5.04, Principles of Inorganic Chemistry II MIT Department of Chemistry Lecture 32: Vibrational Spectroscopy and the IR

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1 5.0, Prncples of Inorganc Chemstry II MIT Department of Chemstry Lecture 3: Vbratonal Spectroscopy and the IR Vbratonal spectroscopy s confned to the cm - spectral regon. The absorpton of a photon characterzes the nfrared technque; the scatterng of a photon characterzes the Raman technque. x E = E 0 cos vt z y Absorpton or scatterng of a photon produces a molecule n a hgher vbratonal state (.e., the molecule s vbratonal ampltude s ncreased). Classcal armonc scllator m CG m x x r r equlbrum dstance: r e = r + r non-equlbrum: r = r e + q (= x + x ) Conservaton of center of gravty (CG) requres: m r = m r m (r + x ) = m (r + x ) for system at equlbrum for dsplaced system Substtutng, x = x, x = x m m m m () 5.0, Prncples of Inorganc Chemstry II Lecture 3 Page of 0

2 Classcally, the chemcal bond s descrbed by a sprng that obeys ooke s law, where the restorng force, f, s expressed as f = -K (x + x ) = -Kq () ndcates restorng force and dsplacement oppose each other Substtuton of () nto () gves the ndvdual forces on each atom, f = ma m + m d x force on atom K = x = m =f (3) fx ( ) fx ( ) m m + m dt d x K = x = m f () = m dt force on atom The total force on the system s the sum of the ndvdual forces f = f + f m m d x d x f = K( x + x ) = + m + m dt dt f = Kq = d q m μ m where μ = dt m + m Solvng the above dfferental equaton gves, mm q = q 0 sn( 0 t + ) where 0 = (5) m + m The total energy, E tot, of the system s, E tot = KE + PE dx dx x μ dq KE = mv = m + m = dt dt dt from (5), dq = 0 q 0 cos( 0 t) + dt KE = 0 μq 0 cos ( 0 t + ) 5.0, Prncples of Inorganc Chemstry II Lecture 3 Page of 0

3 PE = Kq = Kq 0 sn ( 0 t + ) from (5), K = 0 μ PE = 0 μq 0 sn ( 0 t + ) The total energy of the system s therefore, E tot = 0 μq 0 [cos ( 0 t + ) + sn ( 0 t + )] = 0 μq 0 = constant cos a + sn a = The result summarzed graphcally n coordnate q, KE E tot = KE + PE = 0 μq 0 E PE KE PE PE -q o 0 +q 0 q dq turnng pont, = 0 dt PE = 0, q = 0 KE = 0 E tot = KE = 0 μq 0 E tot = PE = 0 μq 0 5.0, Prncples of Inorganc Chemstry II Lecture 3 Page 3 of 0

4 Quantum Mechancal armonc scllator vb = E vb vb + Vq ( ) = E μ q vb vb vb but q = r - r e. Expandng V(q) n a Taylor seres about r e V() r Vr () Vq ( ) = E + r r 0 ( e ) + ( r r e) +... r r but at r = r e, we are at the mnmum of PE and V r ()r = 0 r=re r=re V(r) r e V() r = 0 r Vr () Thus leadng term n the expanson s (r-r e ). If s slowly varyng r functon then can be approxmated as constant K, and V(r) becomes, Vr ()= Kr ( r e) same result as classcal treatment Substtutng nto Schrödnger s equaton and rearrangng, μ vb + E Kr vb ( r e ) = 0 r The soluton to ths amltonan s, h K K E = + = + wth = μ μ where = 0,,, n Note, when = 0, E 0 = energy dfference between and + and E = 5.0, Prncples of Inorganc Chemstry II Lecture 3 Page of 0

5 The graphcal summary of egenvalues s, E v=0 = 0.5 v=0 = <r/0> v=3 E v=0 v= v= r e r The egenfunctons shown above have the general form, = ( ) ( q ) = vb q μk μ e = v! h h where ( q ) s the ermte polynomal of the th degree. The frst four wavefunctons are: 0 = e q 3 for even, q (q) s even = qe for odd, (q) s odd = (q )e q 3 = 3 (q 3 3q)e q 9 5.0, Prncples of Inorganc Chemstry II Lecture 3 Page 5 of 0

