BÀI GIẢNG Inorganic Materials and Ceramics

Inorganic Materials and Ceramics Schedule Introduction • Structures of Crystalline Solids • Syntheses and Fabrications of Inorganic Materials 2.  Structures and Properties of Ceramics Ceramic Phase Diagrams  • Some concepts and Phase rule • One‐component systems • Two‐component condensed systems – Lever rule Physical Techniques for Characterizing Inorganic Materials and Ceramics • X‐Ray Diffraction (Powder Method) • Thermal Analysis (TG/DSC) • Electron Micrograph (SEM, TEM) • X‐Ray Photoelectron Spectroscopy (XPS) 5.  Midterm test (60 min) Contact Pham Anh Son Department of Inorganic Chemistry Email: anhsonhhvc@gmail.com Textbook   Materials Science and Engineering: An Introduction William D. Callister, David G. Rethwisch https://www.dropbox.com/s/sozn9c4rkblmu0v/Materials%20Science%20and%20Enginee ring%20An%20Introduction%208th.pdf?dl=0 Grading: • Attendace (20%) • Midterm test (20%) • Final exam (60%) Introduction Materials Classification of Solid Materials Metals: Ceramics: compounds between metallic  Polymers: large molecule,  • Pure metals: Fe, Al, Cu, … and nonmetallic elements prepared by  composed of many repeated  • Alloys: FeC, AlMg, … heat and subsequent cooling (oxides, …) subunits (plastics,  rubbers,…) Ductile Very hard and strong Extremely ductile Possessing large numbers of  nonlocalized electrons Versatile properties: electrical,  thermal, magnetic, etc Low density, high strength Good electrical and heat  conductivities Excellent resistance to high  temperatures and harsh  environments  Composites Composed of two (or more) individual materials Nonmagnetic, low heat and  electrical conductivities Chemically inert semiconductors Aircraft, rocket Electronic devices Advanced Materials • fiber optical cables • • • New or traditional materials with enhanced properties Semiconductors Biomaterials Smart materials satellite (Utilized in high‐technology applications) catalysts chemicals, fuels artificial body parts sensors Inorganic Materials Metals and Alloys Nonlocalized electrons Ductile Good electrical and heat  conductivities Ceramics (Metal carbide, nitride, boride,  silicide, etc.) Very hard and strong but brittle Resistance to high temp. and harsh  environments Semiconductors Located between metals and insulators Electrical conductivity strongly depends  on the impurity concentration Composites Properties are enhanced  comparing with those of  individual components Inorganic Polymers Skeletal structure does not include C atoms Just few polymers: backbone chain containing Si  (silicates, silicones, silanes), B (borates) Structure of Crystalline Solids Crystalline and Noncrystalline Solids Single crystal long‐range order Polycrystal shorter‐range order Amorphous solid no order • Crystalline solids: Groups of atoms locate in a repeating array over large distances • All metals, most ceramic materials are crystalline solids • Crystals are built up by regular arrangements of atoms in three  dimensions • Unit cell is the basic structural unit of crystal structure Unit cell • smallest repeating unit that shows the full symmetry of crystal structure  • parallelepiped having three sets of parallel faces • xyz coordinate system is established with origin at one of unit cell corners • The orientation and direction of axes: Right hand rule • Each of x, y, and z axis coincides with one of the edges that extend from origin • Lattice parameters (or lattice constants): 6 parameters Geometry of Unit Cell • • Edge lengths: a, b and c • Interaxial angles: , , and  Crystal Systems The volume of unit cell  90° Tetragonal 90° Orthorhombic 90° Cubic Stretch/compress  along one diagonal Cubic Four 3‐axes Three 4‐axes Stretch/compress  along one axis Tetragonal Stretch/compress  along second axis Hexagonal 90° 120° Rhombohedral (Trigonal) Monoclinic 90° Triclinic 90° 90° Trigonal One 3‐axes Orthorhombic Three 2‐axes or  mirror planes Shear one face relative  to opposite face Shear second face  relative to opposite face Triclinic No axis or  mirror  symmetry One 4‐axis Monoclinic One 2‐axis  or mirror  plane Point Coordinates Lattice point is presented by q r s q is fractional multiple of edge length a on x‐axis r is fractional multiple