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## Lecture 8 January 22, 2010

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**Lecture 8 January 22, 2010**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> Ted Yu <tedhyu@wag.caltech.edu>**Course schedule**Monday Jan. 18 holiday, remember MLK, no lecture Wednesday Jan. 20, 2pm L7 regular time Friday Jan. 22, 2pm L8 (regular time) Monday Jan. 25 2pm L9 Wednesday Jan. 27, 2pm L10 regular time Friday Jan. 29, 2pm L11 (regular time) ??**x**Linear molecules, C∞v symmetry z H--C=N 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**Demonstrate that s’ = Rz(a) sxz Rz(-a) is a reflection in**the plane rotated by an angle a from the xz plane y y y y e e e e x x x x Take the z axis out of the plane. The six symmetry operations are: y e a a x -a C-a Ca sxz y a e x s’ s’**x**C∞v symmetry group name 1-e N-e s p d f g z H--C=N The symmetry operators are: Rz(a): counterclockwise rotation by an angle a about the z axis sxz: reflection in the xz plane (this takes +a into –a) s’ = Rz(a) sxz Rz(-a); reflection in a plane rotated by an angle a from the xz plane (there are an infinite number of these) e: einheit (unity) This group is denoted as C∞v,The character table (symmetries) are**Application to FH**The ground state wavefunction of HF is A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](ab-ba)} In C∞v symmetry, the bond pair is s (m=0),while the px and py form a set of p orbitals (m=+1 and m=-1). Consider the case of up spin for both px and py Ψ(1,2) = A{φxaφya}=(φxφy- φyφx) aa Rotating by an angle g about the z axis leads to φa =cosgφx +singφy and φb =cosgφy -singφx This leads to (φaφb- φbφa) = [(cosg)2 +(sing)2] }=(φxφy- φyφx) Thus (φxφy- φyφx) transforms as S.**Continuing with FH**Thus the (px)2(py)2 part of the HF wavefunction A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](ab-ba)} Since both aa and bb transform like S; the total wavefunction transforms as S The symmetry table, demands that we also consider the symmetry with respect to reflection in the xz plane. Here px is unchanged while py changes sign. Since there are two electrons in py the wavefunction is invariant. Thus the ground state of FH has 1S+ symmetry**x**z Next consider the ground state of OH We write the two wavefunctions for OH as 2Px Ψx=A{(sOHbond)2[(pxa)(pya)(pxb)]} Ψy=A{(sOHbond)2[(pxa)(pya)(pyb)]} We saw above that A{(pxa)(pya)} transforms like S. thus we need examine only the transformations of the downspin orbital. But this transforms like p. 2Py Thus the total wavefunction is 2P. Another way of describing this is to note that A{(px)2(py)2} transforms like S and hence one hole in a (p)4shell, (p)3 transforms the same way as a single electron, (p)1**x**z Now consider the ground state of NH A{(NH bond)2(N2pxa)(N2pya)} We saw earlier that up-spin in both x and y leads to S symmetry. Considering now the reflection, sxz, we see that with just one electron in py, we now get S-. Thus the ground state of NH is 3S-.**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**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}**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.**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**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**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**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.**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**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**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**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).**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.**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)**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**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.**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**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**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.**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+**Re-examine 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**The GVB orbitals of Silicon 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**Now reconsider Bonding H atom to all 3 states of C**(2px)(2pz) Bring H1s along z axis to C and consider all 3 spatial states. C 2pz singly occupied. H1s can get bonding Get S= ½ state, Two degenerate states, denote as 2P Just as in our previous analysis (2py)(2pz) (2px)(2py) Now we can get a bond to the lobe orbital just as for BeH This is new**Is the 2P state actually 2P?**The presence of the lobe orbitals might seem to complicate the symmetry Ψyz=A[(sx)(xs)+(xs)(sx)](ab-ba)(ya) (zH bond)2] Ψxz=A[(sy)(ys)+(ys)(sy)](ab-ba)(xa)(zH bond)2)] To see that there is no problem, rewrite in the CI form (and ignore the zH bond) Ψyz=A[(s2 – l x2)](ab-ba)(ya)] Now form a new wavefunction by adding - l y2 to Ψyz Φyz ≡A[s2 – l x2 – l y2](ab-ba)(ya)] But the 3rd term is A[y2](ab-ba)(ya)]= – l A[(ya)(yb)(ya)]=0 Thus Φyz = Ψyz and similarly Φxz = A[s2 – l x2 – l y2](ab-ba)(xa)] = Ψxz Thus the 2s term [s2 – l x2 – l y2] is clearly symmetric about the z axis, so that these wavefunctions have 2P symmetry**Bonding of H to lobe orbital of C, Long R**At large R the lobe orbitals of C are already hybridized Thus the wave function is A{(pxa)(pya)[(sz)(zs)+(zs)(sz)](ab-ba)(Ha)} Unpaired H 2s pair pooched +z and –z xy open shell 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.