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William A. Goddard, III, wag@kaist.ac.kr

Lecture 8, September 29, 2009. Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy. Course number: KAIST EEWS 80.502 Room E11-101 Hours: 0900-1030 Tuesday and Thursday.

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William A. Goddard, III, wag@kaist.ac.kr

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  1. Lecture 8, September 29, 2009 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: KAIST EEWS 80.502 Room E11-101 Hours: 0900-1030 Tuesday and Thursday William A. Goddard, III, wag@kaist.ac.kr WCU Professor at EEWS-KAIST and Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Senior Assistant: Dr. Hyungjun Kim: linus16@kaist.ac.kr Manager of Center for Materials Simulation and Design (CMSD) Teaching Assistant: Ms. Ga In Lee: leeandgain@kaist.ac.kr Special assistant: Tod Pascal:tpascal@wag.caltech.edu EEWS-90.502-Goddard-L08

  2. Schedule changes, reminder There was no lecture on Sept. 22 because of the EEWS conference Goddard will be traveling Oct 2-11 and will not give the lectures scheduled for Oct. 6 and 8 Consequently an extra lecture will be added at 2pm on Wednesday Sept. 30 and another at 2pm Wednesday Oct 14. This will be in the same room, 101 E11 L8: Sept. 29, as scheduled L9: Sept. 30, new lecture replaces Oct 6 L10: Oct. 1, as scheduled L11: Oct. 13, as scheduled L12: Oct. 14, new lecture replaces Oct 8 L13: Oct. 15, as scheduled EEWS-90.502-Goddard-L08

  3. Last time EEWS-90.502-Goddard-L08

  4. Consider the ground state of B: [Be](2p)1 x Ψ(1,2,3,4,5) = A[(1sa)(1sb)(2sa)(2sb)(2pxa)] Ignore the [Be] core then Can put 1 electron in 2px, 2py, or 2pz each with either up or down spin. Thus get 6 states. We will depict these states by simplified contour diagrams in the xz plane, as at the right. Of course 2py is zero on this plane. Instead we show it as a circle as if you can see just the front part of the lobe sticking out of the paper. 2px z 2pz 2py Because there are 3=2L+1 degenerate states we denote this as L=1 or a P state. Because the spin can be +½ or –½, we call it a spin doublet and we denote the overall state as 2P EEWS-90.502-Goddard-L08

  5. Summary ground state for C atom x z (2px)(2pz) x z x z Ψ(1,2,3,4,5,6)xz= A[(1sa)(1sb)(2sa)(2sb)(2pxa)(2pza)] = = A[(1s)2(2s)2(2pxa)(2pza)] =A[(Be)(2pxa)(2pza)] = = A[(xa)(za)] = which we visualize as Ψ(1,2,3,4,5,6)xy = A[(xa)(ya)] which we visualize as Ψ(1,2,3,4,5,6)yz = A[(ya)(za)] which we visualize as Note that we choose to use the xz plane for all 3 wavefunctions, so that the py orbitals look like circles (seeing only the + lobe out of the plane EEWS-90.502-Goddard-L08

  6. Summarizing the energies for C atom E(1S) = 2hpp + Jxx + 2Kxy 3Kxy E(1D) = 2hpp +Jxx - Kxy = 2hpp +Jxy + Kxy 2Kxy E(3P) = 2hpp + Exy – Kxy = 2hpp +Jxy - Kxy EEWS-90.502-Goddard-L08

  7. Energy of the ground state of N: A[(xa)(ya)(za)] = [Axyz]aaa E(2P) 2Kxy E(2D) 4Kxy E(4S) x z Simple product xyz leads to Exyz = 3hpp + Jxy + Jxz + Jyz E(4S) = <xyz|H|A[xyz]>/<xyz|A[xyz]> Denominator = <xyz|A[xyz]> = 1 Numerator = Exyz - Kxy - Kxz – Kyz = E(4S) = 3hpp + (Jxy - Kxy) + (Jxz - Kxz) + (Jyz – Kyz) Pictorial representation of the N ground state =3hpp + 2Jxy + Jxx + 1Kxy TA’s check this =3hpp + 3Jxy + 1Kxy = 3hpp + 2Jxy + Jxx - 1Kxy Since Jxy =Jxz =Jyz and Kxy= Kxz = Kyz =3hpp + 3Jxy - 3Kxy EEWS-90.502-Goddard-L08

