atomic and ionic arrangements

8/10/2019 Atomic and Ionic Arrangements

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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.


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


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


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

atomic and ionic arrangements

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