CHE 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
Form complexes with a variety of geometries
Form kinetically both labile and inert complexes
Form 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. C3v plus i = D3d
b. S6 minus i = C3
c. C5v plus σh= D5h
3. Assign these molecules to point groups
a. (9) All of the dichlorobenzene isomers
Cl
Cl
Cl
Cl
Cl
Cl
C2v
C2v
D2h
b. (3) cis-[PtCl2(NH3)2] (ignore H positions)
c. (3) trans-[PtCl2(NH3)2] (ignore H positions)
NH3
Cl
Pt
NH3
D2h
Cl
Cl
Cl
Pt
NH3
NH3
C2v
1
d. (39) All isomers of tetrafluorocyclooctatetraene (C8H4F4 in a tub conformation)
I found 13 unique isomers. Did I miss any?
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
1,2,3,5-C8H4F4
C1
1,2,3,4-C8H4F4
C2
F
1,2,3,6-C8H4F4
C1
F
F
1,2,3,7-C8H4F4
C1
F
F
F
F
F
F
F
F
F
F
1,2,4,7-C8H4F4
Cs
F
F
F
F
F
1,2,4,6-C 8H4F4
C1
F
1,2,4,5-C8H4F4
C1
1,2,3,8-C8H4F4
Cs
F
F
F
F
F
F
F
F
F
F
F
F
F
1,2,5,6-C8H4F4
C2v
1,3,5,7-C8H4F4
S4
F
1,3,5,8-C8H4F4
C1
1,3,6,8-C8H4F4
C2
F
1,4,5,8-C8H4F4
D2
Picture each isomer in a tub conformation. Here are four examples.
F
F
F
F
F
1,2,3,4-C 8H4F4
C2
F
F
F
F
F
1,2,3,5-C8H4F4
C1
F
F
1,2,3,8-C8H4F4
Cs
2
F
F
FF
1,3,5,7-C8H4F4
S4
e.
(3) SF4
f. (3) POCl3
g. (3) NH4+
h. (3) [B12H12]2-
H
B
F
S
O
F
F
F
P
Cl
H
BH
HB
N
Cl
Cl
H
H
H
BH
BH
C3v
Td
HB
BH
Ih
i. (3)
O
O
O
Zr
O
O
O
D4
O
O
Tetra(acetylacetonato)zirconium(IV) (ignore symmetry of acac)
3
BH
BH
BH
B
H
C2v
2-
j. (18)
Cl
Fe
Cl
1,1'-dichloroferrocene, all eclipsed and staggered rotamers
Cl
Cl
Cl
Cl
Cl
Cl
C2
C2
Cl
Cl
C2h
Cl
Cl
Cl
Cl
C2v
C2
C2
4
k. (9)
[Re2Cl8]2-, eclipsed (shown), staggered (45°) and gauche (any other angle)
Cl
Cl
Cl
Re
Cl
Cl
Cl
Re
Cl
Cl
Cl
Cl
Re
Cl
Cl
Cl
Cl
Re
Cl
Cl
D4d
has 4 σd
Cl
Re
Cl
Cl
D4
lacks 4 σd
(20 points) The interaction of [Mo(CO)6] with triphenylphosphite gives two complexes
with the formula [Mo(CO)4{P(OPh)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 CO stretching frequency, and tell whether
it is IR and/or Raman active.
PR3
CO
OC
Mo
OC
C2v
Cl
Cl
Re
D4h
4.
Cl
Cl
Cl
PR3
OC
PR3
OC
CO
Mo
CO
CO
PR3
cis, C2v
trans, D4h
E
C2 (z)
σv(xz)
5
σv(yz)
linear,
rotations
quadratic
A1
A2
B1
B2
Γ(CO)
1
1
1
1
4
1
1
-1
-1
0
1
-1
1
-1
2
1
-1
-1
1
2
x2, y2, z2
xy
xz
yz
z
Rz
x, Ry
y, Rx
Γ(CO) reduction:
Number of A1 = [(4)(1)(1) + 0 + (2)(1)(1) + (2)(-1)(1)]/4 = 2
Number of B1 = [(4)(1)(1) + 0 + (2)(1)(1) + (2)(1)(1)]/4 = 1
Number of B2 = [(4)(1)(1) + 0 + (2)(-1)(1) + (2)(1)(1)]/4 = 1
The cis isomer has 4 ν(CO) absorptions, all IR and Raman active
D4h
E
2C4 (z)
C2
2C'2
2C''2
i
2S4
σh
2σv
2σd
A1g
A2g
B1g
B2g
Eg
A1u
A2u
B1u
B2u
Eu
Γ(CO)
1
1
1
1
2
1
1
1
1
2
4
1
1
-1
-1
0
1
1
-1
-1
0
0
1
1
1
1
-2
1
1
1
1
-2
0
1
-1
1
-1
0
1
-1
1
-1
0
2
1
-1
-1
1
0
1
-1
-1
1
0
0
1
1
1
1
2
-1
-1
-1
-1
-2
0
1
1
-1
-1
0
-1
-1
1
1
0
0
1
1
1
1
-2
-1
-1
-1
-1
2
4
1
-1
1
-1
0
-1
1
-1
1
0
2
1
-1
-1
1
0
-1
1
1
-1
0
0
linear,
rotations
x2+y2, z2
Rz
(Rx, Ry)
z
(x, y)
Γ(CO) reduction:
Number of A1g = [(4)(1)(1) + 0 + 0 + (2)(2)(1) + 0 + 0 + 0 + (4)(1)(1) + (2)(2)(1) + 0]/16 = 1
Number of B1g = [(4)(1)(1) + 0 + 0 + (2)(2)(1) + 0 + 0 + 0 + (4)(1)(1) + (2)(2)(1) + 0]/16 = 1
Number of Eu = [(4)(2)(1) + 0 + 0 + 0 + 0 + 0 + 0 + (4)(2)(1) + 0 + 0]/16 = 1
The trans isomer has 3 ν(CO) absorptions: Eu is IR active; A1g and B1g are Raman active,
5.
