Chapter 12 The Solid State The Structure of Metals and Alloys

Transcription

1 Chapter 12 The Solid State The Structure of Metals and Alloys

2 The Solid State Crystalline solid a solid made of an ordered array of atoms, ion, or molecules Amorphous solids a solid that lacks long-range order for the atoms, ions or molecules in its structure Molecular solids a solid formed by neutral, covalently bonded molecules held together by intermolecular attractive forces Ionic solids a solid consisting of monatomic or polyatomic ions held together by ionic bonds

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4 Crystal Lattice When allowed to cool slowly, the particles in a liquid will arrange themselves to give the maximum attractive forces. The result will generally be a crystalline solid. The arrangement of the particles in a crystalline solid is called the crystal lattice. The smallest unit that shows the pattern of arrangement for all the particles is called the unit cell.

5 Unit Cells Unit cells are 3-dimensional Unit cells are repeated over and over to give the macroscopic crystal structure of the solid. Each particle in the unit cell is called a lattice point. Lattice planes connect equivalent points in unit cell throughout the lattice.

6 Unit Cells The number of other particles each particle is in contact with is called its coordination number. Higher coordination number stronger attractive forces holding the crystal together Packing efficiency - percentage of volume in the unit cell occupied by particles

7 7 Unit Cells c c c c b a Cubic a = b = c all 90 a b Tetragonal a = c < b all 90 a b Orthorhombic a b c all 90 a b Monoclinic a b c 2 faces 90 c b a Hexagonal a = c < b 2 faces 90 1 face 120 c b a Rhombohedral a = b = c no 90 Triclinic a b c no 90

8 Simple Cubic Cell

9 Simple Cubic Cell all 90 angles between corners of the unit cell the length of all the edges are equal ⅛ of each corner particle is within the cube ½ of each particle on a face is within the cube ¼ of each particle on an edge is within the cube

10 Body-Centered Cubic 9 particles, one at each corner of a cube + one in center 1/8th of each corner particle lies in the unit cell 2 particles in each unit cell 8 corners x 1/8 + 1 center

11 Face-Centered Cubic 14 particles, one at each corner of a cube + one in center of each face 1/8th of each corner particle + 1/2 of face particle lies in the unit cell 4 particles in each unit cell 8 corners x 1/8 + 6 faces x 1/2

12 Structures of Metals

13 Stacking Patterns Crystal lattice Ordered three-dimensional array of particles in a crystalline solid Unit cell basic repeating unit of the arrangement of particles in a crystalline solid Hexagonal closest-packed (hcp) a crystal structure in which the layers of atoms or ions have an ababab stacking pattern Cubic closest-packed (ccp) a crystal structure in which the layers of atoms, ions, have an abcabcabc stacking pattern

14 Stacking Patterns

15 Stacking Patterns Hexagonal closest-packed (hcp) Cubic closest-packed (ccp)

16 aba pattern Hexagonal Unit Cell

17 abc pattern Face-Centered Cubit Unit Cell

18 Body-Centered Cubic Unit Cell

19 Unit Cell Dimensions and Atom Radii Simple cubic (sc) (radius) x 2 = l; r = l/2 r = 0.5 l Face-centered cubic (fcc) (radius) x 4 = l x (2) 1/2 ; r = (l 2)/4 r = l Body-centered cubic (bcc) (radius) x 4 = l x (3) 1/2 ; r = (l 3)/4 r = l

20 1) What is the length of an edge of the unit cell when aluminum (atomic radius 143 pm) crystallizes in a crystal lattice of fcc unit cells? l = 4r/( 2) l = 4(143)/( 2) l = 404 pm

21 2) A crystalline form of molybdenum has a density of g/cc at a temperature at which the radius of a molybdenum atom is 139 pm. Which unit cell is consistent with these data? simple cubic, bcc, or fcc? l = 2r = 2(139 pm) = 278 pm 1 x pm x -10 cm = 1 pm 2.78 x 10-8 cm V = (2.78 x 10-8 cm) 3 = 2.15 x cm 3 l = 2r; 1 atom/cell 1 atom = g/mol x 1 mol 6.02 x atoms = x g/atom D =(1.593 x g)/(2.15 x cm 3 ) = 7.41 g/cm 3

