Download CHE450G Final Exam CP-109 December 11, 2006 10:30

CHE450G Final Exam
CP-109
December 11, 2006
Last name ___________________
First Name ___________________
Score
[_____/151]·100__ = _______ %
10:30-12:30 AM
2
(1)
Construct a physically realistic molecular orbital diagram for CS. Draw all SALC’s, molecular
orbitals, and provide the appropriate Mulliken symbols for each molecular orbital. Show all work
for credit. (10 points)
3
(2)
Metal-ligand bonding is a fundamental concept that is useful in predicting both the properties and
reactivity of a transition metal complex.
(a) Compare and contrast the differences between π-donors and π -acceptor ligands. Give an
example of each ligand type. Show all work for credit. (5 points)
π-donors (π-bases): ligands with filled π symmetry orbitals (p or π) and no low
energy or vacant p or π* orbitals that can engage in π-bonding with the transition
metal orbitals (t2g set in Oh complexes); decrease Δo. (e.g. I-)
π-acceptor (π-acids): ligands that do have empty π symmetry orbitals (p or π*) that
can engage in π-bonding with transition metal orbitals (t2g set in Oh complexes) and
no filled π symmetry orbitals that are close in energy to the metal orbitals; increase
Δo. (e.g. CO)
(b) Use a partial molecular orbital diagram to show how π-donor and π-acceptor ligands differ with
respect to metal-ligand bonding (ligand field splitting). In other words, explain the
spectrochemical series for an octahedral complex. Show all work for credit. (5 points)
π bases: halides (I-, Br-, Cl-)
π acids: CN-, NO+, CO
π*
π*
eg
eg
ΔO
ΔO
E
t2g
t2g
π
π
Metal
Complex
Ligand
Metal
Complex
Ligand
σ-donors: alter eg orbital energies and do not alter the t2g orbital energies (Oh symmetry) in
transition metal complexes. e.g. H- and Meπ-donors/acceptors: alter t2g orbital energies (Oh symmetry) but do not alter the eg orbital
energies in transition metal complexes
4
(3)
Inspection of the Spectrochemical Series reveals an interesting but surprising trend for halide
ligands. It is found that for halides, the ligand field strengths are:
F- > Cl- > Br- > IExplain why this is the case. Hint: It may prove useful to draw a simple molecular orbital diagram
for each M-X. Show all work for credit. (10 points)
Note: Modified version of problem 10-13 in textbook using Table 10-13 (3rd Ed.)
ligand field strengths:
F- > Cl- > Br- > IIII
3For a given [Cr X6] complex, where X- = F, Cl, Br, I:
Ligand
F-
Cl-
Br-
I-
16,600
13,180
12,340
620
_______________________________________________________________________________________
Δo (cm-1)
Decreasing σ and π donation is found when E.N. of X- decreases (see eσ and eπ).
The partial M.O. (octahedral MIIIX63- complex) diagram qualitatively scales as a function of X-:
F- is a generally thought of as a good σ donor while I- is weaker in comparison.
F- is generally thought of as a poor π donor while I- is a better one.
F- gives good M-X orbital overlap while I- is poor in comparison.
So, Δo is largest for F- and smallest for I-
(4)
Construct a molecular orbital diagram for a tetrahedral ML4 complex where L is a hydrogen atom.
Be sure to clearly draw the ligand SALC’s, assign the appropriate Mulliken symbols, and how each
ligand SALC interacts with the transition metal s, p, and d orbitals. Show all work for credit. (10
points)
z
x
Td Complex
(i)
t2
z
t2 (i)
y
4p (t2)
(ii)
transition metal orbitals
a1 (ii)
4s (a1)
t2
x
y
(iii)
a1
t2 (iii)
ΔT
E
t2
(iv)
t2
e
3d (e + t2)
t2 (iv)
e
a1
a1 (v)
(v)
t2 (vi)
a1 (vii)
(vi)
Metal
(vii)
Complex
Ligands
5
(5)
Determine the symmetries (Mulliken symbols) of the d orbitals under D4h symmetry. Show all work
for full credit. (10 points)
d orbitals = xz, yz, xy, z2, and x2 – y2
D4h
E
2C
C
2C
2C
i
2S
σ
2σ
2σ
A1g
A2g
B1g
B2g
Eg
A1u
A2u
B1u
B2u
Eu
1
1
1
1
2
1
1
1
1
2
1
1
-1
-1
0
1
1
-1
-1
0
1
1
1
1
-2
1
1
1
1
-2
1
-1
1
-1
0
1
-1
1
-1
0
1
-1
-1
1
0
1
-1
-1
1
0
1
1
1
1
2
-1
-1
-1
-1
-2
1
1
-1
-1
0
-1
-1
1
1
0
1
1
1
1
-2
-1
-1
-1
-1
2
1
-1
1
-1
0
-1
1
-1
1
0
1
-1
-1
1
0
-1
1
1
-1
0
’
’’
4
2
2
2
4
h
v
d
_________________________________________________________________________________________________________________________________________________________________________
x2 + y2, z2
