The symmetry properties & relative energies of atomic orbitals determine how they react to form molecular orbitals. These molecular orbitals are then
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2 The symmetry properties & relative energies of atomic orbitals determine how they react to form molecular orbitals. These molecular orbitals are then filled with the available electrons according to the same rules used for atomic orbitals. 2
3 5-1 Formation of Molecular Orbitals from Atomic Orbitals HΨ = EΨ (Schrödinger equation ) Approximate solutions to these molecular Schrödinger equations can be constructed from linear combination of atomic orbitals (LCAO), the sums & differences of the atomic wave functions. 3
4 Linear Combination of Atomic Orbitals (LCAO) Ψ = c i Φ i Approximate solution of molecular Schrödinger equation, Ψ, can be represented as the sum of n atomic orbitals, Φ i, each multiplied by a corresponding coefficient, c i. The coefficients are the weights of the contributions of the n atomic orbitals to the molecular orbital. 4
5 For diatomic H 2 molecule, Ψ = c a φ a ± c b φ b Ψ : molecular wave function φ a, φ b : atomic wave functions c a, c b : adjustable coefficients AO MO AO 5
6 Three conditions are essential for overlap to lead to bonding, and such that the atoms must have 1. Same symmetry 2. Similar energy 3. Close contact 6
7 When these conditions are met, the overall energy of the electrons in the occupied molecular orbitals will be lower in energy than the overall energy of the electrons in the original atomic orbitals, & the resulting molecule has a lower total energy than the separated atoms. 7
8 Molecular Orbitals from s Orbitals 8
9 9
10 σ bond : Bonding, antibonding, nonbonding orbitals # of atomic orbitals # of molecular orbitals 10
11 11
12 π bond : When we draw the z axes for the two atoms pointing in the same direction, the p z orbitals subtract to form σ & add to form σ* orbitals, both of which are symmetric to rotation about the z axis, with nodes perpendicular to the line that connects the nuclei. Interactions between p x & p y orbitals lead to π & π* orbitals. The π notation indicates a change in sign of the wave function with C 2 rotation about the bond axis. 12
13 13
14 14
15 5-1-4 Nonbonding orbitals and other factors : 1. There are three atomic orbitals of same symmetry & similar energy, a situation that requires the formation of three molecular orbitals (lowenergy bonding orbitals, high-energy antibonding orbitals, & intermediate energy is nonbonding orbitals). The energy of nonbonding orbitals is essentially that of the original atomic orbitals. 15
16 2. Atomic orbitals whose symmetries do not match, & therefore remain unchanged in the molecule are also called nonbonding. i.e., s & d yz. 3. When the two atomic orbitals have quite different energies, the interaction is weak, & the resulting molecular orbitals have nearly the same (energy & shape) as those of atomic orbitals. (1s v.s. 2s or 2s v.s. 2p). 16
17 17
18 5-2 Homonuclear diatomic molecules Molecular orbitals As we will see, the molecular orbital description is more in agreement with experiment, i.e., magnetic dioxygen molecules. O 2 : paramagnetic molecule 18
19 Electrons fill the molecular orbitals according to the same rules that govern the filling of atomic orbitals. 1. Aufbau principle (from lowest to highest energy). 2. Hund s rule (maximum spin multiplicity). 3. Pauli exclusion principle (no two electrons with identical quantum numbers). AO MO AO B B B O O 2 O 2 19
20 g: gerade u: ungerade symmetric (g) or antisymmetric (u) to inversion. The g or u describes the symmetry of the orbitals without a judgment as to their relative energy. 20
21 Bond order The overall # of bonding and antibonding electrons determines the # of bonds (bond order) : Bond order = ½[(# of electrons in bonding orbitals) -(# of electrons in antibonding orbitals)] For example: O 2 Consider valence electrons only Bond order = ½[6-2] = 2 21
