The fundamental aspects of valence bond theory as outlined
on the previous page are that superposition of wavefunctions
leads to a low energy bonding wavefunction in which the probability
of finding the electrons in the internuclear region has been
substantially increased.
This discussion centred on the H2 molecule, which
contains a single σ bond. To discuss more complicated molecules
it is necessary to extend the theory slightly. For example,
the molecule N2 contains only two atoms, but instead
of a single σ bond has a triple bond joining the two atoms.
The electron configuration of a nitrogen atom is 1s22s22px12py12pz1
where the z axis is, by convention, taken to be the molecular
axis.
Each of the atoms that go to make up an N2
molecule therefore have a 2pz orbital pointing towards
a like orbital on the other atom. These two orbitals can overlap
to form a σ bond, as they have the correct cylindrical
symmetry about the internuclear axis. The spatial wavefunction
for this bond has the following form:

which has the same form as for the σ bond in the H2
molecule. However, in this case A and B refer to nitrogen 2p
orbital wavefunctions on atoms A and B, rather than to hydrogen
1s orbitals.
The 2px and 2py orbitals do not have the
correct cylindrical symmetry to overlap and form two more σ
bonds. Instead, the orbitals overlap with their counterpart
on the other atom to form two π bonds. π bonds arise
from the side-on overlap of p orbitals, and have a nodal plane
(which is the plane in which the atomic nuclei lie). Sausage-shaped
regions of electron density lie above and below the plane:
There are two π bonds in N2, one formed from
overlap and spin pairing between the px orbitals,
and one formed likewise from the py orbitals.
Further extension of the theory is necessary to explain the
bonding in carbon. The ground-state electronic configuration
of a carbon atom is 1s22s22px12py1,
which suggests that carbon should form two covalent bonds, using
its two unpaired electrons. However, experience shows that carbon
is almost invariably tetravalent (forms four bonds) in its stable
compounds.
The explanation for this is that a carbon atom can promote
one of its 2s electrons into its vacant 2pz orbital.
Though energy is required for this, it is an overall energetically
favourable process due to the extra energy recovered from forming
four bonds rather than two.
The electronic configuration of carbon once it has performed
this promotion is 1s22s12px12py12pz1,
which still leaves a problem. It implies that of the four bonds
formed by a carbon atom, three will involve a carbon 2p orbital
and the other one will involve a carbon 2s orbital, so should
show different properties (length, strength) to the other three.
However, in the case of a molecule such as methane, CH4
, the four C-H bonds are entirely equivalent. The explanation
of this lies in a process termed hybridisation.
This is where the four valence orbitals 'mix' (or, more accurately,
their wavefunctions interfere with each other) to produce four
equivalent hybrid orbitals. These hybrid orbitals are called
sp3 orbitals (as they are made from one
s and three p orbitals), and are arranged so that they point
towards the corners of a regular tetrahedron (this being the
arrangement in which all the orbitals occupy an equivalent position
and are separated by the maximum possible amount). Each hybrid
orbital contains one electron, so can pair with another half-filled
orbital to form a σ bond. The spatial arrangement of the
hybrid orbitals explains the tetrahedral shape of molecules
such as methane, and since all four hybrid orbitals are identical,
this explains the equivalence of the four C-H bonds.
Finally we consider unsaturated molecules such as ethene. The
carbons in this molecule have a double bond between them, composed
of one σ and one π bond. This indicates that there
must be one unhybridised p orbital in the valence shell of each
carbon atom (to overlap and form the π bond). The remaining
two 2p orbitals and the 2s orbital hybridise to produce three
identical sp2 orbitals. The lowest energy
arrangement of these three orbitals is when they are pointing
to the corners of an equilateral triangle (this has the maximum
separation between the orbitals). In this situation, the angle
between two adjacent orbitals is 120º. The sp2
orbitals define a plane, and the unhybridised p orbital lies
at 90º to this plane.
This explains well the shape of the ethene molecule. The σ
bonds around each carbon are arranged at 120º to each other,
as they are formed to the three sp2 orbitals.
The hydrogen atoms all lie in the same plane, as in this conformation
the unhybridised p orbitals on the two carbons lie parallel
to one another, allowing maximum overlap and thus maximum strength
of the π bond.
In molecules such as ethyne, which contains a triple bond, there
are two unhybridised p orbitals on each carbon. The remaining
p orbital and the s orbital hybridise to give two sp hybrid
orbitals. The minimum energy arrangement for these is at 180º
to each other. Thus the σ bonds in ethyne are at 180º to
each other, resulting in the molecule's linear shape.
Note throughout the important and
general result that the number of hybrid orbitals produced is
equal to the total number of orbitals that were mixed together
to create them.
All the types of hybrid orbital described above have broadly
similar shapes, which look something like this:
Bonds are usually formed with the major lobe, as the minor
lobe is too small to overlap effectively with other orbitals.
|