Lecture 9 September 30 2009 Nature of the

Lecture 9, September 30, 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 E 11 -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: linus 16@kaist. ac. kr Manager of Center for Materials Simulation and Design (CMSD) Teaching Assistant: Ms. Ga In Lee: leeandgain@kaist. ac. kr Special assistant: copyright. Pascal: tpascal@wag. caltech. edu EEWS-90. 502 -Goddard-L 09 © Tod 2009 William A. Goddard III, all rights reserved 1

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 2 pm on Wednesday Sept. 30 and another at 2 pm Wednesday Oct 14. This will be in the same room, 101 E 11 L 8: Sept. 29, as scheduled, 9 am L 9: Sept. 30, new lecture 2 pm replaces Oct 6 L 10: Oct. 1, as scheduled, 9 am L 11: Oct. 13, as scheduled, 9 am L 12: Oct. 14, new lecture 2 pm, replaces Oct 8 L 13: Oct. 15, as scheduled, 9 am EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 2

Volunteer extra credit assignment For those that will be bored with no lectures, we have a course project for extra credit. The numbers in these notes were mostly put together in 1972 to 1986, when the QM calcualtions were less accurate and the experiments less complete. I would like to update all the numbers using modern QM methods (B 3 LYP DFT with 6 -311 G** basis set) of moderate accuracy. Dr. Hyungjun Kim has a number of systems we would like to update at this level and will work with interested students Satisfactory completion of such a project, with a report summarizing the results can be used in lieu of the class midterm. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 3

Last time EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all

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 R 1 is called a symmetry transformation if it has the property that R 1 (HΨ)=H(R 1Ψ) The set of all possible symmetries transformations of H are collected into what is called a Group. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 5

The definition of a Group 1). Closure: If R 1, R 2 e G (both are symmetry transformations) then R 2 R 1 is also a symmetry transformation, R 2 R 1 e G 2. Identity. The do-nothing operator or identity, R 1 = e e G is clearly is a symmetry transformation 3. Associativity. If (R 1 R 2)R 3 =R 1(R 2 R 3). 4. Inverse. If R 1 e G then the inverse, (R 1)-1 e G , where the inverse is defined as (R 1)-1 R 1 = e. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 6

The degenerate eigenfunctions of H form a representation If HΨ=EΨ then H(R 1Ψ)= E(R 1Ψ) for all symmetry transformations of H. Thus the transformations amount the n denegerate functions, {S=(RiΨ), where Ri Ψi e G} 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. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 7

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 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 8

H 2 O, n example of C 2 v consider the nonlinear H 2 A molecule, with equal bond lengths, e. g. H 2 O, CH 2, NH 2 The symmetry transformations are 1. e for einheit (unity) x x, y y, z z 2. C 2 z, rotation about the z axis by 2 p/2=180º, x -x, y -y, z z 3. sxz, reflection in the xz plane, x x, y -y, z z 4. syz, reflection in the yz plane, x -x, y y, z z 5. Which is denoted as the C 2 v group. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 9

Stereographic projections 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. y x e C 2 z sxz syz EEWS-90. 502 -Goddard-L 09 syz This make relations between the symmetry elements transparent. e. g. C 2 zsxz= syz sxz Start with a general point, denoted as e and follow where it goes on various symmetry operations. Combine these as below to show the relationships C 2 z syz e y © copyright 2009 William A. Goddard III, all rights reserved x sxz C 2 v 10

The character table for C 2 v The basic symmetries (usually called irreducible representations) for C 2 v are given in a table, called the character table My choice of coordinate system follows (Mulliken in JCP 1955). This choice removes confusion about B 1 vs B 2 symmetry (x is the axis for which sxz moves the maximum number of atoms In the previous slide we saw that C 2 zsxz= syz which means that the This group is denoted as C 2 v, symmetries for syz are already which denotes that the implied by C 2 zsxz. Thus we consider generators are C 2 z and a C 2 z and sxz as the generators of the vertical mirror plane 11 group. (containing the C 2 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved axis)