22 5-2-2 Orbital mixing When two molecular orbitals of the same symmetry mix, the one with higher energy moves still higher & the one with lower energy moves lower. This phenomenon is called mixing, which takes into account that molecular orbitals with similar energies interact if they have appropriate symmetry. 22
23 As we will see s-p mixing can have an important influence on the energy of molecular orbitals. In addition, mixing changes the bonding-antibonding nature of some of the orbitals (magnetic properties : high- & low-spin properties). 23
24 s-p mixing changes the relative positions of σ g & π u. With more s-contribution, the energy of σ g (2p) orbital goes higher. same symmetry more s-contribution 24
25 Configuration Interaction (CI) for Pt 2+ (d 8 ) (a) a 2u (b) 6p z a 2u a 1g Energy 5d z 2 a 1g a 2u a 1g free complex free complex free complex MO scheme for a " molecule " consisting of two planar complexes (a) without and (b) with configuration interaction 25
26 higher σ g (2p) energy better s-p mixing small s-p separation large s-p separation electronegativity 26
27 [C C] 2- O 2 27
28 Bond lengths in homonuclear diatomic molecules 28
29 H-B 120 pm; H-C 109 pm, H-N pm; H-O 96 pm; H-F 91.8 pm (effective nuclear charge ) 29
30 5-2-4 Photoelectron Spectroscopy In addition to data on bond distances & energies, specific information about the energies of electrons in orbitals can be determined from photoelectron spectroscopy, one of the more direct methods for determining orbital energies. 30
31 O 2 + hν (UV or X-rays) Ο e - I.E. = hν (photons) K.E. (expelled electron) (binding energy) UV light removes outer electrons, usually from gases; X-rays are more energetic & remove inner electrons as well, from any physical state. 31
32 Photoelectron spectroscopy is related to MO diagram. Electronic levels vibrational hyperfine structure binding energy 32
33 binding energy 33
34 5-2-5 Correlation diagrams A correlation diagram shows the calculated effect of moving two atoms together, from a large interatomic distance on the right, with no interatomic interaction, to zero interatomic distance on the left, where the two nuclei become, in effect, a single nucleus. The simplest example has two hydrogen atoms on the right & a helium atom on the left (hypothesis).
35 The diagram shows how the energies of the orbitals change with the internuclear distance & from the order of atomic orbitals on the left to the order of molecular orbitals of similar symmetry on the right.
36 Correlation diagram for homonuclear diatomic MO As atoms move closer, bonding MOs decrease in energy, while antibonding MOs increase in energy. SYMMETRY is used to connect MOs. 1σ u * is the antibonding of two 1s orbitals (right). It is correlated to the 2p z orbital of united atom (left). Degenerate 1π u from the 2p MOs of the separated atoms have the same symmetry of the p x and p y orbitals of the united atom. 3σ g, 1π u exchange Degenerate 1π g * from the 2p x and 2p y orbitals of the separated atoms is connected to the 3d xz and 3d yz of the united atom. smaller H H distance He (MO) (B 2,C 2,N 2 ) (O 2,F 2 ) 2xH (AO) 36
37 Another consequence of this phenomenon is called the non-crossing rule, which states that orbitals of the same symmetry interact so that their energies never cross. This rule helps in assigning correlations.
38 Concepts and Models of Inorganic Chemistry - Douglas (third ed, P. 449) The energies of two states deviate from what their values would have been in the absence of interaction, bending away from each other instead of displaying the linear behavior. This is referred to as the noncrossing rule (orbital mixing).
39
40 The actual energies of molecular orbitals for diatomic molecules are intermediate between the extremes of this diagram, approximately in the region set off by the vertical lines. Toward the right, closer to the separated atoms, the energy sequence is the normal one of O 2 & F 2 ; further to the left, the order of molecular orbitals is that of B 2, C 2 & N 2, with σ g (2p) above π u (2p).
