1. LINEAR MOLECULE ROTATIONAL TRANSITIONS:  J = 4  J = 3  J = 2  J = 1  J = 0

2. LINEAR MOLECULE ROTATIONAL SPECTRUM:  Intensity J = 4←3  J = 1←0  2B  Absorption Frequencies →

3. EFFECTS OF MASS AND MOLECULAR SIZE:  The slide that follows gives B values for a number of diatomic molecules with different reduced masses and bond distances. What is the physical significance of the very different B values seen for H 35 Cl and D 35 Cl? All data are taken from the NIST site.  http://www.nist.gov/pml/data/molspec.cfm

4. REDUCED MASSES AND BOND DISTANCES: MoleculeBond Distance (Å) B Value (MHz) H 35 Cl1.275312989.3 H 37 Cl1.275312519.1 D 35 Cl1.275161656.2 H 79 Br1.414250360.8 H 81 Br1.414250282.9 D 79 Br1.414127358.1 24 Mg 16 O1.74817149.4 107 Ag 35 Cl2.2813678.04

5. REAL LIFE – WORKING BACKWARDS?  In the real world spectroscopic experiments provide frequency (and intensity) data. It is necessary to assign quantum numbers for the transitions before molecular (chemically useful) information can be determined. Sometimes “all of the data” are not available!

6. SPECTRUM TO MOLECULAR STRUCTURE:  Class Example: A scan of the microwave (millimeter wave!) spectrum of 6 LiF over the range 350 → 550 GHz shows lines at 358856.2 MHz, 448491.1 MHz and 538072.7 MHz. Assign rotational quantum numbers for these transitions. Determine a B value and the bond distance for 6 LiF. Are the “lines” identically spaced?

7. HIGHER ORDER ENERGY TERMS:  The slightly unequal spacing of lines in the 6 LiF spectrum occurs because very rapidly rotating diatomic molecules distort. A “higher order” energy expression accounts for this effect  E J = hBJ(J+1) – hD J J 2 (J+1) 2  D J is the (quartic) centrifugal distortion constant.

8. HIGHER ORDER FREQUENCY EXPRESSION:  The energy expression on the previous slide can be used with the selection rule ΔJ = +1 (for absorption) and ΔE = hν to give:  ν = 2B(J+1) - 4D J (J+1) 3  This expression will be used in the lab (HCl/DCl spectrum). A typical frequency calculation is shown on the next slide.

9. NON-RIGID ROTOR CALCULATION, 7 LIF:  Here B = 40,026.883 MHz & D J = 0.3505 MHz Transition 2B(J+1)4D J (J+1) 3 Freq. Calc.Freq. Obs. J=1←080053.7661.40280052.36 No Data J=2←1160107.5311.216160096.32160096.33 J=3←2240161.3037.854240123.45240123.47 J=4←3320215.0789.728320125.34320125.36 J=5←4400268.83175.2540093.58400093.62 J=6←5480322.60302.83480019.77480019.73

10. NON-RIGID MOLECULES:  Aside: Every spectroscopic constant tells us something. A “small” D J value suggests that a molecule does not distort easily. Comparisons can be made for inertially similar molecules. Explanation? MoleculeB (GHz)D J (kHz) 6 LiF45.23443 13 C 18 O52.36151

11. SPECTRA OF NONLINEAR MOLECULES:  With the particle in the box energy expressions grew more complex as we moved from one to three dimensions.  PIAB one dimension: Energy (eigenvalues!) expression has one term and one quantum number.  PIAB three dimensions: Energy expression has (up to!) three terms and three quantum numbers.

12. ROTATIONS IN THREE DIMENSIONS:  For nonlinear molecules the number of quantum numbers and rotational constant needed to describe rotational energies is greater than one. We also have more than one I value. In general, we have a (3x3) matrix (moment of inertia tensor) that cane be diagonalized to simplify the mathematics.

13. NONLINEAR RIGID ROTORS:  After diagonalization the moment of inertia tensor has three elements with I a ≤ I b ≤ I c.  Types of Rotors:  1. Spherical tops: I a = I b = I c. We need just one quantum number (J again) to describe rotational energies. Examples: CH 4, SF 6 and C 60.

14. NONLINEAR RIGID ROTORS:  2. Symmetric tops: I a = I b ≤ I c (oblate top) and I a ≤ I b = I c (prolate top). Examples:  Oblate top: CHF 3, HSi 79 Br 3.  Prolate top: CH 3 F, CH 3 -CN.  For symmetric tops we need two quantum numbers, J and K, to describe rotational energies.

15. NONLINEAR RIGID ROTORS: 

16. DEGENERACY:  In organic chemistry courses you have discussed NMR spectra – and removal of spin degeneracy using a magnetic field. For the spin case, I = ½, there is a two-fold energy degeneracy in the absence of a magnetic field. In rotational spectroscopy there is, similarly, a (2J+1) fold degeneracy.

17. DIPOLES AND ELECTRIC FIELDS.  From physics, the energy of a linear rod with an electric dipole moment (μ) placed in an electric field can be found as μEcosθ. Similar to NMR, the degeneracy of rotational energy levels can be removed by applying an electric filed to a gas. This enables the size of a molecule’s dipole moment to be determined.

18. DEGENERACY AND DIPOLE MOMENTS:  We will not use the degeneracy of rotational levels for several weeks. It will be useful in calculating the relative intensities of spectral lines (after Boltzmann!). By experiment, it is found that a molecule must have a permanent non-zero electric dipole moment to have a pure rotational spectrum.

19. MOLECULAR STRUCTURE/DIPOLE MOMENTS:  From first year chemistry courses you should be able to take a simple molecular formula and (a) draw a Lewis structure for the molecule, (b) determine a molecule’s shape and (c) predict whether a molecule has net polarity. Review examples on the next slide.

20. MOLECULAR SHAPES AND POLARITY: Molecular FormulaElectrically PolarPure Rotational Spectrum? H2H2 CO HCN CO 2 CH 2 Cl 2 SF 6 O=C=C=C=S CHF=CHF (cis) CHF=CHF (trans)