Midterm 2009
Chem 201/281: Inorganic Chemistry 1
Mar 17, 2009
Chem 201/281: Modern Inorganic Chemistry / Inorganic Chemistry 1 Midterm 2 Answers
Exam version
1. Exercises (please answer directly on the sheet)
A
A. True/False (14 points) False True
Definition or statement 1)
A paramagnetic molecule is a molecule that possesses two unpaired electrons with parallel spins.
2)
Atomic orbitals can only interact if they belong in the same representation
3)
4 electrons interactions lead to a global antibonding interaction because usually the 4 electrons come from different levels of energy. An MO of a molecule does not necessarily have to belong in a representation of the point group of the molecule.
4) 5)
is antibonding because the probability of finding an electron right in the middle of the two atoms is zero.
6)
The intensity of the stabilization and of the destabilization depends solely on the overlap between the interacting atomic orbitals
7)
In the MO diagramme of H He ,
X X X X X X X
is the destabilized orbital.
B. Molecular orbital representations (20 points) Here are 10 molecules and one of their MOs. For each of these molecules and MOs, answer the following questions. Please answer in the table below: a) What is the point group of the molecule? b) What is the representation of the MO? c) What is the dimension of this representation? d) If the representation is of dimension 2 or above, draw the other MO(s) involved in the same representation (limit yourself to the MO(s) of same energy). Otherwise leave blank.
Dr. Audrey H. Moores
1/8
Chem 201/281: Inorganic Chemistry 1 Here is a molecule and one of its MOs 1)
O H
2)
H
H H
3)
B
4)
a)
b)
c)
d)
C2v
B2
1
N/A
D3h
E’
2
D3h
A’’2
1
N/A
C3v
A1
1
N/A
H
H H
Mar 17, 2009
B
H
N
H H
H
5)
H H
D∞h
Σu+
1
N/A
6)
H F
C∞v
Σ+
1
N/A
D∞h
Πu
2
C∞v
Π
2
Td
A1
1
N/A
C2v
A2
1
N/A
7)
8)
9)
O O
C O
H
H H
10)
H O
H
H
2. PROBLEMS (please answer on the McGill booklets) * marked questions are questions that are independent from previous answers.
Dr. Audrey H. Moores
2/8
Chem 201/281: Inorganic Chemistry 1
Mar 17, 2009
A. Be2 and O2 (33 points) x A
A
y
z
1) A2 diagramme a) Draw the AOs you will use. * py px
py s
s
px pz
pz
b) What is the point group of the molecule? * D∞h c) Using overlap considerations, build the MO diagramme of A2, neglecting the interaction between s and p orbitals. Σu+ py px pz
Πg
Πu s
py px pz
Πg
Σg
+
Σu+
Πu s
Σg+
d) Take the second orbital from the bottom. Retrieve the character table of the point group found in b). For each operation of symmetry, apply the operation to the orbital and draw the result. σv i S∞ C’2 Operation E C∞
e) Assign representation to each of the orbitals you found in c). See above c) f) List two pairs of OMs that have a good symmetry to interact. In a new diagramme, picture these two interactions, placing the s OMs on one side and the p ones on the other. Draw the new MOs formed. The two Σg+ MOs and the two Σu+ MOs. They interact in the following way in a secondary interation:
Dr. Audrey H. Moores
3/8
Chem 201/281: Inorganic Chemistry 1
Mar 17, 2009
Σu+
Σu
Σg+
+
Σg+
2) Be2 diagramme a) What is the consequence of the secondary interaction described in 1) f) on the order of the MOs in the case of Be2? Draw the final MO diagramme of Be2, with everything, levels and MOs. The orbital 2σg becomes higher in energy than the two πu.
