Name Ch 1a, Problem Set Eight Section

Name
Section
Rec TA
Side 1 of 5
Ch 1a, Problem Set Eight
Due Friday, November 21, 2014
at 4 PM in the Ch1a Box
Problems 2 and 3 are designated (P) as “no collaboration” problems.
1. (30 Points, 5 points each) Molecular size is governed by factors including bond lengths
and atomic radii. For each pair below, identify which molecule is larger. Briefly explain
each decision on the basis of bond lengths, atomic radii, or both. (Note: you do not need
exact numbers for this part; make qualitative arguments. You do not need to explain any
trends, just cite them; therefore, limit your answer to one sentence.)
a.
b.
c.
d.
e.
f.
O2 or O2+
HCl or HF
NH3 or PH3
H2CCH2 or HCCH
SiBr4 or CCl4
NO or O2 (Hint: you may be able to come up with multiple valid answers and arguments.
Any one of those valid arguments will earn full credit.)
P
2. (40 Points) Determining the structures of large molecules is often difficult due to free
rotations around single bonds. Rotation about double bonds, however, is forbidden.
Because of this, examination of the double and triple bonds in some compounds can help
provide some understanding about their structure.
a. (25 pts) Draw the molecules H2C=C=C=CH2 and H2C=C=C=C=CH2. For each
carbon, indicate the hybridization. Label ALL of the approximate bond angles. Draw
the molecules graphically showing the basic π molecular orbitals. For example:
H
H
C
H
C
H
b. (10 pts) One of these molecules has coplanar H2C groups. Which one does, and why?
(Give the physical reason, not simply the rule. Limit your answer to 3 sentences.)
c. (5 pts) For the molecule that does not have coplanar H2C groups, what is the angle
between the planes of the two H2C groups?
Name
Section
Rec TA
P
Side 2 of 5
Ch 1a, Problem Set Eight
Due Friday, November 21, 2014
at 4 PM in the Ch1a Box
3. (30 Points, 5×5 pts, 2×2.5 pts) The following diagrams illustrate pairs of atomic orbitals
approaching each other to form a molecular orbital (shading indicates the relative sign of
the wavefunction). Sketch the molecular orbital formed by the atomic orbitals, as shown
below (i.e., do not change any of the AOs’ phases or spatial orientation). For
antibonding interactions, draw a line for the antibonding node. (Note: If you determine
the combination is nonbonding, then you do not need to draw anything: simply write that
it is non-bonding.) Finally, indicate the type of molecular orbital (σb, σ*, πb, π*) that
results. As an example:
σb
+
a.
+
b.
+
c.
+
d.
+
e.
f.
g.
+
+
Name
Section
Rec TA
Side 3 of 5
Ch 1a, Problem Set Eight
Due Friday, November 21, 2014
at 4 PM in the Ch1a Box
4. (40 Points) The formula for the conjugated alkene octatetraene is H2C(CH)6CH2.
a. (8 pts) Draw the structure of octatetraene, and indicate the hybridization of each
carbon atom. Try to draw the molecule (line of carbons) as straight as possible,
and make the structure geometrically accurate—that is, the bonding angles you
draw should be those predicted by hybridization theory. (Hint: Put a line
horizontally through the molecule. When drawn geometrically accurately, the carbons should
be equidistant from this line, with half on top and half on the bottom)
b. (32 pts, 4 per MO) The resonance structures and basis atomic orbitals for the π
molecular orbitals for H2CCHCH2 (allyl radical) are given below. Draw all eight π
molecular orbitals for octatetraene in this same fashion. Indicate the phase of the
orbitals and show the nodes. (Hint: the nodes will be somewhat symmetric around the
E
center of the molecule.)
Name
Section
Rec TA
Ch 1a, Problem Set Eight
Due Friday, November 21, 2014
at 4 PM in the Ch1a Box
Side 4 of 5
5. (60 Points) The cycloheptatrienyl radical, C7H7, is cyclic and sometimes found in metal
chemistry.
a. (10 pts) Draw the seven resonance structures for the cycloheptatrienyl radical. What is
the hybridization for the seven carbons?
b. (19 pts) There is a graphic device for cyclic π-bonded π MO systems that can be used to
order the π molecular orbitals in energy. This is called a Frost circle. It involves
inscribing the molecule point-down in a circle. Each vertex then corresponds to a π
molecular orbital. For example, benzene gives the following:
π*
π*
π*
πb
πb
πb
Draw a similar diagram for the cycloheptatrienyl radical. Label the MOs either πb or π*
and fill in the π electrons. (hint: there are three π bonding orbitals)
c. (21 pts) We can use the Frost circle from part b. to draw the shape of the π orbitals for
our cyclic systems. As with atomic orbitals, as you add nodes to the π MOs, the
energy goes up. For our benzene example, we go from zero to three angular nodes in
our MOs and they go up in energy.
Energy
Construct such a diagram for the cycloheptatrienyl radical, showing the nodes and the
sign changes for the MOs. You may draw shaded/unshaded circles at each atom rather
than continuous orbitals if you find it easier, despite that not really being an MO. (Hint: the
easiest way to draw your nodes is with equal angles between the nodes. So, if you had 4 nodes, you could
draw them 45° from each other. Then, fill in your phases. Finally, rotate the nodes by 22.5˚ to get the
degenerate wavefunction.)
d. (10 pts) Use your diagram to predict the bond order (of the π system only) for the
cycloheptatrienyl radical, “tropylium ion” (the cycloheptatrienyl cation), and the
Name
Section
Rec TA
Side 5 of 5
Ch 1a, Problem Set Eight
Due Friday, November 21, 2014
at 4 PM in the Ch1a Box
cycloheptatrienyl anion (sorry, no fun name for that one). For example, benzene has a π
bond order of 3.