CE 610 Spring 2007 Problem Set 1 key Total points = 200 1. (10 points) What distinguishes transition metals from the other elements in the periodic table? List five physical or chemical features that are distinctive properties of transition metals. Many possibilities: Partially filled d orbitals Multiple oxidation states Magnetic behavior (esp. ferromagnetism) Brightly colored complexes orm complexes with a variety of geometries orm kinetically both labile and inert complexes orm a variety of homogenous and heterogeneous catalysts 2. (15) What group is obtained by adding or deleting from each of the following groups the indicated symmetry operation? a. C 3v plus i = D 3d b. S 6 minus i = C 3 c. C 5v plus σ h = D 5h 3. Assign these molecules to point groups a. (9) All of the dichlorobenzene isomers C 2v C 2v D 2h b. (3) cis-[pt 2 (N 3 ) 2 ] (ignore positions) c. (3) trans-[pt 2 (N 3 ) 2 ] (ignore positions) N 3 Pt Pt N 3 N 3 N 3 D 2h C 2v 1
d. (39) All isomers of tetrafluorocyclooctatetraene (C 8 4 4 in a tub conformation) I found 13 unique isomers. Did I miss any? 1,2,3,4-C 8 4 4 1,2,3,5-C 8 4 4 C 2 C 1 1,2,3,6-C 8 4 4 C 1 1,2,3,7-C 8 4 4 C 1 1,2,3,8-C 8 4 4 C s 1,2,4,5-C 8 4 4 1,2,4,6-C 4 4 C 1 C 1 C s 1,2,5,6-C 8 4 4 1,3,5,7-C 8 4 4 1,3,5,8-C 8 4 4 1,3,6,8-C 8 4 4 1,4,5,8-C 8 4 4 C 2v S 4 C 1 C 2 D 2 Picture each isomer in a tub conformation. ere are four examples. 1,2,3,4-C 8 4 4 1,2,3,5-C 8 4 4 C 2 C 1 1,2,3,8-C 8 4 4 C s 1,3,5,7-C 8 4 4 S 4 2
e. (3) S 4 f. (3) P 3 g. (3) N 4 + h. (3) [B 12 12 ] 2- B 2- S P N B B B B B B B B B B C 2v i. (3) C 3v T d I h Zr D 4 Tetra(acetylacetonato)zirconium(IV) (ignore symmetry of acac) 3
j. (18) e 1,1'-dichloroferrocene, all eclipsed and staggered rotamers C 2 C 2 C 2h C 2v C 2 C 2 4
k. (9) [Re 2 8 ] 2-, eclipsed (shown), staggered (45 ) and gauche (any other angle) Re Re Re Re Re Re D 4h D4d has 4 σ d D 4 lacks 4 σ d 4. (20 points) The interaction of [Mo() 6 ] with triphenylphosphite gives two complexes with the formula [Mo() 4 {P(Ph) 3 } 2 ]. Assign the isomers to point groups (treat the phosphites as point ligands i.e., ignore their symmetry). The carbonyl stretching frequencies at about 2000 cm - 1 are well isolated from other vibrations in the molecules. Derive the symmetry (Mulliken label) of each stretching frequency, and tell whether it is IR and/or Raman active. PR 3 C PR 3 Mo C PR 3 C C Mo PR 3 cis, C 2v trans, D 4h C 2v E C 2 (z) σ v (xz) σ v (yz) linear, rotations quadratic 5
A 1 1 1 1 1 z x 2, y 2, z 2 A 2 1 1-1 -1 R z xy B 1 1-1 1-1 x, R y xz B 2 1-1 -1 1 y, R x yz Γ() 4 0 2 2 Γ() reduction: Number of A 1 = [(4)(1)(1) + 0 + (2)(1)(1) + (2)(-1)(1)]/4 = 2 Number of B 1 = [(4)(1)(1) + 0 + (2)(1)(1) + (2)(1)(1)]/4 = 1 Number of B 2 = [(4)(1)(1) + 0 + (2)(-1)(1) + (2)(1)(1)]/4 = 1 The cis isomer has 4 ν() absorptions, all IR and Raman active D 4h E 2C 4 (z) C 2 2C' 2 2C'' 2 i 2S 4 σ h 2σ v 2σ d linear, quadratic rotations A 1g 1 1 1 1 1 1 1 1 1 1 x 2 +y 2, z 2 A 2g 1 1 1-1 -1 1 1 1-1 -1 R z B 1g 1-1 1 1-1 1-1 1 1-1 x 2 -y 2 B 2g 1-1 1-1 1 1-1 1-1 1 xy E g 2 0-2 0 0 2 0-2 0 0 (R x, R y ) (xz, yz) A 1u 1 1 1 1 1-1 -1-1 -1-1 A 2u 1 1 1-1 -1-1 -1-1 1 1 z B 1u 1-1 1 1-1 -1 1-1 -1 1 B 2u 1-1 1-1 1-1 1-1 1-1 E u 2 0-2 0 0-2 0 2 0 0 (x, y) Γ() 4 0 0 2 0 0 0 4 2 0 Γ() reduction: Number of A 1g = [(4)(1)(1) + 0 + 0 + (2)(2)(1) + 0 + 0 + 0 + (4)(1)(1) + (2)(2)(1) + 0]/16 = 1 Number