2σg πu
1σu* 1σg
b) How many valence electrons are available in Be2? Fill out the MO diagramme with them. * 4, see above for filling c) Describe the bonding scheme of the molecule. What can you say about this molecule? The molecule has the following electronic configuration: (1σg)2 (1σu*)2. (1σg) is a bonding interaction and 1σu* is an antibonding interaction: both interactions cancel each other and Be2 does not exist. 3) O2 diagramme a) What is the consequence of the secondary interaction described in 1) f) on the order of the MOs in the case of O2? Draw the final MO diagramme of O2, with everything, levels and MOs. The secondary interaction is not powerful enough in the case of O2 to cause a destabilization of the orbital 2σg to become higher in energy than the two πu. The original order is not affected.
Dr. Audrey H. Moores
4/8
Chem 201/281: Inorganic Chemistry 1
Mar 17, 2009
πg
πu 2σg 1σu* 1σg
b) How many electrons are available in O2? Fill out the MO diagramme with them.* 12, see above c) Describe the bonding scheme of the molecule. What can you say about this molecule? The molecule has the following electronic configuration: (1σg)2 (1σu*)2(2σg)2(πu)2(πu’)2(πg)2(πg’)2. 1σg is a bonding interaction and 1σu* is an antibonding interaction: both interactions cancel each other out and create 2 lone pairs. 2σg is a bonding interaction, σ in nature: this is a σ O-O bond. πu and πu’ are populated with a total of 4 electrons when πg and πg’ with a total of 2. The first orbitals are bonding and the last antibonding thus the last two electrons will cancel two electrons of the πus, creating two lone pairs. The resulting 2 bonding electrons are creating a second bond, π in nature. As a result: A O=O double bond and four lone pairs, two on O O
each atom. Also, it should be noted that O2 is paramagnetic.
B. Methane (33 points): x H H
H z
y
H
1) Methane MO diagramme a) Draw the atomic orbitals at play in methane. * H H
H H
H H
H H
H H
H H
H H
H H
b) What is the point group of methane? * Td c) Assign representation to the AOs of the carbon atom
Dr. Audrey H. Moores
5/8
Chem 201/281: Inorganic Chemistry 1
Mar 17, 2009
T2 A1 H H
H
H H
H
H H
H H
H H
H H
H H
d) Combine the orbitals on Hs to match the representations of the point group. A1
T2
e) Assign representation to the combined AOs found in d) See above f) Build the MO diagramme of methane. 6, 6', 6''
py px pz
5
2, 2', 2'' s 1 2''
2' 2
6''
6'
6
g) How many valence electrons are available? * 4 on C, 1 on each H: 8 h) Fill out the MO diagramme with these electrons. See above
Dr. Audrey H. Moores
6/8
Chem 201/281: Inorganic Chemistry 1
Mar 17, 2009
i) Describe the bonding scheme of the molecule. 1: 2 ē in binding interaction participating in all four C-H bonds: ¼ of the four C-Hs 2, 2’, 2’’: 6 ē in binding interactions participating in all four C-H bonds: ¾ of the four C-Hs. I have formed the four C-H single bonds. 2) Methane deprotonation Let’s consider the following reaction: -
CH3 +
H H
H+
H H
a) Draw the structure of CH3- using Lewis theory and VSEPR. * H H
H
b) Build the MO diagramme of CH3-. Do it step by step following the general method we saw in class and we have been using for methane in 1). Orbitals at play: 2s
2px 2pz 2py
Point group: C3v 2px
2s
Representations assignment:
2pz A1
2py
A1
E
E E
A1 The H AOs need to be combined into: We will mix A1 orbitals together and E together
Dr. Audrey H. Moores
7/8
Chem 201/281: Inorganic Chemistry 1 z
Mar 17, 2009
y x
E
E
2px
E
2py
E
A1 E
2pz A1 E
2s
E
A1 E
A1
A1 A1 C
c) How many valence electrons are available, fill out your MO diagramme and describe the bonding scheme. * Valence electrons: 8. See above for filling d) What MO(s) of CH3- will interact with H+ to form methane? has a good symmetry to interact with an incoming s orbital in the z axis direction. The s on H is also an empty orbital, so it has to interact with a full orbital, which this MO is.
