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Chapter 10 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 10 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 10 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 10 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 10 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula
CHAPTER 10 Molecular structure The concepts developed in Chapter 9, particularly those of orbitals, can be extended to a description of the electronic structures of molecules There are two principal quantum mechanical theories of molecular electronic structure In ‘valence-bond theory’, the starting point is the concept of the shared electron pair 10D Heteronuclear diatomic molecules The MO theory of heteronuclear diatomic molecules introduces the possibility that the atomic orbitals on the two atoms contribute unequally to the molecular orbital As a result, the molecule is polar The polarity can be expressed in terms of the concept of electronegativity 10A Valence-bond theory In this Topic we see how to write the wavefunction for a shared electron pair, and how it may be extended to account for the structures of a wide variety of molecules The theory introduces the concepts of σ and π bonds, promotion, and hybridization that are used widely in chemistry 10B Principles of molecular orbital theory Almost all modern computational work makes use of molecular orbital theory (MO theory), and we concentrate on that theory in this chapter In MO theory, the concept of atomic orbital is extended to that of ‘molecular orbital’, which is a wavefunction that spreads over all the atoms in a molecule The Topic begins with an account of the hydrogen molecule, which sets the scene for the application of MO theory to more complicated molecules 10C Homonuclear diatomic molecules The principles established for the hydrogen molecule are readily extended to other homonuclear diatomic molecules, the principal difference being that more types of atomic orbital must be included to give a more varied collection of molecular orbitals The building-up principle for atoms is extended to the occupation of molecular orbitals and used to predict the electronic structure of molecules 10E Polyatomic molecules Most molecules are polyatomic, so it is important to be able to account for their electronic structure An early approach to the electronic structure of planar conjugated polyenes is the ‘Hückel method’ This procedure introduces severe approximations, but sets the scene for more sophisticated procedures These more sophisticated procedures have given rise to what is essentially a huge and vibrant theoretical chemistry industry in which elaborate computations are used to predict molecular properties In this Topic we see a little of how those calculations are formulated What is the impact of this material? The concepts introduced in this chapter pervade the whole of chemistry and are encountered throughout the text We focus on two biochemical aspects here In Impact I10.1 we see how simple concepts account for the reactivity of small molecules that occur in organisms In Impact I10.2 we see a little of the contribution of computational chemistry to the explanation of the thermodynamic and spectroscopic properties of several biologically significant molecules To read more about the impact of this material, scan the QR code, or go to bcs.whfreeman.com/webpub/chemistry/ pchem10e/impact/pchem-10-1.html 10A Valence-bond theory Contents 10A.1 Diatomic molecules The basic formulation Brief illustration 10.A1: A valence-bond wavefunction (b) Resonance Brief illustration 10A.2: Resonance hybrids (a) 10A.2 Polyatomic molecules Brief illustration 10A.3: A polyatomic molecule (a) Promotion Brief illustration 10A.4: Promotion (b) Hybridization Brief illustration 10A.5: Hybrid structures Checklist of concepts Checklist of equations 400 400 400 401 402 402 402 403 403 403 405 405 406 ➤➤ Why you need to know this material? Valence-bond theory was the first quantum mechanical theory of bonding to be developed The language it introduced, which includes concepts such as spin pairing, σ and π bonds, and hybridization, is widely used throughout chemistry, especially in the description of the properties and reactions of organic compounds ➤➤ What is the key idea? A bond forms when an electron in an atomic orbital on one atom pairs its spin with that of an electron in an atomic orbital on another atom simplification at the outset Whereas the Schrödinger equation for a hydrogen atom can be solved exactly, an exact solution is not possible for any molecule because even the simplest molecule consists of three particles (two nuclei and one electron) We therefore adopt the Born–Oppenheimer approximation in which it is supposed that the nuclei, being so much heavier than an electron, move relatively slowly and may be treated as stationary while the electrons move in their field That is, we think of the nuclei as fixed at arbitrary locations, and then solve the Schrödinger equation for the wavefunction of the electrons alone The Born–Oppenheimer approximation allows us to select an internuclear separation in a diatomic molecule and then to solve the Schrödinger equation for the electrons at that nuclear separation Then we choose a different separation and repeat the calculation, and so on In this way we can explore how the energy of the molecule varies with bond length and obtain a molecular potential energy curve (Fig 10A.1) It is called a potential energy curve because the kinetic energy of the stationary nuclei is zero Once the curve has been calculated or determined experimentally (by using the spectroscopic techniques described in Topics 12C–12E and 13A), we can identify the equilibrium bond length, Re, the internuclear separation at the minimum of the curve, and the bond dissociation energy, D0, which is closely related to the depth, De, of the minimum below the energy of the infinitely widely separated and stationary atoms When more than one molecular parameter is changed in a polyatomic molecule, such as its various bond lengths and angles, we obtain a potential energy surface; the overall equilibrium shape of the molecule corresponds to the global minimum of the surface You need to know about atomic orbitals (Topic 9A) and the concepts of normalization and orthogonality (Topic 7C) This Topic also makes use of the Pauli principle (Topic 9B) Energy ➤➤ What you need to know already? Here we summarize essential topics of valence-bond theory (VB theory) that should be familiar from introductory chemistry and set the stage for the development of molecular orbital theory (MO theory) However, there is an important preliminary point All theories of molecular structure make the same Re Internuclear separation, R –De Figure 10A.1 A molecular potential energy curve The equilibrium bond length corresponds to the energy minimum 400 10 Molecular structure 10A.1 Diatomic molecules e1 We begin the account of VB theory by considering the simplest possible chemical bond, the one in molecular hydrogen, H2 rA1 rA2 A (a) The basic formulation A B (10A.1) if electron is on atom A and electron is on atom B; in this chapter, and as is common in the chemical literature, we use χ (chi) to denote atomic orbitals For simplicity, we shall write this wavefunction as Ψ(1,2) = A(1)B(2) When the atoms are close, it is not possible to know whether it is electron or electron that is on A An equally valid description is therefore Ψ(1,2) = A(2)B(1), in which electron is on A and electron is on B When two outcomes are equally probable, quantum mechanics instructs