5.80 Small-Molecule Spectroscopy and Dynamics MIT OpenCourseWare Fall 2008

MIT OpenCourseWare http://ocw.mit.edu 5.80 Small-Molecule Spectroscopy and Dynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 5.80 Lecture #27 Fall, 2008 Page 1 of 8 pages Lecture #27: Polyatomic Vibrations III: s-Vectors and H2O Last time: F-matrix: too many Fij’s even at quadratic-only level Internal coordinates: types 3N–6 independent ones constraints * translation * rotation * direction of fastest increase s -vectors ∇αSt ≡ s tα    * magnitude resulting from unit displacement in optimum direction N   St ( { ρα } ) = ∑ s tα • ρα α =1    rigid translation ρα = ε for all α ∑ s constraint =0 no center of mass translation      rigid rotation by dΩ ρα ( dΩ) = R α × dΩ    St ( dΩ) = dΩ • ∑ s tα × R α  α e s × R α =0 constraint ∑ tα ECKART α (minimizes vibrational angular momentum) If the normal displacements are built from s tα vectors that satisfy these constraints, then, for  infinitesimal displacements from equilibrium, there is no rotation. For large displacements, or for small displacements away from a non-equilibrium configuration, there is a small vibrational angular momentum. This definition of vibrations embeds a specific partitioning between rotation and vibration. tα α TODAY: G from s tα ’s  Examples of s tα ’s  1. 2. valence bond stretch ∆r valence angle bend ∆φ G matrix using diagrams and tables from WDC pages 304 and 305 H2O FG handout G ≡ DD† recall |S〉 = B|ξ〉 = D|q〉 = DΜ1/2|ξ〉 B = DM1/2 BM−1/2 = D G = DD† = BM−1/2 ( M−1/2 ) B† = BM−1B† † 5.80 Lecture #27 Fall, 2008 3N G tt′ = ∑ i=1 1 BtiB*t′i mi definition of |S〉 = B|ξ〉 → N =∑ α=1 N G tt′ = ∑ α=1 Page 2 of 8 pages ⎛ ∂St ⎞ ⎜ ⎟ ⎝ ∂ξ i ⎠0 ⎛ ∂St ′ ⎞ ⎜ ⎟ ⎝ ∂ξi ⎠0 1 (∇αSt )0 ·(∇αSt′ )0 mα 1 s tα ·s t′α m α   This way to derive G is convenient * locally defined s tα . Easy to compute s tα · s t′α .    * Each St involves small number of stα’s (only the involved atoms). * Small number of topological cases for internal displacements. All analyzed in WDC, pages 303-306. s -Vector Method. WDC pages 54-63.  * start with all atoms at equilibrium positions; * direction of s tα is direction of α’th atom must move to yield maximum increase in St;  * magnitude of s tα is increase in St that results from unit displacement of atom α in optimal  direction; * must verify or impose the 6 constraints (3 Cartesian components for the two vector constraint equations).  ∑ stα = 0 ∑ R αe × s tα = 0 α α 5.80 Lecture #27 Fall, 2008 Page 3 of 8 pages Several possible types of internal displacements. 1. 2. 3. 4. 1. Bond Stretch Valence angle bend  1A2 state of H2CO) defined by 2 bonds angle between a bond and a plane (non-planar A  1Au state of HCCH torsion → trans-bent excited A Bond Stretch St ≡ ∆r st1 1 st2 r12 2 only 2 nonzero s vectors (even in a long linear chain)!     Atom 1 unit displacement s t1 = 1 st1 = e21 = − e12    Atom 2 st 2 = 1 st 2 = −e21 = e12    St { ξα } = ê 21 ( ρ1 − ρ2 ) (displacements of all other atoms have no effect on ∆r12) These are the vector representations of S∆r. ( ) Are the constraints satisfied? ∑ s = s t1 + s t 2 = −ê12 + eˆ 12 = 0!   α    ∑ R αe × s tα = R1e × (−ê12 ) + R e2 × (ê12 ) α   = ( R e2 − R1e ) × ê12 tα 1 2  R1e  R e2 center of mass  e  e  e  R 2 – R1 = R12  e12  e  ∴ R12 × e12 = 0! 5.80 Lecture #27 2. Fall, 2008 Valence Angle Bend 3 ∆φ r31 φ Page 4 of 8 pages St ≡ ∆φ r32 st1 st2 1 2 Exactly 3 atoms are involved. 3 nonzero s tα ’s.  