Symmetries for H 2 O, NH 2, and CH 2 H 2 O NH 2 1 A 2 B 1 1 1 OHz bond A{(N 2 pya)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)](ab-ba)} NHx bond CH 2 1 A A{(O 2 py)2[(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)} OHx bond NH 2 CH 2 NHz bond A{(C 2 py)0[(Cpx )(Hx)+(Hx)(Cpx)](ab-ba)[(Cpz)(Hz)+(Hz)(Cpz)](ab-ba)} EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 12

Now do triplet state of CH 2 A{(C 2 sa)1(2 pxa)1[(Cp. L )(HL)+(HL)(Cp. L)](ab-ba)[(Cp. R)(HR)+(HR)(Cp. R)](ab-ba)} CHL bond CHR bond Since we know that the two CH bonds y are invariant under all symmetry operations, from now on we will write the wavefunction as z A{[(CHL)2(CHR)2](Csa)1(Cpa)1} Here s is invariant (a 1) while p transforms as b 1. Since both s and p are unpaired the ground state is triplet or S=1 Thus the symmetry of triplet CH 2 is 3 B 1 EEWS-90. 502 -Goddard-L 09 s=2 s © copyright 2009 William A. Goddard III, all rights reserved p=2 px 13

Second example, C 3 v, with NH 3 as the prototype A{[(Npy )(Hy)+(Hy)(Npy)](ab-ba)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)] (ab-ba)} NHb bond NHx bond We will consider a system such as NH 3, 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. NHc bond z x The other two NH bonds will be denoted as b and c. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 14

C 3 v, prototype: NH 3 z The symmetry transformations of C 3 v are: Hx 1. e for einheit (unity) x x, y y, z z 2. Cz 3, rotation about the z axis by 2 p/3=120º x Hc Hb 3. (Cz 3)2 rotation about the z axis by 2(2 p/3)=240º 4. sxz, reflection in the xz plane, x x, y -y, z z 5. syz, reflection in the plane through Hb and the z axis 6. syz, reflection in the plane through Hc and the z axis EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 15

Combining the 6 operations for C 3 v leads to Clearly this set of symmetries is closed, which you can see by the symmetry of the diagram. b C 3 sxz. C 32 e sxz Also we see that the generators are sxz and C 3 x Note that the operations do not all commute. Thus sxz. C 3 ≠ sxz. C 3 Instead we get sxz. C 3 c C 32= C 3 -1 EEWS-90. 502 -Goddard-L 09 sxz. C 3 = sxz. C 32 Such groups are called nonabelian and lead to irreducible representations with degree (size) > 1 © copyright 2009 William A. Goddard III, all rights reserved 16

The character table for C 3 v Here the E irreducible representation is of degree 2. This 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(3 P) at a site with C 3 v symmetry, would generally split into 2 levels, {3 Px and 3 Py} of 3 E symmetry 17 and 3 Pz of 3 A 1 symmetry. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved

Application for C 3 v, NH 3 z s. LP We will write the wavefunction for NH 3 as A{(s. LP)2[(NHb bond)2(NHc bond)2(NHx bond)2]} x x c b Where each of the 3 VB bond pairs is written as a pair function and we denote what started as the 2 s pair as slp The rotations and relections interchange pairs of electrons leaving the wavefunction invariant. Thus we get 1 A 1 symmetry EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 18

Linear molecules, C∞v symmetry For a linear molecule (axis along z) the symmetry operators are: e: einheit (unity) Rz(a): counterclockwise rotation by an angle a about the z axis sxz: reflection in the xz plane s’ = Rz(a) sxz Rz(-a); reflection in a plane rotated by an angle a from the xz plane (there an infinite number of these This group is denoted as C∞v EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 19

The character table for C∞v EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A.

