1. Chapter 10 Liquids and Solids

2. QUESTION Following the trend, from right to left , for each of the lines in the graph indicates that boiling point decreases in a group as molar mass decreases—except for the abrupt rise in three of the lines. What is the explanation for the difference in the group 4A line?

3. QUESTION (continued) 1. The intermolecular forces for CH 4 are stronger than the other elements in row two elements. 2. The hydrogen bonding forces are strongest between the compounds in group 4A. 3. The hydrogen bonding forces are strong for row two elements except for carbon. 4. The low electronegativity of carbon compared to N, F and O allows the dipole forces to increase.

4. ANSWER Choice 3 provides the correct explanation. The electronegativity of C is low compared to N, F and O; so no hydrogen bonding is possible for CH 4 . Section 10.1: Intermolecular Forces

5. QUESTION The polarization of electron distributions shown here produce instantaneous, temporary dipoles. Which statement properly implies the importance of this activity?

6. QUESTION (continued) 1. Nonpolar molecules can achieve permanent dipoles through this process. 2. The smaller the electron “cloud,” the greater the opportunity for influence from another electron “cloud” thus producing stronger dipoles. 3. The polarization forces (London forces) shown here are always weaker than dipole – dipole forces. 4. The larger, more polarizable, electron “cloud” of Ar helps explain why its freezing point is higher than Ne.

7. ANSWER Choice 4 correctly connects the result of polarizability with an observation. Larger electron clouds can be influenced by other forces, even other electron clouds, thus creating significant although temporary dipoles. This in turn allows for more inter-atomic attractions that would produce higher freezing points. This is shown in Table 10.2; Argon’s freezing point is –189.4°C whereas Ne freezes at –248.6°C. Section 10.1: Intermolecular Forces

8. QUESTION The electronegativity difference between C and O is fairly high, yet CO 2 is a gas. Which type of intermolecular force is the most likely between CO 2 molecules? 1. Dipole to dipole 2. Dispersion forces 3. Dipole to induced dipole 4. Intramolecular forces

9. ANSWER Choice 2 correctly identifies that molecules with small molar mass that are gases do not have any strong intermolecular forces. The dispersion forces, caused by temporary, instantaneous dipoles are too weak to provide small molecules with enough intermolecular force to overcome their free movements at room temperatures. Section 10.1: Intermolecular Forces

10. QUESTION NaCl and HCl, O 2 and CO 2 are compounds that are important to humans (in fact, are inside us.) Using intermolecular forces, which would have the highest freezing point? 1. O 2 2. HCl 3. NaCl 4. CO 2

11. ANSWER Choice 3 (NaCl) has the highest freezing point because it has the strongest intermolecular forces (ionic.) With strong intermolecular forces it would be the easiest to freeze because the particles (ions in this case) would already have strong attractive forces; thus, not as much energy would need to be removed to cause the phase change to solid. Section 10.1: Intermolecular Forces

12. QUESTION A liquid has its temperature increased slightly. Which of the following sequences correctly reports the effect on viscosity, vapor pressure, and surface tension respectively? 1. Increase, increase, increase 2. Decrease, increase, increase 3. Increase, decrease, decrease 4. Decrease, increase, decrease

13. ANSWER Choice 4 correctly relates the effect of increasing temperature on the three properties of the liquid. An increase in temperature is accompanied by an increase in the average kinetic energy for the molecules. As molecular motion increases, the resistance to flow (viscosity) would decrease since molecules are now flowing faster. The vapor pressure would increase because more molecules will be obtaining the energy needed to change phase. The surface tension will decrease; as the molecules move more, their intermolecular influences will decrease. Section 10.2: The Liquid State

14. QUESTION When reading the volume of water in a graduated cylinder you can notice that the middle portion of the water appears lower than the water on the inside of the glass cylinder. Why is the meniscus lower in the center? 1. The adhesive and cohesive forces in water are both strong enough to allow water to attract to the glass and pull some water with it. 2. The adhesive intermolecular forces of polar water are stronger than its cohesive forces; so it can “wet” the glass. 3. Wetting the glass allows the water to decrease its surface area, which is supported by the cohesive forces’ tendency to increase the surface area. 4. Although I know intermolecular forces are involved, I am not sure about the reason.

