October 5th, 2010
Class 8
Crystal Binding
Now we will concentrate on what keeps the atoms together in a crystal. In all the cases the
reason can be tracked down to an electrostatic interaction of some sort. Magnetic interactions
are usually very weak and gravitational forces are simply negligible at this level.
Cohesive Energy: Energy needed to break apart the crystal into individual neutral atoms at rest
and at infinite separation
Lattice Energy: Energy needed to break apart a crystal into individual ions at rest and at infinite
separation (for ionic crystals)
For ionic crystals, one can be obtained from the other by appropriately adding or subtracting the
energy needed to ionize the atoms (electron affinity to create the anion and ionization potential to
create the cation). Indeed, if the cohesive energy is known, the lattice energy can be obtained by
adding to it the energy needed to remove on electron from the (to be) cation (ionization potential)
and subtracting the energy that it is recovered when the electron is added to the (to be) anion. If
instead the lattice energy is known, the process described above can be reversed to get the
cohesive energy.
Example,
1) The cohesive energy per pair of KF in the crystal is 7.29eV. In kcal/mol, what is the
lattice energy per KF pair? (compare with the value in table 7, page 66)
2) The lattice energy per pair of RbBr in the crystal is 152.6kcal/mol. In eV, what is the
cohesive energy per RbBr pair?
The cohesive energy vary greatly from element to element (distribute table 1), from very small
(inert gases), to intermediate (alkali metals), to large (C, Si, Ge, etc), to very large like for some
transition metals. Cohesive energy is more than just a number, it is related to crystal properties
such us melting temperature, bulk modulus, etc
Different type of crystals are formed depending on the nature of the interaction between the
atoms in the crystal, four different types of bonds are Van der Waals, Ionic, Covalent, and
metallic. Crystal can be formed when atoms bond according to each of these types of bonds,
however most crystals will show a combination of these types of bond.
Inert Gases
Atoms in the last column, known as inert gases, are characterized for having an electronic
structure which forms a closed shell. That means, it is very difficult to add electrons to
(Negative EA) or subtract (Large IP) electrons from the atom and thus they are not very reactive.
From ab initio calculations on a Ne atom, it is obtained that ~21eV need to be added to remove
an electron and ~35eV need to be added to add an electron (the electron has to be pushed in).
These are huge numbers considering that cohesive energies per atom are in the order of 1-10eV
per atom.
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However, even at absolute zero, the electron density would oscillate leading to the instantaneous
formation of dipoles that interact with instantaneous dipoles on other atoms. As a consequence,
there is a dipole-dipole interaction between adjacent atoms that keep them together; this is
known as Van der Waals-London interaction. It can be shown, with a simple model of two
coupled harmonic oscillators, that the strength of this interaction decreases (or approaches zero
since it is negative) with distance from the atom, proportionally to r-6 U=-A/r6. This interaction
is a quantum effect and does not depend on the overlap of electron density.
Thus, when close enough they’ll remain close to each other, however they are not allowed to get
too close. As the nuclei approach each other, the electron density around them overlap, when
this happen, there is a tendency for electron on one atom to occupy energy levels on the other,
however, according to the Pauli Exclusion Principle; two electrons cannot have all their quantum
number equal and so for closed shell atoms or atoms whose electrons have like spins, the only
way their electronic structure can overlap is if some of the electrons are promoted to higher
energy level what cost energy, so they can get closer to each other only if some energy is spent.
To better understand, consider two H atom, each has 1 electron in the 1s level. As they approach
each other, the two atomic levels give rise to two molecular levels, one with lower and the other
with higher energy than the atomic level. Both electrons cannot be accommodated in the lower
level since they have the same spin so one has to go into the higher energy level thus increasing
the energy.
This effect is known to rapidly decrease with the distance between atoms as B/r12. Considering
the two interactions together we get:
U ( R) 
  12    6 
A
B


