10/12/2015 CHAPTER 10: INTERMOLECULAR FORCES AND THE PHYSICAL PROPERTIES OF LIQUIDS AND SOLIDS. Intra- Versus Inter- Molecular Forces

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1 CAPTER 0: INTERMOLECULAR FORCES AND TE PYSICAL PROPERTIES OF LIQUIDS AND SOLIDS The magnitude of intermolecular forces determines whether a substance is a gas, liquid, or solid Gas Liquid Solid Intermolecular forces are attractive forces between particles. (just as interstates run between states and international means between countries) The state of a substance depends largely on the balance between the kinetic energies of the particles and the energies of attraction between particles. Intra- Versus Inter- Molecular Forces Intramolecular forces Ionic bonding (bonding within an ionic compound) and covalent bonding (bonding within a molecule) are two of the strongest forces that hold molecules together. It requires 927 kj to break the two intramolecular covalent O bonds in mol of water. Intermolecular forces Ion Dipole Dipole Dipole It requires 4 kj to overcome the intermolecular attractions between water molecules and convert mol of liquid water to water vapor at 00 C. Induced Dipole Induced Dipole (London Dispersion Forces) Intermolecular Forces Attractive forces that act between atoms or molecules in a pure substance are collectively called van der Waals forces. Dipole-dipole interactions are attractive forces that act between polar molecules. The magnitude of the attractive forces depends on the magnitude of the dipole. Ion-Dipole Forces: Exist between an ion and the partial charge on the end of a polar molecule. Ion Dipole Forces: important for solutions of ionic substances in polar solvents (ex, a solution of NaCl in water)

2 Dipole Dipole Forces: exist between oppositely charged ends of polar molecules. These forces are effective only when polar molecules are very close together. Boiling Point Reflects the Magnitude of Intermolecular Forces The magnitude of the attractive forces depends on the magnitude of the dipole. A substance in which the particles are held together by larger intermolecular attractions will require more energy to separate the particles and will therefore boil at a higher temperature. Temp ( C) Why does this happen? Te Se S 00 2 O Te, Se, S and O are in the same group. We expect their hydrogen compounds to be chemically similar. Extrapolating from the graph, we would expect 2 O to boil at -64 C O?? ydrogen Bonding ydrogen bonding is a special type of dipole-dipole interaction. ydrogen bonding only occurs in molecules that contain bonded to a small, highly electronegative atom such as N, O, or F. F F There is a 64 C difference between the predicted and measured boiling point. Why? ydrogen Bonding: a very strong type of dipole dipole interaction that only occurs in molecules that contain bonded to a small, highly electronegative atom (N, O, or F) Boiling Point Anomaly appens for N, O, and F 00 2 O d + d - O d+ Boiling Point ( C) Te I Sb 3 Sn 4 2 Se Br 2 S Cl As 3 Ge 4 P 3 Si 4 F N 3 C 4 Polar water molecule ydrogen bonding in liquid water

3 Snowflakes and Ice Crystals Reflect the Arrangement of Water Molecules at the Molecular Level! ydrogen Bonding is Expected in Molecules that have O or N Groups d + d - O d + d + d - O d + C 3 d - d + N d + d + Polar Water molecule Polar Methanol molecule Polar Ammonia molecule ydrogen Bonding in Alcohols is Similar to ydrogen Bonding in Water aving more than one O Group increases the number of hydrogen bonds and therefore, increases the boiling point Temp ( C) C 3 C 2 C O C 3 C C 2 O 87 O C 2 O C C 2 O 290 O ydrogen bonding in liquid water ydrogen bonding in an alcohol 0 -propanol,2-propanediol,2,3-propanetriol ydrogen Bonding in DNA In which of the following substances, in the liquid state, would hydrogen bonding occur? a) C C b) Cl N Thymine Adenine 2 hydrogen bonds Cytosine Guanine 3 hydrogen bonds c) N N d) C O C 3