15. ANSWER Choice 1 properly includes the correct explanation. The polar nature of water and its small molar mass allow particularly strong intermolecular forces to hold water molecules to each other while they also are attracted to the slightly polar oxygen atoms in glass. Section 10.2: The Liquid State

16. QUESTION How is it possible that some substances consisting of nonpolar molecules can still have relatively high viscosities? (Consider that this is possible even at room temperature.) 1. The nonpolar nature of the molecules predicts that the boiling point is likely to be relatively low, thus the viscosity would be high. 2. Viscosity is a measure of a liquid’s resistance to flow, so the weaker intermolecular forces of a nonpolar substance would have a higher value. 3. Some nonpolar molecules, if large enough molecules, could develop large enough dispersion forces to have high viscosity. 4. High viscosity for nonpolar liquids does not seem logical, I don’t see the possible connection.

17. ANSWER Choice 3 provides the only way that nonpolar liquids could develop enough intermolecular attractions to cause a significant viscosity (without drastically lowering the temperature.) Some molecules are made from enough atoms and have flexible enough structures that they can have numerous opportunities for dispersion forces to become quite significant. Section 10.2: The Liquid State

18. QUESTION

19. QUESTION (continued) Using the representations here, how many total atoms could be claimed to be part of each of the three unit cells (simple cubic, body-centered cubic, face-centered cubic)? 1. 8, 9,14 respectively 2. 8, 9, 11 respectively 3. 1, 2, 4 respectively 4. It depends on the size of the unit cell.

20. ANSWER Choice 3 provides the correct comparison for the three cubic unit cells. In the third column of the figure, the portion of each atom that is part of the unit cell is displayed. Each vertice of the cube is occupied by only a 1/8 portion of an atom. An atom on the inside of a cube is not shared and therefore can be in its entirety. An atom on the face of a cube is shared between two unit cells; therefore, each may claim only ½. Section 10.3: An Introduction to Structures and Types of Solids Section 10.4: Structure and Bonding in Metals

21. QUESTION Applying the Bragg equation, d = n  /(2 sin  ), to x-ray diffraction data taken from diffraction experiments allows the calculation of several properties of crystals. If a metal was exposed to 1.54 Angstrom X-rays and had an internal spacing of 2.13 Angstroms, what would you calculate as the angle of diffraction ( n = 1) in the experiment? 1. 46.3° 2. 14.2° 3. 21.2° 4. I am not familiar enough with the sine function to do the calculation.

22. ANSWER Choice 3 provides the correct solution to the Bragg equation. The units of measurement, Angstroms, should be consistent and the math work involves solving for sin –1 in order to determine the angle. Section 10.3: An Introduction to Structures and Types of Solids

23. QUESTION Plastics are typically large molecular structures made from smaller chemically bonded repeating units. To which type of atomic solid and bonding type would this be classified? 1. Molecular solid; dipole to dipole 2. Molecular solid; dispersion forces 3. Network solid; covalent bonding 4. Network solid; delocalized covalent bonding

24. ANSWER Choice 3 correctly describes the bonding in plastics. The term “chemically joined” indicates that intramolecular forces are at work (covalent.) Since covalent bonds are directional, the electrons must be localized between bonded atoms. Section 10.3: An Introduction to Structures and Types of Solids

25. QUESTION Representing metal atoms as solid spheres allows us to describe the structures of metals on an atomic scale. Which statement below presents accurate comments about the various possible crystal structures and atomic packing?

26. QUESTION (continued) 1. In abab closest packing, the third layer of atoms is directly over the second layer of atoms. 2. In abca closest packing, the third layer of atoms is directly over the first layer of atoms. 3. In both abca closest packing, and abab packing each atom has 12 “nearest neighbors.” 4. These representations still do not help me visualize the packing.

