Chem 400
Inorganic Chemistry
Exam 1
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6____of 20
7____of 10
Σ_______of 100
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Name _________________________________________________________
(please print)
1. Sketch out the molecular orbital energy level diagram for BH3 applying the LGO
approach. Define the spatial orientation of BH3 using x,y,z directions and identify all
atomic and ligand group orbitals involved in bonding. (10 points)
Planar BH3
ψ7
node
x
ψ5
2pz
2py
2px
Energy
z
ψ6
LGO(3)
LGO(2)
LGO(1)
ψ4
LGO(3)
LGO(2)
y
ψ2
LGO(1)
ψ3
2s
y
H
B
H
ψ1
B
BH3
x
H
H
H
H
All atoms of BH3 lie in the xy plane, x coincides with principal rotation axis. LGO’s of H3
fragment are derived by taking linear combinations of the three H 1 s orbitals.
1
2. Draw schematic representations for all of the bonding and anti-bonding molecular
orbitals (MO’s) of BH3 (use the information derived in Problem 1). (10 points)
ψ1
ψ2
ψ3
ψ4
pz
ψ7
ψ5
ψ6
2
3. Draw Lewis structures for each of the following molecules. (2 points each) Based
on your Lewis structure, develop three-dimensional structures for each compound
using VSEPR theory. (2 points each) Finally, determine the appropriate point group
for each of your three dimensional drawings. (2 points each)
a. BrF4
F
F
F
F
Br
D4h
Br
F
F
F
Lewis Structure
F
Point Group
VSEPR Structure
b. IF7
F
F
F
F
F
F
I
F
F
F
F
F
F
F
I
Point Group
F
Lewis Structure
D5h
VSEPR Structure
c. SiFClBrI
Cl
Cl
Si
F
Br
F
I
d. SF5
C1
Si
Br
I
Lewis Structure
VSEPR Structure
F
F
Point Group
-
F
F
S
F
F
Lewis Structure
F
F
S
F
F
C4v
Point Group
VSEPR Structure
3
e. H2O2
H
O
H
O
H
Lewis Structure
O
O
H
C2h
Point Group
VSEPR Structure
4. How many normal modes of vibration does each of the following molecules possess,
and how many of those are IR active for (2 points each):
a. CCl4
CCl4 is tetrahedral and non-polar; Td symmetry. Four modes of vibrational freedom are
IR active and give rise to two bands in the IR.
b. BCl3
BCl3 has a trigonal planar structure and is non-polar. Therefore, the symmetric stretch
is IR inactive. Five modes of vibrational freedom give rise to three bands in the IR
spectrum.
c. CS2
CS2 is linear and non-polar. Thus the asymteric stretch and the doubly degenerate
deformation are IR active, giving rise to two bands in the IR spectrum.
d. H2S
H2S is polar and possesses a bent molecular shape. All three modes of vibration are
IR active.
e. NF3
NF3 has trigonal pyramidal structure and is polar. There are six modes of vibrational
freedom giving rise to four bands in the IR spectrum.
4
6
5. Give the expected possible covalent MXn compounds ( where X- = Cl- or Br-)
of Sn, At, and Ra. Assume only σ-bonding, and predict the geometry of each
molecule. Hint: To predict possible covalent states, start with the ground
state electron configurations of the atom and obtain the simplest possible
covalent state by utilizing only the initially unpaired electrons in covalent bond
formation. The partial table below indicates would type of answer I expect.
(10 points)
Valence State Configuration
Sn
At
Hybridization
Formula
sp2
SnX2
bent
sp3
SnX4
tetrahedral
sp3
AtX
linear
AtX3
T-shaped
AtX5
square pyramidal
sp3d3
AtX5
pentagonal
bipyramidal
sd
RaX2
bent
3
sp d
sp3d2
Geometry
Ra
6.a Draw an MO energy-level diagram for the MOs that would arise in a diatomic
molecule from the combination of unhybridized d orbitals. Label the AOs being
combined and the resulting MOs. Let the z axis lie along the σ-bond. (10 points)
σ*
Energy
π1* π2*
d x 2- y 2
dxy
dyz
dxz
dz2
δ1*
δ2*
δ1
δ2
π1
π2
Note: d-orbitals
are degenerate,
drawn at different
energies to simplify
figure.
dx2- y2
dxy
dyz
dxz
dz2
σ
AOM
MOM-M
AOM
b. Sketch the shape of the resulting MO’s. (10 points)
y
x
z
π1
σ
dz2 + dz2
δ2
y
π1*
dxy+ dxy
x
dxz - dxz
y
δ2*
dxy - dxy
z
π2*
dx2- y2 + dx2- y2
z
dyz - dyz
x
δ1
dyz + dyz
y
x
x
z
π2
dxz + dxz
y
δ1*
dx2- y2 - dx2- y2
σ*
dz2 - dz2
6
8
7. Assign (1) formal charges, (2) oxidation numbers, and (3) evaluate the relative
importance of the three Lewis structures of OCN-, the cyanate ion. Compare
these (i.e., do the same) to the Lewis structures of the fulminate ion, CNO-.
(10 points)
For cyanate ion:
a)
O
C
N
0
0
+IV
Formal charges
Oxidation
numbers
-II
-
b)
-
c)
O
C
N
0
+1
0
-2
-III
-II
+IV
O
C
N
-1
-1
0
-III
-II
+IV
-
-III
Best charge distribution is in b). Structure b) should be the most important,
with structure a) next. Structure c) should be unimportant because of poor
charge distribution.
For fulminate ion:
a)
Formal charges
Oxidation
numbers
N
0
+1
-2
-1
+IV
0
-II
-II
C
-
O
b)
O
-
c)
O
N
+1 -1
+1
+1 -3
-I
-II
N
C
+II
The best structure is b), the second best a). Structure c) should be
unimportant because of poor charge distribution.
Extra Credit: Explain the observation that the cyanate ion is stable, the
fulminate ion is explosive, and CON- does not exist. (2 points)
The poor charge distribution in the fulminate ion is the source of its
explosiveness.
+III
C
-I
-