10.05.2011 - Lecture 13 Wednesday, October 5, ...

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Unformatted text preview: Lecture 13 Wednesday, October 5, 2011 Today in Chemistry 260/261 •  Electronic spectroscopy •  Applications •  Review of rotation/vibration/electronic spectroscopy Coming up in Chemistry 260/261 • Reading: (Fri) 20.5, 20.7 • Problem set 5 due Monday, October 10 by 3:00 pm • MIDTERM 1: Tuesday, October 11, 8- 10 pm, Chem 1800 Electronic Spectroscopy Anharmonic Oscillator OH stretch of H2O: 3400 cm- 1 OD stretch of HOD: 2500 cm- 1 Anharmonic Oscillator Real molecules are anharmonic, since we know bonds must break 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 ;lattens 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 How can you prove this absorption is due to vibrations? 750 nm two 3- m tubes, one ;illed 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. 1 Water is blue because of anharmonicity! THE RIGID ROTOR Modeling molecular rotations r 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) de[inition: r = r1 + r2 m1 m2 center of mass r1 Balancing equation: m1r1 = m2r2 r2 Moment of inertia Classical de[inition of rotational energy m1m2 2 I= r = µr 2 m1 + m2 K .E . = μ = reduced mass Describes how dif;icult it is to induce rotation in an object 12 Iω 2 angular velocity Depends on the masses and bond lengths … the structure of the molecule! Re Quantum Modeling of Molecular Rotations Rotational Spectroscopy m1 m2 r1 Schrödinger equation for a rigid rotor: 2 − 2I 2 Moment of inertia: I = mred Req + 1 dψ $ dψ ' 1 d 2ψ . sin θ + 0 = Eψ sin θ dθ & dθ ) sin 2 θ dφ 2 / % ( , Energy levels: J (J + 1)h 2 = hBJ (J + 1) 8π 2 I h h B= 2 = 2 8π I 8π µ Req 2 Erot = Looks a lot like our H atom Schrödinger equation with the potential energy and dependence on r shaved off… As before we must introduce a new variable (operator, quantum number) in order to separate and solve this complex equation: rotational constant (dependent on molecule because of its dependence on I) J (J + 1)h 2 Erot = = hBJ (J + 1) 8π 2 I h rotational quantum number, J B= 2 = 8π I 4π I J = 0,1,2,… r2 Selection rules: Gross: molecule must have a permanent dipole ˆ ˆ µ ( x ) = (0 ) + ˆ ˆ dµ ( x) 1 d 2 µ ( x) x+ x2 + … dx x =0 2 dx 2 x =0 Moment of inertia – information on bond lengths and atomic masses – rotational spectroscopy is used to determine structure of molecules. 2 Re Rotational Spectroscopy m1 m2 r1 r2 Combinations of Transitions HCl J=3 vibrational levels: 2888 cm- 1 Moment of inertia: I = mred R 2 eq rotational levels: 65 cm- 1 (for J=0 to J=1) Energy levels: J=2 J=1 J=0 n=2 Vibration- Rotation energy differences à༎ J (J + 1)h 2 = hBJ (J + 1) 8π 2 I h h B= 2 = 2 8π I 8π µ Req 2 J=3 Erot = J=2 J=3 Selection rules: Gross: molecule must have a permanent dipole Δ J = ±1 20Bh Speci[ic: |J [inal - J initial| = 1 Abs. Energy 12Bh 6Bh 2Bh R branch P branch ΔJ=+1 ΔJ = - 1 J=2 J=1 J=0 n=0 Because molecules at room temperature are rotationally excited, we can see transitions from many different initial states 8Bh 4Bh J=1 J=0 n=1 6Bh 2Bh 0 2Bh 4Bh 6Bh 8Bh Energy Combinations of Transitions Pure rotational spectra are evenly spaced where the levels are not. J=3 vibrational levels: 2888 cm- 1 rotational levels: 65 cm- 1 (for J=0 to J=1) Electronic Spectroscopy Selection Rules HCl Transition probabilities: origins of selection rules J=2 n=2 Vibration- Rotation energy differences à༎ J=1 J=0 Lets examine what can occur in UV- Vis spectroscopy (electronic transitions). J=3 J=2 n=1 J=1 J=0 J=3 You can excite an electron in to either an s or a p type orbital in its excited state Electron can retain its spin or take on the opposite spin in its excited state J=2 n=0 J=1 J=0 ∫ψ orb i ψ spin i orb j ˆ µψ ψ spin j dτ orbdτ spin ? For a transition to occur, this integral must not equal zero! 3 Electronic Spectroscopy Selection Rules The Spin Selection Rules Transition probabilities: origins of selection rules φLUMO ∫ψ ? orb i ˆj j ψ ispinµψ orbψ spindτ orbdτ spin Only operates on orbital coordinates; does not effect spin φHOMO ∫ψ orb i ˆj µψ orbdτ orb ∫ψ ispinψ spindτ spin j spin ∫ψ ≠ 0 only if ψ i orb i ˆj j ψ ispinµψ orbψ spindτ orbdτ spin In general the spin does not change Electronic Spectroscopy Selection Rules ? Transition probabilities: origins of selection rules φLUMO …φHOMO1 φ LUMO1 S1 φHOMO φHOMO … φHOMO2 φLUMO ∫ψ ∴ΔS = 0 Electronic Spectroscopy Selection Rules Transition probabilities: origins of selection rules φLUMO = ψ spin j orb i ˆj µψ orbdτ orb ∫ψ ispinψ spindτ spin j In general the spin does not change Transitions where the spin does change tend to be very weak. ∫ψ S0 orb i ˆj µψ orbdτ orb ∫ψ ispinψ spindτ spin j Although we can look at transitions in terms of orbitals – the transition is from one state (electron con[iguration) to another state (a different electron con[iguration). 4 Electronic Spectroscopy Selection Rules Electronic Spectroscopy Selection Rules Lets examine what can occur in UV- Vis spectroscopy (electronic transitions) for the hydrogen atom Lets examine what can occur in UV- Vis spectroscopy (electronic transitions) for the hydrogen atom You can excite an electron in to either an s or a p type orbital in its excited state You can excite an electron in to either an s or a p type orbital in its excited state ∫ψ orb i ˆj µψ orbdτ orb ∫ψ ispinψ spindτ spin j Integral must be ≠ 0! i.e. there must be a polarizing redistribution of electron density in going from ψi and a ψj. ∫ψ ? orb i ˆj µψ orbdτ orb ∫ψ ispinψ spindτ spin j Integral must be ≠ 0! i.e. there must be a polarizing redistribution of electron density in going from ψi and a ψj. 2p 2p 1s 1s Selection Rules 1s 1s 2s Δl = ±1 2s c1 (t )Ψ2 p (q, t ) + c2 (t )Ψ1s (q, t ) Combinations of Transitions Journal of Molecular Spectroscopy 72,p. 36- 43 1978 The extinction coef[icient of benzene vapor in the region 4.6 to 36 eV E. Pantos, J. Philis and A. Bolovinos also observed in electronic spectroscopy 2p à༎ 1s H H H 2s à༎ 1s H 263 nm = 38023 cm- 1 256 nm = 39062 cm- 1 39062 cm- 1 – 38023 cm- 1 ≈ 1000 cm- 1 H H Other vibrational transitions as well … 5 ...
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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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