4

     
r 12 r 6
 r  
 r 
The above is known as the Lennard-Jones potential. Where A=4εσ6 and B=4εσ12. ε is the
energy at the minimum of this potential, and σ is the distance between atoms that makes the
energy to be zero. Distribute the plot of the Lennard-Jones potential
There are other models, particularly for the repulsive part of the interaction, that are sometimes
used, an example is the exponential, e
R

where  is he range of the interaction.
In table 4, the unit for  are 10-16 erg, to convert it to eV, multiply by1x10-16, to account for the
scale factor, now the number is in ergs. Then multiply by 1e-7 Joules/erg and divide by 1.602e-19
Joules/eV (numerically equal to the electron charge).
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He
Ne
Ar
Kr
Xe
8.74e-4 eV
3.12e-2 eV
1.04e-2 eV
1.40e-2 eV
2.00e-2 eV
Equilibrium lattice constant
Considering only contributions from the binding energy to the cohesive energy (i.e. kinetic
energy is ignored). In the calculations below an atomic basis with all identical atoms is assumed.
U tot
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   12
   
1
  
 
 N 4   


2
 j  i  p ij R 
j  i  p ij R  


Where pij is the atomic distance in term of first neighbor’s distance and R is the first neighbor
distance. The cohesive energy is obtained by “picking” one atom, calculating the binding energy
with every other in the lattice and then multiplying by N. This implicitly uses the fact that each
atom is exactly identical to any other atom due to the periodicity of the crystal. The ½ factor
comes to correct for counting each pair twice.
For a fcc lattice the fallowing values are obtained
p
j i
12
ij
p
 12.13188
j i
6
ij
 14.45392
Notice than in an fcc lattice there are 12 first neighbors (for which pij=1) the values shown above
imply that the main contribution to the cohesive energy is by first neighbors (if the summation is
performed ONLY considering the 12 first neighbors, then both summations will give exactly 12,
thus the difference with 12 is the contribution from second, third, and other neighbors. To
compare, notice that there are 6 second neighbors in an fcc lattice at a distance a  2R thus
pij  2 and
p
 
12
ij
p
 0.09375
 
j 2nd
6
ij
 0.75
j 2nd
The equilibrium distance can be found by requiring the total energy to be a minimum
dU tot
dR
R  Ro

 12
6
 2 N 12 12.13188 13  614.45392 7   0
Ro
Ro 

what gives Ro/=1.09
By using this value back into the Lennard-Jones potential, the cohesive energy is found to be.
Utot(R)=-2.15 4N for ALL inert gases.
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Notice how close the theoretical value for Ro/ is to the value for calculated using Ro and 
determined experimentally.

2.74
3.40
3.65
3.98
Ro
3.13
3.76
4.01
4.35
Ne
Ar
Kr
Xe
Ro/
1.14
1.11
1.10
1.09
The difference in Ro/ values between the atoms is due to the kinetic energy not considered in
the description presented above. Let’s consider these atoms as particles in a box (where the
boundaries of the box are determined by the crystal constant). These atoms then have a
DeBroglie wave length determined by the constraint (standing wave) and related to the
momentum by p=h/ and thus the kinetic energy due to the zero point energy (the longest wave
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h
that can fit in the box) is K   . Thus the kinetic energy is inversely proportional to the
2M
mass. In order to partially compensate for this, the equilibrium distance is slightly larger than
predicted by the theory that assumes atoms at rest. By increasing the equilibrium distance, the
wavelength increases and thus the kinetic energy is smaller that it would be if the binding
distance is not altered. The larger the atom, the smaller the kinetic energy and the smaller the
correction and thus the closest the theoretical values from the experimental values. The binding
energy is reduced with respect to the theoretical values by 28%, 10%, 6%, and 4% respectively
for Ne, Ar, Kr, and Xe.
 
