1. Light (electromagnetic radiation or emr) Composed of photons - can be counted travels as spread out wave can interact only at single point E B E = electric field vector B = magnetic field vector absorptionemission scatteringrefractive index rotation of the plane of polarized light Properties of emr interaction with matter

2. nm                    m                                       s -1  rays x rays UV IR  waves radio visiblevisible E = h J = Js  s -1 h = 6.626 x 10 -34 Js high energy low energy c =  = 3.00 x 10 8 ms -1 m s -1 = m  s -1

3. Factors to consider for spectroscopy ….. What are allowed states -  i Original Population: N j /N i = g j /g i exp(-  /kT) The probability of absorption – selection rules Spectroscopy Type Electromagnetic Radiation Frequency s -1 = E/h Wavelength = c/ Typelowhigh low NMRRadio1 x 10 4 1 x 10 9 1000 m50 cm rotationalMicro1 x 10 9 5 x 10 12 50 cm100 mm vibrationalInfrared5 x 10 12 4.3 x 10 14 100mm700 nm electronicVisible4.3 x 10 14 7.5 x 10 14 700 nm400 nm electronicUltraviolet7.5 x 10 14 2 x 10 18 400 nm100 nm X-rays2 x 10 18 1 x 10 20 1000 A5 A  -rays 1 x 10 20 -----500 pm1 pm

4. rot vib  E (eV) wavenumber (cm -1 ) 10 1 10 -1 10 -2 10 -3 10 5 10 4 10 3 100 10 el 300K kT Original Population: N j /N i = g j /g i exp(-  /kT)

5. Selection Rules M = transition dipole moment = probability of transition if M = 0 then “forbidden” if M  0 then “allowed” M =   m *   n d  ^  =  i q i r i ^

6. Selection Rules – 1D Particle in Box  = (2/a) 1/2 sin(n  x/a) M =   m * qx  n d  M = qa/  2 {(cos[(m-n)  ]-1)/(m-n) 2 – (cos[(m+n)  ]-1)/(m+n) 2 }  n =  1,  3,  5,.... any odd # e.g. n = 1  2, 1  4, 1  6, but not n = 1  3, n = 1  5, etc........ sin m ● sin n = ½cos(m-n) - ½cos(m+n) ∫ x cos cx dx = c -2 cos cx + (x/c) sin cx

7. Molecular Energy E = E tr + E int = E tr + E vib + E rot + E e Particle in a box rigid rotor harmonic oscillator AO/MO Nuclear Energy at fixed r H N  N = (K N + V N )  N = (K N + E e )  N = E N  N E e (R e ) + E e ’ (R e )(R - R e ) + 1/2E e ” (R e ) (R-R e ) 2 + negligible …

8. E rot Rigid Rotor (  corresponds to rotation of diatomic molecules B = ħ 2 /(2I) ― rotational constant (J) = h/(8  2Ic) ― rotational constant wavenumber (  = cm -1 ) E rot = J(J+1) ħ 2 /(2I) = BJ(J+1) I =  r 2  reduced mass (m 1 m 2 /(m 1 + m 2 ) Ĥ(  = E(  L 2  = J(J + 1)ħ 2  ^ L z  = M J ħ  |M J | ≤ J ^ Selection Rules — Gross: permanent dipole moment Specific: angular momentum conserved —  J = ±1 if linear —  J = 0, ±1 if symmetric top —  K = 0

9. Pure Rotational Spectra ― E J = BJ(J+1) B (J) =ħ 2 /2I diatomic molecules J max = (0.35T/B) 1/2 Relative population: N j /N i = g j /g i exp(-  /kT) J E  E g 4 20B 8B 9 3 12B 6B 7 2 6B 4B 5 1 2B 2B 3 0 0 --- 1 Selection rule:  J = ±1 (&  M J = 0, ±1) 2B 4B 6B The value of the rotational constant, B, decreases for heavier diatomic molecules B (cm -1 ) = B (J)/hc MoleculeB (cm -1 ) H2H2 60.80 D2D2 30.42 HCl10.59 HBr8.473 N2N2 2.010 NO1.7046 O2O2 1.446 CO1.931

