© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 1 Nature of the Chemical Bond with applications to catalysis, materials.

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 1 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy 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 Course number: KAIST EEWS 80.502 Room E11-101 Hours: 0900-1030 Tuesday and Thursday 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.krleeandgain@kaist.ac.kr Special assistant: Tod Pascal:tpascal@wag.caltech.edu Lecture 10, October 01, 2009

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 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, 9am L9: Sept. 30, new lecture 2pm replaces Oct 6 L10: Oct. 1, as scheduled, 9am L11: Oct. 13, as scheduled, 9am L12: Oct. 14, new lecture 2pm, replaces Oct 8 L13: Oct. 15, as scheduled, 9am

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 3 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 (B3LYP DFT with 6-311G** 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.

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 5 The ground state for C atom Ψ yz = A [(sx)(xs)+(xs)(sx)](  )(y  )(z  )] Instead of the simple (1s) 2 (2s) 2 (2p) 2 form 2s pair pooched ±x yz open shell Ψ yz = A [(2s  s  )(y  )(z  )] pz sx xs py the ground state of C is

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 6 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

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 7 Bonding H atom to all 3 states of C Now we can get a bond to the lobe orbital just as for BeH Bring H1s along z axis to C and consider all 3 spatial states. (2p x )(2p z ) C 2p z singly occupied. H1s can get bonding Get S= ½ state, Two degenerate states, denote as 2  (2p x )(2p y ) (2p y )(2p z )

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 8 Bonding of H to lobe orbital of C, Long R Thus the wave function is A {(p x  )(p y  )[(sz)(zs)+(zs)(sz)](  )(H  At large R the lobe orbitals of C are already hybridized 2s pair pooched +z and –z xy open shell Unpaired H Here the H overlaps slightly more with sz than with zs, but the spin on sz is half the time  Thus at large R we obtain a slightly repulsive interaction.

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 9 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 A{ [(sz)(H)+(H)(sz)](  )(zs  px  py  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= , px=  x, py=  y Since the nonbond orbitals, ,  x,  y are orthogonal to each other the high spin is lowest, get S=3/2 or quartet state We saw for NH that (  x  y –  y  x )(  ) has 3  - symmetry. CH has one additional high spin nonbond  orbital, leading to 4  - Hsz zs px py Sz-H bond pairnonbond orbitals

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 12 The bonding states of CH and SiH 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 2  state is best CHSiH De(2)De(2) 80.070.1 Kcal/mol  ( 2  4  - ) De(4-)De(4-) 62.933.9 17.136.2 p bond 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 reservedEEWS-90.502-Goddard-L10 13 Analysis of bonding in CH and SiH 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 2 nd H to bond to the lobe pair. 104º 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  - 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

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 14 Bonding the 2 nd H to CH and SiH As usual, we start with the ground states of CH or SiH, 2  x and 2  y and bond bring an H along some axis, say x. H sx xs pz py 2y2y 2x2x

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 15 Bonding the 2 nd H to CH and SiH Now we get credible bound states from both components H sx xs pz py 2y2y 2x2x A bond to the sx lobe orbital of CH ( 2  y ) A bond to the p orbital of CH ( 2  x ) This leads to the 1 A 1 state of CH 2 and SiH 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 reservedEEWS-90.502-Goddard-L10 16 The p bond leads to the 1 A 1 state GVB orbitals for SiH( 1 A 1 ) Ψ= A {[(sy)(ys)+(ys)(sy)(  SiH L bond) 2 (SiH R bond) 2 } The wavefunction is Applying C 2z or  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.

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 17 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). The wavefunction becomes Ψ= A {(SiH L bond) 2 (SiH R bond) 2 [(  ℓ  x  } ℓℓ xx Here the two bond pairs and the  ℓ orbital have A 1 while  x 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  ℓ and  x are orthogonal).

