103 Review Exercises. b.) B 2 1σ g 2 1σ u 2 2σ g 2 2σ u 2 1π u 1 2π u. c.) O 2 1σ g 2 1σ u 2 2σ g 2 2σ u 2 1π u 4 3σ g 2 1π g

Transcription

1 103 Review Exercises 1. For the given orbital occupations (configurations) of the following systems, determine all possible states (all possible allowed combinations of spin and space states). There is no need to form the determinental wavefunctions simply label each state with its proper term symbol. One method commonly used is arry Grays "box method" found in Electrons and Chemical Bonding. a.) C 1a 1 a 1 1b 3a 1 1 1b 1 1 b.) B 1σ g 1σ u σ g σ u 1π u 1 π u 1 c.) O 1σ g 1σ u σ g σ u 1π u 4 3σ g 1π g d.) Ti 1s s p 6 3s 3p 6 4s 3d 1 4d 1 e.) Ti 1s s p 6 3s 3p 6 4s 3d Exercises 1. Show that the configuration (determinant) corresponding to the Li + 1s(α)1s(α) state vanishes.. Construct the 3 triplet and 1 singlet wavefunctions for the Li + 1s 1 s 1 configuration. Show that each state is a proper eigenfunction of S and S z (use raising and lowering operators for S ) 3. Construct wavefunctions for each of the following states of C : a.) 1 B 1 (1a 1 a 1 1b 3a 1 1 1b 1 1 ) b.) 3 B 1 (1a 1 a 1 1b 3a 1 1 1b 1 1 ) c.) 1 A 1 (1a 1 a 1 1b 3a 1 ) 4. Construct wavefunctions for each state of the 1σ σ 3σ 1π configuration of N. 5. Construct wavefunctions for each state of the 1s 1 s 1 3s 1 configuration of Li. 6. Determine all term symbols that arise from the 1s s p 3d 1 configuration of the excited N atom. 7. Calculate the energy (using Slater Condon rules) associated with the p valence electrons for the following states of the C atom. i. 3 P(M L =1,M S =1), ii. 3 P(M L =0,M S =0), iii. 1 S(M L =0,M S =0), and iv. 1 D(M L =0,M S =0). 8. Calculate the energy (using Slater Condon rules) associated with the π valence electrons for the following states of the N molecule. i. 1 (M L =, M S =0), ii. 1 Σ (M L =0, M S =0), and iii. 3 Σ (M L =0, M S =0). Problems 1. Let us investigate the reactions: i. C ( 1 A 1 ) + C, and

2 104 ii. C ( 3 B 1 ) + C, under an assumed C v reaction pathway utilizing the following information: C atom: 3 9. kcal/mole P kcal/mole D 1 S C( 3 P) + C ( 3 B 1 ) E = kcal/mole C( 1 D) + C ( 1 A 1 ) E = kcal/mole IP ( ) > IP (s carbon). a. Write down (first in terms of p 1,0,-1 orbitals and then in terms of p x,y,z orbitals) the: i. three Slater determinant (SD) wavefunctions belonging to the 3 P state all of which have M S = 1, ii. five 1 D SD wavefunctions, and iii. one 1 S SD wavefunction. b. Using the coordinate system shown below, label the hydrogen orbitals σ g, σ u and the carbon s, p x, p y, p z, orbitals as a 1, b 1 (x), b (y), or a. Do the same for the σ, σ, σ *, σ *, n, and p π orbitals of C. y z x C c. Draw an orbital correlation diagram for the C + C reactions. Try to represent the relative energy orderings of the orbitals correctly. d. Draw (on graph paper) a configuration correlation diagram for C ( 3 B 1 ) + C showing all configurations which arise from the C( 3 P) + products. You can assume that doubly excited configurations lie much (~100 kcal/mole) above their parent configurations. e. Repeat step d. for C ( 1 A 1 ) + C again showing all configurations which arise from the C( 1 D) + products. f. Do you expect the reaction C( 3 P) + C to have a large activation barrier? About how large? What state of C is produced in this reaction? Would distortions away from C v symmetry be expected to raise of lower the activation barrier? Show how one could estimate where along the reaction path the barrier top occurs. g. Would C( 1 D) + C be expected to have a larger or smaller barrier than you found for the 3 P C reaction?. The decomposition of the ground-state singlet carbene,.., to produce acetylene and 1 D carbon is known to occur with an activation energy equal to the reaction endothermicity. owever, when triplet carbene decomposes to acetylene and ground-state (triplet) carbon, the activation energy exceeds this reaction's endothermicity.

3 105 Construct orbital, configuration, and state correlation diagrams which permit you to explain the above observations. Indicate whether single configuration or configuration interaction wavefunctions would be required to describe the above singlet and triplet decomposition processes. 3. We want to carry out a configuration interaction calculation on at R=1.40 au. A minimal basis consisting of normalized 1s Slater orbitals with ζ=1.0 gives rise to the following overlap (S), one-electron (h), and two-electron atomic integrals: < 1s A 1s B > = S, < 1s A h 1s A > = , < > 1s B h 1s A = , < 1s A 1s A h 1s A 1s A > = 0.65 < AA AA > < AA BB > = 0.33, < AB AB > = 0.504, and < AA AB > = a. The normalized and orthogonal molecular orbitals we will use for this minimal basis will be determined purely by symmetry: σ g = ( +S) -1 ( 1s A + 1s B ), and σ u = ( +S) -1 ( 1s A - 1s B ). Show that these orbitals are indeed orthogonal. b. Evaluate (using the one- and two- electron atomic integrals given above) the unique one- and two- electron integrals over this molecular orbital basis (this is called a transformation from the ao to the mo basis). For example, evaluate < u h u >, < uu uu >, < gu gu >, etc. c. Using the two 1 Σ g + configurations σg, and σ u, show that the elements of the x configuration interaction amiltonian matrix are , 0.140, and d. Using this configuration interaction matrix, find the configuration interaction (CI) approximation to the ground and excited state energies and wavefunctions. e. Evaluate and make a rough sketch of the polarized orbitals which result from the above ground state σ g and σ u CI wavefunction. Review Exercises Solutions 1. a. For non-degenerate point groups one can simply multiply the representations (since only one representation will be obtained): a 1 b 1 = b 1 Constructing a "box" in this case is unnecessary since it would only contain a single row. Two unpaired electrons will result in a singlet (S=0, M S =0), and three triplets (S=1, M S =1; S=1, M S =0; S=1, M S =-1). The states will be: 3 B 1 (M S =1), 3 B 1 (M S =0), 3 B 1 (M S =- 1), and 1 B 1 (M S =0). 1. b. Remember that when coupling non-equivalent linear molecule angular momenta, one simple adds the individual L z values and vector couples the electron spin. So, in this case (1π u 1 π u 1 ), we have M L values of 1+1, 1-1, -1+1, and -1-1 (, 0, 0, and -). The

4 106 term symbol is used to denote the spatially doubly degenerate level (M L =±) and there are two distinct spatially non-degenerate levels denoted by the term symbol Σ (M L =0) Again, two unpaired electrons will result in a singlet (S=0, M S =0), and three triplets (S=1, M S =1;S=1, M S =0;S=1, M S =-1). The states generated are then: 1 (M L =); one state (M S =0), 1 (M L =-); one state (M S =0), 3 (M L =); three states (M S =1,0, and -1), 3 (M L =-); three states (M S =1,0, and -1), 1 Σ (M L =0); one state (M S =0), 1 Σ (M L =0); one state (M S =0), 3 Σ (M L =0); three states (M S =1,0, and -1), and 3 Σ (M L =0); three states (M S =1,0, and -1). 1. c. Constructing the "box" for two equivalent π electrons one obtains: M L 1 0 M S 1 π 1 απ -1 α 0 π 1 απ 1 β π 1 απ -1 β, π -1 απ 1 β From this "box" one obtains six states: 1 (M L =); one state (M S =0), 1 (M L =-); one state (M S =0), 1 Σ (M L =0); one state (M S =0), 3 Σ (M L =0); three states (M S =1,0, and -1). 1. d. It is not necessary to construct a "box" when coupling non-equivalent angular momenta since the vector coupling results in a range from the sum of the two individual angular momenta to the absolute value of their difference. In this case, 3d 1 4d 1, L=4, 3,, 1, 0, and S=1,0. The term symbols are: 3 G, 1 G, 3 F, 1 F, 3 D, 1 D, 3 P, 1 P, 3 S, and 1 S. The L and S angular momenta can be vector coupled to produce further splitting into levels: J = L + S... L - S. Denoting J as a term symbol subscript one can identify all the levels and subsequent (J + 1) states: 3 G 5 (11 states), 3 G 4 (9 states), 3 G 3 (7 states), 1 G 4 (9 states), 3 F 4 (9 states), 3 F 3 (7 states),

