1. Metallic and Ionic Solids Section 13.4 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

2. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Types of Solids Table 13.6 TYPEEXAMPLEFORCE Ionic NaCl, CaF 2, ZnSIon-ion MetallicNa, FeMetallic MolecularIce, I 2 Dipole Ind. dipole NetworkDiamondExtended Graphitecovalent TYPEEXAMPLEFORCE Ionic NaCl, CaF 2, ZnSIon-ion MetallicNa, FeMetallic MolecularIce, I 2 Dipole Ind. dipole NetworkDiamondExtended Graphitecovalent

3. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Types of Solids TypeBuilt fromExamples Amorphous Covalently bonded network with limited ordering Glass, plastics, polymers Ionic+ and – ionsNaCl, CsCl, (NH 4 ) 2 SO 4 MetallicAtoms or metallic ions in sea of e - MolecularMolecules with internal covalent bonds, and intramolcular attractions: dipole-dipole, H- bond, London dispersion H 2, ice, I 2, CH 3 0H NetworkAtoms held in network covalent bondsGraphite, diamond, quartz

4. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Properties of Solids 1. Molecules, atoms or ions locked into a CRYSTAL LATTICE 2. Particles are CLOSE together 3. STRONG IM forces 4. Highly ordered, rigid, incompressible 1. Molecules, atoms or ions locked into a CRYSTAL LATTICE 2. Particles are CLOSE together 3. STRONG IM forces 4. Highly ordered, rigid, incompressible ZnS, zinc sulfide

5. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Crystal Lattices Regular 3-D arrangements of equivalent LATTICE POINTS in space. The lattice points define UNIT CELLS, the smallest repeating internal unit that has the symmetry characteristic of the solid. There are 7 basic crystal systems, but we are only concerned with CUBIC. Regular 3-D arrangements of equivalent LATTICE POINTS in space. The lattice points define UNIT CELLS, the smallest repeating internal unit that has the symmetry characteristic of the solid. There are 7 basic crystal systems, but we are only concerned with CUBIC.

6. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Cubic Unit Cells

7. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Cubic Unit Cells Figure 13.28 Metals have unit cells that are simple cubic (SC)simple cubic (SC) body centered cubic (BCC)body centered cubic (BCC) face centered cubic (FCC)face centered cubic (FCC) Metals have unit cells that are simple cubic (SC)simple cubic (SC) body centered cubic (BCC)body centered cubic (BCC) face centered cubic (FCC)face centered cubic (FCC)

8. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Simple cubic unit cell. Note that each atom is at a corner of a unit cell and is shared among 8 unit cells. Simple Cubic Unit Cell Figure 13.28

9. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Body-Centered Cubic Unit Cell

10. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Body Centered Cubic Unit Cell Has 8 identical atoms or ions at the corners and additional particle at the center of the cubeHas 8 identical atoms or ions at the corners and additional particle at the center of the cube Total of two particles in the unit cellTotal of two particles in the unit cell Ex: alkali metals, iron, K, Na, Ba, Ce, Li, VEx: alkali metals, iron, K, Na, Ba, Ce, Li, V Has 8 identical atoms or ions at the corners and additional particle at the center of the cubeHas 8 identical atoms or ions at the corners and additional particle at the center of the cube Total of two particles in the unit cellTotal of two particles in the unit cell Ex: alkali metals, iron, K, Na, Ba, Ce, Li, VEx: alkali metals, iron, K, Na, Ba, Ce, Li, V

11. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Face Centered Cubic Unit Cell

12. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Face Centered Cubic Unit Cell There is a particle in each of the six faces of the cube in addition to the those at the cornersThere is a particle in each of the six faces of the cube in addition to the those at the corners Each particle in the face of the cube is shared equally between two unit cells (1/2 of particle is within unit cell)Each particle in the face of the cube is shared equally between two unit cells (1/2 of particle is within unit cell) 4 atoms/ions within the unit cell4 atoms/ions within the unit cell There is a particle in each of the six faces of the cube in addition to the those at the cornersThere is a particle in each of the six faces of the cube in addition to the those at the corners Each particle in the face of the cube is shared equally between two unit cells (1/2 of particle is within unit cell)Each particle in the face of the cube is shared equally between two unit cells (1/2 of particle is within unit cell) 4 atoms/ions within the unit cell4 atoms/ions within the unit cell

13. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Crystal Lattices—Packing of Atoms or Ions Assume atoms are hard spheres and that crystals are built by PACKING of these spheres as efficiently as possible. FCC is more efficient than either BC or SC. Assume atoms are hard spheres and that crystals are built by PACKING of these spheres as efficiently as possible. FCC is more efficient than either BC or SC. See Closer Look, pp. 622-623

14. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Packing of C 60 molecules. They are arranged at the lattice points of a FCC lattice. Crystal Lattices—Packing of Atoms or Ions

15. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Unit Cell is the smallest repeating unit that retains the symmetry properties of the Lattice Crystal Latticesand Unit Cells Crystal Lattices and Unit Cells Cubic Cells An Atom in this corner is shared by eight unit cells

16. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved scfccbcc

17. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The cubic crystal lattices are of three types 1.simple cubic (sc) 8 x 1/8 atom per corner = one atom per cell 2. body-centered cubic (bcc) 8 x 1/8 atoms per corner + one atom in center = 2 atoms / cell 3.face-centered cubic (fcc) 8 x 1/8 atom / corner + ½ atom per side x 6 sides = 4 atoms per cell

18. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Density can be used to find atomic radius if type of crystal is known.  2 x cell edge For fcc e 2 + e 2 = l 2 l 2 = 2e 2 l =  2  e l = four atom radii Volume of cell = e 3

19. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved d = m /V density of metal = mass of cell/ Volume Mass of cell is mass of four atoms of metal Mass of cell = atom mass / mol x 1mol/ 6.022 x10 23 atoms x 4 atom/cell Example: the density of Al is 2.699 g/cm 3 the Volume of a unit cell is 6.640 x 10 -23 cm 3 and the atomic radius is then 143 pm l =  2  e l = four atom radii Volume of cell = e 3

20. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved For Iron which has a bcc unit cell the calculation is a little different The Density of Iron is 7.8470 g/cm 3  2 e D 2 = (  2 e) 2 + e 2 = 3e 2 D =  3 e

21. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved There are two atoms per unit cell Volume of cell = e 3 e = D/  3 D = 4 x r atom mass of Fe = 55.85 g/mol This gives cell edge of 287 pm and r = 124 pm  2 e D 2 = (  2 e) 2 + e 2 = 3e 2 D =  3 e e D V = e 3 = m/density m = 2 x 55.85 g/6.022 x 10 23 atoms e 3 = 1.85 x 10 -22 / 7.8470 g/cm 3 e = 2.87 x 10 -8 cm

22. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Finding the Lattice Type To find out if a metal is SC, BCC or FCC, use the known radius and density of an atom to calc. no. of atoms per unit cell. PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 1.Calc. unit cell volume To find out if a metal is SC, BCC or FCC, use the known radius and density of an atom to calc. no. of atoms per unit cell. PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 1.Calc. unit cell volume

23. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Finding the Lattice Type PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 1. Calc. unit cell volume V = (cell edge) 3 Edge distance comes from face diagonal. Diagonal distance = 2 edge Diagonal distance =  2 edge PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 1. Calc. unit cell volume V = (cell edge) 3 Edge distance comes from face diagonal. Diagonal distance = 2 edge Diagonal distance =  2 edge

24. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Finding the Lattice Type PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION V = (cell edge) 3 and face diagonal = 2 edge V = (cell edge) 3 and face diagonal =  2 edge PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION V = (cell edge) 3 and face diagonal = 2 edge V = (cell edge) 3 and face diagonal =  2 edge

25. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Finding the Lattice Type PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION Here diagonal = 4 radius of Al = 572 pm Therefore, edge = 572 pm / 2 = 404 pm Therefore, edge = 572 pm /  2 = 404 pm In centimeters, edge = 4.04 x 10 -8 cm In centimeters, edge = 4.04 x 10 -8 cm So, V of unit cell = (4.04 x 10 -8 cm) 3 V = 6.62 x 10 -23 cm 3 V = 6.62 x 10 -23 cm 3 PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION Here diagonal = 4 radius of Al = 572 pm Therefore, edge = 572 pm / 2 = 404 pm Therefore, edge = 572 pm /  2 = 404 pm In centimeters, edge = 4.04 x 10 -8 cm In centimeters, edge = 4.04 x 10 -8 cm So, V of unit cell = (4.04 x 10 -8 cm) 3 V = 6.62 x 10 -23 cm 3 V = 6.62 x 10 -23 cm 3

26. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Finding the Lattice Type PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 2.Use V and density to calc. mass of unit cell from DENS = MASS / VOL Mass = density volume = (6.62 x 10 -23 cm 3 )(2.699 g/cm 3 ) = (6.62 x 10 -23 cm 3 )(2.699 g/cm 3 ) = 1.79 x 10 -22 g/unit cell = 1.79 x 10 -22 g/unit cell PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 2.Use V and density to calc. mass of unit cell from DENS = MASS / VOL Mass = density volume = (6.62 x 10 -23 cm 3 )(2.699 g/cm 3 ) = (6.62 x 10 -23 cm 3 )(2.699 g/cm 3 ) = 1.79 x 10 -22 g/unit cell = 1.79 x 10 -22 g/unit cell

27. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Finding the Lattice Type PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 3. Calculate number of Al per unit cell from mass of unit cell. PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 3. Calculate number of Al per unit cell from mass of unit cell.

28. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Finding the Lattice Type PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 3. Calculate number of Al per unit cell from mass of unit cell. 1 atom = 4.480 x 10 -23 g, so PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 3. Calculate number of Al per unit cell from mass of unit cell. 1 atom = 4.480 x 10 -23 g, so

29. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Finding the Lattice Type PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 3. Calculate number of Al per unit cell from mass of unit cell. 1 atom = 4.480 x 10 -23 g, so PROBLEM Al has density = 2.699 g/cm 3 and Al radius = 143 pm. Verify that Al is FCC. SOLUTION 3. Calculate number of Al per unit cell from mass of unit cell. 1 atom = 4.480 x 10 -23 g, so

30. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Number of Atoms per Unit Cell How can there be 4 atoms in a unit cell? 1. Each corner Al is 1/8 inside the unit cell. 8 corners (1/8 Al per corner) = 1 net Al 8 corners (1/8 Al per corner) = 1 net Al 2. Each face Al is 1/2 inside the cell 6 faces (1/2 per face) = 3 net Al’s 6 faces (1/2 per face) = 3 net Al’s How can there be 4 atoms in a unit cell? 1. Each corner Al is 1/8 inside the unit cell. 8 corners (1/8 Al per corner) = 1 net Al 8 corners (1/8 Al per corner) = 1 net Al 2. Each face Al is 1/2 inside the cell 6 faces (1/2 per face) = 3 net Al’s 6 faces (1/2 per face) = 3 net Al’s

31. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Number of Atoms per Unit Cell Unit Cell Type Net Number Atoms Unit Cell Type Net Number Atoms FCC4 FCC4 SC 1 SC 1 BCC2 BCC2 Unit Cell Type Net Number Atoms Unit Cell Type Net Number Atoms FCC4 FCC4 SC 1 SC 1 BCC2 BCC2

32. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Simple Ionic Compounds Lattices of many simple ionic solids are built by taking a SC or FCC lattice of ions of one type and placing ions of opposite charge in the holes in the lattice. EXAMPLE: CsCl has a SC lattice of Cs + ions with Cl - in the center.

33. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Simple Ionic Compounds CsCl has a SC lattice of Cs + ions with Cl - in the center. 1 unit cell has 1 Cl - ion plus (8 corners)(1/8 Cs + per corner) = 1 net Cs + ion. = 1 net Cs + ion.

34. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Simple Ionic Compounds Salts with formula MX can have SC structure — but not salts with formula MX 2 or M 2 X

35. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Simple Ionic Compounds Many common salts have FCC arrangements of anions with cations in OCTAHEDRAL HOLES — e.g., salts such as CA = NaCl FCC lattice of anions ----> 4 A - /unit cellFCC lattice of anions ----> 4 A - /unit cell C + in octahedral holes ---> 1 C + at centerC + in octahedral holes ---> 1 C + at center + [12 edges 1/4 C + per edge] = 4 C + per unit cell

36. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Construction of NaCl We begin with a cube of Cl - ions. Add more Cl - ions in the cube faces, and then add Na + ion in the octahedral holes.

37. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Sodium Chloride Lattice Na + ions are in OCTAHEDRAL holes in a face-centered cubic lattice of Cl - ions.

38. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved

39. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Comparing NaCl and CsCl Even though their formulas have one cation and one anion, the lattices of CsCl and NaCl are different. The different lattices arise from the fact that a Cs + ion is much larger than a Na + ion.

40. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Common Ionic Solids Titanium dioxide, TiO 2 There are 2 net Ti 4+ ions and 4 net O 2- ions per unit cell.

41. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Common Ionic Solids Zinc sulfide, ZnSZinc sulfide, ZnS The S 2- ions are in TETRAHEDRAL holes in the Zn 2+ FCC lattice.The S 2- ions are in TETRAHEDRAL holes in the Zn 2+ FCC lattice. This gives 4 net Zn 2+ ions and 4 net S 2- ions.This gives 4 net Zn 2+ ions and 4 net S 2- ions.

42. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Common Ionic Solids Fluorite or CaF 2 FCC lattice of Ca 2+ ions This gives 4 net Ca 2+ ions. F - ions in all 8 tetrahedral holes. This gives 8 net F - ions.

43. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Summary Ionic Solids Compounds with formula MX are commonly either sc or fcc Many salts have NaCl structure (fcc) especially alkali metals Exceptions are CsCl, CsBr, CsI, alkaline oxides and sulfides, and oxides of 4 th row transition metals (MO) Formulas can always be found from unit cell structure

44. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Types of Solids Table 13.6 TYPEEXAMPLEFORCE Ionic NaCl, CaF 2, ZnSIon-ion MetallicNa, FeMetallic MolecularIce, I 2 Dipole Ind. dipole NetworkDiamondExtended Graphitecovalent TYPEEXAMPLEFORCE Ionic NaCl, CaF 2, ZnSIon-ion MetallicNa, FeMetallic MolecularIce, I 2 Dipole Ind. dipole NetworkDiamondExtended Graphitecovalent

45. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Molecular Solids Covalently bonded moleculesCovalently bonded molecules Arrangement depends on shape of molecules and types of IM forcesArrangement depends on shape of molecules and types of IM forces Ex: Ice molecules pack densely and maximize their attractionEx: Ice molecules pack densely and maximize their attraction Covalently bonded moleculesCovalently bonded molecules Arrangement depends on shape of molecules and types of IM forcesArrangement depends on shape of molecules and types of IM forces Ex: Ice molecules pack densely and maximize their attractionEx: Ice molecules pack densely and maximize their attraction

46. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Network Solids Covalently bonded atomsCovalently bonded atoms High mp’s and bp’s, hard, rigid, and high energy required to break bondsHigh mp’s and bp’s, hard, rigid, and high energy required to break bonds Ex: graphite, diamond, elemental silicon, silicaEx: graphite, diamond, elemental silicon, silica Silicon exists in diamond structureSilicon exists in diamond structure Silicates – made of silicon and oxygen (ex: sand, talc, quartz)Silicates – made of silicon and oxygen (ex: sand, talc, quartz) Covalently bonded atomsCovalently bonded atoms High mp’s and bp’s, hard, rigid, and high energy required to break bondsHigh mp’s and bp’s, hard, rigid, and high energy required to break bonds Ex: graphite, diamond, elemental silicon, silicaEx: graphite, diamond, elemental silicon, silica Silicon exists in diamond structureSilicon exists in diamond structure Silicates – made of silicon and oxygen (ex: sand, talc, quartz)Silicates – made of silicon and oxygen (ex: sand, talc, quartz)

47. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Network Solids Diamond Graphite

48. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Network Solids A comparison of diamond (pure carbon) with silicon.

49. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Amorphous Solids PropertiesProperties –Solid lacks regular structures –Has a melting range –Restrictive movement of molecules –Unable to identify unit cell in solid phase –Breaks in random pieces Examples: glass, polymers, nylonExamples: glass, polymers, nylon PropertiesProperties –Solid lacks regular structures –Has a melting range –Restrictive movement of molecules –Unable to identify unit cell in solid phase –Breaks in random pieces Examples: glass, polymers, nylonExamples: glass, polymers, nylon

50. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Properties of Solids Melting point – the crystal lattice of a solid breaks converting to a liquidMelting point – the crystal lattice of a solid breaks converting to a liquid Enthalpy of fusion – energy needed to convert one mole from s  lEnthalpy of fusion – energy needed to convert one mole from s  l –increase in with increasing MW due to strength of IM forces –Increase in ionic compounds due to increase in lattice energy (depends on size and charge) Sublimation – conversion of a solid to a gasSublimation – conversion of a solid to a gas Melting point – the crystal lattice of a solid breaks converting to a liquidMelting point – the crystal lattice of a solid breaks converting to a liquid Enthalpy of fusion – energy needed to convert one mole from s  lEnthalpy of fusion – energy needed to convert one mole from s  l –increase in with increasing MW due to strength of IM forces –Increase in ionic compounds due to increase in lattice energy (depends on size and charge) Sublimation – conversion of a solid to a gasSublimation – conversion of a solid to a gas