Bonding and Moleculal Orbital Theory
Question
i) Construct an MO diagram for the Li2 molecule using the valence electrons only and show the atomic and
molecular orbitals involved.
ii) What is the bond order for the Li2 molecule?
ii) Compare your result with the MO diagram of H 2. Comment on similarities and differences.
Answer
i)
Li
Li
s*2s
Li2
2s
2s
s2s
ii) 1
iii) It looks exactly the same and the orbitals are the same shape. They will be different in energy, though, and
since they are made from 2s orbitals, they will be larger than the corresponding orbitals of H2. In principle,
there are also p-orbitals just like all of the other second row diatomics. All such orbitals will be empty, though.
Question
As carbocations go, allyl carbocations are relatively stable while vinyl carbocations are relatively unstable.
i.) For the allyl carbocation, how is the carbocation carbon hybridized?
ii.) How is the carbocation stabilized by the adjacent double bond?
iii.) For the vinyl carbocation, how is the carbocation carbon hybridized?
iv.) How does the carbocation interact with the double bond?
H
H
C C H
H2C
H
Allyl carbocation
C C
H
H
Vinyl carbocation
Answer
i.) The carbocation carbon is sp2 hybridized and has an empty atomic p orbital. This empty orbital can align
itself parallel to the double bond and be stabilized by overlap with it.
H C
H
C
H
C H
H
iii.) In this case the carbocation carbon is sp hybridized, as predicted by VSEPR. The HCC bond is linear. The
carbocation carbon uses its two hybrid orbitals to form the s-bonds to H and C. It has two unhybridized p
orbitals, the 2py and the 2pz WHICH ARE ORTHOGONAL. One of these orbitals makes up part of the p-bond.
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The other is vacant, and because it is orthogonal to the p-bond IT CANNOT OVERLAP WITH IN ANY
WAY and therefore cannot derive any stabilization.
p-bond in plane of page
H
C
C H
H
C
C H
H
H
p-orbital perpendicular to page
Question (4 marks)
Sketch on the diagram below the atomic orbitals (on the right), and the molecular orbital (on the left) that make
up the C-O p-bond of formaldehyde. Oxygen is more electronegative than carbon and this should be
taken into account in your drawing.
H
H
H
C
O
H
C
O
Answer
H
H
H
C
O
C
H
O
The resulting p -bond is polarized toward the O atom as you'd expect on the basis of O's electronegativity
despite the smaller size of the atomic p-orbital.
Question (3 marks)
The molecule shown below is called allene. It is in a class of molecules known as cumulenes.
i.) What is the hybridization of each of the three carbon atoms?
ii.) Draw a picture of allene (using wedge/dash bonds if necessary) to indicate its structure. You probably
want to build a model of this molecule.
iii.) How are the p-bonds of allene oriented with respect to one another?
H2C C CH2
Answer
i) The central C atom is sp hybridized - the other C atoms are sp2 hybridized.
ii)
H
H
H
C C C H
H
H
C C C H
H
iii) They are at right angles to one another. In the diagram above the p-bond on the right is in the plane of the
page - the one on the right is orthogonal to the page.
2
H
H
H
Csp
H
2
Csp
Csp
2
H
The s-framework.
H
C
C
C
H
H
The atomic orbitals making up
the p-bonds. Note that the two p-orbitals
that result must be orthogonal to each other.
Question (8marks)
The carbon-carbon double bond of ethylene consists of two different types of bond, given the symbols s and p.
H
H
C C
H
H
i.) How are the carbon atoms hybridized?
ii.) Sketch the orbitals (atomic and/or hybrid orbitals) that are used to form the s bond and sketch the bonding
molecular orbital that results from their overlap.
iii.) Sketch the orbitals (atomic and/or hybrid orbitals) that are used to form the p bond and sketch the bonding
molecular orbital that results from their overlap.
iv.) Which of the two types of bonds (s or p) is stronger and why?
v.) Rotation of the carbon-carbon bond in ethylene is not possible. Explain making reference to the bonding.
