Physical Chemistry -Molecular Orbital Theory_lecture28to30, Lecture notes of Physics

Download Physical Chemistry -Molecular Orbital Theory_lecture28to30 and more Lecture notes Physics in PDF only on Docsity! 1 5.61 Physical Chemistry Lecture #28 MOLECULAR ORBITAL THEORY­ PART I At this point, we have nearly completed our crash­course introduction to quantum mechanics and we’re finally ready to deal with molecules. Hooray! To begin with, we are going to treat what is absolutely the simplest molecule we can imagine: H+ . This simple molecule will allow us to work 2 out the basic ideas of what will become molecular orbital (MO) theory. We set up our coordinate system as shown at right, with the electron positioned at r, and the two nuclei positioned at points RA and RB, at a distance R from one HA another. The Hamiltonian is RA easy to write down: H2 Ĥ = − 1 ∇2 − ∇2 A − ∇2 B − 1 − 1 + 1 ˆ2 r 2M A 2M B R̂ − r̂ R − r̂ R̂ − R̂ B A BA e­ HB RB r rArB R + Coordinates Electron HA HB e ­­HA e ­­HB HA ­HB Kinetic Kinetic Kinetic Attraction Attraction Repulsion Energy Energy Energy Now, just as was the case for atoms, we would like a picture where we can separate the electronic motion from the nuclear motion. For helium, we did this by noting that the nucleus was much heavier than the electrons and so we could approximate the center of mass coordinates of the system by placing the nucleus at the origin. For molecules, we will make a similar approximation, called the Born­Oppenheimer approximation. Here, we note again that the nuclei are much heavier than the electrons. As a result, they will move much more slowly than the light electrons. Thus, from the point of view of the electrons, the nuclei are almost sitting still and so the moving electrons see a static field that arises from fixed nuclei. A useful analogy here is that of gnats flying around on the back of an elephant. The elephant may be moving, but from the gnats’ point of view, the elephant is always more or less sitting still. The electrons are like the gnats and the nucleus is like the elephant. 2 5.61 Physical Chemistry Lecture #28 The result is that, if we are interested in the electrons, we can to a good approximation fix the nuclear positions – RA and RB – and just look at the motion of the electrons in a molecule. This is the B­Oppenheimer approximation, which is sometimes also called the clamped­nucleus approximation, for obvious reasons. Once the nuclei are clamped, we can make two simplifications of our Hamiltonian. First, we can neglect the kinetic energies of the nuclei because they are not moving. Second, because the nuclei are fixed, we can replace the operators R̂ A and R̂ B with the numbers RA and RB. Thus, our Hamiltonian reduces to ˆ rH (R ,R ) = − ∇2 − 1 1 1 − + el A B 2 R − r̂ R − r̂ R − R A B A B where the last term is now just a number – the electrostatic repulsion between two protons at a fixed separation. The second and third terms depends only on the position of the electron, r, and not its momentum, so we immediately identify those as a potential and write: ˆ ( ) ∇2 r R A , R