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Lecture_3 - NE 125: Lecture 3 NE Primary and Secondary...

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Unformatted text preview: NE 125: Lecture 3 NE Primary and Secondary Bonding Primary Instructor: Instructor: William K. O’Keefe, P.Eng. wkokeefe@engmail.uwaterloo.ca “Office Hours”: Mondays 1:30 to 2:20 RCH 106 Week 1 Teaching Assistant (TA): Hua (Leanne) Wei h6wei@engmail.uwaterloo.ca Reading: Shackelford 6th Ed, Chapters 1 & 2 Assignment 1 (handout) problem set due: 14 January 2007 Suggested (extra) problems: Shackelford 6th Ed ( 2-2, 2-5, 2-10, 2-11; 2-16, 2-27, 2-28) NE 125: Introduction to Materials Science and Engineering Introduction Review: The relationship between work (energy) and Force If you apply a constant force (F) on an object and cause it to move through a distance “d” the work done on the object is: W = Fxd If the magnitude of the force changes with position then work done on the object moving it from point x1 to x2 by the force is calculated by: W = ∫ F ( x ) dx x1 A charged particle (such as an ion) will create an electric field E The electric field is conservative; the strength of the electric field depends only on the distance r between charges (and magnitude of charges q1 and q2) Both particles at a distance r2 experience the same electrostatic force due to the interaction with particle at the centre This is why ionic bonding is non-directional; an ion will attract charged particles in all directions due to conservative electric field r2 x2 + r1 + + NE 125: Introduction to Materials Science and Engineering Introduction Review: The relationship between work (energy) and Force Consider 2 ions with a +1 valence such as Na+ Any ion with +1 charge and a distance r2 away from p1 will experience the same electrostatic force due to interaction with particle p1 kq1 q 2 ke 2 F c= − =− 2 2 r2 r2 r2 r1 +p + p2 1 The work done on particle p2 due to the electric field of p1 to move particle p2 to the point in space with distance r1 from point with distance r2 is: kq q kq q kq q W = E = ∫ F (r )dr = ∫ − 12 2 dr = + 1 2 − 1 2 r r2 r1 r1 r1 r2 r2 Energy required to move particle p2 from r1 to r2 Work W > 0 for particles with like charge (eg both positive) NE 125: Introduction to Materials Science and Engineering Introduction Review: The relationship between work (energy) and Force The electrical potential energy (coulombic potential) for a system of two charged particles is equivalent to the work that must be done to bring these two particles together from an infinite separation distance to some final separation distance r kq1 q 2 W = E c = ∫ F (r )dr = ∫ − 2 dr r r1 r2 +p + p2 1 kq1 q 2 E c= + r Coulombic Potential Energy If charges q1 and q2 are both of same sign, coulombic force is repulsive; potential energy is increased; system less stable; work Ec is positive, must do work on particle to move them closer together less If charges q1 and q2 are opposite sign; force is attractive; potential energy decreases (becomes more negative) as particles move closer; system more stable; work is negative, particles will move spontaneously closer together from electrical attraction together NE 125: Introduction to Materials Science and Engineering Introduction Review: Ionic Bonding & Electrostatic Forces a 0 = r Na + + r Cl − Cl- Hard Sphere Model (Shackelford 6th Ed. pg. 36) Oppositely charged ions are attracted by the electrostatic force due to net charge on ions Na+ kq1 q 2 r2 When oppositely charged ions get very close, the ions F A= F c= − experience a repulsive force (FR) between them due to repulsion between electrons within each ion (due to overlap of wavefunctions when a0 becomes very small). −r ρ R rClrNa+ a0 F = λe At the equilibrium, the attractive force and repulsive force cancel each other out The equilibrium bond length (a0) is the sum of the two atomic radii (like billiard balls put together; hence the name hard sphere model) The net potential energy E0 associated with the ionic bond with length a0 is known as the bonding energy bonding NE 125: NE Introduction to Materials Science and Engineering Introduction Review: Ionic Bonding: Forces (F) and Potential Energy (E) Review: F A= F c= − kq1 q 2 r2 E F r0 FN r0 E0 (Bonding Energy) attraction 0 r r EN repulsion E N = EA + ER ρ F R = λe −r At the equilibrium bond length (r0) FN = Σ F = F A + F R ∑F = 0 dE N =0 dr NE 125: NE Introduction to Materials Science and Engineering