Spectroscopy in Inorganic Chemistry
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
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
E V =hv (V + 1 ) 2 E V=0 = hv 2 Zero point energy X 4
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Vibrational energy levels in a diatomic molecule continuum Zero point energy D 0 D e 6
first and second anharmonicity constants 7
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
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 )
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
Let s start with the reduced mass of HCl Mass of Cl in kg = 35.453x10-3 /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 ) 2 1.633x10-27 = 485.5 N/m =4.84 10 5 dyne/cm
What would the wavenumber of the corresponding vibration be for DCl, where the proton is replaced with a deuterium atom. 12
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 10 23 H, hydrogen 1/6.02 x 10 23 13 bond absorption region, cm 1 C C, C O, C N 800 1300 C=C, C=O, C=N, N=O 1500 1900 C C, C N 2000 2300 C H, N H, O H 2700 3800
Harmonic oscillator 485.5 N/m =4.84 10 5 dyne/cm Anharmonic oscillator 515.7 N/m =5.157 10 5 dyne/cm 14
overtone 15
Harmonic oscillator 485.5 N/m =4.84 10 5 dyne/cm Anharmonic oscillator 515.7 N/m =5.157 10 5 dyne/cm HCl vibrational spectrum. Transition ṽ obs [cm -1 ] ṽ obs Harmonic [cm -1 ] ṽ obs Anharmonic [cm -1 ] 0 1 (fundamental) 2885.9 2885.9 2885.3 0 2 (first overtone) 5668.0 5771.8 5665.0 0 3 (second overtone) 8347.0 8657.7 8339.0 0 4 (third overtone) 10 923.1 11 543.6 10 907.4 0 5 (fourth overtone) 13 396.5 14 429.5 13 370. 16
force constant K= 2 V r 2 distance potential energy r 0 17
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 10 23 H, hydrogen 1/6.02 x 10 23 18 bond absorption region, cm 1 C C, C O, C N 800 1300 C=C, C=O, C=N, N=O 1500 1900 C C, C N 2000 2300 C H, N H, O H 2700 3800
Stretching and bending vibration 600 4000 cm -1 X-H (NH, OH, CH) over 2900 cm -1 Triplet band 2000-2700 cm -1 Doublet band 1500 1700 cm -1 Fingerprint 400 1500 cm -1 Metal-ligand below 400 cm -1 (far IR) various metal ligand vibrations, ring deformation, rockink and lattice 19
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
Substitution of H by D (deutration) Substitution of 12 C by 13 C (different reduced mass) Metal isotopes 21
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
Vibration in a Polymeric molecule 23
The 3N-6(5) rule Linear molecule of N atoms: normal modes = 3N - 5 Nonlinear molecule of N atoms: normal modes = 3N - 6 24
Effects giving rise to absorption bands SO 2 25
3N-6 (3 3-6=3) 1151 cm -1 519 cm -1 1361 cm -1 A 1 A 1 B 2 antisymmetric stretches > symmetric stretches > bends 26
symbol Stretching vibrations v Bending vibrations δ Out of plane stretching vibrations π Asymmetric as, symmetry a, degenerate d 27
antisymmetric stretches > symmetric stretches > bends 28
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
CO 2 -IR 3N-5 (3 3-5=4) 2349 cm -1 1340 cm -1 667 cm -1 30
CO 2 -Raman 3N-5 (3 3-5=4) 1340 cm -1 31
1340 cm -1 32
CO 2 33
CO 2 34
CO 2 35
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
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
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
Fermi resonance detection: 1- deutration 2- solvent effects (use varies solvents) 39
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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
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800-1200 2860-3000 1400-1470 C=O C-O ±150 43
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NH 3 46
H-C-N & D-C-N 47
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
F 2 O Cl 2 O Br 2 O C=O 1928 cm -1 1827 cm -1 1828 cm -1 K 12.85 mdy/a 12.61 mdy/a 12.83 mdy/a 49
C=O 75% C-C 25% 1750 cm -1 50
Raman 51
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The most common and inexpensive gas laser, the helium-neon laser is usually constructed to operate in the red at 632.8 nm. 53
Ar laser (green - blue) 457.9 nm, 465.8 nm, 476.5 nm, 488.0 nm, 491.5 nm, 514.5 nm 54
Kr laser 568.2 nm, 647.1 nm tunable dye laser 55
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W 57
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