41 5-3 Heteronuclear diatomic molecules Polar bonds X has larger electronegativity. Ψ = c x φ x + c y φ y Y δ+ Y:X δ- X For bonding orbital, C x > C y. φ x contribute more to Ψ, and electrons are populated close to X. 41
42 Predicting bond polarity 1. Atoms with similar electronegativities (ΔEN 0.4) - form nonpolar bonds. 2. Atoms whose electronegativities differ by more than 2 - form ionic bonds. 3. Atoms whose electronegativities differ by less than 2 - form polar covalent bonds. 42
43 Electronegativities are related to the orbital potential energies (Z eff ). These potential energies are negative because they represent the attraction of valence electrons. 43
44 Symmetry of Molecular Orbitals 44
45 Take the MO of C O C O as an example CO CO has C v symmetry, but the p x & p y orbitals have C 2v symmetry if the signs of the orbital lobes are ignored as in the diagram (the signs are ignored only for the purpose of choosing a point group, but must be included for the rest of the process). Using the C 2v point group rather than C v simplifies the orbital analysis by avoiding the infinite rotation axis of C v. 45
46 The s & p z group orbitals have A 1 symmetry & form molecular orbitals with σ symmetry; the p x & p y group orbitals have B 1 & B 2 symmetry, respectively & form π orbitals. The p x & p y orbitals change sign on C 2 rotation & change sign on one σ v reflection character table. C 2v E C 2 (z) σ v (xz) σ v (yz) linear, rotations quadratic A z x 2, y 2, z 2 A R z xy B x, R y xz B y, R x yz 46
47 1π* 2s(C), 2p z (C), 2p z (O). 3σ Lone pair& bonding on C? Frontier orbitals MO versus Conventional Lewis Structure? :C O: HOMO & LUMO are mainly on C, not O. HOMO (3σ) is contributed mainly by 2p z (C) & some 2s(C), while 2p z (O) contributes less because it is divided to 2σ*, 3σ & 3σ*. LUMO (1π*) is mainly contributed by 2p x (C) & 2p y (C). C 2v E C 2 (z) σ v (xz) σ v (yz) A z x 2, y 2, z 2 A R z xy B x, R y xz B y, R x yz 47
48 The Bonding of Metal Carbonyl M-C O HOMO back-bonding LUMO 48
49 HF : C v C 2v Lone pair on F σ bonding with electron density close to F 49
50 5-3-2 Ionic Compounds and Molecular Orbitals Large ΔE = 13.2 ev Electrons on Li are transferred almost completely to F, with little covalent bonding character left. 50
51 Lattice enthalpy is very large for crystal formation. 51
52 5-4 Molecular orbitals for Larger Molecules Application of group theory to construct MOs: 1. Determine the point group of the molecule. Simplify the higher-order of point groups, i.e., D 2h for D h, C 2v for C v. 2. Assign x, y, & z coordinates to the atoms, chosen for convenience, i.e., the highest-order rotation axis is chosen as the z axis. 52
53 3. Find the characters of the representation for the combination of the 2s orbitals on the outer atoms & then repeat the process, finding the representations for each of the other sets of orbitals (p x, p y, & p z ). Later, these will be combined with the appropriate orbital of the central atoms. * Assuming that valence electrons are in 2s & 2p. 53
54 4. Reduce each representation from Step 3 to its irreducible representations. This is equivalent to finding the symmetry of the group orbitals or the symmetry-adapted linear combinations (SALCs) of the orbitals. Ψ = Φ SALC c i i So that Ψ can combine with the AOs of central atom if they have the same symmetry & similar energy. Atomic orbitals of outer atoms. 54