The symmetry functions for C∞v Lower case letters are used to denote one-electron orbitals

Application to FH The ground state wavefunction of HF is A{(F 2 px)2(F 2 py)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 (φaφb- φbφa) = [(cosg)2 +(sing)2] }=(φxφy- φyφx) Thus (φxφy- φyφx) transforms as S. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 22

Continuing with FH Thus the (px)2(py)2 part of the HF wavefunction A{(F 2 px)2(F 2 py)2[(Fpz)(H)+(H)(Fpz)](ab-ba)} 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 1 S+ symmetry EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 23

Next consider the ground state of OH We write the wavefunctions for OH as Ψx=A{(s. OH bond)2[(p xa)(pya)(pxb)]} 2 P x 2 P y Ψy=A{(s. OH bond)2[(pxa)(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. x Thus the total wavefunction is 2 P. z Another way of describing this is to note that A{(px)2(py)2} transforms like S and hence one hole in a (p)4 shell, (p)3 transforms the same way as a single electron, (p)1 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 24

Now consider the ground state of NH A{(NH bond)2(N 2 pxa)(N 2 pya)} We saw earlier that up-spin in both x and y leads to S symmetry. With just one electron in py, we now get S-. x z Thus the ground state of NH is 3 S-. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 25

Now consider Bonding H atom to all 3 states of C (2 px)(2 pz) Bring H 1 s along z axis to C and consider all 3 spatial states. O 2 pz singly occupied. H 1 s can get bonding (2 py)(2 pz) Get S= ½ state, Two degenerate states, denote as 2 P (2 px)(2 py) No singly occupied orbital for H to bond with x z EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 26

Ground state of CH (2 P) The full wavefunction for the bonding state 2 P x A{(2 s)2(OHs bond)2(O 2 pxa)1} 2 P y A{(2 s)2(OHs bond)2(O 2 pya)1} x z EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 27

Bond a 2 nd H atom to the ground state of CH Starting with the ground state of CH, we bring a 2 nd H along the x axis. Get a second covalent bond This leads to a 1 A 1 state. No unpaired orbtial for a second covalent bond. x z EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 28

Analyze Bond in the ground state of CH 2 Ground state has 1 A 1 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 NH 2 (103. 2º) and OH 2 (104. 5º), we expect CH 2 to have bond angle of ~ 102º x z EEWS-90. 502 -Goddard-L 09 θe Re © copyright 2009 William A. Goddard III, all rights reserved 29

But, the Bending potential surface for CH 2 1 B 1 A 9. 3 kcal/mol 1 1 D g 1 3 B 1 3 S g The ground state of CH 2 is the 3 B 1 state not 1 A 1. Thus something is terribly wrong in our analysis of CH 2 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 30

Re-examine the ground state of Be Ground state of Be atom: A[(1 s)2(2 s)2] = A[(1 sa)(1 sb)(2 sa)(2 sb)] 1 s 2 s 1. 06 A 0. 14 A R~0. 14 A Each electron has its maximum amplitude in a spherical torus centered at R 2 s ~ 1. 06 A = 2. 01 bohr 2 a 0 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 e. V EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 31

hybridization of the atom 2 s orbitals of Be. So assumed that the Be wavefunction is A[(1 s)2(2 s)2] = A[(1 sa)(1 sb)(2 sa)(2 sb)] 2 s In fact this is wrong. Writing the wavefunction as A[(1 sa)(1 sb)(φaa)(φbb)] and solving self-consistently (unrestricted Hartree Fock or UHF calculation) for φa and φ2 s + l φpz φb leads to pz φa = φ2 s + l φpz and φb = φ2 s - l φpz where φpz is like the 2 pz orbital of Be+, but φ2 s - l φpz with a size like that of 2 s rather (smaller than a normal 2 p orbital) This pooching or hybridization of the 2 s orbitals in opposite directions leads to a much increased average ee distance, dramatically reducing EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved the ee repulsion. 32

analyze the pooched or hybridized orbitals Pooching of the 2 s orbitals in opposite directions leads to a dramatic increase in ee distance, reducing ee repulsion. φ2 s + l φpz φ2 s - l φpz z z 1 -D x 2 -D z x z Schematic. The line shows symmetric pairing. Notation: sz and sz bar or ℓ and ℓ bar. Cannot type bars. use zs to show the EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved bar case 33

