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Atomic and Ionic Arrangements
Prof. Dr. Hatem AKBULUT
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Objectives of Chapter
To learn classification of materialsbased on atomic/ionic arrangements
To describe the arrangements incrystalline solids based on lattice,basis, and crystal structure
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Chapter Outline
Short-Range Order versus Long-Range
Order Amorphous Materials: Principles and
Technological Applications
Lattice, Unit Cells, Basis, and Crystal
Structures Allotropic or Polymorphic
Transformations
Points, Directions, and Planes in the
Unit Cell Interstitial Sites
Crystal Structures of Ionic Materials
Covalent Structures
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Short-range order (SRO)- The regular and predictablearrangement of the atoms over a short distance - usuallyone or two atom spacings.
Long-range order (LRO)- A regular repetitivearrangement of atoms in a solid which extends over avery large distance.
Bose-Einstein condensate (BEC)- A newlyexperimentally verified state of a matter in which a
group of atoms occupy the same quantum ground state.
Short-Range Order versusLong-Range Order
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Figure. Levels of atomicarrangements inmaterials: (a) Inertmonoatomic gases have
no regular ordering ofatoms. (b,c) Somematerials, includingwater vapor, nitrogengas, amorphous silicon,and silicate glass, haveshort-range order. (d)
Metals, alloys, manyceramics and somepolymers have regularordering of atoms ionsthat extends throughthe material.
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Figure. (a) Basic Si-O tetrahedron in silicateglass. (b) Tetrahedral arrangement of C-Hbonds in polyethylene.
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Unit Cell
In Crystalline Materials;Atoms forms unit cell (or lattice) by joining in 3 dimensions
in a specific arrangements.The places that atoms accommodate called lattice pointsand all the lattice points are identical in crystalline structure.The smallest unit, which the ordered structure is repeatedso called unit cell.
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Figure. (a) Photograph of a silicon single crystal. (b)Micrograph of a polycrystalline stainless steel showinggrains and grain boundaries
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Figure. Liquid crystal display. These materials areamorphous in one state and undergo localizedcrystallization in response to an external electric fieldand are widely used in liquid crystal displays.
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Amorphous materials- Materials, including glasses, thathave no long-range order, or crystal structure.
Glasses- Solid, non-crystalline materials (typically
derived from the molten state) that have only short-range atomic order.
Glass-ceramics- A family of materials typically derivedfrom molten inorganic glasses and processed intocrystalline materials with very fine grain size and
improved mechanical properties.
Amorphous Materials: Principlesand Technological Applications
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Figure. Classification of materials based on the type of atomicorder.
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Figure. This figure shows a schematic of the blow-stretchprocess used for fabrication of a standard two-liter PET(polyethylene terephthalate) bottle from a preform. The stressinduced crystallization leads to formation of small crystals thathelp reinforce the remaining amorphous matrix.
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Figure. Atomic arrangements in crystalline silicon andamorphous silicon. (a) Amorphous silicon. (b) Crystallinesilicon. Note the variation in the inter-atomic distance foramorphous silicon.
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Lattice- A collection of points that divide space intosmaller equally sized segments.
Basis- A group of atoms associated with a lattice point.
Unit cell- A subdivision of the lattice that still retains theoverall characteristics of the entire lattice.
Atomic radius- The apparent radius of an atom, typicallycalculated from the dimensions of the unit cell, usingclose-packed directions (depends upon coordination
number). Packing factor- The fraction of space in a unit cell
occupied by atoms.
Lattice, Unit Cells, Basis, andCrystal Structures
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Cell or Lattice parameters A lattice is a collection of points, called lattice points,
The lattice parameters are the axial lengths or
dimensions of the unit cell.
Defines dimensions and shape of unit cell.
In cubic systems, only one dimension of the cell is
enough to define the all cell.
b
g
a
b
c
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Figure. Thefourteen types ofBravais latticesgrouped in seven
crystal systems.
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Figure. Definition ofthe lattice parameters
and their use in cubic,orthorhombic, andhexagonal crystalsystems.
Prof. Dr. Hatem AKBULUT
Triclinic
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How we determine the number of lattice points per cell in the
cubic crystal systems. If there is only one atom located at eachlattice point, how the number of atoms can be calulated in unitcell.
In the SC unit cell: lattice point / unit cell = (8 corners)1/8 = 1In BCC unit cells: lattice point / unit cell= (8 corners)1/8 + (1 center)(1) = 2
In FCC unit cells: lattice point / unit cell= (8 corners)1/8 + (6 faces)(1/2) = 4
The number of atoms per unit cell would be 1, 2, and 4, for thesimple cubic, body-centered cubic, and face-centered cubic,unit cells, respectively.
