Chapter 7. Periodic Properties of the Elements. Lecture Presentation. 劉玉嬌 ( 中研院化學所 ) 光電學程海洋工程學程 2018/11/09

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1 劉玉嬌 ( 中研院化學所 ) yuchiao@chem.sinica.edu.tw 光電學程海洋工程學程 Lecture Presentation Chapter 7 of the 2018/11/09, Inc. James F. Kirby Quinnipiac University Hamden, CT

2 Outline: 7.1 Development of the Table 7.2 Effective Nuclear Charge 7.3 Sizes of Atoms and Ions 7.4 Ionization Energy 7.5 Electron Affinity 7.6 Metals, Nonmetals, and Metalloids 7.7 Trends for Group 1A and Group 2A Metals 7.8 Trends for Selected Nonmetals (H, 6A 8A), Inc. 1

3 7.1 Development of the Table Dmitri Mendeleev and Lothar Meyer independently came to the same conclusion about how elements should be grouped. Both noted that similar chemical and physical properties recur periodically when the elements are arranged in order of increasing atomic weight. 2 of the

4 Mendeleev and the Table Chemists mostly credit Mendeleev because he also used chemical properties to organize the table and predicted some missing elements and their expected properties, including germanium. 3 of the

5 Atomic Number Mendeleev s table was based on atomic masses. It was the most fundamental property of elements known at the time. About 35 years later, the nuclear atom was discovered by Ernest Rutherford. Henry Moseley developed the concept of atomic number experimentally. The number of protons was considered the basis for the periodic property of elements. For example, the atomic weight of Ar (atomic number 18) is greater than that of K (atomic number 19), yet the chemical and physical properties of Ar are much more like those of Ne and Kr than like of the those of Na and Rb. 4

6 ity ity is the repetitive pattern of a property for elements based on atomic number. The following properties are discussed in this chapter: Sizes of atoms and ions Ionization energy Electron affinity Some group chemical property trends First, we will discuss a fundamental property that leads to many of the trends, effective nuclear charge. 5 of the

7 7.2 Effective Nuclear Charge Many properties depend on attractions between valence electrons and the nucleus. Electrons are both attracted to the nucleus and repelled by other electrons. The forces an electron experiences depend on both factors. of the Na: 1s 2 2s 2 2p 6 3s 1 6

8 Effective Nuclear Charge The effective nuclear charge, Z eff, is found this way: Z eff = Z S where Z is the atomic number and S is a screening constant, usually close to the number of inner electrons. (Electrons in the same valence shell do not screen one another very effectively, but they do affect the value of S slightly) Effective nuclear charge is a periodic property: o It increases across a period. o It increases down a group. of 7 the

9 Screening and Effective Nuclear Charge Atomic number Z effective = Z S core e Li:1S 2 2S 1 ~ 3 2 = 1 Be:1S 2 2S 2 ~ 4 2 = 2 Core ~valence 8 of the

10 (ψ 2 r 2 ) (r) of 9 the

11 Effective Nuclear Charge Increases across a Period Z eff = Z S [7.1] F:1S 2 2S 2 2P (1S = core = 2 e) S = (0.35 * 6) + (0.85 * 2) = 3.8. ~~~ = 5.2 of 10 the

12 of 11 the

13 7.3 Sizes of Atoms and Ions What Is the Size of an Atom? The nonbonding atomic radius or van der Waals radius is half of the shortest distance separating two nuclei during a collision of atoms. 12 of the

14 Sizes of Atoms The bonding atomic radius is half the internuclear distance when atoms are bonded. The bonding atomic radius tends to decrease from left to right across a period (Z eff ). increase from top to bottom of a group (n ). 13 of the

15 small large stable of the 14 Because of the most stable of 8A, 7A is eager to transfer to 8A

16 Sizes of Ions Determined by interatomic distances in ionic compounds Ionic size depends on the nuclear charge. the number of electrons. the orbitals in which electrons reside. 15 of the

17 Sizes of Ions Cations are smaller than their parent atoms: The outermost electron is removed and repulsions between electrons are reduced. Anions are larger than their parent atoms: Electrons are added and repulsions between electrons are increased. 16 of the

