10.03.2011 - Lecture 12 Monday October 3 2011 ...

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Unformatted text preview: Lecture 12 Monday, October 3, 2011 Vibrational and Rotational Spectroscopy Today in Chemistry 260/261 •  •  •  •  •  •  Vibrational (infrared) spectroscopy Rotational (microwave) spectroscopy Quantization condition Selection rules Normal modes Applications This Week in Chemistry 260/261 • Reading: (Wed) 20.3 (Fri) 20.4 • Rotational, vibrational, electronic, magnetic spectroscopy • Problem set 5 due Monday, October 10 by 3:00 pm • MIDTERM 1: Tuesday, October 11, 8- 10 pm The transition dipole moment The transition dipole moment µij = ∫ψ i µψ j dτ Likelihood of making the connection between Ψi and Ψj is given by the transition dipole moment, µij. excited state ψ j ψi ground state measure of the change of the dipole moment associated with the shift of electric charge that accompanies a transition. µij = ∫ψ i µψ j dτ If µij = 0, no absorption occurs. What does this mean? If there is no “oscillating dipole moment” in the classical picture, no transition dipole in the quantum picture, then there is no absorption. Even if the energy is “in resonance.” The details depend on the speciXic kind of transitions. excited state ψ µ = (const )x j ψ2 ( x ) = ψ1 ( x ) = ψi # 2π x & 2 sin % ( $L' L # 1π x & 2 sin % ( $L' L ground state µ12 = ∫ L 0 ψ1 ( x ) x ψ2 ( x ) dx = µ12 = − 8 L2 9π 2 µ13 = 0 2 L ∫ L 0 * $ 1 π x '- * $ 2 π x ')/ x ,sin & )/ dx ,sin & + % L (. + % L (. TDM is non- zero, so light can make transitions from n=1 to n=2 This one does equal zero, so light cannot connect n=1 and n=3 in the same way 1 The interaction of matter and light Selection Rules The transition from 1s to 2s is not observed. Hydrogen atom − RH / 16 − RH / 9 µ1s , 2 s = ∫ψ 1s µψ 2 s dτ = 0 − RH / 4 × ΔE = hν − RH l=0 l =1 l=2 There are two parts to a selection rule: 1. A gross selection rule speciXies the general features that a molecule must have if it is to exhibit a spectrum of a given kind. “Always” true – e.g. oscillating dipole moment 2. A speciXic selection rule states which changes in quantum number may occur in a transition. l=3 In general, the energy of the incident radiation (hν) matches the energy difference between a ground state wave function and an excited state wave function. No oscillating dipole moment to absorb energy from the electromagnetic Xield. µ1s , 2 p ≠ 0 Selection rules in the atmosphere: N2: 78% O2: 21% Ar: 0.93% Ne,He,Kr: 0.0001% A selection rule is a statement about when a transition dipole moment is non- zero, i.e. under what circumstances the transition occurs. SpeciXic to the details A transition permitted by a speciXic and gross selection rule is classiXied as allowed. Transitions that are disallowed by a speciXic selection rule or a gross selection rule are called forbidden. Quantum Modeling of Molecular Vibrations HA RAB HB Consider bonds as harmonic oscillators, in which the potential energy is a parabolic function. No worries! V ( RAB ) = 1 2 k ( RAB − Re ) 2 LETS QUANTIZE! Re Potential energy operator: CO2: 0.0003% Why does CO2 correlate with global warming, but O2 and N2 do not ??? 