EMT 110 Engineering Materials Lecture 2: Atom and Bonding Semester 1 2012/2013
Atomic Structure Fundamental Concept Atoms are the structural unit of all engineering materials! Each atoms consist of nucleus composed of protons and neutron and surrounded by electrons Nucleus = Proton (positively charged particles) + Neutrons (no electrical charge particles) n=1 n=2 n = Quantum number: representing the energy level of the electron ( n, energy) Electrons (negatively charged particles)
Mass, Charge & Charge Unit of the Proton, Neutron, and Electron Particle Mass (g) Charge (C) Charge unit Proton 1.67262X10-24 +1.6022X10-19 +1 Neutron 1.67493X10-24 0 0 Electron 9.10939X10-28 -1.6022X10-19 -1 Source: Smith, W.F. and Hashemi, J., Foundations of Materials Science and Engineering, McGraw-Hill, 2011
Atomic number, Z: Number of protons (p). In a neutral atom the atomic number is equal to the number of electrons (Z=e). Atomic mass, A: Total mass of proton and neutron in the nucleus ( A=Z+N ). Isotope: atoms that have two or more atomic mass. Same number of proton but different number of neutron. One atomic mass unit (amu) = 1/12 of the atomic mass of carbon One mole= 6.023 x 10 23 atoms ( Avogadro s number N A ).
Z Example: Determine the number of electron (e), neutron (N) in fluorin atom A 9 Answer: A=p + N=19 19 Z=p=e= 9 So, N=A - Z=10
Atomic Number and Atomic Mass Atomic Number = Number of Protons in the nucleus Unique to an element Example :- Hydrogen = 1, Uranium = 92 Relative atomic mass = Mass in grams of 6.203 x 10 23 ( Avagadro Number) Atoms. Example :- Carbon has 6 Protons and 6 Neutrons. Atomic Mass = 12. One Atomic Mass unit (amu) is 1/12 th of mass of carbon atom, indicates that the mass of one neutron or one proton very close to one amu. One gram mole = Gram atomic mass of an element. Example :- One gram Mole of Carbon (mol) 12 Grams of Carbon 6.023 x 10 23 Carbon Atoms
Periodic Table Source: Davis, M. and Davis, R., Fundamentals of Chemical Reaction Engineering, McGraw-Hill, 2003.
Example: 1) One mole aluminium have mass of 26.98 g and 6.023 x 10 23 atoms. What is the mass in grams of one atom of aluminium (A=26.98g/mol) 2) How many atom of Copper (Cu) in one gram of Copper?(A=63.54g/mol)
Ans 1: 1 mol = 6.023 x 10 23 atom mass 1 mol Al = 26.98 g mass (g) in 1 atom Al= 26.98 g 6.023 x 10 23 Ans 2: 1 mol Cu= 63.54g 1 mol Cu= 6.023 x 10 23 atom Number of Atom Cu in 1 gram Cu = 6.023 x 10 23 atom 63.54
Example: The Cladding (outside layers) of the US quarter coin consists of an alloy of 75 wt % copper and 25 wt % nickel. What are the atomic percent of Cu and Ni of this materials?
Solutions: Using the basis of 100g of the 75 wt % Cu and 25 wt % Ni alloy, there are 75g Cu and 25g Ni. Thus, the number of gram-moles of copper and nickel is No. of gram-moles of (mol of )Cu= 75g 63.54g/mol 1.1803mol No. of gram moles of (mol of) Ni = 25g 58.69g/mol 0.4260 mol Total gram-moles= mol of Cu + mol of Ni= 1.1803 + 0.4260=1.6063 mol Atomic % Cu= Atomic % Ni= 1.1803mol 1.6063mol 0.4260mol 1.6063mol (100%) 73.5% (100%) 26.5%
Electronic Structure of Atoms Electron rotates at definite energy levels. Energy is absorbed to move to higher energy level. Energy is emitted during transition to lower level. Energy change due to transition = ΔE = hv = Absorb Energy (Photon) Emit Energy (Photon) hc h=planks Constant = 6.63 x 10-34 J.s c = Speed of light λ = Wavelength of light v=frequency of photon Energy levels Energy is released or transmitted in the form of electromagnetic radiation known as photon.
