Chem 104a
Homework 5 Solutions
Fall 2011
Problem Set 5 Solutions
Chemistry 104a
Problem 1
(a) Prepare a molecular orbital energy level diagram for the cyanide ion.
Use sketches to show clearly how the atomic orbitals interact to form
MOs.
E (eV)
C
CN-
N
4σ
2π
V -10.6
2p
-13.2
2p
3σ
]
EV
1π
E]
-19.4
2σ
2s
2s
1,2
-25.6 V E
V E
]
]
1σ
(b) What is the bond order, and how many unpaired electrons does cyanide
have?
We can say that the bond order is approximately 3 because the 1σ
and the two 1π orbitals are filled; 2σ and 3σ are considered
1
non-bonding. (6 − 0) = B.O.
2
Looking at the molecular orbital diagram, you can see that there are
no unpaired electrons.
(c) Which molecular orbital of CN- would you predict to interact most
strongly with a hydrogen 1s orbital to form an H-C bond in the reaction
CN- + H+ −→ HCN? Explain.
1
Chem 104a
Homework 5 Solutions
Fall 2011
The 3σ (HOMO) of CN- would interact most with the LUMO of H+
1s orbital, which has an energy of -13.6 eV, because they match in
symmetry and in energy.
Problem 2
Although KrF+ and XeF+ have been studied, KrBr+ has not yet been prepared. For KrBr+ :
(a) Propose a molecular orbital diagram, showing the interactions of the
valence shell s and p orbitals to form molecular orbitals.
E (eV)
NF
B
Kr
4σ
2π
-12.5
-14.3
2p
2p
3σ
1π
-24.1
2σ
2s
2s
-27.5
1,2
1σ
(b) Toward which atom would the HOMO be polarized? Why?
The HOMO = 2π. The HOMO is more polarized toward the Br 4px
and 4py . These Br 4px and 4py orbitals are closer in energy to the 2π
MO than the Kr 4px and 4py .
2
Chem 104a
Homework 5 Solutions
Fall 2011
(c) Predict the bond order.
Bond order is approximately 1 since the effects of the 2σ and 3σ
cancel each other out just like the 1π and the 2π.
(d) Which is more electronegative, Kr or Br? Explain your reasoning.
Kr is more electronegative because its atomic orbitals are lower in
energy than the corresponding Br atomic orbitals. As a result, you
can see more filled orbitals have Kr character than Br character.
Problem 3
Consider the molecule NF and the ions NF+ and NF- . Write the Lewis
structure and the molecular orbital description of the ground state for each
species. Determine which of the three species would be paramagnetic, and tell
how many unpaired electrons there would be in each paramagnetic molecule.
Predict the bond orders for all three species.
E (eV)
NF
N
F
4σ
11,13
2π
12
-13.2
2p
9,10
3σ
5,7
6,8
-18.6
2p
1π
4
3
-25.6
2s
2σ
1,2
2s
1σ
3
N F
-40.2
-25.6
2s
-13.2
5,7
6,8
3
1,2
4
9,10
2σ
1σ
2s
3σ 6,8
Homework 55,7 Solutions
Chem 104a
-18.6
2p
1π
2p
-25.6
2π
2σ
3σ
-40.2
2s
-18.6
2p Fall 2011
1π
NF
+
1,2
4
3
-25.6
2s
2s
-40.2
1σ
2σ
N F
1,2
11 electrons
-40.2
2s
1σ antibonding. The bond order is
8 electrons are bonding and 3 electrons are
5/2, and the molecule is paramagnetic due to one unpaired electron.
NF
N
N
N
N
N
N
F
FF
F
FF
12 electrons
8 electrons are bonding and 4 electrons are antibonding. The bond order is
2, and the molecule is paramagnetic due to 2 unpaired electrons.
NF-
N F
13 electrons
8 electrons are bonding and 5 electrons are antibonding. The bond order is
3/2, and the molecule is paramagnetic due to one unpaired electron.
4
Chem 104a
Homework 5 Solutions
Fall 2011
Problem 4
The first ionization energies of BF, CO, and N2 are 11.06 eV, 14.01 eV, and
15.57 eV, respectively. Explain the increase in ionization energy for this
isoelectronic series on the basis of atomic-orbital composition of the highest
occupied molecular orbital.
E (eV)
B
BF
F
C
CO
4σ
-8.3
2p
O
N2
N
4σ
2π
2π
-10.6
2p
2π
3σ
-14
N
4σ
3σ
2s
1π
-18.6
2p
1π
3σ
1π
-19.4
2s
2σ
-15.8
2p
-13.2
2p
-13.2
2p
2σ
2σ
-32.3
2s
-25.6
2s
-25.6
2s
1σ
1σ
1σ
2s
-40.2
Note: Not showing s-p mixing for all 3 molecules.
We can answer this question simply by inspecting the HOMO levels of each
molecule because the HOMO level is a minimum value for ionization energy
(IE). The HOMO level of BF is the highest since it is higher than -14.0 eV.
