Lecture 5,6, January 19, 2011 Bonding in NeFONCBBe hydrides

1. Lecture 5,6, January 19, 2011 Bonding in NeFONCBBe hydrides 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 L9 Wednesday January 26: wag L10 and L11 Friday January 28: wag participates in a retreat for our nanotechnology project with UCLA Back on schedule Monday January 31: wag L12

3. Last time

4. Aufbau principle for atoms Uuo, 118 Rn, 86 Particularly stable atoms, closed shells Xe, 54 Kr, 36 Zn, 30 Ar, 18 Ne, 10 He, 2

5. More detailed description of first row atoms Li: (2s) Be: (2s)2 B: [Be](2p)1 C: [Be](2p)2 N: [Be](2p)3 O: [Be](2p)4 F: [Be](2p)5 Ne: [Be](2p)6

6. 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

7. 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

8. 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

9. 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

10. 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)}

11. 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

12. 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.

13. 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

14. 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)

15. 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)}

16. 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

17. 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

18. 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).

19. 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.

20. 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.

21. 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-

22. 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

23. 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

24. 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.

25. New material

26. Now consider Bonding H atom to all 3 states of C x z Bring H1s along z axis to C and consider all 3 spatial states. (2px)(2pz) O 2pz singly occupied. H1s can get bonding Get S= ½ state, Two degenerate states, denote as 2P (2py)(2pz) (2px)(2py) No singly occupied orbital for H to bond with

27. x z Ground state of CH (2P) The full wavefunction for the bonding state 2Px A{(2s)2(OHs bond)2(O2pxa)1} 2Py A{(2s)2(OHs bond)2(O2pya)1}

28. x z Bond a 2nd H atom to the ground state of CH Starting with the ground state of CH, we bring a 2nd H along the x axis. Get a second covalent bond This leads to a 1A1 state. No unpaired orbtial for a second covalent bond.

29. x z θe Re Analyze Bond in the ground state of CH2 Ground state has 1A1 symmetry. 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. As NH2 (103.2º) and OH2 (104.5º), we expect CH2 to have bond angle of ~ 102º

30. But, the Bending potential surface for CH2 1B1 1Dg 1A1 3B1 3Sg- 9.3 kcal/mol The ground state of CH2 is the 3B1 state not 1A1. Thus something is terribly wrong in our analysis of CH2

31. Re-examine bonding to Be, B, and C. Ground state of Be atom: A[(1s)2(2s)2] = A[(1sa)(1sb)(2sa)(2sb)] For the 1s orbital of Be2+,Zeff = 4- 5/16 =3.6875=1/R1s So R1s = 1/3.69 = 0.27 bohr = 0.14 A For the 2s orbital of Be+, Zeff = 4 – 1.72 = 2.28; hence R2s = 4/2.28=1.75 bohr E[Be+ (2s)1] = - ½ Zeff/R2s = 0.65 h0 For the 2s orbital of Be, Zeff = 4 – 1.72 – 0.29 = 1.99 Thus R2s = n2/Zeff = 2.01 bohr = 1.06A E[Be (2s)2] = 2 (- ½ Zeff/R2s )= -0.99 h Thus IP(Be) = 0.99 – 0.65 = .34 h0 = 9.25 eV (exper 9.32 eV) 1s 2s 1.06A 0.14A R~0.07A

32. consider the e-e interactions within the (2s)2 pair 2 a0 2 a0 Each electron has its maximum amplitude in a spherical shell centered at R2s ~ 1.06 A = 2.01 bohr Thus the two electrons will on the average be separated by 2*sqrt(2) = 2.8 bohr leading to an ee repulsion of ~1/2.8 hartree= 9.5 eV. This is BIG, comparable to the predicted IP e1 2.8 a0 2 a0 e2 2 a0

33. hybridization of the atomic 2s orbitals of Be. So far we have assumed the Be wavefunction to be A[(1s)2(2s)2] = A[(1sa)(1sb)(2sa)(2sb)] 2s In fact this is wrong. Writing the wavefunction as A[(1sa)(1sb)(φaa)(φbb)] and solving self-consistently (unrestricted Hartree Fock or UHF calculation) for φa and φbleads to φa = φ2s + lφpz and φb = φ2s - lφpz where φpz is like the 2pz orbital of Be+, but with a size like that of 2s rather (smaller than a normal 2p orbital) This pooching or hybridization of the 2s orbitals in opposite directions leads to a much increased pz φ2s + lφpz φ2s - lφpz average ee distance, dramatically reducing the ee repulsion.