B ( ) 1 H R ,R = − +V r̂ + el A B 2 eff R − R A B This Hamiltonian now only contains operators for the electrons (hence the subscript “el”), but the eigenvalues of this Hamiltonian depend on the distance, R, between the two nuclei. For example, the figure below shows the difference between the effective potentials the electron “feels” when the nuclei are close together versus far apart: R Small R Large Veff(r) Likewise, because the electron feels a different potential at each bond distance R, the wavefunction will also depend on R. In the same limits as above, we will have: R Small R Large ψel(r) 5 5.61 Physical Chemistry Lecture #28 E = 1• E = ∑ an 2 E0 0 0 n ⇒ Eavg − E0 = ∑ an 2 En − ∑ an 2 E0 = ∑ an 2 (E − E ) ≥ 0 since E ≥ En 0 n 0 n n n Where, in the last line we have noted that the sum of terms that are non­ negative must also be non­negative. It is important to note that the equals sign is only obtained if an=0 for all states that have En>E0. In this situation, ψ actually is the ground state of the system (or at least one ground state, if the ground state is degenerate). The variational method does two things for us. First, it gives us a means of comparing two different wavefunctions and telling which one is closer to the ground state – the one that has a lower average energy is the better approximation. Second, if we include parameters in our wavefunction variation gives us a means of optimizing the parameters in the following way. Assume that ψ depends on some parameters c – such as is the case for our LCAO wavefunction above. We’ll put the parameters in brackets ­ψ[c] – in order to differentiate them from things like positions that are inside of parenthesis ­ψ(r).Then the average energy will depend on these parameters: ψ c * Ĥψ c dτ Eavg ( ) c = ∫ψ [ ] [ ] c dτ ∫ [ ] [ ] c *ψ Note that, using the variational principle, the best parameters will be the ones that minimize the energy. Thus, the best parameters will satisfy * ψ c Hψ c dτ = = 0 ∂E ave ( ) c ∂ ∫ [ ] ˆ [ ] * i ψ c τ∂c i ∂c ∫ [ ] c ψ [ ] d Thus, we can solve for the optimal parameters without knowing anything about the exact eigenstates! Let us apply this in the particular case of our LCAO­MO treatment of H2 +. Our trial wavefunction is: ψ [c] = c 1s + c 1s el 1 A 2 B 6 5.61 Physical Chemistry Lecture #28 where c=(c1 c2). We want to determine the best values of c1 and c2 and the variational theorem tells us we need to minimize the average energy to find the right values. First, we compute the average energy. The numerator gives: ∫ψ el * Ĥ el ψ el dτ = ∫ (c 1 1s A + c 2 1s B ) * Ĥ (c 1 1s A + c 2 1s B ) dτ = c * c 1s Ĥ 1s d τ + c * c 1s Ĥ 1s d τ + c * c 1s Ĥ 1s d τ + c * c 1s Ĥ 1s d τ 1 1 ∫ A el A 1 2 ∫ A el B 2 1 ∫ B el A 2 2 ∫ B el B ≡H11 ≡H12 ≡H21 ≡H22 * * * * ≡ c H c + c H c + c H c + c H c 1 11 1 1 12 2 2 21 1 2 22 2 while the normalization integral gives: * ∫ψ el *ψ el dτ = ∫ (c 1 1s A + c 2 1s B ) (c 1 1s A + c 2 1s B ) dτ = c c 1s 1s d τ + c c 1s 1s d τ + c c 1s 1s d τ + c c 1s 1s d τ 1 * 1 ∫ A A 1 * 2 ∫ A B 2 * 1 ∫ B A 2 * 2 ∫ B B ≡S11 ≡S12 ≡S21 ≡S22 * * * * ≡ c S c + c S c + c S c + c S c 1 11 1 1 12 2 2 21 1 2 22 2 So that the average energy takes the form: * * * * c H c + c H c + c H c + c H c E avg ( ) c = 1 * 11 1 1 * 12 2 2 * 21 1 2 * 22 2 c S c + c S c + c S c + c S c 1 11 1 1 12 2 2 21 1 2 22 2 We note that there are some simplifications we could have made to this formula: for example, since our 1s functions are normalized S11=S22=1. By not making these simplifications, our final expressions will be a little more general and that will help us use them in more situations. Now, we want to minimize this average energy with respect to c1 and c2. Taking the derivative with respect to c1 and setting it equal to zero [Note: when dealing with complex numbers and taking derivatives one must treat variables and their complex conjugates as independent variables. Thus d/dc1 * has no effect on c1 ]: ∂E * * c H + c Havg 1 11 2 21 = 0 = * * * * ∂c c S c + c S c + c S c + c S c 1 1 11 1 1 12 2 2 21 1 2 22 2 * * * * − c 1 H 11 c 1 + c 1 H 12 c 2 + c 2 H 21 c 1 + c 2 H 22 c 2 (c * S + c * S )2 1 11 2 21 c S c + c S c + c S c + c S c( 1 * 11 1 1 * 12 2 2 * 21 1 2 * 22 2 ) 7 5.61 Physical Chemistry Lecture #28 * * * * ⇒ 0 = (c * H + c * H ) − c 1 H 11 c 1 + c 1 H 12 c 2 + c 2 H 21 c 1 + c 2 H 22 c 2 (c * S + c * S )1 11 2 21 * * * * 1 11 2 21 c S c + c S c + c S c + c S c 1 11 1 1 12 2 2 21 1 2 22 2 ⇒ 0 = (c 1 * H 11 + c 2 * H 21 ) − Eavg (c 1 * S 11 + c 2 * S 21 ) Applying the same procedure to c2: ∂E c * H + c * Havg = 0 = 1 12 2 22 * * * * ∂c c S c + c S c + c S c + c S c 2 1 11 1 1 12 2 2 21 1 2 22 2 * * * * − c 1 H 11 c 1 + c 1 H 12 c 2 + c 2 H 21 c 1 + c 2 H 22 c 2 2 (c 1 * S 12 + c 2 * S 22 ) (c * S c + c * S c + c * S c + c * S c )1 11 1 1 12 2 2 21 1 2 22 2 * * * * ⇒ 0 = (c * H + c * H ) − c 1 H 11 c 1 + c 1 H 12 c 2 + c 2 H 21 c 1 + c 2 H 22 c 2 (c * S + c * S )1 12 2 22 * * * * 1 12 2 22 c S c + c S c + c S c + c S c 1 11 1 1 12 2 2 21 1 2 22 2 ⇒ 0 = (c 1 * H 12 + c 2 * H 22 ) − Eavg (c 1 * S 12 + c 2 * S 22 ) We notice that the expressions above look strikingly like matrix­vector operations. We can make this explicit if we define the Hamiltonian matrix: ⎛ H11 H12 ⎞ ⎜ ⎛ ∫1sAĤ el 1sAdτ ∫1sAĤ el 1sBdτ ⎟ ⎞ H ≡ ⎜ ⎟ ≡ ⎝ H21 H22 ⎠ ⎜⎜ ⎝ ∫1sBĤ el 1sAdτ ∫1sBĤ el 1sBdτ ⎟⎟ ⎠ and the Overlap matrix: ⎛ S11 S12 ⎞ ⎜ ⎛ ∫1sA1sAdτ ∫1sA1sBdτ ⎟ ⎞ S ≡ ≡⎜ S S ⎟ ⎜ ⎟⎝ 21 22 ⎠ ⎜ ⎝ ∫1s 1s dτ ∫1s 1s dτ ⎟ ⎠B A B B Then the best values of c1 and c2 satisfy the matrix eigenvalue equation: (c1 * c2 * ) ⎜ ⎛ H H11 H H12 ⎟ ⎞ = Eavg (c1 * c2 * ) ⎜ ⎛ S S11 S S12 ⎟ ⎞ ⎝ 21 22 ⎠ ⎝ 21 22 ⎠ Which means that: ∂E avg † † = 0 ⇔ c iH = E c iS Eq. 1 ∂c avg This equation doesn’t look so familiar yet, so we need to massage it a bit. * First, it turns out that if we had taken the derivatives with respect to c1 * and c2 instead of c1 and c2, we would have gotten a slightly different equation: ⎛ H11 H12 ⎞ ⎛ c1 ⎞ ⎛ S11 S12 ⎞ ⎛ c1 ⎞ ⎜ ⎟ ⎜ ⎟ = Eavg ⎜ ⎟ ⎜ ⎟ ⎝ H21 H22 ⎠ ⎝ c2 ⎠ ⎝ S21 S22 ⎠ ⎝ c2 ⎠ or 5.61 Physical Chemistry Lecture #28 10 again guess the form of the antibonding eigenvector, since we know it has the characteristic shape +/­, so that we guess the solution is c1=­c2: ⎛ ε V12 ⎞ ⎛ c1 ⎞ ⎛ 1 S12 ⎞ ⎛ c1 ⎞ ⎜ V ε ⎟ ⎜ −c ⎟ = Eavg ⎜ S 1 ⎟ ⎜ −c ⎟ ⎝ 12 ⎠ ⎝ 1 ⎠ ⎝ 21 ⎠ ⎝ 1 ⎠ ⎛ (ε − V ) c ⎞ ⎛ (1 − S ) c ⎞12 1 12 1⇒ ⎝ ⎜ − (ε − V12 ) c1 ⎠ ⎟ = Eavg ⎝ ⎜ − (1 − S12 ) c1 ⎠ ⎟ ε − V ⇒ Eavg = 12 = Eσ * 1 − S 12 so, indeed the other eigenvector has c1=­c2. The corresponding antibonding orbital is given by: ψσ(r) *ψ σ = c 1s + c 1s = c 1s − c 1s ∝ 1s − 1s el 1 A 2 B 1 A 1 B A B where we note again that c1 is just a normalization constant. Given these forms for the bonding and antibonding orbitals, ψσ(r) we can draw a simple picture for the H2 + MOs (see right). We can incorporate the energies obtained above into a simple MO diagram of H2 +: ε − V Eσ∗ = 12 1 − S Eσ = 12 12 1 S ε + + V 12 E1sA =ε E1sB =ε On the left and right, we draw the energies of the atomic orbitals (1sA and 1sB) that make up our molecular orbitals (σ and σ*) in the center. We note 11 5.61 Physical Chemistry Lecture #28 that when the atoms come together the energy of the bonding and antibonding orbitals are shifted by different amounts: ε − V12 ε − V12 ε (1 − S12 ) εS12 − V12 Eσ * − E1s = − ε = − = 1 − S 1 − S 1 − S 1 − S12 12 12 12 E1s − Eσ * = ε − ε + V12 = ε (1 + S12 ) − ε + V12 = ε S12 − V12 1 + S 1 + S 1 + S 1 + S12 12 12 12 Now, S12 is the overlap between two 1s orbitals. Since these orbitals are never negative, S12 must be a positive number. Thus, the first denominator is greater than the second, from which we conclude εS − V εS − V Eσ* − E1s = 12 12 > 12 12 = E1s − Eσ* 1 − S 1 + S12 12 Thus, the antibonding orbital is destabilized more than the bonding orbital is stabilized. This conclusion need not hold for all diatomic molecules, but it is a good rule of thumb. This effect is called overlap repulsion. Note that in the special case where the overlap between the orbitals is negligible, S12 goes to zero and the two orbitals are shifted by equal amounts. However, when is S12 nonzero there are two effects that shift the energies: the physical interaction between the atoms and the fact that the 1sA and 1sB orbitals are not orthogonal. When we diagonalize H, we account for both these effects, and the orthogonality constraint pushes the orbitals upwards in energy. 1 5.61 Physical Chemistry Lecture #30 MOLECULAR ORBITAL THEORY­ PART II For the simple case of the one­electron bond in H2 + we have seen that using the LCAO principle together with the variational principle led to a recipe for computing some approximate orbitals for a system that would be very difficult to solve analytically. To generalize this to the more interesting case of many electrons, we take our direction from our experience with the independent particle model (IPM) applied to atoms and we build up antisymmetrized wavefunctions out of the molecular orbitals. This is the basic idea behind molecular orbital theory – there are many variations on the central theme, but the same steps are always applied. Rather than go step­ by­step and deal with H2 and then Li2 and then LiH … we will instead begin by stating the general rules for applying MO theory to any system and then proceed to show some illustrations of how this works out in practice. 