Electromagnetic Spectrum
Example Calculate the energy in joules (J) and electron volts (ev) of the photon whose wave length is 121.6nm. (Given 1.00eV=1.60X10-19 J; h= 6.63X10-34 J.s) Answer: ΔE = hc E 34 8 (6.6310 J.s)(3.0010 m/s) -9 (121.6nm)(10 m/nm) 1.6310 1.6310 18-18 J 1eV J 1.6010-19 J 10.2eV
Energy in Hydrogen Atom Hydrogen atom has one proton and one electron Energy of hydrogen atoms for different energy levels is given by E 13. 6 n 2 ev (n=1,2..) principal quantum numbers Example:- If an electron undergoes transition from n=3 state to n=2 state, the energy of photon emitted is 13.6 13.6 E ( ) 1. 89ev 3 2 2 2 Energy required to completely remove an electron from hydrogen atom is known as ionization energy
Energy level diagram for hydrogen This diagram explains the energy level changes for a hydrogen atom, and which transitions cause which type of light. The horizontal lines on the left represent the different energy levels. The red arrows represent the drops an electron can take. Let s say an excited electron is in n=3. From there, that electron can drop down to n=2 or n=1. The transition of n=3 to n=2 produces a red band of light. The transition of n=3 to n=1 produces UV light. These two types of light, along with the others can be detected when electricity is run through a sample of hydrogen gas.
A hydrogen atom exists with its electron in the n= 3 state. The electron undergoes a transition to the n=2 state. Calculate (a) the energy of the photon emitted, (b) its frequency, and (c) its wavelength, (d) energy is absorbed or emitted, and (e) which series it belong to and what type emission does it represent? Ans: (a) Energy of the photon emited is: 13.6eV E 2 n E E E 3-13.6 2 3 1.89eV 2 13.6 2 2 1.6010 1.89eV ev -19 3.0210 J -19 J
b) The frequency of the photon is E hv E v h 3.0210 6.6310 4.5510 c) The wavelength of the photon is hc E 659nm 14 J J.s Hz -19 34 34 (6.6310 J.s)(3.0010 19 3.0210 J -7 6.5910 m 6.5910-7 1nm m -9 10 m 8 m/s)
d) Energy is released as its quantity is positive, and the electron is transitioning from a higher orbit to lower orbit. c) The emission belongs to Balmer series and corresponds to visible red light. Balmer series Visible red light (690 nm)
Quantum Numbers of Electrons of Atoms Principal Quantum Number (n) Represents main energy levels. Range of n from 1 to 7. Larger the n higher the energy. n=2 n=3 n=1 Subsidiary Quantum Number (l) Represents sub energy levels (orbital). Range of l from 0 to n-1. Represented by letters s,p,d and f. n=1 n=2 s orbital (l=0) p Orbital (l=1)
Quantum Numbers of Electrons of Atoms (Cont..) Magnetic Quantum Number m l Represents spatial orientation of single atomic orbital. Permissible values are l to +l. Example:- if l=1, m l = -1,0,+1. I.e. 2l+1 allowed values. No effect on energy. Electron spin quantum number m s Specifies two directions of electron spin. Directions are clockwise or anticlockwise. Values are +1/2 or 1/2. Two electrons on same orbital have opposite spins. No effect on energy. 2-9
Relationship between principal (n), subsidiary (l) and magnetic (m l ) quantum numbers Example: l = 0 to (n -1), e.g: if n=3, then l= 0 to (3-1=2) or l=(0, 1, 2) If l=0-->s orbital, l=1-->p orbital, l=2-->d orbital, l=3-->f orbital ml = -l to +l, if l = 2, so ml = -2 to +2 or ml = (-2, -1, 0, +1, +2)
Electron Structure of Multielectron Atom Maximum number of electrons in each atomic shell is given by 2n 2. Atomic size (radius) increases with addition of shells. Electron Configuration lists the arrangement of electrons in orbitals. Example :- Orbital letters Number of Electrons 1s 2 2s 2 2p 6 3s 2 Principal Quantum Numbers 2-10 For Iron, (Z=26), Electronic configuration is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2
Stable electron configuration Stable electron configurations... have complete s and p subshells tend to be unreactive.
