Chemistry 181
Professor S. Alex Kandel
Fall 2015
Due 9/30/2015
Problem Set 5
Discussion problems:
We will discuss this in class (3:30) on Wednesday the 30th. Be prepared to present
answers to and to discuss any of these up at the blackboard.
1. Use hybrid orbitals and π bonding to draw three-dimensional structures for
the following molecules; pay attention to small details. Determine bond angles
where possible.
(a) CH3 C≡CH
(b) CH3 ON=O
(c) H2 C=C=CH2
This is the most interesting one. Name the carbons:
H2 Ca = Cb = Cc H2
Because the central carbon (=Cb =) needs to make two π bonds, it must
use the π2px MO on one side, and the π2py MO on the other. So, the 2p
orbitals used by carbons a and c must be at right angles to each other.
This means the molecule is not planar: the H2 Ca plane is perpendicular
to the Cb H2 plane.
2. The following are illustrations of the four π molecular orbitals in the NO−
3 anion:
A
B
C
D
(a) Indicate whether each one is bonding, non-bonding, or anti-bonding.
B is bonding, C and D are nonbonding, and A is antibonding.
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(b) Arrange them from lowest in energy to highest. (Two are very close in
energy; as long as they are next to each other, don’t worry which to put
first.) How many electrons are in this π system, and which orbitals are
filled?
lowest → B < C = D < A ← highest
There are 6 electrons, which fill orbitals B, C, and D.
(c) What would you predict for N–O bond order?
There is a σ bond between the N and each O, and one π bonding pair
(in orbital A) distributed among all three N–O bonds. The bond order is
4/3. This is also which is also what you get from Lewis structures (with
resonance).
(d) If you use the same MO diagram for the molecule SO3 , what S–O bond
order do you get?
This is still a 24-electron system, and with the same MOs, you still get
bond order 4/3.
(e) Does the MO description tell you something different about bonding in
NO−
3 or SO3 than what you’d expect from Lewis structures for these
molecules?
If you decide S needs to follow the octet rule, then the predicted bond
order is still 4/3. Expanding the octet on S will result in an S–O bond
order of 2. Problem set 6 will discuss bond order and expanded octets in
more detail.
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Graded problem:
1. Cyanamide, H2 NCN, has the following structure:
(a) Draw Lewis structures for all important resonance structures of cyanamide.
Rank the relative importance of each resonance structure.
There are 16 electrons to allocate. Resonance structures are:
H
N
H
C
N
H
N
C
N
H
major
minor
The structure on the left contributes more, as it has lower formal charges
(i.e., no formal charges) than the structure on the right.
(b) What are the hybridizations for the N, C, and N atoms? Does your answer
change for different resonance structures ... and what does that tell you?
Start with this:
• All single bonds will be sigma bonds, made with hybrid orbitals.
• The second and third bonds in a double or triple bonds will be made
with π molecular orbitals, and require a 2p orbital to be “set aside”
(i.e., not hybridized), by each of the bonded atoms.
• All things being equal, a lone pair would prefer to be in a hybrid orbital
... but only if the pair is localized onto a single atom. Lone pairs may
be participating in π bonding, and it is not always immediately obvious
this is happening.
If you rush ahead and assign all lone pairs to hybrid orbitals, you get:
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major isomer
minor isomer
sp3
sp
sp
sp2
sp
sp2
N (left)
C
N (right)
The hybridization does change between resonance structures, and this
should bother you, a lot. Most notably, you’d predict the left N atom
to be tetrahedral (sp3 ) for the major structure, but planar (sp2 ) for the
minor structure. However, for resonance structures, the geometry can not
change: it’s not like the atoms can be in two places at once.
The minor resonance structure tells you that the left N atom is participating in π bonding, which means one of its 2p orbitals is contributing to
a molecular orbital. Since the resonance structure is part of the overall
description of electrons in the molecule, this statement must hold for both
resonance structures: the left N atom is participating in π bonding, which
means one of its 2p orbitals is contributing to a molecular orbital. It’s not
obvious looking at just the major resonance structure that the lone pair
on the left nitrogen is in a 2p orbital, but it is, and that nitrogen is sp2
and planar. (Similarly, the right N atom is sp hybridized.)
(c) Below are representations of five π molecular orbitals for cyanamide:
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Rank these from lowest to highest energy, label them as bonding, antibonding, or non-bonding, and fill the orbitals with the appropriate number
of electrons.
The top row are bonding, antibonding, and bonding; the bottom row are
nonbonding and antibonding. The bonding orbitals are lowest in energy,
the nonbonding in the middle, and the antibonding highest. It’s likely
that the bonding orbital on the left is lower in energy than the one on the
right, and the antibonding orbital on top higher in energy than the one on
bottom, but this is a detail, and it’s not critical to call it correctly in one
direction or the other: it doesn’t affect the description of bonding in the
molecule.
Counting electrons, there are four σ bonds that contain eight electrons,
and a lone pair (only one) in an sp hybrid orbital on the right nitrogen.
That leaves six electrons to put into π orbitals, and both bonding orbitals
are filled, as well as the nonbonding orbital.
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