start remember last time
we started Chapter twelve solids and
modern materials let's look at what we
have done in fact so far in this chapter
we simply classify the solids according
to the forces that holds the particle
together in a solid and in this respect
we classify them as metallic solids
ionic solids covalent networks and
molecular solids in addition to that we
also have polymers and nano materials
now we also classified the solid into
two as crystalline solids and amorphous
solids crystalline solids are the solids
which have a very regular pattern in
3-dimensional actually structure in
their three-dimensional structure there
is a repeating unit layer actually and
that repeating unit we said is called
the unit cell so the unit cell actually
contains all the structural properties
of the crystalline lattice and also
tells you the atoms and the molecules
whatever in the solid relative positions
of those so the unit cell is very very
important for crystalline solids now
amorphous solids there is no regular
pattern again no order of the atoms in
the solid entry dimensions and they have
different properties and it's very
difficult to simply define more for
solids because every time you have a
different amorphous solids even though
you have the same elements there and we
also said that the crystalline solids
have very sharp melting points but the
amorphous solids do not have sharp
melting points and we also said that we
will simply study the crystalline solids
here not the amorphous solids so we
talked about the unit cells and ledges
and lattice points actually lettuce is
the
three dimensional array of the
crystalline structures and lattice
points defines the lattice structure so
we may have a atom or molecule at the
lattice points or we may not but usually
the lattice points are occupied by the
atoms or the molecules of the solids
again we talked about this how we define
the lettuces and we said that there
simply seven lettuces or I should
probably call unit cell forms these are
called the primitive unit cells and
cubic one is the most we will simply
deal with in this chapter and maybe the
others but the cubic is the most
important then we can derive other unit
cells or lattices in fact from those
primitive lattices by simply placing an
atom and the surface in the middle of
the surface a surface of the unit cell
or in the center of the unit cell or on
middle of edges and so on and for
example if we take the cubic unit cell
there are three types of those one is
that the primitive lattice primitive
cubic actually and the other one is
called body centered and the other one
is called face centred body centered or
face centered this is simply primitive
cubic this is body centered cubic we
have another atom right over here and
face centered is we have a cube we have
the atoms at the corner and also the
atoms in the center of each faces so
these that we should simply be familiar
with because we will be using those in
this chapter so what we said again this
simply says that the lattice points may
be occupied by atoms molecules or group
of other atoms and those groups that
occupies the lattice points
are called motives we talk about that
now when we consider the metallic
structures it would be very easy
actually to consider the solid metal
structures because metallic solids
metals in the solids are crystalline
solids and the unit in that case our
metal atoms right so metal atoms occupy
the lattice points and I mean you may
have simply primitive lattice or you may
have body centered or you may have face
centered type lattices now the metals
are all same sized atoms so you can
simply consider metal as if collections
of the same sized sized spheres right so
when you simply consider as metal atoms
that's fair here the lattice structures
or the unit cell would look like this
again the primitive unit cells and body
centered and face centered and so on now
actually the unit cell is the part that
is inside of the cube drawn here as a
matter of fact so when you have a metal
atom or a sphere at the corner of the
cube that is shared by eight cubes or
eight unit cells only one-eighth of
those is in this and if the in the body
centered actually the one in the middle
is not shared by any others and face
centered ones the ones which in the
middle of the face are shared by two
unit cells so considering that we simply
said that if we try to find out how many
metal atoms belongs to one unit cell or
how many or let's say let's say this way
how many metal atoms corresponds to one
unit cell for the simple cubic unit cell
cases
eight corners we have eight metal atoms
and one over eight of those only in the
unit cell here cubic unit cell so that
simply makes one atom per unit cell now
here in the body centered two atoms per
unit cell one comes from the corner
eight corners and one in the middle and
in the face center case we have simply
one coming from the corners and three
from the faces six faces and that makes
that two actually three four metal atoms
makes a face centred unit cell or face
centered unit cell is made of four atoms