1 1 1 2 ˆ V (r ) = k (r − re ) = kx 2 = µω 2 x 2 2 2 2 Schrödinger equation: ⎡ྎ 2 d 2 1 2 ⎤ྏ + kx ⎥ྏψ ( x) = Eψ ( x) ⎢ྎ− 2 2 ⎣ྏ 2µ dx ⎦ྏ ω= k µ angular frequency Boundary condition: ψ ( x → ±∞ ) = 0 2 The quantum harmonic oscillator − ψ n ( x) = Nn H n ( x) e Transition Dipole Moment µ=Q×r mω 2 x Q+ normalization Hermite polynomial In general, we don’t know the function µ(x), so we expand it about the equilibrium position in a Taylor series: Q- ˆ ˆ µ ( x ) = µ (0 ) + r ˆ µij = ∫ψ i ( x)µ ( x)ψ j ( x) dx Gaussian (bell curve) Keeping only the linear term is the linear dipole approximation: ˆ (1) dµ ( x) Transition dipole moment between states i and j; 1 ⎞ྏ ⎛ྎ En = ⎜ྎ n + ⎟ྏ ω 2 ⎠ྏ ⎝ྎ angular frequency patial ω = 2π ν is related tbo y s2π frequency ˆ ˆ dµ ( x) 1 d 2 µ ( x) x+ x2 + … dx x =0 2 dx 2 x =0 ˆ µ= dx ˆ ˆ µ ( x) ≈ µ (1) x •  Gross selection rule: The molecular dipole moment must change during the vibration (displacement)! O O H N • Homonuclear diatomics cannot absorb infrared light! (the dipole never changes!) • Polyatomics will have some transitions that do and maybe some that do not An example of a resonance in a molecule It’s a number! O • Transition occurs as the molecule moves along the vibrational mode and interacts with the Xield 1 ⎞ྏ ⎛ྎ En = ⎜ྎ n + ⎟ྏ hν 2 ⎠ྏ ⎝ྎ x =0 N O C O H H S C S H Cl H Br H HH C H H Vibrations in Polyatomics: Normal Modes N particles have 3N coordinates (x, y, z for each) unit vectors ⎡ྎ dµ x ⎤ྏ ⎢ྎ ⎥ྏ ≠ 0 ⎣ྏ dQ3 ⎦ྏ z2 z1 x1 y1 antisymmetric stretching vibration z3 y3 ˆ ˆ rn = xni + yn ˆ + zn k j x3 ˆ k x2 y2 ˆ i ˆ j Any position (or motion) can be represented by a sum of these unit vectors µx(t) t Any vibration can be reduced to sum of more simple vibrational motions. These simple vibrational “building blocks” are called normal modes. 3 Vibrations in Polyatomics: Normal Modes Vibrations in Polyatomics: Normal Modes We can pick another “coordinate system”: ˆ α O O γˆ bend C O OC antisymmetric stretch ˆ β Example: H2O Example: CO2 symmetric stretch O C O O C antisymmetric stretch antisymmetric stretch symmetric stretch symmetric stretch in- plane- bend O unit vectors ˆ ˆ r = cα α + cβ β + cγ γˆ Any position (or motion) can be represented by a sum of these unit vectors Any vibration can be reduced to sum of more simple vibrational motions. These simple vibrational “building blocks” are called normal modes. We can describe any complex vibration with a language of more simple motions. Infrared Absorption O C O bend A linear molecule will have 3N- 5 normal modes A nonlinear molecule will have 3N- 6 normal modes Transition Dipole Moment µ=Q×r The normal modes have distinct vibrational frequencies Q+ O C O C O 2350 cm- 1 Stretching (more energy) O C C O O r ˆ µij = ∫ψ i ( x)µ ( x)ψ j ( x) dx 1388 cm- 1 (not by IR) The dipole remains the same during the vibration O Q- O OC O out- of- plane- bend 667 cm- 1 Transition dipole moment between states i and j; In general, we don’t know the function µ(x), so we expand it about the equilibrium position in a Taylor series: ˆ ˆ µ ( x ) = µ (0 ) + ˆ ˆ dµ ( x) 1 d 2 µ ( x) x+ x2 + … dx x =0 2 dx 2 x =0 Keeping only the linear term is the linear dipole approximation: ˆ (1) dµ ( x) ˆ µ= dx (1) ˆ ˆ µ ( x) ≈ µ x x =0 It’s a number! Now we rewrite the transition dipole moment as: Bending 667 cm-1 2350 cm-1 ˆ µij = ∫ψ i ( x)µ ( x)ψ j ( x) dx ˆ µij = ∫ψ i ( x)µ (1) xψ j ( x) dx Substitute approximation ˆ µij = µ (1) ∫ψ i ( x) xψ j ( x) dx Final equation is relatively simple! 