William A. Goddard, III, wag3@kaist.ac.kr

1. Lecture 2, September 3, 2009 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: KIAST EEWS 80.502 Room E11-101 Hours: 0900-1030 Tuesday and Thursday William A. Goddard, III, wag3@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: linus16@kaist.ac.kr Teaching Assistant: Ms. Ga In Lee: leeandgain@kaist.ac.kr EEWS-90.502-Goddard-L02

2. Last time At the beginning of each lecture, Goddard will summarize the key points from the previous lecture. These are the only important parts that you will need to know for the exams. The other material covered plus the material in the course notes that have been passed out are merely to help you understand the material and are not important for the exams EEWS-90.502-Goddard-L02

3. Quantum Mechanics – First postulate Essential postulate of QM is that all properties that can be known about the system are contained in the wavefunction, Φ(x,y,z,t) (for one electron), where the probability of finding the electron at position x,y,z at time t is given by P(x,y,z,t) = | Φ(x,y,z,t) |2 = Φ(x,y,z,t)*Φ(x,y,z,t) where ∫Φ(x,y,z,t)*Φ(x,y,z,t) dxdydz ≡ < Φ| Φ> = 1 EEWS-90.502-Goddard-L02

4. Quantum Mechanics – Second postulate In QM the total energy can be written as EQM = KEQM + PEQM where for a system with a potential energy function, V(x,y,z,t) PEQM=∫Φ(x,y,z,t)*V(x,y,z,t)Φ(x,y,z,t)dxdydz ≡ < Φ| V|Φ> Just like Classical mechancs except weight V with P=|Φ|2 For the H atom PEQM=< Φ| (-e2/r) |Φ> = -e2/ where is the average value of 1/r _ _ R R KEQM = (Ћ2/2me) <(Φ·Φ> where <(Φ·Φ> ≡ ∫[(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz In QM the KE is proportional to the average square of the gradient or slope of the wavefunction Thus KE wants smooth wavefunctions, no wiggles EEWS-90.502-Goddard-L02

5. My form for KE differs from usual one in QM • One dimensional, usual form • KE = - (Ћ2/2me) ∫(Φ*)(d2Φ/dx2) dx = - (Ћ2/2me) <Φ|(d2/dx2)| Φ> • = <Φ| - (Ћ2/2me) (d2/dx2)| Φ> • Where KE operator is - (Ћ2/2me) (d2/dx2) • Now integrate by parts: • ∫u(dv/dx)dx = ∫(du/dx)(v)dx if u,v  0 at boundaries • Let u = Φ* and dv/dx = d2Φ/dx2 then get Goddard form • KE = (Ћ2/2me)<(dΦ/dx)|(dΦ/dx)> Both forms of the KE are correct, I consider the second form more fundamental and more useful It can be used to derive the 1st form by integrating by parts Clearly KE is always positive and decreasing slopes decreases KE EEWS-90.502-Goddard-L02

6. Quantum mechanics description of Hydrogen atom _ _ _ _ R R R R How do PE and KE scale with ? PE ~ -C1/ KE ~ +C2/ 2 Now lets find the optimum electron Charge -e Nucleus Charge +Z EEWS-90.502-Goddard-L02

7. Analysis for optimum Consider very large Here PE is small and negative, but KE is (small)2 but positive, thus PE wins and the total energy is negative Now consider very small Here PE is large and negative, but KE is (large)2 but positive, thus KE wins and the total energy is positive _ _ _ _ R R R R Thus there must be some intermediate for which the total energy is most negative This is the for the optimum wavefunction _ R Conclusion in QM the H atom has a finite size, EEWS-90.502-Goddard-L02

8. Implications In QM KE wants to have a smooth wavefunction but electrostatics wants the electron concentrated at the nucleus. Since KE ~ 1/R2 , KE always keeps the wavefunction finite, leading to the finite size of H and other atoms to the formation of molecules EEWS-90.502-Goddard-L02

9. The wavefunction for H atom _ _ _ _ R R R R The PE = -Ze2/ while KE= Ze2/ 2 Thus the total energy E = -Ze2/2 = PE/2 = -KE This is called the Virial Theorem and is general for all molecules Where = 0.529 A =0.053 nm is the Bohr radius (the average size of the H atom), =a0/Z Here is the normalization constant, <Φ|Φ>=1 Here we use r,Ө,Φspherical coordinates rather than x,y,z EEWS-90.502-Goddard-L02