metal atoms now again they are all same
size same diameters now then we simply
talk about close packed structures what
is the close packed structures that
simply says how can we pack same size of
spheres in a given volume such that we
will put maximum number of spheres in
that volume so there are two ways of
stacking those same size fares or you
can call it metal atoms actually which
belongs to the close packing lattice or
post packing arrays in fact and one is
you have one layer of spheres put
together as close as possible and then
you have the second layer of spheres
again put close as close as possible in
that layer but they put on that sort of
little bit displaced not on the top of
the first then you may have the third
layer as we said right on the top of the
first or you can put it in a different
place actually then the first layer so
if I have this type of stacking I will
have a Bab type close packing structure
and the unit cell in this case is
hexagonal and that is called hexagonal
close
pack an HCP and if I have this type of
packing again ABC ABC ABC that and the
unit cell in this case is face centered
cubic and it's called cubic close
packing type structures this simply
shows the unit cell of this packing and
the this one also now what we said again
in these two cases both are close back
and even though the unit cells are
different the only things is the unit
cells are different I should probably
put it back again they have exactly the
same problem een in terms of packing
maximum number of spheres in a given
volume they are the same exactly the
same now in both cases actually we have
holes between the spheres
holes between the spheres and those are
simply called tetrahedral holes and
octahedral holes let's say holes let me
write it down
pause in closed bag closed packed
letters now one type of hole is formed
by four spheres and it would be like
this this is the second sphere and the
third one here like this and then we
have the fourth one is right on the top
of this just fill in that now there is a
hole between these four spheres in a
closed pack structure in both cases
actually and if you consider the center
of these four spheres they are right on
the corner of a tetrahedron right on the
corner of a tetrahedron therefore this
type of holes is called tetrahedral
tetrahedral holes let's say here
tetrahedral holes or these are simply
sites there sometime called site
tetrahedral site so there is a space
between them that's a tetrahedral holes
now the question here is how many
tetrahedral holes can we obtain if I
packed n spheres and spheres and the
answer again that can be fine
we will have two and tetrahedral holes
so n n closed let's say close packed
structure structure will have two and
tetrahedral holes tetrahedral holes
tetrahedral holes now if you have also
another type of holes here and that is
called octahedral holes and that would
be between the space of six equal sized
fairs now how can we show that we can
draw it again as follow this is one
sphere and this is another same size
another one over here now again this is
just a square like this now I put
another sphere on the top of this and
also below let's say if I put a sphere
on the top here like this
now another sphere in the bottom on the
other side considering that these four
all in the same layer let's say we have
another one on the other side like this
so now I have a whole space between six
atoms or six
Phares you can say same-size fares now
if you consider the center of these six
spheres one here one the other side one
on their side it is they are all on the
corner of an octahedron on the corner of
an octahedron therefore this is simply
called octahedral poles octahedral holes
or sites you can see that in some book
that's called its sides and now if I
have impact closed structure and number
of using M number of atoms if I get a
close packed structure the number of
octahedral holes is equal to the number
of spheres I packed so I will simply
have n both the he drill holes if I
simply pack n number of spheres in a
close-packed structure I will have n
number of octahedral holes octahedral
holes now these are important actually
because when we talk about the ionic
compounds ionic solids we will see that
these holes tetrahedral or octahedral
holes are filled with ions are filled
with ions again you can simply see that
the octahedral holes will be little
larger than the tetrahedral holes right
tetrahedron swaps so both of those close
packed structures in fact have these
holes and the number of holes with
respect to the number of atoms that form
the close pack is 2 n for the
tetrahedral and for the octahedral case
now the other structure I guess
character property of the close packing
if I take one atom here somewhere and
then find out how many other atoms are
in the closest neighborhood or if I put
it
different way how many atoms are
touching to one atom in the close-packed
structures and we will find out that
there are 12 actually atoms and now if
you simply consider an atom here or a
sphere in the middle right here there
