Chemistry 481 Answer Set #1 Question #1 Identify the symmetry elements and operations present in each of the following molecules:

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1 Chemistry 81 Answer Set #1 Question #1 Ientify the symmetry elements an operations present in each of the following molecules: Molecule Structure Point group Symmetry Elements Symmetry Operations (a) Chloroform C 3v C 3, 3σ v E, C 3, 3σ v (b) 1,3,5-trichlorobenzene D 3h C 3, 3C, 3σ v, σ h, S 3 E, C 3, 3C, 3σ v, σ h, S 3 (c) chair cyclohexane D 3 S 6, C 3, i, 3C, 3σ E, 3S 6, C 3, i, 3C, 3σ (c) boat cyclohexane C v C, σ v E, C, σ v () B H 6 (iborane) D h 3C, 3σ, i E, 3C, 3σ, i (e) planar, trans-h O C h C, σ h, i E, C, σ h, i (f) Re Cl 8 - D h C, C, σ v, σ,σ h, C ', C '', i, S E, C, C, σ v, σ, σ h, C ', C '', i, S

2 (g) Co(NH 3 ) 5 (H O) 3+ (without the hyrogens) C v C, C, σ v ', σ v '' E, C, C, σ v ', σ v '' (h) S 8 D S 8, C, C, E, S 8, S 8, C ', σ C, C, C ', σ Note: The re atoms are all in the same plane, an the blue atoms are all in the same plane. Question #Ientify the single symmetry operations that gives the same result. a..σz (a,b,c) = (a,b,-c) = (-a,-b,-c) = i(a,b,c) b..i(a,b,c) = (-a,-b,-c) = (a,b,-c)=σ z (a,b,c) c..σy (a,b,c) = (a,-b,c) = (-a,b,c) = σ x (a,b,c)..c z (a,b,c) = (b,-a,c) = (-b,a,c) = ( ) 3 (a,b,c) e..c x (a,b,c) = (a,-b,-c) = (-b,-a,-c) = C x=-y x=-y (Note: C is a C rotation about the line x=-y.) f..σx (a,b,c) = (-a,b,c) = (b,a,c) = σ x=y (a,b,c) (Note: σ x=y is a mirror plane containing the line x=y an the z-axis.) g. σ x. (a,b,c) = σ x (b,-a,c) = (-b,-a,c) = σ x=-y (a,b,c) (Note: σ x=-y is a mirror plane containing the line x=-y an the z-axis.) h. S z.σz (a,b,c) = S z (a,b,-c) = (b,-a,c) = z (a,b,c) (Recall that S = z C.σz ) i. Two C axes at an angle of π/ 3. For simplicity, we will choose one of the axes as the x-axis (enote C a ); the other axis is enote as C b.

3 C a.c b C b.c a C a.c b = C 3 z C b.c a = (C 3 z ) They o not commute. Question #3constructing group multiplication tables, fin the groups with following symmetry operations. Note: For the group multiplication tables, the operation liste in the top row was one first, followe by the operation in the first column. Remember that orer is important. 3a C v E C z σ x σ y 3b C h E C z i σ z E E σ x σ y E σ y σ x σ x σ x σ y E σ y σ y σ x E E E i σ z E σ z i i i σ z E σ z σ z i E

4 3c D h E C x C y C z i σ x σ y σ z E E C x C y C z i σ x σ y σ z C x C x E C y σ x i σ z σ y C y C y E C x σ y σ z i σ x C y C x E σ z σ y σ x i i i σ x σ y σ z E C x C y σ x σ x i σ z σ y C x E C y σ y σ y σ z i σ x C y E C x σ z σ z σ y σ x i C y C x E 3 C v E C z C z (C z ) 3 σ x σ y σ (1) σ () E E C z C z (C z ) 3 σ x σ y σ (1) σ () ( ) 3 E σ () σ (1) σ x σ y ( ) 3 E σ y σ x σ () σ (1) ( ) 3 ( ) 3 E σ (1) σ () σ y σ x σ x σ x σ (1) σ y σ () E ( ) 3 σ y σ y σ () σ x σ (1) E ( ) 3 σ (1) σ (1) σ x σ () σ y ( ) 3 E σ () σ () σ y σ (1) σ x ( ) 3 E Note: All of the rotations were one clockwise. σ (1) is a mirror plane that contains the () line x=y an the z-axis, while σ is a mirror plane that contains the line x=-y an the z-axis.

5 3e S E S z C z (S z ) 3 E E S z (S z ) 3 S z S z (S z ) 3 E (S z ) 3 E S z (S z ) 3 (S z ) 3 E S z 3f D 3 E C 3 z (C 3 z ) C a C b C c E E 3 ( 3 ) C a C b C c 3 3 ( 3 ) E C c C a C b ( 3 ) ( 3 ) E 3 C b C c C a C a C a C b C c E 3 ( 3 ) C b C b C c C a ( 3 ) E 3 C c C c C a C b C 3 z (C 3 z ) E Question a. See column 3 of Question 1 answers. b. See top left cell of Question 3 answers. c. CCl : T. Benzene: D 6h e. Pyriine: C v f. Fe(CO) 5 D 3h g. Ferrocene Staggere: D 5 ; Eclipse: D 5h h. IrCl 3 (PMe 3 ) 3 fac: C 3v mer: C v i. [Ni(CN) ] - D h Question #5 a. cyclopropane, D 3h b. SO, C v c. CO, D h. B H 6, D h e. P, T f. Cl C=C=CCl, D g. BF 3, D 3h h. PH 3, C 3v i. OSCl, C S j. O SCl, C v 1 k. B(OH) 3, C 3h l. P I, C h

