INTERMOLECULAR FORCES AND CONDENSED STATES

Transcription

1 INTERMOLECULAR FORCES AND CONDENSED STATES I. Intermolecular Forces A. Types of Intermolecular Forces. 1. Van der Waals forces = attractive forces that exist between neutral molecules. a. Are much weaker than the intramolecular force of the covalent bond.( kj/mol) b. Weaker than ion-ion forces (Lattice energies, kj/mol) 2. Dipole-dipole forces. a. Exist between polar molecules. Due to the attraction of the negative end of one dipole and the positive end of another. b. Vary as 1/d 2 ( d = distance ). Fairly strong ( 5-25 kj/mol), should depend on the dipole moments of the molecules. c. Especially important in understanding the behavior of small polar molecules. 3. Dipole-induced dipole forces. a. The permanent dipole in one molecule can distort, or polarize, the electron density in a neighboring molecule creating, or inducing, a dipole. The resulting force of attraction is called dipole-induced dipole attraction. It can take place between a polar and a nonpolar molecule or between two polar molecules. - + polar molecule nonpolar molecule Polarization - dipole induced dipole dipole - induced dipole! attraction b. These forces of attraction will increases as the polarizability of the molecule increases. The polarizability is a measure of the ease of distortion of the electron density of a molecule. c. The polarizability increases as 1

2 1) the number of electrons increases. 2) the size of the molecule increases d. Forces 1/r 6 and are fairly weak (2-10 kj/mol). They are important in that they are responsible for the solubility of nonpolar gases, such as O 2 in polar solvents, such as water. 4. Induced dipole-induced dipole forces (London dispersion forces). a. A dipole created by an instantaneous asymmetry in the electron density of one molecule can induce a dipole in a neighboring molecule, giving rise to a force of attraction. b. Example. Consider two He atoms. Each He has two electron in spherically symmetric 1s orbitals. However, it is possible to have an instantaneous asymmetry in the electron density of one of the He atoms. This will induce an asymmetry in neighboring He atoms. If the electrons move in a synchronous manner, there will always be a weak attractive force between the He atoms. +2 Symmetric He 1s Instantaneous Dipole Induced dipole Induced Dipole - Induced Dipole Attraction This type of attractive forces are called London Dispersion Forces or Induced Dipole- Induced Dipole forces. 2

3 c. Such forces should increase as the polarizabilities of the molecules increase. It is found in all molecules, polar and nonpolar. It is the only type of attractive forces that exist between nonpolar molecules. Forces 1/d 6 ( kj/mol). 5. Hydrogen Bonding. a. Special force of attraction that exists between unshielded nuclear charge on a H, when it is covalently bonded to an O, N or F and the lone pair of electrons on a neighboring O, N, or F. Range is kj/mol. This is in addition to the other van der Waals forces. b. F, O, and N are the three most electronegative elements in the Periodic Table and are also the smallest members of their respective groups. Therefore any lone pairs on these atoms will be in small concentrated orbitals. 1) When these electronegative elements form covalent bonds with hydrogen, the electron density in the bonds will be greatly polarized away from the H atom. 2) Since H has only one electron (no core electrons) the polarization of the bonds will mean that the side of the H opposite a bond to F, O, or N will be greatly depleted of electron density, leaving an unshielded, "bare", nucleus. 3) The hydrogen bond arises from and attraction of a lone pair on F, O, or N with the unshielded nuclear charge on hydrogen (due to its covalent bonding to another F, O, or N). c. Hydrogen bonding is unique for H bonded to F, O, or N. It is in addition to any dipole- dipole, dipole-induced dipole or London forces. There are geometric constraints on the atoms involved in hydrogen bonding. Hydrogen Bond X H : X (X = F, O, or N) Lone Pair d. Some examples of molecules that undergo hydrogen bonding in their pure states (the hydrogen atoms capable of undergoing hydrogen bonding are in bold). O H O CH 3 C-O-H CH 3 N-H H-C-O-H H 2 O HF NH 3. 3