Answer
i) They are sp2 hybridized.
ii)
H
H
C
C
H
H
1s atomic orbital
H
C
sp2 hybrid
C
H
H
H
iii)
p-bond
Unhybridized atomic p orbitals
H
H
C
C
H
H
H
H
C
C
H
H
iv) Because s-bonds result in enhanced electron density in the region of space directly between the two positive
nuclei, they will be stronger. The enhancement of electron density in the p-bonds is above and below the plane
- less direct, therefore a p-bond is not quite as strong. Do not confuse the strength of a double vs. a single bond
with this question. Double bonds are stronger than single bonds because they are composed of a s- and a pbond.
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v) Rotation of one end of ethylene destroys the overlap of the atomic p-orbitals necessary for the formation of a
p-bond. Effectively, the bond must be broken for rotation to occur.
Question (8marks)
The carbon-carbon triple bond of acetylene consists of two different types of bond, given the symbols s and p.
H C C H
i.) What is the hybridization of the carbon atoms?
ii.) Sketch and label the atomic orbitals of carbon that are used to form the two hybrid orbitals and sketch the
hybrid orbital(s) that result from them.
iii.) Sketch the C-C s bonding molecular orbital that results from the overlap of the two hybrid orbitals.
iv.) Sketch the atomic orbitals that are used to form one of the p bonds and sketch the bonding molecular
orbital that results from their overlap.
Answer
i) They are sp hybridized.
ii)
Usually omitted
for clarity.
2s
C
C
2p
C
2 sp hybrids
C
2p
2s
iii)
C
C
C
C
s-bond
2 sp hybrids
iv)
p-bond
Unhybridized atomic p orbitals
H
C
C
H
H
C
C
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H
Question
Shown below is the MO diagram for the p-bond in cis-2-butene.
i.) On the left diagram, sketch the two molecular p-orbitals and the atomic orbitals from which they are
derived.
ii.) Add the electrons to the diagram.
iii.) If cis-2-butene is irradiated with UV light, then a facile bond rotation is possible:
H
H
hn
H
C C
H3C
CH3
C C
CH3
H3C
H
The light converts cis-2-butene into its first excited state (an electron excited state). In the right diagram, show
what has happened to the electron distribution in the p-MOs upon excitation.
v.) Why is bond rotation facilitated by irradiation?
Answer
ii and iii)
C
C
p*
p*
p
C
p
Ground state:
p bond order = 1
C
Excited state, p bond
order = 0
v) This excitation does not effect the s-bond. But the p-bond order becomes zero since there are now equal
numbers of bonding and antibonding electrons. The C-C bond effectively becomes single in character.
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Question
Sketch the bonding molecular orbital (MO) formed from the overlap of a N sp3 hybrid and hydrogen 1s atomic
orbital.
N
H
Answer
N
H
Question (four marks)
Metal carbonyls such as Fe(CO)5 are compounds made from transition metals and carbon monoxide. It is
believed that part of the metal-carbon bond results from the overlap of a filled (with electrons) dxy Fe orbital
with the p* orbital of carbon monoxide. These orbitals are shown below.
CO
OC
Fe
CO
CO
Fe
C
O
Fe
C
O
CO
Fe(CO)5
a.)
b.)
c.)
d.)
Sketch the bonding molecular orbital that results from the overlap of these two orbitals.
Does this MO resemble more closely a p-type or a s-type orbital? Why?
Draw lines to show the position of all nodes.
Which is longer: the C-O bond length in free carbon monoxide or the C-O bond length in Fe(CO)5 ?
Explain your answer.
Answer
a)
Fe
C
O
Fe
C
O
Nodes
b) It is not radially symmetrical and has a node in the plane of the Fe-C bond. It is a p-type of orbital.
c) Shown above.
d) The orbital drawn above is bonding with respect to the Fe and C atoms, but anti-bonding with respect to the
C and O atoms. If this orbital is populated with electrons, then you'd expect to see a weakening and lengthening
of the CO bond.
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