27. ANSWER Choice 3 accurately notes a similar feature of both aba and abca closest packing for metals. A selected central atom will have 6 neighbors surrounding it in the same plane plus three each above and below in the adjacent layers. Section 10.5: Carbon and Silicon: Network Atomic Solids

28. QUESTION The expensive element platinum represents the close packing structure with a face-centered unit cell similar to silver. From X-ray data it has been determined that the length of an edge on the unit cell is 392.4 pm? From this information, what would be the radius of an atom of Pt? 1. 138.7 pm 2. 196.2 pm 3. 1109 pm 4. 392.4 pm

29. ANSWER Choice 1 correctly predicts the radius of Pt using the Pythagorean theorem. Note, that in the face centered cubic unit cell, the diagonal across the face of the unit cell crosses the radius of 4 atoms. The dimensions of the arms of the right triangle are the edges of the unit cell. Thus, the hypotenuse, once determined must be divided by 4 to get the radius of one atom. edge 2 + edge 2 = (4r) 2 Section 10.4: Structure and Bonding in Metals

30. QUESTION Lightweight, strong, and relatively inert, titanium makes an excellent metal for some body replacement parts. Titanium forms in a body centered unit cell. An atom has a radius of 142 pm. Using these measurements, what is the density, in g/cm 3 , of Ti? (Atomic mass of Ti = 47.88 g/mol) 1. Approximately 9.04 g/cm 3 2. Approximately 2.46 g/cm 3 3. Approximately 4.52 g/cm 3 4. I know that I need both mass and volume for the cell, but cannot get both from this data.

31. ANSWER Choice 3 is the best choice for the density of Ti. The mass can be obtained by determining the mass of 2 atoms of Ti (because 2 atoms make up a body-centered unit cell). The radius of an atom of titanium can be used to determine the length of the hypotenuse through the body of the cell. (radius  4) This can be used to determine the length of the edge (use Pythagoras theorem). Once the edge length is cubed it provides the volume, which can be divided into the mass of the unit cell. Section 10.4: Structure and Bonding in Metals

32. QUESTION The diagram here shows a treatment of energy levels for a crystal of magnesium. Explain how this diagram supports the property of conductivity found in Mg.

33. QUESTION (continued) 1. The filled MO’s represent electrons that are free to move downward to the 2 p and lower levels. 2. The small energy differences between the empty and filled MO’s for 3 s and 3 p enable electrons to freely move throughout the crystal. 3. Keeping the core electrons localized enables a great many electrons in that area to move freely and conduct electricity. 4. The purple and brown areas at the top of the diagram represent closely spaced energy bands that form when two magnesium atoms are metallically bonded.

34. ANSWER Choice 2 provides the accurate way to describe the connection between band theory and electrical conductivity. Many atoms of a metal are needed to produce the closely spaced bands that enable some electrons to move freely, and thus carry a current, in a crystal. Section 10.4: Structure and Bonding in Metals

35. QUESTION The graphite and diamond forms of carbon are both classified as network solids. However, diamonds do not conduct an electrical current whereas graphite does. What important difference accounts for this distinction? 1. Graphite forms ionic bonds, which do conduct electrical current. 2. The band gap in diamond is smaller than the band gap in graphite. 3. The unhybridized p orbital found in the carbon atoms in graphite create a weaker bond, which allows graphite sheets to slide and conduct an electrical current.   delocalized electron orbitals, with small band gaps, are not present in diamonds

36. ANSWER Choice 4 points out the critical difference that accounts for the lack of conductivity of diamonds compared to graphite. In graphite the carbon atoms have sp 2 hybridization (compared to sp 3 in diamonds). This allows the unhybridized p orbital of one atom to overlap with those nearby forming a delocalized pi bonding system with a very small energy gap between filled and unfilled MOs. Section 10.5: Carbon and Silicon: Network Atomic Solids

37. QUESTION The figure shown here presents both a P type and N type semiconductor. Silicon is a good semiconductor material but diamond, which has a similar bonding type, is not a good semiconductor. Which explanation offers the best reason for this?