  8. Summary ground state for O atom x z x x z z Ψ(1,2,3,4,5,6)xz= A[(1sa)(1sb)(2sa)(2sb)(2pya)(2pyb)(2pxa)(2pza)] = A[(1s)2(2s)2(2pya)(2pyb)(2pxa)(2pza)] = = A[(ya)(yb)(xa)(za)] = which we visualize as (2py)2(2px)(2pz) Ψ(1,2,3,4,5,6)yz = A[(xa)(xb)((ya)(za)] which we visualize as (2px)2(2py)(2pz) Ψ(1,2,3,4,5,6)xy = A[(za)(zb)(((xa)(ya)] which we visualize as (2pz)2(2px)(2py) We have 3 = 2L+1 equivalent spin triplet (S=1) states that we denote as L=1 orbital angular momentum, leading to the 3P state EEWS-90.502-Goddard-L08

  9. Calculating energies for O atom (2py)2(2px)(2pz) Ψ(1,2,3,4,5,6)xz= A[[Be](ya)(yb)(xa)(za)] Energy is Exz = E(Be) + 2hyy + hxx + hzz + Jyy+ 2Jxy+ 2Jyx+ Jxz – Kxy – Kyz – Kxz Check: 4 electrons, therefore 4x3/2 = 6 coulomb interactions 3 up-spin electrons, therefor 3x2/2 = 3 exchange interactions Other ways to group energy terms Exz = 4hpp + Jyy + (2Jxy – Kxy) + (2Jyx – Kyz) + (Jxz– Kxz) Same energy for other two components of 3P state EEWS-90.502-Goddard-L08

  10. Comparison of O states with C states x x z z Ne (2p)61S C (2p)23P O (2p)43P Compared to Ne, we have Hole in x and z Hole in y and z Hole in x and y Compared to Be, we have Electron in x and z Electron in y and z Electron in x and y Thus holes in O map to electrons in C EEWS-90.502-Goddard-L08

  11. Comparison of F states with B states x x z z Ne (2p)61S B (2p)12P F (2p)52P Compared to Ne, we have Hole in z Hole in x Hole in y Compared to Be, we have Electron in z Electron in x Electron in y Thus holes in F map to electrons in B EEWS-90.502-Goddard-L08

  12. Calculating energies for F atom (2px)2(2py)2(2pz) The energy of Ψ(1-9)z= A{[Be](2pxa)(2pxb)(2pya)(2pyb)(2pza)] is Exz = 5hpp + Jxx+ Jyy+ (4Jxy – 2Kxy) + (2Jxz – Kxz) + (2Jyx– Kyz) Check: 5 electrons, therefore 5x4/2 = 10 coulomb interactions 3 up-spin electrons, therefore 3x2/2 = 3 exchange interactions 2 down-spin electrons, therefore 2x1/2 = 1 exchange interaction Other ways to group energy terms Same energy for other two components of 2P state EEWS-90.502-Goddard-L08

  13. Calculating energy for Ne atom (2px)2(2py)2(2pz)2 The energy of Ψ(1-9)z= A{[Be](2pxa)(2pxb)(2pya)(2pyb)(2pza)(2pzb)} is Exz = 6hpp +Jxx+Jyy+ Jzz + (4Jxy – 2Kxy) + (4Jxz – 2Kxz) + (4Jyx– 2Kyz) Check: 6 p electrons, therefore 6x5/2 = 15 coulomb interactions 3 up-spin electrons, therefore 3x2/2 = 3 exchange interactions 3 down-spin electrons, therefore 3x2/2 = 3 exchange interaction Since Jxx = Kxx we can rewrite this as Exz = 6hpp +(2Jxx-Kxx ) +(2Jyy-Kyy ) +(2Jzz-Kzz ) + 2(2Jxy – Kxy) + 2(2Jxz – Kxz) + 2(2Jyx– Kyz) Which we will find later to be more convenient for calculating the wavefunctions using the variational principle EEWS-90.502-Goddard-L08