will be six of those around in the same
layer and there will be three on the top
and three in the bottom so one sphere
actually is touching twelve spheres in a
close-packed structures and this number
12 is simply called the coordination
number of that atom coordination number
of atom so when we say coordination
number it is the number of atoms that
actually touching to that atom or I
should say the number of atoms which are
fine in the closest neighborhood of that
atom so in both cases coordination
number of the atoms in fact in HCP or
CCP is 12 and 12 is almost the highest
coordination number heis coordination
number this also indicates that we pack
those atoms as close as possible in a
given volume in a given volume okay so
for example if somebody tells you that
the unit cell of a metal atom solid
metal atom is body centered cubic and
ask you is this a closed pack structure
is it a closed pack structure if the
unit cell off the metal is body centered
cubic and the answer will be no why
because close packed structures the unit
cell has to be face centered or
hexagonal inner-self
right face centered or exactly if the
structure if the unit cell actually is
not cubic face-centered cubic or
hexagonal that
that structure is not closed pack
structure closed pack structure so we
have to keep that in mind again now by
using actually these unit cells in fact
let's say these three unit cells last
time remember we calculated the percent
efficiency packing or efficiency of
packing in fact in these three unit cell
structures or unit cell lettuces in fact
and we find out that the packing
efficiency in the primitive unit cell
cubic unit cell calculated to be 52%
what does that means that means that in
the three dimensional lattice which have
that unit cell 52 percent of the space
is full and 48 percent is empty right
and what is the empty means that means
those are they simply the spaces between
the spheres you can think about that
between the same size spheres
now when we calculated this one we find
out it is 68% for the body center and
for this one which is the face center we
calculated it to be 74% so you see again
this one if the unit cells face centered
cubic that means that that lattice
structure is closed packing lattice that
means we will have the highest packing
efficiency and we have actually 74%
which is the highest packing
efficiencies now let me before I go I
will solve you another problem example
problem how we use these unit cells and
unit cells dimension to calculate the
density for example of the solids or if
you know the density to calculate the
edge of the unit cell or if you know the
edge of the unit cell we can calculate
the radius of the
atoms actually radiate off the atoms now
let me just go through this example
actual problem at the end of the chapter
and this is problem 12:35 from the book
chapter 12 problem 35 now this simply
says that we have aluminum solid
aluminum this is a metal right solid and
it has face centered cubic unit cell so
phase centered cubic unit cell this is
the information given now the first
question it says how many aluminum atoms
are present in one unit cell
so we have simply corners here and also
at the middle of the faces so how many
you know how many aluminum atoms belongs
to one unit cell in this structure it
would be actually this right right there
so it is four so the answer to that for
aluminum atoms in one unit cell let's
say four aluminum atoms in one unit cell
okay in the second part it says what is
the coordination number of aluminum atom
now again remember coordination number
is the number of atoms that are touching
in contact with that so for the
close-packed structure this this since
this is a face centered cubic unit cell
that means that the aluminum lattice is
close packed cubic close pack lattice
right so what does that means that means
that the number of atoms that touch in
one atom or coordination number is 12 12
is actually the maximum as I said 12 so
as here is coordination number
coordination number of aluminum atom
aluminum atom is 12 and then the third
it says can you calculate the density of
aluminum metal provided that you're
given the radius of aluminum and you're
also given the atomic mass
of aluminum right so aluminum let's say
our aluminum the radius of aluminum is
given as 1 point 43 angstrom and the
molar mass of aluminum is given as let's
say molar mass aluminum of aluminum is
this is again from periodic table right
create a table you just look at the
atomic weight and 26 point 98 grams per
mole okay these are given and now we're
asked to calculate what is the density
see here what is the density of aluminum
solid right how again these type of
problems are very common usually appears
in the exam how can we calculate it once
you learn how to do that for a problem
like this you can simply apply it to all
others now what we simply need here is
this we will calculate density of
aluminum to calculate density I have to
know the mass of a given volume of the
substance right so that means that I
should simply have a mass let's say here