10. What is the bond length of 1 H 80 Br? Verify J max What is B? A pure rotational spectrum of 1 H 80 Br gives the 1 st 4 energy levels at 0, 16.95, 50.84, and 101.7 cm -1. 14.8? B = 16.95/2 = 8.475 cm -1 = 1.68 x 10 -22 J. I =  r 2 = ħ 2 /2B = 3.30 x 10 -47 need to find reduced mass, . r = (3.30 x 10 -47 /1.64 x 10 -27 ) ½ = 1.42 Å B (J) =ħ 2 /2I need to find I  (kg) = (80 ∙ 1)/(80 + 1) ÷ 6.022 x 10 26. J E rel pop g 5 30B 11 4 20B 9 3 12B 7 2 6B 5 1 2B 3 0 0 1 1 J max = (0.35T/B) 1/2 = 3.5 Relative population: N j /N i = g j /g i exp(-  /kT) at 298K, kT = 4.11 x 10 -21 J B (cm -1 ) = B (J)/hc B (J) = ħ 2 /2I need to find I 2.76 3.91 4.29 3.23 3.97

11. 196 Ag 1 Hr (Ǻ)r (m)m (kg)IB (J)B (cm -1 )JE (cm -1 ) 1.6171.62E-101.65E-274.32E-471.29E-226.47377.6 238.8 112.9 00 J E rel pop g 3 12B 7 2 6B 5 1 2B 3 0 0 1 1 14.8 197 Ag 1 H bond length = 1.617Å Predict E for 1 st 4 energy levels

12. 10S2S2 r ( m )  (kg) IB (J)B (cm) 1.88E-102.66E-269.39E-465.92E-24 0.298 E rot = J(J + 1)BB = ħ 2 /(2I) I =  r 2 B (cm -1 ) = B (J)/(hc)

13. 11r ( m )  (kg) IB (J)B (cm)book SF 6 1.56E-101.98E-264.84E-461.15E-230.581.80E-24 UF 6 2.00E-102.92E-261.16E-454.77E-240.241.10E-24  (kg) IB (J) 1.58E-261.54E-453.60E-241.80E-24 1.58E-262.51E-452.21E-241.10E-24 Ignore central atom – use 2Fs at 2 x bond divide by 2 after determining B (J) for both bonds Is the bond length of a diatomic molecule influenced by its rotational energy? Why? Does this influence the energy levels and spectrum?

14. Centrifugal Distortion R e  as E rot  E rot = BJ(J+1) – D J J 2 (J + 1) 2 D = centrifugal distortion constant (>0) MoleculeB (cm -1 )D J (cm -1 ) H2H2 60.804.64 x 10 -2 D2D2 30.421.159 x 10 -2 HCl10.595.32 x 10 -4 HBr8.4733.72 x 10 -4 N2N2 2.0105.8 x 10 -6 NO1.70465 x 10 -6 O2O2 1.4464.95 x 10 -6 CO1.9316.12 x 10 -6 This correction is usually small and often omitted from the total E equation however, as J↑, the correction becomes more significant. 14.27 and 14.28 D J ~ 4B 3 /ΰ 2 ΰ Wavenumber for vibration

15. Centrifugal Distortion R e  as E rot  E rot = BJ(J+1) – D J J 2 (J + 1) 2 D = centrifugal distortion constant (>0) Will centrifugal distortion alter largest band of HBr spectrum? J = 3 → 4 J E ignore CDwith CD 5 30B 4 20B169.46 - 1.49 = 169.31 3 12B 101.68 -.054 = 101.63 2 6B 1 2B 0 MoleculeB (cm -1 )D J (cm -1 ) HBr8.4733.72 x 10 -4 J = 3 → 4 67.69 cm -1 vs. 67.68 cm -1 This correction is usually small and often omitted from the total E equation however, as J↑, the correction becomes more significant.