6 Infrared Experment The nfrared experment s an absorpton experment where the ncdent radaton s absorbed at the frequences of dscrete, quantzed vbratonal levels. The harmonc potental works here because most nfrared spectra arse from transtons between = 0 and =. ( ) ( ) ( ) ( ) = = vb n n fundamental: ( 0) j () ( 0) overtone: ( 0) n j( ) ( 0) n > j combnaton: ( 0) j ( n ) k ( m) ( 0) n, m = 0,, j jk v= v=0 v=3 v= overtones fundamental Lke any absorpton experment, the ncdent radaton after passng through the sample s measured and referenced to an ncdent beam, I 0 IR source reference sample d e t e c t o r T = I s I r (= I 0 ) 5.0, Prncples of Inorganc Chemstry II Lecture 3 Page 6 of 0

7 Let s consder z z y z 3 y x y3 x x 3 Ths bass has translaton, rotaton and vbraton ncluded: 3N degrees of freedom 3N - 6 vbratonal degrees of freedom Can determne rreducble representaton of these degrees of freedom by smple applcaton of group theory C v xyz unmoved atoms tot E C (xz) (yz) a + a + b + 3b tot = trans + rot + vb xyz R x R y R z (3a + a + b + 3b ) = (a + b + b ) + (a + b + b ) + vb vb = a + b these are the symmetres of the normal modes of vbraton The probablty for the radatve, IR, transton s P t E0 ex μ gs t a b ()= where μ = (e x + e y + e z ) x coordnate of charge on th atom 5.0, Prncples of Inorganc Chemstry II Lecture 3 Page 7 of 0

8 so the rate of transton from gs to es s the probablty per unt tme, a b 0 ex gs es 0 gs R = E μ = μ E Ths ntegral s non zero ff ( vb es ) x μ x ( vb gs ) contans the totally symmetrc representaton. Snce ( q = 0 ) s a totally symmetrc functon, gs es ( vb ) a g ( vb ) ( q ), whch s the symmetry of the vbraton the transton s allowed f ( es vb ) x μ contans a g, whch wll only be the case f ( es vb ) = μ. a Therefore for the example vb b s allowed and z-polarzed, vb s allowed and y-polarzed. 5. Symmetry Coordnates (Vbratonal SALCs) Molecular vbratons are superpostons of the normal modes of vbraton, whch are dagonalzed or symmetry adapted lnear combnatons (SALCs) of nternal dsplacement coordnates. The symmetry coordnates may be projected from an arbrtrary set of nternal coordnates usng a projecton operator, P () order = * h [ x() R ] Rˆ R dmenson of the th rreducble representaton operator Character of symmetry operator Rˆ n Choosng three arbrtrary coordnates z r r x y 5.0, Prncples of Inorganc Chemstry II Lecture 3 Page 8 of 0

9 r r r r r C v E C v (xz) v (yz) r r r r r r 0 0 a + b a Thus, the a and b SALC s must be projected from the r bass and a SALC from. P a ( r = xe E r + xc C r + x xz xz r + x yz ) ( ) ( ) ( ) ( ) ( v ( )) v( )( ) v ( ) v ( yz)( r ) Normalzng, = [ r + r + r + r ] = ( r + r ) ( r + r ) c r + r r + r = j c = snce r r = 0 c = ( a ) = ( r r ) S + [ r r r P b ( r ) = r + ( ) r + ( ) r + r ] S ( b) = ( r ) For bass, = S 3 (a ) = Summarzng P a ( ) [ ] lkewse for these SALCs smply multplyng SALC by (-) S (a ) S (b ) S 3 (a ) 5.0, Prncples of Inorganc Chemstry II Lecture 3 Page 9 of 0

10 Stretchng Mode Analyss ften a partcular mode of vbraton occurs n a characterstc regon. Can determne or analyze these modes wth the approprate choce of bass sets. For example consder the C stretchng frequences n Mo(C) 6. C C C Mo C C C Typcally, the carbonyl stretch n transton metal complexes occurs n the cm - regon. Usng the above bass, we fnd stretch h E 8C 3 6C 6C 3C I 6S 8S 6 3 h 6 d Decomposng, we expect to see IR actve mode, stretch = a g + e g + t u IR actve By droppng n symmetry to Mn(C) 5 I, the story changes sgnfcantly C v stretch E 5 C C v 3 d a + b + e IR IR actve Now expect to see 3 bands n the C stretchng regon 5.0, Prncples of Inorganc Chemstry II Lecture 3 Page 0 of 0