Ionic Crystals
Ionic crystals are formed for atomic basis with two different atoms such that one looses one
electron and the other gains one electron so both become ions. Particularly, ionic bonds occur
between atoms that have a closed shell+1 electron with atoms that have a close shell minus an
electron, those are atoms in the first column and atoms in the last but one columns respectively,
like Na and Cl, or Cs and Cl, or Li and F. Atoms in the first column have a very small IP, what
means that it takes very little energy to remove one electron from them. On the other hand,
elements in the last but one column are one electron from being inert gases and as that they have
a very large EA what means a lot of energy is released if an extra electron is added to it, in other
words the negative ion (anion) is very stable.
For ionic metals, almost all the binding energy is due to the coulombic attraction between ions
(there is a minimal contribution from Van der Waals forces what accounts for 1 or 2% only).
The coulomb law indicates that the potential energy when two ions are separated by a distance r
is q2/4or where the “+” sign if for pairs with the same charge and the “–” is for pairs of
different charge. (Notice that Kittel uses the CGS unit system where the coulombic potential
energy is simply q2/r. I chose to use the SI units. To go back and forth between these two units
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systems do the following transformation SI
q2
 q 2 CGS where q is in coulombs in the SI
4o
but in electrostatic units in the CGS
Equilibrium distance will be determined by considering a coulombic attraction/repulsion and a
repulsion of the same nature of the Van der Waals repulsion. The interaction energy between
two ions is then given by
U ij  e

rij


q2
4o rij
Where an exponential model for the repulsion term is used. Notice that for a crystal with N ionic
pairs, the total crystal energy is
 R 
q2

e

for nearest neighbors

4o Ro

U tot  N  U ij where U ij  
q2
j i
 1
otherwise
 pij 4o Ro
Notice that it was implicitly assumed that there is no contribution to the repulsion from not
nearest neighbors. This is a more than appropriate assumption since there will be no overlap
between the electron density for two atoms that are not near each other.

1
q 2 
R 

 where z is the number of nearest neighbors and    
is the
U tot  N  ze 
pij
4o Ro 
j i

Madelung Constant and is an indication of the effect of all neighbors, not only the first neighbor.
Consider a linear lattice with ions with alternate charges at a distance R from each other. The
Madelung constant can be written as the series:
 1 1 1

  2 1     
 2 3 4

 x 2 x3 x 4

Using the following identity ln 1  x    x     
2
3 4


Then =2 ln 2 = 1.386 for a linear ionic crystal.
To find the equilibrium distance, we need to take the derivative of the energy and make it equal
R
 o
 q 2
to zero. By doing that the following condition is obtained: ze  
Ro 4 o Ro
What inserted back into the energy equation leads to:
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U tot  
Nq 2
4o Ro

 
1  
 Ro 
Nq 2
is the Madelung energy. For most ionic crystals ~0.1Ro. Remember that N is
4o Ro
the number of ionic pairs which of course it depends on the crystal size, thus in general you will
be solving problems considering the energy “per ionic pair”.
where 
Covalent Crystals
Covalent bonds are formed between two atoms of the same type. Each bond is populated by two
electrons, equally shared between the atoms.
The two types of crystals defined above, Van der Waals and Ionic crystals, are characterized by
non-directional bonds, this type of bonds allow for closed packing with up to 12 nearest
neighbors. Covalent bonds are highly directional and allow for only a small number of neighbors
leading to less compact structures. For instance, C atoms can only form up to 4 bonds, when
they do; they bind on a tetrahedral configuration to only 4 first neighbors leading to a diamond
structure. The packing fraction of the diamond structure is 0.34 compared to the packing
fraction for hcp or fcc which is 0.74.
Between ionic and covalent, there is a range of crystals where bonds have partially ionic
and partially covalent nature.
Metallic Bond
Metallic bond are characterized by a lattice of ions held together by electrons that are (almost)
free to move around the structure and cannot be associated with any of the ions in particular, but
belong to all of them. Each atom usually contributes with 1 or 2 electrons. Transition metals
have rather complicated structures with electrons in d levels and are characterized by a large
binding energy. The metallic bond is also non-directional allowing for high coordination and
compact structures such us hcp and fcc, although bcc structures with a packing fraction of 0.68
are also commonly observed.
Hydrogen bond
According to the electronic structure of H, bonding should be limited to diatomic molecules.
However when H bonds to an atom other than H, H usually partially loses its electron and
becomes positive, actually when bonded to highly electronegative atoms like F it practically
looses the electron entirely and become a proton. Now it can bond to second atom mainly by an
ionic bond. Usually only two atoms can simultaneously bond to H.
H bond is what allow water to be liquid and it is extremely important in DNA and other
biological molecules interactions.
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