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Vibrational Theory in Molecules: Bonds, Springs, and Energy Levels

Explore molecular vibrations, bond characteristics, energy levels, and rotational transitions. Learn about Hooke's law, simple harmonic oscillators, vibrational frequencies, and molecular structures.

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Vibrational Theory in Molecules: Bonds, Springs, and Energy Levels

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  1. VIBRATIONAL THEORY p.55 bonds ~ springs C H E = ½ kx2 m1 m2 x is + or - x2 is +

  2. Graph ½ kx2 gives a parabola p. 55 SIMPLE HARMONIC OSCILLATOR

  3. p. 55 for a SHO k = force constant m = reduced mass m(CH) = 12/13 = 0.923 m(OH) = 16/17 = 0.941 O------H C------H 3000cm-1 3400 cm-1 when m1 is very large, m does not change much however n(OH) > n(CH) so k(OH) > k(CH)

  4. However, if the light element is different: p. 56 C----H C-----D 1 2 = 1.7 = 0.92 k’s are same or nCH (0.92/1.7)1/2 = 2200 cm-1 Huge difference

  5. A spring (SHO) has all energies possible, MOLECULES DO NOT p. 57 Ev = (v+ ½)hn only one fundamental frequency (spacings equal, hn)

  6. normally, only a one level jump allowedp. 57 hot lines, overtones weak in IR, only lowest level is populated

  7. REAL MOLECULES: Not a simple parabola! p. 58 Dissociation energy Ev ≈ (v + ½)hno E3 = 7/2 hno - a ‘bit’ more v=3 E2 = 5/2 hno - a ‘bit’ v=2 hot line, n<no n<2no E0→E2 1st overtone E1 = 3/2 hno v=1 fundamental frequency of vibration, DE = hno Eo = 1/2 hno v=0 ground vibrational state Internuclear separation r average bond length o

  8. p. 59 so in CO n0 = 2143 cm-1, n1 = 4250 cm-1not 4286 cm-1 Reminder: Intensity of band depends on change in dipole during stretch or bend

  9. p. 59 Polyatomics X-----Y has one stretching vibration ONLY What about X---Y---Z ? Separate atoms EACH need an x, y, z co-ordinate So N atoms require 3N coordinates to specify position BUT not all movements of atoms in space correspond to a vibration: 3 possible translations along x, y or z AND 3 possible rotations around x, y and z do NOT change the relative positions of the atoms

  10. In general, if molecule has N atoms there are only 3N-6possible fundamental vibrations [3N-5 if molecule is linear] p. 59/60 look at this one

  11. p. 60 During the vibration, the bond dipole changes - + but it is the VECTOR SUM that is important, If this changes during vibration =IR active

  12. p.60

  13. p. 60 3750 asym 3650 sym 1600 bend (overlapping) Structure is ‘hairy’ due to rotational fine structure: more on this shortly

  14. CO2 p. 61 Dm = 0 Dm = 0 Dm = 0

  15. p. 61 CO2

  16. p. 62 SUMMARY Reduced mass effect: Force constant effect:

  17. p. 62 (1) C≡X C=X C-XX=C,O,N ~2200 cm-1 ~1650 cm-1 ~1100 cm-1 (2) C-F C-Cl C-Br C-I 1050 cm-1 725 cm-1 650 cm-1 550 cm-1 • M-F M-Cl M-Br M-I where Metal is more massive, e.g. Sn, Pb 600 cm-1 350 cm-1 225 cm-1 150 cm-1 From these, you can predict many others!