Problem with UHF wavefunction A[(φaa)(φbb)] = [φa(1)][(φb(2)] – [φb(1)] [φa(2)] 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 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 34

The GVB orbitals for the (3 s)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 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 35

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 = φ2 s + l φpz and φb = φ2 s - l φpz into the spatial factor leads to (ab+ba) = (s+lz)(s-lz)+(s-lz)(s+lz) = [s(1)s(2) - l 2 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) -l 2 z(R) which is smaller than s(R) while the probability of being on opposite sides is s(R)s(-R)- l 2 z(R)z(-R) = s(R)+l 2 36 z(R) which is increased. 2009 William A. Goddard III, all rights reserved EEWS-90. 502 -Goddard-L 09 © copyright

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 = φ2 s + l φpz and φb = φ2 s - l φpz 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) - l 2 z(1)z(2)]/sqrt(1+l 4) We see that the wave function is still 86% (2 s)2 character. This is expected since promotion of 2 s to 2 p costs a significant amount in the one electron energy. This promotion energy limits the size of l. Normalizing the GVB orbitals leads to φa = (φ2 s + l φpz)/sqrt(1+l 2) Thus the overlap of the GVB pair is < φa | φa > = (1 -l 2)/(1+l 2)=0. 752, similar to a bond pair EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 37

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 = φ2 s + l φpz and φb = φ2 s - l φpz 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 2 s electron to 2 pz, which is very high (xx e. V). 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 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 38

Problem with the GVB wavefunction A problem with this simple GVB wavefunction is that it does not have the spherical symmetry of the 1 S ground state of Be. This problem is easily fixed in the CI form by generalizing to {s(1)s(2) - m 2 [z(1)z(2)+x(1)x(2)+y(1)y(2)]} which does have 1 S symmetry. Here the value of m 2 ~ l 2/3 This form of CI wavefunction can be solved for selfconsistently, referred to as MC-SCF for multiconfiguration self-consistent field. But the simple GVB description is obscured. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 39

Role of pooched or hybridized atomic lobe orbitals in bonding of Be. H+ Consider the bonding of H to Be+ In fact optimizing the wavefunction for Be. H+ leads The simple VB combination of to pooching of the 2 s toward H 1 s with the 2 s orbital of Be+ the H 1 s with much improved leads to a very small overlap and contragradience EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 40

Role of pooched or hybridized atomic lobe orbitals in bonding of Be. H neutral At small R the H can overlap At large R the orbitals of Be are significantly more with sz than already hybridized with z. H, so that we can form a bond pair just like in Be. H+. This leads to the wavefunction A{[(sz)(H)+(H)(sz)](ab-ba)(zsa)} zs sz H In which the zs hybrid must Thus the wave function is A{[(sz)(zs)+(zs)(sz)](ab-ba)(Ha)} now get orthogonal to the sz and H bond pair. This where sz≡(s+lz) and zs ≡(s-lz) weakens the bond from that Here the H overlaps slightly more of Be. H+ by ~ 1 e. V with sz than with zs, but the spin on sz is half the time a Thus at large R we obtain a zs sz slightly repulsive interaction. William A. Goddard III, all rights reserved EEWS-90. 502 -Goddard-L 09 © copyright 2009 H 41

Short range Attractive interaction sz with H Be. H Compare bonding in Be. H+ and Be. H Long range Repulsive interaction with H TA’s check numbers, all from memory 2 e. V Be. H+ has long range attraction no short range repulsion 3 e. V 1 e. V Be. H+ EEWS-90. 502 -Goddard-L 09 1 e. V Repulsive orthogonalization of 42 © copyright 2009 William A. Goddard III, all rights reserved sz H zs with