4 London Dispersion Forces: result from the Coulombic attractions between instantaneous dipoles of nonpolar molecules You might think that two nearby nonpolar molecules A and B would be unaffected by each other. The electric field of a molecule fluctuates rapidly. For the briefest of moments, the electrons may be on one side of the molecule. When that happens, molecule B can be considered to have a temporary separation of charge. A B A B The neighboring molecules A and C feel the electric field of B and undergo a spontaneous adjustment in their electrons positions, giving them a complementary temporary separation of charge (remember like charges repel, opposites attract ). London Dispersion Forces are Weak A B C Dispersion forces or London dispersion forces result from the Coulombic attractions between instantaneous dipoles of non-polar molecules. The Number and Kind of London Dispersion Forces Affect the Boiling Point eavier Molecules ave igher Boiling Points 600. Relative atomic weight of the atoms involved 500 I Number of atoms in the molecule. Shape of the molecule Boiling point (K) F 2 Cl 2 Br 2 Ar Kr Xe 0 e Ne -00 4

5 Intermolecular Forces Boiling Points of Alkanes Increases with the number of carbon atoms due to the induced dipole induced dipole attractions (London Dispersion forces) 400 Boiling point ( C) Room temperature Number of carbons Shape Affects Boiling Point The more compact the molecule, the less surface area there is for interactions, and therefore the lower the boiling point. A Comparison of Intermolecular Forces C C C C C pentane bp 36. C C C C C C 2-methylbutane bp 28 C C C C C C 2,2-dimethylpropane bp 9.5 C What kinds of intermolecular forces exist in the following molecules? a) Ar b) Cl c) F d) C 4 Worked Example 2. What kind(s) of intermolecular forces exist in (a) CCl 4 (l), (b) C 3 COO(l), (c) C 3 COC 3 (l), and (d) 2 S(l). Strategy Draw Lewis dot structures and apply VSEPR theory to determine whether each molecule is polar or nonpolar. Nonpolar molecules exhibit dispersion forces only. Polar molecules exhibit dipole-dipole interactions and dispersion forces. Polar molecules with N, F, or O bonds exhibit dipoledipole interactions (including hydrogen bonding) and dispersion forces. e) CaCl 2 f) CCl 4 g) C 3 COO (a) (b) (c) (d) h) C 3 COC 3 i) 2 S 5

6 Worked Example 2. (cont.) Properties of Liquids (a) (b) (c) (d) Solution (a) CCl 4 is nonpolar, so the only intermolecular forces are dispersion forces. (b) C 3 COO is polar and contains an O bond, so it exhibits dipole-dipole interactions (including hydrogen bonding) and dispersion forces. (c) C 3 COC 3 is polar but does not contain N, F, or O bonds, so it exhibits dipole-dipole interactions and dispersion forces. (d) 2 S is polar but does not contain N, F, or O bonds, so it exhibits dipole-dipole interactions and dispersion forces. Think About It Being able to draw correct Lewis structures is, once again, vitally important. Review, if you need to, the procedure for drawing them. Surface tension Viscosity Vapor Pressure Surface tension: the amount of energy required to stretch or increase the surface of a liquid The molecules on the surface feel a net inward pull. Molecules with large IMF, like polar molecules, tend to have high surface tension. Capillary action: movement of a liquid up a narrow tube Two types of forces bring about capillary action: Cohesion is the attraction between like molecules Adhesion is the attraction between unlike molecules Adhesive forces are greater than cohesive forces Cohesive forces are greater than adhesive forces Viscosity: a measure of a fluid s resistance to flow Factors that Affect the Viscosity of a Liquid 3000 The higher the viscosity the more slowly a liquid flows. Liquids that have high IMFs have higher viscosities. Viscosity (cst) Viscosity decreases with increasing temperature. Viscosity increases with increasing molecular weight Temp ( C) 6

7 Glycerol is a thick, syrupy liquid. In terms of IMFs, explain why glycerol has a higher viscosity than water. Phase Changes Since it is a bigger molecule, glycerol has greater London dispersion forces that water. Glycerol can make three hydrogen bonds whereas water can make only one hydrogen bond. Therefore, glycerol molecules have more attraction for one another than water molecules have for one another and so have a higher viscosity. If a molecule at the surface of the liquid has enough kinetic energy, it can escape to the gas phase in a process called evaporation or vaporization. In a sealed vessel, vapor pressure increases until the rate of evaporation equals the rate of condensation 2 O(l) 2 O(g) Number of molecules Low temp igh temp Energy needed to escape liquid Evaporation: Condensation: 2 O(l) 2 O(g) 2 O(l) 2 O(g) When the forward process and reverse process are occurring at the same rate, the system is in dynamic equilibrium. Kinetic energy If the container is open, evaporation will continue until the liquid evaporates away. In a sealed vessel, vapor pressure increases until the rate of evaporation equals the rate of condensation The vapor pressure of a liquid is the pressure exerted by its vapor when the liquid and vapor states are in equilibrium. A vapor is a gas that exists at a temperature and pressure at which it ordinarily would be thought of as a liquid or solid. 7