6 Question #6 The symmetry elements in an octaheron are: four C 3 axes (there are eight faces--the C 3 axes pass through opposite faces); three C axes (there are six vertices--the C axes pass through opposite vertices); six C axes (there are twelve eges--the C axes pass through opposite eges); three aitional C axes that are colinear with the three C axes; a center of symmetry, i; three S axes that are colinear with the three C axes; four S 6 axes that are colinear with the four C 3 axes; three planes of symmetry that are perpenicular to the three C axes; six planes of symmetry that bisect the twelve eges. Question #7 :What are vibrational moes in BF3? Nitrogen trifluorie, a nonlinear molecule, will have six vibrational egrees of freeom (3 N - 6). It has C 3v symmetry. The character table for this point group shows three classes of symmetry operations: E, C 3, an 3σ v. The number of atoms that are unshifte when these operations are carrie out are counte an multiplie by the contribution per atom to give a reucible representaion: E C 3 3σ Unshifte atoms 1 Contributions per atom Γ total Next we fin the irreucible components: N(A 1 ) = (1/6)[1(1)(1) + 0(1)() + (1)(3)] = 3 N(A ) = (1/6)[1(1)(1) + 0(1)() + (-1)(3)] = 1 N E = (1/6)[1(1)() + 0(-1)() + (0)(3)] = Thus Γ total = 3A 1 + A + E. Subtracting the translational moes (A 1 + E) leaves A 1 + A + 3E. Subtracting the rotational moes (A + E) gives A 1 + E. The symmetric stretching moe an the symmetric bening moe are both nonegenerate aan belong to the totally symmetric representation, A 1. The asymmetric stretch an the asymmetric ben are both oubly egenerate an belong to the irreucible representation E. What are vibrational moes in OF? OF has the same number of atoms an the same symmetry as H O.

7 E C σ v (xz) σ v (yz) Unshifte atoms Contribution per atom Γ total The irreucible components (page 63 of the text) are 3A 1 + A + B 1 + 3B. Subtracting the translational an rotational moes (page 68) gives the irreucible representations of the vibrational moes, A 1 + B. The symmetric stretch an the symmetric ben both transform as the totally symmetric representation, A 1, an the asymmetric stretch belongs to the irreucible representation B. Question #8 :What are vibrational moes in XeF? The molecule has D h symmetry. A reucible representation can be obtaine as follows: E C C C C i S σ h σ v σ Unshifte Atoms Contribution per atom Γ total The irreucible representaions are foun as follows: N(A 1g ) = (1/16)[(15)(1)(1) + (1)(1)() + (-1)(1)(1) + (-3)(1)() + (-1)(1)() + (-3)(1)(1) + (-1)(1)() + (5)(1)(1) + (3)(1)() + (1)(1)()] = 1 N(A g ) = (1/16)[(15)(1)(1) + (1)(1)() + (-1)(1)(1) + (-3)(-1)() + (-1)(-1)() + (-3)(1)(1) + (-1)(1)() + (5)(1)(1) + (3)(-1)() + (1)(-1)()] = 1 N(B 1g ) = (1/16)[(15)(1)(1) + (1)(-1)() + (-1)(1)(1) + (-3)(1)() + (-1)(-1)() + (-3)(1)(1) + (-1)( -1)() + (5)(1)(1) + (3)(1)() + (1)(-1)()] = 1 N(B g ) = (1/16)[(15)(1)(1) + (1)(-1)() + (-1)(1)(1) + (-3)(-1)() + (-1)(1)() + (-3)(1)(1) + (-1)(-1)() + (5)(1)(1) + (3)(-1)() + (1)(1)()] = 1 N(E g ) = (1/16)[(15)()(1) + (1)(0)() + (-1)(-)(1) + (-3)(0)() + (-1)(0)() + (-3)()(1) + (-1)(0)() + (5)(-)(1) + (3)(0)() + (1)(0)()] = 1

8 N(A 1u ) = (1/16)[(15)(1)(1) + (1)(1)() + (-1)(1)(1) + (-3)(1)() + (-1)(1)() + (-3)(-1)(1) + (-1)(-1)() + (5)(-1)(1) + (3)(-1)() + (1)(-1)()] = 0 N(A u ) = (1/16)[(15)(1)(1) + (1)(1)() + (-1)(1)(1) + (-3)(-1)() + (-1)(-1)() + (-3)(-1)(1) + (-1)(-1)() + (5)(-1)(1) + (3)(1)() + (1)(1)()] = N(B 1u ) = (1/16)[(15)(1)(1) + (1)(-1)() + (-1)(1)(1) + (-3)(1)() + (-1)(-1)() + (-3)(-1)(1) + (-1)(1)() + (5)(-1)(1) + (3)(-1)() + (1)(1)()] = 0 N(B u ) = (1/16)[(15)(1)(1) + (1)(-1)() + (-1)(1)(1) + (-3)(-1)() + (-1)(1)() + (-3)(-1)(1) + (-1)(1)() + (5)(-1)(1) + (3)(1)() + (1)(-1)()] = 1 N(E u ) = (1/16)[(15)()(1) + (1)(0)() + (-1)(-)(1) + (-3)(0)() + (-1)(0)() + (-3)(-)(1) + (-1)(0)() + (5)()(1) + (3)(0)() + (1)(0)()] = 3 Subtracting rotational moes A g an E g, an translational moes A u an E u, leaves the following vibrational moes: A 1g + B 1g + B g + A u + B u + +E u. All of the gerae moes are Raman active but not IR active. The B u moe is neither Raman nor IR active. The A u an E u moes are IR active but not Raman active. Note that none of the funamental vibrations are simultaneously Raman an IR active, consistent with the rule of mutual exclusion which tells us that a molecule with a center of symmetry cannot have funamental vibrations that are both Raman an IR active.