mass divided by the volume that means
this is the volume of this mass and that
that will give you simply the density
now what mass I should take in this case
and what volume I should take in this
case now the mass that I should take is
the number of aluminum atoms in one unit
cell right and the volume should be the
volume of the unit cell
right the volume after yourself now how
can I find the volume of the unit cell
let's see actually the mass let's say
over here the mass of unit cell will be
equal to four times the mass of aluminum
atoms right why because it's face
centered cubic unit cell in phase
centered cubic unit cell for atoms
belongs to one unit cell so that means
that this will be multiplied by the mass
of one aluminum atom now the mass of one
aluminum atom in grams one aluminum atom
in grams would be equal to the molar
mass which is simply 2698 2698 divided
by what Avogadro's number right
Avogadro's number 6.02 times 10 to the
23 and if you simply work that out this
will give you the mass of the unit cell
the mass of the unit cell and that will
be again if I work it out it will be one
point seven nine I did work it out times
10 to the minus 22 in grams
it has to be in grams because this is
gram per mole this is number of atoms
per mole so that cancels out that
becomes the grams per four atoms
actually this becomes the mass of the
unit cell now how about the volume of
the unit cell now I can find out the
volume of the unit cell again knowing
the unit cell structure it is face
centered so if I simply take one face
off the unit cell it's a cubic unit cell
right and I know that face centered unit
cells I have simply one atom at the
center and now all the others are simply
touching to this they're all the same
size like this now this is actually the
edge of the unit cell let's call that a
a is edge off the unit cell right edge
of the unit cell so if I can calculate
that I will calculate the volume because
a to the cube will be the volume of the
unit cell now how can I calculate the
Valley a actually from the radius of the
atoms the radius of the atoms are simply
1.48 and we will consider this right
angle triangle here and this part here
is the phase diagonal and that phase
diagonal is equals to 4 R right and then
this is the edge of the unit cell so I
can simply write down this is phase
diagonal this d square would be equal to
a squared plus a squared that means a
squared plus another way square that's
equal to 2a squared now from here if you
replace the D by R and make it as for R
to the square
is equal to 2a squared and you can find
out that the edge of the unit cell which
is a will be equal to two times square
root two times the greatest radius of
aluminum radius of the aluminum and if I
put the values here radius of aluminum
as 1.43 and the edge of the unit cell
will turn out to be as 4.0 for let's say
five here angstrom
clear so that is the edge now if I would
like to calculate the density in units
of gram per cubic centimeter it would be
wise to convert this actually to the
centimeters right centimeter and if I
convert it to the centimeter this is
going to be simply 4.0 five times ten to
the what centimeter one angstrom is ten
to the eighth right we have to learn
these these are very important know
nanometers angstroms and millimeters and
so on but these are simple so it is 10
to the minus 8 centimeters 10 to the
minus 8 centimeters now if I simply
calculate the volume of the unit cell
and that will be simply a to the cube
and I replace that a here and the volume
of the unit cell would be simply equals
to 6 6 2 times 10 to the minus 23 cubic
centimeter cubic centimeter now I am
ready to calculate the density density
of aluminum in fact would be equal to
mass of one unit cell which is 1.7
9 times 10 to the minus 22 divided by
the volume this is graham and that's
cubic centimeter six point six two times
ten to the minus 23 cubic centimeter and
that is equals to if you work it out
turns out to be two point seven Oh gram
per cubic centimeter is the answer for
aluminum so you see again by using the
relationship between the radius of the
atoms and the edges of the unit cell and
the unit cell we can calculate the
density if the information here that
like the mass of the atoms and the
radius of the atoms are given again for
example I could have given you the
density of aluminum and then I would
have asked you what is the radius of
aluminum assuming that all aluminum
atoms are touching each other as in
close packed structures or we may ask
some other things for example we can
give everything here and we can ask you
can give you the density we can give you
the radius we can give you the mass of
the aluminum actually atomic mass there
and we can ask you to calculate the
Avogadro's number right you can do that
because in this calculation you know if
you don't know how a god Rose number and
if you know all the others you can
calculate it actually the Avogadro's
number calculated from these crystalline
solids turns out to be very very close
to 6.0 to 10 to the 23 that's very
important very important as a matter of