8 The exponential rise in vapor pressure with increasing temperature allows us to use natural logarithms to express the nonlinear relationship as a linear relationship Clausius Clapeyron Equation lnp vap - + RT ln P = natural log of vapor pressure Δ vap = the molar heat of vaporization R = the gas constant (8.34 J/K mol) T = the kelvin temperature C is an experimentally determined constant C Clausius Clapeyron Equation: A Plot of lnp Versus /T(K) Gives a Straight Line with a Slope of vap /R Clausius Clapeyron Equation: Can be used to calculate the vap of a liquid from its measured vapor pressure at two temperatures R = 8.34 J mol K y = m x + b y = mx +b Diethyl ether is a volatile, highly flammable organic liquid that today is used mainly as a solvent. (It was used as an anesthetic during the nineteenth century and as a recreational intoxicant early in the twentieth century during prohibition, when ethanol was difficult to obtain.) The vapor pressure of diethyl ether is 40 mmg at 8 C, and its molar heat of vaporization is 26 kj/mol. Calculate its vapor pressure at 32 C. Strategy Given the vapor pressure at one temperature, P, use the equation below to calculate the vapor pressure at a second temperature, P 2. P ln Δ = vap P 2 R T 2 T Temperature must be expressed in kelvins, so T = 29.5 K and T 2 = K. Because the molar heat of vaporization is given in kj/mol, we will have to convert it to J/mol for the units of R to cancel properly: Δ vap = J/mol. The inverse function of ln x is e x. Solution P ln = 4 J/mol P J/K mol K 29.5 K = P P 2 = e = P = P 2 P 2 = 40 mmg = mmg Think About It It is easy to switch P and P 2 or T and T 2 accidentally and get the wrong answer to a problem such as this. One way to help safeguard against this common error is to verify that the vapor pressure is higher at the higher temperature. 8

9 What is the boiling point of water at a pressure of atm? You need two points: (P,T ) and (P 2,T 2 ) P =.00 atm (the normal boiling point) T = 00 C = 373 K (the normal boiling point) P 2 = atm vap = 40,670 J/mol R = 8.34 J/mol K D T 2 = ln P = vap T æ æ ö ö ç ç è P 2 ø + D vap + RT ln P = 365K = 92 C æ ö ç ç è P 2 ø çæ D vap ö T ç ç èè R ø ø Using the following data, determine vap for hexafluorobenzene (C 6 F 6 ) T(K) P(torr) Use Excel s slope function to determine the slope of the line: /T ln P y = ln P m x + b /T ln P known_x s known_y s =SLOPE(known_y s, known_x s) D vap = -slope R /T = 35, 700 J mol = 35.7 kj mol = 35, 700 J mol = 35.7 kj mol Determine the normal boiling point for hexachlorobenzene given that vap = 35,700 J/mol T (K) P (torr) T 2 = æ æ ln P ö ö ç ç è P 2 ø + ç çæ D vap ö T çç èè R ø ø /T ln P y = mx +b x = y - b m ln P /T ln P -intercept = T slope slope T = ln(760)- intercept 9

10 Use Excel s slope and intercept functions: =SLOPE(known_y s, known_x s) =INTERCEPT(known_y s, known_x s) Consider the following vapor pressure versus temperature plot for three different substances A, B, and C. If the three substances are C 4, Si 4, and N 3, match each curve to the correct substance. x = y - b m ln P -intercept = T slope slope T = ln(760)- intercept Normal boiling point = K = 79. C Vapor pressure (torr) A B C Temperature ( C) When a substance goes from one phase to another phase, it has undergone a phase change. Phase changes are generally caused by the addition or removal of energy, usually in the form of heat. Example Freezing Evaporation (vaporization) Melting (fusion) Condensation Sublimation of dry ice Deposition of iodine Phase Change 2 O(l) 2 O(s) 2 O(l) 2 O(g) 2 O(s) 2 O(l) 2 O(g) 2 O(l) CO 2 (s) CO 2 (g) I 2 (g) I 2 (s) Vapor Pressure Increases as Temperature Increases The boiling point of a liquid is the temperature at which the vapor pressure becomes equal to the external atmospheric pressure exerted on the liquid. External Pressure Affects Boiling Point The normal boiling point of a liquid is the temperature at which a liquid boils at a pressure of atm. 0