  14. Summary of ground states of Li-Ne Li 2S (2s)1 N 4S (2p)3 Be 1S (2s)2 O 3P (2p)4 Ignore (2s)2 B 2P (2p)1 F 2P (2p)5 C 3P (2p)2 Ne 1S (2p)6 EEWS-90.502-Goddard-L08

  15. Bonding H atom to He Start with the ground state of He, (1s)2 = A(He1sa)(He1sb) and bring up an H atom (H1sa), leads to HeH: A[(He1sa)(He1sb)(H1sa)] But properties of A (Pauli Principle) tell us that the H1s must get orthogonal to the He 1s since both have an a spin. A[(He1sa)(He1sb)(θa)] Where θ = H1s – S He1s Consequently θ has a nodal plane, increasing its KE. Smaller R  larger S  larger increase in KE. Get a repulsive interaction, no bond R EEWS-90.502-Goddard-L08

  16. Bonding H atom to Ne Start with the ground state of Ne, (1s)2(2s)2(2p)6 Ψ(Ne)= A{(2pxa)(2pxb)(2pya)(2pyb)(2pza)(2pzb)} (omitting the Be) and bring up an H atom (H1sa) along the z axis, leads to A{(2px)2(2py)2(Ne2pza)(Ne2pzb)(H1sa)} Where we focus on the Ne2pz orbital that overlaps the H atom The properties of A (Pauli Principle) tell us that the H1s must get orthogonal to the Ne 2pz since both have an a spin. θ = H1s – S Ne2pz R θ has a nodal plane, increasing its KE. Smaller R  larger S  larger increase in KE. Get a repulsive interaction, no bond EEWS-90.502-Goddard-L08

  17. Mixture of old and New material EEWS-90.502-Goddard-L08

  18. x z Now consider Bonding H atom to all 3 states of F Bring H1s along z axis to F and consider all 3 spatial states. F 2pz doubly occupied, thus H1s must get orthogonal  repulsive A{(2pxa)1(2py)2(F2pza)(F2pzb)(H1sa)} F 2pz doubly occupied, thus H1s must get orthogonal  repulsive A{(2px)2(2pya)1(F2pza)(F2pzb)(H1sa)} F 2pz singly occupied, Now H1s need not get orthogonal if it has opposite spin, can get bonding R EEWS-90.502-Goddard-L08

  19. energy x (Pz H - H Pz) z R (Pz H + H Pz) Now consider Bonding H atom to x2y2z1 state of F Focus on 2pz and H1s singly occupied orbitals 3S+ Antibonding state (S=1, triplet) [φpz(1) φH(2) - φH(1) φpz(2)](aa) Just like H2. Bonding state (S=0, singlet) [φpz(1) φH(2) + φH(1) φpz(2)](ab-ba) 1S+ Full wavefunction for bond becomes A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](ab-ba)} Full wavefunction for antibond becomes A{(F2px)2(F2py)2[(Fpz)(H)-(H)(Fpz)](aa)} EEWS-90.502-Goddard-L08

  20. Schematic depiction of HF Denote the ground state of HF as Where the line connecting the two singly occupied orbitals  covalent bonding We will not generally be interested in the antibonding state, but if we were it would be denoted as EEWS-90.502-Goddard-L08

  21. Bond a 2nd H atom to the ground state of HF? Starting with the ground state of HF as can a 2nd H can be bonded covalently, say along the x axis? antibond bond A{(F2pxa)(F2pxb)(Hxa)(F2py)2[(Fpz)(Hz)+(Hz)(Fpz)](ab-ba)} This leads to repulsive interactions just as for NeH. Since all valence orbitals are paired, there are no other possible covalent bonds and H2F is not stable. EEWS-90.502-Goddard-L08