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 18 Analyze Bond angles of CH 2 and SiH 2 θeθe ReRe 1 A 1 state. Optimum bonding,  p z orbital points at H z while p x orbital points at H x, 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. θeθe ReRe 3 B 1 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  ℓ orbital must get orthogonal to the two bond pairs  a bond angle decrease. The lone pair affect dominates for SiH 2 decreasing the bond angle by 10º to 118º while the HH affect dominates for CH 2, increasing the bond angle by 5º to 133º ℓℓ

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 20 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 arises from the exchange terms. For C or Si A [(2s) 2 (pz  )(px  )] leads to a term in the energy of the form (J xz –K xz ) 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)(  )(px  )}. Now the pz and px orbitals have the same spin only have the time, so that this exchange term is decreased to - ½ K xz. 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 C8090 Si7076.2 This is close to the observed differences. TA’s check numbers

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 21 Analysis of bond energies of CH 2 and SiH 2 state 22 4-4- CH 62.9ℓ SiH 3P3P 80.0p 22 4-4- 33.9ℓ 70.1p 3B13B1 1A11A1 90.0p 99.1ℓ 3B13B1 1A11A1 56.9ℓ 76.1p CH 2 SiH 2 CH Lobe bonds: 63 and 99 50% increase SiH Lobe bonds: 35 and 57 50% increase Assume 50% in lobe bond is from the first p bond destabilizing the lone pair

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 22 The Bending potential surface for CH 2 3B13B1 1A11A1 1B11B1 3g-3g- 1g1g 9.3 kcal/mol The ground state of CH 2 is the 3 B 1 state not 1 A 1.

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 23 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 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 SiH 2 ( 3 B 1 ) state is 19 kcal/mol higher than SiH 2 ( 1 A 1 ) but SiH( 4  - ) is 35 kcal/mol higher than Si( 2  ), 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.

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 25 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-of- plane p  orbital but this would lead to large HH repulsions. However the possibility of favorable out-of- plane bonding leads to an extremely flat potential curve for CH 3. Since the lobe orbital is already unpaired, we get a very strong bond energy of 109 kcal/mol. 120º 133º

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 26 Add 3 rd H to 1 A 1 SiH2 to form SiH 3. For SiH 2 we start with the 1 A 1 state and add H to The lobe pair. Clearly this leads to a pyramidal SiH 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 reservedEEWS-90.502-Goddard-L10 27 Add 4 th H to form CH 4 and SiH 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 SiH 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 SiH4 molecule. No unpairing is required so that we get a strong bond, 92 kcal/mol (the 3 rd bond was 72)

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 31 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 reservedEEWS-90.502-Goddard-L10 33 The ground state for B atom Ψ yz = A [(sx)(xs)+(xs)(sx)](  )(z  )] which we visualize as z x Based on our study of C, we expect that the ground state of B is 2s pair pooched ±x (or ±y) z open shell pz sx xs 2s pair pooched ± z x open shell px szzs Ψ yx = A [(sz)(zs)+(zs)(sz)](  )(x  )] which we visualize as Ψ xz = A [(sz)(zs)+(zs)(sz)](  )(y  )] which we visualize as 2s pair pooched ± z y open shellsz py zs

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 34 form BH and AlH by bonding along the z axis Ψ yz = A [(sx)(xs)pair](z  z  ] Bonding to the pz state of B we obtain 2s pair pooched ±x (or ±y) H-z covalent bond pz sx xs H-sz covalent bond open shell px sz zs Ψ yx = A [(sz)(H)+(H)(sz)](  )(x  )(zs  ] Ψ xz yx = A [(sz)(H)+(H)(sz)](  )(y  )(zs  ] open shell szpy zs 128º H H H H-sz covalent bond 1+1+ 3x3x 3y3y

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 35 The bonding states of BH and AlH The ground state of BH and AlH is obtained by bonding to the p orbital (leading to the 1  + state. However the bond to the lobe orbital (leading to the 3  state  is also quite strong. The bond to the lobe orbital is weaker than the p, but it is certainly significant 1+1+ 33 11 3+3+ 2 P+ H

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 36 BH 2 and AlH 2 128º Starting from the ground state of BH or AlH, the second bond is to a lobe orbital, to form the 2 A 1 state. Just as for the 3 B 1 state of CH 2 and SiH 2 the bond for BH 2 opens by several degrees to 131º while the bond to the AlH 2 closes down by ~9º. θeθe ReRe

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 37 BH 3 and AlH 3 Bonding the 3 rd H to the 2A1 state of BH2, leads to planar BH3 or AlH3. But there is no 4 th bond since there remain no additional unpaired orbitals to bond to.