10. Atomic Units = 0.529 A =0.053 nm is the Bohr radius For Z ≠ 1, = a0/Z _ _ R R E = -Ze2/2 Z=1 for Hydrogen atom = - Z2 e2/2a0 = - meZ2 e4/2 Ћ2 = Z2 h0/2 where h0 = e2/a0 = me e4/ Ћ2 = Hartree = 27.2116 eV = 627.51 kcal/mol = 2625.5 kJ/mol Atomic units: me = 1, e = 1, Ћ = 1 leads to unit of length = a0 and unit of energy = h0 In atomic units: KE= <Φ.Φ>/2 (leave off Ћ2/me) PE = <Φ|-1/r|Φ>/2 (leave off e2) EEWS-90.502-Goddard-L02

11. 2 views of orbitals contour plot of orbital in xz plane, adjacent contours differ by 0.05 au Iine plot of orbital along z axis EEWS-90.502-Goddard-L02

12. Now consider H2+ molecule e Bring a proton up to an H atom to form H2+ Is the molecule bound a lower energy at finite R than at R = ∞ Two possibilities Electron is on the left proton, L Electron is on the right proton, R Or we could combine them rL rR L R At R = ∞ these are have the same energy, but not for finite R In QM we always want the wavefunction with the lowest energy. Question: which combination is lowest? EEWS-90.502-Goddard-L02

13. Combine Atomic Orbitals for H2+ molecule Symmetric combination Two extreme possibilities Antisymmetric combination Which is best (lowest energy)? the Dg = Sqrt[2(1+S)] and Du = Sqrt[2(1-S)] factors above are the constants needed to ensure that <Φg|Φg> =∫ Φg|Φg dxdydz= 1 (normalized) <Φu|Φu> =∫ Φu|Φu dxdydz= 1 (normalized) I will usually writing such factors, but they are understood EEWS-90.502-Goddard-L02

14. The change in electron density for molecular orbitals The densities rg and ru for the g and u LCAO wavefunctions of H2+ compared to superposition of rL + rR atomic densities (all densities add up to one electron) Adding the two atomic orbitals to form the g molecular orbital increases the electron density in the bonding region, as expected. This is because in QM, one adds amplitudes and the squares to get probability density EEWS-90.502-Goddard-L02

15. Compare change in density with local PE function The local PE for the electron is lowered at the bond midpoint from the value of a single atom PE(r) = -1/ra – 1/rb But the best local PE is still near the nucleus Thus the Φg = L + R wavefunction moves charge to the bond region AT THE EXPENSE of the charge near the nuclei, causing an increase in the PE, and opposing bonding EEWS-90.502-Goddard-L02

16. Potential energy for Molecular Orbitals of H2+ The total PE of H2+ for the Φg = L + R and Φu = L - R wavefunctions (relative to the values of Vg = Vu = -1 h0 at R = ∞) EEWS-90.502-Goddard-L02

17. If the bonding is not due to the PE, then it must be KE We see a dramatic decrease in the slope of the g orbital along the bond axis compared to the atomic orbital. This leads to a dramatic decrease in KE compared to the atomic orbital This decrease arises only in the bond region. The shape of the Φg = L + R and Φu = L - R wavefunctions compared to the pure atomic orbital (all normalized to a total probability of one). It is this decrease in KE that is responsible for the bonding in H2+ EEWS-90.502-Goddard-L02

18. The KE of g and u wavefunctions Use top part of 2-7 The change in the KE as a function of distance for the g and u wavefunctions of H2+ (relative to the value at R=∞ of KEg=KEu=+0.5 h0 Comparison of the g and u wavefunctions of H2+ (near the optimum bond distance for the g state, showing why g is so bonding and u is so antibonding EEWS-90.502-Goddard-L02

19. KE dominates PE Changes in the total KE and PE for the g and u wavefunctions of H2+ (relative to values at R=∞ of KE :+0.5 h0 PE: -1.0 h0 E: -0.5 h0 The g state is bound between R~1.5 a0 and ∞ (starting the atoms at any distance in this range leads to atoms vibrating forth and back. Exciting to the u state leads to dissociation Ungood state: u Good state: g EEWS-90.502-Goddard-L02

20. Why does KEg has an optimum? R too short leads to a big decrease in slope but over a very short region,  little bonding R is too large leads to a decrease in slope over a long region, but the change in slope is very small  little bonding Optimum bonding occurs when there is a large region where both atomic orbitals have large slopes in the opposite directions (contragradient). This leads to optimum bonding EEWS-90.502-Goddard-L02