5 107 3 F (5 states), 1 F 3 (7 states), 3 D 3 (7 states), 3 D (5 states), 3 D 1 (3 states), 1 D (5 states), 3 P (5 states), 3 P 1 (3 states), 3 P 0 (1 state), 1 P 1 (3 states), 3 S 1 (3 states), and 1 S 0 (1 state). 1. e. Construction of a "box" for the two equivalent d electrons generates (note the "box" has been turned side ways for convenience): M S 1 0 M L 4 d αd β 3 d αd 1 α d αd 1 β, d βd 1 α d αd 0 α d αd 0 β, d βd 0 α, d 1 αd 1 β 1 d 1 αd 0 α, d αd -1 α d 1 αd 0 β, d 1 βd 0 α, d αd -1 β, d βd -1 α 0 d αd - α, d 1 αd -1 α d αd - β, d βd - α, d 1 αd -1 β, d 1 βd -1 α, d 0 αd 0 β The term symbols are: 1 G, 3 F, 1 D, 3 P, and 1 S. The L and S angular momenta can be vector coupled to produce further splitting into levels: 1 G 4 (9 states), 3 F 4 (9 states), 3 F 3 (7 states), 3 F (5 states), 1 D (5 states), 3 P (5 states),

6 108 3 P 1 (3 states), 3 P 0 (1 state), and 1 S 0 (1 state). Exercises 1. Constructing the Slater determinant corresponding to the "state" 1s(α)1s(α) with the rows labeling the orbitals and the columns labeling the electron gives: 1sα1sα = 1! 1sα(1) 1sα() 1sα(1) 1sα() = 1 ( 1sα(1)1sα() - 1sα(1)1sα() ) = 0. Starting with the M S =1 3 S state (which in a "box" for this M L =0, M S =1 case would contain only one product function; 1sαsα ) and applying S - gives: S - 3 S(S=1,M S =1) = 1(1 + 1) - 1(1-1) h 3 S(S=1,M S =0) = h 3 S(S=1,M S =0) = ( S - (1) + S - () ) 1sαsα = S - (1) 1sαsα + S - () 1sαsα = h + h sβsα sαsβ = h ( 1sβsα + 1sαsβ ) So, h 3 S(S=1,M S =0) = h ( 1sβsα + 1sαsβ ) 3 S(S=1,M S =0) = 1 ( 1sβsα + 1sαsβ ) The three triplet states are then: 3 S(S=1,M S =1)= 1sαsα, 3 S(S=1,M S =0) = 1 ( 1sβsα + 1sαsβ ), and 3 S(S=1,M S =-1) = 1sβsβ. The singlet state which must be constructed orthogonal to the three singlet states (and in particular to the 3 S(S=1,M S =0) state) can be seen to be: 1 S(S=0,M S =0) = 1 ( 1sβsα - 1sαsβ ).

7 109 Applying S and S z to each of these states gives: S z 1sαsα = ( S z (1) + S z () ) 1sαsα = S z (1) 1sαsα + S z ()) 1sαsα = h 1 1sαsα + h 1 1sαsα S 1sαsα = h 1sαsα = (S - S + + S z + h S z ) 1sαsα = S - S + 1sαsα + S z 1sαsα + h S z 1sαsα = 0 + h 1sαsα + h 1sαsα = h 1sαsα S 1 z ( 1sβsα + 1sαsβ ) = ( S z (1) + S z () 1 ) ( 1sβsα + 1sαsβ ) = 1 ( S z (1) + S z () ) 1sβsα + 1 ( S z (1) + S z () ) 1sαsβ = 1 h - 1 1sβsα + 1 h 1 + h - 1sαsβ = 0h 1 ( 1sβsα + 1sαsβ ) S 1 ( 1sβsα + 1sαsβ ) = (S - S + + S z + h S z ) 1 ( 1sβsα + 1sαsβ ) = S - S 1 + ( 1sβsα + 1sαsβ ) = 1 ( S - (S + (1) + S + ()) 1sβsα + S - (S + (1) + S + ()) 1sαsβ ) = 1 S - h 1sαsα + S - h 1sαsα = h 1 ( (S - (1) + S - ()) 1sαsα ) = h 1 h 1sβsα + h 1sαsβ = h 1 ( 1sβsα + 1sαsβ ) S z 1sβsβ = ( S z (1) + S z () ) 1sβsβ = S z (1) 1sβsβ + S z ()) 1sβsβ = h - 1 1sβsβ + h - 1 1sβsβ

8 110 S 1sβsβ = -h 1sβsβ = (S + S - + S z - h S z ) 1sβsβ = S + S - 1sβsβ + S z 1sβsβ - h S z 1sβsβ = 0 + h 1sβsβ + h 1sβsβ = h 1sβsβ S 1 z ( 1sβsα - 1sαsβ ) = ( S z (1) + S z () 1 ) ( 1sβsα - 1sαsβ ) = 1 ( S z (1) + S z () ) 1sβsα - 1 ( S z (1) + S z () ) 1sαsβ = 1 h - 1 1sβsα - 1 h 1 + h - 1sαsβ = 0h 1 ( 1sβsα - 1sαsβ ) S 1 ( 1sβsα - 1sαsβ ) = (S - S + + S z + h S z ) 1 ( 1sβsα - 1sαsβ ) = S - S 1 + ( 1sβsα - 1sαsβ ) = 1 ( S - (S + (1) + S + ()) 1sβsα - S - (S + (1) + S + ()) 1sαsβ ) = 1 S - h 1sαsα - S - h 1sαsα = 0 h 1 ( (S - (1) + S - ()) 1sαsα ) = 0 h 1 h 1sβsα - h 1sαsβ = 0 h 1 ( 1sβsα - 1sαsβ ) 3. a. Once the spatial symmetry has been determined by multiplication of the irreducible representations, the spin coupling is identical to exercise and gives the result: 1 ( 3a 1 α1b 1 β - 3a 1 β1b 1 α ) 3. b. There are three states here (again analogous to exercise ): 1.) 3a 1 α1b 1 α,.) 1 ( 3a 1 α1b 1 β + 3a 1 β1b 1 α ), and 3.) 3a 1 β1b 1 β 3. c. 3a 1 α3a 1 β 4. As shown in review exercise 1c, for two equivalent π electrons one obtains six states:

9 111 1 (M L =); one state (M S =0), 1 (M L =-); one state (M S =0), 1 Σ (M L =0); one state (M S =0), and 3 Σ (M L =0); three states (M S =1,0, and -1). By inspecting the "box" in review exercise 1c, it should be fairly straightforward to write down the wavefunctions for each of these: 1 (M L =); π 1 απ 1 β 1 (M L =-); π -1 απ -1 β 1 Σ (M L =0); 1 ( π 1 βπ -1 α - π 1 απ -1 β ) 3 Σ (M L =0, M S =1); π 1 απ -1 α 3 Σ (M L =0, M S =0); 1 ( π 1 βπ -1 α + π 1 απ -1 β ) 3 Σ (M L =0, M S =-1); π 1 βπ -1 β 5. We can conveniently couple another s electron to the states generated from the 1s 1 s 1 configuration in exercise : 3 S(L=0, S=1) with 3s 1 (L=0, S= 1 ) giving: L=0, S= 3, 1 ; 4 S (4 states) and S ( states). 1 S(L=0, S=0) with 3s 1 (L=0, S= 1 ) giving: L=0, S= 1 ; S ( states). Constructing a "box" for this case would yield: M L 0 M S 3 1 1sαsα3sα 1sαsα3sβ, 1sαsβ3sα, 1sβsα3sα One can immediately identify the wavefunctions for two of the quartets (they are single entries): 4 S(S= 3,M S= 3 ): 1sαsα3sα 4 S(S= 3,M S=- 3 ): 1sβsβ3sβ Applying S - to 4 S(S= 3,M S= 3 ) yields:

