Page 87 - Electrical Properties of Materials

Bond types 69

For the purpose of making some rough calculations, the characteristic
curve of Fig. 5.1 may be approximated by the following simple mathematical
expression,
A B
E(r)= – , (5.8)
r n r m
where the first term on the right-hand side represents repulsion and the second
term attraction. By differentiating eqn (5.8) we can get E c the minimum of the
E(r) curve at the equilibrium distance r = r 0 ,inthe form
B m
E c = m –1 . (5.9)
r n
0
For a stable bond, E c < 0, which can be satisfied only if The repulsive force has a higher
index than the attractive one.
m < n. (5.10)

5.3 Bond types
There is no sharp distinction between the different types of bonds. For most
bonds, however, we may say that one or the other mechanism dominates. Thus,
a classification is possible; the four main types are: (i) ionic, (ii) metallic,
(iii) covalent, and (iv) van der Waals. ∗ ∗ Johannes Diderik Van der Waals,
Nobel Prize, 1910.
5.3.1 Ionic bonds
A typical representative of an ionic crystal is NaCl, which we have already
discussed in some detail. The crystal structure is regular and looks exactly like
the one shown in Fig. 1.1. We have negatively charged Cl ions and positively
charged Na ions. We may now ask the question, what is the cohesive energy of
this crystal? Cohesive energy is what we have denoted by E c in Fig. 5.1, that is
the energy needed to take the crystal apart. How could we calculate this? If the
binding is due mainly to electrostatic forces, then all we need to do is to sum
the electrostatic energy due to pairs of ions.
Let us start with an arbitrary Na ion. It will have six Cl ions at a distance,
a, giving the energy,
e 2 6
– . (5.11)
4π 0 a
√
There are then 12 Na ions at a distance a 2 contributing to the energy by the
amount
e 2 12
√ . (5.12)
4π 0 a 2
√
Next come eight chlorine atoms at a distance a 3, and so on. Adding up the
contributions from all other ions, we have an infinite sum (well, practically
infinite) of the form
e 2 6 12 8
– – √ + √ ··· . (5.13)
4π 0 a a 2 a 3
We have to add together sums such as eqn (5.13) for every Na and Cl ion to get
the cohesive energy. It would actually be twice the cohesive energy, because