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 38 Re-examine bonding in NH, OH, and FH Why did we ignore hybridization of the (2s) pair for NH, OH, and FH? The reason is that the ground state of N atom A [(2s  )(2s  )(x  )(y  )(z  )] Already has occupied px,py,pz orbitals. Thus Pauli annihilates any hybridization in the 2s orbital.

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 39 Re-examine bonding in NH, OH, and FH However, the doublet excited state of N can have hybridization, eg A (2s) 2 (y) 2 (z  )  A [(sx)(xs)+(xs)(sx)](  )(y) 2 (z  ) which leads to the 2 A 1 excited states of NH 2 of the form θeθe ReRe

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 41 Bonding to halides (AXn, for X=F,Cl,… The ground state of F has just one singly-occuped orbital and hence bonding to C is in many ways similar to H, leading to CF, CF2, CF3, and CF4 species. However there are significant differences. Thus CF leads to two type so bonds, p and lobe just like CH 22 44 p bond lobe bond Covalent bond expect (CH) 80 kcal/mol 63 kcal/mol actual bond (CF) 120 kcal/mol 63 kcal/mol

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 42 How can CF lead to such a strong bond, 120 vs 80 kcal/mol? Consider the possible role of ionic character in the bonding In the extreme limit + CFC+C+ F-F- E (R=∞) = 0E (R=∞) =IP (C) – EA (F) =11.3 – 3.4 =7.9 eV IP (C) = 11.3 eV = 260 kcal/mol EA (F) = 3.40 eV = 78.4 kcal/mol Can Coulomb attraction make up this difference?

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 43 Estimate energy of pure ionic bond for CF Covalent limit (C + F) Ionic limit (C + + F - ) 7.9 eV 14.4/R R e =1.27A 14.4/1.27 = 11.3 eV Net bond = 11.3-7.9 = 3.4 eV= 78 kcal/mol Units for electrostatic interactions E=Q1*Q2/(  0 *R) where  0 converts units (called permittivity of free space) E(eV) = 14.4 Q1(e units)*Q2(e units)/R(angstrom) E(kcal/mol) = 332.06 Q1(e units)*Q2(e units)/R(angstrom) Ionic estimate ignores shielding and pauli repulsion for small R. Thus too large

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 44 CF has strong mixture of covalent and ionic character + Pure covalent bond ~ 80 kcal/mol (based on CH) Pure ionic bond ~ 78 kcal/mol (ignore Pauli and shielding) Net bond = 120 kcal/mol is plausible for 2  state But why is the bond for the 4  - state only 63, same as for covalent bond?

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 45 Consider ionic contribution to 4    state + CFC+C+ F-F- E (R=∞) = 0E (R=∞) =IP (C) – EA (F) =16.6 – 3.4 =13.2 eV To mix ionic character into the 4   state the electron must be pulled from the sz lobe orbital. This leads to the (2s) 1 (2p) 2 state of C + rather than the ground state (2s) 2 (2p) 1 which is 123 kcal/mol = 5.3 eV higher Thus ionic bond is NEGATIVE (78-123=-45 kcal/mol)

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 46 Bonding the 2 nd F to CF With the 4  - state at 57 kcal/mol higher than  , we need only consider bonding to 2 , leading to the 1 A 1 state. Bad Pauli repulsion increases FCF angle to 105º 1A11A1 3B13B1 57 kcal/mol

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 47 Now bond 3 rd F to form CF 3 Get pyramidal CF 3 (FCF angle ~ 112º) In sharp contrast to planar CH 3 The 3 rd CF bond should be much weaker than 1 st two. This strong preference for CF to use p character makes conjugated flourocarbons much less stabe than corresponding saturated compounds. Thus C4H6 prefers butadiene but C4F6 prefers cylcobutene Of course the 4 th bond to form CF 4 leads to a tetrahedral structure

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 48 Summary bonding to form hydrides General principle: start with ground state of AH n and add H to form the ground state of AH n+1 Thus use 1 A 1 AH 2 for SiH 2 and CF 2 get pyramidal AH 3 Use 3 B 1 for CH 2 get planar AH 3. For less than half filled p shell, the presence of empty p orbitals allows the atom to reduce electron correlation of the (ns) pair by hybridizing into this empty orbital. This has remarkable consequences on the states of the Be, B, and C columns.

© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 1 Nature of the Chemical Bond with applications to catalysis, materials.

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© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L10 1 Nature of the Chemical Bond with applications to catalysis, materials.

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