4 Harmonic Oscillator Transitions Harmonic Oscillator Transitions Transition dipole moment for the harmonic oscillator: Transition dipole moment for the harmonic oscillator: ˆ µij = µ (1) ∫ψ ( x) xψ i j ˆ µij = µ (1) ∫ψ i ( x) xψ j ( x) dx ( x) dx Use parity (symmetry): As x is odd;ψi (x) x ψj (x) must be odd or even integrand; Integral is non zero if Δn =± 1 − ψ n ( x) = Nn H n ( x) e mω 2 x 1 ⎞ྏ ⎛ྎ En = ⎜ྎ n + ⎟ྏ ω 2 ⎠ྏ ⎝ྎ Use parity (symmetry): As x is odd;ψi (x) x ψj (x) must be odd or even integrand; Integral is non zero if Δn =± 1 − mω 2 x ψ n ( x) = Nn H n ( x) e 1 ⎞ྏ ⎛ྎ En = ⎜ྎ n + ⎟ྏ ω 2 ⎠ྏ ⎝ྎ 1→ 0 ψ 0 ( x) Gross selection: Physically it means that for a normal mode to be spectroscopically active (IR active) the dipole must change during the vibration. SpeciTic selection of the HO: Δn =± 1 mω 2 x − Investigate graphically: ψ 1 ( x) x xψ 0 ( x )ψ 1 ( x ) x Integrand is positive everywhere, so the integral is necessarily positive. This transition is allowed. Harmonic Oscillator Transitions Harmonic Oscillator Transitions Transition dipole moment for the harmonic oscillator: Transition dipole moment for the harmonic oscillator: ˆ µij = µ (1) ∫ψ ( x) xψ i j ˆ µij = µ (1) ∫ψ i ( x) xψ j ( x) dx ( x) dx Use parity (symmetry): As x is odd;ψi (x) x ψj (x) must be odd or even integrand; Integral is non zero if Δn =± 1 − Investigate graphically: ψ n ( x) = Nn H n ( x) e 2→0 1 ⎞ྏ ⎛ྎ En = ⎜ྎ n + ⎟ྏ ω 2 ⎠ྏ ⎝ྎ ψ 0 ( x) mω 2 x Use parity (symmetry): As x is odd;ψi (x) x ψj (x) must be odd or even integrand; Integral is non zero if Δn =± 1 Investigate graphically: 2 →1 ψ n ( x) = Nn H n ( x) e 1 ⎞ྏ ⎛ྎ En = ⎜ྎ n + ⎟ྏ ω 2 ⎠ྏ ⎝ྎ ψ 1 ( x) ψ 2 ( x) x ψ 2 ( x) x x 0 ( x )ψ 2 ( x ) xψ 1 ( x )ψ 2 ( x ) x This integrand has equal negative and positive areas. This transition is forbidden. x This integrand is mostly positive. This transition is allowed. 5 Harmonic Oscillator Transitions Anharmonic Oscillator Transition dipole moment for the harmonic oscillator: Real molecules are anharmonic, since we know bonds must break ˆ µij = µ (1) ∫ψ ( x) xψ i j ( x) dx Use parity (symmetry): As x is odd;ψi (x) x ψj (x) must be odd or even integrand; Integral is non zero if Δn =± 1 − Investigate graphically: ψ n ( x) = Nn H n ( x) e mω 2 x 1 ⎞ྏ ⎛ྎ En = ⎜ྎ n + ⎟ྏ ω 2 ⎠ྏ ⎝ྎ 4 →1 ψ 1 ( x) The speciTic selection rule for a harmonic oscillator is that the quantum number must change by one unit: ψ 4 ( x) x xψ 1 ( x )ψ 4 ( x ) x ni − n j = 1 Though it is not obvious, this integrand has equal negative and positive areas. This transition is forbidden. Δn = 1 OH stretch of H2O: 3400 cm- 1 OD stretch of HOD: 2500 cm- 1 Anharmonic Oscillator Main differences from the HO: 1) Only a harmonic oscillator has equally spaced energy levels. By measuring the “anharmonicity” it is possible to determine how the potential curve Tlattens out. 2) Also, |ni- nj|=1 selection rule no longer holds! One photon may connect two levels that differ by more than one quantum absorption spectrum of water Water is blue because of anharmonicity! How can you prove this absorption is due to vibrations? 