Dr. Audrey H. Moores
8/8
Mar 17, 2009
Chem 201/281: Modern Inorganic Chemistry / Inorganic Chemistry 1 Midterm 2 Answers
Exam version
1. Exercises (please answer directly on the sheet)
A
A. True/False (14 points) False True
Definition or statement 1)
A paramagnetic molecule is a molecule that possesses two unpaired electrons with parallel spins.
2)
Atomic orbitals can only interact if they belong in the same representation
3)
4 electrons interactions lead to a global antibonding interaction because usually the 4 electrons come from different levels of energy. An MO of a molecule does not necessarily have to belong in a representation of the point group of the molecule.
4) 5)
is antibonding because the probability of finding an electron right in the middle of the two atoms is zero.
6)
The intensity of the stabilization and of the destabilization depends solely on the overlap between the interacting atomic orbitals
7)
In the MO diagramme of H He ,
X X X X X X X
is the destabilized orbital.
B. Molecular orbital representations (20 points) Here are 10 molecules and one of their MOs. For each of these molecules and MOs, answer the following questions. Please answer in the table below: a) What is the point group of the molecule? b) What is the representation of the MO? c) What is the dimension of this representation? d) If the representation is of dimension 2 or above, draw the other MO(s) involved in the same representation (limit yourself to the MO(s) of same energy). Otherwise leave blank.
Dr. Audrey H. Moores
1/8
Chem 201/281: Inorganic Chemistry 1 Here is a molecule and one of its MOs 1)
O H
2)
H
H H
3)
B
4)
a)
b)
c)
d)
C2v
B2
1
N/A
D3h
E’
2
D3h
A’’2
1
N/A
C3v
A1
1
N/A
H
H H
Mar 17, 2009
B
H
N
H H
H
5)
H H
D∞h
Σu+
1
N/A
6)
H F
C∞v
Σ+
1
N/A
D∞h
Πu
2
C∞v
Π
2
Td
A1
1
N/A
C2v
A2
1
N/A
7)
8)
9)
O O
C O
H
H H
10)
H O
H
H
2. PROBLEMS (please answer on the McGill booklets) * marked questions are questions that are independent from previous answers.
Dr. Audrey H. Moores
2/8
Chem 201/281: Inorganic Chemistry 1
Mar 17, 2009
A. Be2 and O2 (33 points) x A
A
y
z
1) A2 diagramme a) Draw the AOs you will use. * py px
py s
s
px pz
pz
b) What is the point group of the molecule? * D∞h c) Using overlap considerations, build the MO diagramme of A2, neglecting the interaction between s and p orbitals. Σu+ py px pz
Πg
Πu s
py px pz
Πg
Σg
+
Σu+
Πu s
Σg+
d) Take the second orbital from the bottom. Retrieve the character table of the point group found in b). For each operation of symmetry, apply the operation to the orbital and draw the result. σv i S∞ C’2 Operation E C∞
e) Assign representation to each of the orbitals you found in c). See above c) f) List two pairs of OMs that have a good symmetry to interact. In a new diagramme, picture these two interactions, placing the s OMs on one side and the p ones on the other. Draw the new MOs formed. The two Σg+ MOs and the two Σu+ MOs. They interact in the following way in a secondary interation:
Dr. Audrey H. Moores
3/8
Chem 201/281: Inorganic Chemistry 1
Mar 17, 2009
Σu+
Σu
Σg+
+
Σg+
2) Be2 diagramme a) What is the consequence of the secondary interaction described in 1) f) on the order of the MOs in the case of Be2? Draw the final MO diagramme of Be2, with everything, levels and MOs. The orbital 2σg becomes higher in energy than the two πu.
