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Rotational Spectra. Simplest Case: Diatomic or Linear Polyatomic molecule Rigid Rotor Model: Two nuclei joined by a weightless rod. J = Rotational quantum number (J = 0, 1, 2, …) I = Moment of inertia = m r 2 = reduced mass = m 1 m 2 / (m 1 + m 2 ) r = internuclear distance. m 2.
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Rotational Spectra Simplest Case: Diatomic or Linear Polyatomic molecule Rigid Rotor Model:Two nuclei joined by a weightless rod • J = Rotational quantum number (J = 0, 1, 2, …) • I = Moment of inertia = mr2 • = reduced mass = m1m2 / (m1 + m2) r = internuclear distance m2 m1 r
Rigid Rotor Model In wavenumbers (cm-1): Separation between adjacent levels: F(J) – F(J-1) = 2BJ
Rotational Energy Levels Selection Rules: Molecule must have a permanent dipole. DJ = 1 J. Michael Hollas, Modern Spectroscopy, John Wiley & Sons, New York, 1992.
Rotational Spectra J. Michael Hollas, Modern Spectroscopy, John Wiley & Sons, New York, 1992.
Intensity of Transitions %T cm-1 J. Michael Hollas, Modern Spectroscopy, John Wiley & Sons, New York, 1992.
Are you getting the concept? Calculate the most intense line in the CO rotational spectrum at room temperature and at 300 °C. The rigid rotor rotational constant is 1.91 cm-1. Recall: k = 1.38 x 10-23 J/K h = 6.626 x 10-34 Js c = 3.00 x 108 m/s
The Non-Rigid Rotor Account for the dynamic nature of the chemical bond: DJ = 0, 1 D is the centrifugal distortion constant (D is large when a bond is easily stretched) Typically, D < 10-4*B and B = 0.1 – 10 cm-1
More Complicated Molecules Still must have a permanent dipole DJ = 0, 1 K is a second rotational quantum number accounting for rotation around a secondary axis A.
Vibrational Transitions Simplest Case: Diatomic Molecule Harmonic Oscillator Model:Two atoms connected by a spring. in Joules in cm-1 v = vibrational quantum number (v = 0, 1, 2, …) n = classical vibrational frequency k = force constant (related to the bond order).
Vibrational Energy Levels • Selection Rules: • Must have a change in dipole moment (for IR). • 2) Dv = 1 J. Michael Hollas, Modern Spectroscopy, John Wiley & Sons, New York, 1992.
Anharmonicity Selection Rules: Dv = 1, 2, 3, … Dv = 2, 3, … are called overtones. Overtones are often weak because anharmonicity at low v is small. Ingle and Crouch, Spectrochemical Analysis
Rotation – Vibration Transitions The rotational selection rule during a vibrational transition is: DJ = 1 Unless the molecule has an odd number of electrons (e.g. NO). Then, DJ = 0, 1 Bv signifies the dependence of B on vibrational level
Rotation – Vibration Transitions If DJ = -1 P – Branch If DJ = 0 Q – Branch If DJ = +1 R – Branch Ingle and Crouch, Spectrochemical Analysis
Rotation – Vibrational Spectra Why are the intensities different? J. Michael Hollas, Modern Spectroscopy, John Wiley & Sons, New York, 1992.
Are you getting the concept? In an infrared absorption spectrum collected from a mixture of HCl and DCl, there are eight vibrational bands (with rotational structure) centered at the values listed below. Identify the cause (species and transition) for each band. Atomic masses H → 1.0079 amu D → 2.0136 amu 35Cl → 34.9689 amu 37Cl → 36.9659 amu
Raman Spectra Selection Rule: DJ = 0, 2 J. Michael Hollas, Modern Spectroscopy, John Wiley & Sons, New York, 1992.
Polyatomics If linear (3N – 5) vibrational modes (N is the # of atoms) If non-linear (3N – 6) vibrational modes Only those that have a change in dipole moment are seen in IR. http://jchemed.chem.wisc.edu/JCEWWW/Articles/WWW0001/index.html
Linear Polyatomic How many vibrational bands do we expect to see? J. Michael Hollas, Modern Spectroscopy, John Wiley & Sons, New York, 1992.
Nonlinear Polyatomic (Ethylene) J. Michael Hollas, Modern Spectroscopy, John Wiley & Sons, New York, 1992.
Infrared Spectroscopy • Near Infrared: 770 to 2500 nm • 12,900 to 4000 cm-1 • Mid Infrared: 2500 to 50,000 nm (2.5 to 50 mm) • 4000 to 200 cm-1 • Far Infrared: 50 to 1000 mm • 200 to 10 cm-1
Infrared Spectroscopy: Vibrational Modes Ingle and Crouch, Spectrochemical Analysis
Group Frequencies Estimate band location: Pretsch/Buhlmann/Affolter/ Badertscher, Structure Determination of Organic Compounds
Are you getting the concept? Estimate the stretching vibrational frequency for a carbonyl group with a force constant, k, of 12 N/cm. If a C=S bond had the same force constant, where would its stretching band appear in the infrared absorption spectrum? Recall: 1 amu = 1.6605 x 10-27 kg 1N = 1 kg*m*s-2 Atomic masses C → 12.000 amu O → 15.995 amu S → 31.972 amu
Infrared Spectroscopy • Near Infrared: 770 to 2500 nm • 12,900 to 4000 cm-1 • *Overtones • * Combination tones • * Useful for quantitative measurements • Mid Infrared: 2500 to 50,000 nm (2.5 to 50 um) • 4000 to 200 cm-1 • *Fundamental vibrations • * Fingerprint region 1300 to 400 cm-1 • (characteristic for molecule as a whole) • Far Infrared: 2.5 to 1000 um • 200 to 10 cm-1 • *Fundamental vibrations of bonds with heavy • atoms (useful, e.g., for organometallics)
Example of an Overtone • Wagging vibration at 920 cm-1. • Overtone at approximately 2 x 920 cm-1 = 1840 cm-1.
Fermi Resonance N.B. Colthup et al., Introduction to Infrared and Raman Spectroscopy, Academic Press, Boston, 1990.
Example of a Fermi Resonance • Stretching vibration of C-C=(O) at 875 cm-1. • Overtone at approximately 2 x 875 cm-1 = 1750 cm-1 • coincides with C=O stretch