of B 1g = [(4)(1)(1) + 0 + 0 + (2)(2)(1) + 0 + 0 + 0 + (4)(1)(1) + (2)(2)(1) + 0]/16 = 1 Number of E u = [(4)(2)(1) + 0 + 0 + 0 + 0 + 0 + 0 + (4)(2)(1) + 0 + 0]/16 = 1 The trans isomer has 3 ν() absorptions: E u is IR active; A 1g and B 1g are Raman active, 5. (24 points) The interaction of trigonal bipyramidal [e() 5 ] with P 3 gives several products with the formula [e() 3 (P 3 )} 2 ]. Sketch each isomer and assign its point group. The carbonyl stretching frequencies at about 2000 cm - 1 are well isolated from other vibrations in the molecules. Derive the symmetry (Mulliken label) of each stretching frequency, and tell whether it is IR and/or Raman active. 6
P 3 P 3 C e P 3 C e P 3 P 3 3 P e D 3h C 2v C s D 3h E 2C 3 3C' 2 σ h 2S 3 3σ v linear, quadratic rotations A' 1 1 1 1 1 1 1 x 2 +y 2, z 2 A' 2 1 1-1 1 1-1 R z E' 2-1 0 2-1 0 (x, y) (x 2 -y 2, xy) A'' 1 1 1 1-1 -1-1 A'' 2 1 1-1 -1-1 1 z E'' 2-1 0-2 1 0 (R x, R y ) (xz, yz) Γ() 3 0 1 3 0 1 Number of A' 1 = [(3)(1)(1) + 0 + (1)(3)(1) + (3)(1)(1) + 0 + (1)(3)(1)]/12 = 1 Number of E' = [(3)(1)(1) + 0 + 0 + (3)(1)(2) + 0 + 0]/12 = 1 The D 3h isomer has 2 ν() absorptions: E' is IR and Raman active A' 1 is only Raman active, C 2v E C 2 (z) σ v (xz) σ v (yz) linear, quadratic rotations A 1 1 1 1 1 z x 2, y 2, z 2 A 2 1 1-1 -1 R z xy B 1 1-1 1-1 x, R y xz B 2 1-1 -1 1 y, R x yz Γ() 3 1 1 3 Number of A 1 = [(3)(1)(1) + (1)(1)(1) + (1)(1)(1) + (3)(1)(1)]/4 = 2 Number of B 2 = [(3)(1)(1) + (1)(1)(-1) + (1)(1)(-1) + (3)(1)(1)]/4 = 1 The C 2v isomer has 3 ν() absorptions, all IR and Raman active, C s E σ h linear, rotations quadratic A' 1 1 x, y, Rz x2, y2, z2, xy A'' 1-1 z, Rx, Ry yz, xz Γ() 3 1 7
Number of A' = [(3)(1)(1) + (1)(1)(1)]/2 = 2 Number of A'' = [(3)(1)(1) + (1)(1)(-1)]/2 = 1 The C s isomer has 3 ν() absorptions, all IR and Raman active, 6 (5 each = 20) Name these compounds, both a line formula (where not given) and a full name a. K 2 [ebr 4 ] = Potassium tetrabromoferrate(ii) b. Λ-cis- [Co( 2 NC 2 C 2 N 2 ) 2 ( 2 )( κ-n 2 )] 2+ N 2 2 N 2 Co 2 N N N 2 2+ = Λ-cis-aquabis(ethane-1,2-diamine) (nitritoκ)cobalt(iii) ion Many variations are possible. ethylenediamine and 1,2-diaminoethane are widely accepted names in fact, I think the latter is more proper than the version in the IUPAC guide. Nitrito- is in common use. Indicating the 2+ charge on the ion instead of the oxidation state is accepted. c. 3 N 3 N N 3 Co N 3 N 3 Co N 3 N 3 N 3 4+ [Co 2 (N 3 ) 8 ( µ ) 2 ] 4+ or [{Co(N 3 ) 4 } 2 ( µ ) 2 ] 4+ octaamminedi-µ-hydroxodicobalt(4+). Possible variants: drop one a between octa and ammine, use hydroxy instead of hydroxo, indicate oxidation state III instead of the ionic charge. 8
d. trans, trans-[mn 2 () 8 (PPh 3 ) 2 ] or trans, trans-[{mn() 4 (PPh 3 )} 2 ] Ph 3 P C Mn C Mn PPh 3 trans, trans-octacarbonylbis(triphenylphosphine- 1κP,2κP)dimanganese (Mn Mn). Possible variants: phosphane instead of phosphine, indicate the distribution of carbonyls as (octacarbonyl-1κ 4 C,2κ 4 C), or trans, transbis(tetracarbonyltriphenylphosphinemanganese) (Mn Mn) would be understandable. 7. (5 each = 15) Draw structures. Draw all isomers if a single isomer is not specified a. carbonylchlorodihydridobis(trimethylphosphane)rhodium C Me 3 P Me 3 P C 2 pairs of enantiomers b. sodium hexafluoromanganate(iv) 4- Na + 2 Mn c. (µ-ethane-1,1-diyl)bis(pentacarbonyltechnetium) C Tc C C 3 C Tc C 9