4 Liquid State I. Properties of Liquids. A. Kinetic Molecular Theory and its Application to Liquids. 1. In liquid state molecules cannot move independently of one another. Molecules condense together and can move only in the body of the liquid. a. Because of the motion collisions take place and there is a distribution of kinetic energies much like that in gases. b. The average kinetic energy per mole is much less than that required to overcome attractive forces and go into the gas phase. However, at any instant in time a certain fraction of molecules will have sufficient kinetic energy to escape into the gaseous state If such a high energy molecule is on the surface, it can escape from the liquid and go into the gas phase. c. This process is called evaporation. 1) Evaporation is a surface phenomenon. 2) Evaporation is a cooling process. The high kinetic energy molecules leave which lowers the average kinetic energy per mole, and hence the temperature. 2. Evaporation in a closed container--vapor Pressure. a. In a closed container, as evaporation puts more molecules in the vapor phase above the liquid, the probability that a gas molecule, through random collision, will end up back in the liquid state increases. b. Therefore, as evaporation continues, the rate of condensation will increase. A point will be reached when the rate of evaporation = rate of condensation. At this point a dynamic equilibrium is established and the concentration of vapor molecules above the liquid will reach a constant value. c. The partial pressure of vapor above the liquid will therefore reach a constant value, called the equilibrium vapor pressure of the liquid (or the vapor pressure). This is the same pressure that is required to liquefy the gas at the particular temperature. It is independent of the size of the container but increases as the temperature increases. d. Boiling point = the temperature at which the vapor pressure equals atmospheric pressure. At that temperature evaporation can take place in the body of the liquid and the phenomenon of boiling takes place. 4

5 Normal boiling point = temperature at which the vapor pressure is equal to 1 atm (760 Torr) e. Substances having strong intermolecular forces of attraction (having large values of a in the van der Waals equation, high critical temperatures and critical pressures) will have low vapor pressures and high normal boiling points. B. Variation of Vapor Pressure with Temperature -- The Clausius-Clapeyron Equation. 1. Consider the equilibrium at constant pressure a. ΔH for this process = Molar heat of vaporization = ΔH vap b. The partial pressure of the vapor in equilibrium with the liquid is lnp = - ΔH vap RT + B where B is a temperature independent constant. This equation is called the Clausius-Clapeyron Equation. 2. ΔH vap is always positive (evaporation is endothermic), therefore, a plot of lnp vs. 1 T will be a straight line with a slope of - ΔH vap R a. A convenient form of the equation for calculations is ln ( P 2 P 1 ) = - ΔH vap R [ 1 T 2 -, that is, a negative slope. 1 T 1 ] b. Example. The molar heat of vaporization of H 2 O is kj/mol and its normal boiling point is C. Calculate the vapor pressure of H 2 O at 27.0 C. P 2 = 760 Torr T 2 = = K P 1 =? ln( 760 P 1 ) = - Liquid T 1 = = K x10-3 [ 760 P 1 = e = 24.5 P 1 = ] = = 31.0 Torr. Vapor The actual vapor pressure of H 2 O at 27 C = 27.0 Torr. The discrepancy is due to the fact that ΔH vap is not exactly temperature independent. C. Other Properties. 5

6 1. Surface Tension, γ. a. A molecule in the bulk liquid is stabilized by the attraction of the other liquid molecules. Therefore, a molecule on the surface will be at a higher energy than one in the bulk of the liquid. b. Surface tension = excess energy per unit area of the surface = force required to stretch the surface of the liquid per unit area. Dimensions of γ are Newton/meter. c. Surface tension is responsible for the rise of a liquid such as H 2 O in a thin glass capillary (water "wets" or adheres to glass). This also gives rise to a meniscus when water or an aqueous solution is in a pipette or a burette. d. Surface tension should be large for a liquid having high attractive forces. 2. Viscosity. a. Viscosity is the resistance to flow of a liquid. b. When a liquid flows down a tube, the molecules slide over one another. The higher the forces of attraction, the less of a tendency the liquid will have to flow, and the higher will be the viscosity. c. Viscosity will also depend on the size and shape of the molecules. 3. The following Table lists some properties as some common liquids. Properties of Liquids. Liquid NBP, C ΔH vap (kj/mol) γ (N/m) Viscosity (centipoise) Ethyl Ether x (CH 3 OCH 3 ) Methanol x (CH 3 OH) Ethanol x (CH 3 CH 2 OH) Benzene x (C 6 H 6 ) Water x (HOH) II. Intermolecular Attraction. A. General intermolecular attractive forces. 1. The intermolecular forces that exist in the liquid state are the same as found among gases. a. Dipole-dipole forces. b. Dipole-induced dipole forces. c. Induced dipole-induced dipole (London dispersion) forces 2. The first two are important in polar liquids while the last one exists in all liquids. 6