  22. x z Now consider Bonding H atom to all 3 states of O Bring H1s along z axis to O and consider all 3 spatial states. O 2pz doubly occupied, thus H1s must get orthogonal  repulsive O 2pz singly occupied. Now H1s need not get orthogonal if it has opposite spin, can get bonding Get S= ½ state, Two degenerate states, denote as 2P R EEWS-90.502-Goddard-L08

  23. x z Bonding H atom to x1y2z1 and x2y1z1 states of O The full wavefunction for the bonding state 2Py bond A{(O2px)2(O2pya)1[(Opz)(H)+(H)(Opz)](ab-ba)} 2Px bond A{(O2pxa)1(O2py)2[(Opz)(H)+(H)(Opz)](ab-ba)} R 2P (Pz H + H Pz) EEWS-90.502-Goddard-L08

  24. x z Bond a 2nd H atom to the ground state of OH Starting with the ground state of OH, we can ask whether a 2nd H can be bonded covalently, say along the x axis. Bonding a 2nd H along the x axis to the 2Py state leads to repulsive interactions just as for NeH. No bond. 2Py bond antibond A {(O2pxa)(O2pxb)(Hxa)(O2pya)1[(Opz)(H)+(H)(Opz)](ab-ba)} Bonding a 2nd H along the x axis to the 2Px state leads to a covalent bond 2Px bond bond A{(O2py)2[(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)} EEWS-90.502-Goddard-L08

  25. x z θe Re Analize Bond in the ground state of H2O bond bond A{(O2py)2[(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)} This state of H2O is a spin singlet state, which we denote as 1A1. For optimum bonding, the pz orbital should point at the Hz while the px orbital should point at the Hx Thus the bond angle should be 90º. In fact the bond angle is far from 90º for H2O, but it does approach 90º for S  Se  Te EEWS-90.502-Goddard-L08

  26. x z What is origin of large distorsion in bond angle of H2O A{(O2py)2[(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)} OHx bond OHz bond Bonding Hz to pz leaves the Hz orbital orthogonal to px and py while bonding Hx to px leaves the Hx orbital orthogonal to py and pz, so that there should be little interference in the bonds, except that the Hz orbital can overlap the Hx orbital. Since the spin on Hx is a half the time and b the other half while the same is true for Hz, then ¼ the time both are a and ¼ the time both are b. Thus the Pauli Principle (the antisymmetrizer) forces these orbitals to become orthogonal. This increases the energy as the overlap of the 1s orbitals increases. Increasing the bond angle reduces this repulsive interaction EEWS-90.502-Goddard-L08

  27. x z Testing the origin of bond angle distorsion in H2O If the increase in bond angle is a response to the overlap of the Hz orbital with the Hx orbital, then it should increase as the H---H distance decreases. In fact: θe Re Thus the distortion increases as Re decreases, becoming very large for R=1A (H—H of 1.4A, which leads to large overlap ~0.5). To test this interpretation, Emily Carter and wag (1983) carried out calculations of the optimum θe as a function of R for H2O and found that increasing R from 0.96A to 1.34A, decreases in θe by 11.5º (from 106.5º to 95º). EEWS-90.502-Goddard-L08

  28. Validation of concept that the bond angle increase is due to H---H overlap Although the driving force for distorting the bond angle from 90º to 104.5º is H—H overlap, the increase in the bond angle causes many changes in the wavefunction that can obscure the origin. Thus for the H’s to overlap the O orbitals best, the pz and px orbtials mix in some 2s character, that opens up the angle between them. This causes the O2s orbital to build in p character to remain orthogonal to the bonding orbitals. EEWS-90.502-Goddard-L08

  29. Bond a 3rd H atom to the ground state of H2O? Starting with the ground state of H2O We can bring a 3rd H along the y axis. This leads to A{(O2pya)(O2pyb)(Hya) [(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)} OHx bond OHz bond OHy antibond This leads to repulsive interactions just as for NeH. Since all valence orbitals are paired, there are no other possible covalent bonds and H3O is not stable. EEWS-90.502-Goddard-L08