1) Define a basis of atomic orbitals For H2 + our atomic orbital basis was simple: we used the 1s functions from both hydrogen atoms and wrote our molecular orbitals as linear combinations of our basis functions: ψ = c 1s + c 1s1 A 2 B Note that the AO basis determines the dimension of our MO vector and also determines the quality of our result – if we had chosen the 3p orbitals instead of the 1s orbitals, our results for H2 + would have been very wrong! For more complicated systems, we will require a more extensive AO basis. For example, in O2 we might want to include all the 2s and 2p orbitals on both oxygens, in which case our MOs would take the form ψ = c1 2s A + c2 2 p xA + c3 2 p yA + c4 2 p zA + c5 2s B + c6 2 p xB + c7 2 p yB + c8 2 p zB Meanwhile, for methane we might want to include the 1s functions on all four hydrogens and the 2s and 2p functions on carbon: ψ = c11s1 + c21s2 + c31s3 + c41s4 + c5 2s + c6 2 p x + c7 2 p y + c8 2 p z In the general case, we will write: N AO ψ = ∑ciφi i=1 and represent our MOs by column vectors: 4 5.61 Physical Chemistry Lecture #30 is usually best to ask a computer to solve the generalized eigenvalue problem for you. 4) Occupy the orbitals according to a stick diagram At this point, we must depart from the H2 + model and begin to account for the fact that we have multiple electrons. To do so, we follow the prescription of the independent particle model and build a Slater determinant out of our orbitals. However, whereas for atoms we built the determinant out of atomic orbitals, for molecules we will build the determinant out of molecular orbitals: ( ) 1 1ψ ↑ ( ) 1 1ψ ↓ ... ( ) 1Nψ ↓ Ψ ≡ ( ) 1 2ψ ↑ ( ) 1 2ψ ↓ ... ( ) 2Nψ ↓ ... ... ... ... ( )1 Nψ ↑ ( )1 Nψ ↓ ... ( )N Nψ ↓ As was the case for atoms, it is much easier to reason in terms of stick diagrams, rather than write out all of the orbitals in determinant form. So, for example, we would associate a stick diagram like this ψ2 ψ1 with a determinant: ψ 1 ψ 1 ψ 1 ψ 1 1↑ ( ) 1↓ ( ) 2↑ ( ) 2↓ ( ) 1↑ ( ) 1↓ ( ) 2↑ ( ) 2↓ ( ) ψ 2 ψ 2 ψ 2 ψ 2 Ψ (1, 2,3, 4 ) ≡ 1↑ ψ 1↓ 3 2↑ ( ) ψ 2↓ 3ψ ( ) 3 ( ) ψ 3 ( ) 1↑ ( ) 1↓ ( ) 2↑ ( ) 2↓ ( ) ψ 4 ψ 4 ψ 4 ψ 4 But all the information we would need is contained in the stick diagram and, of course, the MOs. 5) Compute the energy There are a variety of ways to compute the energy once the MOs have been obtained. The simplest is to use the non­interacting particle picture we used for atoms. Here, the energy of N electrons is just given by the sum of the energies of the N orbitals that are occupied: � * * 5 5.61 Physical Chemistry Lecture #30 N E = ∑E i i=1 A more accurate way is to use the independent particle model to add an average electron­electron repulsion to the energy: N N E = Ĥ = ∑Ei +∑J� ij − K� ij i=1 i< j Where now the Coulomb and exchange integrals use molecular orbitals rather than atomic orbitals: Jij ≡ ∫∫ ψ i (1)ψ j (2) r − 1 r ψ i (1)ψ j (2) dr 1 dr 2 dσ 1 dσ 2 1 2 K� ij ≡ ∫∫ ψ i * (1)ψ * j (2) r − 1 r ψ i (2)ψ j (1) dr 1 dr 2 dσ 1 dσ 2 1 2 In fact, as we will see later on, there