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The Solid State
Crystalline and Amorphous Solids
Ionic Crystals
“The universe consists only of atoms and the void: all else is opinion and
illusion.”–Edward Robert Harrison
Exam 3 Post-Mortem—Last time I taught 107
average score=81.2
actually rather high, although 3 points below exams 1&2
high score=98
the biggest problem area:
difficulty completing a calculation with no errors
Exam 3 Post-Mortem—April 2005
average score=82.1
actually rather high, although 3 points below exams 1&2
high score=100
the biggest problem area:
difficulty completing a calculation with no errors
A note on attendance policy. The syllabus says:
“The lowest score of the four exams (3 semester, 1 final) will be dropped.
However, you will not be allowed to drop the final exam unless
you attend at least 2/3 of the scheduled class meetings after
exam 3. If you attend less than 2/3 of the scheduled class meetings during
that time, I will drop the lowest of the three in-semester exam scores.”
This means exactly what it says. If you attend at least 2/3 of the
final nine class meetings and skip the final exam, a “0” goes in
your final exam score, and will be the dropped score.
If you attend less than 2/3 of the final nine class meetings and
skip the final exam, a “0” goes in your final exam score. The
dropped score will be the lowest of your 3 exams. The “0” for
the final will kill your grade!
Chapter 10
The Solid State
10.1 Crystalline and Amorphous Solids
Crystalline and amorphous are the two major categories into
which solids are divided.
Crystalline solids exhibit long-
range order in their atomic
arrangements.
Silicon crystal surface.
http://www.aip.org/history/einstein/atoms.htm
The order in crystals is usually three dimensional, but
lower dimensionality order is possible.
Bonds in crystalline solids are more or less the same in
energy, and crystalline solids have distinct melting
temperature.
mercury
(Don’t try
this at
home!)
Amorphous solids exhibit only short-range order in their atomic
arrangements. They lack long-range order.
Their bonds vary in energy and are weaker; they have no
distinct melting temperature.
A good example is B2O3. Here’s figure 10.1b, crystalline B203.
All crystals have atoms which are
“equivalent;” the symmetry of all
equivalent atoms is the same.
For example, these two boron atoms
are equivalent; everything around
them looks the same.
Imagine you are a nanohuman. If you build a house on some
boron atom and look out any window in some direction…
You’ll have exactly the same
view if you build your house on
an equivalent boron atom and
look out in an equivalent
direction.
Note that “equivalent direction”
does not mean “the same
direction.”
Here’s how you tell if a solid is crystalline:
See Figure 10.1a for amorphous B2O3.
Every B is surrounded by a
“triangle” of 3 O’s, just as for
crystalline B2O3.
But you can’t find two B atoms
that give you an identical “view.”
Eventually the short-range order
breaks down, and every “view” is
different.
“Amorphous” means “formless” or “shapeless” but amorphous
materials still have short-range order. The appropriate definition
of “amorphous” for us is “lacking a distinct crystalline structure.”
(www.dictionary.com)
Defects in Crystals
Image “borrowed” from http://comp.uark.edu/~pjansma/geol3513_25_defmechs1_04.ppt who got
it from “Davis & Reynolds 1996.”
It's easy to think of real crystals as having these ideal
structures.
In fact, no crystals are perfect; all crystals have defects.
Crystals can have:
• The "right" atoms in "wrong" places.
• "Wrong" atoms in "right" or "wrong" places.
• Missing atoms.
• Etc.
One type of defect is the point defect.
There are three basic kinds of point defects.
• (1) A vacancy.
Actually, a vacancy would probably look more like this:
• (2) An interstitial.
There is likely a vacancy somewhere else in the crystal, which
supplied the interstitial atom.
• (3) An impurity,
substitutional interstitial
• (3) An impurity, which could be either substitutional• (3) An impurity, which could be either substitutional or
interstitial.
Point defects makes diffusion in solids possible.
Either vacancies or interstitial atoms can migrate
through a crystal.
Diffusion is strongly temperature dependent.
Higher dimensional defects include edge and screw
dislocations. A dislocation occurs when a line of atoms is in the
wrong place.
Dislocations are important but more difficult to deal with
than point defects.