(24 points) The interaction of trigonal bipyramidal [Fe(CO)5] with PF3 gives several
products with the formula [Fe(CO)3(PF3)}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 CO
stretching frequency, and tell whether it is IR and/or Raman active.
6
quadratic
x2-y2
xy
(xz, yz)
PF3
OC
CO
CO
Fe
OC
Fe
CO
PF3
PF3
F3P
Fe
PF3
CO
CO
PF3
CO
CO
D3h
C2v
Cs
D3h
E
2C3
3C'2
σh
2S3
3σv
A'1
A'2
E'
1
1
2
1
1
-1
1
-1
0
1
1
2
1
1
-1
1
-1
0
A''1
A''2
E''
Γ(CO)
1
1
2
3
1
1
-1
0
1
-1
0
1
-1
-1
-2
3
-1
-1
1
0
-1
1
0
1
linear,
rotations
quadratic
x2+y2, z2
Rz
(x, y)
(x2-y2,
xy)
z
(Rx, Ry)
(xz, yz)
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 D3h isomer has 2 ν(CO) absorptions: E' is IR and Raman active A'1 is only Raman active,
C2v
E
C2 (z)
σv(xz)
σv(yz)
A1
A2
B1
B2
Γ(CO)
1
1
1
1
3
1
1
-1
-1
1
1
-1
1
-1
1
1
-1
-1
1
3
linear,
rotations
z
Rz
x, Ry
y, Rx
quadratic
x2, y2, z2
xy
xz
yz
Number of A1 = [(3)(1)(1) + (1)(1)(1) + (1)(1)(1) + (3)(1)(1)]/4 = 2
Number of B2 = [(3)(1)(1) + (1)(1)(-1) + (1)(1)(-1) + (3)(1)(1)]/4 = 1
The C2v isomer has 3 ν(CO) absorptions, all IR and Raman active,
Cs
A'
A''
Γ(CO)
E
1
1
3
linear, rotations
x, y, Rz
z, Rx, Ry
σh
1
-1
1
7
quadratic
x2, y2, z2, xy
yz, xz
Number of A' = [(3)(1)(1) + (1)(1)(1)]/2 = 2
Number of A'' = [(3)(1)(1) + (1)(1)(-1)]/2 = 1
The Cs isomer has 3 ν(CO) 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. K2[FeBr4] = Potassium tetrabromoferrate(II)
b. Λ-cis- [Co(H2NCH2CH2NH2)2(H2O)( κO-NO2)]2+
2+
NH2
H2N
OH2
Co
H2 N
O
NH2
N
O
= Λ-cis-aquabis(ethane-1,2-diamine) (nitritoκO)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. “NitritoO” is in common use. Indicating the 2+ charge on the ion instead of the oxidation state is
accepted.
c.
NH3
H3 N
Co
H3N
NH3
H
O
Co
O
H
4+
NH3
NH3
NH3
NH3
[Co2(NH3)8( µ−OH)2]4+ or
[{Co(NH3)4}2( µ−OH)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-[Mn2(CO)8(PPh3)2] or trans, trans-[{Mn(CO)4(PPh3)}2]
CO
Ph3P
CO
CO
Mn
CO
Mn
PPh3
OC
OC
CO
CO
trans, trans-octacarbonylbis(triphenylphosphine1κP,2κP)dimanganese (Mn–Mn). Possible variants: phosphane instead of phosphine, indicate
the distribution of carbonyls as (octacarbonyl-1κ4C,2κ4C), 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
PMe3
OC
PMe3
H
H
CO
Rh
H
Cl
Rh
H
PMe3
Cl
CO
PMe3
H
H
Rh
H
PMe3
PMe3
Rh
H
Cl
CO
Me3P
H
OC
CO
F
Mn
F
F
F
c. (µ-ethane-1,1-diyl)bis(pentacarbonyltechnetium)
CO
CH3
CO
CH
OC
CO
Tc
Tc
CO OC
OC
CO
CO
CO
9
H
Rh
H
Cl
2 pairs of enantiomers
F
PMe3
Rh
b. sodium hexafluoromanganate(IV)
4F
Na+2
PMe3
H
PMe3
PMe3
Cl
H
Rh
Cl
Me3P
PMe3
H
PMe3
CO
Cl