us to describe the true state of the system as a superposition of the wavefunctions for each possibility (Topic 7C), so a better description of the molecule than either wavefunction alone is one of the (unnormalized) linear combinations Ψ(1,2) = A(1)B(2) ± A(2)B(1) The combination with lower energy is the one with a + sign, so the valence-bond wavefunction of the electrons in an H2 molecule is A valence-bond wavefunction Ψ (1, 2) = A(1)B(2) + A(2)B(1) (10A.2) The reason why this linear combination has a lower energy than either the separate atoms or the linear combination with a negative sign can be traced to the constructive interference between the wave patterns represented by the terms A(1)B(2) and A(2)B(1), and the resulting enhancement of the probability density of the electrons in the internuclear region (Fig 10A.2) Brief illustration 10.A1 A valence-bond wavefunction The wavefunction in eqn 10A.2 might look abstract, but in fact it can be expressed in terms of simple exponential functions Thus, if we use the wavefunction for an H1s orbital (Z = 1) given in Topic 9A, then, with the radii measured from their respective nuclei (1), A(1) B (2) A(2) 1 Ψ (1, 2) = e − rA1 /a0 × e − rB /a0 + e − rA /a0 1/2 1/2 (πa0 ) (πa0 ) (πa03 )1/2 B (1) × R rB1 e2 rB2 B The spatial wavefunction for an electron on each of two widely separated H atoms is Ψ (1, 2) = χ H1s (r1 ) χ H1s (r2 ) r12 1 e − rB1 /a0 = {e −(rA1 +rB )/a0 + e −(rA +rB1 )/a0 } (πa03 )1/2 πa0 Self-test 10A.1 Express this wavefunction in terms of the Cartesian coordinates of each electron given that the inter nuclear separation (along the z-axis) is R Answer: rAi = (xi2 + yi2 + zi2 )1/2 , rBi = (xi2 + yi2 + (zi − R)2 )1/2 A(1)B(2) A(1)B(2) + A(2)B(1) A(2)B(1) Enhanced electron density Figure 10A.2 It is very difficult to represent valencebond wavefunctions because they refer to two electrons simultaneously However, this illustration is an attempt The atomic orbital for electron is represented by the purple shading, and that of electron is represented by the green shading The left illustration represents A(1)B(2), and the right illustration represents the contribution A(2)B(1) When the two contributions are superimposed, there is interference between the purple contributions and between the green contributions, resulting in an enhanced (two-electron) density in the internuclear region The electron distribution described by the wavefunction in eqn 10A.2 is called a σ bond A σ bond has cylindrical symmetry around the internuclear axis, and is so called because, when viewed along the internuclear axis, it resembles a pair of electrons in an s orbital (and σ is the Greek equivalent of s) A chemist’s picture of a covalent bond is one in which the spins of two electrons pair as the atomic orbitals overlap The origin of the role of spin, as we show in the following Justification, is that the wavefunction in eqn 10A.2 can be formed only by a pair of spin-paired electrons Spin pairing is not an end in itself: it is a means of achieving a wavefunction and the probability distribution it implies that corresponds to a low energy Justification 10.A1 Electron pairing in VB theory The Pauli principle requires the overall wavefunction of two electrons, the wavefunction including spin, to change sign when the labels of the electrons are interchanged (Topic 9B) The overall VB wavefunction for two electrons is 10A Valence-bond theory Ψ (1, 2) = {A(1)B(2) + A(2)B(1)}σ(1, 2) + where σ represents the spin component of the wavefunction When the labels and are interchanged, this wavefunction becomes Internuclear axis – = {A(1)B(2) + A(2)B(1)}σ(2,1) – The Pauli principle requires that Ψ(2,1) = −Ψ(1,2), which is satisfied only if σ(2,1) = −σ(1,2) The combination of two spins that has this property is σ − (1, 2) = (1/ 21/2 ){α(1)β(2) − β(1)α(2)} Nodal plane + Ψ (2,1) = {A(2)B(1) + A(1)B(2)}σ(2,1) 401 which corresponds to paired electron spins (Topic 9C) Therefore, we conclude that the state of lower energy (and hence the formation of a chemical bond) is achieved if the electron spins are paired The VB description of H2 can be applied to other homonuclear diatomic molecules For N2, for instance, we consider the valence electron configuration of each atom, which is 2s2 2p1x 2p1y 2p1z It is conventional to take the z-axis to be the internuclear axis, so we can imagine each atom as having a 2pz orbital pointing towards a 2pz orbital on the other atom (Fig 10A.3), with the 2px and 2py orbitals perpendicular to the axis A σ bond is then formed by spin pairing between the two electrons in the two 2pz orbitals Its spatial wavefunction is given by eqn 10A.2, but now A and B stand for the two 2pz orbitals The remaining N2p orbitals cannot merge to give σ bonds as they not have cylindrical symmetry around the internuclear axis Instead, they merge to form two π bonds A π bond arises from the spin pairing of electrons in two p orbitals that approach side-by-side (Fig 10A.4) It is so called because, viewed along the inter-nuclear axis, a π bond resembles a pair of electrons in a p orbital (and π is the Greek equivalent of p) There are two π bonds in N2, one formed by spin pairing in two neighbouring 2px orbitals and the other by spin pairing in two neighbouring 2py orbitals The overall bonding pattern in Figure 10A.4 A π bond results from orbital overlap and spin pairing between electrons in p orbitals with their axes perpendicular to the internuclear axis The bond has two lobes of electron density separated by a nodal plane + – – + + + – – + – – Figure 10A.5 The structure of bonds in a nitrogen molecule, with one σ bond and two π bonds The overall electron density has cylindrical symmetry around the internuclear axis N2 is therefore a σ bond plus two π bonds (Fig 10A.5), which is consistent with the Lewis structure :N ≡ N: for nitrogen (b) Resonance Another term introduced into chemistry by VB theory is resonance, the superposition of the wavefunctions representing different electron distributions in the same nuclear framework To understand what this means, consider the VB description of a purely covalently bonded HCl molecule, which could be written as Ψ = A(1)B(2) + A(2)B(1), with A now a H1s orbital and B a Cl2p orbital However, this description is physically unlikely: it allows electron to be on the H atom when electron is on the Cl atom, and vice versa, but it does not allow for the possibility that both electrons are on the Cl atom (Ψ = B(1)B(2), representing H+Cl−) or even on the H atom (Ψ = A(1)A(2), representing the much less likely H−Cl+) A better description of the wavefunction for the molecule is as a superposition of the covalent and ionic descriptions, and we write (with a slightly simplified notation, and ignoring the less likely H−Cl+ possibility) ΨHCl =ΨH−Cl + λΨH Cl with λ (lambda) some numerical coefficient In general, we write + Figure 10A.3 The orbital overlap and spin pairing between electrons in two collinear p orbitals that results in the formation of a σ bond Ψ =Ψcovalent + λΨionic − (10A.3) 402 10 Molecular structure If an arbitrary wavefunction is used to calculate the energy, then the value calculated is never less than the true energy Variation principle where Ψcovalent is the two-electron wavefunction for the purely covalent form of the bond and Ψionic is the two-electron wavefunction for the ionic form of the bond The approach summarized by eqn 10A.3, in which we express a wavefunction as the superposition of wavefunctions corresponding to a variety of structures with the nuclei in the same locations, is called resonance In this case, where one structure is pure