Rz
x2 – y2
xy
(xz, yz)
(Rx, Ry)
z
(x, y)
____________________________________________________________________________________________________________________________________________________________________________
z
x
y
χ
d(xy)
E
E
1
χ
d(yz)
E
1
χ
d(x2-y2)
E
1
-1
C4
yz
C4
xz
C4
C2
1
C2
-1
C2
-1
C2
C2'
-1
C2 '
C2'
-1
C2"
C2
1
i
σv
σv
-1
σd
σd
1
yz
1
C2 '
1
"
C2
1
"
1
-1
1
σv
1
S4
1
σh
1
σv
1
σd
xz
i
1
σh
-1
σd
1
C2
1
S4
-1
σv
1
1
C4
-1
i
-xz
σh
-1
E
-1
1
S4
-yz
C2
-1
i
1
σh
1
C2"
1
S4
-1
σh
"
i
1
S4
1
χ
d(z2)
1
C4
C2'
(6)
χ
d(xz)
1
σd
-1
1
For each molecule and ion below give the appropriate (i) electron pair geometry and (ii)
molecular geometry/structure names. Show all work for credit. (6 points)
(a) [ICl2]Cl
(b)
-
SeCl4
Se
Cl
Cl
Cl
Cl
-
6-
Cl
I
P46-
(c)
P
P
P
P
O
F
F
Cl
F
F
total # valence electrons:
3(7) + 1 = 22
1(6) + 4(7) = 34
4(5) + 6 = 26
5(7) + 1(6) + 1 = 42
6
# nonbonding pairs on central atom:
7 - 2(1) +1 = 3
2
6 - 4(1) + 0 = 1
2
5 - 3(3) + 6 = 1
2
7 - 4(1) -1(2) + 1 = 1
2
(i) electron pair geometry considers all electron pairs around central atom (i)
trigonal bipyramidal
trigonal bipyramidal
tetrahedral
octahedral
(ii) molecular geometry considers only bonding electron pairs (ii)
linear
(7)
see-saw
square pyramidal
The following regular polyhedra are members of what are known collectively as the five Platonic
solids. To which point group(s) do each belong? (6 points)
(a)
(b)
(c)
Oh
Oh
(8)
trigonal pyramidal
Td
Assign the proper point group for the ions below. Assume an idealized (highest
symmetry) structure whenever possible. (12 points)
en = H2NC2H4NH2 ; Tp* = HB(3,5-dimethylpyrazol-yl)borate or [HB(Me2C3HN2)3](a) [NbIV(OEt)2(CN)4]2-
C2h
(9)
(b) [MnII(en)3]2+
D3
(c) [Tp*VII(CN)3]2-
C3v
(d) [NbV(O)(CN)6]3(linear CN- ligands)
C5v
Explain why the electronic configuration for Ti is [Ar]4s2 3d2 but for Cr2+ it is [Ar] 3d4. (5 points)
Short answer:
4s electrons have lower energies than 3d electrons in Ti
In ions, 3d levels move to lower energies and electron-electron pairing acts to increase the
energy of the ion (going from left to right)
In Cr, 3d and 4s levels are very close in energy
7
Second electron in Cr2+ goes to 3d shell to minimize electron-electron repulsion
Long answer:
To answer this question you must consider electron-electron repulsion and Zeff for the 4s and 3d
electrons since titanium(0) and CrII are isoelectronic. As Zeff increases, electrons become strongly
bound to the transition metal center (more negative energies) and the energy levels decrease in
energy. The energy of the d orbitals decrease more rapidly than do the s orbitals as a function of
increasing Z since they are shielded less than the s orbitals. The intraorbital repulsion interactions
become greater than the subshell energies and consequently CrII does not have 4s electrons.
(10) Determine the number of infrared active νCO stretching absorptions for fac-Mo(CO)3(NCMe)3. Use
vectors, symmetry labels, and show all work to justify your answers. Show all work for credit. (12
points)
8
(11)
(i) Draw the infrared active stretching and deformation modes for ammonia. (ii) Assign each
using the correct terminology and Mulliken symbol appropriate for the point group designation.
(iii) Use vectors to indicate apparent motion of atoms where necessary. (20 points)
9
(12)
(1a) Plot the radial probability functions for the 3d, 3s, and 3p orbitals. (5 points)
(1b) Arrange the 3d, 3s, and 3p orbitals as a function of increasing energies. (5 points)
3d < 3p < 3s
(13)
(1c) Arrange the orbitals as a function of increasing shielding. (5 points)
E3s < E3p < E3d
(1d) Arrange the orbitals as a function of increasing Zeff. (5 points)
3d < 3p < 3s
Solid CrF3 contains a CrIII ion surrounded by six F- ions in an octahedral geometry. All of the
Cr-F distances are 190 pm in length. However, MnF3 adopts a distorted geometry, with Mn-F
distances of 179, 191, and 209 pm (two of each), respectively. Provide an explanation for these
observations. (10 points)
CrIII (t2g3) is not subject to Jahn-Teller distortion while MnIII (t2g3eg1) exhibits tetragonal elongation
that is typical of Jahn-Teller distorted ions.
(14)
Complete the missing portions of the periodic table. (10 points)