  30. x z Now consider Bonding H atom to the ground state of N Bring H1s along z axis to O and N 2pz singly occupied, forms bond to Hz R A{(N2pxa)(N2pya)[(Npz)(Hz)+(Hz)(Npz)](ab-ba)} NHz bond Two unpaired spins, thus get S=1, triplet state Denote at 3S- EEWS-90.502-Goddard-L08

  31. x A{(N2pya)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)](ab-ba)} z bond bond Bond a 2nd H atom to the ground state of NH Starting with the ground state of NH, bring a 2nd H along the x axis. This leads to a 2nd covalent bond. Denote this as 2B1 state. Again expect 90º bond angle. Indeed as for H2O we find big deviations for the 1st row, but because N is bigger than O, the deviations are smaller. Re θe EEWS-90.502-Goddard-L08

  32. A{[(Npy )(Hy)+(Hy)(Npy)](ab-ba)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)] (ab-ba)} NHx bond NHz bond NHy bond Bond a 3rd H atom to the ground state of H2N Starting with the ground state of H2N We can a 3rd H along the y axis. This leads to Denote this as 1A1 state. Again expect 90º bond angle. Indeed as for H2N we find big deviations for the 1st row, but because there are now 3 bad H---H interactions the deviations are larger. Re θe EEWS-90.502-Goddard-L08

  33. Bond a 4th H atom to the ground state of H3N? The ground state of H3N has all valence orbitals are paired, there are no other possible covalent bonds and H4N is not stable. EEWS-90.502-Goddard-L08

  34. New material EEWS-90.502-Goddard-L08

  35. More on symmetry We saw in previous lectures that symmetry often has a profound effect on the solutions of the Schrödinger Equation For example inversion symmetry  g or u states Permutational symmetry in electrons  symmetric and antisymmetric wavefunctions for transposition of space, spin, and space-spin coordinates. More general discussions make use of group theory or more correctly group representation theory. Here we will use outline some of the essential elements relevant to this course EEWS-90.502-Goddard-L08

  36. The symmetry operations for the Schrödinger equation form a group Consider some symmetry operator, R1, of the Hamiltonian, for example R1 = I (inversion) for H2 or the transposition, t12, for any two electron system. Then If HΨ=EΨ if follows that R1(HΨ)=H(R1Ψ)= E(R1Ψ) so that R1Ψ is an eigenfunction with the same E. The set of symmetry operators {R1 , R2 , … Rn} = G forms a Group. This follows since: 1). Closure: If R1,R2 e G (that is both are symmetry elements) then R2 R1is also a symmetry element, R2 R1e G(we say that the set of symmetry operations is closed). This follows since (R2 R1) (HΨ)= R2 H(R1Ψ)= R2E(R1Ψ)= E(R2 R1Ψ)}, that is (R2 R1Ψ) is also an eigenstate of H with the same energy E 2. Identity. Also R1 = e (identity) e G. Clearly eΨis a symmetry element EEWS-90.502-Goddard-L08

  37. The symmetry operations for the schrodinger equation form a group, continued 3. Associativity. If {R1,R2,R3 } e G then (R1R2)R3 =R1(R2R3). This follows since (R1R2)R3HΨ = (R1R2)ER3Ψ = E (R1R2)R3Ψ and R1(R2R3)HΨ = R1H(R2R3)Ψ = R1E (R2R3)Ψ= ER1(R2R3)Ψ 4. Inverse. If R1e G then the inverse, (R1)-1e G,where the inverse is defined as (R1)-1R1=e. This follows for any finite set that is closed since, there must be some integer p, such that (R)p = e.Thus [(R)p-1]R = e = R[(R)p-1] where (R)p-1 ≡(R1)-1 The above four conditions are necessary and sufficient to define what the Mathematicians call a group. The theory of the properties of such a group is called Group Theory. This is a vast field, but the only part important to QM and materials is group representation theory. EEWS-90.502-Goddard-L08