covalent and the other pure ionic, it is called ionic–covalent resonance The interpretation of the wavefunction, which is called a resonance hybrid, is that if we were to inspect the molecule, then the probability that it would be found with an ionic structure is proportional to λ2 If λ2 is very small, the covalent description is dominant If λ2 is very large, the ionic description is dominant Resonance is not a flickering between the contributing states: it is a blending of their characteristics, much as a mule is a blend of a horse and a donkey It is only a mathematical device for achieving a closer approximation to the true wavefunction of the molecule than that represented by any single contributing structure alone A systematic way of calculating the value of λ is provided by the variation principle which is proved in Topic 10C: The arbitrary wavefunction is called the trial wavefunction The principle implies that, if we vary the parameter λ in the trial wavefunction until the lowest energy is achieved (by evaluating the expectation value of the hamiltonian for the wavefunction), then that value of λ will be the best and through λ2 represents the appropriate contribution of the ionic wavefunction to the resonance hybrid Brief illustration 10A.2 Resonance hybrids Consider a bond described by eqn 10A.3 We might find that the lowest energy is reached when λ = 0.1, so the best description of the bond in the molecule is a resonance structure described by the wavefunction Ψ = Ψ covalent + 0.1Ψ ionic This wavefunction implies that the probabilities of finding the molecule in its covalent and ionic forms are in the ratio 100:1 (because 0.12 = 0.01) 10A.2 Polyatomic molecules Each σ bond in a polyatomic molecule is formed by the spin pairing of electrons in atomic orbitals with cylindrical symmetry around the relevant internuclear axis Likewise, π bonds are formed by pairing electrons that occupy atomic orbitals of the appropriate symmetry Brief illustration 10A.3 A polyatomic molecule The VB description of H2O will make this approach clear The valence-electron configuration of an O atom is 2s2 2p2x 2p1y 2p1z The two unpaired electrons in the O2p orbitals can each pair with an electron in an H1s orbital, and each combination results in the formation of a σ bond (each bond has cylindrical symmetry about the respective O eH internuclear axis) Because the 2px and 2py orbitals lie at 90° to each other, the two σ bonds also lie at 90° to each other (Fig 10A.6) We predict, therefore, that H2O should be an angular molecule, which it is However, the theory predicts a bond angle of 90°, whereas the actual bond angle is 104.5° H1s H1s O2px O2py Figure 10A.6 In a primitive view of the structure of an H2O molecule, each bond is formed by the overlap and spin pairing of an H1s electron and an O2p electron Self-test 10A.2 Use VB theory to suggest a shape for the ammonia molecule, NH3 Answer: Trigonal pyramidal with HNH bond angle 90°; experimental: 107° Resonance plays an important role in the valence-bond description of polyatomic molecules One of the most famous examples of resonance is in the VB description of benzene, where the wavefunction of the molecule is written as a superposition of the many-electron wavefunctions of the two covalent Kekulé structures: ψ=ψ ( ) ( ) +ψ (10A.4) The two contributing structures have identical energies, so they contribute equally to the superposition The effect of resonance , in (which is represented by a double-headed arrow, this case) is to distribute double-bond character around the ring and to make the lengths and strengths of all the carbon–carbon bonds identical The wavefunction is improved by allowing resonance because it allows for a more accurate description of the location of the electrons, and in particular the distribution can adjust into a state of lower energy This lowering is called the resonance stabilization of the molecule and, in the context of VB theory, is largely responsible for the unusual stability of 10A Valence-bond theory aromatic rings Resonance always lowers the energy, and the lowering is greatest when the contributing structures have similar energies The wavefunction of benzene is improved still further, and the calculated energy of the molecule is lowered further still, if we allow ionic–covalent resonance too, by allow+ – ing a small admixture of structures such as (a) Promotion As pointed out in Brief illustration 10A.3, simple VB theory predicts a bond angle of 90°, whereas the actual bond angle is 104.5° Another deficiency of this initial formulation of VB theory is its inability to account for carbon’s tetravalence (its ability to form four bonds) The ground-state configuration of C is 2s2 2p1x 2p1y , which suggests that a carbon atom should be capable of forming only two bonds, not four This deficiency is overcome by allowing for promotion, the excitation of an electron to an orbital of higher energy In carbon, for example, the promotion of a 2 s electron to a 2p orbital can be thought of as leading to the configuration 2s1 2p1x 2p1y 2p1z, with four unpaired electrons in separate orbitals These electrons may pair with four electrons in orbitals provided by four other atoms (such as four H1s orbitals if the molecule is CH4), and hence form four σ bonds Although energy was required to promote the electron, it is more than recovered by the promoted atom’s ability to form four bonds in place of the two bonds of the unpromoted atom Promotion, and the formation of four bonds, is a characteristic feature of carbon because the promotion energy is quite small: the promoted electron leaves a doubly occupied 2s orbital and enters a vacant 2p orbital, hence significantly relieving the electron–electron repulsion it experiences in the former However, it is important to remember that promotion is not a ‘real’ process in which an atom somehow becomes excited and then forms bonds: it is a notional contribution to the overall energy change that occurs when bonds form Brief illustration 10A.4 Promotion Sulfur can form six bonds (an ‘expanded octet’), as in the molecule SF6 Because the ground-state electron configuration of sulfur is [Ne]3s23p , this bonding pattern requires the promotion of a 3s electron and a 3p electron to two different 3d orbitals, which are nearby in energy, to produce the notional configuration [Ne]3s13p33d with all six of the valence electrons in different orbitals and capable of bond formation with six electrons provided by six F atoms Self-test 10A.3 Account for the ability of phosphorus to form five bonds, as in PF5 Answer: Promotion of a 3s electron from [Ne]3s23p3 to [Ne]3s13p33d1 403 (b) Hybridization The description of the bonding in CH4 (and other alkanes) is still incomplete because it implies the presence of three σ bonds of one type (formed from H1s and C2p orbitals) and a fourth σ bond of a distinctly different character (formed from H1s and C2s) This problem is overcome by realizing that the electron density distribution in the promoted atom is equivalent to the electron density in which each electron occupies a hybrid orbital formed by interference between the C2s and C2p orbitals of the same atom The origin of the hybridization can be appreciated by thinking of the four atomic orbitals centred on a nucleus as waves that interfere destructively and constructively in different regions, and give rise to four new shapes As we show in the following Justification, the specific linear combinations that give rise to four equivalent hybrid orbitals are h1 = s + p x + p y + pz h2 = s − p x − p y + pz h3 = s − p x + p y − pz h4 = s + p x − p