22 2) A crystalline form of molybdenum has a density of g/cc at a temperature at which the radius of a molybdenum atom is 139 pm. Which unit cell is consistent with these data? simple cubic, bcc, or fcc? l = 4r/( 3) = 321 pm 1 x pm x -10 cm = 1 pm 3.21 x 10-8 cm V = (3.21 x 10-8 cm) 3 = 3.31 x cm 3 l = 4r/ 3; 2 atoms/cell 1 atom = g/mol x 1 mol 6.02 x atoms = x g/atom D =(3.186 x g)/(3.31 x cm 3 ) = 9.63 g/cm 3

23 2) A crystalline form of molybdenum has a density of g/cc at a temperature at which the radius of a molybdenum atom is 139 pm. Which unit cell is consistent with these data? simple cubic, bcc, or fcc? l = 4r/( 2) = 393 pm 1 x pm x -10 cm = 1 pm 3.93 x 10-8 cm V = (3.93 x 10-8 cm) 3 = 6.07 x cm 3 l = 4r/ 2 ; 4 atoms/cell 1 atom = g/mol x 1 mol 6.02 x atoms = x g/atom D =(6.372 x g)/(6.07 x cm 3 ) = 10.5 g/cm 3

24 Alloys Alloy a blend of a host metal and one or more other elements, which may or may not be metals, that are added to change the properties of the host metal Homogenous alloys a solid solution in which atoms of host and added elements are randomly and uniformly distributed Heterogenous alloys a matrix of host metal atoms with islands of atoms of added elements interspersed

25 Substitutional Alloy Substitutional alloy atoms of nonhost metal replace host atoms in the crystal lattice Bronze

26 Interstitial Alloy Interstitial alloy atoms of added element occupy the spaces between atoms of the host Carbon Steel

27 3) Nitinol, a memory alloy with the formula NiTi, crystallizes in a bcc structure. ( Atomic radii for Ni and Ti are 124 pm and 147 pm.) What is the density of the alloy? body diagonal = 2r Ni +2r Ti body diagonal = 542 pm = 5.42 x10-8 cm (body diagonal) 2 = (edge) 2 + (face diagonal) 2 (2r Ni +2r Ti ) 2 = l 2 + (l 2) 2 1 Ti per cell 1 Ni per cell (5.42 x10-8 cm) 2 = l 2 + (l 2) x cm 2 = 3l 2 l = x 10-8 cm V = l 3 = x cm 3

28 3) Nitinol, a memory alloy with the formula NiTi, crystallizes in a bcc structure. ( Atomic radii for Ni and Ti are 124 pm and 147 pm.) What is the density of the alloy? 1 atom Ti = g/mol x 1 mol 6.02 x atoms = 7.95 x g/atom 1 atom Ni = g/mol x 1 mol 6.02 x atoms 1 Ti per cell 1 Ni per cell = 9.75 x g/atom = 1.77 x g/unit cell D =(1.77 x g)/(3.06 x cm 3 ) = 5.78 g/cm 3

29 X-ray Diffraction

30 Determing Crystal Structures Crystalline solids have a regular geometric arrangement of particles. The arrangement of particles can be determined by x-ray diffraction. A crystal is struck by beams of x-rays which are then reflected. The wavelength is adjusted to result in an interference pattern, at which point the wavelength is an integral multiple of the distances between the particles.

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32 When the interference between x-rays is constructive, the distance between the two paths (a) is an integral multiple of the wavelength: nλ = 2a Braggs Law Combining equations, rearranging, to give nλ = 2d sinθ The angle of reflection is therefore related to the distance (d) between two layers of particles: sinθ = a/d

33 Problem: An x-ray beam at λ=154 pm striking an iron crystal results in the angle of reflection θ = 32.6º. Assuming n = 1, calculate the distance between layers.

34 Four Circle Diffractometer