Determining the Number of Lattice Pointsin Cubic Crystal Systems
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Simple Cubic Structure, SC
Number of the atoms in anunit cell= the number ofthe atoms at the corners/8
Nat-Corner= 8/8=1 atom
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Each corner atom shared by 8unit cells
Face centered atom (in fccunit cell) shared by two unitcells
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Body centered cubic structure (BCC)
Cr, -Fe, Mo
NBCC= Number of atoms at corners/8 + atoms in the cell
NBCC= 8/8 + 1= 1 + 1= 2 atoms
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Face centered cubic structure (FCC),g-Fe, Al, Cu
NFCC= Number of atoms at corners/8 +number of atoms at the surfaces/2
NFCC= 8/8 + 6 1/2= 1+ 3 = 4 atoms
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The hexagonal close-packed structure (HCP),Cd, Mg, Zn, Ti
NHCP= Number of atoms at corners/6+number of atoms atsurface/2+atoms in the cell
NHCP= 12/6 + 2 + 3= 2+1+ 3 = 6 atoms
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How we determine the relationship between the atomic radius and
the lattice parameter in SC, BCC, and FCC structures when oneatom is located at each lattice point.
Determining the Relationship betweenAtomic Radius and Lattice Parameters
Figure A. The relationships between the atomic radius and theLattice parameter in cubic systems.
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SOLUTION
Referring to Figure A, we find that atoms touch alongthe edge of the cube in SC structures.
3
40
ra
In FCC structures, atoms touch along the face diagonalof the cube. There are four atomic radii along this
lengthtwo radii from the face-centered atom and oneradius from each corner, so:
2
40
ra
ra 20
In BCC structures, atoms touch along the body diagonal.There are two atomic radii from the center atom and oneatomic radius from each of the corner atoms on the body
diagonal, so
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Figure. Illustration of coordinations in (a) SC and (b) BCCunit cells. Six atoms touch each atom in SC, while the eightatoms touch each atom in the BCC unit cell.
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APF = ( number of atoms/cell)( volume of eachatom)/volume of unit cell
Prof. Dr. Hatem AKBULUT
The packing factor or atomic packing fraction is the fraction
of space occupied by atoms, assuming that the atoms are hardspheres. The general expression for the packing factor is
Atomic Packing Factor (APF)
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Example A.Calculating the Packing Factor
74.018)2/4(
)3
4(4)(
FactorPacking
24r/cells,unitFCCforSince,
)3
4)(atoms/cell(4
FactorPacking
3
3
0
3
0
3
r
r
r
a
a
Calculate the packing factor for the FCC cell.Example A. SOLUTION
In a FCC cell, there are four lattice points per cell; if there isone atom per lattice point, there are also four atoms per cell.The volume of one atom is 4r3/3 and the volume of the unitcell is .3
0a
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Calculate the planar density and planar packing fractionfor the (010) and (020) planes in simple cubic polonium,which has a lattice parameter of 0.334 nm.
Example BCalculating the Planar Density and Packing
Fraction
Figure. The planerdensities of the(010) and (020)planes in SC unitcells are not
identical.
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Example B. SOLUTION
The total atoms on each face is one. The planar densityis:
2142
2
atoms/cm1096.8atoms/nm96.8
)334.0(faceperatom1
faceofareafaceperatom(010)densityPlanar
The planar packing fraction is given by:
79.0
)2(
)(
)(atom)1(
faceofarea
faceperatomsofarea(010)fractionPacking
2
2
20
2
r
r
r
a
However, no atoms are centered on the (020) planes.Therefore, the planar density and the planar packingfraction are both zero. The (010) and (020) planes arenot equivalent!
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Coordination Number
The coordination number is the number of atoms
touching a particular atom, or the number of nearestneighbors for that particular atom
Interstitials
The vacancies between the interatomic vacancies in thecrystal structures can be filled small sized atoms. These
regions are called interstitials (e.g.. C in FCC Fe). If an atom settled in an interstitial it touches two or more
atoms. The coordination number of the interstitial atomequals the number of atoms that it touches.