7 3. In some liquids, such as H 2 O, hydrogen bonding is also important. B. Effects of Hydrogen Bonding on the Properties of Water. 1. Abnormally high melting points, boiling points, and ΔH vap. a. Compared to the other group 16 hydrides, H 2 O has unusually high melting and boiling points. b. NH 3 and HF also show these abnormalities. 2. The density of H 2 O(s) is less than that of H 2 O(l). a. In H 2 O(s) each O is tetrahedrally surrounded by four H's; two are covalently bonded and two are hydrogen bonded. b. Because of the orientation requirements imposed by hydrogen bonding, the H 2 O molecules are not packed together in ice in the spatially most efficient way. c. When ice melts, the hydrogen bonds are broken and the ice lattice structure collapses in on itself. Therefore, the density increases. 3. Between 0 C and 4 C, the density of H 2 O(l) increases as the temperature increases. a. For most substances density decreases as the temperature increases ( things expand on heating). b. In H 2 O(l) between 0 and 4 C there are still large domains of water molecules involved in ice-like hydrogen bonded structures. As the temperature increases from 0 to 4 C,these domains are progressively broken down and the liquid contracts. The maximum density occurs at 3.98 C (1.000 g/ml), as the temperature increases from this point, the natural tendency of a substance to expand on heating dominates and the density decreases. 4. Other Effects. a. Protein Structure. Hydrogen bonding is one of the main forces stabilizing the α-helix structure of proteins. b. Genetic Code. Base pair recognition in DNA and RNA is through hydrogen bonding. 7

8 Phase Diagrams I. Phase Diagram. A. Diagrams on a pressure vs. temperature scale that show the conditions under which a phase will be stable or in equilibrium with other phases. B. Phase diagram of H 2 O (not to scale). Consists of three areas that are divided by lines (A-O, O-B, and O-C) that intersect at a point (point O). 760 C B Pressure, Torr Solid Liquid 4.58 O Vapor A Temperature, o C 1. Areas. At any temperature and pressure that falls in an area, only one phase is stable. Any other phase will spontaneously convert to the particular stable phase. 2. Lines. Delineate temperatures and pressures in which two phases are in equilibrium. a. Line A-O. H 2 O(s) H 2 O(g) 1) A plot of the sublimation pressure of the solid as a function of temperature. 2) Governed by a Clausius-Clapeyron type of equation. lnp = -!H sub RT +!S sub R ΔH sub = ΔH vap + ΔH fus 8