fact and that that's one way of you know
experimentally shown that the Avogadro's
number is right a correct number yes
any questions
how close we how close we get it
actually it says estimate the it is
pretty close I mean the value of the
aluminum I can't remember exactly but
it's somewhere about 2.7 3 maybe 75 and
so on yes no no I don't I don't have to
consider the empty spaces in this case
right you can you can probably set up
your problem such that since this is a
closed pack structure the packing
efficiency is 76% what percent is empty
and no no no because in that case you
have to know the density of one aluminum
atomosphere which is not possible
actually right that would be very very
heavy yes how I found the coordination
number okay
if you have a close-packed lettuce now
school clothes packed letters made off
from the same size atoms same size atoms
coordination number is always 12 where
do I know it's a closed pack because the
unit cell is face centered cubic and
only the close packed structures will
have the phase centered cubic unit cell
again assuming that all atoms are
touching each other
okay now the next we will talk about
ellos
and alloys are also very important
remember alloys are the mixtures of the
elements which have metallic properties
so all alloys should have a metallic
properties actually another way of
defining alloys is called the solid
solutions of metal solid solutions of
metals are simply alloys or that that
means that the alloys are homogeneous
mixtures of the elements right
homogeneous mixtures of the elements now
by simply alloying and metal in fact you
can change the properties of that metal
you can increase the strength you can
increase the resistance to corrosion and
you can increase the boiling point and
so on so therefore by simply preparing
alloys mixing the elements to form
alloys you may simply obtain the solid
which have particular properties that
you want and that's possible now the
alloys are are also classified here or
this simply gives the some of the
examples of the alloys if you go through
here as you see again we have for
example bronze and we have brass and
those are the copper and zinc and copper
and tin but the most known alloys that
we come across everyday and use it are
stainless steel right stainless steels
are alloys of iron and carbon and
chromium and depending on again nickel
you may have nickel there too but almost
all items that is used in industry are
alloys of iron this way you increase the
strength of iron
you increase the corrosion resistance to
the corrosion stainless steel actually
it's simply that says that is very
corrosion resistant substance corrosion
resistant substance again sometimes we
want a very strong substance but low
density and sometimes we want substance
that will have very high melting points
and so on so all of these are done by
alloying simply the elements types of
alloys we classify the alloys one is
called the substitutional alloys and the
other one is called interstitial alloys
and then we have heterogeneous alloys
now in addition to that we also have
compounds alloys which are the compounds
of metal actually these are metal
compounds and they have very distinct
properties as we will talk about that
but let's look at again those three
types of alloys here substitutional
alloys it says a second element takes
the place of a metal atom that means you
simply mix the two element together and
you consider that the second element is
replaces some of the atoms of the first
element in the lattice right in the
lattice for example as we have here
usually the silver and gold alloys are
like that
then you mix silver and gold silver
replaces some of the alloys in the
lattice points for example it is forming
if it is forming a cubic face-centered
cubic lattice gold and the silver goes
and replaces one of the gold atom in
fact in that structure in that letters
and those kind of alloys are called
substitutional alloys as you say you
substitute gold by silver
example in gold and silver alloys now
for this we should have the size of the
metal atoms close to each other if the
sizes are very different then we can't
really do that right the size has to be
about the same so that one atom will
replace the other atom in the lattice
without disturbing the lattice structure
actually gold pure gold is very soft to
make jewelry and so on most of the gold
jewelry's are alloys of gold with silver
with copper with certain percent again
that's what we simply have when we buy
gold jewelry's actually not they're not
pure gold now interstitial alloys in
this case the second element atom do not
replaces in the crystal letters of the
first element atom but it goes between
the layers between the empty spaces like
it may go to tetrahedral or octahedral
holes right and that kind of alloys are
called
interstitial alloys for example if an
atom second atom goes into the
octahedral holes then it will simply
coordinate to six other atoms and
increase the bond strength and this way
actually you increase the bond strength
of the solids and what does that mean is