The HOMO level of CO is slightly higher than the O 2p orbital (-15.8 eV)
from s-p mixing; therefore, its HOMO level is lower than the HOMO level
of BF, so its IE is larger. The HOMO level of N2 is the lowest because its
3σ bond is strongly bonding despite weak s-p mixing. Therefore, diatomic
nitrogen has the largest IE.
5
4nb
-18.6
3nb
1σ
Chem 104a
Homework 5 Solutions
Fall 2011
Problem 5
2nb
1nb
The species -40.2
of HF2 - is linear and symmetrical,
F-H-F- , and has one of the
strongest known systems of “hydrogen bonding”.
(a) Draw atomic overlap representations for each of the σ and π orbitals
of HF2 - .
F
H
F
σ
π
π
(b) Draw a molecular-orbital energy-level scheme for HF2 - . Place the H
atom 1s orbital on the left, the fluorine 2s and 2p atomic orbitals (or
the appropriate SALCs) on the right, and the orbitals of HF2 - in the
center.
HF2-
H
F2-
E (eV)
2σ
-13.6
5nb
4nb
-18.6
3nb
1σ
2nb
-40.2
1nb
6
F
σ
H
F
Chem 104a
Homework 5 Solutions
Fall 2011
(c) Given the energy-level scheme derived in part (b), would you expect
HF2 - to be diamagnetic or paramagnetic?
diamagnetic (all electrons are paired up)
(d) What is the H-F bond order in HF2 - ?
1
The total bond order = (2 − 0) = 1, which represents the bond order
2
over two H-F bonds.
A single BOH-F would then be 1/2
Problem 6
Consider the borohydride anion, BH4 - , to be formed in the reaction
H- + BH3 −→ BH4 D3h
BH3
B
H3
E (eV)
2E’
2E’
2A1’
2A1’
-8.3
px
py
pz
E’
E’
A2’’
1A2”
1A2”
-13.6
-14.0
E’
s
A1 ’
A1’
1E’
1E’
1A1’
1A1’
Which orbital will the H- nucleophile attack on the BH3 molecule?
Following the symmetry rules for chemical reactions laid out in pg. 324 328 of DG, we can argue that H- will act as a nucleophile and attack the
LUMO on BH3 . The LUMO is a non bonding a2 ” orbital, which
corresponds to the 2pz orbital of boron.
7
Chem 104a
Homework 5 Solutions
Fall 2011
Problem 7
Diborane, B2 H6 , has the structure shown. Using molecular orbitals (and
showing appropriate orbitals on B and H from which the MOs are formed),
explain how hydrogen can form “bridges” between two B atoms. (This type
of bonding is discussed in Chapter 8 of MT)
H
H
H
H
H
B
H
H
B
H
B
H
H
B
H
H
You can look at this problem as 2Hmolecules
of BH3 , labeled (1) and (2),
H
H
reacting with each other to make
B2 H6B.
B
H
H
H
H
B
H
(1) H
B
HH (1)
H
B
B
H
H
(2)
H
H
H
HH
H
(2)
B
H
H
H
B LUMO of each BH molecule
Problem #6 shows us that the HOMO
and
3
H
will look like:
(1)
(2)
H
B
H
H
H
H HOMO
B
LUMO
H
2 interactions
LUMO
(1)
(1)
H
The goal now is to orient the HOMO Hon (1) with the LUMO on (2) to
2 interactions
create one interaction. To create another
interaction, we can orient the
H
H
(1)
(1)
HOMO on (2) with the H
LUMO on (1) to create another interaction
while
H
the molecules areHOMO
still in the same relative
position.
This
can
create
aH
H LUMO
(2)
(2)
H
“bridged” molecule.
H
HOMO
H
H
(1)
2 interactions
(2)
(2)
(1)
H
H
H
H
(2)
(2)
8
H
H
Chem 104a
Homework 5 Solutions
Fall 2011
Problem 8
In the bent geometry, we should consider s-p mixing. The Walsh diagram
is made with s-p mixing. Please note that the position of 2a1 is system
dependent.
(a)
NH2-
N
(c)
E (eV)
H-H
E (eV)
2b2
2b2
3a1
-13.2
-13.6
(px) b2
(py) b1
(pz) a1
3a1
1b1
b2
a1
1b1
-13.2
-13.6
2a1
2a1
nonbonding (py)
2a1 (slightly bonding)
1b2
1b2
1b2
-25.6
PES
-25.6
a1 (s)
1a1
1a1
1a1
(b)
E (eV)
2σu
2b2
3a1
2σg
1b1
px
py
2a1
1b2
1σu
1σg
1a1
C2v
D∞h
C2v is more stable because it provides the greater energy
stabilization for 8 valence electrons
Problem 9
Consider the hypothetical square-planar molecule XeH4 in the following coordinate system:
9
Chem 104a
Homework 5 Solutions
Fall 2011
Z
Hc
Xe
Hd
Y
Hb
Ha
X
(a) Write appropriate SALCs of the hydrogen 1s atomic orbitals which
will overlap with each of the following xenon orbitals that can form
molecular orbitals: 5s, 5px , 5py , and 5dx2 −y2 .