9 b. Line O-B. H 2 O(l) H 2 O(g) 1) A plot of the vapor pressure of the liquid as a function of temperature, or a plot of the boiling point of the liquid as a function of pressure. 2) The plot is governed by the Clausius-Clepeyron equation. The line O-B stops at the critical temperature and critical pressure of the substance. c. Line O-C. H 2 O(s) H 2 O(l) 1) A plot of the freezing point of the liquid, or melting point of the solid, as a function of pressure. 2) For H 2 O the freezing point decreases as the temperature increases. This is because the density of H 2 O(s) is less than the density of H 2 O(l). This is predicted from Le Chatelier's Principle (when a stress is placed on a system in equilibrium the equilibrium will shift so as to counteract the stress). The molar volume of H 2 O(s) is greater than that of H 2 O(l). Therefore, when pressure is applied the equilibrium H 2 O(s) H 2 O(l) will shift to the side having the smaller molar volume, that is, to the liquid and the temperature would have to be decreased in order to reestablish equilibrium. 3. Point O. Triple Point. a. At this temperature and pressure all three phases coexist in equilibrium, that is, the melting point and the boiling point coincide. This is a unique point for a substance. b. Note that the liquid phase can not exist at pressures below that of the triple point. c. For H 2 O the pressure of the triple point is quite low, 4.58 Torr. However, the triple point for CO 2 is at 5.2 atm and -57 C. Therefore, while ice melts on heating at normal atmospheric pressure, CO 2 (s) will sublime, and hence the name "dry ice". Crystalline State I. Structure of Solids. A. Crystal lattice. 1. Crystalline solids (true solids) are characterized by both short range and long range order. a. The particles that make up the solid (ions, molecules, or atoms) are arranged in definite repeating geometric patterns. b. This internal order gives rise to an external order in that crystalline solids possess plane faces that intersect at definite angles. 2. The internal geometric order can be shown by the crystal lattice (or space lattice). a. The lattice is a network of lines connecting the centers of the particles that make up the solid. b. Shown below is a two dimensional array of spheres forming a monolayer "solid" 9

10 and its corresponding lattice. Solid Lattice 3. Unit Cell. a. It is not necessary to consider the complete space lattice. We are only interested in the smallest part of the space lattice which, when repeated, will generate the complete lattice. This is called the unit cell. b. The unit cell is the smallest part of the space lattice (that is the solid) that has all the properties of the solid. That is, it has 1) the same geometry or morphology as the solid. 2) the same stoichiometry as the solid. 3) the same density as the solid. c. The unit cell can be determined experimentally from X-ray diffraction. 1) X-rays are scattered by the electrons of the atoms in a solid. The process acts like the X-rays were being scattered (or diffracted) by the lines on a grating ( the "lines" are the rows of atoms in the solid). 2) Depending on the wavelength of the X-rays, the angle of incidence of the X-ray 10

11 beam, and the geometry of the solid, the diffracted beam could be in phase and are detectable, or out of phase and not detectable. This gives rise to a diffraction pattern. 3) By analyzing the diffraction pattern of the X-rays of a particular wavelength, the structure of the solid, that is the unit cell, can be determined. B. Systems of Unit Cells. 1. There are seven systems of unit cells. These are described in the following Figure. We will be concerned only with the Cubic System. 2. In the cubic system the basic shape of the unit cell is that of a cube (all sides of equal length and meeting at right angles). There are three unit cells in the cubic system. a. Simple Cubic Unit Cell (Primitive Cubic). 1) The unit cell consists of eight particles, one at each of the corners of the unit cell. 2) Since each of the eight particles belong to eight unit cells, only 1/8 of each particle is contained within a particular unit cell. The total number of particles per simple cubic unit cell = 1/8(8) = 1 3) Therefore, there is one unit cell per particle. If one had a mole of a solid that crystallized in a simple cubic unit cell, there would be 6.03x10 23 unit cells in the solid. 4) In a simple cubic unit cell the particles are in contact along the edge of the unit cell. The radius of the particle, r, and the length of the unit cell, a, are related by 2r = a or r = a 2 5) The simple cubic unit cell is rather rare in nature. The only metal crystallizing in this unit cell is Po. b. Body Centered Cubic (bcc) Unit Cell. 1) The unit cell consists of the eight particles at the corners plus another particle directly in the center of the unit cell. 2) There are 1/8(8) + 1 = 2 particles per body centered cubic unit cell. 3) The particles are in contact along the diagonal of the cube. There are three particles stacked along this diagonal. Let d = the length of the diagonal; r = the radius of a particle and a = the length of the edge of the unit cell. Then d = a 2 + a 2 + a 2 = 3 a d = 4r 3! 4r = 3 a or r = a 4 11