10 11 S - 4 S(S= 3,M S= 3 ) = h 3 (3 + 1) - 3 (3-1) 4 S(S= 3,M S= 1 ) = h 3 4 S(S= 3,M S= 1 ) S - 1sαsα3sα = h ( 1sβsα3sα + 1sαsβ3sα + 1sαsα3sβ ) So, 4 S(S= 3,M S= 1 ) = 1 3 ( 1sβsα3sα + 1sαsβ3sα + 1sαsα3sβ ) Applying S + to 4 S(S= 3,M S=- 3 ) yields: S + 4 S(S= 3,M S=- 3 ) = h 3 (3 + 1) - -3 (-3 + 1) 4 S(S= 3,M S=- 1 ) = h 3 4 S(S= 3,M S=- 1 ) S + 1sβsβ3sβ = h ( 1sαsβ3sβ + 1sβsα3sβ + 1sβsβ3sα ) So, 4 S(S= 3,M S=- 1 ) = 1 3 ( 1sαsβ3sβ + 1sβsα3sβ + 1sβsβ3sα ) It only remains to construct the doublet states which are orthogonal to these quartet states. Recall that the orthogonal combinations for systems having three equal components (for example when symmetry adapting the 3 sp hybrids in C v or D 3h symmetry) give results of + + +, + - -, and Notice that the quartets are the combinations and therefore the doublets can be recognized as: S(S= 1,M S= 1 ) = 1 6 ( 1sβsα3sα + 1sαsβ3sα - 1sαsα3sβ ) S(S= 1,M S= 1 ) = 1 ( 1sβsα3sα - 1sαsβ3sα + 0 1sαsα3sβ ) S(S= 1,M S=- 1 ) = 1 6 ( 1sαsβ3sβ + 1sβsα3sβ - 1sβsβ3sα ) S(S= 1,M S=- 1 ) = 1 3 ( 1sαsβ3sβ - 1sβsα3sβ + 0 1sβsβ3sα ) 6. As illustrated in this chapter a p configuration (two equivalent p electrons) gives rise to the term symbols: 3 P, 1 D, and 1 S. Coupling an additional electron (3d 1 ) to this p configuration will give the desired 1s s p 3d 1 term symbols: 3 P(L=1,S=1) with D(L=,S= 1 ) generates; L=3,,1, and S= 3, 1 with term symbols 4 F, F, 4 D, D, 4 P, and P, 1 D(L=,S=0) with D(L=,S= 1 ) generates; L=4,3,,1,0, and S= 1 with term symbols G, F, D, P, and S, 1 S(L=0,S=0) with D(L=,S= 1 ) generates; L= and S= 1 with term symbol D. 7. The notation used for the Slater Condon rules will be the same as used in the text: (a.) zero (spin orbital) difference;

11 113 < F + G > = i < φ i f φ i > + i>j = f ii + i i>j ( g ijij - g ijji ) (b.) one (spin orbital) difference (φ p φ p' ); < > φ p f φ p' + j p;p' = f pp' + ( g pjp'j - g pjjp' ) j p;p' (c.) two (spin orbital) differences (φ p φ p' and φ q φ q' ); F + G = < > F + G = < φ p φ q g φ p' φ q' > - < φ p φ q g φ q' φ p' > < > = g pqp'q' - g pqq'p' (d.) three or more (spin orbital) differences; < F + G > = 0 7. i. 3 P(M L =1,M S =1) = p 1 αp 0 α < p 1 αp 0 α p 1 αp 0 α > = Error!. Using the Slater Condon rule (a.) above (SCa): < > = f 11 + f 00 + g g ii. 3 P(M L =0,M S =0) = 1 ( p 1 αp -1 β + p 1 βp -1 α ) < φ i φ j g φ i φ j > - < φ i φ j g φ j φ i > < φ p φ j g φ p' φ j > - < φ p φ j g φ j φ p' > < 3 P(M L =0,M S =0) 3 P(M L =0,M S =0) > = 1 ( < p 1 αp -1 β p 1 αp -1 β > + < p 1 αp -1 β p 1 βp -1 α > + < p 1 βp -1 α p 1 αp -1 β > + < > Evaluating each matrix element gives: < p 1 αp -1 β p 1 αp -1 β > = f 1α1α + f -1β-1β + g 1α-1β1α-1β - g 1α-1β-1β1α (SCa) = f 11 + f g < p 1 αp -1 β p 1 βp -1 α > = g 1α-1β1β-1α - g 1α-1β-1α1β (SCc) = 0 - g < p 1 βp -1 α p 1 αp -1 β > = g 1β-1α1α-1β - g 1β-1α-1β1α (SCc) = 0 - g p 1 βp -1 α p 1 βp -1 α ) < p 1 βp -1 α p 1 βp -1 α > = f 1β1β + f -1α-1α + g 1β-1α1β-1α - g 1β-1α-1α1β (SCa) = f 11 + f g Substitution of these expressions give: < 3 P(M L =0,M S =0) 3 P(M L =0,M S =0) > = 1 (f 11 + f g g g f 11 + f g )

12 114 = f 11 + f g g iii. 1 S(M L =0,M S =0); 1 3 ( p 0αp 0 β - p 1 αp -1 β - p -1 αp 1 β ) < 1 S(M L =0,M S =0) 1 S(M L =0,M S =0) > = 1 3 ( < p 0 αp 0 β p 0 αp 0 β > - < p 0 αp 0 β p 1 αp -1 β > - < p 0 αp 0 β p -1 αp 1 β > - < p 1 αp -1 β p 0 αp 0 β > + < p 1 αp -1 β p 1 αp -1 β > + < p 1 αp -1 β p -1 αp 1 β > - < p -1 αp 1 β p 0 αp 0 β > + < p -1 αp 1 β p 1 αp -1 β > + < p -1 αp 1 β p -1 αp 1 β > ) Evaluating each matrix element gives: < p 0 αp 0 β p 0 αp 0 β > = f 0α0α + f 0β0β + g 0α0β0α0β - g 0α0β0β0α (SCa) = f 00 + f 00 + g < p 0 αp 0 β p 1 αp -1 β > = < > p 1 αp -1 β p 0 αp 0 β = g 0α0β1α-1β - g 0α0β-1β1α (SCc) = g < p 0 αp 0 β p -1 αp 1 β > = < > p -1 αp 1 β p 0 αp 0 β = g 0α0β 1α1β - g 0α0β1β 1α (SCc) = g < p 1 αp -1 β p 1 αp -1 β > = f 1α1α + f -1β-1β + g 1α-1β1α-1β - g 1α-1β-1β1α (SCa) = f 11 + f g < p 1 αp -1 β p -1 αp 1 β > = < > p -1 αp 1 β p 1 αp -1 β = g 1α-1β-1α1β - g 1α-1β1β-1α (SCc) = g < p -1 αp 1 β p -1 αp 1 β > = f -1α 1α + f 1β1β + g -1α1β 1α1β - g -1α1β1β 1α (SCa) = f f 11 + g Substitution of these expressions give: < 1 S(M L =0,M S =0) 1 S(M L =0,M S =0) > = 1 3 (f 00 + f 00 + g g g g f 11 + f g g g g f f 11 + g ) = 1 3 (f 00 + f 11 + f g g g g ) 7. iv. 1 D(M L =0,M S =0) = 1 6 ( p 0 αp 0 β + p 1 αp -1 β + p -1 αp 1 β ) Evaluating < > 1 D(M L =0,M S =0) 1 D(M L =0,M S =0) we note that all the Slater Condon matrix elements generated are the same as those evaluated in part iii. (the signs for the