Compare bonding in Be. H and Be. H 2 Be. H+ Mg. H+ 3. 1 e. V R=1. 31 A 2. 1 e. V R=1. 65 A 2. 03 e. V R=1. 34 A 1. 34 e. V R=1. 73 A ~3. 1 e. V ~2. 1 e 1 S + 2 S + 1 S + linear Expect linear bond in H-Be-H and much stronger than the 1 st bond Expect bond energy similar to Be. H+, maybe stronger, because the zs orbital is already decoupled from the sz. TA’s check numbers, Cannot bind 3 rd H because no singly all from memory EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved occupied orbitals left. 43

New material EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all

The ground state for C atom Based on our study of Be, we expect that the ground state of C is Modified from the simple (1 s)2(2 p)2 form Ψyz=A[(2 sa)(2 sb)(ya)(za)] Consider first pooching the 2 s orbitals in the z direction Ψyz=A[(2 s+lz)(2 s-lz)+(2 s-lz)(2 s+lz)](ab-ba)(ya)(za)] 2 s pair pooched +z and –z Expanding in the CI form leads to Ψyz=A[(2 s)-l 2(z)(z)](ab-ba)(ya)(za)] yz open shell =A[(2 sa)(2 sb)](ya)(za)] -l 2 A[(za)(zb)(ya)(za)] But the 2 nd term is zero since the za is already occupied sx Thus the 2 s can only pooch in the x direction pz Ψyz=A[(sx)(xs)+(xs)(sx)](ab-ba)(ya)(za)] EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved py xs 45

The GVB orbitals of Silicon atom EEWS-90. 502 -Goddard-L 09 Long dashes indicate zero amplitude, solid lines indicate positive amplitude while short dashes indicate negative amplitude. The spacing between contours is 0. 05 in 46 © copyright 2009 William A. Goddard III, all atomic units rights reserved

The ground state for C atom 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 2 s pair pooched yz open shell pz +x and –x xs Ψyx=A[(sz)(zs)+(zs)(sz)](ab-ba)(ya)(xa)] which we visualize as px py 2 s 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 x xz open shell 2 s pair pooched pz +y and –y z EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. sy, ys III, all rights reserved Goddard 47

Now reconsider Bonding H atom to all 3 states of C (2 px)(2 pz) Bring H 1 s along z axis to C and consider all 3 spatial states. C 2 pz singly occupied. H 1 s can get bonding (2 py)(2 pz) Get S= ½ state, Two degenerate states, denote as 2 P (2 px)(2 py) Now we can get a bond to the lobe orbital just as for Be. H EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 48

Is the 2 P state actually 2 P? The presence of the lobe orbitals might seem to complicate the symmetry Ψyz=A[(sx)(xs)+(xs)(sx)](ab-ba)(ya) (z. H bond)2] Ψxz=A[(sy)(ys)+(ys)(sy)](ab-ba)(xa)(z. H bond)2)] To see that there is no problem, rewrite in the CI form (and ignore the z. H bond) Ψyz=A[(s 2 – l x 2)](ab-ba)(ya)] Now form a new wavefunction by adding - l y 2 to Ψyz Φyz ≡ A[s 2 – l x 2 – l y 2](ab-ba)(ya)] But the 3 rd term is A[y 2](ab-ba)(ya)]= – l A[(ya)(yb)(ya)]=0 Thus Φyz = Ψyz and similarly Φxz = A[s 2 – l x 2 – l y 2](ab-ba)(xa)] = Ψxz But the 2 s terms [s 2 – l x 2 – l y 2] are clearly symmetric about the 49 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved z axis. Thus these wavefunctions have 2 P 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)} xy open shell 2 s pair pooched +z and –z Unpaired H 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. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 50

Bonding of H to lobe orbital of C, small R At small R the H can overlap significantly more with sz than with z. H, so that we can form a bond pair just like in Be. H+. This leads to the wavefunction A{[(sz)(H)+(H)(sz)](ab-ba)(zsa)(pxa)(pya)} Sz-H bond pair py zs px sz H nonbond orbitals But now the zs hybrid must now get orthogonal to the sz and H bond pair. This destabilizes the bond by ~ 1 e. V 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, get S=3/2 or quartet state We saw for NH that (pxpy –pypx)(aa) has 3 S- symmetry. CH has one additional high spin nonbond s orbital, leading to 4 SEEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 51

GVB orbitals of Si. H 4 S- state EEWS-90. 502 -Goddard-L 09 © copyright

GVB orbitals of Si. H 2 P state sx pz H py xs EEWS-90.