11 Frozen Fruit Juice Concentrate: At a Reduced Pressure, the Boiling Point is Lower Pressure Cooker: At Increased Pressure, the Boiling Point is igher For every 0 C increase in temperature, food cooks twice as fast (takes half the amount of time). Some of the water in the fruit juice is boiled away at a reduced pressure, thus concentrating the juice without heating it to a high temperature (which spoils the taste of the juice and reduces its nutritional value). With an internal pressure of 2.02 atm in the pressure cooker, water boils at 2 C. Food cooks four times faster in the pressure cooker. The Molar eat of Vaporization ( vap )is the amount of heat required to vaporize a mole of substance at its boiling point. The Molar eat of Fusion ( vap ) is the energy required to melt one mole of a solid. The boiling point increases as vap increases. A Typical eating Curve The Molar Enthalpy of Sublimation is the energy required to sublime one mole of a solid. Temperature Boiling point Melting point Solid Solid and liquid in equilibrium Liquid and vapor in equilibrium Liquid Vapor Δ sub = Δ fus + Δ vap Solid I 2 in equilibrium with its vapor Sublimation is the process by which molecules go directly from the solid phase to the vapor phase. Time Deposition is the reverse process of sublimation.

12 (a) Calculate the amount of heat deposited on the skin of a person burned by.00 g of liquid water at 00.0 C and (b) the amount of heat deposited by.00 g of steam at 00.0 C. (c) Calculate the amount of energy necessary to warm 00.0 g of water from 0.0 C to body temperature and (d) the amount of heat required to melt 00.0 g of ice 0.0 C and then warm it to body temperature. (Assume that body temperature is 37.0 C.) Strategy For the purpose of following the sign conventions, we can designate the water as the system and the body as the surroundings. (a) eat is transferred from hot water to the skin in a single step: a temperature change. (b) The transfer of heat from steam to the skin takes place in two steps: a phase change and a temperature change. (c) Cold water is warmed to body temperature in a single step: a temperature change. (d) The melting of ice and the subsequent warming of the resulting liquid water takes place in two steps: a phase change and a temperature change. In each case, the heat transferred during a temperature change depends on the mass of the water, the specific heat of the water, and the change in temperature. For the phase changes, the heat transferred depends on the amount of water (in moles) and the molar heat of vaporization (Δ vap ) or molar heat of fusion (Δ fus ). In each case, the total energy transferred or required is the sum of the energy changes for the individual steps. The specific heat is 4.84 J/g C for water and.99 J/g C for steam. Δ vap is kj/mol and Δ fus is 6.0 kj/mol. Note: The Δ vap of water is the amount of heat required to vaporize a mole of water. owever, we want to know how much heat is deposited when water vapor condenses, so we use kj/mol. Solution (a) ΔT = 37.0 C 00.0 C = 63.0 C q = msδt =.00 g 4.84 J 63.0 C = g C 2 J = kj Thus,.00 g of water at 00.0 C deposits kj of heat on the skin. (The negative sign indicates that heat is given off by the system and absorbed by the surroundings.).00 g (b) = mol water 8.02 g/mol kj q = nδ vap = mol = 2.26 kj mol 4.84 J q 2 = msδt =.00 g 63.0 C = g C 2 J = kj The overall energy deposited on the skin by.00 g of steam is the sum of q and q 2 : 2.26 kj + ( kj) = 2.53 kj Solution (c) ΔT = 37.0 C 0.0 C = 37.0 C q = msδt =.00 g 4.84 J 37.0 C =.55 0 g C 4 J = 5.5 kj The energy required to warm 00.0 g of water from 0.0 C to 37.0 C is 5.5 kj. Think 00.0 g About It In problems that include phase changes, the q (d) = 5.55 mol water 8.02 values g/mol corresponding to the phase-change steps will be the largest contributions to the total. 6.0 If you kj find that this is not the case in your q = nδ fus = 5.55 mol = 33.4 kj solution, check to see if you mol have made the common error of neglecting to convert the 4.84 q values J corresponding q 2 = msδt = 00.0 g 37.0 C =.55 0 to temperature g C 4 J = 5.5 kj changes from J to kj. The energy rquired to melt 00.0 g of ice at 0.0 C and warm it to 37.0 C is the sum of q and q 2 : 33.4 kj kj = 48.9 kj A phase diagram summarizes the conditions at which a substance exist as a s, l, or g. Phase Diagram for CO 2 2