13 115 wavefunction components and the multiplicative factor of two for one of the components, however, are different). < 1 D(M L =0,M S =0) 1 D(M L =0,M S =0) > = 1 6 (4f f g g g g f 11 + f g g g g f f 11 + g ) = 1 6 (8f 00 + f 11 + f g g g g ) 8. i. 1 (M L =,M S =0) = π 1 απ 1 β < 1 (M L =,M S =0) 1 (M L =,M S =0) > = < π 1 απ 1 β π 1 απ 1 β > = f 1α1α + f 1β1β + g 1α1β1α1β - g 1α1β1β1α (SCa) = f 11 + f 11 + g = f 11 + g ii. 1 Σ(M L =0,M S =0) = 1 ( π 1 απ -1 β - π 1 βπ -1 α ) < 3 Σ(M L =0,M S =0) 3 Σ(M L =0,M S =0) > = 1 ( < π 1 απ -1 β π 1 απ -1 β > - < π 1 απ -1 β π 1 βπ -1 α > - < π 1 βπ -1 α π 1 απ -1 β > + < > Evaluating each matrix element gives: π 1 βπ -1 α π 1 βπ -1 α ) < π 1 απ -1 β π 1 απ -1 β > = f 1α1α + f -1β-1β + g 1α-1β1α-1β - g 1α-1β-1β1α (SCa) = f 11 + f g < π 1 απ -1 β π 1 βπ -1 α > = g 1α-1β1β-1α - g 1α-1β-1α1β (SCc) = 0 - g < π 1 βπ -1 α π 1 απ -1 β > = g 1β-1α1α-1β - g 1β-1α-1β1α (SCc) = 0 - g < π 1 βπ -1 α π 1 βπ -1 α > = f 1β1β + f -1α-1α + g 1β-1α1β-1α - g 1β-1α-1α1β (SCa) = f 11 + f g Substitution of these expressions give: < 3 Σ(M L =0,M S =0) 3 Σ(M L =0,M S =0) > = 1 (f 11 + f g g g f 11 + f g ) = f 11 + f g g iii. 3 Σ(M L =0,M S =0) = 1 ( π 1 απ -1 β + π 1 βπ -1 α ) < 3 Σ(M L =0,M S =0) 3 Σ(M L =0,M S =0) > = 1 ( < π 1 απ -1 β π 1 απ -1 β > + < π 1 απ -1 β π 1 βπ -1 α >

14 116 + < π 1 βπ -1 α π 1 απ -1 β > + < > Evaluating each matrix element gives: π 1 βπ -1 α π 1 βπ -1 α ) < π 1 απ -1 β π 1 απ -1 β > = f 1α1α + f -1β-1β + g 1α-1β1α-1β - g 1α-1β-1β1α (SCa) = f 11 + f g < π 1 απ -1 β π 1 βπ -1 α > = g 1α-1β1β-1α - g 1α-1β-1α1β (SCc) = 0 - g < π 1 βπ -1 α π 1 απ -1 β > = g 1β-1α1α-1β - g 1β-1α-1β1α (SCc) = 0 - g < π 1 βπ -1 α π 1 βπ -1 α > = f 1β1β + f -1α-1α + g 1β-1α1β-1α - g 1β-1α-1α1β (SCa) = f 11 + f g Substitution of these expressions give: < 3 Σ(M L =0,M S =0) 3 Σ(M L =0,M S =0) > Problems = 1 (f 11 + f g g g f 11 + f g ) = f 11 + f g g a. All the Slater determinants have in common the 1sα1sβsαsβ "core" and hence this component will not be written out explicitly for each case. 3 P(M L =1,M S =1) = p 1 αp 0 α 3 P(M L =0,M S =1) = p 1 αp -1 α = 1 (p x + ip y ) α(p z )α = 1 ( p x αp z α + i p y αp z α ) = 1 (p x + ip y ) α 1 (p x - ip y ) α = 1 ( p x αp x α - i p x αp y α + i p y αp x α + p y αp y α ) = 1 ( 0 - i p x αp y α - i p x αp y α + 0 ) = 1 ( -i p x αp y α ) = -i p x αp y α 3 P(M L =-1,M S =1) = p -1 αp 0 α = 1 (p x - ip y ) α(p z )α = 1 ( p x αp z α - i p y αp z α ) As you can see, the symmetries of each of these states cannot be labeled with a single irreducible representation of the C v point group. For example, p x αp z α is xz (B 1 ) and

15 117 p y αp z α is yz (B ) and hence the 3 P(M L =1,M S =1) state is a combination of B 1 and B symmetries. But, the three 3 P(M L,M S =1) functions are degenerate for the C atom and any combination of these three functions would also be degenerate. Therefore we can choose new combinations which can be labeled with "pure" C v point group labels. 3 P(xz,M S =1) = p x αp z α 3 P(yx,M S =1) = p y αp x α = 1 ( 3 P(M L =1,M S =1) + 3 P(M L =-1,M S =1) ) = 3 B 1 = 1 i ( 3 P(M L =0,M S =1) ) = 3 A 3 P(yz,M S =1) = p y αp z α = 1 i ( 3 P(M L =1,M S =1) - 3 P(M L =-1,M S =1) ) = 3 B Now we can do likewise for the five degenerate 1 D states: 1 D(M L =,M S =0) = p 1 αp 1 β 1 D(M L =-,M S =0) = p -1 αp -1 β = 1 (p x + ip y ) α 1 (p x + ip y ) β = 1 ( p x αp x β + i p x αp y β + i p y αp x β - p y αp y β ) = 1 (p x - ip y ) α 1 (p x - ip y ) β = 1 ( p x αp x β - i p x αp y β - i p y αp x β - p y αp y β ) 1 D(M L =1,M S =0) = 1 ( p 0 αp 1 β - p 0 βp 1 α ) = 1 (p z )α 1 (p x + ip y )β - (p z )β 1 (p x + ip y )α = 1 ( p z αp x β + i p z αp y β - p z βp x α - i p z βp y α ) 1 D(M L =-1,M S =0) = 1 ( p 0 αp -1 β - p 0 βp -1 α ) = 1 (p z )α 1 (p x - ip y )β - (p z )β 1 (p x - ip y )α = 1 ( p z αp x β - i p z αp y β - p z βp x α + i p z βp y α ) 1 D(M L =0,M S =0) = 1 6 ( p 0 αp 0 β + p 1 αp -1 β + p -1 αp 1 β ) = 1 6 ( p zαp z β + 1 (p x + ip y )α 1 (p x - ip y )β + 1 (p x - ip y ) α 1 (p x + ip y ) β ) = 1 6 ( p zαp z β

16 ( p x αp x β - i p x αp y β + i p y αp x β + p y αp y β ) + 1 ( p x αp x β + i p x αp y β - i p y αp x β + p y αp y β ) ) = 1 6 ( p z αp z β + p x αp x β + p y αp y β ) ) Analogous to the three 3 P states we can also choose combinations of the five degenerate 1 D states which can be labeled with "pure" C v point group labels: 1 D(xx-yy,M S =0) = p x αp x β - p y αp y β = ( 1 D(M L =,M S =0) + 1 D(M L =-,M S =0) ) = 1 A 1 1 D(yx,M S =0) = p x αp y β + p y αp x β = 1 i ( 1 D(M L =,M S =0) - 1 D(M L =-,M S =0) ) = 1 A 1 D(zx,M S =0) = p z αp x β - p z βp x α = ( 1 D(M L =1,M S =0) + 1 D(M L =-1,M S =0) ) = 1 B 1 1 D(zy,M S =0) = p z αp y β - p z βp y α = 1 i ( 1 D(M L =1,M S =0) - 1 D(M L =-1,M S =0) ) = 1 B 1 D(zz+xx+yy,M S =0) = 1 6 ( p z αp z β + p x αp x β + p y αp y β ) ) = 1 D(M L =0,M S =0) = 1 A 1 The only state left is the 1 S: 1 S(M L =0,M S =0) = 1 3 ( p 0 αp 0 β - p 1 αp -1 β - p -1 αp 1 β ) = 1 3 ( p zαp z β - 1 (p x + ip y )α 1 (p x - ip y )β - 1 (p x - ip y ) α 1 (p x + ip y ) β ) = 1 3 ( p zαp z β - 1 ( p x αp x β - i p x αp y β + i p y αp x β + p y αp y β ) - 1 ( p x αp x β + i p x αp y β - i p y αp x β + p y αp y β ) ) = 1 3 ( p z αp z β - p x αp x β - p y αp y β ) ) Each of the components of this state are A 1 and hence this state has A 1 symmetry. 1. b. Forming SALC-AOs from the C and atomic orbitals would generate the following:

17 119 C C 1s + 1s = σ g = a 1 1s - 1s = σ u = b C C C C C s = a 1 C p = a 1 C p = b C p = b 1 z The bonding, nonbonding, and antibonding orbitals of C can be illustrated in the following manner: y x C C C C σ = a 1 σ = b n = a 1 p π = b 1 C C σ* = a 1 σ* = b 1. c.

