Revisit vibrational Spectroscopy Frequency: Infrared Near-IR 4000-14000 cm-1 Mid-IR 400 - 4000 cm-1 Far-IR 10-400 cm-1 Model: Harmonic oscillator Hamiltonian: Wave function: Energy level: Transition energy: Selection rule: E= h

Anharmonic oscillator Energy curve: Harmonic oscillator Anharmonic oscillator E 4 D0 3 De 2 1 1 h at n=0, E0  0 Normal modes: 3N-6 (non-linear), 3N-5 (linear)

Revisit Rotational Spectroscopy Frequency: Microwave 1 - 100 cm-1 Model: Rigid rotor Hamiltonian: Wave function: Energy level: Degeneracy: 2J+1 values mJ = 0, 1, 2,.. J Transition energy: Selection rule:

Rotational angular momentum: Rotational spectral lines:

Rovibrational Spectroscopy Outlines Operator, wavefunctions, energy Rovibrational Energy of rigid diatomic molecules Rotational and vibrational energy levels Rovibrational transitions and selection rule Transition energy of rovibration Rovibrational spectrum Intensity of a spectral line: Boltzmann distribution Rovibration of non-rigid rotation

Operator, wavefunctions, energy Rovibration = Rotation + Vibration - Energy operator of rovibration motion : Summation of H - Wavefunctions of the system : Product of  Total energy of rovibration motion: Erovib = Erot + Evibr

Rovibrational Energy of rigid diatomic molecules Absorption of infrared region, molecules can change vibrational and rotational states because vibrational transitions can couple with rotational transitions to give rovibrational spectra. By Treating as harmonic oscillator and rigid rotor, energy of vibration and rotation can expressed as: n = 0, 1, 2, … J = 0, 1, 2, … J unit

From previous slide: n = 0, 1, 2, … J = 0, 1, 2, … In unit of Wave number (cm-1) n = 0, 1, 2, … J = 0, 1, 2, …

Rotational and vibrational energy levels A series of J rotational states in vibrational states.

n = + 1 Vibrational Transition Rovibrational transitions and selection rule Selection rules: n =  1 and J =  1 Absorption: n = + 1 Vibrational Transition  Rotational Transition J = - 1 J = + 1 Typically at room temperature, only ground vibrational state populated but several rotational levels may be populated.

Transition energy of rovibration Rotational Transition Erot ; Selection rules: J =  1 For J = + 1 : Vibrational Transition Evib ; Selection rules: n = + 1 For J = - 1 :

n = + 1 Vibrational Transition Rovibrational spectrum n = + 1 Vibrational Transition  Rotational Transition J = - 1 J = + 1 Q branch  P branch (lower frequency)    R branch (higher frequency)  cm-1 2B Decreases as J increases increases as J increases

Ex. Rovibrational transitions J = - 1 J = + 1

Ex. Rovibrational transitions J = - 1 (P branch) J = + 1 (R branch)

Simulated 300. K infrared absorption spectrum and energy diagram for HCl.

Ex. Infrared absorption by HI gives rise to an R-branch from n = 0 Ex. Infrared absorption by HI gives rise to an R-branch from n = 0.What is the wavenumber of the line originating from the rotational state with J = 4? (Assuming HI as rigid rotor model where k =314 N m-1 and its bond length = 163 pm) Absorption lines in R-branch region correspond to: Calculate the vibrational transition energy at n=0 =1.66 10-27kg = 6.92 1013 Hz = 4.59 10-20 J

Calculate the vibrational transition energy at J=4 the wavenumber of the line originating from the rotational state with J = 4 is = 2371.95 cm-1

Rovibrational spectrum of HCl  

Intensity of a spectral line: Boltzmann distribution Depends on the number of molecules in the energy level from which the transition originates. Boltzmann distribution A high-resolution spectrum is shown for CO in which the P and R branches are resolved into the individual rotational transitions.

