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Chapter Atomic Structure and Interatomic Bonds (Jif 104 Chpter 1
1. Materials science is the study of relationships between the structure and properties of materials. It relates how the atomic and molecular structure of a material influences its properties. 2. A material's properties determine how it responds to external forces and the environment. Key properties include mechanical, electrical, thermal, optical, and chemical properties. Mechanical properties describe response to forces like strength and toughness. 3. There are three main classes of materials: metals, ceramics, and polymers. Metals are strong, ductile, and conductive. Ceramics are brittle but heat resistant. Polymers are lightweight and insulating. Materials science helps understand materials and design new components.
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Chapter Atomic Structure and Interatomic Bonds (Jif 104 Chpter 1
- 1. MATERIALS SCIENCE MENJANA MINDAKREATIF DAN INOVATIF
- 2. Introduction What is materialsscience ? Relationship between structures and properties of materials Relates to the arrangements of electrons surrounding of the atom which influence the + Cr atomic bonding Examples 1. Fe +C Heat Treatment SiC particles 2. Al + Si + Heat treatment Jasmi 2011
- 3. Introduction Properties are theway the material responds to the environment and external forces Mechanical Properties Electrical & Magnetic Properties Response to mechanical forces, strength, etc Response to electrical and magnetic fields, conductivity, etc Thermal Properties Related to transmission of heat and heat capacity Optical Properties Include to absorption, transmission and scattering of light Chemical Properties In contact with the environment eg : corrosion resistance Jasmi 2011
- 4. Introduction Properties are theway the material responds to the environment and external forces Mechanical Properties : Response to mechanical forces, such as Strength (……….) Toughness Hardness Ductility Elasticity, Fatigue, Creep…etc Jasmi 2011
- 5. Introduction Why study MaterialsScience ? 1 Important to understand capabilities and limitation of materials Lack of fundamental understanding of materials and their properties will cause catastrophic failure Jasmi 2011
- 6. Introduction 2 An understandingof Materials Science helps us to design better components, parts , devices, etc. How do you make something stronger or lighter? How do elements come together to form alloys ? Why …….. 3 It is interesting and helps to make you more informed person Jasmi 2011
- 7. Introduction There are 3major classes 1. Metal Strong, ductile, High thermal & electrical conductivity, Opaque, reflective Pure metallic elements or combination of metallic elements (alloys) . Air frame, landing gear, engine components 2. Ceramic Brittle, glassy, elastic, Non-conducting (insulators) Molecules based on bonding between metallic and non-metallic elements. Typically insulating and refractory – coating on high temp engine components 3. Polymers Ductile, low strength, low density, Thermal & electrical insulators Optically translucent or transparent Many are organic compound Chemically based on C , H other non-metals – windows , cabin interior Jasmi 2011
- 8. Introduction Sub-classes of Materials i.Semiconductor (ceramics), Intermediate electrical properties ii. Composite (all three classes), combination iii. Bio Materials (all three classes), Compatible with body tissue Jasmi 2011
- 9. Introduction Engineering Materials Metals Ferrous Non-Metals Non ferrous Ceramics Aluminium Glass Thermoplastics PMC CarbonSteel Copper Graphite Thermosetting MMC Alloy Steel Titanium Diamond Elastomers CMC …... …... Irons Polymers Composites Jasmi 2011
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- 15. Li Be Na Ca Rb Sr Cs Ba Fr inert gases accept 1e- He Mg K giveup 3e- H accept 2e- THE PERIODIC TABLE give up 2e- give up 1e- Introduction O F Ne S Cl Ar Sc Se Br Kr Y Te I Xe Po At Rn Ra • Columns: Similar Valence Structure 6 Jasmi 2011