Most elements: Electron configuration not stable! Electron configuration 1s 1 1s 2 1s 2 2s 1 1s 2 2s 2 1s 2 2s 2 2p 1 1s 2 2s 2 2p 2... (stable) 1s 2 2s 2 2p 6 (stable) 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 3s 2 1s 2 2s 2 2p 6 3s 2 3p 1... 1s 2 2s 2 2p 6 3s 2 3p 6 (stable)... 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4 6 (stable) Why? Valence (outer) shell usually not filled completely.
Example Write the electron configuration for the following atoms by using conventional spdf notation. a) Fe atom (Z=26) and the Fe 2+ and Fe 3+ ions
remember!!!!! 7s 7p 7d 6s 6p 6d 6f 5s 5p 5d 5f 4s 4p 4d 4f 3s 3p 3d 2s 2p 1s
Electrons fill up in this order The order for writing the orbitals
Electron Structure and Chemical Activity (Cont..) Electronegative elements accept electrons during chemical reaction. Some elements behave as both electronegative and electropositive. Electronegativity is the degree to which the atom attracts electrons to itself Measured on a scale of 0 to 4.1 Example :- Electronegativity of Fluorine is 4.1 Electronegativity of Sodium is 1. Electropositive Na Te N O Fl 0 K 1 W 2 H Se 3 4 Electronegative
Atomic and Molecular Bonds Ionic bonds :- Strong atomic bonds due to transfer of electrons Covalent bonds :- Large interactive force due to sharing of electrons Metallic bonds :- Non-directional bonds formed by sharing of electrons Permanent Dipole bonds :- Weak intermolecular bonds due to attraction between the ends of permanent dipoles. Fluctuating Dipole bonds :- Very weak electric dipole bonds due to asymmetric distribution of electron densities. 2-12
Ionic Bonding Ionic bonding is due to electrostatic (Coulombic) force of attraction between cations and anions. It can form between metallic and nonmetallic elements. Electrons are transferred from electropositive to electronegative atoms Electropositive Element Electron Transfer Electronegative Atom Cation +ve charge Electrostatic Attraction Anion -ve charge IONIC BOND 2-14
Ionic Bonding Lattice energies and melting points of ionically bonded solids are high. Lattice energy decreases when size of ion increases. Multiple bonding electrons increase lattice energy. Sodium Atom Na Sodium Ion Na + I O N I C B O N D Chlorine Atom Cl Chlorine Ion Cl -
Ionic Force for Ion Pair Nucleus of one ion attracts electron of another ion. The electron clouds of ion repulse each other when they are sufficiently close. Force versus separation Distance for a pair of oppositely charged ions Figure 2.11 2-16
Ion Force for Ion Pair (Cont..) Z ez e 1 2 F Z1Z 2 attractive 2 2 4 Z 1,Z 2 = Number of electrons removed or 0 added during ion formation e = Electron Charge a = Interionic seperation distance ε = Permeability of free space (8.85 x 10-12 c 2 /Nm 2 ) e 4 a 0a (n and b are constants) 2 F repulsive nb a n1 F net F attractive F repulsive F net Z 1 Z 2 e nb 2 4 0a a 2 n1 2-17
Interionic Force - Example Force of attraction between Na+ and Cl - ions Z 1 = +1 for Na +, Z 2 = -1 for Cl - e = 1.60 x 10-19 C, ε 0 = 8.85 x 10-12 C 2 /Nm 2 a 0 = Sum of Radii of Na + and Cl - ions = 0.095 nm + 0.181 nm = 2.76 x 10-10 m Na + Cl- a 0 F attraction Z 1 4 2 Z 2e 2 0a 4 (8.85 x 10 ( 1)( 1)(1.60 10-12 C 2 19 C) /Nm2)(2.76 x 10 2-10 m) 3.02 10 9 N 2-18
Interionic Energies for Ion Pairs Net potential energy for a pair of oppositely charged ions = E net Z 1 Z 2 e b 2 4 0a a 2 Attraction Energy n Repulsion Energy Energy Released Energy Absorbed E net is minimum when ions are at equilibrium seperation distance a 0 2-19
Ion Arrangements in Ionic Solids Ionic bonds are Non Directional Geometric arrangements are present in solids to maintain electric neutrality. Example:- in NaCl, six Cl- ions pack around central Na+ Ions Ionic packing In NaCl and CsCl Figure 2.13 CsCl NaCl As the ratio of cation to anion radius decreases, fewer anion surround central cation. 2-20