34. analyze the pooched or hybridized orbitals x x z z Pooching of the 2s orbitals in opposite directions leads to a dramatic increase in the ee distance, reducing ee repulsion. φ2s + lφpz φ2s - lφpz z z 1-D 2-D Schematic. The line shows symmetric pairing. Notation: sz and sz bar or ℓ and ℓ bar. Cannot type bars. use zs to show the bar case

35. Problem with UHF wavefunction for Be A[(φaa)(φbb)] = [φa(1)a(1)][(φb(2)b(2)] – [φb(1)b(1)] [φa(2)a(2)] Interchange s1 and s2: get A[(φab)(φba)] = [φa(1)b(1)][(φb(2)a(2)] – [φb(1)a(1)] [φa(2)b(2)] Now φahas b spin rather than a. Does not have proper spin or space permutation symmetry. Combine to form proper singlet and triplet states. 1Ψ(1,2) = [φa(1)φb(2)+φb(1)φa(2)][a(1)b(2)–b(1)a(2)] 3Ψ(1,2) = [φa(1)φb(2)-φb(1)φa(2)][a(1)b(2)+b(1)a(2)] and aa, bb The Generalized Valence Bond (GVB) method was developed to optimize wavefunctions of this form. The result is qualitatively the same as UHF, but now the wavefunction is a proper singlet. I do not have handy a plot of these GVB orbitals for Be but there are similar to the analogous orbtials for Si, which are shown next

36. The GVB orbitals for the (3s)2 pair of Si atom Long dashes indicate zero amplitude, solid lines indicate positive amplitude while short dashes indicate negative amplitude. The spacing between contours is 0.05 in atomic units

37. Analysis of the GVB singlet wavefunction 1Ψ(1,2) = [φa(1)φb(2)+φb(1)φa(2)][a(1)b(2)–b(1)a(2)] Substituting φa = φ2s + lφpz and φb = φ2s - lφpz into the spatial factor leads to (ab+ba) = (s+lz)(s-lz)+(s-lz)(s+lz) = [s(1)s(2) - l2 z(1)z(2)] (ignoring normalization), which we will refer to as the CI form (for configuration interaction).

38. Analysis of the GVB singlet wavefunction 1Ψ(1,2) = [φa(1)φb(2)+φb(1)φa(2)][a(1)b(2)–b(1)a(2)] Substituting φa = φ2s + lφpz and φb = φ2s - lφpz into the spatial factor leads to (ab+ba) = (s+lz)(s-lz)+(s-lz)(s+lz) = [s(1)s(2) - l2 z(1)z(2)] (ignoring normalization), which we will refer to as the CI form (for configuration interaction). In the GVB wavefunction it is clear from the shape of the sz and zs wavefunctions that the average distance between the electrons is dramatically increased. This is a little more complicated to see in the CI form. Consider two electrons a distance R from the nucleus. Then the probability for the two electrons to be on the same side is s(R)s(R)-l2 z(R)z(R) which is smaller than s(R)s(R) while the probability of being on opposite sides is s(R)s(-R)- l2 z(R)z(-R) = s(R)s(R)+l2 z(R)z(R) which is increased.

39. Analysis of the GVB triplet wavefunction 3Ψ(1,2) = [φa(1)φb(2)-φb(1)φa(2)][a(1)b(2)+b(1)a(2)] and aa, bb Substituting φa = φ2s + lφpz and φb = φ2s - lφpz into the spatial factor leads to (ab-ba) = (s+lz)(s-lz)-(s-lz)(s+lz) = [s(1)z(2)-z(1)s(2)] (ignoring normalization). This is just the wavefunction for the triplet state formed by exciting the 2s electron to 2pz, which is very high (xx eV). Thus we are interested only in the singlet pairing of the two lobe or hybridized orbitals. This is indicated by the line pairing the two lobe functions Bottom line: 2s orbitals of (1s)2(2s)2 state of Be hybridize in ±z direction in order to reduce electron repulsion (as the cost of promotion to the 2p orbital 16% of the time)

40. Role of pooched or hybridized atomic lobe orbitals in bonding of BeH+ In fact optimizing the wavefunction for BeH+ leads to pooching of the 2s toward the H1s with much improved overlap and contragradience. Consider the bonding of H to Be+ The simple VB combination of H1s with the 2s orbital of Be+ leads to a very small overlap and contragradience

41. Role of pooched or hybridized atomic lobe orbitals in bonding of BeH neutral At large R the the orbitals of Be are already hybridized zs sz H Thus the wave function is A{[(sz)(zs)+(zs)(sz)](ab-ba)(Ha)} where sz≡(s+lz) and zs ≡(s-lz) Here the H overlaps slightly more with sz than with zs, but the spin on sz is half the time a Thus at large R we obtain a slightly repulsive interaction.