38. QUESTION (continued) 1. As the diagram shows, a semiconductor must have band gap between conducting and non-conducting MO’s that is large so that “doping” atoms can fit. 2. The band gap in silicon must be slightly smaller than in diamond but not too large that some electrons may be excited. 3. The “doping” of silicon makes it possible to become a semiconductor, but the bonding of diamond is too stable for doping. 4. The p - n junction is not possible with the tetrahedron shape of diamonds.

39. ANSWER Choice 2 accurately states that band gaps for semiconductors must be, in some way, like goldilocks and the three bears – just right. If the gap is too large the substance is an insulator. If it is small, it will be a conductor. Semiconductors have a gap that allows some electrons to make the transition from non-conducting MO’s to conducting MOs. Section 10.5: Carbon and Silicon: Network Atomic Solids

40. QUESTION

41. QUESTION (continued) Using the information presented here, would C 2 H 5 OH (ethanol) have a higher or lower heat of vaporization than water? What would be the calculated value for  H vaporization? The slope of the ethanol line is approximately –4175. 1. Ethanol  H vap is higher than water;  H vap = 34 700 J/mol 2. Ethanol  H vap is higher than water;  H vap = 4 175 J/mol 3. Ethanol  H vap is lower than water;  H vap = 34 700 J/mol 4. I do not know how to determine this value.

42. ANSWER Choice 3 provides the correct response to both questions. The slope of the ethanol line is slightly less steep than water so it will have a slightly lower value for heat of vaporization. Using the relationship for these graphs, slope = –  H / R , yields the value of approximately 34 700 J/mol. Section 10.8: Vapor Pressure and Changes of State

43. QUESTION The apparatus depicted here allows the comparison of vapor pressures of solids and liquids at the same temperature. In order for the diagram to depict the freezing point, what must be true?

44. QUESTION (continued) 1. The vapor pressure of the liquid must exceed that of the solid so that liquid can form a solid at the freezing point. 2. The vapor pressure of the solid must be greater than that of the liquid present so that the liquid will evaporate more to reach equilibrium. 3. The vapor pressure of the liquid must equal the vapor pressure of the solid. 4. All of the liquid must be converted to the solid phase.

45. ANSWER Choice 3 states the correct relationship between the vapor pressure of the liquid and solid at the freezing point. The apparatus should show some of both phases present in an equilibrium. Section 10.8: Vapor Pressure and Changes of State

46. QUESTION The phase diagram for CO 2 is shown here. Which statement offers accurate and related observations?

47. QUESTION (continued) 1. At one atmosphere, it is not possible to form liquid CO 2 because the triple point is shown above the melting point. 2. The density of solid CO 2 is less than liquid CO 2 so, unlike water, the solid will sink in its own liquid. 3. Regardless of pressure, liquid CO 2 will not form higher than –56.6°C. This is because the critical point is not found until 72.8 atm 4. Solid CO 2 is known as dry ice because the solid form is very cold, and it cannot liquefy at room temperatures.

48. ANSWER Choice 1 is correct. At one atmosphere of pressure CO 2 will either be a solid or a gas, depending on the temperature, but not a liquid. Liquid CO 2 is not found until a minimum of – 56.6°C, and only then if the pressure is 5.1 atm. Section 10.9: Phase Diagrams

49. QUESTION Bringing water to boil is very pressure dependent. If a recipe called for placing your food in boiling water for 10 minutes at sea level, would your food be undercooked or overcooked if you did that on a mountain camping trip in Montana? Which of the following provides the correct response and reason? 1. Overcooked; at high altitudes water comes to a boil faster, so ten minutes would provide more cooking time in boiling water. 2. Undercooked; at high altitudes water boils at a lower temperature, so ten minutes in boiling water would not provide as much heat. 3. Overcooked; at high altitudes the pressure is less, so the the water will be hotter as it boils, thus overcooking food. 4. I don’t know; on camping trips I’m not that picky.

50. ANSWER Choice 2 is correct about both the cooking and the reason. At higher mountain altitudes where the air pressure is less, water will boil at lower temperatures. If food is placed in less hot water for the same amount of time as hotter water, it will not be as cooked. Cleaning with the cooler water may not be as sanitary on the campout either. Section 10.9: Phase Diagrams