How to move each atom to increase φ by maximum amount? s tan∆ φ ≈ ∆ φ = t1 r31 How to define a UNIT VECTOR pointing in correct direction? ê × ê ê st1 = ê 31 × 31 32 Recall ê 31 × ê 32 = sin φ  sin φ right hand rule ⊥ to plane, up out of board Rules for vector triple product cos φ ê st1  ê st1  1 ê 31 ·ê 32 ) ê 31 − ( ê 31 ·ê 31 ) ê 32 ( = sin φ cos φê 31 − ê 32 = sin φ Now, how much does unit displacement of atom 1 in ê st1 direction increase St?  unity 1 tan ∆ St ≈ ∆ St = = = s t1 r31 r31 ∴ s t1 = s t1 ê st1 =    cos φê 31 − ê 31 r31 sin φ this is a vector of specified length and direction 5.80 Lecture #27 Fall, 2008 Page 5 of 8 pages similarly for atom 2 cos φê 32 − ê 31 st 2 =  r32 sin φ now for the hard one: atom 3! Easy way: impose constraint ∑ s tα = 0 α ∴ s t 3 = − (s t1 + s t 2 ) =    (r31 − r32 cos φ) ê 31 + (r32 − r31 cos φ) ê 32 r31r32 sin φ Hard way: move atom 3 1 unit in optimal direction, then translate deformed molecule rigidly to put atom 3 back at its original position. This evidently leaves atoms 1 and 2 displaced by s t1 and s t 2   respectively. move atom 3 3 2 1 translate distorted structure back to put atom 3 in original location s t2  s t1  This obviously satisfies s t 3 = − ( s t1 + s t 2 )    −s t 3  ∑ s tα =0 α  It is harder to show that it also satisfies 0 = ∑R αe × s tα .  Grind out the algebra! (see Non-Lecture on next page) 5.80 Lecture #27 Fall, 2008 Page 6 of 8 pages Alternative definition of S∆θ as a linear displacement rather than an angular displacement is possible. e.g. r31∆φ, r32∆φ, or (r31r32)1/2∆φ. Then S∆φ would have dimension of length and all bending force constants would have same units as stretching ones. The derivation of S∆φ would follow same path, but each s tα gets multiplied by the  relevant length factor, r31 or r32 or (r31r32)1/2. NON-LECTURE Proof that s tα ’s satisfy Eckart Condition  s t1 =  cos φê 31 − ê 32 r31 sin φ s t 2 =  s t 3 =  cos φê 32 − ê 31 r32 sin φ (r31 − r32 cos φ) ê 31 + (r32 − r31 cos φ) ê 32 r31r32 sin φ 0 ? R1e × s t1 + R e2 × s t 2 + R e3 × s t 3    R3 R1 R2      0 ? ( R 3 − R13 ) × s t1 + ( R 3 − R 23 ) × s t 2 + R 3 × s t 3       0 ? R 3 × (s t1 + s t 2 + s t 3 ) − ( R13 × s t1 + R 23 × s t 2 )       =0  R13 × eˆ 31 = 0  R 23 × sˆ 32 = 0  ⎛ −ê 32 ⎞  ⎛ −ê 31 ⎞ 0 ? R13 × ⎜ ⎟ − R 23 × ⎜ ⎟ ⎝ r32 sin φ ⎠ ⎝ r31 sin φ ⎠  R13 × ê 32 = r31ê13 × ê 32 = −r31ê 31 × ê 32  R 23 × ê 31 = r32 ê 23 × ê 31 QED    R 3 = R 2 + R 23    R 3 = R1 + R13 5.80 Lecture #27 Fall, 2008 Page 7 of 8 pages G-Matrix N 1 s tα × s t′α  mα  G tt′ = ∑ α=1 Could compute directly from s tα ’s, but easier to use diagrams from WDC page 304 and table on  WDC, page 305. DIAGRAMS 2 1 diagonal stretch-stretch an atom involved in both t and t′ # of atoms common to both t and t′ G 2rr stretch 2 1 off-diagonal stretch-adjacent stretch G′rr 3 3 2 1 diagonal bend-bend 3 G φφ 2 1 off-diagonal bend-internal stretch G 2rφ 3 etc. TABLE G 2rr = µ1 + µ 2 1 mα cφ ≡ cos φ µα ≡ G1rr = µ1cφ 3 2 2 2 G φφ = ρ12 µ1 + ρ223 µ 3 + (ρ12 + ρ23 − 2ρ12ρ23cφ ) µ 2 ρij ≡ ( rije ) sφ = sin φ G 2rφ = −ρ23 µ 2sφ 1 cis bent acetylene 4 3 2 1 2 3 4 5 6 −1 s ∆ r12 s ∆ r23 s ∆ r34 s ∆ φ123 s ∆ φ234 s ∆ τ bend 5.80 Lecture #27 Fall, 2008 Page 8 of 8 pages 6×7 G matrix has 2 = 21 independent elements G: 1 2 3 4 5 6 G 2 rr G 1 rr G 2rr 0 G 2 rφ ⎛1⎞ G ⎜ ⎟ G 2rτ ⎝2⎠ 1 rφ G1rr G 2 rr 3 G φφ 3 G φφ G See J. C. Decius Journal of Chemical Physics 16 1025 (1948)! for torsion and out of plane bend distortions! 4 ττ

5.80 Small-Molecule Spectroscopy and Dynamics MIT OpenCourseWare Fall 2008

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