Edge dislocation are relatively easy to draw and visualize. See
Figure 10.3.
http://uet.edu.pk/dmems/edge_dislocation.htm dead link spring 2005
Atoms in crystals are not static, and neither are dislocations.
Screw dislocation are more difficult to draw. See Figures 10.4
and 10.5.
edge dislocation
edge dislocation
http://www.techfak.uni-kiel.de/matwis/amat/def_en/kap_5/illustr/a5_1_1.html
screw dislocation
http://uet.edu.pk/dmems/screw_dislocation.htm
Work hardening (you have probably tried this at home).
Work hardening occurs when so many dislocations are
formed in a material that they impede each others'
motion.
Hard materials are usually brittle.
Annealing.
Heating (annealing) a crystal can remove
dislocations. The edge dislocation animation
from Germany we saw showed an edge
dislocation being annealed out.
Annealing makes metals more ductile, and can be used
to remove secondary phases in crystals.
10.2 Ionic Crystals
An atom with a low ionization energy can give up an electron to
another atom with a high electron affinity. The result is an ionic
crystal.
To calculate the stability of an ionic crystal, we need to consider
all of the energies involved in its formation.
Positive energy is required to ionize an atom.
Energy is released when a highly electronegative atom
gains an electron (energy becomes more -).
There are ± contributions to the energy from the
Coulomb force between charged ions.
There is a + contribution to the energy from the overlap
of core atomic electrons (the Pauli exclusion principle at
work).
+
+
±
-
Remember that negative energies mean stable systems. If you
add up all the above energies and get a more negative energy
than for the separate, isolated atoms, then the ionic crystal is
stable.
We’ll go through that exercise soon.
Ionic crystals are generally close-packed, because nature
"wants" as many ions of different charge squeezed together as
possible.
Like-charged ions never come in contact.
There are two primary structures for ionic crystals.
Face-centered cubic (fcc).
http://members.tripod.com/shiner17/Crystals/BravaisLattice/
Cubic/FaceCenteredCubic.htm
An example of the fcc structure is sodium chloride (NaCl).
NaCl NaCl
a bit of the
front sliced off
http://sbchem.sunysb.edu/msl/nacl.html
The other main ionic crystal structure is body-centered cubic
(bcc).
An example of the bcc structure is cesium chloride (CsCl). Click
here for a model you can manipulate.
It is not too difficult to calculate the energies involved in ionic
bonding, so let’s do it.
We begin by defining the cohesive energy of an ionic crystal
as “the reduction in the energy per ion of the ionic crystal
relative to the neutral atoms,” or “the energy per ion needed to
break the crystal up into individual atoms.”
Later Beiser implicitly generalizes this definition to all
crystals.
Let's calculate the contribution to the cohesive energy from the
Coulomb potential energy. Let's do it for an fcc structure, NaCl
for example.
Let’s add up all contributions to the Coulomb energy from ion-
ion interactions. Let's take a Na+
ion as the reference* ion (NaCl
is made of Na+
Cl-
ions). We get the same result if we take a Cl-
as the reference.
Here’s the reference ion.
Each Na+ ion has six Cl-
nearest
neighbors a distance r away. Here are
four of them.
Where are the other two Cl’s?
r
showing these 4 Na’s just to help
you get your bearings
here
and
here
*Remember, the energy is
calculated per ion.
The contribution to the Coulomb potential from these six nearest
neighbors is
( ) ( )
π π
2
1
0 0
6 +e -e 6 e
U = = - .
4ε r 4 ε r
This represents a negative (more stable) contribution to the total
energy.
r 2
Each Na+
has 12 Na+
next-nearest
neighbors at a distance of 21/2
r.
Each Na+
has 12 Na+
next-nearest
neighbors at a distance of 21/2
r. These
are + ions, so the interaction is repulsive,
and the contribution to the total energy is
positive. This figure shows four of the
next-nearest neighbors.
There are four more (red) in the plane aboveThere are four more (red) in the plane above and four more in
the plane below.
“original” plane
“above”
“below”
The contribution from the twelve next-nearest neighbors is
π
2
2
0
12 e
U =+ .
4ε r 2
This represents a positive (less stable) contribution to the total
energy.
