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Lecture 11 January 28, 2011 Symmetry, Homonuclear diatomics

Lecture 11 January 28, 2011 Symmetry, Homonuclear diatomics. Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy. William A. Goddard, III, wag@wag.caltech.edu 316 Beckman Institute, x3093

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Lecture 11 January 28, 2011 Symmetry, Homonuclear diatomics

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  1. Lecture 11 January 28, 2011 Symmetry, Homonuclear diatomics Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy William A. Goddard, III, wag@wag.caltech.edu 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants: Wei-Guang Liu <wgliu@wag.caltech.edu> Caitlin Scott <cescott@caltech.edu>

  2. Course schedule Friday January 14: L3 and L4 Monday January 17: Caltech holiday (MLKing) Wednesday January 19: wag L5 and L6 Friday January 21: wag L7 and L8, caught up Monday January 24: wag L7 and L8 Wednesday January 26: wag L9 and L10 Friday January 28: wag participates in a retreat for our nanotechnology project with UCLA Friday January 28: wag L11 Back on schedule Monday January 31: wag L12 Wag rotator cuff operation

  3. Last time

  4. Reconstruction of (110) surface, side view along [-1,1,0] Surface As has 3 covalent bonds to Ga, with 2 e in 3s lone pair, relaxes upward until average bond angle is 95º Surface Ga has 3 covalent bonds leaving 0 e in 4th orbital, relaxes downward until average bond angle is 119º. GaAs angle 0º 26º Si has dangling bond electron at each surface atom 54.7º 54.7º As Ga 54.7º [110] Si (110) GaAs (110) [001]

  5. Reconstruction of (110) GaAs

  6. Reconstruction of GaAs(110) surface, discussion We consider that bulk GaAs has an average of 3 covalent bonds and one donor acceptor (DA) bond. But at the surface can only make 3 bonds so the weaker DA bond is the one broken to form the surface. The result is that GaAs cleaves very easily compared to Si. No covalent bonds to break. As has 3 covalent bonds, leaving 2 electrons in 3s lone pair. AsH3 has average bond angle of 92º. At the GaAs surface As relaxes upward until has average bond angle of 95º Ga has 3 covalent bonds leaving 0 eletrons in 4th orbital. GaH3 has average bond angle of 120º. At the GaAs surface Ga relaxes downward until has average bond angle of 119º. This changes the surface Ga-As bond from 0º (parallel to surface to 26º. Observed in LEED experiments and QM calculations

  7. Analysis of charges Ga As Ga As Ga As Bulk structure: each As has 3 covalent bonds and one Donor-accepter bond(Lewis base – Lewis acid). This requires 3+2=5 electrons from As and 3+0=3 electrons from Ga. We consider that each bulk GaAs bond has 5/4 e from As and ¾ e form Ga. Each surface As has 5/4+1+1+2 = 5.25e for a net charge of -0.25 each surface Ga has ¾+1+1+0= 2.75 e for a net charge of +0.25 Thus considering both surface Ga and As, the (110) is neutral 5.25e 2.75e Net Q =0 0 2 0 2 0 2 1 1 1 1 1 1 1 1 1 1 5/4 3/4 3/4 3/4 5/4 5/4 5/4 3/4 3/4 5/4 5/4 3/4 5/4 3/4 3/4 5/4 5/4 3/4 a g a g a g 5/4 3/4 3/4 5/4 5/4 3/4 5/4 3/4 3/4 5/4 5/4 3/4

  8. The GaAs (100) surface, unreconstructed Every red surface atom is As bonded to two green 2nd layer Ga atoms, but the other two bonds were to two Ga that are now removed. This leaves three non bonding electrons to distribute among the two dangling bond orbitals sticking out of plane (like AsH2) 1st Layer  RED 2nd Layer  GREEN 3rd Layer  ORANGE 4th Layer  WHITE

  9. GaAs(100) surface reconstructed (side view) For the perfect surface, As in top layer, Ga in 2nd layer, As in 3rd layer, Ga in 4th layer etc. For the unreconstructed surface each As has two bonds and hence three electrons in two nonbonding orbitals. Expect As atoms to dimerize to form a 3rd bond leaving 2 electrons in nonbonding orbitals. Surface As-As bonds As Ga As Ga As Ga As Ga

  10. Charges for 2x1 GaAs(100) 2 2 2 2 2nd layer ga has 3 e 1 1 Top layer, As 2nd layer, ga 5/4 5/4 3/4 3/4 3/4 3/4 3/4 3/4 3/4 3/4 5/4 2e As-ga bond 5/4 3rd layer, as 1 1 2e As LP 5/4 5/4 Each surface As has extra 0.5 e  dimer has extra 1e Not stable 3/4 3/4 3/4 3/4 3/4 1st layer As has 5.5 e 3/4 3/4 2e As-As bond 3/4

  11. Now consider a missing row of As for GaAs(100) 0 0 0 0 1 1 Top layer, As 2nd layer, ga 5/4 5/4 3/4 3/4 3/4 3/4 3/4 3/4 ga empty LP 3rd layer, as 2nd layer ga has 2.25e Each 2nd layer ga next to missing As is deficient by 0.75e extra 0.5 e  4 ga are missing 3e 3/4 3/4 3/4 1st layer As has 5.5 e 3/4 3/4 3/4