Introduction Review: Ionic Bonding: Forces (F) and Potential Energy (E) Review: Homework assignment #1 Problem 3(c) You were asked to calculate the coulombic potential energy associated with the You attractive force EA associated with coulombic force of attraction Fc between Mg+2 and Oattractive 2 ions kq1 q 2 kZ 1Z 2e 2 E c= + =+ a0 a0 You will find that EA=Ec is negative. This means that a system consisting of an Mg+2 ion and a O-2 ion can reduce the total system energy by an amount EA by Mg coming together to form an ion-cation pair coming In nature, thermodynamic diving forces move systems in a direction which In minimizes their energy NE 125: NE Introduction to Materials Science and Engineering Introduction Review: Ionic Bonding: Forces (F) and Potential Energy (E) Review: Sample Problem 2-27 from Shackelford 6th Edition For the 1 dimensional ionic crystal of NaCl, the net coulombic bonding force is a simple For multiple of the force of attraction between an adjacent ion pair F=AFc multiple a0 + -+ + + - By summing up the coulombic potential energies due to every electrostatic interaction By with the reference ion, we can show that the potential energy of the reference ion is lowered by a factor of A=2ln2 relative to the anion-cation pair A=Madelung constant lowered kq 2 kq 2 kq 2 kq 2 E N = −2 +2 −2 +2 + ... a0 2a 0 3a 0 4a 0 kq 2 E N = −2 a0 kq 2 EN= −A a0 1 1 1 1 1 − + − + + ... 2 3 4 5 A = 2ln2 = Madelung constant The potential energy of the reference ion is reduced by a factor of A relative to the anion-cation pair i.e. the 1D crystal is more stable NE 125: NE Introduction to Materials Science and Engineering Introduction Review: Ionic Bonding: Forces (F) and Potential Energy (E) Review: Sample Problem 2-28 from Shackelford 6th Edition In problem 2.28 of Shackelford 6th Edition, you are given A for NaCl is 1.748 for the three dimensional crystal and asked to calculate the net coulombic force of attraction on a reference ion ion For 1D crystal A = 2ln2 = 1.386 For For 3D crystal A = 1.748 For Therefore, the potential energy of an ion in a 3D lattice is lower (and therefore in a more stable Therefore, state) than in a 1D crystal or in an anion-cation pair state) Eg. a Na+ ion in a 3D NaCl crystal is bonding to every Cl- ion in the lattice through long range electrostatic interactions electrostatic Important Points from problems 2-27 and 2-28: The electrostatic force is a long range interaction which extends beyond nearest neighbour The interactions ( unlike covalent bonding) interactions In an ionic solid, must consider electrostatic interactions between all ions In Ionic crystals are stable: salts have very high melting points Ionic NE 125: NE Introduction to Materials Science and Engineering Introduction Review: The Madelung constant A Review: EN= ∑ i≠ j kq i q j a ij r 1 A=∑ i ≠ j a ij To calculate the net coulombic potential energy for an ion within a lattice associated with interactions with all ions in the lattice, sum up the individual potentials The Madelung constant is characteristic of the crystal structure (distances between ions) For an ionic solid with 2 kinds of ions with equal and opposite charge (Z1=-Z2) such as such NaCl or MgO NaCl 2 2 EN= − Examples NaCl MgF2 Ak ( Z ) (e) r (fcc structure) (fcc (tetragonal structure) A=1.748 A=2.381 Tmp = 801 °C Tmp = 1261 °C NE 125: Introduction to Materials Science and Engineering Introduction Covalent Bonding Another way individual atoms may achieve a closed outer shell (kernel) i.e. a complete octet; is by sharing electrons Covalent bonding occurs between atoms with similar electronegativities A rule of thumb: if a bond has less than 20% ionic character it is classified as a covalent bond % = 100% 1 − exp ( − 0.25) ( X A− X B) 2 [ { }] NE 125: Introduction to Materials Science and Engineering Introduction Review: Quantum Mechanics of atomic orbitals The energy of electrons in atoms are quantized; only certain quantum states with specific The energies are allowed. A quantum state is defined by 4 quantum numbers energies The first 3 quantum numbers define the spatial distribution of electrons (i.e. the size and shape of the orbitals) 1. Principle quantum number n 1. Determines the size of the orbital. Defines the shells in the atom n= 1,2,3… Determines 2. Orbital Angular Momentum (l) Defines the sub-shells (type of orbital; shape); Defines l: 0, 1,2,3,4… Old notation s,p,d,f,g…. 3. Magnetic Quantum Number ml ml: l, l-1, l-2….