  38. Group representation Theory If G is a group with n elements, consider the set of functions S={Ψ1= R1Ψ, Ψ2= R2Ψ, …Ψn= RnΨ}. From the properties of a group, any operation R e G on any ΨieS leads to a linear combination of functions in S. This leads to a set of nxn matrices that multiple in exactly the same way as the elements of G , so the Mathematicians say that S is a basis for the group G and that these matrices form a representation of G. The mathematicians went on to show that one could derive a set of irreducible reorientations from which one can construct any possible representation. This theory was worked out ~1905 mostly in Germany but it had few real practical applications until QM. In QM these irreducible representations are important because they constitute the possible symmetries of all possible eigenfunctions of the Hamiltonian H. EEWS-90.502-Goddard-L08

  39. An example, C2v The point of going through these definitions is to make sure that the students understand that the use of Group Theory in QM involves very simple concepts. No need to get afraid of the complex notations and nomenclature used sometimes. For those that want to see my views on Group Theory with some simple applications. Some notes are available for this course, denoted as Volume V, chapter 9. This dates from ~1972 when I used to teach a full year course on QM and from 1976 when I included such materials in my then full year course on chemical bonding Lets consider an example, system the nonlinear H2A molecule, with equal bond lengths, e.g. H2O, CH2, NH2 EEWS-90.502-Goddard-L08

  40. C2v The symmetry operators are And the group {e, C2z, sxz, syz} is denoted as C2z Czn denotes a rotation of 2p/n about the z axis (C for cyclisch, German for cyclic) sxz denotes a reflection or mirror in the xz plane (s for spiegel, German for mirror) EEWS-90.502-Goddard-L08

  41. Stereographic projections e e y y x x C2z C2z syz sxz sxz syz syz C2z C2z syz sxz sxz Consider the stereographic projection of the points on the surface of a sphere onto a plane, where positive x are circles and negative x are squares. Start with a general point, denoted as e and follow where it goes on various symmetry operations. This make relations between the symmetry elements transparent. e.g. C2zsxz= syz Combine these as below to show the relationships C2v EEWS-90.502-Goddard-L08

  42. The character table for C2v The basic symmetries (usually called irreducible representations) for C2v are given in a table, called the character table My choice of coordinate system follows (Mulliken in JCP 1955). This choice removes confusion about B1 vs B2 symmetry (x is the axis for which sxz moves the maximum number of atoms In the previous slide we saw that C2zsxz= syz which means that the symmetries for syz are already implied by C2zsxz. Thus we consider C2z andsxz as the generators of the group. This group is denoted as C2v, which denotes that the generators are C2z and a vertical mirror plane (containing the C2 axis) EEWS-90.502-Goddard-L08

  43. More on C2v Each of the symmetry elements are of order two, That is Rp = e where p=2. We saw earlier, with R = I (Inversion) and t (transposition) That with p=2 the eigenfunctions of the H must transform as symmetric (+) or antisymmetric (-) under the operation. The character table reflects this. Also since C2zsxz= syz , we see that the symmetry of syz is determined by those of C2z andsxz The general naming convention is A and B are symmetric and antisymmetric with respect to the principal rotation axis. Subscripts 1 and 2 are symmetric and antisymmetric with respect to the mirror plane generator. EEWS-90.502-Goddard-L08

  44. Applications for C2v Consider that an atom with a single electron in a p orbital (say B or Al) is placed at a site in a crystal with C2v symmetry. The character table tells us that in general, the px, py, and pz states will all have different energies. On the other hand if the symmetry were that of a square (D4h), px and py would be degenerate, but pz might be different, and in the symmetry of an octahedron (Oh) or tetrahedron (Td), the three p states will be degenerate. We will get to such issues later in the course but for now we will use the symmetry only to provide names for the states N electron wavefunction, use A1,A2 etc one electron orbitals: use a1,a2 etc name 1-e N-e a1 a2 b1 b2 EEWS-90.502-Goddard-L08

  45. Symmetries for H2O, NH2, and CH2 CH2 NH2 H2O A{(O2py)2[(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)} H2O OHx bond OHz bond Previously we discussed the wavefunctions for H2O, NH2, and CH2 from bonding H to two p orbitals NH2 A{(N2pya)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)](ab-ba)} NHx bond NHz bond CH2 A{(C2py)0[(Cpx )(Hx)+(Hx)(Cpx)](ab-ba)[(Cpz)(Hz)+(Hz)(Cpz)](ab-ba)} EEWS-90.502-Goddard-L08