You could keep on like this for shell after shell. Pretty soon a
pattern would emerge. After “many” shells, you get
 
 π π 
2 2
0 0
e 12 e
U= - 6 - +... = -1.748
4ε r 4 ε r2
α
π
2
0
e
U= -
4ε r
The constant α (which is constant only for a given type of
structure) is known as the Madelung constant. Beiser gives
values of α for a couple of other structure types.
( )
 
α  
 
12
= 6 - +... = 6 - 8.485281+...
2
The convergence of this series is very poor. You have to find a
clever way to do this series if you want to calculate α with a
reasonable amount of effort.
We've accounted for the Coulomb attraction. Beiser calls this
Ucoulomb (using U instead of V as in the previous edition) so let’s
make it official:
α
π
2
coulomb
0
e
U = -
4ε r
This isn’t the whole story!
α
π
2
coulomb
0
e
U = -
4ε r
Note the - sign in the equation for U. The net coulomb
interaction is attractive.
In fact, the closer the ions, the more negative the energy. The
more negative, the more stable. What is the logical conclusion
from this observation?
As r gets smaller and smaller, electron shells start to overlap,
and electrons from different atoms share the same potential.
What does Pauli have to say about that?
Electrons would have to be promoted to higher energies to
allow atoms to come closer together. The result is a more
positive energy, i.e., a less stable crystal.
So we need to account for the repulsive forces that take over
when electron shells start to overlap, and different electrons
share the same set of quantum numbers.
We model this repulsive force with a potential of the form
repulsive n
B
U = ,
r
where n is some exponent. The exact value of n isn't too critical
(see the figure on the next page).
A simple scale factor change could make 1/r9
and 1/r10
look
nearly the same.
With the electron overlap repulsive energy accounted for, the
potential energy of interaction of our Na+
reference ion with all
the other ions is
α
π
2
total coulomb repulsive n
0
e B
U = U + U = - + .
4ε r r
Even though this says Utotal, we are not done! Do not use this
equation in a test/quiz problem!
For one thing, we have accounted for the total energy of
interaction of the ions (that’s what “total” means), but we haven’t
accounted for the energy required to ionize the neutral atoms.
For another thing, we have this adjustable parameter, B. (You
do remember adjustable parameters?)
Ucoulomb was called Vlattice in the previous text edition, so you’ll see that terminology in
homework solutions and prior exams.
At equilibrium the energy is minimized, which allows us to find B
in terms of α and the equilibrium near-neighbor separation r0.
∂ α 
 ∂ π  0
2
2 n+1
0 0 0r =r
U e nB
0 = = -
r 4ε r r
α
π
2
2 n+1
0 0 0
e nB
=
4ε r r
α
π
2
n-1
0
0
e
B= r
4ε n
In a sense, we have “adjusted” B to get the right answer
(energy minimum at equilibrium).
Plugging B back into the expression for U gives the total
potential energy at equilibrium separation. Beiser used to call it
the lattice energy, Vlat. Now he calls it U0.
 
α  π  
2
0
0 0
e 1
U = - 1- .
4ε r n
U0 is the reduction in the energy of the ionic crystal relative to
the ions at infinite separation.
We set out to calculate the cohesive energy, so we are still not
done yet.
The cohesive energy, Ecohesive, takes into account the energy
needed to create the ions.
Beiser doesn’t give an equation for the cohesive energy, so I’ll
make one up. The energy of the ions is U0. The energy to
create a Na+
ion is the ionization energy of Na. I’ll call that Uion.
When a Cl atom combines with an electron to form Cl-
, energy
is actually released. Chemists call this energy the electron
affinity. I’ll write that Uea.
coh, pair 0 ion eaE = U + U + U .
You’ll see what “pair” means in a minute. Also, the numerical
value of Ecoh had better be negative, right?
Uion is positive, because it is energy put into the crystal. Uea is
negative, because it is energy released from the crystal.
Part of your homework/quiz/exam responsibility is to get the
right signs on these energies!
Oh, and one more thing. We’ve calculated the coulomb energy
for our reference Na+
with all other ions in the crystal, one pair
at a time. Thus, U0 is the energy per ion pair.