18 10 Orbital-correlation diagram for the reaction C > C (bent) σ u (b ) b (antibonding) a 1 (antibonding) p x (b 1 ) p y (b ) p z (a 1 ) s(a 1 ) σ g (a 1 ) b 1 (p π ) a 1 (non-bonding) b (bonding) a 1 (bonding) C + C (bent) 1. d. - e. It is necessary to determine how the wavefunctions found in part a. correlate with states of the C molecule: 3 P(xz,M S =1); 3 B 1 = σ g s p x p z σ n p π σ* 3 P(yx,M S =1); 3 A = σ g s p x p y σ n p π σ 3 P(yz,M S =1); 3 B = σ g s p y p z σ n σσ* 1 D(xx-yy,M S =0); 1 A 1 σ n p π - σ n σ 1 D(yx,M S =0); 1 A σ n σp π 1 D(zx,M S =0); 1 B 1 σ n σ*p π 1 D(zy,M S =0); 1 B σ n σ*σ 1 D(zz+xx+yy,M S =0); 1 A 1 σ n σ* + σ n p π + σ n σ Note, the C + state to which the lowest 1 A 1 (σ n σ ) C state decomposes would be σ g s p y. This state (σ g s p y ) cannot be obtained by a simple combination of the 1 D states. In order to obtain pure σ g s p y it is necessary to combine 1 S with 1 D. For example, σ g s p y = 1 6 ( 6 1 D(0,0) S(0,0) ) - 1 ( 1 D(,0) + 1 D(-,0) ). This indicates that a CCD must be drawn with a barrier near the 1 D asymptote to represent the fact that 1 A 1 C correlates with a mixture of 1 D and 1 S carbon plus hydrogen. The C + state to which the lowest 3 B 1 (σ nσ p π ) C state decomposes would be σ g sp y p x.

19 11 3 B1 (σ n σ p π ) 3 B1 σ g spy px C( 1 D) + 3 B1 3 B1 ( 1 B 1 A1 1 A1 1 A 1 B1 ) 3 B (σ σn σ ) 9. Kcal/mole 3 B C( 3 P) + ( 3 B 1 3 B 3 A ) 3 A 3 A (σ σn p π ) 78.8 Kcal/mole 1 A1 1 A1 (σ σ n ) 97.0 Kcal/mole 3 B1 (σ σ np π ) 1. f. If you follow the 3 B 1 component of the C( 3 P) + (since it leads to the groundstate products) to 3 B 1 C you must go over an approximately 0 Kcal/mole barrier. Of course this path produces 3 B 1 C product. Distortions away from C v symmetry, for example to C s symmetry, would make the a 1 and b orbitals identical in symmetry (a'). The b 1 orbitals would maintain their identity going to a'' symmetry. Thus 3 B 1 and 3 A (both 3 A'' in C s symmetry and odd under reflection through the molecular plane) can mix. The system could thus follow the 3 A component of the C( 3 P) + surface to the place

20 1 (marked with a circle on the CCD) where it crosses the 3 B 1 surface upon which it then moves and continues to products. As a result, the barrier would be lowered. You can estimate when the barrier occurs (late or early) using thermodynamic information for the reaction (i.e. slopes and asymptotic energies). For example, an early barrier would be obtained for a reaction with the characteristics: Energy Progress of Reaction and a late barrier would be obtained for a reaction with the characteristics: Energy Progress of Reaction This relation between reaction endothermicity or exothermicity is known as the ammond postulate. Note that the C( 3 P 1 ) + --> C reaction of interest here (see the CCD) has an early barrier. 1. g. The reaction C( 1 D) + ---> C ( 1 A 1 ) should have no symmetry barrier (this can be recognized by following the 1 A 1 (C( 1 D) + ) reactants down to the 1 A 1 (C ) products on the CCD).. This problem in many respects is analogous to problem 1. The 3 B 1 surface certainly requires a two configuration CI wavefunction; the σ σ np x (π p y sp x ) and the σ n p x σ* (π s p x p z ). The 1 A 1 surface could use the σ σ n (π s p y ) only but once again there is no combination of 1 D determinants which gives purely this configuration (π s p y ). Thus mixing of both 1 D and 1 S determinants are necessary to

21 13 yield the required π s p y configuration. ence even the 1 A 1 surface would require a multiconfigurational wavefunction for adequate description. C C + C y C: x z Orbital-correlation diagram for the reaction C + C -----> C 3 π*(b ) b (antibonding) a 1 (antibonding) σ CC σ CC p x (b 1 ) p y (b ) p z (a 1 ) s(a 1 ) b 1 (p π ) a 1 (non-bonding) n π(a 1 ) b (bonding) a 1 (bonding) σ CC σ CC C + C C 3 Configuration correlation diagram for the reaction C + C ---> C 3.

22 14 π sp y px 3 B1 3 B1 (σ n σ p π ) 3 B (σ σn σ ) C( 1 D) + C 3 A (σ σn p π ) π s p y 1 A1 E a 3 B 1 C( 3 P) + C π s 3 p x p z B1 π s 3 p x p y A π s 3 p y p z B E a > E (for 3 B 1 ) E 3 B 1 3 B1 (σ σ np π ) E a = E (for 1 A 1 ) 1 A1 (σ σ n ) 3. a. < σ g σ g > = < ( +S) -1 ( 1s A + 1s B ) ( +S) -1 ( ) = ( ) 1s A + 1s B > +S -1 (< 1s A 1s A > + < 1s A 1s B > + < 1s B 1s A > + < > = (0.85)((1.000) + (0.753) + (0.753) + (1.000)) = < σ g σ u > = < ( +S) -1 ( 1s A + 1s B ) ( -S) -1 ( 1s A - 1s B ) > 1s B 1s B ) = ( +S) -1 ( -S) -1 (< 1s A 1s A > + < 1s A 1s B > + < 1s B 1s A > + < > = (1.43)(0.534)((1.000) - (0.753) + (0.753) - (1.000)) = 0 < σ u σ u > = < ( -S) -1 ( 1s A - 1s B ) ( -S) -1 ( ) = ( ) 1s A - 1s B > 1s B 1s B ) -S -1 (< 1s A 1s A > - < 1s A 1s B > - < 1s B 1s A > + < > = (.04)((1.000) - (0.753) - (0.753) + (1.000)) = s B 1s B )

23 15 3. b. < σ g h σ g > = < ( +S) -1 ( 1s A + 1s B ) h ( +S) -1 ( ) = ( +S ) -1 (< 1s A h 1s A > + < 1s A h 1s B > + < 1s B h 1s A > + < > 1s A + 1s B > 1s B h 1s B ) = (0.85)((-1.110) + (-0.968) + (-0.968) + (-1.110)) = < σ u h σ u > = < ( -S) -1 ( 1s A - 1s B ) h ( -S) -1 ( ) = ( -S ) -1 (< 1s A h 1s A > - < 1s A h 1s B > - < 1s B h 1s A > + < > 1s A - 1s B > 1s B h 1s B ) = (.04)((-1.110) + (0.968) + (0.968) + (-1.110)) = S -1. < σ g σ g h σ g σ g > < gg gg > = ( +S ) -1 ( ) < ( 1s A + 1s B ) ( 1s A + 1s B ) ( 1s A + 1s B ) ( 1s A + 1s B ) > = ( +S ) -. (< AA AA > + < AA AB > + < AA BA > + < AA BB > + < AB AA > + < AB AB > + < AB BA > + < AB BB > + < BA AA > + < BA AB > + < BA BA > + < BA BB > + < BB AA > + < BB AB > + < BB BA > + < BB BB > ) = (0.081) ( (0.65) + (0.46) + (0.46) + (0.33) + (0.46) + (0.504) + (0.33) + (0.46) + (0.46) + (0.33) + (0.504) + (0.46) + (0.33) + (0.46) + (0.46) + (0.65) ) = S -1. < uu uu > = ( -S ) -1 ( ) < ( 1s A - 1s B ) ( 1s A - 1s B ) ( 1s A - 1s B ) ( 1s A - 1s B ) > = ( -S ) -. (< AA AA > - < AA AB > - < AA BA > + < AA BB > - < AB AA > + < AB AB > + < AB BA > - < AB BB > - < BA AA > + < BA AB > + < BA BA > - < BA BB > + < BB AA > - < BB AB > - < BB BA > + < BB BB > ) = (4.100) ( (0.65) - (0.46) - (0.46) + (0.33) - (0.46) + (0.504) + (0.33) - (0.46) - (0.46) + (0.33) + (0.504) - (0.46) + (0.33) - (0.46) - (0.46) + (0.65) ) = S -1. < gg uu > = ( +S ) -1 ( ) < ( 1s A + 1s B ) ( 1s A + 1s B ) ( 1s A - 1s B ) ( 1s A - 1s B ) >