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- 20. . Introduction Angstrom =1Å = 1/10,000,000,000 meter = 10-10m Nanometer = 10nm = 1/1,000,000,000 meter = 10-9m Micrometer = 1μm = 1/1,000,000 meter = 10-6m Millimeter = 1mm = 1/1,000 meter = 10-3m Jasmi 2011
- 21. Atomic Structure Atoms = nucleus(protons and neutron) + electrons Electrons, protons have negative and positive charges of the same magnitude Neutron are electrically neutral Proton and neutron have the same mass 1.67 x 10 –27 kg Mass of an electron is much smaller (9.11x 10 –31 kg and can be neglected in calculation The atomic mass (A) = mass of proton mass of neutron Atomic number (Z) = number of proton Jasmi 2011
- 22. Atomic Structure The atomicmass unit(amu) is often used to express atomic weight. The number of atom in a mole is called the Avogadro number, (Nav), Nav = 6.023 x 10 23 Nav = 1 gram/amu Example : Atomic weight of iron =55.85 amu/atom = 55.85 g/mol Valence electrons Valence electrons – those in unfilled shells Valence electrons determine all of the following properties : Chemical, Electrical, Thermal , Optical Filled shells more stable Valence electrons are most available for bonding and tend to control the chemical properties, etc.. Jasmi 2011
- 23. Atomic Bonding 1. Ionicbonding 2. Covalent bonding Strong interaction among negative atom (have an extra electron ) and positive atom (lost an electron) Electrons are shared between the molecules to saturate the valence Large interactive force due to sharing of electrons Strong atomic bonds due to transfer of electrons 3. Metallic bonding The atoms are ionized, loosing some electrons from the valence band. Those electrons form a electron sea, which binds the charged nuclei in place. Jasmi 2011
- 24. Atomic Bonding 1. IonicBonding • Occurs between + and - ions. • Requires electron transfer. • Large difference in electronegativity required. • Example: NaCl Ionic bond : Metal donates electrons + Non-metal accepts electrons Jasmi 2011
- 25. Atomic Bonding Ionic bondingin NaCl 3s1 Sodium Atom Na Sodium Ion Na+ 3p6 I O N I C Chlorine Atom Cl Chlorine Ion Cl - B O N D Jasmi 2011
- 26. Atomic Bonding EXAMPLES: IONICBONDING • Predominant bonding in Ceramics NaCl MgO CaF2 CsCl H 2.1 Li 1.0 Be 1.5 Na 0.9 Mg 1.2 K 0.8 Ca 1.0 Rb 0.8 Cs 0.7 Fr 0.7 He - O F 3.5 4.0 Cl 3.0 Ne - Br 2.8 Kr - Sr 1.0 I 2.5 Ba 0.9 At 2.2 Xe Rn - Ra 0.9 Ti 1.5 Cr 1.6 Fe 1.8 Ni 1.8 Zn 1.8 As 2.0 Ar - Jasmi 2011
- 27. Atomic Bonding 2. CovalentBonding Electrons are shared between the molecules to saturate the valence Electron Pair H + H 1s1 Electrons H H H Hydrogen Molecule Jasmi 2011
- 28. Atomic Bonding Requires sharedelectrons Example : CH4 C : has 4 valence e, needs 4 more H : has 1 valence e, needs 1 more Si with electron valense : 4 Four covalent bonds must be formed Jasmi 2011
- 29. Atomic Bonding EXAMPLES: COVALENTBONDING H2 H 2.1 Li 1.0 Na 0.9 Ca 1.0 Rb 0.8 Sr 1.0 Cs 0.7 Ba 0.9 Fr 0.7 Ra 0.9 C(diamond) Be 1.5 Mg 1.2 K 0.8 column IVA H2O SiC Ti 1.5 Cr 1.6 Fe 1.8 F2 He O 2.0 C 2.5 Si 1.8 Ni 1.8 Zn 1.8 Ga 1.6 Ge 1.8 As 2.0 Sn 1.8 Pb 1.8 F 4.0 Cl 3.0 Ne - Br 2.8 Kr - I 2.5 Xe - At 2.2 Cl2 Rn - Ar - GaAs Jasmi 2011
- 30. Atomic Bonding 3. MetallicBonding Arises from a sea of donated valence electrons (1, 2, or 3 from each atom). Primary bond for metals and their alloys Valence electrons are detached from atoms, and spread in an electron sea that “glues’ the ions together Jasmi 2011
- 31. • Pure metalsare significantly more malleable than ionic or covalent networked materials. • Strength of a pure metal can be significantly increased through alloying. • Pure metals are excellent conductors of heat and electricity. Jasmi 2011
- 32. Crystalline Solid Levels ofatomic arrangements in materials 1. Gas 2. Water 3. glass 3. Solid metal or alloy - crystal Jasmi 2011
