COVALENT BONDING
29
four electrons are accommodated one in each hybrid
orbital and one in the remaining 2p orbital.
The sp 2 hybrid orbitals are distributed in a
planar array around the atom; this spacing minimizes
any interactions. The 2p orbital is then located
perpendicular to this plane. Such information is
again obtained from the mathematical analysis, but
simple logic would lead us to predict that this is
the most favourable arrangement to incorporate the
components. The sp
2 orbital will be similar in shape
to the sp
3 orbital, but somewhat shorter and fatter,
in that it has more s character and less p character
(Figure 2.14).
The bonding in ethylene is based initially on one
C–C σ bond together with four C–H σ bonds, much
as we have seen in ethane. We are then left with a p
orbital for each carbon, each carrying one electron,
and these interact by side-to-side overlap to produce
a π bond (Figure 2.15). This makes the ethylene
molecule planar, with bond angles of 120
◦ , and the
π bond has its electron density above and below
this plane. The combination of the C–C σ bond and
the C–C π bond is what we refer to as a double
bond; note that we cannot have π bond formation
without the accompanying σ bond. You will observe
that it becomes progressively more difficult to draw a
combination of σ and π molecular orbitals to illustrate
the bonding that constitutes a double bond. We often
resort to a picture that illustrates the potential overlap
of p orbitals by means of a dotted line or similar
device. This cleans up the picture, but leaves rather
more to the imagination. The properties of an alkene
(like ethylene) are special, in that the π bond is more
reactive than the σ bond, so that alkenes show a range
of properties that alkanes (like ethane) do not (see
Chapter 8).
We can only get overlap of the p orbitals if their
axes are parallel. If their axes were perpendicular,
then there would be no overlap and, consequently,
no bonding (Figure 2.16). This situation might arise
if we tried to twist the two parts of the ethylene
molecule about the C–C link. This is not easily
achieved, and would require a lot of energy (see
Section 3.4.3). It can be achieved by absorbing
sufficient energy to promote an electron to the
antibonding π ∗ orbital. This temporarily destroys the
π bond, allows rotation about the remaining σ bond,
and the π bond may reform as the electron is restored
1/3 s
2/3 p
s p
2
+
top view:
three sp
2 orbitals in
planar array
side view:
three sp
2 hybrid orbitals +
one p orbital
120˚
Figure 2.14 sp
2 hybrid orbitals
π bond
π bond
formation of one
and
four
σ bonds, plus one
π bond in ethylene
for clarity, overlap of p
orbitals is represented
by the dotted lines
C
C
H
H
H
H
π molecular orbital
in ethylene
C C
C H
C C
Figure 2.15 Bonding in ethylene
29
four electrons are accommodated one in each hybrid
orbital and one in the remaining 2p orbital.
The sp 2 hybrid orbitals are distributed in a
planar array around the atom; this spacing minimizes
any interactions. The 2p orbital is then located
perpendicular to this plane. Such information is
again obtained from the mathematical analysis, but
simple logic would lead us to predict that this is
the most favourable arrangement to incorporate the
components. The sp
2 orbital will be similar in shape
to the sp
3 orbital, but somewhat shorter and fatter,
in that it has more s character and less p character
(Figure 2.14).
The bonding in ethylene is based initially on one
C–C σ bond together with four C–H σ bonds, much
as we have seen in ethane. We are then left with a p
orbital for each carbon, each carrying one electron,
and these interact by side-to-side overlap to produce
a π bond (Figure 2.15). This makes the ethylene
molecule planar, with bond angles of 120
◦ , and the
π bond has its electron density above and below
this plane. The combination of the C–C σ bond and
the C–C π bond is what we refer to as a double
bond; note that we cannot have π bond formation
without the accompanying σ bond. You will observe
that it becomes progressively more difficult to draw a
combination of σ and π molecular orbitals to illustrate
the bonding that constitutes a double bond. We often
resort to a picture that illustrates the potential overlap
of p orbitals by means of a dotted line or similar
device. This cleans up the picture, but leaves rather
more to the imagination. The properties of an alkene
(like ethylene) are special, in that the π bond is more
reactive than the σ bond, so that alkenes show a range
of properties that alkanes (like ethane) do not (see
Chapter 8).
We can only get overlap of the p orbitals if their
axes are parallel. If their axes were perpendicular,
then there would be no overlap and, consequently,
no bonding (Figure 2.16). This situation might arise
if we tried to twist the two parts of the ethylene
molecule about the C–C link. This is not easily
achieved, and would require a lot of energy (see
Section 3.4.3). It can be achieved by absorbing
sufficient energy to promote an electron to the
antibonding π ∗ orbital. This temporarily destroys the
π bond, allows rotation about the remaining σ bond,
and the π bond may reform as the electron is restored
1/3 s
2/3 p
s p
2
+
top view:
three sp
2 orbitals in
planar array
side view:
three sp
2 hybrid orbitals +
one p orbital
120˚
Figure 2.14 sp
2 hybrid orbitals
π bond
π bond
formation of one
and
four
σ bonds, plus one
π bond in ethylene
for clarity, overlap of p
orbitals is represented
by the dotted lines
C
C
H
H
H
H
π molecular orbital
in ethylene
C C
C H
C C
Figure 2.15 Bonding in ethylene