750 nm two 3- m tubes, one Tilled with H2O, the other with D2O. • Isotope accounts for the frequency shift! • What is the color of these two absorptions? http://www.dartmouth.edu/~etrnsfer/water.htm Once symmetry (equally spaced energy levels) is broken, the selection rule is relaxed and forbidden transitions become allowed, albeit weak. 6 THE RIGID ROTOR Modeling molecular rotations r Schrödinger equation for a rigid rotor: The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. (corresponds to a linear molecule) deXinition: r = r1 + r2 m1 − m2 center of mass r1 m1m2 2 r = µr 2 m1 + m2 Balancing equation: m1r1 = m2r2 K .E . = . 0 = Eψ / As before we must introduce a new variable (operator, quantum number) in order to separate and solve this complex equation: Classical deXinition of rotational energy μ = reduced mass Describes how difTicult it is to induce rotation in an object 2 + 1 dψ $ dψ ' 1 d 2ψ & sin θ dθ ) + sin 2 θ dφ 2 2 I , sin θ dθ % ( Looks a lot like our H atom Schrödinger equation with the potential energy and dependence on r shaved off… r2 Moment of inertia I= Quantum Modeling of Molecular Rotations J (J + 1)h 2 = hBJ (J + 1) 8π 2 I h rotational quantum number, J B= 2 = 8π I 4π I J = 0,1,2,… Erot = 12 Iω 2 angular rotational constant (dependent on molecule because of its dependence on I) velocity Depends on the masses and bond lengths … the structure of the molecule! Moment of inertia – information on bond lengths and atomic masses – rotational spectroscopy is used to determine structure of molecules. Re Rotational Spectroscopy Re m1 m2 r1 Rotational Spectroscopy m1 r2 m2 r1 2 Moment of inertia: I = mred Req 2 Moment of inertia: I = mred Req Energy levels: r2 Energy levels: J (J + 1)h 2 Erot = = hBJ (J + 1) 8π 2 I h h B= 2 = 2 8π I 8π µ Req 2 J (J + 1)h 2 = hBJ (J + 1) 8π 2 I h h B= 2 = 2 8π I 8π µ Req 2 Erot = Selection rules: Selection rules: Gross: molecule must have a permanent dipole Gross: molecule must have a permanent dipole Δ J = ±1 20Bh SpeciXic: |J Xinal - J initial| = 1 Because molecules at room temperature are rotationally excited, we can see transitions from many different initial states ˆ ˆ dµ ( x) 1 d 2 µ ( x) + x+ x2 + … dx x =0 2 dx 2 x =0 12Bh Abs. ˆ ˆ µ ( x) = µ (0 ) Energy 8Bh 6Bh 4Bh 2Bh 6Bh 2Bh 0 2Bh 4Bh Energy 6Bh 8Bh Pure rotational spectra are evenly spaced where the levels are not. 7 Combinations of Transitions Combinations of Transitions HCl J=3 vibrational levels: 2888 cm- 1 rotational levels: 65 cm- 1 (for J=0 to J=1) J=2 n=2 Vibration- Rotation energy differences à༎ J=1 J=0 J=3 HCl J=3 vibrational levels: 2888 cm- 1 rotational levels: 65 cm- 1 (for J=0 to J=1) J=2 n=2 Vibration- Rotation energy differences à༎ J=3 J=2 n=1 J=1 J=0 J=2 n=1 J=3 P branch ΔJ = - 1 R branch ΔJ=+1 J=1 J=0 J=1 J=0 J=3 J=2 n=0 J=1 J=0 J=2 n=0 J=1 J=0 8 ...
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This note was uploaded on 01/19/2012 for the course CHEM 260 taught by Professor Staff during the Fall '08 term at University of Michigan.

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