APTER 9 OVALET BODIG: ORBITALS 323 2 3 2 2 2 3 3 2 2 3 2 3 O * * 2 o; most of the carbons are not in the same plane since a majority of carbon atoms exhibit a tetrahedral structure (19.5 bond angles). 73. a. 2 has 5 + 4 + 5 + 2 = 16 valence electrons. 2-2- 2-2 has 2(1) + 5 + 4 + 5 = 16 valence electrons. +1-1 favored by formal charge ( 2 ) 2 has 5 + 4 + 5 + 4 + 2(5) + 4(1) = 32 valence electrons. -1 +1 favored by formal charge
324 APTER 9 OVALET BODIG: ORBITALS Melamine ( 3 6 6 ) has 3(4) + 6(5) + 6(1) = 48 valence electrons. b. 2 : is sp hybridized. Each resonance structure predicts a different hybridization for the atom. Depending on the resonance form, can be sp, sp 2, or sp 3 hybridized. or the remaining compounds, we will give hybrids for the favored resonance structures as predicted from formal charge considerations. 2 sp 3 2 sp sp 2 sp 3 sp Melamine: in 2 groups are all sp 3 hybridized. Atoms in ring are all sp 2 hybridized. c. 2 : 2 σ and 2 π bonds; 2 : 4 σ and 2 π bonds; dicyandiamide: 9 σ and 3 π bonds; melamine: 15 σ and 3 π bonds d. The π-system forces the ring to be planar just as the benzene ring is planar. e. The structure: is the most important since it has three different bonds. This structure is also favored on the basis of formal charge. 74. One of the resonance structures for benzene is:
APTER 9 OVALET BODIG: ORBITALS 325 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 = 6 D + 3 D = + 3 D Δ = 6(413 kj) + 3(614 kj) + 3(347 kj) = 5361 kj The question asks for f for 6 6 (g), which is Δ for the reaction: 6 (s) + 3 2 (g) 6 6 (g) Δ = Δ f, 6 6 (g) To calculate Δ for this reaction, we will use ess s law along with the value and the bond energy value for 2 ( D 2 = 432 kj/mol). 6 (g) + 6 (g) 6 6 (g) Δ 1 = 5361 kj Δ f for (g) 6 (s) 6 (g) Δ 2 = 6(717 kj) 3 2 (g) 6 (g) Δ 3 = 3(432 kj) 6 (s) + 3 2 (g) 6 6 (g) Δ = Δ 1 + Δ 2 + Δ 3 = 237 kj; Δ = 237 kj/mol f, 6 6 (g) The experimental f for 6 6 (g) is more stable (lower in energy) by 154 kj as compared to 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 as all bonds in benzene are equivalent. We say the π electrons in benzene are delocalized over the entire surface of 6 6 (see Section 9.5 of the text). The large discrepancy between f values is due to the delocalized π electrons, whose effect was not accounted for in the calculated 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. 75. a. E = hc λ 34 (6.6261 J s)(2.9981 m/s) = 7.9 9 251 m 8 18 1 J 7.9 18 1 J 6.221 mol 23 1kJ 1J = 48 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 142 kj/mol.
326 APTER 9 OVALET BODIG: ORBITALS 941kJ mol 1J mol 23 6.221 kj = 1.56 18 1 J/photon λ = hc E 34 (6.6261 J s)(2.9981 m/s) = 1.27 18 1.561 J 142kJ mol 1J mol 23 6.221 1 kj = 2.328 18 1 J/photon λ = hc E 34 (6.6261 1 J s)(2.99791 m/s) = 8.533 18 2.3281 J 8 8 7 1 m = 127 nm 8 1 m = 85.33 nm Light with wavelengths in the range of 85.33 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. Since the electron removed from 2 is lower in energy than the electron in, the ionization energy of 2 is greater than that for. 76. The π bonds between two S atoms and between and S atoms are not as strong. The orbitals do not overlap with each other as well as the smaller atomic orbitals of and O overlap. 77. O= l: The bond order of the O bond in Ol is 2 (a double bond). O: rom molecular orbital theory, the bond order of this O bond is 2.5. Both reactions apparently involve only the breaking of the l bond. owever, in the reaction Ol O + l, some energy is released in forming the stronger O bond, lowering the value of Δ. Therefore, the apparent l bond energy is artificially low for this reaction. The first reaction involves only the breaking of the l bond. 78. 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. One of the sigma bonding orbitals forms from in phase 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 MO diagram for Be 2 is:
APTER 9 OVALET BODIG: ORBITALS 327 Be 2 * s * p 2p 2s 2px 2p y 1s 1s p s Bond Order = (4 )/2 = 2; The MO diagram predicts Be 2 to be a stable species and also predicts that Be 2 is diamagnetic. ote: The σ s MO is a mixture of the two hydrogen 1s orbitals with the 2s orbital from beryllium and the σ p MO is a mixture of the two hydrogen 1s orbitals with the 2p z orbital from beryllium. The MOs are not localized between any two atoms; instead, they extend over the entire surface of the three atoms. 79. a. The O bond is polar with the negative end around the more electronegative oxygen atom. We would expect metal cations to be attracted to and bond to the oxygen end of O on the basis of electronegativity. b. O (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 O are usually transition metals, all of which have outer electrons in the d orbitals. The only molecular orbitals of O that have proper symmetry to overlap with d orbitals are the π 2p * orbitals, whose shape is similar to the d orbitals (see igure 9.34). Since the antibonding molecular orbitals have more carbon character (carbon is less electronegative than oxygen), one would expect the bond to form through carbon.