The bonding states of CH and Si. H The low-lying state of Si. H are shown at the right. Similar results are obtained for CH. The bond to the p orbital to form the 2 P state is best Si. H Kcal/mol CH De(2 P) 80. 0 70. 1 p bond D(2 P-4 S-) 17. 1 36. 2 D e ( 4 S -) 62. 9 33. 9 sz bond The bond to the lobe orbital is weaker than the p, but it is certainly significant © copyright 2009 William A. Goddard III, all rights reserved EEWS-90. 502 -Goddard-L 09 54

Analysis of bonding in CH and Si. H Bond to p orbital is still the best for C and Si but the lobe bond is also quite strong. Thus hydridization in the atom due to electron correlation leads naturally to the new 4 S- 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 Si. H 180º 104º The reason is that as the p. H bond is formed, the incoming H orbital overlaps the 2 s part of the lobe orbital. To remain orthogonal, the 2 s 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 55 easier for the 2 nd EEWS-90. 502 -Goddard-L 09 H to bond to the lobe pair. all rights reserved © copyright 2009 William A. Goddard III,

Bonding the 2 nd H to CH and Si. H As usual, we start with the ground states of CH or Si. H, 2 Px and 2 Py and bond bring an H along some axis, say x. sx pz H 2 P y py xs 2 P EEWS-90. 502 -Goddard-L 09 x © copyright 2009 William A. Goddard III, all rights reserved 56

Bonding the 2 nd H to CH and Si. H Now we get credible bound states from both components A bond to the p A bond to the sx lobe orbital of CH (2 Px) orbital of CH (2 Py) sx 2 P pz 2 P x H y py xs This leads to the 1 A 1 state of CH 2 and Si. H 2 that has already been discussed. This leads to the 3 B 1 state that is the ground state of CH 2 © copyright 2009 William A. Goddard III, all rights reserved EEWS-90. 502 -Goddard-L 09 57

The p bond leads to the 1 A 1 state GVB orbitals for Si. H(1 A 1) The wavefunction is Ψ=A{[(sy)(ys)+(ys)(sy)(ab-ba)](Si. HLbond)2(Si. HRbond)2} Applying C 2 z or s in the plane interchanges (sy) and (ys) but the (sy)(ys) pair is symmetric under this interchange. Thus the total symmetry is 1 A. 1 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 58

Bonding the 2 nd H to the lobe orbital 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). sℓ px The wavefunction becomes Ψ=A{(Si. HLbond)2(Si. HRbond)2 [(sℓa)(pxa)]} Here the two bond pairs and the sℓ orbital have A 1 while px has b 1 symmetry so that the total spatial symmetry is B 1. This leads to both 3 B 1 and 1 B 1 states, but triplet is lower (since sℓ and px are orthogonal). 59 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved

Analyze Bond angles of CH 2 and Si. H 2 1 A θe Re 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 NH 2 and 14. 5º for OH 2 and indeed it increases by 12. 4º. But for Si the increase from 90º is only 2º as for P and As. θe Re 3 B state. Optimum bonding, sℓ 1 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 Si. H 2 decreasing the bond angle by 10º to 118º while the HH affect 60 dominates for CH 2, increasing the bond angleall rights reserved EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, by 5º to 133º 1

The GVB orbitals for Si. H 2 (3 B 1) EEWS-90. 502 -Goddard-L 09

Analysis of bond energies of 1 A 1 state Consider the first two p bonds. Ignoring the affect of the bonds on the lobe orbitals, The main difference arrise from the exchange terms. For C or Si A[(2 s)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{(2 s)2[(pz. H+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 2 nd bond should be stronger than the first by 7 kcal/mol for C and by 4. 5 kcal/mol for Si. E(kcal/mol)1 st bond 2 nd bond C 80 90 This is close to the observed Si 70 76. 2 differences. 62 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, allnumbers TA’s check rights reserved