of edge length b on y‐axis s is fractional multiple of edge length c on z‐axis Location of point having specified coordinates    M Crystallographic Directions Crystallographic direction [u v w] is determined by two point • A suitable vector is positioned so that it passes through  the origin of the coordinate system • Length of vector projection on each axis is determined by  a, b, and c • u, v, and w are reduced to smallest integer values • There will exist both positive and negative coordinates • Changing sign of all indices produces an opposite  Attention direction, such as  110 and   110 • The direction goes through two point q1 r1 s1 and q2 r2 s2 is  (reduced if necessary) Direction  • Vector passes through the origin of coordinate system ‐‐‐‐‐‐‐‐> no translation is necessary x Projections Fractional multiple Reduction Enclosure 2 y z 1 [120] Determine the indices for the direction  and  shown in below picture Direction  • Vector does not pass the origin, the translation is necessary in order to the translated vector passes origin x y Determine the indices for the direction  and  shown in below picture z Projections Fractional multiple Reduction Enclosure ′ Hexagonal System • Having 4 axes: a1, a2, and a3 in basal plane that is  perpendicular to z axis • The direction will be denoted by four indices [u v w t] Conversion from three‐index to four‐index → ′ ′ ′ In three‐index system • Vector passes the origin, no translation is necessary a1 a2 z Determine the indices in four‐index  system for the direction  Projections Fractional multiple Reduction Enclosure Conversion from three‐index to four‐index 3 ′ ′ Crystallographic Planes • • Crystallographic planes are specified by three Miller  indices as (h k l) Any two parallel planes are equivalent and have same  indices Determine Miller indices (h k l) • If plane passes through origin ‐‐‐> translation is necessary • Finding the length of planar intercept for each axis (ma, nb, pc) • Taking the reciprocals of these fractional numbers (1/m, 1/n, 1/p) • Changing to set of smallest integer numbers by multiplication or division by a common factor • Enclosing three numbers within parentheses (hkl) Miller Indices in Hexagonal Crystal Systems • Using four‐index system (h k i l) • Four‐index system used for clearly identifying the orientation of planes in a hexagonal crystal • The fourth index i is determined by sum of h and k Therefore Determine Miller indices for below plane • • Plane passes through origin Move plane (or coordination system) to appropriate position x y Intercepts ∞ Fractional  multiple ∞ Reciprocals Reduction Not necessary Enclosure 012 z 2 Determine Miller indices for below plane  in hexagonal unit cell • In three‐index system a1 a2 z Intercepts Fractional multiple Reciprocals Reduction Enclosure • In four‐index system for hexagonal crystal Face‐Centered Cubic (fcc) Crystal Structure • Atoms are located at each of the corners and center of all faces of cubic • Each corner occupies 1/8 atom • Each face contains 1/2 atom • Each fcc unit cell contains • Coordination number is … 4 atoms Relation between edge length a and atomic radius R: 2 Atomic Packing Factor (APF) 0.74 Body‐Centered Cubic (bcc) Crystal Structure • Atoms are located at each of the corners and center unit cell • Each corner occupies 1/8 atom • atom is at center of cubic • Each fcc unit cell contains • Coordination number is … 2 atoms Relation between edge length a and atomic radius R: Atomic Packing Factor (APF) 0.68 Hexagonal Close‐Packed (hcp) Crystal Structure • Atoms are located at each of the corners and center unit cell • Each corner occupies 1/8 atom • atom is at center of cubic • 1.633 • Each hpc unit cell contains • Coordination number is … 2 atoms Atomic Packing Factor (APF) 0.74 Calculate density of copper (in g/cm3) with given data below: Density computation () n number of atoms in each unit cell A atomic weight • fcc crystal structure • atom radius • atomic weight 0.128 nm 63.5 g/mol VC volume of the unit cell NA Avogadro’s number (6.022 10 atoms/mol) Synthesis and Fabrication of  Inorganic Materials Fabrication of Inorganic Materials • • • • Solid State  Low‐temperature  Reactions Methods Oldest, simplest, most  widely Press solids