Also, Uion is the energy to make Na+
and Uea is the energy
released on making Cl-
. Thus Uion + Uea represents the energy
needed to make a Na+
Cl-
pair.
Beiser defines the cohesive energy as “per ion.” Thus finally
(and officially):
0 ion ea
coh
U + U + U
E = .
2
Beiser on page 342 calculates the cohesive energy for
NaCl.
A “typical” value for n is 9. The equilibrium distance r0 between
Na+
and Cl-
ions is 0.281 nm.
Beiser calculates numerical values for the different energies,
and then combines them to get the cohesive energy.
We know it’s better to do the problem algebraically first, and get
numerical results only at the very end.
Let’s do it Beiser’s way here.
 
α  π  
2
0
0 0
e 1
U = - 1-
4ε r n
( )
( )
( )( )
×  
 
π × ×  
2
-19
0 -12 -9
1.6 10 1
U = - 1.748 1-
94 8.85 10 0.281 10
× -18
0U = -1.27 10 joules = -7.96 eV .
only for NaCl
structure!
The ionization energy of Na is 5.14 eV and the electron affinity
of Cl is 3.61 eV. With the correct signs, the energies are +5.14
eV and -3.61 eV. Make sure you understand why! Many of you
will lose several points because of sign errors.
0 ion ea
coh
U + U + U
E =
2
coh
-7.96 eV + 5.14 eV -3.61 eV
E =
2
cohE = -3.21 eV
Great news! We got a – sign for the cohesive energy. You can
sleep well at night knowing that NaCl is stable.
Wonder how it compares with experiment? The measured
value is -3.28 eV. That’s a 2% error. For this kind of
calculation, the agreement is good.
Why did I bother with this rather lengthy calculation to show that
NaCl is stable, when we already know it is stable?
Stupid?
Physicists may do stupid things, but we are not stupid.
Designer molecules! If you have an application that needs a
compound with special properties, do you spend millions in the
lab and hope you stumble across it…
Designer molecules! If you have an application that needs a
compound with special properties, do you spend millions in the
lab and hope you stumble across it… or do you spend a few
hundred thousand on computational modeling that tells you
what kind of compound to make?
The calculation we did is your first step towards making
designer molecules.
Some properties of ionic crystals, which Beiser mentions briefly:
• They have moderately high
melting points.
• They are brittle due to
charge ordering in planes.
• They are soluble in polar
fluids.

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Solid

  • 1. The Solid State Crystalline and Amorphous Solids Ionic Crystals “The universe consists only of atoms and the void: all else is opinion and illusion.”–Edward Robert Harrison
  • 2. Exam 3 Post-Mortem—Last time I taught 107 average score=81.2 actually rather high, although 3 points below exams 1&2 high score=98 the biggest problem area: difficulty completing a calculation with no errors Exam 3 Post-Mortem—April 2005 average score=82.1 actually rather high, although 3 points below exams 1&2 high score=100 the biggest problem area: difficulty completing a calculation with no errors
  • 3. A note on attendance policy. The syllabus says: “The lowest score of the four exams (3 semester, 1 final) will be dropped. However, you will not be allowed to drop the final exam unless you attend at least 2/3 of the scheduled class meetings after exam 3. If you attend less than 2/3 of the scheduled class meetings during that time, I will drop the lowest of the three in-semester exam scores.” This means exactly what it says. If you attend at least 2/3 of the final nine class meetings and skip the final exam, a “0” goes in your final exam score, and will be the dropped score. If you attend less than 2/3 of the final nine class meetings and skip the final exam, a “0” goes in your final exam score. The dropped score will be the lowest of your 3 exams. The “0” for the final will kill your grade!
  • 4. Chapter 10 The Solid State 10.1 Crystalline and Amorphous Solids Crystalline and amorphous are the two major categories into which solids are divided. Crystalline solids exhibit long- range order in their atomic arrangements. Silicon crystal surface. http://www.aip.org/history/einstein/atoms.htm
  • 5. The order in crystals is usually three dimensional, but lower dimensionality order is possible. Bonds in crystalline solids are more or less the same in energy, and crystalline solids have distinct melting temperature. mercury (Don’t try this at home!)
