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Quantum Chemistry “ Physical Chemistry”, Atkins, 6 th ed., chap. 14.

Quantum Chemistry “ Physical Chemistry”, Atkins, 6 th ed., chap. 14. Revision of third year Exact solution for the simplest molecule H 2 + Approximate Methods LCAO-MOT Hückel theory Calculations, applications Beyond Hückel: Semi-empirical, ab initio methods

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Quantum Chemistry “ Physical Chemistry”, Atkins, 6 th ed., chap. 14.

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  1. Quantum Chemistry“Physical Chemistry”, Atkins, 6th ed., chap. 14. • Revision of third year • Exact solution for the simplest molecule H2+ • Approximate Methods LCAO-MOT • Hückel theory • Calculations, applications • Beyond Hückel: Semi-empirical, ab initio methods • Molecular mechanics / force field methods Copies of these overheads are available in Secretariat in Concourse or download from http://www.nuigalway.ie/chem/degrees.htm

  2. 3rd Year Quantum Chemistry Revisited • The probablitity of finding a particle (or system) is given by the square of the wavefunction. • The wavefunction is a solution of the SWE. • The SWE for a particle in 1-dimension is: • y must be single-valued, finite & continuous. • Given m (what particle) & V (the environment), solving SWE gives allowed energies, E, & ys

  3. Exact solutions of SWE • Particle in a box • V = 0 for 0£ x£ a • Particle in parabolic V • V = kx2/2 • vibrational energy • Particle on ring • V = 0 • rotational energy • Particle in coulombic V • V = -e2/(4pe0r) • H-atom, electronic energy

  4. SWE; molecular ion H2+ • Bates et al. (1953) • De = 269 kJ mol-1 • re = 106 pm • Excellent agreement with experiment

  5. Approximate solution for H2+ Guess form of wavefunction • y = c1 f1 + c2 f2 Linear combination of atomic orbitals Solve variant of SWE Þ E(approx.) ³ E(exact) Pick c’s such that E(approx.) is a minimum c1 (H11-E) + c2 (H21-S21E) = 0 c1 (H12-S12E) + c2 (H22-E) = 0 • Solution of these simultaneous equations for c1 & c2 requires that their determinant be zero

  6. Two solutions • Go back to simultaneous equations insert the appropriate value of E & solve for c’s Eg = (H11+H12)/(1+S12)yg = c1 (f1 + f2) Eu = (H11- H12)/(1-S12) yu = c1 (f1 - f2)

  7. Hydrogen molecular ion Atomic units a0 = 52.9 pm • r = R / a0 • S = e-r (1+ r + r2/3) • J = e-2r (1 + 1/r) • K = (S / r) - e-r (1 + r) E+ = [(J+K)/(1+S)] + (1/r) – (1/2) Find: • Re = 132 pm (106 pm) • De = 170 kJ/mol (268)

  8. s-MOs from 1s AOs sg su

  9. s-MOs & p-MOs from 2p AOs • 2s AOs generate 1sg & 2su* MOs as previously • 2p AOs generate 3sg & 3su* also two 3pu & 4pg* MOs • Above true for O2 & F2; order 3sg & 3pureversed for N2, etc.

  10. Electron spin? • Only considered y (x,y,z) not spin y. Quite simple, either a or b (up or down) • For H2a(1)a(2), b(1)b(2), a(1)b(2) ±a(2)b(1) • for spin ½ particles y must be antisymmetric wrt exchange • y = [a(1)b(2) + a(2)b(1)][a(1)b(2) - a(2)b(1)] • y = [a(1)b(2) – a(2)b(1)][a(1)a(2)] or [b(1)b(2)] or[a(1)b(2) + a(2)b(1)] Ground state—singlet; excited—state triplet

  11. Hückel theory: p bonded Cs Solve secular equation: |H-ES|=0 • All Hii have the same value, a (-ve value!) • If i bonded to j then all Hij =b (-ve value!) • If i not bonded to j then all Hij = 0 • All Sij = 0 (and all Sii = 1)

  12. Ethene or ethylene Quadratic equation Two solutions • E1 =a + b • E2 =a - b • For E = E1 • Y1 = c1 f1 + c2 f2 • ForE = E2 • Y2 = c1 f1-c2 f2