13 Phase Diagram for Water Using the following phase diagram, (a) determine the normal boiling point and the normal melting point of the substance, (b) determine the physical state of the substance at 2 atm and 0 C, and (c) determine the pressure and temperature that correspond to the triple point of the substance. Strategy Each point on the phase diagram corresponds to a pressuretemperature combination. The normal boiling and melting points are the temperatures at which the substance undergoes phase changes. These points fall on the phase boundary lines. The triple point is where the three phase boundaries meet. Solution By drawing lines corresponding to a given pressure and/or temperature, we can determine the temperature at which a phase change occurs, or the physical state of the substance under specified conditions. (a) Solution (b) The normal boiling and melting points are ~40 C and ~205 C, respectively. At 2 atm and 0 C the substance is a solid. Use the phase diagram for carbon to answer the following questions: Solution (c). ow many triple points are in the phase diagram? The triple point occurs at ~0.8 atm and ~5 C. Think About It The triple point of this substance occurs at a pressure below atmospheric pressure. Therefore, it will melt rather than sublime when it is heated under ordinary conditions. 2. What phases coexist at each triple point? 3. What happens if graphite is subjected to very high pressures at room temperature? 4. If we assume that density increases with an increase in pressure, which is more dense, graphite or diamond? 3

14 A crystalline solid possess rigid and long-range order; its atoms, molecules, or ions occupy specific positions. A unit cell is the basic repeating structural unit of a crystalline solid. There are seven types of unit cells. The coordination number is defined as the number of atoms surrounding an atom in a crystal lattice. The value of the coordination number indicates how tightly the atoms are packed together. The basic repeating unit in the array of atoms is called a simple cubic cell. There are three types of cubic cells. In a body-centered cubic cell (bcc) the spheres in each layer rest in the depressions between spheres in the previous layer. The coordination number is 8. In a face-centered cubic cell (fcc) the coordination number is 2. 4

15 Most of a cell s atoms are shared by neighboring cells. A simple cubic cell has the equivalent of only one complete atom contained within the cell. 8 atoms at corners equivlent atom 8 A corner atom is shared by eight unit cells. An edge atom is shared by four unit cells. A face-centered atom is shared by two unit cells. A body-centered cubic cell has two equivalent atoms: 8 atoms at corners 8 equivalent atom + atom in the center 2 equivalent atoms total exagonal close-packed (hcp) structure: Site directly over an atom in layer A A face-centered cubic cell contains four complete atoms: 8 atoms at corners equivalent atom atoms on faces 3 equivalent atoms 2 4 equivalent atoms total Close packing starts with a layer of atoms (A) Atoms in the second layer (B) fit into the depressions of the first layer exagonal closepacked structure. Cubic close-packed (ccp) structure: Closest packing: Site directly over an atom in layer A (hcp) Site NOT directly over an atom in layer A (ccp) Cubic close-packed structure exagonal close-packing (hcp) Cubic close-packing (ccp) corresponds to a facecentered cubit cell. 5