  • 6. Amorphous solids exhibit only short-range order in their atomic arrangements. They lack long-range order. Their bonds vary in energy and are weaker; they have no distinct melting temperature. A good example is B2O3. Here’s figure 10.1b, crystalline B203. All crystals have atoms which are “equivalent;” the symmetry of all equivalent atoms is the same. For example, these two boron atoms are equivalent; everything around them looks the same.
  • 7. Imagine you are a nanohuman. If you build a house on some boron atom and look out any window in some direction… You’ll have exactly the same view if you build your house on an equivalent boron atom and look out in an equivalent direction. Note that “equivalent direction” does not mean “the same direction.” Here’s how you tell if a solid is crystalline:
  • 8. See Figure 10.1a for amorphous B2O3. Every B is surrounded by a “triangle” of 3 O’s, just as for crystalline B2O3. But you can’t find two B atoms that give you an identical “view.” Eventually the short-range order breaks down, and every “view” is different. “Amorphous” means “formless” or “shapeless” but amorphous materials still have short-range order. The appropriate definition of “amorphous” for us is “lacking a distinct crystalline structure.” (www.dictionary.com)
  • 9. Defects in Crystals Image “borrowed” from http://comp.uark.edu/~pjansma/geol3513_25_defmechs1_04.ppt who got it from “Davis & Reynolds 1996.”
  • 10. It's easy to think of real crystals as having these ideal structures. In fact, no crystals are perfect; all crystals have defects. Crystals can have: • The "right" atoms in "wrong" places. • "Wrong" atoms in "right" or "wrong" places. • Missing atoms. • Etc. One type of defect is the point defect.
  • 11. There are three basic kinds of point defects. • (1) A vacancy. Actually, a vacancy would probably look more like this:
  • 12. • (2) An interstitial. There is likely a vacancy somewhere else in the crystal, which supplied the interstitial atom.
  • 13. • (3) An impurity, substitutional interstitial • (3) An impurity, which could be either substitutional• (3) An impurity, which could be either substitutional or interstitial.
  • 14. Point defects makes diffusion in solids possible. Either vacancies or interstitial atoms can migrate through a crystal. Diffusion is strongly temperature dependent. Higher dimensional defects include edge and screw dislocations. A dislocation occurs when a line of atoms is in the wrong place. Dislocations are important but more difficult to deal with than point defects.
  • 15. Edge dislocation are relatively easy to draw and visualize. See Figure 10.3. http://uet.edu.pk/dmems/edge_dislocation.htm dead link spring 2005
  • 16. Atoms in crystals are not static, and neither are dislocations. Screw dislocation are more difficult to draw. See Figures 10.4 and 10.5.
  • 20. Work hardening (you have probably tried this at home). Work hardening occurs when so many dislocations are formed in a material that they impede each others' motion. Hard materials are usually brittle. Annealing. Heating (annealing) a crystal can remove dislocations. The edge dislocation animation from Germany we saw showed an edge dislocation being annealed out. Annealing makes metals more ductile, and can be used to remove secondary phases in crystals.
  • 21. 10.2 Ionic Crystals An atom with a low ionization energy can give up an electron to another atom with a high electron affinity. The result is an ionic crystal. To calculate the stability of an ionic crystal, we need to consider all of the energies involved in its formation. Positive energy is required to ionize an atom. Energy is released when a highly electronegative atom gains an electron (energy becomes more -). There are ± contributions to the energy from the Coulomb force between charged ions. There is a + contribution to the energy from the overlap of core atomic electrons (the Pauli exclusion principle at work). + + ± -
  • 22. Remember that negative energies mean stable systems. If you add up all the above energies and get a more negative energy than for the separate, isolated atoms, then the ionic crystal is stable. We’ll go through that exercise soon. Ionic crystals are generally close-packed, because nature "wants" as many ions of different charge squeezed together as possible. Like-charged ions never come in contact.
  • 23. There are two primary structures for ionic crystals. Face-centered cubic (fcc).
  • 25. An example of the fcc structure is sodium chloride (NaCl). NaCl NaCl a bit of the front sliced off http://sbchem.sunysb.edu/msl/nacl.html
  • 26. The other main ionic crystal structure is body-centered cubic (bcc).