12 4) Iron, among other metals, crystallize in a bcc unit cell. c. Face Centered Cubic (fcc) Unit Cell. 1) The unit cell consists of the eight particles at the corners of the unit cell plus a particle in the center of each of the six faces of the cube. Each particle on a face is in common with two unit cells and contributes 1/2 per unit cell. 2) The unit cell contains 1/8(8) + 1/2(6) = 4 particles. 3) Contact between the particles is along the diagonal of the face of the cube. Let d = the diagonal of a face; r = the radius of the particle; and a = the length of the edge of the unit cell. Then d = a 2 + a 2 = 2 a 4r = d! 4r = 2 2 a or r = a 4 4) Al, Cu, Au, Ag, and many other substances, crystallize in a fcc unit cell. C. Calculations on Cubic unit cells. Aluminum (At. wt. = 27.0) crystallizes in a fcc unit cell and has a density of 2.7 g/cc. Calculate, 1. the molar volume V g/mol V = 2.7 g/cc = 10.0 cm3 /mol 2. the cm 3 per atom of Al. cm cm 3 /mol per atom = 6.02x10 23 atoms/mol = 1.67x10-23 cm 3 /atom 3. the cm 3 per unit cell, that is the volume of the unit cell, V uc. V uc = (1.67x10-23 cm 3 /atom)(4 atoms/uc) = 6.7x10-23 cm 3 /uc 4. the length of the edge of the unit cell, a. V uc = a 3! a = 3 V uc 5. the radius of an Al atom, r. In a fcc unit cell, a = 3 6.7x10-23 cm 3 = 4.05x10-8 cm = 405 pm r = 2 2 a = (405 pm) = 142 pm

13 II. Close Packing. Many common metals and salts crystallize in structures that can be understood by considering how spheres of equal size can be packed together in the spatially most efficient way possible. This is called close packing. A. They will stack in Layers. 1. First Layer (layer a). a. Each sphere in the layer is touching six other spheres in that layer. b. This layer is shown in Figure Second Layer (layer b). a. Each sphere in this layer will come to rest in the crevice formed by the juncture of three spheres in layer a (see Figure 2). b. Each sphere is then in contact with 12 other spheres, 6 in the same layer + 3 in the layer below + 3 in the layer above. 1) The number of nearest neighbor spheres (the number of spheres touching a particular sphere) is called the coordination number (CN). 2) In close packing the coordination number = 12. c. All space is not occupied by the spheres. There are holes, or cavities, between the two layers. There are two types of such cavities. 1) Tetrahedral Cavities are formed by the juncture of three spheres in one layer and one sphere in an adjacent layer (see Figure 3). - If a particle were in this cavity it would be tetrahedrally coordinated by the four spheres. - They are the smaller of the two types of cavities. - They are more numerous than the other type of cavity. The number of tetrahedral holes twice is equal to the number of spheres in the close packed array (one cavity above and below each sphere) 2) Octahedral Cavities are formed by the juncture of three spheres in one layer and three spheres in an adjacent layer (see Figure 4). - If a particle were in this cavity it would be octahedrally surrounded by the six spheres. - They are larger but less numerous than tetrahedral cavities. - The number of octahedral cavities is equal to the number of spheres in the close packed array. 13

14 First Layer Sphere Second Layer Sphere Figure 1. Layer a in a close packed array. Figeue 2. Layers a and b in a close packed array. Figure 3. Tetrahedral cav ities. Figure 4. An octahedral cavity. 14