18 Size of Ions Isoelectronic Series In an isoelectronic series, ions have the same number of electrons. Ionic size decreases with an increasing nuclear charge. An Isoelectronic Series (10 electrons) Note increasing nuclear charge with decreasing ionic radius as atomic number increases O 2 F Na + Mg 2+ Al Å 1.19 Å 1.16 Å 0.86 Å 0.68 Å 17 of the

19 Sample Exercise 7.2 Predicting Relative Sizes of Atomic Radii Referring to the periodic table, arrange (as much as possible) the atoms B, C, Al, and Si in order of increasing size. Solution Analyze and Plan We are given the chemical symbols for four elements and told to use their relative positions in the periodic table to predict the relative size of their atomic radii. We can use the two periodic trends just described to help with this problem. Solve C and B are in the same period, with C to the right of B. Therefore, we expect the radius of C to be smaller than that of B because radii usually decrease as we move across a period. Likewise, the radius of Si is expected to be smaller than that of Al. Al is directly below B, and Si is directly below C. We expect, therefore, the radius of B to be smaller than that of Al and the radius of C to be smaller than that of Si. Thus, so far we can say C < B, B < Al, C < Si, and Si < Al. We can therefore conclude that C has the smallest radius and Al has the largest radius, and so can write C <? <? < Al. Our two periodic trends in atomic size do not supply enough information to allow us to determine, however, whether B or Si (represented by the two question marks) has the larger radius. Going from B to Si in the periodic table, we move down (radius tends to increase) and to the right (radius tends to decrease). It is only because Figure 7.7 provides numerical values for each atomic radius that we know that the radius of Si is greater than that of B. If you examine the figure carefully, you will discover that for the s- and p-block elements the increase in radius moving down a column tends to be the greater effect. There are exceptions, however. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc. 18

20 Sample Exercise 7.2 Predicting Relative Sizes of Atomic Radii Continued Check From Figure 7.7, we have C(0.76 Å) < B(0.84 Å) < Si(1.11 Å) < Al(1.21 Å). Comment Note that the trends we have just discussed are for the s- and p-block elements. Figure 7.7 shows that the transition elements do not show a regular decrease moving across a period. Practice Exercise 1 By referring to the periodic table, but not to Figure 7.7, place the following atoms in order of increasing bonding atomic radius: N, O, P, Ge. (a) N < O < P < Ge (b) P < N < O < Ge (c) O < N < Ge < P (d) O < N < P < Ge (e) N < P < Ge < O Practice Exercise 2 Arrange Be, C, K, and Ca in order of increasing atomic radius. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc. 19

21 Sample Exercise 7.3 Predicting Relative Sizes of Atomic and Ionic Radii Arrange Mg 2+, Ca 2+, and Ca in order of decreasing radius. Solution Cations are smaller than their parent atoms, and so Ca 2+ < Ca. Because Ca is below Mg in group 2A, Ca 2+ is larger than Mg 2+. Consequently, Ca > Ca 2+ > Mg 2+. Practice Exercise 1 Arrange the following atoms and ions in order of increasing ionic radius: F, S 2, Cl, and Se 2. (a) F < S 2 < Cl < Se 2 (b) F < Cl < S 2 < Se 2 (c) F < S 2 < Se 2 < Cl (d) Cl < F < Se 2 < S 2 (e) S 2 < F < Se 2 < Cl Practice Exercise 2 Which of the following atoms and ions is largest: S 2, S, O 2? Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc. 20

22 Sample Exercise 7.4 Ionic Radii in an Isoelectronic Series Arrange the ions K +, Cl, Ca 2+, and S 2 in order of decreasing size. Solution This is an isoelectronic series, with all ions having 18 electrons. In such a series, size decreases as nuclear charge (atomic number) increases. The atomic numbers of the ions are S 16, Cl 17, K 19, Ca 20. Thus, the ions decrease in size in the order S 2 > Cl > K + > Ca 2+. Practice Exercise 1 Arrange the following atoms and ions in order of increasing ionic radius: Br, Rb +, Se 2, Sr 2+, Te 2. (a) Sr 2+ < Rb + < Br < Se 2 < Te 2 (b) Br < Sr 2+ < Se 2 < Te 2 < Rb + (c) Rb + < Sr 2+ < Se 2 < Te 2 < Br (d) Rb + < Br < Sr 2+ < Se 2 < Te 2 (e) Sr 2+ < Rb + < Br < Te 2 < Se 2 Practice Exercise 2 In the isoelectronic series Ca 2+, Cs +, Y 3+, which ion is largest? Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc. 21