  • 27.
  • 28. An example of the bcc structure is cesium chloride (CsCl). Click here for a model you can manipulate. It is not too difficult to calculate the energies involved in ionic bonding, so let’s do it. We begin by defining the cohesive energy of an ionic crystal as “the reduction in the energy per ion of the ionic crystal relative to the neutral atoms,” or “the energy per ion needed to break the crystal up into individual atoms.” Later Beiser implicitly generalizes this definition to all crystals. Let's calculate the contribution to the cohesive energy from the Coulomb potential energy. Let's do it for an fcc structure, NaCl for example.
  • 29. Let’s add up all contributions to the Coulomb energy from ion- ion interactions. Let's take a Na+ ion as the reference* ion (NaCl is made of Na+ Cl- ions). We get the same result if we take a Cl- as the reference. Here’s the reference ion. Each Na+ ion has six Cl- nearest neighbors a distance r away. Here are four of them. Where are the other two Cl’s? r showing these 4 Na’s just to help you get your bearings here and here *Remember, the energy is calculated per ion.
  • 30. The contribution to the Coulomb potential from these six nearest neighbors is ( ) ( ) π π 2 1 0 0 6 +e -e 6 e U = = - . 4ε r 4 ε r This represents a negative (more stable) contribution to the total energy. r 2 Each Na+ has 12 Na+ next-nearest neighbors at a distance of 21/2 r. Each Na+ has 12 Na+ next-nearest neighbors at a distance of 21/2 r. These are + ions, so the interaction is repulsive, and the contribution to the total energy is positive. This figure shows four of the next-nearest neighbors.
  • 31. There are four more (red) in the plane aboveThere are four more (red) in the plane above and four more in the plane below. “original” plane “above” “below”
  • 32. The contribution from the twelve next-nearest neighbors is π 2 2 0 12 e U =+ . 4ε r 2 This represents a positive (less stable) contribution to the total energy. You could keep on like this for shell after shell. Pretty soon a pattern would emerge. After “many” shells, you get    π π  2 2 0 0 e 12 e U= - 6 - +... = -1.748 4ε r 4 ε r2 α π 2 0 e U= - 4ε r
  • 33. The constant α (which is constant only for a given type of structure) is known as the Madelung constant. Beiser gives values of α for a couple of other structure types. ( )   α     12 = 6 - +... = 6 - 8.485281+... 2 The convergence of this series is very poor. You have to find a clever way to do this series if you want to calculate α with a reasonable amount of effort. We've accounted for the Coulomb attraction. Beiser calls this Ucoulomb (using U instead of V as in the previous edition) so let’s make it official: α π 2 coulomb 0 e U = - 4ε r
  • 34. This isn’t the whole story! α π 2 coulomb 0 e U = - 4ε r Note the - sign in the equation for U. The net coulomb interaction is attractive. In fact, the closer the ions, the more negative the energy. The more negative, the more stable. What is the logical conclusion from this observation? As r gets smaller and smaller, electron shells start to overlap, and electrons from different atoms share the same potential. What does Pauli have to say about that?
  • 35. Electrons would have to be promoted to higher energies to allow atoms to come closer together. The result is a more positive energy, i.e., a less stable crystal. So we need to account for the repulsive forces that take over when electron shells start to overlap, and different electrons share the same set of quantum numbers. We model this repulsive force with a potential of the form repulsive n B U = , r where n is some exponent. The exact value of n isn't too critical (see the figure on the next page).
  • 36. A simple scale factor change could make 1/r9 and 1/r10 look nearly the same.
  • 37. With the electron overlap repulsive energy accounted for, the potential energy of interaction of our Na+ reference ion with all the other ions is α π 2 total coulomb repulsive n 0 e B U = U + U = - + . 4ε r r Even though this says Utotal, we are not done! Do not use this equation in a test/quiz problem! For one thing, we have accounted for the total energy of interaction of the ions (that’s what “total” means), but we haven’t accounted for the energy required to ionize the neutral atoms. For another thing, we have this adjustable parameter, B. (You do remember adjustable parameters?) Ucoulomb was called Vlattice in the previous text edition, so you’ll see that terminology in homework solutions and prior exams.