16 Edge length (a) and radius (r) are related: Gold crystallizes in a cubic close-packed structure (face-centered cubic unit cell) and has a density of 9.3 g/cm 3. Calculate the atomic radius of an Au atom in angstroms (Å). Strategy Using the given density and the mass of gold continued within a facecentered cubic unit cell, determine the volumes of the unit cell. Then, use the volume to determine the value of a, and use the equation a = 8r to find r. Be sure to use consistent units for mass, length, and volume. The face-centered cubic unit cell contains a total of four atoms of gold [six faces, each shared by two unit cells, and eight corners, each shared by eight unit cells]. d = m/v and V = a 3. Solution First, we determine the mass of gold (in grams) contained within a unit cell: 4 atoms mol 97.0 g Au m = = g/unit cell unit cell atoms mol Au Simple cubic Body-centered cubic Face-centered cubic Types of Crystals Solution Then we calculate the volume of the unit cell in cm 3 : m.3 0 V = = -2 = cm d g 3 Using the calculated volume and the 9.3 relationship g/cm 3 V = a 3 (rearranged to solve for a), we determine the length of a side of a unit cell: a = 3 3 Think About It Atomic V = radii tend to -23 be cmon 3 the = order of -8 cm Å, so this answer is reasonable. Using the relationship provided a = 8r (rearranged to solve for r), we determine the radius of a gold atom in centimeters. a r = 8 = -8 = cm cm Finally, we convert centimeters to angstroms: cm -2 Å =.44 Å m 0-0 cm m Ionic crystals are composed of charged ions that are held together by Coulombic attraction. The unit cell of an ionic compound can be defined be either the positions of the anions or the positions of the cations. Types of Crystals Crystal structures of three ionic compounds: ow many of each ion are contained within a unit cell of ZnS? Strategy Determine the contribution of each ion in the unit cell based on its position. Referring to the figure, the unit cell has four Zn 2+ ions completely contained within the unit cell, and S 2- ions at each of the eight corners and at each of the six faces. Interior ions (those completely contained within the unit cell) contribute one, those at the corners each contribute one-eighth, and those on the faces contribute Think About one-half. It Make sure that the ratio of cations to anions that you determine for a unit cell matches the ratio expressed in the compound s empirical formula. CsCl Simple cubic lattice ZnS Zincblende structure (based on FCC) CaF 2 fluorite structure (based on FCC) Solution The ZnS unit cell contains four Zn 2+ ions (interior) and four S 2- ions [8 (corners) and 6 (faces)] 8 2 6

17 The edge length of the NaCl unit cell is 564 pm. Determine the density of NaCl in g/cm 3. Strategy Use the number of Na + and Cl - ions in a unit cell (four of each) to determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is mass divided by volume (d = m/v). Be careful to use units consistently. The masses of Na + and Cl - ions are amu and amu, respectively. The conversion factor from amu to grams is g so the masses of the Na + and Cl - amu ions are g and g, respectively. The unit cell length is pm -2 cm = cm m 0-2 pm m Solution The mass of the unit cell is g ( g g). The volume of a unit cell is cm 3 [( cm) 3 ]. Therefore, the density is given by d = -22 g = 2.6 g/cm cm 3 3 Think About It If you were to hold a cubic centimeter ( cm 3 ) of salt in your hand, how heavy would you expect it to be? Common errors in this type of problem include errors of unit conversion especially with regard to length and volume. Such errors can lead to results that are off by many orders of magnitude. Often you can use common sense to gauge whether or not a calculated answer is reasonable. For instance, simply getting the centimeter-meter conversion upside down would result in a calculated density of g/cm 3! You know that a cubic centimeter of salt doesn t have a mass that large. (That s billions of kilograms!) If the magnitude of a result is not reasonable, go back and check your work. Types of Crystals In covalent crystals, atoms are held together in an extensive threedimensional network entirely by covalent bonds. Types of Crystals In molecular crystals, the lattice points are occupied by molecules; the attractive forces between them are van der Waals forces and/or hydrogen bonding. The metal iridium (Ir) crystallizes with a face-centered cubic unit cell. Given that the length of the edge of a unit cell is 383 pm, determine the density of iridium in g/cm 3. Strategy A face-centered metallic crystal contains four atoms per unit cell [8 (corners) and 6 (faces)]. Use the number of atoms per cell and the 8 2 atomic mass to determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is then mass divided by volume (d = m/v). Be sure to make all necessary unit conversions. Solution The mass of the unit cell is g ( g). The volume of a unit cell is cm 3 [( cm) 3 ]. Therefore, the density is given by d = -2 g = 22.7 g/cm cm 3 3 Think About It Metals typically have high densities, so common sense can help you decide whether or not your calculated answer is reasonable. The mass of an Ir atom is 92.2 amu. The conversion factor from amu to grams is g so the mass of an Ir atom is amu g. The unit cell length is pm -2 cm = cm m 0-2 pm m 7

18 Types of Crystals In metallic crystals, every lattice point is occupied by an atom of the same metal. Electrons are delocalized over the entire crystal. Delocalized electrons make metals good conductors. Large cohesive force resulting from delocalization makes metals strong. Types of Crystals Amorphous Solids Amorphous solids lack a regular three-dimensional arrangement of atoms. Glass is an amorphous solid. Glass is a fusion product. SiO 2 is the chief component. Na 2 O and B 2 O 3 are typically fused with molten SiO 2 and allowed to cool without crystallizing. Amorphous Solids Amorphous Solids Crystalline quartz Noncrystalline (amorphous) quartz glass 8