together  and heat at high  temperature Slow reaction Stoichiometry • • • • • Reactants are mixed at  atomic scale Easily diffuse Reduce time and  temperature Contaminations are  formed easily Expensive and difficult  to handle for large scale Mixing and  grind High‐pressure  Gas‐phase Methods • • • Reactants (precursors)  are vaporized and  unstable Use for developing new  materials, thinfilms,  coating  Expensive, require  complicated  equipments Compression Methods • Using very high pressure  (collaborating with high  temperature) Calcination Solid State Reactions • Solid state reactions are oldest, simplest, most widely used methods to prepare  inorganic materials • Reactants are pressed together and heated at high temperature for a long time • Solid state reactions occur very slowly due to the limitation of diffusion • It is very easy to handle and prepare the reaction with stoichiometric ratios Solid State Reactions: Nucleation and growth MgO Al2O3 MgAl2O4 fcc hpc fcc Position of Mg2+ Octahedral ‐ Tetrahedral Position of Al3+ ‐ Octahedral Octahedral Structure of O2‐ lattice • At the interface between MgO and Al2O3, the first MgAl2O4 crystals are formed as nuclei (nucleation) • Mg2+ changes from octahedral hole to tetrahedral hole • O2‐ lattice in Al2O3 changes from hpc to fcc • The growth of nuclei on MgO structure is easy than on Al2O3 because of the similarity of MgO and spinel MgO + Al2O3 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐> MgAl2O4 (spinel) (AB: close‐packed stacking sequence for hpc ABC: close‐packed stacking sequence for fcc) Counter‐diffusion process • In the growth stage, the counter‐ diffusion process is required • Mg2+ ions diffuse away from and  Al3+ ions diffuse towards Low‐temperature methods • • • • • Reactants can be mixed at atomic scale Diffusion rate is enhanced Reaction temperature and time are reduced Contaminations are formed easily Expensive and difficult to handle for large scale Sol‐gel method Single‐source  Hydrothermal  precursors synthesis Sol‐gel method • Alkoxides as precursors: Mg(OCH3)2, Al(OC4H9)3 M(OR)n + mOH‐ ‐‐‐‐‐‐‐‐‐‐> (OH)mM(OR)n‐m + mOR‐ {‐M‐OH + HO‐M’‐} ‐‐‐‐‐‐‐‐‐‐‐‐‐‐> {‐M‐O‐M’‐} + H2O Fabrication of MgAl2O4, SiO2, ITO (Indium tin oxide), YSZ, etc • Oxyhydroxides as precursors for preparation of many oxides: Al2O3, SiO2, Fe2O3, ZrO2, Y2O3, etc Co‐precipitation method Hydrothermal methods • Reaction in water/steam • High pressure, high temperature: controllable • Using autoclave: Teflon‐lined steel reactors • Syntheses of zeolites, mesoporous materials (SBA,  MCM, etc.) and other mixed oxides (BaFe12O19, BaTiO3,  SiO2, TiO2, etc.) Fabrication of Materials using Template Surfactant (2) (3) (4) (5) H 2O (1) Micelle Condensation + + (1) + + + (2) (3) HO HO HO HO Si O Si O Si OH O O O HO Si O Si O Si OH O O O + + + + + + + + + - O HO Si O HO Si HO HO HO Si O HO Si O + - + H 2O HO HO HO Si OH HO Si OH HO Si OH O O O HO Si OH HO Si OH HO Si OH O O O + + + - O O O HO Si O Si O Si OH O O O HO Si O Si O Si OH H O HO HO + HO - - O O O Si O Si OH O O O Si O Si OH H O HO H O HO O Si O Si OH O O O Si O Si OH O O + + - (4) 1.    Show that the atomic packing factor (APF) of bcc is 0.68 2.    Compute density of iron that has a bcc crystal structure, an atomic  radius is 0.124 nm and an atomic weigh of 55.85 g/mol.  3.    Give unit cell as Fig. 1 a.Towhichcrystalsystemdoesthisunitcellbelong? Fig.1 b.Whatwouldthiscrystalstructurebecalled? c.Indicatethelocationoftheẵ1ẵandẳẵắpoint 4.Withinacubicunitcell,sketchfollowingdirections: b. 111 c. 012 d.  133 5.    What are the indices for the directions indicated by two vectors (Fig.2) Fig. 2 6.    Within a cubic cell, sketch  102 and  011 planes Determine the Miller indices for the planes shown in Fig. 3 What are advantages and dis advantages of methods as follows: a. Conventional method b. Coprecipitation method Why do spinel nuclei grow on MgO more easily than on Al2O3? - - O O O HO Si O Si O Si O O O HO Si O Si O Si O O O H HO HO Si O Si O Si O O O HO Si O Si O Si O O O + + + Assignment  1 a.  110 Calcination + - Further condensation + SiO2 Precursor Fig. 3 OH - OH OH OH (5) - - O O O HO Si O Si O Si OH O O O HO Si O Si O Si OH O O O HO Si O Si O Si OH O O O HO Si O Si O Si OH O O O