Chap15 solutions

Physical Chemistry

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15 Molecular symmetry Solutions to exercises Discussion questions E15.1(b) Symmetry operations Symmetry elements 1. Identity, E 1. The entire object 2. n -fold rotation 2. n -fold axis of symmetry, C n 3. ReFection 3. Mirror plane, σ 4. Inversion 4. Centre of symmetry, i 5. n -fold improper rotation 5. n -fold improper rotation axis, S n E15.2(b) A molecule may be chiral, and therefore optically active, only if it does not posses an axis of improper rotation, S n . An improper rotation is a rotation followed by a reFection and this combination of operations always converts a right-handed object into a left-handed object and vice-versa ; hence an S n axis guarantees that a molecule cannot exist in chiral forms. E15.3(b) See Sections 15.4(a) and (b). E15.4(b) The direct sum is the decomposition of the direct product. The procedure for the decomposition is the set of steps outlined in Section 15.5(a) on p. 471 and demonstrated in Illustration 15.1. Numerical exercises E15.5(b) CCl 4 has 4 C 3 axes (each C–Cl axis), 3 C 2 axes (bisecting Cl–C–Cl angles), 3 S 4 axes (the same as the C 2 axes), and 6 dihedral mirror planes (each Cl–C–Cl plane). E15.6(b) Only molecules belonging to C s , C n , and C n v groups may be polar, so ... (a) CH 3 Cl (C 3v ) may be polar along the C–Cl bond; (b) HW 2 ( CO ) 10 (D 4h ) may not be polar (c) SnCl 4 (T d ) may not be polar E15.7(b) The factors of the integrand have the following characters under the operations of D 6h E 2 C 6 2 C 3 C 2 3 C 0 2 3 C 0 2 i 2 S 3 2 S 6 σ h 3 σ d 3 σ v p x 21 1 200 2 11200 z 111 1 1 1 1 1 1 p z 1 1 1 1 1 1 Integrand 2 1 1 2 The integrand has the same set of characters as species E 1 u , so it does not include A 1g ; therefore the integral vanishes E15.8(b) We need to evaluate the character sets for the product A 1g E 2u q , where q = x, y ,or z E 2 C 6 2 C 3 C 2 3 C 0 2 3 C 0 2 i 2 S 3 2 S 6 σ h 3 σ d 3 σ v A 1g E 2u 2 1 12 0 0 211 (x, y) 1 2 Integrand 4 11 400 4
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244 INSTRUCTOR S MANUAL To see whether the totally symmetric species A 1g is present, we form the sum over classes of the number of operations times the character of the integrand c( A 1g ) = ( 4 ) + 2 ( 1 ) + 2 ( 1 ) + ( 4 ) + 3 ( 0 ) + 3 ( 0 ) + ( 4 ) + 2 ( 1 ) + 2 ( 1 ) + ( 4 ) + 3 ( 0 ) + 3 ( 0 ) = 0 Since the species A 1g is absent, the transition is forbidden for x -or y -polarized light. A similar analysis leads to the conclusion that A 1g is absent from the product A 1g E 2u z ; therefore the transition is forbidden. E15.9(b) The classes of operations for D 2 are: E , C 2 (x) , C 2 (y) , and C 2 (z) . How does the function xyz behave under each kind of operation? E leaves it unchanged. C 2 (x) leaves x unchanged and takes y to y and z to z , leaving the product unchanged. C 2 (y) and C 2 (z) have similar effects, leaving one axis unchanged and taking the other two into their negatives. These observations are summarized as follows EC 2 (x) C 2 (y) C 2 (z) 11 1 1 A look at the character table shows that this set of characters belong to symmetry species A 1 E15.10(b) A molecule cannot be chiral if it has an axis of improper rotation. The point group
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Chap15 solutions - 15 Molecular symmetry Solutions to...

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