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CRYSTAL STRUCTURE. Crystal Structure. Objectives Relationships between structures-bonding-properties of engineering materials. Arrangements in crystalline solids Give examples of each: Lattice, Crystal Structure, Unit Cell and Coordination Numbers

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## CRYSTAL STRUCTURE

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**Crystal Structure**Objectives • Relationships between structures-bonding-properties of engineering materials. • Arrangements in crystalline solids • Give examples of each: Lattice, Crystal Structure, Unit Cell and Coordination Numbers • Describe hard-sphere packing and identify cell symmetry. • Define directions and planes (Miller indices) for crystals Outline • Structure of the Atom and Atomic Bonding • Electronic Structure of the Atom • Lattice, Unit Cells, Basis, and Crystal Structures • Points, Directions, and Planes in the Unit Cell • Crystal Structures of Ionic Materials • Covalent Structures**Crystal Structure**Crystal Structure • Lattice- A collection of points that divide space into smaller equally sized segments. • Unit cell - A subdivision of the lattice that still retains the overall characteristics of the entire lattice. • Atomic radius - The apparent radius of an atom, typically calculated from the dimensions of the unit cell, using close-packed directions (depends upon coordination number). • Packing factor - The fraction of space in a unit cell occupied by atoms. Types of Crystal Structure • Body centered cubic (BCC) • Face centered cubic (FCC) • Hexagonal close packed (HCP)**Crystal Structure**A number of metals are shown below with their room temperature crystal structure indicated. There are substances without crystalline structure at room temperature; for example, glass and silicone. All metals and alloys are crystalline solids, and most metals assume one of three different lattice, or crystalline, structures as they form: body-centered cubic (BCC), face-centered cubic (FCC), or hexagonal close-packed (HCP).**Number of Lattice Points in Cubic Crystal Systems**Crystal Structure • In the SC unit cell: point / unit cell = (8 corners)1/8 = 1 • In BCC unit cells: point / unit cell = (8 corners)1/8 + (1 center)(1) = 2 • In FCC unit cells: point / unit cell = (8 corners)1/8 + (6 faces)(1/2) = 4 Relationship between Atomic Radius and Lattice Parameters In SC, BCC, and FCC structures when one atom is located at each lattice point.**Crystal Structure**Packing Factor • In a FCC cell, there are four lattice points per cell; if there is one atom per lattice point, there are also four atoms per cell. The volume of one atom is 4πr3/3 and the volume of the unit cell is a0 3**Crystal Structure**Density Density of BCC iron, which has a lattice parameter of 0.2866 nm. • Atoms/cell = 2, a0 = 0.2866 nm = 2.866 10-8 cm • Atomic mass = 55.847 g/mol • Volume of unit cell = = (2.866 10-8 cm)3 = 23.54 10-24 cm3/cell • Avogadro’s number NA = 6.02 1023 atoms/mol**Crystal Structure**Points, Directions, and Planes in the Unit Cell Geometry Unit Cell: The basic structural unit of a crystal structure. A unit cell is the smallest component of the crystal that reproduces the whole crystal when stacked together with purely translational repetition. R R R R a Atomic configuration in Face-Centered-Cubic Atomic configuration in Simple Cubic**Crystal Structure**Bravais Lattices**Crystal Structure**Unit Cells Types A unit cell is the smallest component of the crystal that reproduces the whole crystal when stacked together with purely translational repetition. • Primitive (P) unit cells contain only a single lattice point. • Internal (I) unit cell contains an atom in the body center. • Face (F) unit cell contains atoms in the all faces of the planes composing the cell. • Centered (C) unit cell contains atoms centered on the sides of the unit cell. Face-Centered Primitive Body-Centered End-Centered Crystal Classes (cubic, tetragonal, orthorhombic, hexagonal, monclinic, triclinic, trigonal) with 4 unit cell types (P, I, F, C) symmetry allows for only 14 types of 3-D lattice.**Crystal Structure**Counting Number of Atoms Per Unit Cell • Counting Atoms in 3D Cells • Atoms in different positions are shared by differing numbers of unit cells. • • Vertex atom shared by 8 cells => 1/8 atom per cell. • • Edge atom shared by 4 cells => 1/4 atom per cell. • • Face atom shared by 2 cells => 1/2 atom per cell. • • Body unique to 1 cell => 1 atom per cell. Simple Cubic • 8 atoms but shared by 8 unit cells. So, • 8 atoms/8 cells = 1 atom/unit cell How many atoms/cell for Body-Centered Cubic? And, Face-Centered Cubic?**= 0.74**Crystal Structure Atomic Packing Fraction for FCC Face-Centered-Cubic Arrangement APF = vol. of atomic spheres in unit cell total unit cell vol. No. of atoms per unit cell = Volume of one atom= Volume of cubic cell = “R” related to “a” by 4/cell 4R3/3 a3**Crystal Structure**APF for BCC Again, For BCC • BCC: a = b = c = a and angles a = b =g= 90°. • 2 atoms in the cubic cell: (0, 0, 0) and (1/2, 1/2, 1/2).