Spectroscopy in Inorganic Chemistry. Vibration and Rotation Spectroscopy
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1 Spectroscopy in Inorganic Chemistry
2 Vibrational energy levels in a diatomic molecule f = k r r V = ½kX 2 Force constant r Displacement from equilibrium point 2 X= r=r-r eq
3 V = ½kX 2 Fundamental Vibrational frequency 0 1 E V =hv (V + 1 ) 2 Vibrational quantum number 0,1,2, X The energy of V th state 3
4 E V =hv (V + 1 ) 2 E V=0 = hv 2 Zero point energy X 4
5 5
6 Vibrational energy levels in a diatomic molecule continuum Zero point energy D 0 D e 6
7 first and second anharmonicity constants 7
8 Selection rules * Change in direction or magnitude of the dipole N-C-H bending mode Change in direction of the dipole * V = ±1 first overtone Second overtone 8
9 Force constant Force constant E = h 2π k μ ½ Reduce mass E = hv = hcv As the mass of the atoms increases, the vibration frequency decreases 9 C-H (3000 cm -1 ) C-C (1000 cm -1 ) C-Cl (800 cm -1 ) C-Br (550 cm -1 ) C-I (about 500 cm -1 )
10 Gaseous HCL exhibits an IR absorption at 2890 cm -1 due to the H-Cl stretching vibration. Calculate the force constant of this bond. What would the wavenumber of the corresponding vibration be for DCl, where the proton is replaced with a deuterium atom. 10
11 Let s start with the reduced mass of HCl Mass of Cl in kg = x10-3 /6.0x10 23 = 5.9x10-26 kg Mass of H in kg = 1.008x10-3 /6.0x10 23 = 1.68x10-27 kg reduced mass of molecule (ì)= (5.9x10-26 kg 1.68x10-27 kg)/(5.9x10-26 kg +1.68x10-27 kg)=1.633x10-27 kg 11 Using the equation: 2890 = 5.3x10-12 (K/1.633x10-27 ) 1/2 2890/5.3x10-12 = (K/1.633x10-27 ) 1/2 (2890/5.3x10-12 ) 2 = K/1.633x10-27 K = (2890/5.3x10-12 ) x10-27 = N/m = dyne/cm
12 What would the wavenumber of the corresponding vibration be for DCl, where the proton is replaced with a deuterium atom. 12
13 single bond 5 x 10 5 dyne/cm double bond 10 x 10 5 dyne/cm triple bond 15 x 10 5 dyne/cm C, carbon 12/6.02 x H, hydrogen 1/6.02 x bond absorption region, cm 1 C C, C O, C N C=C, C=O, C=N, N=O C C, C N C H, N H, O H
14 Harmonic oscillator N/m = dyne/cm Anharmonic oscillator N/m = dyne/cm 14
15 overtone 15
16 Harmonic oscillator N/m = dyne/cm Anharmonic oscillator N/m = dyne/cm HCl vibrational spectrum. Transition ṽ obs [cm -1 ] ṽ obs Harmonic [cm -1 ] ṽ obs Anharmonic [cm -1 ] 0 1 (fundamental) (first overtone) (second overtone) (third overtone) (fourth overtone)
17 force constant K= 2 V r 2 distance potential energy r 0 17
18 single bond 5 x 10 5 dyne/cm double bond 10 x 10 5 dyne/cm triple bond 15 x 10 5 dyne/cm C, carbon 12/6.02 x H, hydrogen 1/6.02 x bond absorption region, cm 1 C C, C O, C N C=C, C=O, C=N, N=O C C, C N C H, N H, O H
19 Stretching and bending vibration cm -1 X-H (NH, OH, CH) over 2900 cm -1 Triplet band cm -1 Doublet band cm -1 Fingerprint cm -1 Metal-ligand below 400 cm -1 (far IR) various metal ligand vibrations, ring deformation, rockink and lattice 19
20 C-H C-D E = h k 2π μ Variations in force constant (little) and reduced mass (magnificent) ½ C-D 1/ 2 C-H v v = 1.04 v =0.76 C-H C-D 20 C-H=3000 cm-1 and C-D=2100 cm-1 v C-H v C-D =1.4
21 Substitution of H by D (deutration) Substitution of 12 C by 13 C (different reduced mass) Metal isotopes 21
22 C-H (3000 cm -1 ) C-C (1000 cm -1 ) C-Cl (800 cm -1 ) C-Br (550 cm -1 ) C-I (about 500 cm -1 ) 22
23 Vibration in a Polymeric molecule 23
24 The 3N-6(5) rule Linear molecule of N atoms: normal modes = 3N - 5 Nonlinear molecule of N atoms: normal modes = 3N
25 Effects giving rise to absorption bands SO 2 25
26 3N-6 (3 3-6=3) 1151 cm cm cm -1 A 1 A 1 B 2 antisymmetric stretches > symmetric stretches > bends 26
27 symbol Stretching vibrations v Bending vibrations δ Out of plane stretching vibrations π Asymmetric as, symmetry a, degenerate d 27
28 antisymmetric stretches > symmetric stretches > bends 28
29 overtone: 2v 1 = 2305 cm -1 Combination bands: 1871, 2499 cm -1 v 2 + v 3 = 1871 cm -1 v 1 + v 3 = 2499 cm -1 Difference band: 606 cm -1 v 1 - v 2 = 606 cm -1 29
30 CO 2 -IR 3N-5 (3 3-5=4) 2349 cm cm cm -1 30
31 CO 2 -Raman 3N-5 (3 3-5=4) 1340 cm -1 31
32 1340 cm -1 32
33 CO 2 33
34 CO 2 34
35 CO 2 35
36 CO 2 v 3 Asymetric Stretch 2349 cm -1 v 1 Symetric Stretch (not IR active) 1340 cm -1 Vertical Bend (scissoring) 667 cm -1 v 2 Π u Horizontal Bend (scissoring) (A degenerate mode with same motion as above but rotated by 90 o ) 667 cm -1 36
37 Fermi resonance split the band at 1340 cm -1 to two bands at 1286 cm -1 and 1388 cm -1 overtone: 2v 2 = 1334 cm -1 (2 667) almost v 1 = 1340 cm -1 Wave function of excited state of 2v 2 and v 1 mixed with each other and made two new bands. 37
38 Fermi resonance v 1 = 1340 cm -1 2v 2 = 1334 cm -1 v= 1388 cm -1 v= 1286 cm -1 v 2 = 667 cm -1 38
39 Fermi resonance detection: 1- deutration 2- solvent effects (use varies solvents) 39
40 40
41 Group vibrations and the limitations of this idea Group vibration concept the most useful information obtained from an IR spectrum is what functional groups are present within the molecule 41
42 42
43 C=O C-O ±150 43
44 44
45 45
46 NH 3 46
47 H-C-N & D-C-N 47
48 H-C-N & D-C-N v 1 =2089 cm -1 (C-N stretching) 1096 cm -1 v 2 =714 cm -1 (doubly degenerate bending) v 3 =3312 cm -1 (C-H stretching) 2629 cm -1 48
49 F 2 O Cl 2 O Br 2 O C=O 1928 cm cm cm -1 K mdy/a mdy/a mdy/a 49
50 C=O 75% C-C 25% 1750 cm -1 50
51 Raman 51
52 52
53 The most common and inexpensive gas laser, the helium-neon laser is usually constructed to operate in the red at nm. 53
54 Ar laser (green - blue) nm, nm, nm, nm, nm, nm 54
55 Kr laser nm, nm tunable dye laser 55
56 56
57 W 57
58 58
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