu -1 —2 V0: _1_ 1550.Nrn x 1kgms =5-627x 1013 5-1 2 7: 1.2398 x 10‘26 kg N = 2.9979 x 1010 cms ‘1 1:3 =5.328 x 104 cm v 5.627x 1013s-1 _ 1 _ 1 __ -1 wave number ~ —— ——-—————— — 1877 cm 9» 5.328x10‘4cm 45. a. AE=2hB(Ji+1)=hv=l‘f, %=ZB(L+1) 8 -1 3 =23 (0+ 1)=2B=——————~—2'998x 10 “‘5 = 1.15 X10119l 9» 2.60x10“3m -11 -1 B: S 10105-1 2 ~34 I: h = 6.626x10 Js =1_46X1046kgm2 8 n2 (5.75 ><1010 s '1) 8 1:2 B m -27 “R32, [1: 1m2 = 12.000(15.995)amu x In1 + In2 12.000 + 15.995 amu I ll = 1.1385 X 10'26 kg = 1.46x 10‘46 kg m2 1 26 k =1.28 ><10‘2° m2, Re=bond length= 1.13 X 10'1°m=113 Pm .1385 X 10— g Re2=l ,U. a. my: CHAPTER 14 COVALENT BONDING: ORBITALS 473 47. 48. C. CHZCHs c b a H H c CH3CH2 CH2CH3 a b b a H c The three different groups of equivalent hydrogen atoms are labeled a, b, and c.. The H-atoms marked c will not exhibit spin-spin coupling, but the H-atoms marked a and b will. Since the H—atoms marked a in the — CH3 groups neighbor — CH2 groups, then a triplet line pattern will result with intensities of 1:2:1. The H—atoms marked b in the — CH2 groups neighbor -— CH3 groups, so a quartet peak should result with intensities of 1 :3 :3 : 1. The relative area ratios of the _ three different H—atoms (a:b:c) should be 9:63 (or 3:221). The —CH2 group neighbors a —CH3 group so a quartet of peaks should result (iv). The —CH3 H—atoms are separated by more than three sigma bonds from other H—atoms, so no spin-spin coupling should occur. A singlet peak should result (i). The -CH2 H—atoms neighbor two H-atoms, so a triplet peak should result (iii). (1. The 2 H—atoms in the —CH2F group neighbor one H—atom, so a doublet peak should result (ii). C6Hl2 NMR spectrum: Since only one peak is present, then all of 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 CsHu. Following the “organic rules” in Exercise 14.60 and knowing a double bond is present, the only possibility to explain the NMR pattern is: ‘EH3 CH3—C|:C_C H3 CH3 Here, all of the H—atoms have equivalent neighboring atoms and there would be no spin-spin coupling since all H-atoms are separated by more than three sigma bonds. ...
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This note was uploaded on 04/14/2008 for the course CHEM 20A taught by Professor Scerri during the Fall '05 term at UCLA.

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20ac14k - CHAPTER 14 COVALENT BONDING: ORBITALS 457 11. 12....

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