D4h
Γred
E
4
2C4
0
C2
0
2C20
2
2C200
0
i 2S4
0 0
σh
4
Applying the reduction formula:
2σv
2
2σd
0
Z
Γred = A1g + B1g + Eu
Hc
We can now apply the Projection
operator
to Hgenerate LGOs.
Xe
Hd
b
Don’t pay too much attention to the shading. The sign is arbitrary.
Just know when the sign flips on a symmetry.
Ha
Y
X
P̂ A1g (Ha ) = 4(Ha + Hb + Hc + Hd )
Eu
A1g
B1g
P̂ B1g (Ha ) = 4(Ha − Hb + Hc − Hd )
10
5py with Eu
A1g
5px with Eu
2
c
Xe
Hd
Y
Hb
Ha
X
Homework 5 Solutions
Z
Chem 104a
Fall 2011
Hc
Y
Hb
Eu
Z
X
B1g
A1g
Eu
Hc
P̂ Eu (Ha ) = 4(Ha − Hc )
Xe
Hd
Y
Hb
5py with Eu
Z
Ha
Eu
X
A1g
B1g
Xe
Hd
Eu
Hb
Ha
X
Eu
B1g
5dx^2-y^2 with B1gEu Eu
5px with
E
u
(b) Draw atomic-orbital overlaps with
the appropriate Xe orbital for each
of these SALCs.
B1g
A1g
Eu
5py with Eu
with A1g
A1g
Y
P̂ Eu (Hb ) = 4(Hb − Hd )
5dz^2 with A1g
5py with Eu
A1g
5dx^2-y^2 with B1g
5px with Eu
Hc
A1g E
5px with
u
5px with Eu
5py with Eu
5dx^2-y^2 with B1g
5dx^2-y^2 with
B1g
5dz^2 with A1g
5dz^2 with (c)
A1g Sketch a molecular-orbital energy-level diagram for XeH , showing the
4
Xe atomic orbitals on the left, the H 1s orbitals on the right, and the
XeH4 orbitals in the center.
11
Chem 104a
Homework 5 Solutions
D4h
Fall 2011
XeH4
Xe
H4
E (eV)
2B1g
3A1g
5d
xy
yz
xz
B2g
Eg
Eg
x2-y2
z2
A1g
B1g
1B2g 1Eg
2Eu
5p
px
py
pz
Eu
Eu
A2u
1A2u
B1g
2A1g
-13.6
Eu
A1g
1B1g
1Eu
5s
A1g
1A1g
(d) Indicate which molecular orbitals are occupied, and calculate the net
Xe-H bond order, based on the molecular-orbital model.
ab1g , ebu , bb1g and a∗1g are occupied.
Bond Order =
1 1
· (8 − 2) = 3/4
4 2
Problem 10
Consider the molecule SiCl4 :
(a) Draw the MO diagram for the molecule in a planar configuration.
Use isolobal analogy so that we can use only s-orbitals for our LGO’s.
This will make life easier. These orbitals will be take on the energy of
the cholorine 3p (13.7 eV) orbitals.
D4h
Γred
E
4
2C4
0
C2
0
2C20
2
12
2C200
0
i 2S4
0 0
σh
4
2σv
2
2σd
0
Cl
Cl
Chem 104a
Si
Homework 5 Solutions
Cl
Fall 2011
Cl
Γred = A1g + B1g + Eu
Cl
Don’t pay too much attention to the shading. The sign is arbitrary.
Just know when the sign flips on a symmetry.
Cl
Si
Cl
P̂ A1g (σ1 ) = 4(σ1 + σ2 + σ3 + σ4 )
Cl
Cl
Cl
Si
Eu
Cl
B1g
A1g
Cl
P̂ B1g (H1 ) = 4(σ1 − σ2 + σ3 − σ4 )
Cl
Cl
Si
5py with Eu
Cl
E
5pux with Eu
A1g
Cl
B1g
A1g
Eu
P̂ Eu (H1 ) = 4(σ1 − σ3 )
Eu
A1g
A1g
B1g
P̂ Eu (H2 ) = 4(σ2 − σ4 )
5px with Eu
Eu
A1g
B1g
5py with Eu
Eu
5px with Eu
13
5py with Eu
5px with Eu
5py with Eu
Eu
Chem 104a
Homework 5 Solutions
D4h
Fall 2011
SiCl4
Si
Cl4
E (eV)
D4h
SiCl4
Si
E (eV)
-7.7
-7.7
px
py
Eu
Eu
px
py
Eu
Eu
pz
1A2u
A2u
pz
A2u
Cl4
2Eu
2Eu
1A2u
2A1g
B1g
-13.7
s
A1g
-14.9
-13.7
-14.9
Eu
1B1g
A1g
2A1g
B1g
1Eu
Eu
1B1g
A1g
s
A1g
1Eu
(b) It will
be tetrahedral.
E (eV)
1A1g
1A1g
E (eV)
2Eu
2Td
2Eu
2Td
1A2u
1A2u
2A1
2A1g2A1g
2A1
1B1g
1B1g
1Eu
1Td
1Td
1Eu
1A1g
1A1
1A1g
D4h
Td
D4h
1A1
Td
14