16. Types of molecules ― (based on 3 mutually  moments of inertia) linear same as diatomic but reduced mass is difficult to calc. spherical topI a = I b = I c symmetric top (oblate)I a = I b < I c ≥ C 3 symmetric top (prolate)I a < I b = I c ≥ C 3 asymmetric topI a < I b < I c Pure Rotational Spectra ― polyatomic molecules N ≡ C – C H H H J = total angular momentum quantum # (same for all rotations) M J = z-component of the angular momentum quantum # K = figure axis component of angular momentum IaIa IbIb z axis K is larger M J is small K = 0 but J ~ M J

17. Types of molecules ― (based on 3 mutually  moments of inertia) linear same as diatomic but reduced mass is difficult to calc. spherical topI a = I b = I c symmetric top (oblate)I a = I b < I c ≥ C 3 symmetric top (prolate)I a < I b = I c ≥ C 3 asymmetric topI a < I b < I c 14.9 a)CH 3 -C≡C-CH 3 b) SF 6 c) PO 4 3- d) H 2 N-CH 2 -COOH e) cis-1,2-dichloroethylene f) trans-1,2-dichloroethylene g) Hexamethylbenzene h) HC≡C-C≡CH i) CN - Pure Rotational Spectra ― polyatomic molecules a dimethylacetyleneprolate b SF6sph top c phosphate ionsph top d glycineasym top e cis C2H2Cl2asym top f trans C2H2Cl2asym top g hexamethylbenzeneoblate h diacetylenelinear i CN linear

18. Molecular symmetryIE rot spherical topI a = I b = I c BJ(J + 1) Oblate symmetric top Principal axis -  length I a = I b < I c BJ(J + 1) + (C - B)K 2 C < B Prolate symmetric top Principal axis - length I a < I b = I c BJ(J + 1) + (A - B)K 2 A > B Asymmetric topI a < I b < I c Rotaional ConstantJcm -1 Aħ 2 /(2I a ) h/(8  2 I a c) Bħ 2 /(2I b ) h/(8  2 I b c) Cħ 2 /(2I c ) h/(8  2 I c c) c = 3.00 x 10 10 cm s -1 14.12 K = 0, ±1, …±J K = figure axis (unique) component of total angular momentum. Pure Rotational Spectra ― polyatomic molecules

19. GivenIaIbIc PH 3 5.48E-47 6.65E-47 ABC21 29.71 calcuate 1.02E-22 8.37E-23 22 27.02 10 10.20 11 9.31 00 0.00 JKE (cm -1 ) 14.12 & 14.14 ― rotation of PH 3 oblate symmetric top ─ E rot = BJ(J + 1) + (C - B)K 2 Selection Rules — Gross: permanent dipole moment Specific:  J = ±1  J = 0, ±1  K = 0 (symmetric top) 14.17 Rotaional ConstantJcm -1 Aħ 2 /(2I a ) h/(8  2 I a c) Bħ 2 /(2I b ) h/(8  2 I b c) Cħ 2 /(2I c ) h/(8  2 I c c) c = 3.00 x 10 10 cm s -1

20. Vibrational spectroscopy Normal modes of vibration – Molecules tend to vibrate in sync – i.e. all the atoms in a vibrating molecule have the same frequency of motion. The quantized energy levels indicate the different energy/frequency of the same normal mode of vibration. Different normal modes of vibration correspond to the different patterns of motion of the molecule. The number of Normal modes of vibration for a molecule – Linear = 3N – 5 (where N is the # of atoms in the molecule) Nonlinear = 3N – 6 HF # normal modes = 6 – 5 = 1 O = C = O # normal modes = 9 – 5 = 4 H 2 O# normal modes = 9 – 6 = 3 Diatomic molecule have only 1 normal mode and therefore their spectra are fairly simple. They are reasonably approximated by the harmonic oscillator.

21. E rot Rigid Rotor (  I =  r 2 B = ħ 2 /(2I) E rot ~ J(J+1)ħ 2 /(2I) = BJ(J+1) E vib Harmonic Oscillator (diatomic molecules) E vib = (  + 1/2)h e e = 1/(2  ) (k e /  ) 1/2 k = d 2 V/dx 2 Selection rules ―  = ±1 &  J = ± 1 Table III – Boltzmann Distribution N i /N o = (g i /g o ) exp(-  E/kT) (values below assume g i /g o = 1) kT at Room Temp = 4.14 x 10 -21 J emr spectra  E = hc/ N i /N 0 UV280 nmelectronic7.10 x 10 -19 3.3 x 10 -75 vis700 nmelectronic2.84 x 10 -19 1.6 x 10 -30 IR 3.45  m vibrational5.76 x 10 -20 9.1 x 10 -7 microwave1 cmrotational2.00 x 10 -23 0.995 Radio wave1 mNMR2.00 x 10 -25.99995