y − pz sp3 hybrid orbitals (10A.5) As a result of the interference between the component orbitals, each hybrid orbital consists of a large lobe pointing in the direction of one corner of a regular tetrahedron (Fig 10A.7) The angle between the axes of the hybrid orbitals is the tetrahedral angle, arccos(−1/3) = 109.47° Because each hybrid is built from one s orbital and three p orbitals, it is called an sp3 hybrid orbital Justification 10A.2 Determining the form of tetrahedral hybrid orbitals We begin by supposing that each hybrid can be written in the form h = as + bx px + by py + bz pz The hybrid h1 that points to the (1,1,1) corner of a cube must have equal contributions from all three p orbitals, so we can set the three b coefficients equal to each other and write h1 = as + b(px + py + pz) The other three hybrids have the same composition (they are equivalent, apart from their direction in space), but are orthogonal to h1 This orthogonality is achieved by choosing different signs for the p-orbitals but the same overall composition For instance, we might choose h 2 = as + b(−px − p y + pz), in which case the orthogonality condition is ∫ h h dτ = ∫ (as + b(p + p x =a ∫ s dτ − b ∫ 2 ∫ − b p x p y dτ + y + pz ))(as + b(− p x − p y + pz ))dτ p2x dτ − ∫ − ab sp x dτ − = a − b2 − b2 + b2 = a − b2 = We conclude that a solution is a = b (the alternative solution, a = −b, simply corresponds to choosing different absolute 404 10 Molecular structure phases for the p orbitals) and the two hybrid orbitals are the h1 and h2 in eqn 10A.3 A similar argument but with h3 = as + b (−px + py − pz) or h4 = as + b(px − py − pz) leads to the other two hybrids in eqn 10A.3 109.47° A hybrid orbital has enhanced amplitude in the internuclear region, which arises from the constructive interference between the s orbital and the positive lobes of the p orbitals As a result, the bond strength is greater than for a bond formed from an s or p orbital alone This increased bond strength is another factor that helps to repay the promotion energy Hybridization is used to describe the structure of an ethene molecule, H2CaCH2, and the torsional rigidity of double bonds An ethene molecule is planar, with HCH and HCC bond angles close to 120° To reproduce the σ bonding structure, each C atom is regarded as promoted to a 2 s12p3 configuration However, instead of using all four orbitals to form hybrids, we form sp2 hybrid orbitals: h1 = s + 21/2 p y h2 = s + ( 23 ) p x − ( 12 ) p y 1/2 Figure 10A.7 An sp3 hybrid orbital formed from the superposition of s and p orbitals on the same atom There are four such hybrids: each one points towards the corner of a regular tetrahedron The overall electron density remains spherically symmetrical It is now easy to see how the valence-bond description of the CH4 molecule leads to a tetrahedral molecule containing four equivalent CeH bonds Each hybrid orbital of the promoted C atom contains a single unpaired electron; an H1s electron can pair with each one, giving rise to a σ bond pointing in a tetrahedral direction For example, the (un-normalized) two-electron wavefunction for the bond formed by the hybrid orbital h1 and the 1sA orbital (with wavefunction that we shall denote A) is Ψ (1, 2) = h1 (1)A(2) + h1 (2)A(1) (10A.6) As for H2, to achieve this wavefunction, the two electrons it describes must be paired Because each sp3 hybrid orbital has the same composition, all four σ bonds are identical apart from their orientation in space (Fig 10A.8) 1/2 h3 = s − ( 23 ) p x − ( 12 ) p y 1/2 sp2 hybrid orbitals (10A.7) 1/2 These hybrids lie in a plane and point towards the corners of an equilateral triangle at 120° to each other (Fig 10A.9 and Problem 10A.3) The third 2p orbital (2pz) is not included in the hybridization; its axis is perpendicular to the plane in which the hybrids lie The different signs of the coefficients, as well as ensuring that the hybrids are mutually orthogonal, also ensure that constructive interference takes place in different regions of space, so giving the patterns in the illustration The sp2-hybridized C atoms each form three σ bonds by spin pairing with either the h1 hybrid of the other C atom or with H1s orbitals The σ framework therefore consists of CeH and CeC σ bonds at 120° to each other When the two CH2 groups lie in the same plane, the two electrons in the unhybridized p orbitals can pair and form a π bond (Fig 10A.10) The formation of this π bond locks the framework into the planar arrangement, for any rotation of one CH2 group relative to the other leads to a weakening of the π bond (and consequently an increase in energy of the molecule) H C 120° (a) Figure 10A.8 Each sp3 hybrid orbital forms a σ bond by overlap with an H1s orbital located at the corner of the tetrahedron This model accounts for the equivalence of the four bonds in CH4 (b) Figure 10A.9 (a) An s orbital and two p orbitals can be hybridized to form three equivalent orbitals that point towards the corners of an equilateral triangle (b) The remaining, unhybridized p orbital is perpendicular to the plane 10A Valence-bond theory 405 Table 10A.1 Some hybridization schemes Coordination number Arrangement Composition Linear sp, pd, sd Angular sd Trigonal planar sp2, p2d Unsymmetrical planar spd Trigonal pyramidal pd2 Tetrahedral sp3, sd3 Irregular tetrahedral spd2, p3d, dp3 Square planar p2d2, sp2d Trigonal bipyramidal sp3d, spd3 Tetragonal pyramidal sp2d2, sd4, pd4, p3d2 Pentagonal planar p2d3 Octahedral sp3d2 Trigonal prismatic spd4, pd5 Trigonal antiprismatic p3d3 Figure 10A.10 A representation of the structure of a double bond in ethene; only the π bond is shown explicitly A similar description applies to ethyne, HC ≡ CH, a linear molecule Now the C atoms are sp hybridized, and the σ bonds are formed using hybrid atomic orbitals of the form h1 = s + pz h2 = s − pz sp hybrid orbitals (10A.8) These two hybrids lie along the internuclear axis The electrons in them pair either with an electron in the corresponding hybrid orbital on the other C atom or with an electron in one of the H1s orbitals Electrons in the two remaining p orbitals on each atom, which are perpendicular to the molecular axis, pair to form two perpendicular π bonds (Fig 10A.11) Other hybridization schemes, particularly those involving d orbitals, are often invoked in elementary descriptions of mol ecular structure to be consistent with other molecular geo metries (Table 10A.1) The hybridization of N atomic orbitals Figure 10A.11 A representation of the structure of a triple bond in ethyne; only the π bonds are shown explicitly The overall electron density has cylindrical symmetry around the axis of the molecule always results in the formation of N hybrid orbitals, which may either form bonds or may contain lone pairs of electrons Brief illustration 10A.5 Hybrid structures For example, sp3d hybridization results in six equivalent hybrid orbitals pointing towards the corners of a regular octahedron; it is sometimes invoked to account for the structure of octahedral molecules, such as SF6 (recall the promotion of sulfur’s electrons in Brief illustration 10A.4) Hybrid orbitals not always form bonds: they may also contain lone pairs of electrons For example, in the hydrogen peroxide molecule, H2O2, each O atom can be regarded as sp3 hybridized Two of the hybrid orbitals form bonds, one OeO bond and one OeH bond at approximately 109° (the experimental value is much less, at 94.8°) The remaining two hybrids on each atom accommodate lone pairs of electrons Rotation around the O e O bond is possible, so the molecule is conformationally mobile Self-test 10A.4 