Name Ch 1a, Problem Set Seven Section

Name
Section
Rec TA
Ch 1a, Problem Set Seven
Due Friday, Nov. 14, 2014
at 4 PM in the Ch1a Box
Side 1 of 4
Problems 3 and 6 are designated P as no collaboration problems.
1. (50 points) Consider the neutral, anion (–1), and cation (+1) species for each of the
following homonuclear diatomics. Using the general MO diagram given for
homonuclear diatomics, list the orbital configuration (e.g., 1sσb2 1sσ*2, …, 2px,yπb4 2pσ
2
b …) and bond order (e.g., 0.5, 1.5, 2, etc.) for each of the following molecules and ions.
Also label which species are paramagnetic and which are diamagnetic. (For Li and
higher, you may omit MOs formed from the 1s orbitals) Use a table for your responses
(species, config, bond order, P/D). The orbitals in your configuration must be in the
proper filling order.
a. Hydrogen (i.e. H+2 , H2 , H−2 )
e. Nitrogen (…)
b. Helium (i.e. He+2 , He2 , He−2 )
f. Oxygen (…)
g. Fluorine (…)
c. Beryllium (…)
€
d. Carbon (…)
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For oxygen and fluorine, remember to put
the 2pσb below the 2pxyπb MO.
2p
2pσ*
2pxyπ*
2p
2pσb
2pxyπb
2s
2sσ *
2s
2sσ b
1s
1sσ *
1sσ b
1s
Name
Section
Rec TA
2.
Ch 1a, Problem Set Seven
Due Friday, Nov. 14, 2014
at 4 PM in the Ch1a Box
Side 2 of 4
(40 points) Consider the series of homonuclear dihalogens: F2, Cl2, Br2, I2.
Molecule
Bond
Energy1
eV
Bond
Length1
pm
F2
1.610
141.2
Cl2
Br2
I2
2.488
1.972
1.540
198.8
228.1
266.6
1
2
Ionization
Potential2
eV
data not
provided
11.6
10.56
9.4
Electron
Affinity2
eV
data not
provided
2.4
2.6
2.6
Color (phase)
nearly colorless (g)
(absorbs UV)
green/yellow (g)
red/brown (ℓ)
violet (s)
A. Haaland, Molecules and Models, Oxford University Press (2008), p. 58.
G. Wulfsberg, Inorganic Chemistry, University Science Books (2000), p. 521.
a) (7 pts) Explain the trends in the data for bond length of the series of dihalogens.
Support your explanation with quantitative evidence – use atomic radii to predict
bond lengths. Atomic radii values can be found in Gray.
b) (10 pts) Explain the trends in the data for bond energy of the series of dihalogens.
Also explain why the bond energy of F2 breaks the general trend. Hint: Something
was slightly off for F2 in part a, and this may help you explain what’s going on.
c) (15 pts) Identify and sketch the shape of the Lowest Unoccupied Molecular
Orbital (LUMO). Label the node(s). To a first approximation, homonuclear
dihalogens all have the same shape LUMO. Why?
d) (8 pts) When a molecule absorbs energy with the correct wavelength, an electron
will move from the HOMO to the LUMO. This HOMO-LUMO gap determines
the color of homonuclear dihalogens. Use the color data in the table to
qualitatively describe the trend for the HOMO-LUMO gap (energy difference and
what wavelength that difference corresponds to) going down the column.
Remember, the perceived color is the complement of the color absorbed. Page
836 of OGC has a depiction of the color wheel.
P
3. (20 points) Consider the molecules NO and N2.
a) (14 points) Draw populated molecular orbital diagrams for each molecule using
proper relative orbital energies. Use the table of valence orbital energies on the next
page. The electron configuration of NO is available on page 100 of Gray. (Note that
πb,2px,2py are lower in energy than σb,2pz.)
b) (6 points) Which molecule should have the higher ionization energy?
Guidelines for your diagrams:
• Label all AOs and MOs with the proper designation (1s, 1σb, etc.)
• Show which AOs mix to form each MO.
• Fill in the electrons on both the AOs and the MOs
• Relative orbital energies are important!
• Basically, it should look similar in style to the diagram on the first page
Name
Section
Rec TA
Ch 1a, Problem Set Seven
Due Friday, Nov. 14, 2014
at 4 PM in the Ch1a Box
Side 3 of 4
4. (40 points) Carbon monoxide forms numerous complexes as a ligand to transition metals.