Ex. Using the Boltzmann distribution equation to calculate the number of CO molecules in J = 2, 5, 9 and 18 relative to the ground state (J=0) at 300K. Giving the rotational constant = 1.9313 cm-1 kB = 1.38 X 10-23 m2 kg s-2 K-1 J 2J+1 BJ(J+1) NJ/N0 2 5 2.3019E-22 4.73 11 1.1509E-21 8.33 9 19 3.4528E-21 8.25 18 37 1.3121E-20 1.56

If we calculate NJ/N0 with various J values, we would obtain a graph below. At 300K At 700K 1 2.945 2.98 2 4.730 4.88 3 6.264 6.67 4 7.478 8.31 5 8.331 9.77 6 8.810 11.00 9 8.255 13.29 12 5.894 13.46 15 3.357 11.96 18 1.558 9.52 21 0.596 6.87 24 0.189 4.53 27 0.050 2.74 30 0.011 1.52 33 0.002 0.78

Rovibration of non-rigid rotation Two effects: Vibration-Rotation Coupling : As the molecule vibrates more, bond stretches The effective rotational constant Be : the rotational constant for a rigid rotor αe : the rotational-vibrational coupling constant 2) Centrifugal Distortion As a molecule spins faster, the bond is pulled apart

Centrifugal Distortion : non-rigid rotation When a molecule rotates faster and faster (J increases), the centrifugal force pulls the atoms apart. As a result, its bond length increases and its moment of inertia (I) increases, thus decreasing B Centrifugal force Rotational energy of non-rigid rotor: rigid rotot D = the centrifugal distortion constant

Real spectrum Ideal spectrum Rovibrational energy of non-rigid rotor: As energy increases, the R-branch lines move closer and as energy decreases, the P-branch lines move farther apart. This is attributable to two phenomena: rotational-vibrational coupling and centrifugal distortion.

Ex. Using the rovibrational spectrum of HCl to answer the questions below (consider as non rigid molecule) A D E C F G 2906 B 2926 H 2865 Spectral region with J = -1 …….. Spectral region with J = +1 …….. R-branch …….. P-branch …….. Q-branch …….. 6) J=7  J=8 transition …….. 7) J=5  J=4 transition …….. 8) Spectral lines belong to 1H35Cl …….. 9) Spectral lines belong to 1H37Cl …….. 10) 𝐵 𝑒 = ………cm-1, 𝐷 = ………… cm-1

Transition Selection rule J En,J E0,0 E0,1 E1,0 E1,1 E1,2 2 1 1 1 Observed freq. Transition Selection rule n = 0  n =1 J = 0  J = 1 n = 1 J = +1 n = 0  n =1 J = 1  J = 0 n = 1 J = -1 n = 0  n =1 J = 1  J = 2 n = 1 J = +1

Rovibrational energy of non-rigid rotor: 2926 2865 2906

(1) - (2) 2906-2865 = 41 (3) - (1) 2926-2906 = 20

Ex. Using the rovibrational spectrum of HCl to answer the questions below 2906 2926 H F C G E D A B Spectral region with J = -1 …A….. Spectral region with J = +1 … B .. R-branch … B …. P-branch … A .. Q-branch … D….. 6) J=7  J=8 transition … E ….. 7) J=5  J=4 transition … C ….. 8) Spectral lines belong to 1H35Cl … F .. 9) Spectral lines belong to 1H37Cl … G….. 10) 𝐵 𝑒 = …10.3 cm-1, 𝐷 = …0.02…… cm-1

Rigid rotor wavefunctions The full wavefunctions Reduce to Consider only the angular part of the wavefunctions Set the Schrödinger Equation equal to zero and

From the previous slide Rearranging the differential equation separating the θ-dependent terms from the -dependent terms: Only  Only  the Schrödinger equation can be solved using separation of variables.

Use the substitution method (similar to the previous one) For the J=0 → J=2 transition, Consider Use the substitution method (similar to the previous one) Replace x with  and integrate from 0 to , we get: Do the same for

From the previous derivation: For the J=0 → J=2 transition, Thus: From the previous derivation: Therefore: Thus, the J=0 → J=2 transition is forbidden.

Home Work 2 1. Spectral line spacing of rotational microwave spectrum of OH radical is 37.8 cm-1 . Determine the OH bond length (in pm unit) and moment of inertia (in kg m2) mO = 15.994 amu Spectrum (cm-1) mH = 1.008 amu 37.8 cm-1 2. Use the bond length of diatomic molecules in Table to predict line spacing (in cm-1 unit) of rotational microwave spectrum. molecule bond length (pm) HF 91.7 HI 161 HCl 128 HBr 141

3. Determine the bond length of these diatomic gases in Table and arrange them in order of increasing the bond length. Bond length (pm) OH 37.80 ICl 0.11 ClF 1.03 AlH 12.60 4. Using the information in the Table to calculate the ratio between the transition energy of rotation from J=0 to J =1 and vibration from n=0 to n=1 for H2 Atomic mass 1.008 amu Bond length of H2 74.14 pm Force constant of H2 575 N/m