**Bonding of H to lobe orbital of C, small R**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 px py zs sz H A{[(sz)(H)+(H)(sz)](ab-ba)(zsa)(pxa)(pya)} Sz-H bond pair nonbond orbitals But now the zs hybrid must now get orthogonal to the sz and H bond pair. This destabilizes the bond by ~ 1 eV The symmetries of the nonbond orbitals are: zs=s, px=px, py=py Since the nonbond orbitals, s, px, py are orthogonal to each other the high spin is lowest (S=3/2 or quartet state) We saw for NH that (pxpy –pypx)(aa) has 3S- symmetry. CH has one additional high spin nonbond s orbital, leading to 4S-**GVB orbitals of SiH 2P state**sx pz H py xs**The bonding states of CH and SiH**CH SiH Kcal/mol p bond De(2P) 80.0 70.1 D(2P-4S-) 17.1 36.2 De(4S-) 62.9 33.9 sz bond The low-lying state of SiH are shown at the right. Similar results are obtained for CH. The bond to the p orbital to form the 2P state is best The bond to the lobe orbital is weaker than the p, but it is certainly significant**Analysis of bonding in CH and SiH**Bond to p orbital is still the best for C and Si but the lobe bond is also quite strong, especially for CH Thus hydridization in the atom due to electron correlation leads naturally to the new 4S- bonding state. Note that although the (sx)(xs) lobe pair for the atom are at 180º in the atom, they bend to ~104º for CH and SiH 104º 180º The reason is that as the pH bond is formed, the incoming H orbital overlaps the 2s part of the lobe orbital. To remain orthogonal, the 2s orbital builds in some –z character along with the x character already there. This rotates the lobe orbital away from the incoming H. This destabilizes the lobe pair making it easier for the 2nd H to bond to the lobe pair.**Bonding the 2nd H to CH and SiH**sx pz H py xs As usual, we start with both components of the ground state of CH or SiH, 2Px and 2Py and bond bring an H along some axis, say x. 2Py 2Px**Bonding the 2nd H to CH and SiH**Now we get credible bound states from both components A bond to the p orbital of CH (2Px) A bond to the sx lobe orbital of CH (2Py) sx 2Px pz 2Py H py xs This leads to the 1A1 state of CH2 and SiH2 that has already been discussed. This leads to the 3B1 state that is the ground state of CH2**The p bond leads to the 1A1 state**GVB orbitals for SiH(1A1) The wavefunction is Ψ=A{[(sy)(ys)+(ys)(sy)(ab-ba)](SiHLbond)2(SiHRbond)2} Applying C2z or s in the plane interchanges (sy) and (ys) but the (sy)(ys) pair is symmetric under this interchange. Thus the total symmetry is 1A1.**Bonding the 2nd H to the lobe orbital**sℓ px At large distances the bond to the lobe orbital will be slightly repulsive and at an angle of 128 to the already formed p bond. However at short distances, we form a strong bond. After forming the bond, each bond pair readjusts to have equivalent character (but an average of lobe bond and p bond). The wavefunction becomes Ψ=A{(SiHLbond)2(SiHRbond)2 [(sℓa)(pxa)]} Here the two bond pairs and the sℓ orbital have A1 while px has b1 symmetry so that the total spatial symmetry is B1. This leads to both 3B1 and 1B1 states, but triplet is lower (since sℓ andpx are orthogonal).**θe**Re Analyze Bond angles of CH2 and SiH2 1A1 state. Optimum bonding, pz orbital points at Hz while px orbital points at Hx, leading to a bond angle of 90º. We expect that HH repulsion increases this by slightly less than the 13.2º for NH2 and 14.5º for OH2 and indeed it increases by 12.4º. But for Si the increase from 90º is only 2º as for P and As. θe Re sℓ 3B1 state. Optimum bonding, the two bonds at ~128º. Here HH orthogonalization should increase this a bit but C much less than 12º since H’s are much farther apart. However now the sℓ orbital must get orthogonal to the two bond pairs a bond angle decrease. The lone pair affect dominates for SiH2 decreasing the bond angle by 10º to 118º while the HH affect dominates for CH2, increasing the bond angle by 5º to 133º**Analysis of bond energies of 1A1 state**Consider the first two p bonds. Ignoring the affect of the bonds on the lobe orbitals, the main difference arises from the exchange terms. For C or Si A[(2s)2(pza)(pxa)] leads to a term in the energy of the form (Jxz –Kxz) since the x and z orbitals have the same spin. But upon bonding the first H to pz, the wavefunction becomes A{(2s)2[(pzH+Hpx)(ab-ba)(pxa)}. Now the pz and px orbitals have the same spin only have the time, so that this exchange term is decreased to - ½ Kxz. However in forming the second bond, there is no additional correction. Since Kxz ~ 14 kcal/mol for C and ~9 kcal/mol for Si. This means that the 2nd bond should be stronger than the first by 7 kcal/mol for C and by 4.5 kcal/mol for Si. E(kcal/mol)1st bond 2nd bond C 80 90 Si 70 76.2 This is close to the observed differences.**Analysis of bond energies of CH2 and SiH2 state**3P 4S- 33.9ℓ 4S- 62.9ℓ 2P 2P 70.1p 80.0p SiH CH 3B1 56.9ℓ 1A1 76.1p 1A1 90.0p SiH2 3B1 99.1ℓ CH2 SiH Lobe bonds: 35 and 57 50% increase CH Lobe bonds: 63 and 99 50% increase We find that forming the first bond to the p orbital, destabilizes the lobe pair so that bonding to the lobe orbital of XH is 50% stronger than bonding to lobe orbital of X**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.**Analysis of bond energies of 3B1 state**For CH the lobe bond is 17 kcal/mol weaker than the p bond while for SiH it is 37 kcal/mol weaker. Forming the lobe bond requires unpairing the lobe pair which is ~1 eV for the C row and ~1.5 eV for the Si row. This accounts for the main differences suggesting that p bonds and lobe bonds are otherwise similar in energy. Forming a lobe bond to CH or SiH should be easier than to C or Si, because the first p bond has already partially destabilized the lobe pair. Since the SiH2(3B1) state is 19 kcal/mol higher than SiH2(1A1) but SiH(4S-) is 35 kcal/mol higher than Si(2P), we conclude that lobe bond has increased in strength by ~16 kcal/mol Indeed for CH the 3B1 state is 9.3 kcal/mol lower than 1A1 implying that the lobe bond has increased in strength relative to the p bond by 26 kcal/mol.