Now that you have learned about molecular orbital (M.O.) theory, we can look at CO
from that perspective and compare it to the Lewis structure.
a. (8 pts) Construct an M.O. diagram of CO and use it to ascertain the bond order.
When combining the p orbitals, the πb molecular orbital has a lower energy than
the σb molecular orbital.
b. (8 pts) Draw the Lewis dot structure of CO and assign formal charges based on
the bond order from part a. In terms of electronegativity, why is this unusual?
c. (8 pts) Based on your answer to part b, would positive metal ions bind to the lone
pairs of carbon (Mn+-CO) or the lone pairs of oxygen (Mn+-OC)? Explain.
Although CO has “lone pairs” in the Lewis sense, the more rigorous molecular orbital
theory assigns it only bonding and antibonding orbitals. These Lewis “lone pairs” are in
fact molecular orbitals formed from the 2s atomic orbitals from carbon and oxygen.
Due to their different energies, these 2s atomic orbitals combine poorly. The lowest σb
retains most of the character of one of the 2s atomic orbitals, while the lowest σ* retains
most of the character of the other 2s atomic orbital. In this sense, these are the molecular
orbitals that correspond to the Lewis “lone pairs”.
d. (8 pts) Using the table of valence orbital energies, predict which atom donates the
majority of its character to the σb orbital and which atom contributes to the σ*. Be
sure to explain your prediction!
e. (8 pts) CO forms a bond to the metal by donating “lone pair” electrons from the
MOs arising from the 2s orbitals of oxygen and carbon into empty d orbitals of
the metal. Energetically, would you expect this bonding arrangement to be metalCO or metal-OC? Explain. Compare this to the bonding arrangement (Mn+-CO, or
Mn+-OC) acquired with the Lewis model from part c. (Remember, the energy
levels are given in the table below!)
This is a table of valence orbital energies (given in eV). Use this for relative energies in your MO diagrams, but
only as a qualitative guide to the relative energies: you do not need to draw your molecular orbital diagrams to
scale. Only parts 4d and 4e require you to quantitatively compare energies. The electron configuration of CO is
given in Gray on page 100.
Atom
H
He
Li
Be
B
C
N
O
F
Ne
metals
1s
-13.6 (eV)
-24.6
2s
2p
-5.4
-9.3
-14
-19.4
-25.6
-32.3
-40.2
-48.5
-8.3
-10.6
-13.2
-15.8
-18.6
-21.6
3d
about -15
Name
Section
Rec TA
Side 4 of 4
Ch 1a, Problem Set Seven
Due Friday, Nov. 14, 2014
at 4 PM in the Ch1a Box
5. (30 points, 5 pts each) The hydroxyl radical, OH•, is important not only in atmospheric
chemistry but also in cell biology, as it participates in biochemical reactions as a free
radical. (For an atmospheric chemistry example, check out Professor Mitchio Okumura’s
work on OH• + NO2•)
a. Formulate the electronic structure of OH• in terms of M.O. theory. Use only the
oxygen 2p and hydrogen 1s atomic orbitals.
b. Which type of M.O. (σb, σ*, πb, π*, nonbonding) contains the unpaired electron?
c. Why don’t the other oxygen and hydrogen atomic orbitals play a role in the M.O.
diagram for OH•?
d. Is the orbital you specified in part (b.) associated with both constituent atoms or is
it localized on a single atom (if so, which one)?
e. Based on your M.O. diagram what is the bond order of OH•?
f. Compare your answers in (d.) and (e.) to the Lewis dot structure of OH•.
P
6. (20 points) Problem Five teaches us that when atomic orbitals combine poorly the new
molecular orbitals retain much of their original atomic orbital character. That is to say,
these molecular orbitals closely resemble Lewis lone pairs.
a. (10 pts) Consider fluorine as it forms bonds in a Lewis dot structure. Usually
fluorine is written as having one bond and three lone pairs. Would you expect
any of the molecular orbitals in F2 to behave as lone pairs as written in a Lewis
dot structure? Explain why.
b. (10 pts) Repeat this treatment for HF. Refer to the table of valence energies on
the previous page (and maybe even look in one of the textbooks for an MO
diagram for HF). Explain the possibility that any of the molecular orbitals in HF
behave like fluorine lone pairs in the Lewis dot structure.