Account for the structure of methylamine, CH3NH2 Answer: C, N both sp3 hybridized; a lone pair on N Checklist of concepts ☐ 1 The Born–Oppenheimer approximation treats the nuclei as stationary while the electrons move in their field ☐ 2 A molecular potential energy curve depicts the variation of the energy of the molecule as a function of bond length 406 10 Molecular structure ☐ 3 The equilibrium bond length is the internuclear separation at the minimum of the curve ☐ 4 The bond dissociation energy is the minimum energy need to separate the two atoms of a molecule ☐ 5 A bond forms when an electron in an atomic orbital on one atom pairs its spin with that of an electron in an atomic orbital on another atom ☐ 6 A σ bond has cylindrical symmetry around the internuclear axis ☐ 7 A π bond has symmetry like that of a p orbital perpendicular to the internuclear axis ☐ 8 Promotion is the notional excitation of an electron to an empty orbital to enable the formation of additional bonds ☐ 9 Hybridization is the blending together of atomic orbitals on the same atom to achieve the appropriate directional properties and enhanced overlap ☐ 10 Resonance is the superposition of structures with different electron distributions but the same nuclear arrangement Checklist of equations Property Equation Valence-bond wavefunction Ψ = A(1)B(2) + A(2)B(1) Resonance Ψ = Ψcovalent + λΨionic Hybridization h= ∑c χ i i i Comment Equation number 10A.2 Ionic–covalent resonance 10A.3 All atomic orbitals on the same atom; specific forms in the text 10A.5 10A.6 10B Principles of molecular orbital theory Contents 10B.1 Linear combinations of atomic orbitals The construction of linear combinations Example 10.B1: Normalizing a molecular orbital Brief illustration 10B.1: A molecular orbital (b) Bonding orbitals Brief illustration 10B.2: Molecular integrals (c) Antibonding orbitals Brief illustration 10B.3: Antibonding energies (a) 10B.2 Orbital notation Checklist of concepts Checklist of equations 407 407 408 408 409 410 411 411 412 412 412 ➤➤ Why you need to know this material? Molecular orbital theory is the basis of almost all descriptions of chemical bonding, including that of individual molecules and of solids It is the basis of almost all computational techniques for the prediction and analysis of the properties of molecules ➤➤ What is the key idea? Molecular orbitals are wavefunctions that spread over all the atom in a molecule and each one can accommodate up to two electrons ➤➤ What you need to know already? You need to be familiar with the shapes of atomic orbitals (Topic 9B) and how an energy is calculated from a wavefunction (Topic 7C) The entire discussion is within the framework of the Born–Oppenheimer approximation (Topic 10A) In molecular orbital theory (MO theory), electrons not belong to particular bonds but spread throughout the entire molecule This theory has been more fully developed than valence-bond theory (Topic 10A) and provides the language that is widely used in modern discussions of bonding To introduce it, we follow the same strategy as in Topic 9B, where the one-electron H atom was taken as the fundamental species for discussing atomic structure and then developed into a description of many-electron atoms In this chapter we use the simplest molecular species of all, the hydrogen molecule-ion, H2+ , to introduce the essential features of bonding and then use it to describe the structures of more complex systems 10B.1 Linear combinations of atomic orbitals The hamiltonian for the single electron in H2+ is H =− 2me ∇12 + V V = − 1 e2 + − πε rA1 rB1 R (10B.1) where rA1 and rB1 are the distances of the electron from the two nuclei A and B (1) and R is the distance between the two nuclei In the expression for V, the first two terms in parentheses are the attractive contribution from the interaction between the electron and the nuclei; the remaining term is the repulsive interaction between the nuclei The collection of fundamental constant e2/4πε0 occurs widely throughout this chapter, and we shall denote it j0 e rB1 rA1 A R B The one-electron wavefunctions obtained by solving the Schrödinger equation Hψ = Eψ are called molecular orbitals (MOs) A molecular orbital ψ gives, through the value of |ψ|2, the distribution of the electron in the molecule A molecular orbital is like an atomic orbital, but spreads throughout the molecule (a) The construction of linear combinations The Schrödinger equation can be solved analytically for H2+ (within the Born–Oppenheimer approximation), but the wavefunctions are very complicated functions; moreover, the solution cannot be extended to polyatomic systems Therefore, we adopt a simpler procedure that, while more approximate, can be extended readily to other molecules 10E Polyatomic molecules (b) Benzene and aromatic stability In butadiene it is E π = 2(α + 1.62β ) + 2(α + 0.62β ) = 4α + 4.48β Therefore, the energy of the butadiene molecule lies lower by 0.48β (about 110 kJ mol−1) than the sum of two individual π bonds This extra stabilization of a conjugated system compared with a set of localized π bonds is called the delocalization energy of the molecule A closely related quantity is the π-bond formation energy, Ebf, the energy released when a π bond is formed Because the contribution of α is the same in the molecule as in the atoms, we can find the π-bond formation energy from the π-electron binding energy by writing E bf = E π − N Cα Definition π-Bond formation energy (10E.12) where NC is the number of carbon atoms in the molecule The π-bond formation energy in butadiene, for instance, is 4.48β Example 10E.2 Estimating the delocalization energy Use the Hückel approximation to find the energies of the π orbitals of cyclobutadiene, and estimate the delocalization energy Method Set up the secular determinant using the same basis as for butadiene, but note that atoms A and D are also now neighbours Then solve for the roots of the secular equation and assess the total π-bond energy For the delocalization energy, subtract from the total π-bond energy the energy of two π-bonds Answer The hamiltonian matrix is α β β β α β 0 H = β α β β β α 0 1 = α 1+ β 0 1 431 1 0 1 1 2 Diagonalize → 1 0 0 0 0 0 0 0 0 0 0 −2 The most notable example of delocalization conferring extra stability is benzene and the aromatic molecules based on its structure In elementary accounts, benzene, and other aromatic compounds, is often expressed in a mixture of valence-bond and molecular orbital terms, with typically valence-bond language used for its σ framework and molecular orbital language used to describe its π electrons First, the valence-bond component The six C atoms are regarded as sp2 hybridized, with a single unhybridized perpendicular 2p orbital One H atom is bonded by (Csp2,H1s) overlap to each C carbon, and the remaining hybrids overlap to give a regular hexagon of atoms (Fig 10E.3) The internal angle of a regular hexagon is 120°, so sp2 hybridization is ideally suited for forming σ bonds We see that the hexagonal shape of benzene permits strain-free σ bonding Now consider the molecular orbital component of the description The six C2p orbitals overlap to give six π orbitals that spread all round the ring Their energies are calculated within the Hückel approximation by diagonalizing the hamiltonian matrix α β 0 β β α β 0 0 β α β 0 H = 0 β α β 0 0 β α β β 0 β α 0 1 0 =α 1+ β 0 0 1 0 1 2 0 0 0 1 0 Diagonalize → 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 −1 0 −2 The MO energies, the eigenvalues of this matrix, are simply E = α ± 2β , α ± β , α ± β (10E.13) Diagonalization gives the energies of the