  12. Consider 1 missing As row out of 4 Extra 1e missing 3e -1-1-1+3=0 net charge Extra 1e Thus based on electron counting expect simplest surface reconstruction to be 4x2. This is observed Extra 1e Extra 1e missing 3e

  13. Different views of GaAs(100)4x2 reconstruction -1.0e +1.5e Two missing As row plus missing Ga row Exposes 3rd row As Agrees with experiment Previous page, 3 As dimer rows then one missing Hashizume et al Phys Rev B 51, 4200 (1995)

  14. summary • Postulate of surface electro-neutrality • Terminating the bulk charges onto the surface layer and considering the lone pairs and broken bonds on the surface should lead to: • the atomic valence configuration on each surface atom. For example As with 3 covalent bonds and a lone pair and Ga with 3 covalent bonds and an empty fourth orbital • A neutral surface • This leads to the permissible surface reconstructions

  15. Excitation energies semiconductors

  16. To be added – band states

  17. To be added – band states

  18. Semiconducting properties

  19. Semiconducting properties

  20. To be added – band states IP(P)=4.05 eV 0.054 eV Remove e from P, add to conduction band = 4.045-4.0 = 0.045 eV Thus P leads to donor state just 0.045eV below LUMO or CBM

  21. To be added – band states EA(Al)=5.033 eV 0.045 eV Add e to Al, from valence band = 5.1 -5.033 = 0.067 eV Al leads to acceptor state just 0.067eV above HOMO or VBM

  22. New material

  23. The role of symmetry in QM In this course we are concerned with the solutions of the Schrodinger equation, HΨ=EΨ, but we do actually want to solve this equation. Instead we want to extract the maximum information about the solutions without solving it. Symmetry provides a powerful tool for doing this. Some transformation R1is called a symmetry transformation if it has the property that R1(HΨ)=H(R1Ψ) The set of all possible symmetries transformations of H are collected into what is called a Group.

  24. The definition of a Group 1). Closure: If R1,R2 e G (both are symmetry transformations) then R2 R1is also a symmetry transformation, R2 R1e G 2. Identity. The do-nothing operator or identity, R1 = ee G isclearly is a symmetry transformation 3. Associativity. If (R1R2)R3 =R1(R2R3). 4. Inverse. If R1e G then the inverse, (R1)-1e G,where the inverse is defined as (R1)-1R1 = e.

  25. The degenerate eigenfunctions of H form a representation If HΨ=EΨ then H(R1Ψ)= E(R1Ψ) for all symmetry transformations of H. Thus the transformations amount the n denegerate functions, {S=(RiΨ), where RiΨieG} lead to a set of matrices that multiply in the same way at the group operators. The Mathematicians say that these functions form a basis for a representation of G. Of course the functions in S may not all be different, so that this representation can be reduced. The mathematicians went on to show that one could derive a set of irreducible representations that give all possible symmetries for the H. reorientations from which one can construct any possible.

  26. Example, an atom. For an atom any rotations about any axis passing through the nucleus is a symmetry transformation. This leads to the group denoted as SO(3) by the mathematicians [O(3) indicates 3 three-dimensional real space, S because the inversion is not included). The irreducible representations of O(3) are labeled as S (non degenerate) and referred to as L=0 P (3 fold degenerate) and referred to as L=1 D (5 fold degenerate) and referred to as L=2 F (7 fold degenerate) and referred to as L=3 G (9 fold degenerate) and referred to as L=4

  27. H2O, an example of C2v consider the nonlinear H2A molecule, with equal bond lengths, e.g. H2O, CH2, NH2 • The symmetry transformations are • e for einheit (unity) xx, yy, zz • C2z, rotation about the z axis by 2p/2=180º, x-x, y-y, zz • sxz, reflection in the xz plane, xx, y-y, zz • syz, reflection in the yz plane, x-x, yy, zz • Which is denoted as the C2v group.

  28. 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

  29. The character table for C2v Since (C2z)2 = e, (sxz)2 = e, (syz)2 = e We expect wavefunctions to be ±1 under each operation name 1-e N-e a1 a2 b1 b2 Since C2zsxz= syz the symmetries for syz are already implied by C2zsxz. Thus there are only 4 possible symmetries.

  30. Symmetries for NH2 Ψ1 = A{(N2pya)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)](ab-ba)} NHx bond NHz bond Applying sxz to the wavefunction leads to Ψ2 = A{(N2py)0[(NpR)(HR)+(HR)(NpR)](ab-ba)[(NpL)(HL)+(HL)(NpL)](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 find that sxzΨ1 = Ψ2 = Ψ1 Thus interchanging a bond pair leaves Ψ invariant Also the N(1s)2 and (2s)2 pairs are invariant under all operations

  31. NH2 symmetry continued Consider now the singly occupied Npx orbital C2z changes the sign but, sxz does not. Thus Npx transforms as b1 and The total wavefunction transforms as B1. Include the S= 1/2 , leads to The 2B1 state 1-e N-e a1 a2 b1 b2

  32. Symmetries for H2O 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 CH2 A{(C2py)0[(Cpx )(Hx)+(Hx)(Cpx)](ab-ba)[(Cpz)(Hz)+(Hz)(Cpz)](ab-ba)} Since have enen number electrons in 2py, wavefunction is invarient under all symmetry transformations, thus must be 1A1.