55 5. Find the atomic orbitals of the central atom with the same symmetries (irreducible representations) as those found in Step Combine the atomic orbitals of the central atom & those of the SALC group orbitals with the same symmetry & similar energy to form Molecular orbitals. Ψ MO = aψ a + bψ SALC 55
56 In summary, the process used in creating molecular orbitals is to match the symmetries of the group orbitals using their irreducible representations with the symmetries of the central atom orbitals. 56
57 5-4-1 FHF - (D h ) D 2h 1. Group orbital : 2 x F (2s, 2p) As usual, we need to consider only the valence atomic orbitals (2s & 2p). The symmetry of each group orbital (SALC) can be found by comparing its behavior with each symmetry operation with the irreducible representations of the character table. F-H-F 57
58 D 2h E C 2 (z) C 2 (y) C 2 (x) i σ (xy) σ (xz) σ (yz) A g x 2, y 2, z 2 B 1g R z xy B 2g R y xz B 3g R x yz A u B 1u z B 2u y B 3u x Γs Reducible representation Γp Reducible representation F-H-F y x z 58
59 Reducible to irreducible representations For Γs: N(A g ) = (1/8)x[2x1+2x1+2x1+2x1] = 1 N(B 1u ) = (1/8)x[2x1+2x1+2x1+2x1] = 1 All the rest are zero. So that, Γs = A g + B 1u For Γp: N(A g ) = (1/8)x[6x1+(-2)x1+2x1+2x1] = 1 N(B 2g ) = (1/8)x[6x1+(-2)x(-1)+2x1+2x(-1)] = 1 All the rest follow the same procedure. So we have Γp = A g + B 2g + B 3g + B 1u + B 2u + B 3u However, it is a little hard to arrange p x, p y and p z as shown in Figure Therefore, we can process p x, p y & p z separately. 59
60 Derive the reducible representations for p x, p y, & p z Γpx Γpy Γpz Following the same procedure as previous page, we have Γpx = B 2g + B 3u Γpy = B 3g + B 2u Γpz = A g + B 1u Plus what we obtained earlier: Γs = A g + B 1u 60
61 Ψ = ( 1/ 2)[ φ 1 ± φ2] For two outer AOs, there are only two possibilities: + or - 61
62 H : 1s (A g ) 62
63 a g Antibonding with aψ H ( 1s) bψ (2 p) cψ (2s) SALC SALC Small contribution of F 2s orbitals makes its energy higher ev a g non-bonding orbitals (symmetry does not match) 2p x,y a g 2p z a g a g ev 3-center, 2-electron bond The concept of MO differs from conventional Lewis approach. a g ev 63
64 MO versus Lewis Structure The Lewis approach to bonding requires two electrons to represent a single bond between two atoms & would result in four electrons around the hydrogen atom of FHF -. The molecular orbital picture is more successful, with a 2-electron bond delocalized over 3 atoms (a 3-center, 2-electron bond : 2 electrons occupy a low-energy orbital formed by the interaction of all 3 atoms (a central atom & a two-atom group orbital)). 64
65 5-4-2 CO 2 (D h ) D 2h Although the group orbitals for the oxygen atoms are identical to the group orbitals for the fluorine atoms in FHF -, the central carbon atom in CO 2 has both s & p orbitals capable of interacting with the 2p group orbitals on the oxygen atoms. 65
66 The SALC group orbitals of outer atoms (the above procedure for 2s, 2p x, 2p y, & 2p z ) 66
67 Find the symmetry of central orbitals 67
68 Match the symmetries of central & outer orbitals 68
69 Symmetry allowed but energies poorly matched -19.4eV -32.4eV -10.7eV -32.4eV 69
70 Symmetry allowed & energies better matched eV -15.9eV -10.7eV -15.7eV 70
71 5.2 ev 3.5 ev better 16.5 ev not good (poor)
72 Both symmetry & energy match Symmetry does not match In cases 6 & 8, symmetry does not match, forming nonbonding MO. 72
73 In summary MO of CO 2 antibonding (σ or π?) nonbonding Bonding but more like nonbonding 73
74 5-4-3 H 2 O (C 2v ) 74