22. Vibration/Rotational Spectra – diatomic molecules  = ± 1 &  J = ± 1 R branch P branch   J = 3 J = 2 J = 4 Note  E vib And  E rot not to scale

23.  E  = 0 to  =1 R P 1-00-1 Diatomic Vib-Rot Spectrum e

24. E rot Rigid Rotor (  I =  r 2 B = ħ 2 /(2I) E rot ~ J(J+1)ħ 2 /(2I) = BJ(J+1) E vib Harmonic Oscillator (diatomic molecules) E vib = (  + 1/2)h e e = 1/(2  ) (k e /  ) 1/2 k = d 2 V/dx 2 Vibrational Energy Corrections Vibrations are not harmonic (anharmonic corrections) Vibrations effect Rotations (vibration-rotation interaction) Rotors are not rigid (centrifugal distortion) As k↑ the width of the potential energy well decreases Selection rules ―  = ±1 &  J = ± 1

25. EeEe r Anharmonic Correction harmonic actual ReRe DeDe E vib = h e (  + ½) - h e x e (  + ½) x e = e /(4D e ) D e = D 0 + ½h moleculeB (cm -1 )DJDJ e (cm -1 ) r (Ǻ)  (kg) I  e  e (cm -1 ) D e (cm-1) HF21.082.19E-034138.520.91711.58E-271.33E-4790.0747539

26. E vib (with anharmonic correction) E vib = (  + ½)h e  h e x e (  + ½) 2 0 < e x e < e E rot with vibration-rotation interaction E rot = B e hJ(J+1) -h  e (  + 1/2) J(J+1)  e > 0 vib-rot coupling constant Vibrations are not harmonic (anharmonic corrections) Vibrations effect Rotations (vibration-rotation interaction) Rotors are not rigid (centrifugal distortion) E int  E el + h e (  +½) - h e x e (  +½)² + B e hJ(J+1) -h  e (  +½)J(J+1) -hDJ²(J+1)² E vib anharmonicity E rot vib/rot interaction centrifugal distortion

27. molecule B (cm -1 )DJDJ e (cm -1 ) r (Ǻ)  (kg) I  e  e (cm -1 ) D e (cm-1) HCl 10.595.32E-042989.741.2751.61E-272.62E-4752.0542932 HBr 8.4733.73E-042649.741.4131.64E-273.27E-4745.2138825 NO 1.70465E-061904.031.15081.24E-261.64E-4613.9764877 HF 21.082.19E-034138.520.91711.58E-271.33E-4790.0747539 CO 1.936.11E-062170.211.12811.14E-261.45E-4613.4687478 OH 19.001.97E-033735.210.97061.56E-271.47E-4782.8142120 LiH 7.571.56E-031055.121.59491.45E-273.70E-4713.2221053 E int  E el + h e (  +½) - h e x e (  +½)² + B e hJ(J+1) -h  e (  +½)J(J+1) -hDJ²(J+1)² E v anharmonicity E rot vib/rot interaction centrifugal distortion

28. Polyatomic molecules These have more complex spectra due to multiple vibrational modes These include stretching and bending vibrational modes. Stretching modes are only IR active if the dipole moment changes during the stretch. Individual bonds stretch similar to diatomic model. The concept of reduced mass is complex for bending modes. O = C = O O ← C → O symmetric stretch – not IR active O → C → O asymmetric stretch – is IR active ↕C ↕ bending mode – is IR active O ↕ O

29. H2OH2O H2OH2O CH 4 CO 2 O3O3 Greenhouse Gases 71016  m Emission intensity

30. Raman Spectroscopy Rayleigh Scattering ─ about 1 in 10,000 photons will scatter at an angle from a sample without being absorbed while keeping the same frequency. Raman Scattering ─ about 1 in 10 7 photons will scatter at an angle from a sample and change frequency. The loss of energy by the scattered photons correspond to the vibrational (and/or rotational) energy levels of the molecule. Selection Rule ─ Laser Raman spectra depend on a change in the polarizability of a molecule (how easily an electric field can induce a dipole moment). The transition moment for polarizability is  rather than . Because of this some vibrations that are inactive in vibrational or rotational spectra can be active in laser Raman spectra. Homonuclear diatomic molecules display Raman spectra. It can also be used to eliminate interference due to water for molecules in aqueous solution.