23 In Section 7.2 we said that Z eff generally increases when you move down a column of the periodic table, whereas in Chapter 6 we saw that the size of an orbital increases as the principal quantum number n increases. With respect to atomic radii, do these trends work together or against each other? Which effect is larger? a. Trends work with each other but trend in orbital size is more important. b. Trends work with each other but trend in effective nuclear charge is more important. c. Trends work against each other but trend in orbital size is more important. d. Trends work against each other but trend in effective nuclear charge is more important. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc. 22

24 7.4 Ionization Energy (I) The ionization energy is the minimum energy required to remove an electron from the ground state of a gaseous atom or ion. The first ionization energy (I 1 ) is that energy required to remove the first electron. The second ionization energy (I 2 ) is that energy required to remove the second electron, etc. Na(g) --> Na + (g) + e - Na + (g) --> Na 2+ (g) + e - I 1 =496 kj/mol I 2 =4562 kj/mol Note: the higher the ionization energy, the more difficult it is to remove an electron! 23 of the

25 Ionization Energy It requires more energy to remove each successive electron. When all valence electrons have been removed, it takes a great deal more energy to remove the next electron. 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 3s 2 1s 2 2s 2 2p 6 3s 2 3p 1 1s 2 2s 2 2p 6 3s 2 3p 2 Highest second IE? lowest second IE? 24 of the

26 Trends in First Ionization Energy (I 1 ) 1) I 1 generally increases across a period. 2) I 1 generally decreases down a group. 3) The s- and p-block elements show a larger range of values for I 1. (The d-block generally increases slowly across the period; the f-block elements show only small variations.) 25 of the

27 Factors that Influence Ionization Energy Smaller atoms have higher I values. I values depend on effective nuclear charge and average distance of the electron from the nucleus. 26 of the

28 Sample Exercise 7.5 Trends in Ionization Energy Three elements are indicated in the periodic table in the margin. Which one has the largest second ionization energy? Solution Analyze and Plan The locations of the elements in the periodic table allow us to predict the electron configurations. The greatest ionization energies involve removal of core electrons. Thus, we should look first for an element with only one electron in the outermost occupied shell. Solve The red box represents Na, which has one valence electron. The second ionization energy of this element is associated, therefore, with the removal of a core electron. The other elements indicated, S (green) and Ca (blue), have two or more valence electrons. Thus, Na should have the largest second ionization energy. Check A chemistry handbook gives these I 2 values: Ca, 1145 kj mol; S, 2252 kj mol; Na, 4562 kj mol. of 27 the

29 Irregularities in the General Trend The trend is not followed when the added valence electron in the next element enters a new sublevel (higher energy sublevel); is the first electron to pair in one orbital of the sublevel (electron repulsions lower energy). Why is it easier to remove a 2p electron from an oxygen atom than from a nitrogen atom? 28 of the

30 Sample Exercise 7.6 Trends in Ionization Energy Referring to the periodic table, arrange the atoms Ne, Na, P, Ar, K in order of increasing first ionization energy. Solution Analyze and Plan We are given the chemical symbols for five elements. To rank them according to increasing first ionization energy, we need to locate each element in the periodic table. We can then use their relative positions and the trends in first ionization energies to predict their order. Solve Ionization energy increases as we move left to right across a period and decreases as we move down a group. Because Na, P, and Ar are in the same period, we expect I 1 to vary in the order Na < P < Ar. Because Ne is above Ar in group 8A, we expect Ar < Ne. Similarly, K is directly below Na in group 1A, and so we expect K < Na. From these observations, we conclude that the ionization energies follow the order K < Na < P < Ar < Ne 29 of the