are even more elaborate ways to obtain the energy from an MO calculation. When we work things out by hand, the non­interacting picture is easiest and we will usually work in that approximation when dealing with MOs. Diatomic molecules As a first application of MO theory, it is useful to consider first­row diatomic molecules (B2, C2, N2,O2, CO,CN, NO, etc.), which actually map rather nicely on to an MO picture. We’ll go step­by step for the generic “AB” diatomic to show how this fits into the MO theory framework. 1) Define a basis of atomic orbitals. To begin with, one would consider a set consisting of 10 atomic orbitals – 5 on A and 5 on B: ψ = c11sA + c21sB + c3 2sA + c4 2sB + c5 2 pzA + c6 2 pzB + c7 2 pyA + c8 2 pyB + c9 2 pxA + c10 2 pxB However, for all the diatomics above, the 1s orbitals on both atoms will be doubly occupied. Since we will primarily be interested in comparing the MO descriptions of different diatomics the eternally occupied 1s orbital will have no qualitative effect on our comparisons. It is therefore customary to remove the 1s orbitals from the expansion: ψ = c12sA + c2 2sB + c3 2 pzA + c4 2 pzB + c5 2 pyA + c6 2 pyB + c7 2 pxA + c8 2 pxB The latter approximation is referred to as the valence electron or frozen core approximation. The advantage is that it reduces the length of our vectors from 10 to 8. 5.61 Physical Chemistry Lecture #30 6 2) Compute the Matrix Representations. Here, we have the rather daunting task of computing two 8­by­8 matrices. As mentioned above, we won’t be concerned in this class about filling in precise values for matrix elements here. However, we will be very interested in obtaining the proper shape of the matrix by determining which matrix elements are zero and which are not. The Hamiltonian takes the shape: s Hs s Hs s Hp s Hp s Hp s Hp s Hp s Hp ⎛ ∫ A ˆ A ∫ A ˆ B ∫ A ˆ zA ∫ A ˆ zB ∫ A ˆ yA ∫ A ˆ yB ∫ A ˆ xA ∫ A ˆ xB ⎞ ⎜ ⎟ ⎜ ˆ s Hs ˆ s Hp s Hp ˆ s Hp s ˆ s Hp ˆ s ˆ ⎟ ⎜ ∫ sB Hs A ∫ B B ∫ B ˆ zA ∫ B zB ∫ B ˆ yA ∫ B Hp yB ∫ B xA ∫ B Hp xB ⎟ ⎜ ⎟ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ⎜ ∫ pzA Hs A ∫ pzA Hs B ∫ pzA Hp zA ∫ pzA Hp zB ∫ pzA Hp yA ∫ pzA Hp yB ∫ pzA Hp xA ∫ pzA Hp xB ⎟ ⎜ ⎟ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ⎜ ∫ pzB Hs A ∫ pzB Hs B ∫ pzB Hp zA ∫ pzB Hp zB ∫ pzB Hp yA ∫ pzB Hp yB ∫ pzB Hp xA ∫ pzB Hp xB ⎟ H ≡ ⎜ ⎟ ˆ⎜ pyA Hs ˆ A pyA ˆ B pyA Hp zA pyA ˆ zB pyA Hp yA pyA ˆ yB pyA Hp xA pyA ˆ xB ⎜ ∫ ∫ Hs ∫ ˆ ∫ Hp ∫ ˆ ∫ Hp ∫ ∫ Hp ⎟ ⎟ ⎜ ⎟ p Hs p Hs p Hp p Hp p Hp p Hp p Hp p Hp ⎜ ∫ yB ˆ A ∫ yB ˆ B ∫ yB ˆ zA ∫ yB ˆ zB ∫ yB ˆ yA ∫ yB ˆ yB ∫ yB ˆ xA ∫ yB ˆ xB ⎟ ⎜ ⎟ Hs ˆ Hp ˆ Hp ˆ Hp ⎜ ∫ pxA Hs ˆ A ∫ pxA ˆ B ∫ pxA Hp zA ∫ pxA ˆ zB ∫ pxA Hp yA ∫ pxA ˆ yB ∫ pxA Hp xA ∫ pxA ˆ xB ⎟ ⎜ ⎟ ⎜ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ⎟ ⎝ ∫ pxB Hs A ∫ pxB Hs B ∫ pxB Hp zA ∫ pxB Hp zB ∫ pxB Hp yA ∫ pxB Hp yB ∫ pxB Hp xA ∫ pxB Hp xB ⎠ We assume, for simplicity, that the AB­bond lies along the z­axis. Then it is relatively easy A B y to see that the molecule is symmetric upon reflection around the x and y axes. As a result, the Hamiltonian for AB is also z symmetric (even) with