Bonding Energies Lattice energies and melting points of ionically bonded solids are high. Lattice energy decreases when size of ion increases. Multiple bonding electrons increase lattice energy. Example :- NaCl CsCl BaO Lattice energy = 766 KJ/mol Melting point = 801 o C Lattice energy = 649 KJ/mol Melting Point = 646 o C Lattice energy = 3127 KJ/mol Melting point = 1923 o C 2-21
Covalent Bonding In Covalent bonding, outer s and p electrons are shared between two atoms to obtain noble gas configuration. Takes place between elements with small differences in electronegativity and close by in periodic table. In Hydrogen, a bond is formed between 2 atoms by sharing their 1s 1 electrons Electron Pair H + H H H 1s 1 Electrons Hydrogen Molecule Overlapping Electron Clouds
Covalent Bonding - Examples In case of F 2, O 2 and N 2, covalent bonding is formed by sharing p electrons Fluorine gas (Outer orbital 2s 2 2p 5 ) share one p electron to attain noble gas configuration. F + F F F H F F Bond Energy=160KJ/mol Oxygen (Outer orbital - 2s 2 2p 4 ) atoms share two p electrons O + O O O O = O Bond Energy=28KJ/mol 2-23 H H Nitrogen (Outer orbital - 2s 2 2p 3 ) atoms share three p electrons N + N N N N N Bond Energy=54KJ/mol
Covalent Bonding in Benzene Chemical composition of Benzene is C 6 H 6. The Carbon atoms are arranged in hexagonal ring. Single and double bonds alternate between the atoms H H H C C C C C H H H Structure of Benzene Simplified Notations
Metallic Bonding Atoms in metals are closely packed in crystal structure. Loosely bounded valence electrons are attracted towards nucleus of other atoms. Electrons spread out among atoms forming electron clouds. These free electrons are reason for electric conductivity and thermal conductivity Since outer electrons are shared by many atoms, metallic bonds are Non-directional Positive Ion Valence electron charge cloud
Metallic Bonds (Cont..) Overall energy of individual atoms are lowered by metallic bonds Minimum energy between atoms exist at equilibrium distance a 0 Fewer the number of valence electrons involved, more metallic the bond is. Example:- Na Bonding energy 108KJ/mol, Melting temperature 97.7 o C Higher the number of valence electrons involved, higher is the bonding energy. Example:- Ca Bonding energy 177KJ/mol, Melting temperature 851 o C 2-29
Secondary Atomic Bonding Secondary bonds are due to attractions of electric dipoles in atoms or molecules. Dipoles are created when positive and negative charge centers exist. Bonding result from the columbic attraction between positive end of one dipole and the negative region of an adjacent one Sometimes called Van der Waals bond +q Dipole moment=μ =q.d q= Electric charge d = separation distance d There two types of bonds permanent and fluctuating.
Fluctuating Dipole Bond Weak secondary bonds in noble gasses. Dipoles are created due to asymmetrical distribution of electron charges. Electron cloud charge changes with time. Symmetrical distribution of electron charge Asymmetrical Distribution (Changes with time)
Permanent Dipole Bond Secondary bond created by the attraction of molecules that have permanent dipole. That is, each molecule has positive and negative charge center separated by distance. CH 4 Symmetrical Arrangement Of 4 C-H bonds No Dipole moment CH 3 Cl Asymmetrical Tetrahedral arrangement Creates Dipole
Hydrogen Bond Hydrogen bonds are Dipole-Dipole interaction between polar bonds containing hydrogen atom. Special type of intermolecular permanent dipole attraction that occur between hydrogen atom bonded to a highly electronegative element (F, O, N or Cl) and another atom of a highly electronegative element. H O H 105 0 Hydrogen Bond
Mixed Bonding Chemical bonding of atoms or ions can involve more than one type of primary bond and can also involve secondary dipole bond. 1. Ionic - covalent 2. Metallic covalent 3. Metallic ionic 4. Ionic covalent - metallic