For the COORDINATION number, there is critical and or
minimum radius ratio (ri/ror rK/rA).ri: radius of interstitial atom, r:atom radius in the cell
point. rK: radius of cation, rA: radius of anion
C N r /r or r /r
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C.N. ri/r or rK/rA
2
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For simple cubic cell (BCC) COORDINATION NUMBER is 6
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For body centered cubic
(BCC) cell COORDINATIONNUMBER is 8
For face centered cubic (FCC)
cell COORDINATION NUMBERis 12
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Radius ratio for C.N. 3:
Radius ratio for C.N. 6 :
r
rri
Cos 30 = r / r+ri, 0,866 = r /
r+ri
0,866 ri= (1-0,866)r
ri / r = 0,155
irrr2
)ir(r22r2
i2r2ra2
1,414 r= r+ri, 0,41 r =ri
ri / r = 0,414
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O,O,O
1,O,O
0,1,O
1,1,O
1,O,1 1,1,1
0,0,1 0,1,1
x
y
z
Coordinates
Coordination number 8 SC
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Coordination number 8, SC
, ,
Example C
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Example CDetermining the Density of BCC Iron
Determine the density of BCC iron, which has a latticeparameter of 0.2866 nm.
Example C SOLUTION
Atoms/cell = 2, a0= 0.2866 nm = 2.866 10-8cm
Atomic mass = 55.847 g/mol
Volume of unit cell = = (2.866 10-8cm)3 = 23.54 10-24cm3/cell
Avogadros number NA= 6.02 1023atoms/mol
3
0a
3
2324/882.7
)1002.6)(1054.23(
)847.55)(2(
number)sadro'cell)(Avogunitof(volume
iron)ofmass)(atomicatoms/cellof(number
Density
cmg
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(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Figure. The hexagonal close-packed (HCP) structure (left)and its unit cell.
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Allotropy- The characteristic of an element being able toexist in more than one crystal structure, depending ontemperature and pressure.
Polymorphism- Compounds exhibiting more than onetype of crystal structure.
Allotropic or PolymorphicTransformations
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Figure. Oxygen gas sensors used in carsand other applications are based on
stabilized zirconia compositions
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Calculate the percent volume change as zirconia transformsfrom a tetragonal to monoclinic structure. The lattice constantsfor the monoclinic unit cells are: a= 5.156, b= 5.191, and c=5.304 , respectively. The angle for the monoclinic unit cell is98.9. The lattice constants for the tetragonal unit cell are a=5.094 and c= 5.304 , respectively. Does the zirconia expandor contract during this transformation? What is the implicationof this transformation on the mechanical properties of zirconiaceramics?
Example DCalculating Volume Changes in Polymorphs
of Zirconia
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Example D SOLUTION
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Example D SOLUTION
The volume of a tetragonal unit cell is given byV= a2c= (5.094)2 (5.304) = 134.33 3.
The volume of a monoclinic unit cell is given byV= abcsin = (5.156) (5.191) (5.304) sin(98.9) = 140.25 3.
Thus, there is an expansion of the unit cell as ZrO2 transformsfrom a tetragonal to monoclinic form.
The percent change in volume= (final volume initial volume)/(initial volume) 100
= (140.25 - 134.33 3)/140.25 3* 100 = 4.21%.
Most ceramics are very brittle and cannot withstand morethan a 0.1% change in volume. The conclusion here is that ZrO
2
ceramics cannot be used in their monoclinic form since, whenzirconia does transform to the tetragonal form, it will most likelyfracture. Therefore, ZrO2is often stabilized in a cubic form usingdifferent additives such as CaO, MgO, and Y2O3.
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Example EDesigning a Sensor to
Measure Volume Change
To study how iron behaves at elevated temperatures, wewould like to design an instrument that can detect (with a1% accuracy) the change in volume of a 1-cm3iron cubewhen the iron is heated through its polymorphictransformation temperature. At 911oC, iron is BCC, with alattice parameter of 0.2863 nm. At 913oC, iron is FCC, with alattice parameter of 0.3591 nm. Determine the accuracyrequired of the measuring instrument.
Example E SOLUTION
The volume of a unit cell of BCC iron before transforming is:
VBCC= = (0.2863 nm)3= 0.023467 nm33
0a
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Example E SOLUTION (Continued)
The volume of the unit cell in FCC iron is:VFCC= = (0.3591 nm)
3= 0.046307 nm3
But this is the volume occupied by fourironatoms, as there are four atoms per FCC unit cell.Therefore, we must compare two BCC cells (with avolume of 2(0.023467) = 0.046934 nm3) with each FCC
cell. The percent volume change during transformationis:
3
0a
%34.11000.046934
0.046934)-(0.046307changeVolume
The 1-cm3cube of iron contracts to 1 - 0.0134 =0.9866 cm3after transforming; therefore, to assure 1%accuracy, the instrument must detect a change of:
V= (0.01)(0.0134) = 0.000134 cm3