orbitals as E = α + 2β , α , α , α − 2β Four electrons must be accommodated Two occupy the lowest orbital (of energy α + 2β), and two occupy the doubly degenerate orbitals (of energy α) The total energy is therefore 4α + 4β Two isolated π bonds would have an energy 4α + 4β; therefore, in this case, the delocalization energy is zero H C Self-test 10E.3 Repeat the calculation for benzene (use software!) Answer: See next subsection Figure 10E.3 The σ framework of benzene is formed by the overlap of Csp2 hybrids, which fit without strain into a hexagonal arrangement 432 10 Molecular structure – + + + + – + + such as to be able to accommodate all the electrons in bonding orbitals, and the delocalization energy is large b2g – + e2u e1g + + + – – – + – – + + – + – – + – + + – – Example 10E.3 Judging the aromatic character of a molecule + a2u Figure 10E.4 The Hückel orbitals of benzene and the corresponding energy levels The symmetry labels are explained in Topic 11B The bonding and antibonding character of the delocalized orbitals reflects the numbers of nodes between the atoms In the ground state, only the bonding orbitals are occupied as shown in Fig 10E.4 The orbitals there have been given symmetry labels that are explained in Topic 11B Note that the lowest energy orbital is bonding between all neighbouring atoms, the highest energy orbital is antibonding between each pair of neighbours, and the intermediate orbitals are a mixture of bonding, nonbonding, and antibonding character between adjacent atoms The simple form of the eigenvalues in eqn 10E.13 suggests that there is a more direct way of determining them than by using mathematical software That is in fact the case, for symmetry arguments of the kind described in Topic 11B show that the 6 × 6 matrix can be factorized into two 1 × 1 matrices and two 2 × 2 matrices, which are very easy to deal with We now apply the building-up principle to the π system There are six electrons to accommodate (one from each C atom), so the three lowest orbitals (a2u and the doubly-degenerate pair e1 g) are fully occupied, giving the ground-state configuration a 22u e1g A significant point is that the only molecular orbitals occupied are those with net bonding character (the analogy with the very stable N2 molecule, Topic 10B, should be noted) The π-electron energy of benzene is Decide whether the molecules C H4 and the molecular ion C H2+ are aromatic when planar Method Follow the procedure for benzene Set up and solve the secular equations within the Hückel approximation, assuming a planar σ framework, and then decide whether the ion has nonzero delocalization energy Use mathematical software to diagonalize the hamiltonian (in Topic 11B it is shown how to use symmetry to arrive at the eigenvalues more simply.) Answer The hamiltonian matrix is 0 α β β 1 β α β = α 1+ β H = 0 0 β α β 1 β β α The matrix multiplying β diagonalizes as follows: 0 1 0 1 1 −2 Diagonalize → 1 0 0 0 0 0 0 0 0 0 It follows that the energy levels of the two species are E = α ± 2β, α, α There are four π electrons to accommodate in C 4H4, so the total π-bonding energy is 2(α + 2β ) + 2α = 4(α + β ) The energy of two localized π-bonds is 4(α + β ) Therefore, the delocalization energy is zero, so the molecule is not aromatic There are only two π electrons to accommodate in C H2+ , so the total π-bonding energy is 2(α + 2β ) = 2α + 4β The energy of a single localized π-bond is 2(α + β ), so the delocalization energy is 2β and the molecular-ion is aromatic Self-test 10E.4 What is the total π-bonding energy of C 3H3− ? Answer: 4α + 2β E = 2(α + 2β ) + 4(α + β ) = 6α + 8β If we ignored delocalization and thought of the molecule as having three isolated π bonds, it would be ascribed a π-electron energy of only 3(2α + 2β) = 6α + 6β The delocalization energy is therefore 2β ≈ −460 kJ mol−1, which is considerably more than for butadiene The π-bond formation energy in benzene is 8β This discussion suggests that aromatic stability can be traced to two main contributions First, the shape of the regular hexagon is ideal for the formation of strong σ bonds: the σ framework is relaxed and without strain Second, the π orbitals are 1 0 1 0 10E.3 Computational chemistry The severe assumptions of the Hückel method are now easy to avoid by using a variety of software packages that can be used not only to calculate the shapes and energies of molecu lar orbitals but also to predict with reasonable accuracy the 10E Polyatomic molecules (a) Semi-empirical and ab initio methods In a semi-empirical method, many of the integrals are estimated by appealing to spectroscopic data or physical properties such as ionization energies, and using a series of rules to set certain integrals equal to zero A primitive form of this procedure is used in Brief illustration 10D.1 of Topic 10D where we identify the integral α with a combination of the ionization energy and electron affinity of an atom In an ab initio method an attempt is made to calculate all the integrals, including overlap integrals Both procedures employ a great deal of computational effort and, along with cryptanalysts and meteorologists, theoretical chemists are among the heaviest users of the fastest computers The integrals that are required involve atomic orbitals that in general may be centred on different nuclei It can be appreciated that, if there are several dozen atomic orbitals used to build the molecular orbitals, then there will be tens of thousands of integrals of this form to evaluate (the number of integrals increases as the fourth power of the number of atomic orbitals in the basis) Some kind of approximation scheme is necessary One severe semi-empirical approximation used in the early days of computational chemistry was called complete neglect of differential overlap (CNDO), in which all molecular integrals are set to zero unless A and B are the same orbitals centred on the same nucleus, and likewise for C and D The surviving integrals are then adjusted until the energy levels are in good agreement with experiment or the computed enthalpy of 1 A more complete account with detailed examples will be found in our companion volume, Physical chemistry: Quanta, matter, and change (2014) formation of the compound is in agreement with experiment More recent semi-empirical methods make less draconian decisions about which integrals are to be ignored, but they are all descendants of the early CNDO technique Commercial packages are also available for ab initio calculations Here the problem is to evaluate as efficiently as possible thousands of integrals that arise from the Coulombic interaction between two electrons and have the form ( AB | CD) = j0 ∫ ∫ A(1)B(1) r 12 C(2)D(2)dτ 1dτ Notation Molecular integral (10E.14a) with j0 = e2/4πε0 and the possibility that each of the atomic orbitals A, B, C, D is centred on a different atom, a so-called ‘four-centre integral’ This task is greatly facilitated by expressing the atomic orbitals used in the LCAOs as linear combinations of Gaussian orbitals A Gaussian type orbital (GTO) is a function of the form e −r The advantage of GTOs over the correct orbitals (which for hydrogenic systems are proportional to exponential functions of the form e−r) is that the product of two Gaussian functions is itself a Gaussian function that lies between the centres of the two contributing functions (Fig 10E.5) In this way, the four-centre integrals like these become two-centre integrals of the form ( AB | CD) = j0 ∫ ∫ X(1) r 12 Y (2)dτ 1dτ (10E.14b) where X is the Gaussian corresponding to the product AB and Y is the corresponding Gaussian from CD Integrals