  18. p. 62 Problem: POCl3 shows IR bands at 1290, 582, 486, & 267 cm-1 Cl Cl > O, so P-O > P-Cl, so P-O is likely the highest = 1290 (plus P-O has some double bond character) O=P Cl Cl stretches > bends asym str > sym str 582 = asym str; 486 sym str; 267 is a bend Note: 3N-6 = 9, so there are 5 other fundamentals *** not always obvious what these are, i.e. Raman (IR inactive) or degenerate or combination bands

  19. p. 63 Methane, CH4 likewise only shows 4 bands but has 9 {3N-6} fundamentals (others degenerate) sym bend NOT IR active bend IR ACTIVE sym str if atoms same NOT IR active dipoles cancel asym str always IR active 3020 cm-1 2914 (Raman) 1520 (Raman) 1305

  20. p. 63 3020 cm-1 2914 (Raman) 1520 (Raman) 1305

  21. Vibrational Mode Assignments Tabulated for Many Common Geometries Manual pages 65-68

  22. p. 63 ROTATIONAL LINES: the hairy bits 3020 cm-1 is n0 but many lines caused by many rotational energy levels, much closer spaced than vibrational levels

  23. p. 69 Compared to stretching the bond, rotation around the bond axes takes relatively little energy

  24. p. 69 From quantum mechanics I = Moment of inertia I = mr2 • = reduced mass = m1m2 /(m1+m2) B = constant for a particular bond note: E = hc/l

  25. p. 70 Erot = B J (J+1) J = 0, 1, 2, 3, 4,.... J=0, E0=0J=1, E1=2BJ=2, E2=6BJ=3, E3=12B DE = 8B J0→J1 J1→J2 J2→J3 J3→J4 DE = 6B DE = 4B DE = 2B Rotational energy level spacing increases with J

  26. p. 71 In spectrum of CO, lines are equally spaced Notes: Missing ‘middle’ Right side slightly larger ‘grass’ due to 13C

  27. p. 71 C-------O r Spacing can give r

  28. p. 71 Why this shape?

  29. p. 72

  30. P Branch R Branch – high energy side p. 71

  31. p. 71 Why the difference in intensities? Energy levels are Boltzmann distributed more molecules in lower energy levels But also, level J has a degeneracy of 2J+1 (from quantum) 2J+1 sum up = 1 J 0

  32. p. 71 Real molecules: as bond stretches, r increases so I = mr2, increases and B decreases: spacing (2B) decreases as go to higher J as I gets large (in large molecules), B and hence spacing gets smaller and smaller, so only see rotational lines for small molecules In larger molecules, there is an I value for each axis (x, y, z) giving rise to 3 sets of overlapping rotational lines

  33. When we see the Q (middle) branch: p. 73 ASSIGNMENT 3

  34. Summary: Rotational fine structure • During vibrational transitions, changes in rotational state can also occur • Rotational state changes have a selection rule ΔJ = ± 1 or ΔJ = 0, ±1 depending on molecular symmetry and type of vibration • Rotational levels are spaced increasingly far apart according to E = B(J)(J+1) • Many J levels are occupied at room temperature and the number of equal energy levels (the degeneracy) of a given J is actually 2J+1 • Thus transition V=0 to V=1 is accompanied by a change in J but since there are many starting J states occupied, you get an equally spaced progression of absorptions to the higher energy side (R branch) if ΔJ = +1, to the lower energy side (P branch) if ΔJ = -1 and right at the fundamental frequency ν0 (Q branch) if ΔJ = 0 • The Q branch will NOT be observed if the vibration is along the principle axis of rotation (bond axis) of a linear molecule

  35. p. 75 INORGANIC APPLICATIONS Group frequencies used less, e.g. P=O ~ 1140-1300cm-1 GEOMETRY information from IR, consider the following: Hg Symmetric stretch is only IR active if bent Cl----Hg----Cl Cl Cl Active in IR Inactive in IR In fact we find: 413 (IR, asym stretch), 360 (Raman, sym stretch), 70 (bend) so this suggests the molecule is linear

  36. p. 75 Raman only manual, p 75

  37. p. 76 Structure of complexes

  38. p. 77 METAL CARBONYLS n0 (CO) = 2143 cm-1, but Cr(CO)6 has 2100, 2000, 1985 cm-1 : empty full : s-bonding orbital Donates electron density to metal … BUT

  39. p. 77 empty antibonding (pp*) filled dp puts electron density back on carbon because this electron density is in an antibonding orbital, bond weakens, frequency decreases

  40. p. 78 2100 – 2000 typically sometimes, CO can bridge two metals then n ~ 1850 cm-1

  41. p. 78 bridging CO terminal CO

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