-l (2l+1) (depends on value of l) defines number of orbitals Eg. In s sub shell (l=0) there is only one orbital which can hold 2 electrons eg. 1s2 In p sub shell there are 3 orbitals (l=1,0,-1) (px, py, pz) In NE 125: Introduction to Materials Science and Engineering Introduction Review: Quantum Mechanics of atomic orbitals Pauli Exclusion Principle (Lecture 1) Pauli The Pauli Exclusion Principle states that no two electrons can exist in the same quantum state at the same time. Therefore, electrons in the same orbital must have opposite “spins” An “up spin” and a “down spin” 4. Spin quantum number (ms) 4. (m An electron has either ms = ± 1/2 An (no two electrons in an atom have the exact same 4 quantum numbers: No more than two electrons can be found in an orbital) No Fermi-Dirac Distribution (Lecture 1) The distribution of electrons in the various quantum states is governed by the Fermi-Dirac distribution. f ( Ei ) = e ( E i − E F ) / kT 1 +1 Fermi-Dirac Distribution NE 125: Introduction to Materials Science and Engineering Introduction Review: Quantum Mechanics of atomic orbitals Shapes of atomic orbitals1 The principal quantum number determines the size of the orbital Heisenberg Uncertainty Principle (1927) Can not simultaneously specify the position and momentum of an electron h=∆ x∆ p The orbitals define regions of space for which an electron can be found within a certain confidence bounds (eg. 95% probability) [1] Illustration courtesy of Wikimedia; used by permission NE 125: Introduction to Materials Science and Engineering Introduction Review: Quantum Mechanics of atomic orbitals To understand atomic orbitals and the nature of the covalent bond, we need to think of an electron as a wave rather than a particle De Broglie (1924) Particles have wave properties. The wavelength of a particle is inversely proportional to its momentum. A wavefunction ψ gives the amplitude of the wave at a point (x,y,z) in space Born (1926) Probability Density The probability of finding an electron at some point in space is proportional to the square of the normalized wavefunction p=|ψ |2 The Schrodinger Equation (1927) Solution of the Schrodinger Equation gives and expression for ψ (x,y,z) 2 2 2 − h ∂ ψ ∂ ψ ∂ ψ 2 2 + 2 + 2 + V (ψ ) = E (ψ ) ∂y ∂z 8π m ∂x NE 125: Introduction to Materials Science and Engineering Introduction Covalent Bonding Valence electrons: The electrons in unfilled shell furthest from the nucleus determine the chemical behaviour of atoms What is a covalent bond? When valence electrons are shared between atoms How? When two atoms move close enough together so that the wavefunctions of the valence electrons can overlap Constructive Interference: stable, reduces energy of system, a covalent bond is formed covalent Destructive Interference: unstable, increases energy of system, repulsion occurs repulsion NE 125: Introduction to Materials Science and Engineering Introduction Covalent Bonding When a single covalent bond (eg. C-C) is formed, the probability of finding an electron |ψ |2 in the region between the nuclei of the two atoms is high nucleus ψ = ψ1 + ψ2 H2 The molecular σ bond Eg. Hydrogen H: 1s1 H H Atomic s orbitals have Spherical symmetry Example of overlap of two s orbitals to create a σ bond (Eg. H-H) Example NE 125: Introduction to Materials Science and Engineering Introduction Covalent Bonding Energy σ *1s (antibonding orbital) 1s 1s σ 1s molecular orbital The system energy is reduced when the two electrons in the 1s The atomic orbitals of hydrogen combine to form a hybrid molecular orbital NE 125: Introduction to Materials Science and Engineering Introduction Covalent Bonding When a single covalent bond is formed, the probability of finding an electron |ψ |2 in the region between the nuclei of the two atoms is high The “head on” overlap of pz orbitals can result in a covalent bond The x x x z x z σ p molecular orbital Overlap of 2 pz orbitals to form a σ p molecular orbital NE 125: Introduction to Materials Science and Engineering Introduction Covalent Bonding The Pz and Py orbitals can also overlap but the electron density is very low in the The internuclear region in the π p molecular orbital. The electron density is concentrated molecular above and below the internuclear region above z x π p molecular orbital NE 125: Introduction to Materials Science and Engineering Introduction Covalent Bonding Multiple bonds can occur if several atomic orbitals are overlapping + π p molecular