of this form are much easier and faster to evaluate numerically than the original four-centre integrals Although more GTOs have to be used to simulate the atomic orbitals, there is an overall increase in speed of computation G1G2 (magnified) G1 y(x) structure and reactivity of molecules The full treatment of molecular electronic structure has received an enormous amount of attention by chemists and has become a keystone of modern chemical research However, the calculations are very complex, and all this section seeks to is to provide a brief introduction.1 In every case, the procedures focus on the calculation or estimation of integrals like HJJ and HIJ rather than setting them equal to the constants α or β, or ignoring them entirely In all cases the Schrödinger equation is solved iteratively and self-consistently, just as for the self-consistent field (SCF) approach to atoms (Topic 9B) First, the molecular orbitals for the electrons present in the molecule are formulated as LCAOs One molecular orbital is then selected and all the others are used to set up an expression for the potential energy of an electron in the chosen orbital The resulting SchrÖdinger equation is then solved numerically to obtain a better version of the chosen molecular orbital and its energy The procedure is repeated for all the molecular orbitals and used to calculate the total energy of the molecule The process is repeated until the computed orbitals and energy are constant to within some tolerance 433 G2 x Figure 10E.5 The product of two Gaussian functions on different centres is itself a Gaussian function located at a point between the two contributing Gaussians The scale of the product has been increased relative to that of its two components 434 10 Molecular structure (c) Graphical representations Brief illustration 10E.2 Gaussian type orbitals Suppose we consider a one-dimensional ‘homonuclear’ sys2 tem, with GTOs of the form e −ax located at and R Then one of the integrals that would have to be evaluated would include the term χ A (1)χ B (1) = e − ax e − a( x −R ) = e −2 ax 2 +2 axR −aR Next we note that −2a(x − 12 R)2 = −2ax + 2axR − 12 aR , so we can write χ A (1)χ B (1) = e 1 −2 a( x − R )2 − aR 2 =e 1 −2 a( x − R )2 − aR 2 e which is proportional to a single Gaussian (the term in blue) centred on the mid-point of the internuclear separation, at x = 12 R Self-test 10E.5 Repeat the analysis for a heteronuclear species 2 with GTOs of the form e −ax and e −bx Answer: χ A (1)χ B (1) = e −(cx −bR/c ) −a2 R /c , c = (a + b)1/2 (b) Density functional theory A technique that has gained considerable ground in recent years to become one of the most widely used techniques for the calculation of molecular structure is density functional theory (DFT) Its advantages include less demanding computational effort, less computer time, and—in some cases (particularly d-metal complexes)—better agreement with experimental values than is obtained from other procedures The central focus of DFT is the electron density, ρ, rather than the wavefunction, ψ The ‘functional’ part of the name comes from the fact that the energy of the molecule is a function of the electron density, written E[ρ], the electron density is itself a function of position, ρ(r), and in mathematics a function of a function is called a ‘functional’ The occupied orbitals are used to construct the electron density from ρ(r ) = ∑ m ,occupied ψ m (r ) Electron probability density (10E.15) and are calculated from modified versions of the Schrödinger equation known as the Kohn–Sham equations The Kohn–Sham equations are solved iteratively and selfconsistently First, the electron density is guessed For this step it is common to use a superposition of atomic electron densities Next, the Kohn–Sham equations are solved to obtain an initial set of orbitals This set of orbitals is used to obtain a better approximation to the electron density and the process is repeated until the density and the computed energy are constant to within some tolerance One of the most significant developments in computational chemistry has been the introduction of graphical representations of molecular orbitals and electron densities The raw output of a molecular structure calculation is a list of the coefficients of the atomic orbitals in each molecular orbital and the energies of these orbitals The graphical representation of a molecular orbital uses stylized shapes to represent the basis set, and then scales their size to indicate the coefficient in the linear combination Different signs of the wavefunctions are represented by different colours Once the coefficients are known, it is possible to construct a representation of the electron density in the molecule by noting which orbitals are occupied and then forming the squares of those orbitals The total electron density at any point is then the sum of the squares of the wavefunctions evaluated at that point (as in eqn 10E.15) The outcome is commonly represented by an isodensity surface, a surface of constant total electron density (Fig 10E.6) As shown in the illustration, there are several styles of representing an isodensity surface, as a solid form, as a transparent form with a ball-and-stick representation of the molecule within, or as a mesh A related representation is a solvent-accessible surface in which the shape represents the shape of the molecule by imagining a sphere representing a solvent molecule rolling across the surface and plotting the locations of the centre of that sphere One of the most important aspects of a molecule other than its geometrical shape is the distribution of charge over its surface, which is commonly depicted as an electrostatic potential surface (an ‘elpot surface’) The potential energy, Ep, of an imaginary positive charge Q at a point is calculated by taking into account its interaction with the nuclei and the electron density throughout the molecule Then, because Ep = Qϕ, where ϕ is the electric potential, the potential energy can be interpreted as a potential and depicted as an appropriate colour (Fig 10E.7) Electron-rich regions usually have negative potentials and electron-poor regions usually have positive potentials (a) (b) (c) Figure 10E.6 Various representations of an isodensity surface of ethanol: (a) solid surface, (b) transparent surface, and (c) mesh surface 10E Polyatomic molecules – + 435 Representations such as those we have illustrated are of critical importance in a number of fields For instance, they may be used to identify an electron-poor region of a molecule that is susceptible to association with or chemical attack by an electron-rich region of another molecule Such considerations are important for assessing the pharmacological activity of potential drugs Figure 10E.7 An elpot diagram of ethanol; the molecule has the same orientation as in Fig 10E.6 Red denotes regions of negative electrostatic potential and blue regions of positive potential (as in δ−O–Hδ+) Checklist of concepts ☐ 1 The Hückel method neglects overlap and interactions between atoms that are not neighbours ☐ 2 The Hückel method may be expressed in a compact manner by introducing matrices ☐ 3 The π-bond formation energy is the energy released when a π bond is formed ☐ 4 The π-electron binding energy is the sum of the energies of each π electron ☐ 5 The delocalization energy is the difference between the π-electron energy and the energy of the same molecule with localized π bonds ☐ 6 The highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) form the frontier orbitals of a molecule ☐ 7 The stability of benzene arises from the