  • 38. At equilibrium the energy is minimized, which allows us to find B in terms of α and the equilibrium near-neighbor separation r0. ∂ α   ∂ π  0 2 2 n+1 0 0 0r =r U e nB 0 = = - r 4ε r r α π 2 2 n+1 0 0 0 e nB = 4ε r r α π 2 n-1 0 0 e B= r 4ε n In a sense, we have “adjusted” B to get the right answer (energy minimum at equilibrium).
  • 39. Plugging B back into the expression for U gives the total potential energy at equilibrium separation. Beiser used to call it the lattice energy, Vlat. Now he calls it U0.   α  π   2 0 0 0 e 1 U = - 1- . 4ε r n U0 is the reduction in the energy of the ionic crystal relative to the ions at infinite separation. We set out to calculate the cohesive energy, so we are still not done yet. The cohesive energy, Ecohesive, takes into account the energy needed to create the ions.
  • 40. Beiser doesn’t give an equation for the cohesive energy, so I’ll make one up. The energy of the ions is U0. The energy to create a Na+ ion is the ionization energy of Na. I’ll call that Uion. When a Cl atom combines with an electron to form Cl- , energy is actually released. Chemists call this energy the electron affinity. I’ll write that Uea. coh, pair 0 ion eaE = U + U + U . You’ll see what “pair” means in a minute. Also, the numerical value of Ecoh had better be negative, right? Uion is positive, because it is energy put into the crystal. Uea is negative, because it is energy released from the crystal.
  • 41. Part of your homework/quiz/exam responsibility is to get the right signs on these energies! Oh, and one more thing. We’ve calculated the coulomb energy for our reference Na+ with all other ions in the crystal, one pair at a time. Thus, U0 is the energy per ion pair. Also, Uion is the energy to make Na+ and Uea is the energy released on making Cl- . Thus Uion + Uea represents the energy needed to make a Na+ Cl- pair. Beiser defines the cohesive energy as “per ion.” Thus finally (and officially): 0 ion ea coh U + U + U E = . 2
  • 42. Beiser on page 342 calculates the cohesive energy for NaCl. A “typical” value for n is 9. The equilibrium distance r0 between Na+ and Cl- ions is 0.281 nm. Beiser calculates numerical values for the different energies, and then combines them to get the cohesive energy. We know it’s better to do the problem algebraically first, and get numerical results only at the very end. Let’s do it Beiser’s way here.
  • 43.   α  π   2 0 0 0 e 1 U = - 1- 4ε r n ( ) ( ) ( )( ) ×     π × ×   2 -19 0 -12 -9 1.6 10 1 U = - 1.748 1- 94 8.85 10 0.281 10 × -18 0U = -1.27 10 joules = -7.96 eV . only for NaCl structure! The ionization energy of Na is 5.14 eV and the electron affinity of Cl is 3.61 eV. With the correct signs, the energies are +5.14 eV and -3.61 eV. Make sure you understand why! Many of you will lose several points because of sign errors.
  • 44. 0 ion ea coh U + U + U E = 2 coh -7.96 eV + 5.14 eV -3.61 eV E = 2 cohE = -3.21 eV Great news! We got a – sign for the cohesive energy. You can sleep well at night knowing that NaCl is stable. Wonder how it compares with experiment? The measured value is -3.28 eV. That’s a 2% error. For this kind of calculation, the agreement is good.
  • 45. Why did I bother with this rather lengthy calculation to show that NaCl is stable, when we already know it is stable? Stupid? Physicists may do stupid things, but we are not stupid. Designer molecules! If you have an application that needs a compound with special properties, do you spend millions in the lab and hope you stumble across it… Designer molecules! If you have an application that needs a compound with special properties, do you spend millions in the lab and hope you stumble across it… or do you spend a few hundred thousand on computational modeling that tells you what kind of compound to make? The calculation we did is your first step towards making designer molecules.
  • 46. Some properties of ionic crystals, which Beiser mentions briefly: • They have moderately high melting points. • They are brittle due to charge ordering in planes. • They are soluble in polar fluids.