31 Electron Configurations of Ions Cations: The electrons are lost from the highest energy level (n value). Li + is 1s 2 (losing a 2s electron). Fe 2+ is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 (losing two 4s electrons). Anions: The electron configurations are filled to ns 2 np 6 ; e.g., F is 1s 2 2s 2 2p 6 (gaining one electron in 2p). of 30 the

32 Sample Exercise 7.7 Electron Configurations of Ions!!! Write the electron configurations for (a) Ca 2+, (b) Co 3+, and (c) S 2. Solution Analyze and Plan We are asked to write electron configurations for three ions. To do so, we first write the electron configuration of each parent atom, and then remove or add electrons to form the ions. Electrons are first removed from the orbitals having the highest value of n. They are added to the empty or partially filled orbitals having the lowest value of n. Solve (a) Calcium (atomic number 20) has the electron configuration [Ar]4s 2. To form a 2+ ion, the two outer 4s electrons must be removed, giving an ion that is isoelectronic with Ar: Ca 2+ : [Ar] (b) Cobalt (atomic number 27) has the electron configuration [Ar]4s 2 3d 7. To form a 3+ ion, three electrons must be removed. As discussed in the text, the 4s electrons are removed before the 3d electrons. Consequently, we remove the two 4s electrons and one of the 3d electrons, and the electron configuration for Co 3+ is Co 3+ : [Ar]3d 6 of 31 the

33 Sample Exercise 7.7 Electron Configurations of Ions!!!! Continued (c) Sulfur (atomic number 16) has the electron configuration [Ne]3s 2 3p 4. To form a 2 ion, two electrons must be added. There is room for two additional electrons in the 3p orbitals. Thus, the S 2 electron configuration is S 2 : [Ne]3s 2 3p 6 = [Ar] Comment Remember that many of the common ions of the s- and p-block elements, such as Ca 2+ and S 2, have the same number of electrons as the closest noble gas. (Section 2.7) Practice Exercise 1 The ground electron configuration of a Tc atom is [Kr]5s 2 4d 5. What is the electron configuration of a Tc 3+ ion? (a) [Kr]4d 4 (b) [Kr]5s 2 4d 2 (c) [Kr]5s 1 4d 3 (d) [Kr]5s 2 4d 8 (e) [Kr]4d 10 Practice Exercise 2 Write the electron configurations for (a) Ga 3+, (b) Cr 3+, and (c) Br. 32 of the

34 7.5 Electron Affinity Electron affinity is the energy change accompanying the addition of an electron to a gaseous atom: Cl(g) + e Cl (g) ΔE= 349 kj/mol It is typically exothermic, so, for most elements, it is negative!!! of 33 the

35 General Trend in Electron Affinity Not much change in a group. Across a period, it generally increases. Three notable exceptions include the following: 1) Group 2A: s sublevel is full! 2) Group 5A: p sublevel is halffull! 3) Group 8A: p sublevel is full! Note: the electron affinity for many of these elements is positive (X is unstable). of 34 the

36 Why are the electron affinities of the Group 4A elements more negative than those of the Group 5A elements? a. Group 5A has a greater affinity for electrons than Group 4A. b. Group 4A elements would prefer to lose electrons to have a p 2 configuration. c. Electrons added to Group 4A result in a half-filled p 3 configuration. of 35 the

37 7.6 Metals, Nonmetals, and Metalloids of 36 the

38 Metals Differ from Nonmetals Metals tend to form cations. Nonmetals tend to form anions. of 37 the

39 Metals Most of the elements in nature are metals. of metals: Shiny luster Conduct heat and electricity Malleable and ductile Solids at room temperature (except mercury) Low ionization energies/form cations easily of 38 the

40 Metal Chemistry Compounds formed between metals and nonmetals tend to be ionic. Metal oxides tend to be basic. 39 of the

41 Would you expect NiO to dissolve in an aqueous solution of NaNO 3? 40 of the

42 Nonmetals Nonmetals are found on the right hand side of the periodic table. of nonmetals include the following: Solid, liquid, or gas (depends on element) Solids are dull, brittle, poor conductors Large negative electronegativity/form anions readily 41 of the

43 Nonmetal Chemistry Substances containing only nonmetals are molecular compounds. Most nonmetal oxides are acidic. 42 of the

