CHEM 3340 Physical Chemistry II PDF

CHEM 3340 Physical Chemistry II Molecular Spectroscopy I: rotational and vibrational spectroscopy Molecular Spectroscopy Spectroscopy: analysis of electromagnetic radiation emitted, absorbed, or scattered by atoms and molecules Energy of photons gives information about energies of molecule Energy of...

CHEM 3340 Physical Chemistry II Molecular Spectroscopy I: rotational and vibrational spectroscopy Molecular Spectroscopy Spectroscopy: analysis of electromagnetic radiation emitted, absorbed, or scattered by atoms and molecules Energy of photons gives information about energies of molecule Energy of molecule gives information about bonds (strengths, lengths, angles) and important molecular properties (dipole moment, shapes) Molecular spectrum is more complicated then atomic: ◊ Electronic structure is more complicated ◊ Molecules can vibrate and rotate as well General Features of Spectroscopy Emission spectroscopy: molecule undergoes transition from a high-energy state (E1) to a low-energy state (E2) and emits the excess energy as a photon Absorption spectroscopy: the net absorption of nearly monochromatic (single frequency) incident radiation is monitored as the radiation is swept over a range of frequencies Energy of photon absorbed or emitted (Bohr frequency): hν = |E1 - E2| Often instead of frequency wavelength (λ=c/ν) or wavenumber is used ⇥˜ = 1/ = ⇥/c Absorption Spectrometer Source: light bulb for given frequency range (mercury, quartz, tungsteniodine) Grating: dispersing element that separates radiation into di erent frequencies Di raction grating: constructive interference occurs only at a frequency dependent angle Detector: convert radiation into current or potential (e.g., photodiode: conducts electricity when photons hit the surface) Reference is used to determine the incident intensity (I0) ff ff Goal: obtain transmitted intensity (I) as a function of frequency Lambert-Beer Law ffi What is the amount of ‘absorbed’ light and how it relates to the length of the sample, l, (cuvette length) and the concentration of the chemical species (c, in mol/dm3)? I = I010-εcl ε: molar absorption coe cient [dm3 cm-1 mol-1] Absorbance, A: A = log (I0/I) Simpli ed form of Lambert-Beer Law: A = ε c l Transmittance: T = I/I0 fi Usually ε is given vs. frequency in a spectrum Example ffi The transmittance of an aqueous solution that contained 0.1 mol/dm3 Cu++ was measured as 0.3 at 600 nm in cell length of 5mm. Calculate the molar absorption coe cient and the absorbance! Linewidth in the spectrum Reasons: ◊Lifetime broadening: Time dependent Schrödinger equation results in a similar relationship to the Heisenberg uncertainty, but with the energy and the lifetime of the state τ: δE τ ≈ ħ Shorter lifetime of the state, less de ned the energy ◊ Doppler broadening: molecules moving toward the detector have higher frequency, molecules moving away the detector lower frequency: Doppler e ect. Depends on temperature, can be compensated In liquids: the molecules interact with each other; this changes the energies and broadens the lines fi The spectrum is in theory composed of lines, however, actual lines often have widths: ff Selection Rules Classical idea: molecule to be able to interact with the electromagnetic eld and absorb or create a photon, it must possess, at least transiently, a dipole oscillating at that frequency For each spectroscopy, we will give ◊Gross selection rule: general features a molecule needs to have to exhibit transitions (e.g., permanent electric dipole moment) ◊ Speci c selection rules: what transitions are allowed in terms of changes in quantum numbers fi There are many possible energy levels fi Rotational Spectrum Molecules can rotate; the rotation is quantized Energy depends on two factors: E = Jz2/(2I) J: Angular momentum around some axis, e.g., z-> will be quantized I: Moment of inertia corresponding to rotation around a certain axis Moment of inertia I = Σi mi ri2 mi: mass of the i-th atom ri: distance from the axis of rotation