  46. First do CH2 singlet HL y z pL HR pR 2s Ψ1=A{(C2py)0[(CpL )(HL)+(HL)(CpL)](ab-ba)[(CpR)(HR)+(HR)(CpR)](ab-ba)} CHL bond CHR bond We used x and z for the bond directions before but now we want new x,y,z related to the symmetry group. Thus we denote bonds as L and R for left and right. Note that the C (1s) and (2s) pairs are invariant under all operations Applying sxz to the wavefunction leads to Ψ2=A{(C2py)0[(CpR)(HR)+(HR)(CpR)](ab-ba)[(CpL)(HL)+(HL)(CpL)](ab-ba)} Each term involves transposing two pairs of electrons, e.g., 13 and 24 as interchanging electrons. Since each interchange leads to a sign change we get that sxzΨ1 = Ψ2 = Ψ1 Thus interchanging a bond pair leaves Ψ invariant EEWS-90.502-Goddard-L08

  47. Finish CH2 singlet HL y z pL HR pR 2s Ψ1=A{(C2py)0[(CpL )(HL)+(HL)(CpL)](ab-ba)[(CpR)(HR)+(HR)(CpR)](ab-ba)} CHL bond CHR bond Applying C2z to the wavefunction also interchanges two bond pairs so that C2zΨ1 = Ψ2 = Ψ1  A1 or A2 symmetry Also sxzΨ1 = Ψ2 = Ψ1  A1 or B1 symmetry Thus we conclude that the symmetry of this state of CH2 is 1A1, where the superscript 1  spin singlet or S=0 EEWS-90.502-Goddard-L08

  48. Next consider NH2 HL y z pL NHL bond NHR bond HR pR 2s The wavefunction for NH2 differs from that of CH2 only in having a singly occupied px orbital A{(N2pxa)1[(NpL )(HL)+(HL)(NpL)](ab-ba)[(NpR)(HR)+(HR)(NpR)](ab-ba)} Since the symmetry operation simultaneously operates on all electrons, we can consider the effects on Npx separately. Here we see that C2z changes the sign but, sxz does not. Thus Npx transforms as b1. (note it is not necessary to examine sxz since it is not a generator). With one unpaired spin this state is S= ½ or doublet Thus the symmetry of NH2 is 2B1 EEWS-90.502-Goddard-L08

  49. Now do OH2 OHL bond OHR bond The wavefunction for OH2 differs from that of NH2 only in having px orbital doubly occupied A{(O2pxa)2[(OpL )(HL)+(HL)(OpL)](ab-ba)[(OpR)(HR)+(HR)(OpR)](ab-ba)} Since the symmetry operation simultaneously operates on all electrons, we can consider the effects on Opx separately. Since C2z changes the sign of Opx twice, the wavefunction is invariant. Thus (Opx)2 transforms as A1. With no unpaired spin this state is S= 0 or singlet HL y z pL HR pR 2s Thus the symmetry of OH2 is 1A1 EEWS-90.502-Goddard-L08

  50. Now do triplet state of CH2 A{(C2sa)1(2pxa)1[(CpL )(HL)+(HL)(CpL)](ab-ba)[(CpR)(HR)+(HR)(CpR)](ab-ba)} CHL bond CHR bond Soon we will consider the triplet state of CH2 in which one of the 2s nonbonding electrons (denoted as s to indicate symmetric with respect to the plane of the molecule) is excited to the 2px orbital (denoted as p to indicate antisymmetric with respect to the plane) Since we know that the two CH bonds are invariant under all symmetry operations, from now on we will write the wavefunction as y z A{[(CHL)2(CHR)2](Csa)1(Cpa)1} Here s is invariant (a1) while p transforms as b1. Since both s and p are unpaired the ground state is triplet or S=1 p=2px s=2s Thus the symmetry of triplet CH2 is 3B1 EEWS-90.502-Goddard-L08

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