24 16 = ( +S ) -1 ( -S ) -1. (< AA AA > - < AA AB > - < AA BA > + < AA BB > + < AB AA > - < AB AB > - < AB BA > + < AB BB > + < BA AA > - < BA AB > - < BA BA > + < BA BB > + < BB AA > - < BB AB > - < BB BA > + < BB BB > ) = (0.85)(.04) ((0.65) - (0.46) - (0.46) + (0.33) + (0.46) - (0.504) - (0.33) + (0.46) + (0.46) - (0.33) - (0.504) + (0.46) + (0.33) - (0.46) - (0.46) + (0.65)) = < gu gu > = ( +S ) -1 ( -S ) -1. < ( 1s A + 1s B ) ( 1s A - 1s B ) ( 1s A + 1s B ) ( 1s A - 1s B ) > = ( +S ) -1 ( -S ) -1. (< AA AA > - < AA AB > + < AA BA > - < AA BB > - < AB AA > + < AB AB > - < AB BA > + < AB BB > + < BA AA > - < BA AB > + < BA BA > - < BA BB > - < BB AA > + < BB AB > - < BB BA > + < BB BB > ) = (0.85)(.04) ((0.65) - (0.46) + (0.46) - (0.33) - (0.46) + (0.504) - (0.33) + (0.46) + (0.46) - (0.33) + (0.504) - (0.46) - (0.33) + (0.46) - (0.46) + (0.65)) = Note, that < gg gu > = < uu ug > = 0 from symmetry considerations, but this can be easily verified. For example, < gg gu > = ( +S) -1 ( -S) -3. < ( 1s A + 1s B ) ( 1s A + 1s B ) ( 1s A + 1s B ) ( 1s A - 1s B ) > = ( +S) -1 ( -S) -3. (< AA AA > - < AA AB > + < AA BA > - < AA BB > + < AB AA > - < AB AB > + < AB BA > - < AB BB > + < BA AA > - < BA AB > + < BA BA > - < BA BB > + < BB AA > - < BB AB > + < BB BA > - < BB BB > ) = (0.534)(.880) ((0.65) - (0.46) + (0.46) - (0.33) + (0.46) - (0.504) + (0.33) - (0.46) + (0.46) - (0.33) + (0.504) - (0.46) + (0.33) - (0.46) + (0.46) - (0.65)) = c. We can now set up the configuration interaction amiltonian matrix. The elements are evaluated by using the Slater-Condon rules as shown in the text. 11 = < σ g ασ g β σ g ασ g β >

25 17 = f σg σ g + g σg σ g σ g σ g = (-1.184) = = 1 = < σ g ασ g β σ u ασ u β > = g σg σ g σ u σ u = = < σ u ασ u β σ u ασ u β > = f σu σ u + g σu σ u σ u σ u = (-0.575) = d. Solving this eigenvalue problem: ε ε = 0 ( ε)( ε) - (0.140) = ε ε + ε = 0 ε +.37ε = 0 ε = -.37 ± (.37) - 4(1)(1.005) ()(1) = ± = -1.80, and Solving for the coefficients: ε C ε = 0 C 0 For the first eigenvalue this becomes: C C = C C = 0 0 (0.140)(C 1 ) + (1.5)(C ) = 0 C 1 = C C 1 + C = 1 (from normalization) ( C ) + C = C = 1 C = 0.111, C 1 = For the second eigenvalue this becomes: C C = C C = 0 0 (-1.5)(C 1 ) + (0.140)(C ) = 0

26 18 C 1 = 0.11 C C 1 + C = 1 (from normalization) (0.11 C ) + C = C = 1 C = 0.994, C 1 = e. The polarized orbitals, R ±, are given by: R ± = σ g ± R ± = σ g ± C C σ u σ u R ± = σ g ± σ u R + = σ g σ u (left polarized) R - = σ g σ u (right polarized) Section 4 Exercises, Problems, and Solutions Exercises: 1. Consider the molecules CCl 4, CCl 3, and C Cl. a. What kind of rotor are they (symmetric top, etc; do not bother with oblate, or near-prolate, etc.) b. Will they show pure rotational spectra? c. Assume that ammonia shows a pure rotational spectrum. If the rotational constants are 9.44 and 6.0 cm -1, use the energy expression: E = (A - B) K + B J(J + 1), to calculate the energies (in cm -1 ) of the first three lines (i.e., those with lowest K, J quantum number for the absorbing level) in the absorption spectrum (ignoring higher order terms in the energy expression).

27 19. The molecule 11 B 16 O has a vibrational frequency ω e = 1885 cm -1, a rotational constant B e = 1.78 cm -1, and a bond energy from the bottom of the potential well of D 0 e = 8.8 ev. Use integral atomic masses in the following: a. In the approximation that the molecule can be represented as a Morse oscillator, calculate the bond length, R e in angstroms, the centrifugal distortion constant, D e in cm -1, the anharmonicity constant, ω e x e in cm -1, the zero-point corrected bond energy, D 0 0 in ev, the vibration rotation interaction constant, α e in cm -1, and the vibrational state specific rotation constants, B 0 and B 1 in cm -1. Use the vibration-rotation energy expression for a Morse oscillator: E = h _ ω e (v + 1/) - h _ ω e x e (v + 1/) + B v J(J + 1) - D e J (J + 1), where B v = B e - α e (v + 1/), α e = -6B e + 6 B e 3 h _ ω e x e, and D e = 4B e 3. _ h ωe h ωe h ωe b. Will this molecule show a pure rotation spectrum? A vibration-rotation spectrum? Assume that it does, what are the energies (in cm -1 ) of the first three lines in the P branch ( v = +1, J = -1) of the fundamental absorption? 3. Consider trans-c Cl. The vibrational normal modes of this molecule are shown below. What is the symmetry of the molecule? Label each of the modes with the appropriate irreducible representation.

28 130 Problems: 1. Suppose you are given two molecules (one is C and the other is C - but you don't know which is which). Both molecules have C v symmetry. The C bond length of molecule I is 1.11 Å and for molecule II it is Å. The bond angle of molecule I is 104 and for molecule II it is 136. z R C y θ a. Using a coordinate system centered on the C nucleus as shown above (the molecule is in the YZ plane), compute the moment of inertia tensors of both species (I and II). The definitions of the components of the tensor are, for example: I xx = m j (y j + z j ) - M(Y + Z ) j m j x j y j - MXY I xy = - j ere, m j is the mass of the nucleus j, M is the mass of the entire molecule, and X, Y, Z are the coordinates of the center of mass of the molecule. Use Å for distances and amu's for masses ḃ. Find the principal moments of inertia Ia < I b < I c for both compounds ( in amu Å units) and convert these values into rotational constants A, B, and C in cm -1 using, for example, A = h(8π ci a ) -1. c. Both compounds are "nearly prolate tops" whose energy levels can be well approximated using the prolate top formula: E = (A - B) K + B J(J + 1), if one uses for the B constant the average of the B and C valued determined earlier. Thus, take B and C values (for each compound) and average them to produce an effective B constant to use in the above energy formula. Write down (in cm -1 units) the energy formula for both species. What values are J and K allowed to assume? What is the degeneracy of the level labeled by a given J and K? d. Draw a picture of both compounds and show the directions of the three principle axes (a,b,c). On these pictures show the kind of rotational motion associated with the quantum number K. e. Given that the electronic transition moment vector µ connecting species I and II is directed along the Y axis, what are the selection rules J and K?