  33. 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

  34. Second example, C3v, with NH3 as the prototype z x A{[(Npy )(Hy)+(Hy)(Npy)](ab-ba)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)] (ab-ba)} NHx bond NHc bond NHb bond We will consider a system such as NH3, with three equal bond lengths. Here we will take the z axis as the symmetry axis and will have one H in the xz plane. The other two NH bonds will be denoted as b and c.

  35. Symmetry elements for C3v b b b b b b x x x x x x z x c c c c c c Take the z axis out of the plane. The six symmetry operations are: C3 e C32= C3-1 sxzC32 sxz sxzC3

  36. The C3v symmetry group b sxzC32 C3 e x sxz sxzC3 C32= C3-1 c The sxzC3 transformation corresponds to a reflection in the bz plane (which is rotated by C3 from the xz plane) and the sxzC32transformation corresponds to a reflection in the cz plane (which is rotated by C32 from the xz plane). Thus these 3 reflections are said to belong to the same class. Since {C3 and C3-1 do similar things and are converted into each other by sxz we say that they are in the same class.

  37. The character table for C3v The E symmetry (irreducible representation) is of degree 2, which means that if φpx is an eigenfunction of the Hamiltonian, the so is φpy and they are degenerate. This set of degenerate functions would be denoted as {ex,ey} and said to belong to the E irreducible representation. The characters in this table are used to analyze the symmetries, but we will not make use of this until much later in the course. Thus an atom in a P state, say C(3P) at a site with C3v symmetry, would generally split into 2 levels, {3Px and 3Py} of 3E symmetry and 3Pz of 3A1 symmetry.

  38. Application for C3v, NH3 z x sLP We will write the wavefunction for NH3 as x =A{(sLP)2[(NHb bond)2(NHc bond)2(NHx bond)2]} c b where we combined the 3 Valence bond wavefunctions in 3 pair functions and we denote what started as the 2s pair as sLP Consider first the effect of sxz. This leaves the NHx bond pair invariant but it interchanges the NHb and NHc bond pairs. Since the interchange two pairs of electrons the wavefunction does not change sign. Also the sLP orbital is invariant.

  39. Consider the effect of C3 Next consider the C3 symmetry operator. It does not change sLP. It moves the NHx bond pair into the NHb pair, moves the NHb pair into the NHc pair and moves the NHc pair into the NHx pair. A cyclic permutation on three electrons can be written as (135) = (13)(35). For example. φ(1)φ(3)φ(5) φ(3)φ(5)φ(1) (say this as e1 is replaced by e3 is replaced by e5 is replace by 1) This is the same as φ(1)φ(3)φ(5) φ(1)φ(5)φ(3)  φ(3)φ(5)φ(1) The point is that this is equivalent to two transpostions. Hence by the PP, the wavefunction will not change sign. Since C3 does this cyclic permutation on 6 electrons,eg (135)(246)= (13)(35)(24)(46), we still get no sign change. (135) (13) (35) Thus the wavefunction for NH3 has 1A1 symmetry

  40. x Linear molecules, C∞v symmetry z O=C=O Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, r = sqrt(x2+y2), a, z (axis along z) Since the wavefunction has period of 2p in a, the a dependence of any wavefunction can be expanded as a Fourier series, φ(r,a,z)=f(r,z){a0 +S m=1m=∞ [am cos ma + bm sin ma]

  41. x Linear molecules, C∞v symmetry z O=C=O Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, r = sqrt(x2+y2), a, z (axis along z) Since the wavefunction has period of 2p in a, the a dependence of any wavefunction can be expanded as a Fourier series, φ(r,a,z)=f(r,z){a0 +S m=1m=∞ [am cos ma + bm sin ma] Clearly the kinetic energy will increase with m, so that for the samef(r,z), we expect m=0 lowest, then m=±1, then m=±2, m=±3, etc

  42. x Linear molecules, C∞v symmetry z O=C=O Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, r = sqrt(x2+y2), a, z (axis along z) Since the wavefunction has period of 2p in a, the a dependence of any wavefunction can be expanded as a Fourier series, φ(r,a,z)=f(r,z){a0 +S m=1m=∞ [am cos ma + bm sin ma] Clearly the kinetic energy will increase with m, so that for the samef(r,z), we expect m=0 lowest, then m=±1, then m=±2, m=±3, etc Also if we rotate the molecule about the z axis by some angle b, the states with the same m get recombined [cos m(a+b)] = (cos ma)(cosmb) – (sin ma)(sin mb) [sin m(a+b)] = (sin ma)(cosmb) + (cos ma)(sin mb) Which means that the wavefunctions with the same m are degenerate

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