2σg πu
1σu* 1σg
b) How many valence electrons are available in Be2? Fill out the MO diagramme with them. * 4, see above for filling c) Describe the bonding scheme of the molecule. What can you say about this molecule? The molecule has the following electronic configuration: (1σg)2 (1σu*)2. (1σg) is a bonding interaction and 1σu* is an antibonding interaction: both interactions cancel each other and Be2 does not exist. 3) O2 diagramme a) What is the consequence of the secondary interaction described in 1) f) on the order of the MOs in the case of O2? Draw the final MO diagramme of O2, with everything, levels and MOs. The secondary interaction is not powerful enough in the case of O2 to cause a destabilization of the orbital 2σg to become higher in energy than the two πu. The original order is not affected.
Dr. Audrey H. Moores
4/8
Chem 201/281: Inorganic Chemistry 1
Mar 17, 2009
πg
πu 2σg 1σu* 1σg
b) How many electrons are available in O2? Fill out the MO diagramme with them.* 12, see above c) Describe the bonding scheme of the molecule. What can you say about this molecule? The molecule has the following electronic configuration: (1σg)2 (1σu*)2(2σg)2(πu)2(πu’)2(πg)2(πg’)2. 1σg is a bonding interaction and 1σu* is an antibonding interaction: both interactions cancel each other out and create 2 lone pairs. 2σg is a bonding interaction, σ in nature: this is a σ O-O bond. πu and πu’ are populated with a total of 4 electrons when πg and πg’ with a total of 2. The first orbitals are bonding and the last antibonding thus the last two electrons will cancel two electrons of the πus, creating two lone pairs. The resulting 2 bonding electrons are creating a second bond, π in nature. As a result: A O=O double bond and four lone pairs, two on O O
each atom. Also, it should be noted that O2 is paramagnetic.
B. Methane (33 points): x H H
H z
y
H
1) Methane MO diagramme a) Draw the atomic orbitals at play in methane. * H H
H H
H H
H H
H H
H H
H H
H H
b) What is the point group of methane? * Td c) Assign representation to the AOs of the carbon atom
Dr. Audrey H. Moores
5/8
Chem 201/281: Inorganic Chemistry 1
Mar 17, 2009
T2 A1 H H
H
H H
H
H H
H H
H H
H H
H H
d) Combine the orbitals on Hs to match the representations of the point group. A1
T2
e) Assign representation to the combined AOs found in d) See above f) Build the MO diagramme of methane. 6, 6', 6''
py px pz
5
2, 2', 2'' s 1 2''
2' 2
6''
6'
6
g) How many valence electrons are available? * 4 on C, 1 on each H: 8 h) Fill out the MO diagramme with these electrons. See above
Dr. Audrey H. Moores
6/8
Chem 201/281: Inorganic Chemistry 1
Mar 17, 2009
i) Describe the bonding scheme of the molecule. 1: 2 ē in binding interaction participating in all four C-H bonds: ¼ of the four C-Hs 2, 2’, 2’’: 6 ē in binding interactions participating in all four C-H bonds: ¾ of the four C-Hs. I have formed the four C-H single bonds. 2) Methane deprotonation Let’s consider the following reaction: -
CH3 +
H H
H+
H H
a) Draw the structure of CH3- using Lewis theory and VSEPR. * H H
H
b) Build the MO diagramme of CH3-. Do it step by step following the general method we saw in class and we have been using for methane in 1). Orbitals at play: 2s
2px 2pz 2py
Point group: C3v 2px
2s
Representations assignment:
2pz A1
2py
A1
E
E E
A1 The H AOs need to be combined into: We will mix A1 orbitals together and E together
Dr. Audrey H. Moores
7/8
Chem 201/281: Inorganic Chemistry 1 z
Mar 17, 2009
y x
E
E
2px
E
2py
E
A1 E
2pz A1 E
2s
E
A1 E
A1
A1 A1 C
c) How many valence electrons are available, fill out your MO diagramme and describe the bonding scheme. * Valence electrons: 8. See above for filling d) What MO(s) of CH3- will interact with H+ to form methane? has a good symmetry to interact with an incoming s orbital in the z axis direction. The s on H is also an empty orbital, so it has to interact with a full orbital, which this MO is.
Dr. Audrey H. Moores
8/8