29 131 f. Suppose you are given the photoelectron spectrum of C -. In this spectrum Jj = J i + 1 transitions are called R-branch absorptions and those obeying J j = J i - 1 are called P-branch transitions. The spacing between lines can increase or decrease as functions of J i depending on the changes in the moment of inertia for the transition. If spacings grow closer and closer, we say that the spectrum exhibits a so-called band head formation. In the photoelectron spectrum that you are given, a rotational analysis of the vibrational lines in this spectrum is carried out and it is found that the R-branches show band head formation but the P-branches do not. Based on this information, determine which compound I or II is the C - anion. Explain you reasoning. g. At what J value (of the absorbing species) does the band head occur and at what rotational energy difference?. Let us consider the vibrational motions of benzene. To consider all of the vibrational modes of benzene we should attach a set of displacement vectors in the x, y, and z directions to each atom in the molecule (giving 36 vectors in all), and evaluate how these transform under the symmetry operations of D 6h. For this problem, however, let's only inquire about the C- stretching vibrations. a. Represent the C- stretching motion on each C- bond by an outward-directed vector on each atom, designated r i : r 1 r r 6 r 3 r 5 r 4 These vectors form the basis for a reducible representation. Evaluate the characters for this reducible representation under the symmetry operations of the D 6h group. b. Decompose the reducible representation you obtained in part a. into its irreducible components. These are the symmetries of the various C- stretching vibrational modes in benzene. c. The vibrational state with zero quanta in each of the vibrational modes (the ground vibrational state) of any molecule always belongs to the totally symmetric representation. For benzene the ground vibrational state is therefore of A 1g symmetry. An excited state which has one quantum of vibrational excitation in a mode which is of a given symmetry species has the same symmetry species as the mode which is excited (because the vibrational wave functions are given as ermite polynomials in the stretching coordinate). Thus, for example, excitation (by one quantum) of a vibrational mode of A u symmetry gives a wavefunction of A u symmetry. To resolve the question of what vibrational modes may be excited by the absorption of infrared radiation we must examine

30 13 the x, y, and z components of the transition dipole integral for initial and final state wave functions ψ i and ψ f, respectively: < ψ f x ψ i >, < ψ f y ψ i >, and < ψ f z ψ i >. Using the information provided above, which of the C- vibrational modes of benzene will be infrared-active, and how will the transitions be polarized? ow many C- vibrations will you observe in the infrared spectrum of benzene? d. A vibrational mode will be active in Raman spectroscopy only if one of the following integrals is nonzero: < ψ f xy ψ i >, < ψ f xz ψ i >, < ψ f yz ψ i >, < ψ f x ψ i >, < ψ f y ψ i >, and < ψ f z ψ i >. Using the fact that the quadratic operators transform according to the irreducible representations: (x + y, z ) A 1g (xz, yz) E 1g (x - y, xy) E g Determine which of the C- vibrational modes will be Raman-active. e. Are there any of the C- stretching vibrational motions of benzene which cannot be observed in either infrared of Raman spectroscopy? Give the irreducible representation label for these unobservable modes. 3. In treating the vibrational and rotational motion of a diatomic molecule having reduced mass µ, equilibrium bond length r e and harmonic force constant k, we are faced with the following radial Schrödinger equation: -h d µr dr r dr dr + J(J + 1)h - µr R + 1 k(r - r e) R = E R a. Show that the substitution R = r -1 F leads to: -h µ F'' + J(J + 1)h - µr F + 1 k(r - r e) F = E F b. Taking r = r e + r and expanding (1 + x) - = 1 - x + 3x +..., show that the so-called vibration-rotation coupling term J(J + 1)h - µr can be approximated (for small r) by J(J + 1)h - µr e 1 - r r + 3 r. Keep terms only through order r e r e. c. Show that, through terms of order r, the above equation for F can be rearranged to yield a new equation of the form: -h- µ F'' + 1 k - (r - re - ) F = E - J(J + 1)h - + µr F e Give explicit expressions for how the modified force constant k-, bond length r- e, and energy shift depend on J, k, r e, and µ. d. Given the above modified vibrational problem, we can now conclude that the modified energy levels are:

31 133 E = h- k- µ v J(J + 1)h - -. µr e Explain how the conclusion is "obvious", how for J = 0, k = k-, and = 0, we obtain the usual harmonic oscillator energy levels. Describe how the energy levels would be expected to vary as J increases from zero and explain how these changes arise from changes in k and r e. Explain in terms of physical forces involved in the rotating-vibrating molecule why r e and k are changed by rotation. Exercises: Solutions 1. a. CCl 4 is tetrahedral and therefore is a spherical top. CCl 3 has C 3v symmetry and therefore is a symmetric top. C Cl has C v symmetry and therefore is an asymmetric top. b. CCl 4 has such high symmetry that it will not exhibit pure rotational spectra. CCl 3 and C Cl will both exhibit pure rotation spectra. c. N 3 is a symmetric top (oblate). Use the given energy expression, E = (A - B) K + B J(J + 1), A = 6.0 cm -1, B = 9.44 cm -1, selection rules J = ±1, and the fact that µ 0 lies along the figure axis such that K = 0, to give: E = B (J + 1) = B, 4B, and 6B (J = 0, 1, and ). So, lines are at cm -1, cm -1, and cm -1.. To convert between cm -1 and energy, multiply by hc = (6.6618x10-34 J sec)(.99795x10 10 cm sec -1 ) = x10 3 J cm. Let all quantities in cm -1 be designated with a bar, e.g. B e = 1.78 cm -1. h _ a. hcb e = µr e _ h R e =, µhcb e m µ = B m O m B + m = (11)(16) O ( ) x x10-7 kg = 1.084x10-6 kg. hcb e = hc(1.78 cm -1 ) = x10-3 J x10 R e = -34 J sec ()1.084x10-6 kg x10-3 J R e = 1.05x10-10 m = 1.05 Å

32 134 D e = 4B e 3 4B, D _ e = e 3 = (4)(1.78 cm-1 ) 3 h ωe ω e (1885 cm -1 ) = 6.35x10-6 cm -1 ω e x e = h_ ω e ω 4D 0, ω e x e = e = (1885 cm -1 ) e 4D 0 (4)(6678. cm -1 ) = cm-1. e D 0 0 = D0 e - h_ ω e + h_ ω e x e 4, D 0 0 = D 0 ω e - e + ω ex e 4 = = cm -1 = 8.16 ev. α e = -6B e + 6 B e 3 h _ ω e x e h ωe h ωe -6B α e = e + 6 B e 3 ω e x e ω e ω e (-6)(1.78) α e = (1885) + 6 (1.78)3 (13.3) (1885) = cm -1. B 0 = B e - α e (1/), B 0 = Be - αe(1/) = / = 1.77 cm -1 B 1 = B e - α e (3/), B 1 = Be - αe(3/) = (1.5) = 1.75 cm -1 b. The molecule has a dipole moment and so it should have a pure rotational spectrum. In addition, the dipole moment should change with R and so it should have a vibration rotation spectrum. The first three lines correspond to J = 1 0, J = 1, J = 3 E = h _ ω e (v + 1/) - h _ ω e x e (v + 1/) + B v J(J + 1) - D e J (J + 1) E = _ h ω e - h _ ω e x e - B 0 J(J + 1) + B 1 J(J - 1) - 4D e J 3 E = ω e - ωe x e - B0 J(J + 1) + B1 J(J - 1) - 4De J 3 E = (13.3) J(J + 1) J(J - 1) - 4(6.35x10-6 )J 3 = J(J + 1) J(J - 1) -.54x10-5 J 3 E (J = 1) = cm -1 E (J = ) = cm -1 E (J = 3) = cm The C Cl molecule has a σ h plane of symmetry (plane of molecule), a C axis ( to plane), and inversion symmetry, this results in C h symmetry. Using C h symmetry labels

33 135 the modes can be labeled as follows: ν 1, ν, ν 3, ν 4, and ν 5 are a g, ν 6 and ν 7 are a u, ν 8 is b g, and ν 9, ν 10, ν 11, and ν 1 are b u. Problems: 1. z R C y θ Molecule I Molecule II R C = 1.11 Å R C = Å C = 104 C = 136 y = R Sin (θ/) = ± y = ± z = R Cos (θ/) = z = Center of Mass(COM): clearly, X = Y = 0, 1(0) - RCos(θ/) Z = 14 = Z = a. I xx = m j (y j + z j ) - M(Y + Z ) j m j x j y j - MXY I xy = - j I xx = (1.11) - 14( ) I xx = (1.076) - 14( ) =.377 =.69 I yy = (0.690) - 14( ) I yy = (0.4031) - 14( ) = = I zz = (0.8834) I zz = (0.9976) = = I xz = I yz = I xy = 0 b. Since the moment of inertia tensor is already diagonal, the principal moments of inertia have already been determined to be (I a < I b < I c ): I yy < I zz < I xx I yy < I zz < I xx < < < <.69 h Using the formula: A = 8π = 6.66x10-7 ci a 8π (3x10 10 X 6.0x103 )I a (1x10-8 ) A = I a cm -1