orbital σ molecular orbital Example Carbon C-C CC sigma bond sigma plus pi bond C C sigma plus two pi bonds NE 125: Introduction to Materials Science and Engineering Introduction Covalent Bonding Eg. Aromatic compounds The overlap of p orbitals in benzene results in a “pi cloud” of delocalized electrons, which The gives rise to the unique characteristics of aromatic compounds Graphic courtesy of Wikepedia, used by permission NE 125: Introduction to Materials Science and Engineering Introduction Covalent Bonding Lewis Dot Diagrams Lewis diagrams are useful constructs to help us visualize how electrons are shared Lewis between atoms between Eg. Chlorine Cl2 Cl Cl: [Ne] 3s2 3p5 Cl-Cl Eg. Oxygen O2 O: [He]2s2 2p4 + Cl Cl Cl Why are electrons grouped in pairs? Why an octet? O O+ O O O=O NE 125: Introduction to Materials Science and Engineering Introduction Covalent Bonding is Directional The covalent bond angles are determined by electronic repulsion of lone The pairs of electrons Eg methane CH4 has 4 sp3 hybrid molecular orbitals the molecule has Eg tetrahedral shape with bond angles of 109.5°C tetrahedral H atoms move as far apart as atoms possible to minimize repulsion possible Illustration courtesy of Wikipedia; used by permission NE 125: Introduction to Materials Science and Engineering Introduction Secondary Bonding: Van der Waals Bonding (a.k.a. London Dispersion Forces) Secondary bonding refers to bonding between molecules Secondary Relatively weak compared to primary bonding Relatively Occurs as a result of random fluctuation of electron distribution in an atom resulting in a Occurs temporary dipole temporary Electrons are not transferred Dipole-dipole interactions occur between atoms with closed shell configurations Fc -+ -+ Two neighboring atoms from separate molecules Temporary dipole in one atom induces dipole in neighbour and pulls molecules together NE 125: Introduction to Materials Science and Engineering Introduction Secondary Bonding: Van der Waals Bonding The Lennard-Jones 6-12 potential The potential energy curve for secondary bonding is very similar to ionic bonding σ σ E ( r ) = 4ε − r r 12 6 E r0 E0 (Bonding Energy) r The term raised to the 12 power accounts for repulsive force (overlapping wave functions of electrons in full shells ) th EN EN = E A + E R The term raised to the 6th power accounts for Van der Waals force of attraction Is the Van der Waals interaction a short range or long range interaction ? NE 125: Introduction to Materials Science and Engineering Introduction Secondary Bonding: Van der Waals Bonding (a.k.a. London Dispersion Forces) The effect of Van der Waals interactions becomes more pronounced with increasing The molecular weight molecular Secondary bonding can be very significant in macromolecules (polymers, proteins) Secondary Can significantly influence material properties of plastics, rubbers Can Illustration courtesy of Prof. L. Simon; used by permission NE 125: Introduction to Materials Science and Engineering Introduction Secondary Bonding: Hydrogen Bonding Permanent Dipole When hydrogen forms a covalent bond with a very electronegative atom (O,N,F) a permanent dipole is created Eg. H2O NH3, δ+ δ- O δ+ δ+ δ- O δ+ H2O Tbp = 100 °C H2S Tbp = -60.7 °C Hydrogen Bridge (eg. water) Water molecules are highly polar Water Electron deficient hydrogen atoms (essentially protons) interact via electrostatic force with electron rich oxygen force Hydrogen bonding is much stronger than Van der Waals interaction NE 125: Introduction to Materials Science and Engineering Introduction Summary of Week 1 Monday Classification of materials by atomic bonding and elemental constituents Classification (metals, ceramics, composites, semiconductors, polymers) (metals, Classification of materials by morphology (level of molecular order-disorder): (Crystalline, semi-crystalline, amorphous, liquid crystal, fractal) Metallic Bonding vs Ionic bonding Band Theory: Band explains conductivity in metals, insulators and semiconductors explains Explains light transmission in insulators Wednesday Ionic bonding: Coulomb’s law, potential energy, Madelung constant Sample problems 2-16, 2-27, 2-28 Friday Ionic bonding, Covalent Bonding, secondary bonding ...
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This note was uploaded on 10/08/2010 for the course NE 125 taught by Professor Simon during the Spring '10 term at Waterloo.

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