21. H2 molecule, independent atoms Start with non interacting H atoms, electron 1 on H on earth, E(1) the other electron 2 on the moon, M(2) What is the total wavefunction, Ψ(1,2)? Maybe Ψ(1,2) = E(1) + M(2) ? Since the motions of the two electrons are completely independent, we expect that the probability of finding electron 1 somewhere to be independent of the probability of finding electron 2 somewhere. Thus P(1,2) = PE(1)*PM(2) This is analogous to the joint probability, say of rolling snake eyes (two ones) in dice P(snake eyes)=P(1 for die 1)*P(1 for die 2)=(1/6)*(1/6) = 1/36 Question what wavefunction Ψ(1,2) leads to P(1,2) = PE(1)*PM(2)? EEWS-90.502-Goddard-L02

22. Answer: product of amplitudes Ψ(1,2) = E(1)M(2) leads to P(1,2) = |Ψ(1,2)|2 = Ψ(1,2)*Ψ(1,2) = = [E(1)M(2)]* [E(1)M(2)] = = [E(1)* E(1)] [M(2)* M(2)] = = PE(1) PM(2) Conclusion the wavefunction for independent electrons is the product of the independent orbitals for each electron Back to H2, ΨEM(1,2) = E(1)M(2) But ΨME(1,2) = M(1)E(2) is equally good since the electrons are identical Also we could combine these wavefunctions E(1)M(2)  E(1)M(2) Which is best? EEWS-90.502-Goddard-L02

23. Beginning of Lecture 2 Consider H2 at R=∞ Two equivalent wavefunctions ΦLR ΦRL At R=∞ these are all the same, what is best for finite R EEWS-90.502-Goddard-L02

24. Plot of two-electron wavefunctions along molecular axis x z EEWS-90.502-Goddard-L02

25. Plot of two-electron wavefunctions along molecular axis x z EEWS-90.502-Goddard-L02

26. Plot of two-electron wavefunctions along molecular axis x z Lower KE (good) Higher PE (bad) Any other terms? EEWS-90.502-Goddard-L02

27. E(H2+)1 = KE1 + PE1a + PE1b + 1/Rab for electron 1 E(H2+)2 = KE2 + PE2a + PE2b + 1/Rab for electron 2 E(H2) = (KE1 + PE1a + PE1b) + (KE2 + PE2a+PE2b) + 1/Rab+EE EE = <Φ(1,2)|1/r12| Φ(1,2)> assuming <Φ(1,2)|Φ(1,2)>=1 Note that EE = ∫dx1dy1dz1 ∫dx2dy2dz2[| Φ(1,2)|2/r12] > 0 since all terms in integrand > 0 EEWS-90.502-Goddard-L02

28. Plot of two-electron wavefunctions along molecular axis x z Lower KE (good) Higher PE (bad) EE bad EEWS-90.502-Goddard-L02

29. Plot two-electron wavefunctions along molecular axis x z Lower KE (good) Higher PE (bad) EE poor higher KE (bad) Lower PE (good) EE good EEWS-90.502-Goddard-L02

30. Valence Bond wavefuntion of H2 Valence bond: start with ground state at R=∞ and build molecule by combining best state of atoms EEWS-90.502-Goddard-L02

31. Molecular Orbitals: Alternative way to view states of H2 Valence bond: start with ground state at R=∞ and build molecule by bonding atoms Molecular orbitals (MO): start with optimum orbitals of one electron molecule at R=Re and add electrons H2: two MO’s u: antibonding g: bonding EEWS-90.502-Goddard-L02

32. Put 2 elecrons into the 2 MO’s, get four states worst In between Best EEWS-90.502-Goddard-L02

33. Analyze Φgu = φg(1)φu(2) wavefunction in 2-electron space EEWS-90.502-Goddard-L02

34. Analyze gu and ug states in 2 electron space All four have one nodal plane and lead to same KE and same PE except for the electron-electron repulsion term <Φ(1,2)|1/r12| Φ(1,2)> Worst case if for z1=z2, along diagonal Never have z1=z2 great EE Maximum at have z1=z2 terrible EE EEWS-90.502-Goddard-L02

35. Any other states? Ionic: two electrons on left Or on the righ EEWS-90.502-Goddard-L02

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49. E = <Φ(1,2)| | Φ(1,2)>/ <Φ(1,2)|Φ(1,2)> E = KE1 + KE2 + PE1 + PE2 + 1/r12 EEWS-90.502-Goddard-L02

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