geometry of the ring and the high delocalization energy ☐ 8 Semi-empirical calculations approximate integrals by estimating integrals using empirical data; ab initio methods evaluate all integrals numerically ☐ 9 Density functional theories develop equations based on the electron density rather than the wavefunction itself ☐ 10 Graphical techniques are used to plot a variety of surfaces based on electronic structure calculations Checklist of equations Property Equation LCAO ψ= ∑c χ o o Comment Equation number χ0 are atomic orbitals 10E.1 Hückel approximation: S = 0 except between neighbours 10E.9 o Hückel equations Hc = ScE Diagonalization c−1Hc = E π-Bond formation energy Ebf = Eπ − NCα Definition; NC is the number of carbon atoms 10E.12 Molecular integrals ( AB | CD) = j0 ∫ ∫ A(1)B(1)(1/ r12 )C(2)D(2) dτ 1dτ A, B, C, D are atomic orbitals 10E.14a Electron probability density ρ(r ) = Sum over occupied molecular orbitals m 10E.15 ∑ψ m,occ 10E.10 m (r ) 436 10 Molecular structure CHAPTER 10 Molecular structure TOPIC 10A Valence-bond theory Discussion questions 10A.1 Discuss the role of the Born–Oppenheimer approximation in the calculation of a molecular potential energy curve or surface hybridization explain that in allene, CH2 = C = CH2, the two CH2 groups lie in perpendicular planes? 10A.2 Why are promotion and hybridization invoked in valence-bond theory? 10A.4 Why is spin-pairing so common a features of bond formation (in the 10A.3 Describe the various types of hybrid orbitals and how they are used to describe the bonding in alkanes, alkenes, and alkynes How does context of valence-bond theory)? 10A.5 What are the consequences of resonance? Exercises 10A.1(a) Write the valence-bond wavefunction for the single bond in HF 10A.1(b) Write the valence-bond wavefunction for the triple bond in N2 10A.4(a) Describe the bonding in 1,3-butadiene using hybrid orbitals 10A.4(b) Describe the bonding in 1,3-pentadiene using hybrid orbitals 10A.2(a) Write the valence-bond wavefunction for the resonance hybrid 10A.5(a) Show that the linear combinations h1 = s + px + py + pz and HF ↔ H+F− ↔ H−F+ (allow for different contributions of each structure) 10A.2(b) Write the valence-bond wavefunction for the resonance hybrid N2 ↔ N+N− ↔ N2–N2+ ↔ structures of similar energy h2 = s − px − py + pz are mutually orthogonal 10A.5(b) Show that the linear combinations h1 = (sin ζ )s + (cos ζ )p and h2 = (cos ζ )s − (sin ζ )p are mutually orthogonal for all values of the angle ζ (zeta) 10A.3(a) Describe the structure of a P2 molecule in valence-bond terms 10A.6(a) Normalize the sp2 hybrid orbital h = s + 21/2p given that the s and p 10A.3(b) Describe the structures of SO2 and SO3 in terms of valence bond 10A.6(b) Normalize the linear combinations in Exercise 10A.5b given that the Why is P4 a more stable form of molecular phosphorus than P2? theory orbitals are each normalized to s and p orbitals are each normalized to Problems 10A.1 An sp2 hybrid orbital that lies in the xy plane and makes an angle of 120° to the x-axis has the form ψ = 1/2 31/2 s − 21/2 p x + 21/2 p y 10A.2 Confirm that the hybrid orbitals in eqn 10A.5 make angles of 120° to each other 10A.3 Show that two equivalent hybrid orbitals of the form spλ make an angle θ to each other, then λ = −1/cos θ Plot a graph of λ against θ and confirm that θ = 180° when no s orbital is included and θ = 120° when λ = 2 Use hydrogenic atomic orbitals to write the explicit form of the hybrid orbital Show that it has its maximum amplitude in the direction specified TOPIC 10B Principles of molecular orbital theory Discussion questions 10B.1 What feature of molecular orbital theory is responsible for bond formation? 10B.2 Why is spin-pairing so common a features of bond formation (in the context of molecular orbital theory)? Exercises 10B.1(a) Normalize the molecular orbital ψ = ψA + λψB in terms of the parameter λ and the overlap integral S 10B.1(b) A better description of the molecule in Exercise 10B.1(a) might be obtained by including more orbitals on each atom in the linear combination Normalize the molecular orbital ψ = ψA + λψB + λ′ψB′ in terms of the parameters λ and λ′ and the appropriate overlap integrals S, where ψB and ψB′ are mutually orthogonal orbitals on atom B 10B.2(a) Suppose that a molecular orbital has the (unnormalized) form 0.145A + 0.844B Find a linear combination of the orbitals A and B that is orthogonal to this combination and determine the normalization constants of both combinations using S = 0.250 Exercises and problems 437 10B.2(b) Suppose that a molecular orbital has the (unnormalized) form where Eh = 27.2 eV, a0 = 52.9 pm, and EH = − 12 Eh 10B.3(b) The same data as in Exercise 10B.3(a) may be used to calculate the molecular potential energy curve for the antibonding orbital, which is given by eqn 10B.7 Plot the curve 10B.3(a) The energy of H2+ with internuclear separation R is given by 10B.4(a) Identify the g or u character of bonding and antibonding π orbitals formed by side-by-side overlap of p atomic orbitals 10B.4(b) Identify the g or u character of bonding and antibonding δ orbitals formed by face-to-face overlap of d atomic orbitals 0.727A + 0.144B Find a linear combination of the orbitals A and B that is orthogonal to this combination and determine the normalization constants of both combinations using S = 0.117 eqn 10B.4 The values of the contributions are given below Plot the molecular potential energy curve and find the bond dissociation energy (in electronvolts) and the equilibrium bond length R/a0 j/Eh 1.000 0.729 0.472 0.330 0.250 k/Eh 1.000 0.736 0.406 0.199 0.092 S 1.000 0.858 0.587 0.349 0.189 Problems 10B.1 Calculate the (molar) energy of electrostatic repulsion between two hydrogen nuclei at the separation in H2 (74.1 pm) The result is the energy that must be overcome by the attraction from the electrons that form the bond Does the gravitational attraction between them play any significant role? Hint: The gravitational potential energy of two masses is equal to −Gm1m2/r; G is listed inside the front cover 10B.2 Imagine a small electron-sensitive probe of volume 1.00 pm3 inserted into an H2+ molecule-ion in its ground state Calculate the probability that it will register the presence of an electron at the following positions: (a) at nucleus A, (b) at nucleus B, (c) half way between A and B, (c) at a point 20 pm along the bond from A and 10 pm perpendicularly Do the same for the molecule-ion the instant after the electron has been excited into the antibonding LCAO-MO 10B.3 Derive eqns 10B.4 and 10B.7 by working with the normalized LCAO-MOs for the H2+ molecule-ion Proceed by evaluating the expectation value of the hamiltonian for the ion Make use of the fact that A and B each individually satisfy the Schrödinger equation for an isolated H atom 10B.4 Examine whether occupation of the bonding orbital with one electron (as calculated in the preceding problem) has a greater or lesser bonding effect than occupation of the antibonding orbital with one electron Is that true at all internuclear separations? 10B.5‡ The LCAO-MO approach described in the text can be used to introduce numerical methods needed in quantum chemistry In this problem we evaluate the overlap, Coulomb, and resonance integrals numerically and compare the results with the analytical equations (eqns 10B.5) (a) Use the LCAO-MO wavefunction and the H2+ hamiltonian to derive equations for the relevant integrals and use mathematical software or an electronic spreadsheet to evaluate the overlap, Coulomb, and resonance integrals numerically, and the total energy for the 1sσg MO in the range a0