From the rotational spectrum we will be able to determine I From I we shall be able to predict bond angles and lengths Moment of Inertia r I = mr2 R I = μR2 m1 m2 Equivalent to rotation of a ball of reduced mass μ on a circle of radius R 1/μ = 1/m1 + 1/m2 μ = m1 m2 /(m1+ m2) For more complicated examples look in Table 11B.1 of the book (p 431) Energy of Rotation Energy of rotation: E = J2/(2I) According to quantum mechanics, the angular momentum is quantized (see rotation on a sphere): J2 = l(l+1)ħ2 , l =0,1,2,... For general case we replace l with J J: rotational quantum number E = J(J+1) ħ2/(2I), J=0,1,2,.... It will be simpler to use this formula in another form: E = J(J+1)B hc where B is rotational constant (wavenumber) hcB = ħ2/(2I) B is the energy expressed as the wavenumber of a corresponding photon Typical values of B for molecules: 0.1 to 10 1/cm fi Selection Rules Gross selection rule: the molecule must be polar Rotationally active molecules: HCl, NH3, CO, H2O Rotationally inactive molecules: O2, N2, CO2, CH4 Speci c selection rule: ∆J = ± 1 (the photon takes some angular momentum, and only these transitions can provide that) Transition will occur from J level to J+1 (absorption), or J to J-1 (emission) Notation: (J+1 ← J) Energy Levels and Allowed Transitions Energy Di erence: ∆E (J+1 ← J) = E(J+1) - E(J) = hcB(J+2)(J+1) - hcB(J+1)(J) = 2hcB(J+1) Corresponding photon frequency: hν With wavenumber: Therefore E = hc˜ 2hcB(J + 1) = hc˜ ˜ = 2B(J + 1) ff Wavenumbers of the allowed transitions J=0,1,2... Example: What is the wavenumber of the line in the rotational spectrum of 1H35Cl corresponding to transition J=1 ←0? (bond length: R = 127 pm) Rotational Spectrum Each line corresponds to a transition with wavenumber 2B(J+1) Easy estimation of B: two lines are separated by 2B(J+1) -2B(J) = 2B The intensity of the lines is determined by the number of molecules (N) at levels J and J+1(for transition J+1← J) I ∝ N(J+1)/N(J) Boltzmann Distribution Average energy of the molecules is related to temperature (thermal motion has energy proportional to kT, where k is the Boltzmann constant) The number of molecules (NJ) on each energy level EJ is obtained from Boltzmann distribution NJ N gJ e EJ /(kT ) N: Total number of molecules gJ: degeneracy of level E For rotational energies: EJ = 2hcB(J+1) gJ = 2J+1 (Magnetic quantum number M, M=0, ±1,...,±J) Intensity of Spectral Lines At each level the number of molecules NJ N (2J + 1)e 2BhcJ(J+1)/(kT ) This gives a maximum: at T=298.15 K, J=30 Microwave oven: By microwave absorption Jmax is increased -> temperature is increased The intensity for transition J+1← I J (2J + 3)e 2Bhc(J+1)(J+2)/(kT ) NJ+1 = NJ (2J + 1)e 2BhcJ(J+1)/(kT ) The maximum of I is not necessarily at J=30 at room temperature! Vibration of Molecules Potential energy vs. internuclear distance often approximated by a parabola close to the bottom of the well V=1/2 k x2 x=R-Re, displacement k: force constant of the bond (N/m) (F = -kx) Molecule will vibrate Force constant and reduced mass Large k -> steep potential energy-> sti Selection Rules for Vibrational Spectroscopy fi c selection rule: ∆v = ±1 +1: absorption, -1: emission Infrared Spectroscopy Energy of transition v+1 ← v: (same as spacing) hν = Each transition has the same energy: one line in spectrum (in real spectra there are small variations because the potential energy in not perfectly symmetric parabola) Most molecules at T=289.15 K are at v=0. Strongest transition: 1 ← 0 1H35Cl, Infrared (IR): wavelengths between 750 nm (visible) and 1mm (microwave) Corresponding wavenumber: 300 -3000 1/cm Infrared Spectroscopy: vibrational energy levels hc˜ k=516 N/m, ν = 89.7 THz, λ=3.5 μm Vibration-Rotation Spectra Molecules can both rotate and vibrate For each vibrational energy level (e.g., v=0) there are corresponding rotations (J=0,1,2,...) Total energy: sum of vibration and rotation E = hcJ(J + 1)B + hc(v + 1/2)˜ When there is an absorption due to vibration, the rotational quantum number might change Speci ∆v = ±1