- 33. Crystalline Solid How doatoms arrange themselves to form solid? Crystalline materials • atoms pack in periodic, 3D arrays • typical of: -metals -many ceramics -some polymers Crystalline SiO2 Non-crystalline materials... • atoms have no periodic packing • occurs for: -complex structures -rapid cooling "Amorphous" = Noncrystalline Non- crystalline SiO2 Jasmi 2011
- 34. Crystal Structure Crystal structure(microstructure) affects the mechanical properties of materials such as tensile strength, ductility..etc Crystal structure is describe as: i. lattice : a 3-D of point in space. Each point must have identical surrounding. ii. Unit cell : the simplest repeating unit in a lattice Jasmi 2011
- 35. Crystal Structure Crystal structure: Atoms arranged in repetitive 3-D pattern, in long range order (LRO) Properties of solids depends upon crystal structure and bonding force. Space lattice : An imaginary network of lines, with atoms at intersection of lines, representing the arrangement of • atoms is called. • Unit cell : block of atoms which repeats itself to form space lattice. Jasmi 2011
- 36. Crystal Structure Assume atomsas being hard spheres with well-defined radii a = Lattice parameter or lattice constant The unit cell is the smallest structural unit or building block than can describe the crystal structure. Repeating of the unit cell generates the entire crystal Jasmi 2011
- 37. Crystal Structure Unit cell: Smallestrepetitive volume which contains the complete lattice pattern of a crystal. 7 crystal systems 14 crystal lattices a, b, and c are the lattice constants Jasmi 2011
- 38. Crystal Structure Only 7different types of unit cells are necessary to create all point lattices. According to Bravais (1811-1863) 14 standard unit cells can describe all possible lattice networks Bravais Lattice : 7 crystal systems give 14 lattice 1. Cubic 2. Tetragonal 3. Hexagonal 4. Orthorombic 5. Rhombohendral 6. Monoclinic 7. Triclinic Jasmi 2011
- 39. Crystal Structure 1. CubicUnit Cell a=b=c α = β = γ = 900 ii. Simple iii. Body Centered i. Face centered 2. Tetragonal a =b ≠ c α = β = γ = 900 i. Simple ii. Body Centered Jasmi 2011
- 40. Crystal Structure 3. Orthorhombic a≠ b≠ c α = β = γ = 900 iii. Base Centered i. Face Centered ii. Simple iv. Body Centered 4. Rhombohedral a =b = c α = β = γ ≠ 900 Simple Jasmi 2011
- 41. Crystal Structure 5. Hexagonal a≠ b≠ c α = β = γ = 900 Simple 6. Monoclinic a≠ b≠ c α = β = γ = 900 i. Simple ii. Base Centered 7. Triclinic a≠ b≠ c α = β = γ = 900 Simple Jasmi 2011
- 42. Crystal Structure Most ofengineering metals have one of the following crystal structure i. Body-centered cubic (BCC) ii. Face-centered cubic (FCC) iii. Hexagonal close packed (HCP) Jasmi 2011
- 43. FCC 1. Face-Centered Cubic(FCC) Atoms are located at each of the corners and on the centers of all the faces of cubic unit cell Cu, Al, Ag, Au, Pb, Ni, Pt, ….. have this crystal structure Good ductility Jasmi 2011
- 44. FCC Number of atomsper unit cell, n = 4 Fraction of volume occupied by hard sphere, APF = 0.74 Jasmi 2011
- 45. BCC 2. Body-Centered Cubic(BCC) Atom at each corner and at center of cubic unit cell Cr, Fe-, Mo, Li, W have this crystal structure Jasmi 2011
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- 47. HCP 3. Hexagonal Close-Packed(HCP) Atom at each corner and at center of unit cell Be, Cd, Co, Mg, Ti, Zn have this crystal structure Jasmi 2011
- 48. Crystal Structure APF fora body-centered cubic structure R a 3 8.373R 0.68 3 12.32R Jasmi 2011
- 49. Crystal Structure APF fora Face-centered cubic 16 R 3 V s 3 3 0.74 Vc 16R 2 APF for a Hexagonal Close-Pack a APF = 0.74 Jasmi 2011
- 50. Crystal Structure THEORETICAL DENSITY, atoms/unitcell r nA Volume/unit cell VcNA (cm3/unit cell) r Atomic weight (g/mol) Avogadro's number (6.023x 1023 atoms/mol) Jasmi 2011
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