Analysis of bond energies of CH 2 and Si. H 2 state 3 P 4 S 2 P CH 1 A 1 3 B 1 CH 2 90. 0 p 99. 1ℓ 33. 9ℓ 62. 9ℓ 70. 1 p 80. 0 p 4 S 2 P Si. H 56. 9ℓ 76. 1 p 3 B 1 1 A 1 Si. H 2 CH Lobe bonds: 63 and 99 Si. H Lobe bonds: 35 and 57 50% increase Assume 50% in lobe bond is from the first p bond destabilizing 63 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved the lone pair

Analysis of bond energies of 3 B 1 state We saw for CH that the lobe bond is 17 kcal/mol weaker than the p bond while it is 37 kcal/mol weaker for Si. Forming the lobe bond requires unpairing the lobe pair which is ~1 e. V for the C row and ~1. 5 e. V 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 Si. H should be easier than to C or Si, because the first p bond has already partially destabilized the lobe pair. Since the Si. H 2(3 B 1) state is 19 kcal/mol higher than Si. H 2(1 A 1) but Si. H(4 S-) is 35 kcal/mol higher than Si(2 P), we conclude that lobe bond has increased in strength by ~16 kcal/mol Indeed for CH the 3 B 1 state is 9. 3 kcal/mol lower than 1 A 1 implying that the lobe bond has increased in strength relative to the p bond by 26 kcal/mol. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 64

CH 2 GVB orbitals EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard

Add 3 rd H to form CH 3 For CH 2 we start with the 3 B 1 state and add H. Clearly the best place is in the plane, leading to planar CH 3, as observed. As this 3 rd bond is formed, each bond pair readjusts to a mixture of p and lobe character to become equivalent We could also make a bond to the out-ofplane pp orbital but this would lead to large HH repulsions. However the possibility of favorable out-ofplane bonding leads to an extremely flat potential curve for CH 3. Since the lobe orbital is no longer paired, we get a very strong bond energy of 109 kcal/mol. EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved 66

Add 3 rd H to form Si. H 3. For Si. H 2 we start with the 1 A 1 state and add H to The lobe pair. Clearly this leads to a pyramidal Si. H 3. Indeed the optimum bond angle is 111º. Since we must unpair the lobe orbital this 3 rd bond is relatively weak, 72 kcal/mol. © copyright 2009 William A. Goddard III, all rights reserved EEWS-90. 502 -Goddard-L 09 67

Add 4 th H to form CH 4 and Si. H 4. For CH 3 we start with the planar molecule and bring the H up to the out of plane p orbital. As the new bond forms all four bonds readjust to become equivalent leading to a tetrahedral CH 4 molecule. This bond is 105 kcal/mol, slightly weaker than the 3 rd 109 kcal/mol) since it is to a p orbital and must interact with the other H’s. For Si. H 2 we start with the pyramidal geometry (111º bond angle) and add to the remaining lobe orbital. As the new bond forms all four bonds readjust to become equivalent leading to a tetrahedral Si. H 4 molecule. No unpairing is required so that we get a strong bond, 92 kcal/mol (the EEWS-90. 502 -Goddard-L 09 rd © copyright 2009 William A. Goddard III, all rights reserved 68

EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III, all rights reserved

Si. H 3 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III,

Si. H 4 EEWS-90. 502 -Goddard-L 09 © copyright 2009 William A. Goddard III,

Hybridization of GVB Orbitals Idealized case. Tetrahedral: sp 3 (s+x+y+z)/2 (s-x-y+z)/2 (s-x+y-z)/2 (s+x-y-z)/2 Orthogonal and point to vertices of a tetrahedron. Rationalizes bonding in CH 4. Assumes 75% p character (atom is 57% p) GVB 70% p copyright 2009 William A. Goddard III, all rights reserved EEWS-90. 502 -Goddard-L 09 © 72