15 3. Third Layer. There are two possibilities. a. Spheres in the third layer cover tetrahedral holes. 1) When one third layer comes to rest over a tetrahedral hole, all other spheres in that layer must cover tetrahedral holes. Every sphere in the third layer will be directly over a sphere in the first layer. Therefore, the third layer is just a repeat of the first layer, layer a. 2) When each succeeding layer covers tetrahedral cavities, the layer sequence is ababab 3) This type of close packing is called Hexagonal Close Packing (hcp). The unit cell of the resulting solid is in the hexagonal system. 4) Examples of hcp solids are Zn, Cd, and H 2 (s). b. Spheres in the third layer cover octahedral holes. 1) When one sphere comes to rest over an octahedral cavity, all other spheres in that layer must cover octahedral cavities. The spheres in the third layer are not over spheres in either the first or the second layer. This in a unique layer, called layer c. 2) When octahedral holes are covered by each succeeding layers, the layer sequence is abcabcabc 3) This type of close packing is called Cubic Close Packing (ccp). The unit cell is face centered cubic. 4) Examples: Al, Cu, Ag, Au, and many others. B. Structures of Ionic Compounds. 1. The structures of many ionic compounds can be understood in terms of one ion forming a close packed array and the other ion occupying the one of the cavities in that array. a. Since anions are larger than cations, most of the time, but not all the time, the anions form the close packed array and the smaller cations are in the cavities. b. In ionic compounds, The coordination number of an ion is defined as the number of nearest neighbor counter-ions. 1) The coordination number of an ion in a tetrahedral cavity = 4. 2) The coordination number of an ion in a octahedral cavity = Examples. a. NaCl (rock salt) 1) The Cl - ions form a ccp array and the Na + ions occupy all the octahedral cavities in the Cl - close packed array. 2) The coordination number of Na + = 6 Since there are just as many Na + ions as Cl - ions, there also must six Na + ions around each Cl -. Therefore, the coordination number of Cl - = 6. (NaCl could 15

16 also be described as an array of close packed Na + ions with the Cl - ions occupying octahedral holes.) 3) Per unit cell there are 4 Cl - ions (the Cl - ions crystallize in a fcc unit cell) and 4 Na + ions. 4) Many MX type salts crystallize in the NaCl structure. This is called the rock salt structure or a 6:6 structure. The 6:6 refers to the coordination numbers of the cation and anion, respectively. b. Li 2 O 1) The O 2- ions form a ccp array with the Li + ions occupying all the tetrahedral cavities in the O 2- array. 2) The coordination number of Li + = 4. Since there are twice as many Li + ions as there are O 2- ions, there must be twice as many Li + 's around an O 2- as there are O 2- 's around a Li +. Therefore the coordination number of O 2- = 8. 3) Per unit cell there are 4 O 2- ions (O 2- forms a fcc unit cell) and 8 Li + ions. 4) This structure is called an antifluorite structure. The mineral fluorite, CaF 2, has a structure in which the Ca 2+ ions are ccp and the F - ions occupy the tetrahedral cavities. c. MgAl 2 O 4 (Spinel). 1) The O 2- ions form a ccp array with the Mg 2+ ions in the tetrahedral holes and the Al 3+ ions in octahedral holes. 2) Not all the cavities are occupied. Per formula unit there are four O 2- ions which gives rise to eight tetrahedral and four octahedral holes. Therefore, the Mg 2+ ion occupy only 1 8 of the tetrahedral and the Al 3+ ions occupy 1 2 of the octahedral cavities. 3) Per unit cell there are 4 O 2- ions, 1 Mg 2+ ion and 1 Al 3+ ion. 4) This structure is called the spinel structure. There are a number of minerals of the formula AB 2 O 4, where A is a divalent ion and B is a trivalent ion, that have the spinel structure. IV. Solid State Defects. A. Defects in chemically pure substances. 1. Vacancies. Some of the lattice sites are not occupied. 2. Interstitials. Due to thermal motion, a particle can move into a type of cavity that is not ordinarily occupied, an interstitial position. The introduction of these types of defects on heating of a solid is important in determining the melting point of the solid. 3. Dislocations. The planes of atoms are not perfectly aligned. 16