respect to reflection about x and y. Similarly, the s and p orbitals ­x all have definite reflection symmetries: Hamiltonian s pz py px X Reflection Y Reflection + + + + + + + ­ ­ + Further, we note that if we perform an integral, if the integrand is odd with respect to reflection about either x or y the integrand will be zero. 9 5.61 Physical Chemistry Lecture #30 ⎛ H H H H ⎞ ⎛ c1 α ⎞ ⎛ S S S S ⎞ ⎛ c1 α ⎞ ⎜ 11 12 13 14 ⎟ ⎜ α ⎟ ⎜ 11 12 13 14 ⎟ ⎜ α ⎟ ⎜ H21 H22 H23 H24 ⎟ ⎜ c2 ⎟ = Eα ⎜ S21 S22 S23 S24 ⎟ ⎜ c2 ⎟ ⎜ H31 H32 H33 H34 ⎟ ⎜ ⎜ c3 α ⎟ ⎟ ⎜ S31 S32 S33 S34 ⎟ ⎜ ⎜ c3 α ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ H41 H42 H43 H44 ⎠ ⎜⎝ c4 α ⎟ ⎠ ⎝ S41 S42 S43 S44 ⎠ ⎜⎝ c4 α ⎟ ⎠ which will give us four molecular orbitals that can be written as linear combinations of the first four AOs (sA, sB, pzA, pzB) ψ α = c1 α 2s A + c2 α 2s B + c3 α 2 p zA + c4 α 2 p zB Because these orbitals are symmetric with respect to reflection about both x and y, they will look something like the H2 + bonding and antibonding orbitals, and so they are referred to as σ­orbitals. For example, we can make the +/­ combinations of the 2s orbitals to obtain one bonding orbital and one antibonding: + σ1­orbital σ1∗­orbital ­ A B A B A B we can make the similar linear combinations of the 2pz orbitals to obtain: σ2∗­orbital A B A B A B + σ2­orbital ­ where we label the upper orbital σ* because of the nodes between the nuclei, whereas the σ orbital has no nodes between the nuclei. Note 10 5.61 Physical Chemistry Lecture #30 that these +/­ combinations are just to illustrate what the orbitals will look like; in order to get the actual molecular orbitals we would need to diagonalize the 4­by­4 and get the eigenvectors. However, if we do that for a molecule like N2 we actually get orbitals that look  strikingly similar to the ones above: σ2 * σ2 σ1 * σ1 B) The second eigenvalue problem to be solved is a 2­by­2: ⎛ H55 H56 ⎞ ⎛ c5 α ⎞ α ⎛ S55 S56 ⎞ ⎛ c5 α ⎞ ⎜ ⎟ ⎜ ⎟ = E ⎜ ⎟ ⎜ ⎟ ⎝ H65 H66 ⎠ ⎜⎝ c6 α ⎟ ⎠ ⎝ S65 S66 ⎠ ⎜⎝ c6 α ⎟ ⎠ which will give us two molecular orbitals that can be written as linear combinations of the next two AOs (pyA,pyB): ψ α = c5 α 2 pyA + c6 α 2 pyB These orbitals get “­“ signs upon reflection about y, so we designate them πy orbitals. We can again make the +/­ combinations to get an idea what these orbitals look like: 11 5.61 Physical Chemistry Lecture #30 πy­orbital A B + ­ A B A B πy * ­orbital Here again, the use of +/­ combinations is only for illustration purposes. Unless we have a homonuclear diatomic, the coefficients will not be ±1. However, for a molecule like N2, the orbitals look strikingly similar again: πy πy * C) The last eigenvalue problem is also 2­by­2: ⎛ H77 H78 ⎞ ⎛ c7 α ⎞ α ⎛ S77 S78 ⎞ ⎛ c7 α ⎞ ⎜ ⎟ ⎜⎜ α ⎟ ⎟ = E ⎜ ⎟ ⎜⎜ α ⎟ ⎟⎝ H87 H88 ⎠ ⎝ c8 ⎠ ⎝ S87 S88 ⎠ ⎝ c8 ⎠ which will give us two molecular orbitals that can be written as linear combinations of the last two AOs (pxA,pxB): ψ α = c5 α 2 p xA + c6 α 2 p xB These orbitals get “­“ signs upon reflection about x, so we designate them πx orbitals. The qualitative picture of the πx orbitals is the same