34 136 similarly, B = I b cm -1, and C = I c cm -1. So, Molecule I Molecule II y A = 0.6 y A = z B = z B = 8.46 x C = 7.08 x C = 7.4 c. Averaging B + C: B = (B + C)/ = 8.94 B = (B + C)/ = 7.94 A - B = A - B = 5.51 Using the prolate top formula: E = (A - B) K + B J(J + 1), Molecule I Molecule II E = 11.68K J(J + 1) E = 5.51K J(J + 1) Levels: J = 0,1,,... and K = 0,1,... J For a given level defined by J and K, there are M J degeneracies given by: 1 for K = 0 for K 0 d. Molecule I Molecule II (J + 1) x z => I b z => I b C y => I a C y => I a e. Since µ is along Y, K = 0 since K describes rotation about the y axis. Therefore J = ±1 f. Assume molecule I is C - and molecule II is C. Then, E = E Jj (C ) - E Ji (C - ), where: E(C ) = 5.51K J(J + 1), and E(C - ) = 11.68K J(J + 1) For R-branches: J j = J i + 1, K = 0: E R = E Jj (C ) - E Ji (C - ) = 7.94(J i + 1)(J i ) J i (J i + 1) = (J i + 1){7.94(J i ) J i } = (J i + 1){( )J i + (7.94)} = (J i + 1){-J i } For P-branches: J j = J i - 1, K = 0: E P = E Jj (C ) - E Ji (C - )

35 137 = 7.94(J i - 1)(J i ) J i (J i + 1) = J i {7.94(J i - 1) (J i + 1)} = J i {( )J i } = J i {-J i } This indicates that the R branch lines occur at energies which grow closer and closer together as J increases (since the J i term will cancel). The P branch lines occur at energies which lie more and more negative (i.e. to the left of the origin). So, you can predict that if molecule I is C - and molecule II is C then the R-branch has a band head and the P-branch does not. This is observed therefore our assumption was correct: molecule I is C - and molecule II is C. g. The band head occurs when d( E R) dj = 0. d( E R ) dj = d dj [(J i + 1){-J i }] = 0 = d dj (-J i - J i J i ) = 0 = -J i = 0 J i = 7.44, so J = 7 or 8. At J = 7.44: E R = (J + 1){-J } E R = ( ){ } = (8.44)(8.44) = 71. cm -1 above the origin.. a. D 6h E C 6 C 3 C 3C ' 3C " i S 3 S 6 σ h 3σ d 3σ v A 1g x +y,z A g R z B 1g B g E 1g (R x,r y ) (xz,yz) E g (x -y,xy A 1u A u z B 1u B u E 1u (x,y) E u Γ C b. The number of irreducible representations may be found by using the following formula:

36 n irrep = 1 g χ red (R)χ irrep (R), R where g = the order of the point group (4 for D 6h ). n A 1g = 4 1 Γ C- (R). A 1g (R) R = 1 4 {(1)(6)(1)+()(0)(1)+()(0)(1)+(1)(0)(1) = 1 +(3)(0)(1)+(3)()(1)+(1)(0)(1)+()(0)(1) +()(0)(1)+(1)(6)(1)+(3)()(1)+(3)(0)(1)} n A g = 1 4 {(1)(6)(1)+()(0)(1)+()(0)(1)+(1)(0)(1) +(3)(0)(-1)+(3)()(-1)+(1)(0)(1)+()(0)(1) +()(0)(1)+(1)(6)(1)+(3)()(-1)+(3)(0)(-1)} = 0 n B 1g = 1 4 {(1)(6)(1)+()(0)(-1)+()(0)(1)+(1)(0)(-1) +(3)(0)(1)+(3)()(-1)+(1)(0)(1)+()(0)(-1) +()(0)(1)+(1)(6)(-1)+(3)()(1)+(3)(0)(-1)} = 0 n B g = 1 4 {(1)(6)(1)+()(0)(-1)+()(0)(1)+(1)(0)(-1) +(3)(0)(-1)+(3)()(1)+(1)(0)(1)+()(0)(-1) +()(0)(1)+(1)(6)(-1)+(3)()(-1)+(3)(0)(1)} = 0 n E 1g = 1 4 {(1)(6)()+()(0)(1)+()(0)(-1)+(1)(0)(-) +(3)(0)(0)+(3)()(0)+(1)(0)()+()(0)(1) +()(0)(-1)+(1)(6)(-)+(3)()(0)+(3)(0)(0)} = 0 n E g = 1 4 {(1)(6)()+()(0)(-1)+()(0)(-1)+(1)(0)() +(3)(0)(0)+(3)()(0)+(1)(0)()+()(0)(-1) +()(0)(-1)+(1)(6)()+(3)()(0)+(3)(0)(0)} = 1 n A 1u = 1 4 {(1)(6)(1)+()(0)(1)+()(0)(1)+(1)(0)(1) +(3)(0)(1)+(3)()(1)+(1)(0)(-1)+()(0)(-1) +()(0)(-1)+(1)(6)(-1)+(3)()(-1)+(3)(0)(-1)} = 0 n A u = 1 4 {(1)(6)(1)+()(0)(1)+()(0)(1)+(1)(0)(1) +(3)(0)(-1)+(3)()(-1)+(1)(0)(-1)+()(0)(-1) +()(0)(-1)+(1)(6)(-1)+(3)()(1)+(3)(0)(1)} = 0 n B 1u = 1 4 {(1)(6)(1)+()(0)(-1)+()(0)(1)+(1)(0)(-1) +(3)(0)(1)+(3)()(-1)+(1)(0)(-1)+()(0)(1) +()(0)(-1)+(1)(6)(1)+(3)()(-1)+(3)(0)(1)} = 0 138

37 139 n B u = 1 4 {(1)(6)(1)+()(0)(-1)+()(0)(1)+(1)(0)(-1) +(3)(0)(-1)+(3)()(1)+(1)(0)(-1)+()(0)(1) +()(0)(-1)+(1)(6)(1)+(3)()(1)+(3)(0)(-1)} = 1 n E 1u = 1 4 {(1)(6)()+()(0)(1)+()(0)(-1)+(1)(0)(-) +(3)(0)(0)+(3)()(0)+(1)(0)(-)+()(0)(-1) +()(0)(1)+(1)(6)()+(3)()(0)+(3)(0)(0)} = 1 n E u = 1 4 {(1)(6)()+()(0)(-1)+()(0)(-1)+(1)(0)() +(3)(0)(0)+(3)()(0)+(1)(0)(-)+()(0)(1) +()(0)(1)+(1)(6)(-)+(3)()(0)+(3)(0)(0)} = 0 We see that Γ C- = A 1g E g B u E 1u c. x and y E 1u, z A u, so, the ground state A 1g level can be excited to the degenerate E 1u level by coupling through the x or y transition dipoles. Therefore E 1u is infrared active and polarized. d. (x + y, z ) A 1g, (xz, yz) E 1g, (x - y, xy) E g,so, the ground state A 1g level can be excited to the degenerate E g level by coupling through the x - y or xy transitions or be excited to the degenerate A 1g level by coupling through the xz or yz transitions. Therefore A 1g and E g are Raman active.. e. The B u mode is not IR or Raman active. 3. a. d dr (F r-1 ) = F' r -1 - r - F r d dr (F r-1 ) = r F' - F d dr r d dr (F r-1 ) = F' - F' + r F'' So, -h- d µr dr r d dr (Fr-1 ) = -h - F'' µ r. Rewriting the radial Schrödinger equation with the substitution: R = r -1 F gives: -h d µr dr r d(fr-1 ) dr + J(J + 1)h - µr (Fr -1 ) + 1 k(r - r e) (Fr -1 ) = E (Fr -1 ) Using the above derived identity gives: -h- F'' µ r + J(J + 1)h - µr (Fr -1 ) + 1 k(r - r e) (Fr -1 ) = E (Fr -1 ) Cancelling out an r -1: -h- µ F'' + J(J + 1)h - µr F + 1 k(r - r e) F = E F