17 a. Edge dislocations. A plane of particles will not go all the way through the crystal. b. Screw dislocation. The planes of particles are warped about an axis such that they form a spiral like the pitch of a screw. B. Chemical Impurities. 1. Nonstoichiometric solids. a. These defects can occur in salts where the metal cation can exist in several oxidation states. Some examples are TiO and Fe 0.95 O. Small amounts of the higher oxidation state metal could replace the lower oxidation state metal in the lattice. In order to maintain electrical neutrality, some metal sites must be vacant. For example, Fe 0.95 O is due to the presence of Fe 3+ being in some of the Fe 2+ sites. There are also some metal rich nonstiochiometric compounds, such as, Cu 1.7 S and Cu 1.65 Te. b. Many of the heavier metal hydrides can have varying amounts of hydrogen. An example is VH 0.56 Other metals, such as, Fe, Pd, and Pt are permeable to hydrogen. It is thought that hydrogen atoms diffuse through these metals by occupying interstitial sites in the metal lattice. 2. Chemical doping - semiconductors. a. n-type semiconductor. 1) Formed when small amounts of a group 15 element are introduced into a group 14 element's lattice. The group 14 solid is said to be doped with the group 15 element. An example is Si doped with P 2) The Si is sp 3 hybridized and forms covalent bonds with four other silicons. When a P replaces a Si in this lattice, four of its valence electrons are involved in covalent bonding while the extra P electron is free to move in the crystal. The doped Si then becomes a good conductor of electricity. b. p-type semiconductor. 1) Formed when a group 14 crystal is doped with a group 13 element. An example is Si doped with B. 2) The B has only three valence electrons and when it replaces a Si atom in the Si crystal there will be a vacancy of one electron or a "hole" into which a electron from a neighboring bond could migrate. In this way the vacancy, or hole, could move through the crystal in a direction opposite to that of electron motion; this will have the effect of a "positive hole" moving through the crystal giving rise to p-type semiconduction. 17

18 V. Types of Solids. It is possible to classify solids not according to their structures, but according to the particles occupying the lattice sites. Molecular _Ionic Covalent Metallic Units that occupy Molecules Cations Atoms Metal cations Anions in electron sea Binding Forces Van der Waals Electrostatic Covalent Electrical (dipole-dipole, attraction attraction between London) + ions and - electrons Properties Very soft Hard and Very Hard Hard to soft brittle Low melting High melting Very High Very variable points points points melting points Volatile Good Insula- Good insula- Nonconduc- Good tors tors tors conductors Examples H 2, H 2 O, NaCl, KBr, C(diamond), Na, K, Al, CO 2 Fe(NO 3 ) 2 Carborundum Fe (SiC), Quartz (SiO 2 ) 18

19 Condensed Phases Problem Set 1. Consider the following diagram of the logarithm of the vapor pressure vs. the reciprocal of the temperature for several substances. ( Note (l) = liquid, (s) = solid ) 12 C(l) 10 B(l) ln P (P in torr) 8 A(l) C(s) / T, K - 1 a. Which liquid is the most volatile at 127 C? (liquid C) b. At what temperature will liquid A abd B have the same vapor pressure? ( 526 K) c. Estimate the normal boiling point of A. (350 K) d. Which liquid has the highest value of ΔH vap? (liquid A) e. Estimate the temperature and pressure of the triple point of C.(303 K, 3.63x10 4 Torr) 2. The vapor pressure of a liquid obeys the relation 4000 ln P = T where P is the vapor pressure in torr. a. What is the value of ΔH vap for this liquid? (3.33x10 4 J/mol) b. What is the normal boiling point of the liquid? (3206 K) 19