  13. For E = E1 = a + b Y1 = c1 f1 + c2 f2 c1 (a-E) + c2b = 0 c1 (a-a-b) + c2b = 0 \ c1 = c2 Y1 = c1 (f1 + f2) Normalised Y1 = 0.707 (f1 + f2) For E = E2 = a - b Y2 = c1 f1 + c2 f2 c1 b + c2 (a-E) = 0 c1 b+ c2 (a-a+b) = 0 \ c1 = - c2 Y2 = c1 (f1 - f2) Normalised Y2 = 0.707 (f1 - f2) Complete solutions

  14. Energy level diagram • a and b are both negative energies • hence (a + b) lies lower in energy than (a - b) • So Y1 is the bonding MO and Y2 is the anti-bonding MO

  15. Butadiene; secular determinant1-2-3-4Simplifying determinants:(i) divide through by b, (ii) replace (a-E)/b by xMultiply out 4´4 and solve for x; replace x in En = a - xnb:E1= a + 1.62b E2= a + 0.62bE3= a - 0.62b E4= a - 1.62b

  16. Butadiene: MO energies • MO # 1; lowest • E1 = a+1.62b • MO #2 (neutral, HOMO) • E2 = a+0.62b • MO #3 (neutral, LUMO) • E3 = a–0.62b • MO #4; highest • E4 = a–1.62b

  17. Butadiene H2C=CH-CH=CH2 Four solutions • Note energy solutions are paired about a Symmetry of wavefunctions • Coefficients of pair are same magnitude • But starred set differ in sign C*¾C¾C*¾C even alternant

  18. Delocalisation energy • For neutral butadiene the total p-energy is: Ep = 2 (a + 1.618 b) + 2 (a + 0.618 b) • For the structure C=C–C=C the predicted e is: Ep = 4 (a + b) • So real butadiene is more stable by DEp = (4 – 4)a + (4.472 – 4)b = 0.472b • Delocalisation e per electron a measure of stability of species: • benzene > azulene > pentalene > cyclobutadiene

  19. Summary of solutions Let (a–E)/b= x , so that En =a + xnb • Sum of the roots: Sx= 0 • No root ³ 3 or £ –3 Draw a circle radius 2b, inscribe cyclic pointing down, roots occur where figure touches circle. • Alternant hydrocarbons • energy levels paired • paired e levels differ only in sign of one set of c’s • for an odd AH one MO is non-bonding & is not paired

  20. Exercises • Set up & solve determinants (by hand) for the energies for: • linear & cyclic C3 and C4 • try application “Huckel-c” on Win-95 PC in C217 for the above & more complicated cases such as benzene • Sketch the resulting wavefunctions, identifying the HOMO and LUMO in each case

  21. Using symmetry • Solutions are: • 2(x-1) = 0 or x = + 1  = c (1 - 3) • 2(x+1)(x) – 4 = 0 or 2x2 + 2x – 4 = 0 or x2 + x – 2 = 0 • (x – 1)(x + 2) = 0 or x = -2, +1 • Minors: M1 = x, M2 = -2 • x = - 2 then  = - 2 (1 + 3) – 2 2 • x = + 1 then  = + 1 (1 + 3) – 2 2

  22. Butadiene; numerical data

  23. Charge density, qR • The total p-electron density at atom R is defined as: qR = S{nj c 2(R,j)} where nj is the no. of electrons in MO no. j c(R,j) is the coefficient of atom R in the jth MO • Useful reactivity index

  24. Unpaired electron density, rR The spin density at atom R is: c2(R) • in the MO that the unpaired electron resides (sum if more than 1) • Eg in H2C·CH=CH2 (allyl radical) the unpaired electron is in the non-bonding MO y2= 0.707 f1 – 0.707 f3 • Thus, r1 = r3 = (0.707)2 = 0.50 and r2 = 0.0

  25. Problem • The benzene radical anion has 2 degenerate MOs: • y4= [1/2] {f2 – f3 + f5 – f6} • y5= [1/(2Ö3)] {2 f1 – f2 – f3 + 2 f4 – f5 – f6} Calculate the spin densities at atoms 1 thru 6. • If e in MO#4 then r2=r3=r5=r6= (1/4) = 0.25 • If e in MO#5 then r2=r3=r5=r6= (1/12) = 0.083... and r1=r4= (1/3) = 0.333... • Do these answers make sense?

  26. Bond order, pRS • A measure of the “electronic cement” between two adjacent atoms R and S is given by: p(R,S) =Snj c(R,j) c(S,j) • For ovalene below the X-ray bond lengths, d, are ranked from shortest (1) to longest (7) versus bond order from largest (1) to smallest (7)

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