42. Role of pooched or hybridized atomic lobe orbitals in bonding of BeH neutral At small R the H can overlap significantly more with sz than with zH, so that we can form a bond pair just like in BeH+. This leads to the wavefunction At large R the the orbitals of Be are already hybridized A{[(sz)(H)+(H)(sz)](ab-ba)(zsa)} zs sz H In which the zs hybrid must now get orthogonal to the sz and H bond pair. This weakens the bond from that of BeH+ by ~ 1 eV Thus the wave function is A{[(sz)(zs)+(zs)(sz)](ab-ba)(Ha)} where sz≡(s+lz) and zs ≡(s-lz) Here the H overlaps slightly more with sz than with zs, but the spin on sz is half the time a Thus at large R we obtain a slightly repulsive interaction. zs sz H

43. More analysis of the GVB singlet wavefunction 1Ψ(1,2) = [φa(1)φb(2)+φb(1)φa(2)][a(1)b(2)-b(1)a(2)] where φa = φ2s + lφpz and φb = φ2s - lφpz The optimum value of l~0.4 (it is 0.376 for Si) which leads to a significant increase in the average ee distance, but from the CI expansion [s(1)s(2) - l2 z(1)z(2)]/sqrt(1+l4) We see that the wave function is still 86% (2s)2 character. This is expected since promotion of 2s to 2p costs a significant amount in the one electron energy. This promotion energy limits the size of l. Normalizing the GVB orbitals leads to φa = (φ2s + lφpz)/sqrt(1+l2) Thus the overlap of the GVB pair is < φa | φa > = (1-l2)/(1+l2)=0.752, similar to a bond pair

44. Problem with the GVB wavefunction A problem with this simple GVB wavefunction is that it does not have the spherical symmetry of the 1S ground state of Be. This problem is easily fixed in the configuration interaction (CI) form by generalizing to {s(1)s(2) - m2 [z(1)z(2)+x(1)x(2)+y(1)y(2)]} which does have 1S symmetry. Here the value of m2 ~ l2/3 This CI wavefunction can be solved for self-consistently. It is referred to as MC-SCF for multiconfiguration self-consistent field. But the simple GVB description of independent electrons in each orbital is obscured.

45. Contragradience хL хR We saw that the bonding in H2+ and H2 was dominated by the decrease in KE as the atoms were brought together. This decrease in the KE due to overlapping orbitals is dominated by skip ½ [< (хL). ((хR)> Which is large and negative in the shaded region between atoms, where the L and R orbitals have opposite slope (congragradience)

46. Role of pooched or hybridized atomic lobe orbitals in bonding of BeH neutral At small R the H can overlap significantly more with sz than with zH, so that we can form a bond pair just like in BeH+. This leads to the wavefunction At large R the the orbitals of Be are already hybridized A{[(sz)(H)+(H)(sz)](ab-ba)(zsa)} zs sz H In which the zs hybrid must now get orthogonal to the sz and H bond pair. This weakens the bond from that of BeH+ by ~ 1 eV Thus the wave function is A{[(sz)(zs)+(zs)(sz)](ab-ba)(Ha)} where sz≡(s+lz) and zs ≡(s-lz) Here the H overlaps slightly more with sz than with zs, but the spin on sz is half the time a Thus at large R we obtain a slightly repulsive interaction. zs sz H

47. Short range Attractive interaction sz with H Compare bonding in BeH+ and BeH Long range Repulsive interaction with H BeH TA’s check numbers, all from memory 2 eV BeH+ has long range attraction no short range repulsion 3 eV 1 eV BeH+ 1eV Repulsive orthogonalization of zs with sz H

48. Now bond 2nd H to BeH ~3.1 eV 1S+ Expect linear bond in H-Be-H and much stronger than the 1st bond Expect bond energy similar to BeH+, maybe stronger, because the zs orbital is already decoupled from the sz. Cannot bind 3rd H because no singly occupied orbitals left.

49. Compare bonding in BeH and BeH2 BeH+ MgH+ 1S+ 3.1 eV R=1.31A 2.1 eV R=1.65 A 1.34 eV R=1.73A 2.03 eV R=1.34A 2S+ ~3.1 eV 1S+ linear ~2.1 e Expect linear bond in H-Be-H and much stronger than the 1st bond Expect bond energy similar to BeH+, maybe stronger, because the zs orbital is already decoupled from the sz. TA’s check numbers, all from memory Cannot bind 3rd H because no singly occupied orbitals left.

50. The ground state for C atom x z Based on our study of Be, we expect that the ground state of C is Ψyz=A[(sx)(xs)+(xs)(sx)](ab-ba)(ya)(za)] which we visualize as sx py 2s pair pooched +x and –x yz open shell pz xs Ψyx=A[(sz)(zs)+(zs)(sz)](ab-ba)(ya)(xa)] which we visualize as px py 2s pair pooched +z and –z xy open shell zs sz Ψxz=A[(sy)(ys)+(ys)(sy)](ab-ba)(xa)(za)] which we visualize as px xz open shell 2s pair pooched +y and –y pz sy,ys