75 Molecular orbitals of nonlinear molecules 1. H 2 O - a simple triatomic bent molecule with C 2v point group 2. The C 2 axis is chosen as the z axis & the xz plane as the plane of the molecule (no assigned axis for H atoms) 75
76 3. Because the hydrogen atoms determine the symmetry of the molecule, we will use their orbitals as a starting point. 4. Γ = A 1 (+ : constructive) + B 1 (- : destructive) (1s±1s) 5. 2s : A 1 2p x : B 1 2p y : B 2 2p z : A 1 6. The atomic & group orbitals with the same symmetry are combined into molecular orbitals. 76
77 Or 1s atomic orbitals of H atoms A1 B1 77
78 Check symmetry match for central & outer orbitals 78
79 Ψ MO = aψ a + bψ SALC 79
80 1b 2 : 2p y (nonbonding) 3 a1 : 2p z 1b 1 : 2px 2a 1 : 2s -15.9eV b 1 a 1 B 1 A 1 B 1 b 2 A eV Nonbonding Nonbonding but slightly bonding a 1 b 1 Lewis Structure -32.4eV A 1 2a 1 should be higher in energy (poor energy match). a 1 80
81 5-4-4 NH 3 (C 3v ) 81
82 82
83 Γ = H N 3 bonding 2 antibonding 1 antibonding + 1 nonbonding (=1+1) (=?) (=1-1) Γ = A 1 + E 83
84 mixed Lone pair on N slightly bonding 84
85 5-4-5 BF 3 (D 3h ) Derive reducible representations : Γ σ & Γ π Lewis Acid π-interactions? Acceptor! z-axis is perpendicular to the plane σ could be s-orbitals or p y orbitals on F atoms. 85
86 Γ σ = A 1 + E Symmetry for SALC group orbitals (3F) 86
87 3 bonding 2 antibonding 1 antibonding + 1 nonbonding + p y + p x 1 ( σ + σ + ) 1 ' = σ Ψ a SALC ' Ψ e ' SALC = ( 2σ σ σ ) = ( σ σ ) Ψ e SALC σ could be s-orbitals or p y orbitals on F atoms. How these Ψ SALC were obtained? Projection Operator Method 87
88 Projection Operator Method For cyclic systems, reduce D 3h to C 3 since its transformation properties will be preserved. Γ σ = A 1 + E becomes Γ σ = A + E Projection Operator p l h () i i () i = x ( R) R [ ] R l i : dimension h: order R: operator (ex. E, C 3, (C 3 ) 2 etc) x (i) : character of Γ i under operator R 88
89 Note: clockwise system is used here. (1) (2) Taking appropriate linear combination and normalization: } (1) +(2) [(1)-(2)]/i (1) +(2) [(1)-(2)]/i a 1 e 89
90 Symmetry match of central & outer orbitals Group orbitals Central atom s p x p y 90
91 Group orbitals 2p z Group orbitals 2p x G pz or G p A 2 + E G px A 2 + E Applying projection operator method, similar results can be obtained. 1 Ψ a SALC = ( φ + φ + φ ) Ψ e SALC = ( 2φ φ φ ) = ( φ φ ) Ψ e SALC
92 Symmetry match of SALC group orbitals for outer atoms (3F) & the atomic orbitals of the central atom (B) 92
93 LUMO (electron-pair acceptor) The > 10 ev separation between the B & F 2p orbitals energies means this π-bond is weak. HOMO Nonbonding orbitals Slightly bonding orbitals Weak π interactions Bonding orbitals π bonds 93
94 The qualitative methods described do not allow us to determine the energies of the molecular orbitals, but we can place them in approximate order from their shapes & the expected overlap. s & p : 345 kj/mol (avg) [63 : (C 6 H 5 ) 3 C C(C 6 H 5 ) 3 ; 628 : H C C C C H]
95 5-4-7 Hybrid orbitals CH 4, NH 3, H 2 O s Γ = A 1 + T 2 Therefore, C atom uses 2s and 2p (x,y,z) for hybridization. 95
96 96
97 1. Determine the molecular shape by VSEPR. 2. Find the reducible representation (Γ). 3. The atomic orbitals that fit the irreducible representations are those used in the hybrid orbitals. 97
98 Γ = A 1 + E A 1 : s E : p x, p y ; d x 2-y 2, d xy 2s, 2p x, 2p y (sp 2 ) The 2p z orbital is not involved in the bonding & serves as an acceptor in acid-base reactions (LUMO). 98
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