20 3. The vapor pressure of a liquid is 5.2 torr at 25.0 C and 35.6 torr at 40.0 C. Calculate a. the value of ΔH vap of the liquid.(9.95x10 4 J/mol) b. The vapor pressure of the liquid at 60.0 C. (353 Torr) 4. Consider the following liquids: C 6 H 6, CH 4, C 6 H 5 Cl, CH 3 Cl. Arrange the liquids in order of a. increasing normal boiling points. (CH 4 <C 6 H 6 <CH 3 Cl<C 6 H 5 Cl) b. increasing surface tension. (CH 4 <C 6 H 6 <CH 3 Cl<C 6 H 5 Cl) c. decreasing ΔH vap. (C 6 H 5 Cl>CH 3 Cl>C 6 H 6 >CH 4 ) d. increasing vapor pressure at 25 C. (C 6 H 5 Cl<CH 3 Cl<C 6 H 6 <CH 4 ) 5. Of the two isomers of dichlorobenzene shown below, which one will have the higher viscosity? ( Justify your answer) Cl Cl Cl Cl 1,4-dichlorobenzene 1,2-dichlorobenzene (1,2-dichlorobenzene) 6. Of the two compounds CH 3 OCH 3 and CH 3 CH 2 OH, which one will have the highest normal boiling point?( Justify your answer ) (CH 3 CH 2 OH) 7. Silver crystallizes in a face centered cubic unit cell, the length of the edge of the unit cell is pm. Calculate the density of silver. (10.5 g/cc) 8. Gold has a radius of 144 pm and crystallizes in a body centered cubic unit cell. Calculate the density of gold. (17.79 g/cc) 9. Barium metal has a density of 3.50 g / cm 3 and an atomic radius of 220 pm. In what type of cubic unit cell does Barium crystallize? (bcc) 10. The salt CuI was studied and the following facts were obtained: a) The I - ions form a cubic close packed array. b) The Cu + ions are in cavities that give rise to a coordination number of 4. a. What type of cavities in the I - array do the Cu + ions occupy? What fraction of these cavities are occupied by the Cu + ions? b. The arrangement of layers of I - ions is described by the letter sequence. c. In forming the particular lattice the third layer of I - ions will cover what type of cavity? d. What is the coordination number of I -? e. If the coordination number of Cu + had been 6, what type of cavity would it occupy in the I - array? What would be the coordination number I - under these conditions? f. Of the two types of cavities in the I - array, is the larger and 20

21 is the more numerous one. g.per unit cell, there are I - ions and Cu + ions. 11. Shown below is a representation of the first two layers in a close packed array. (The solid circles are in the first layer.) a. Two types of cavities exist in this lattice. Identify several of these using "O' for octahedral and "T" for tetrahedral. b. Over what type of cavity would the third layer be located if the lattice were cubic? If the lattice were hexagonal close packed? 12. The usual geometric picture drawn of the octahedral arrangement of six groups about a central point is shown below. For one of the octahedral holes in the above diagram, label the surrounding spheres a, b, c, d, e,and f to indicate how they fit the drawing below. e c b hole d a f 21

22 13. Consider the following substances in their solid state. NO 2 N 2 CaO Cu SiO 2 Listed below are a number of statements. By each statement write the formula of the above substance that is best described by the statement. a. Is a good conductor in the liquid state but not in the solid state. b. Has the lowest melting point. c. Has the highest melting point. d. At any particular temperature will have the highest vapor pressure. e. Crystal lattice is stabilized by dipole and London forces only. f. Lattice points occupied by ions. g. Crystal lattice stabilized by covalent bonds. 14. The mineral perovskite has a structure in which Ca 2+ ions occupy the corners of a cubic unit cell, O 2- ions are in the centers of each of the faces, and a Ti 4+ is at the center of the cell. What is the formula of perovskite. 15. On the folowing page is a phase diagram for a hypothetical substance. a. On the diagram label the regions in which the solid, liquid, and vapor phases would be stable. b. Give the temperature and pressure of the triple point. c. Which line represents a plot of the sublimation pressure of the solid as a function of temperature? d. Which line represents a plot of the vapor pressure of the liquid as a function of temperature? e. What is the normal boiling point of the substance? f. Suppose you have a sample of this substance at 0.8 atm and 20 C. Describe what would happen if you heated the substance to 120 C at a constant pressure of 0.8 atm. Give the temperatures of any phase changes that may occur during the heating process. g. Do the same as in (f) when the pressure is 0.4 atm. h. Which would be greater, the density of the solid or the density of the liquid? Justify your answer. 22

23 C 1.0 B Pressure, atm 0.5 O A Temperature, o C 120 Answers a. Tetrahedral b. abcabcabc... c. Octahedral d. 4 e. Octahedral, 6 f. Octahedral, Tetrahedra; g. 4,4 11. a. see notes. b. Octahedral. Tetrahedral. 13. a. CaO b. N 2 c. SiO 2 d. N 2 e. NO 2 f. CaO & Cu g. SiO CaTiO b. triple point: 32 C & 0.5 atm c. AO d. OB e. 120 C f. melts at 32 C, boils at 96 C g. sublimes at 23 C h. d solid = d liquid since OC is essentially vertical. 23