328 APTER 9 OVALET BODIG: ORBITALS 8. 2p 2s O 2 O 2 O 2 + O The order from lowest IE to highest IE is: O 2 < O 2 < O 2 + < O. The electrons for O 2, O 2, and O + 2 that are highest in energy are in the π MOs. But for O 2, these electrons are paired. O 2 should have the lowest ionization energy (its paired π * 2p electron is easiest to remove). The species O + 2 has an overall positive charge, making it harder to remove an electron from O + 2 than from O 2. The highest energy electrons for O (in the 2p atomic orbitals) are lower in energy than the π * 2p electrons for the other species; O will have the highest ionization energy because it requires a larger quantity of energy to remove an electron from O as compared to the other species. 81. The electron configurations are: * 2p 2 : 2 * 2 4 2 ( σ ) (σ ) (π ) (σ 2 s 2s 2p 2p) O 2 : 2 * 2 2 4 * 2 ( σ ) (σ ) (σ ) (π ) (π 2 2 : 2 * 2 4 2 * 2 ( σ ) (σ ) (π ) (σ ) (π 2 : 2 * 2 4 2 * 1 ( σ ) (σ ) (π ) (σ ) (π O + 2 : 2 * 2 2 4 * 1 ( σ ) (σ ) (σ ) (π ) (π ote: the ordering of the 2p and 2p orbitals is not important to this question. The species with the smallest ionization energy has the electron which is easiest to remove. rom the MO electron configurations, O 2, 2 2, + 2, and O 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. 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 O 2 and O 2 +, the electron in O 2 should be easier
APTER 9 OVALET BODIG: ORBITALS 329 to remove. O 2 has one more electron than O 2 +, and one would expect the fewer electrons in O 2 + to be better attracted to the nuclei (and harder to remove). Between 2 2 and O 2, both have 16 electrons; the difference is the number of protons in the nucleus. Because 2 2 has two fewer protons than O 2, one would expect the 2 2 to have the easiest electron to remove which translates into the smallest ionization energy. 82. 2 : 2 * 2 2 4 * 4 ( σ ) (σ ) (σ ) (π ) (π B.O. (8 6)/2 = 1 2 : 2 * 2 2 4 * 4 * 1 ( σ ) (σ ) (σ ) (π ) (π ) (σ B.O. = (8 7)/2 =.5 2 s 2s 2p 2p 2p 2p) MO theory predicts that 2 should have a stronger bond than 2 because 2 has the larger bond order. Let s compare the 2 bond energy in Table 8.4 (154 kj/mol) to the calculated 2 bond energy. 2 (g) (g) + (g) = 2 bond energy Using ess s law: 2 (g) 2 (g) + e = 29 kj (IE for 2 ) 2 (g) 2 (g) = 154 kj (BE for 2 ) (g) + e (g) = 327.8 kj (EA for from Table 7.7) 2 (g) (g) + (g) = 116 kj As predicted from the MO theory, the bond energy of 2 is smaller than the bond energy of 2. Integrative Problems 83. a. Li 2 : s 2 ( σ 2 ) B.O. = (2 )/2 = 1 B 2 : 2 * 2 2 ( σ ) (σ ) (π B.O. = (4 2)/2 = 1 2 s 2s 2p) Both have a bond order of 1. b. B 2 has four more electrons than Li 2 so four electrons must be removed from B 2 to make it isoelectronic with Li 2. The isoelectronic ion is B 2 4+. 1g 1molB2 6455kJ c. 1.5 kg B 2 = 4.5 1 5 kj 1kg 21.62g B molb 2 2
33 APTER 9 OVALET BODIG: ORBITALS 84. a., 1 + 7 = 8 e Sb 5, 5 + 5(7) = 4 e linear, sp 3 (if is hybridized) Sb trigonal bipyramid, dsp 3 2 +, 2(1) + 7 1 = 8 e Sb 6, 5 + 6(7) + 1 = 48 e + Sb V-shaped, sp 3 octahedral, d 2 sp 3 - b. 2.93 ml.975g 1mol =.143 mol ml 2.1g 3.1g Sb 5 1molSb 5 1. ml =.143 mol Sb 5 ml 216.8 g Sb 5 The balanced equation requires a 2:1 mol ratio between and Sb 5. Because we have the same amount (moles) of the reactants, is limiting. 1mol [ 256.8 g 2 ] [Sb 6].143 mol = 18.4 g [ 2 ] + [Sb 6 ] 2 mol mol[ ] [Sb ] 85. Element X has 36 protons which identifies it as Kr. Element Y has one less electron than Y, so the electron configuration of Y is 1s 2 2s 2 2p 5. This is. Kr 3 +, 8 + 3(7) 1 = 28 e 2 6 + Kr T-shaped, dsp 3