that increases the melting point
increases the hardness of the metal
actually the initial metal now the third
is called a heterogeneous alloy this is
the count components not dispersed
uniformly actually if we assume that the
if you define a loyal as solids
solutions of metals and this is not
there right because at your juniors
solutions is always homogeneous
you can't have a genius solutions but
there are arrows where you have for
example like this one this is again
steel iron and carbon alloy and you have
here this carbon and iron and here we
have pure iron so that's the sort of a
homogeneous heterogeneous type alloys
now inter metallic compounds these are
simply compounds that simply means that
you can only have constant fixed ratio
of the atoms because to have a compound
remember the ratio of the elements
should be the same it shouldn't change
in other alloys you can increase or
decrease the percentages of one element
on still it is the alloy even though the
properties will change a little bit but
it will be the alloy but in this case
for the inter metallic compounds the
ratio will never change for example this
one over here three nickel and one
aluminum and this is the structure of
this compound in the solid state you see
how the aluminum and the nickels in fact
are stacked together this is like a
compound you know just compound and then
you have this one here and then you have
that one here now what is it using these
have very very different properties for
example this one is used to make
permanent magnets permanent magnet for
magnetic circular magnetic resonance
instruments or EMR instruments mr images
remember those instruments
now this one is superconductor below 19
degrees centigrade and it's also used as
a superconductor in fact and you know
they have very different properties as a
matter of fact and again you see
distinct properties definite composition
this is important to be a compound you
have to have a definite composite
ordered rather than randomly distributed
what does that means that means that
they are crystalline structures ordered
in the solid means it's a crystalline
structures okay now let's talk about
this metallic bond in a little bit now
remember again the metals are considered
as if the positively charged metal atoms
are placed in an electron sea and these
electrons C are actually the valence
electrons of the metal atom for example
sodium if you consider the sodium sodium
has one valence electrons so if you look
at the structure crystalline structure
of the sodium it would be something like
this you have the sodium plus charged
ions like this and you have the electron
cloud and these electrons are
delocalized that means they move all
over the solid sodium and now why the
metal simply accepts or forms this kind
of structure the reason is because the
metals do not have enough valence
electrons to make octet rule to make
bonds with the other metals so in order
to get the highest sharing of the
valence electrons
they simply form a structure like this
and this way all the valence electrons D
localizes and they belong to all of the
positively charged metal atoms this
actually increases the sharing of
electron between the metal ions that
means it increases the bond strength the
more electrons shared the stronger the
bonds as a matter of fact all metals
except few of those are solid at room
temperature what does that means that
means that the metallic bond which is
the
electrostatic attraction between this
electron cloud and positively charged
metal ions is pretty strong and they
keep the metal together but now if you
look at some of the properties of the
metals for example look at how the
mountain points changes as you go from
left to the right in 3-d or 4d in the
d-block
what you expect that the more electron
valence electron a metal will have there
will be more electrons will be in this
cloud and more electron in that cloud
means as stronger attractions and that
means a higher melting point but if you
simply look at the melting points of the
transition metals
d-block from left to the right it
increases and becomes maximum in the
middle but when you have 5d electrons
for example and then it decreases again
this really cannot be explained with
this model because you expect that more
valence electrons means stronger bond
right but that doesn't happens now in
order to explain the metallic properties
another theory came alone and that's
called the band theory and the band
theory as I will show you next time
based on the formations of molecular
orbitals which extend over all metal and
the electron goes into the molecular
orbital we will talk about that next
time ok let's stop here now let me make
an announcement Tuesday this coming
Tuesday that will be what today is what
twenty four twenty five twenty six
twenty seven twenty eight twenty eighth
of November Tuesday in the afternoon I
will be available in my office to show
you your
papers exam papers so those who are
interested to see their papers please
come and see me in my office Tuesday
afternoon
Tuesday afternoon