**A**B C Crystal Structure FCC Stacking Highlighting the stacking Highlighting the faces**Crystal Structure**FCC Stacking • ABCABC.... repeat along <111> direction gives Cubic Close-Packing (CCP) • Face-Centered-Cubic (FCC) is the most efficient packing of hard-spheres of any lattice. • Unit cell showing the full symmetry of the FCC arrangement : a = b =c, angles all 90° • 4 atoms in the unit cell: (0, 0, 0) (0, 1/2, 1/2) (1/2, 0, 1/2) (1/2, 1/2, 0)**A**Crystal Structure HCP Stacking Highlighting the stacking B A Highlighting the cell Layer A Layer B Layer A**Crystal Structure**HCP Stacking • ABABAB.... repeat along <111> direction gives Hexagonal Close-Packing (HCP) • Unit cell showing the full symmetry of the HCP arrangement is hexagonal • Hexagonal: a = b, c = 1.633a and angles a = b = 90°,g = 120° • 2 atoms in the smallest cell: (0, 0, 0) and (2/3, 1/3, 1/2).**pt.**Crystal Structure Crystallographic Points, Directions, and Planes. To define a point within a unit cell…. Express the coordinates uvw as fractions of unit cell vectors a, b, and c (so that the axes x, y, and z do not have to be orthogonal). pt. coord. x (a) y (b) z (c) 0 0 0 origin 1 0 0 1 1 1 1/2 0 1/2**Crystal Structure**Crystallographic Points, Directions, and Planes.**c**b a • 5. Designate negative numbers by a bar • Pronounced “bar 1”, “bar 1”, “zero” direction. 6. “Family” of [110] directions is designated as <110>. DIRECTIONS will help define PLANES (Miller Indices or plane normal). Crystal Structure Crystallographic Points, Directions, and Planes. • Procedure: • Any line (or vector direction) is specified by 2 points. • The first point is, typically, at the origin (000). • Determine length of vector projection in each of 3 axes in units (or fractions) of a, b, and c. • X (a), Y(b), Z(c) • 1 1 0 • Multiply or divide by a common factor to reduce the lengths to the smallest integer values, u v w. • Enclose in square brackets: [u v w]: [110] direction.**Crystal Structure**Crystallographic Points, Directions, and Planes.**Crystal Structure**Miller Indices for HCP Planes 4-index notation is more important for planes in HCP, in order to distinguish similar planes rotated by 120o. t As soon as you see [1100], you will know that it is HCP, and not [110] cubic! Find Miller Indices for HCP: Find the intercepts, r and s, of the plane with any two of the basal plane axes (a1, a2, or a3), as well as the intercept, t, with the c axes. Get reciprocals 1/r, 1/s, and 1/t. Convert reciprocals to smallest integers in same ratios. Get h, k, i , l via relation i = - (h+k), where h is associated with a1, k with a3, i with a2, and l with c. Enclose 4-indices in parenthesis: (h k i l) . r s**The plane’s intercept a1, a3 and c at r=1, s=1 and t= ,**respectively. The reciprocals are 1/r = 1, 1/s = 1, and 1/t = 0. They are already smallest integers. We can write (h k i l) = (1 ? 1 0). Using i = - (h+k) relation, k=–2. Miller Index is Crystal Structure Miller Indices for HCP Planes What is the Miller Index of the pink plane?**Number of atoms centered on a direction vector**LD = Length of the direction vector Crystal Structure Linear Density in FCC Example: Calculate the linear density of an FCC crystal along [1 1 0]. ASK a. How many spheres along blue line? b. What is length of blue line? ANSWER 2 atoms along [1 1 0] in the cube. Length = 4R XZ = 1i + 1j + 0k = [110]**Area of atoms centered on a given plane**PD = Area of the plane Crystal Structure Planar Packing Density in FCC Example: Calculate the PPD on (1 1 0) plane of an FCC crystal. • Find area filled by atoms in plane: 2R2 • Find Area of Plane: 8√2 R2 Hence, Always independent of R!**Theoretical Density, **Crystal Structure • crystal structure = FCC: 4 atoms/unit cell • atomic weight = 63.55 g/mol (1 amu = 1 g/mol) • atomic radius R = 0.128 nm**Crystal Structures of Ionic Materials**Crystal Structure • Factors need to be considered in order to understand crystal structures of ionically bonded solids are: • Ionic Radii • Electrical Neutrality • Connection between Anion Polyhedra • Visualization of Crystal Structures Using Computers The cesium chloride structure, a SC unit cell with two ions (Cs+ and CI-) per lattice point. (b) The sodium chloride structure, a FCC unit cell with two ions (Na+ + CI-) per lattice point**Crystal Structures of Ionic Materials**Crystal Structure Fluorite unit cell, (b) plan view. The zinc blende unit cell, (b) plan view.**Crystal Structures of Ionic Materials**Crystal Structure The perovskite unit cell showing the A and B site cations and oxygen ions occupying the face-center positions of the unit cell.**Covalent Structures**Crystal Structure Tetrahedron and (b) the diamond cubic (DC) unit cell. This open structure is produced because of the requirements of covalent bonding The silicon-oxygen tetrahedron and the resultant β-cristobalite form of silica**Covalent Structures**Crystal Structure Tetrahedron and (b) the diamond cubic (DC) unit cell. This open structure is produced because of the requirements of covalent bonding

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