44 Sample Exercise 7.9 Reactions of Nonmetal Oxides Write a balanced chemical equation for the reaction of solid selenium dioxide, SeO 2 (s), with (a) water, (b) aqueous sodium hydroxide. Solution Analyze and Plan We note that selenium is a nonmetal. We therefore need to write chemical equations for the reaction of a nonmetal oxide with water and with a base, NaOH. Nonmetal oxides are acidic, reacting with water to form an acid and with bases to form a salt and water. Solve (a) The reaction between selenium dioxide and water is like that between carbon dioxide and water (Equation 7.13): SeO 2 (s) + H 2 O(l) H 2 SeO 3 (aq) (It does not matter that SeO 2 is a solid and CO 2 is a gas under ambient conditions; the point is that both are water-soluble nonmetal oxides.) (b) The reaction with sodium hydroxide is like that in Equation 7.15: SeO 2 (s) + 2 NaOH(aq) Na 2 SeO 3 (aq) + H 2 O(l) 43 of the

45 Recap of a Comparison of the of Metals and Nonmetals of 44 the

46 Metalloids Metalloids have some characteristics of metals and some of nonmetals. Several metalloids are electrical semiconductors (computer chips). of 45 the

47 Group Trends in a group have similar properties. Trends also exist within groups. Groups Compared: Group 1A: The Alkali Metals Group 2A: The Alkaline Earth Metals Group 6A: The Oxygen Group Group 7A: The Halogens Group 8A: The Noble Gases 7.7 Metals 7.8 Nonmetals 46 of the

48 Alkali Metals Alkali metals are soft, metallic solids. They are found only in compounds in nature, not in their elemental forms. Typical metallic properties (luster, conductivity) are seen in them. of 47 the

49 Alkali Metal They have low densities and melting points. They also have low ionization energies. 48 of the

50 Alkali Metal Chemistry Their reactions with water are famously exothermic. Because they also have low ionization energies Would you expect rubidium metal to be more or less 49 reactive with water than potassium metal? of the

51 Differences in Alkali Metal Chemistry Lithium reacts with oxygen to make an oxide: 4 Li + O 2 2 Li 2 O Sodium reacts with oxygen to form a peroxide: 2 Na + O 2 Na 2 O 2 K, Rb, and Cs also form superoxides: M + O 2 MO 2 50 of the

52 Sample Exercise 7.10 Reactions of an Alkali Metal Write a balanced equation for the reaction of cesium metal with (a) Cl 2 (g), (b) H 2 O(l), (c) H 2 (g). Solution Analyze and Plan Because cesium is an alkali metal, we expect its chemistry to be dominated by oxidation of the metal to Cs + ions. Further, we recognize that Cs is far down the periodic table, which means it is among the most active of all metals and probably reacts with all three substances. Solve The reaction between Cs and Cl 2 is a simple combination reaction between a metal and a nonmetal, forming the ionic compound CsCl: 2 Cs(s) + Cl 2 (g) 2 CsCl(s) From Equations 7.18 and 7.16, we predict the reactions of cesium with water and hydrogen to proceed as follows: 2 Cs(s) + 2 H 2 O(l) 2 CsOH(aq) + H 2 (g) 2 Cs(s) + H 2 (g) 2 CsH(s) All three reactions are redox reactions where cesium forms a Cs + ion. The Cl, OH, and H are all 1 ions, which means the products have 1:1 stoichiometry with Cs of the

53 Flame Tests Qualitative tests for alkali metals include their characteristic colors in flames. of 52 the

54 Alkaline Earth Metals Compare to Alkali Metals Alkaline earth metals have higher densities and melting points than alkali metals. Their ionization energies are low, but not as low as those of alkali metals. 53 of the

55 Alkaline Earth Metals Beryllium does not react with water, and magnesium reacts only with steam, but the other alkaline earth metals react readily with water. Reactivity tends to increase as you go down the group. of the Because they also have low ionization energies 54

56 7.8 Trends for Selected Nonmetals (H, 6A 8A) Group 6A Increasing in Metallic Character down the Group Oxygen, sulfur, and selenium are nonmetals. Tellurium is a metalloid. The radioactive polonium is a metal. of 55 the

57 Group 7A Halogens The halogens are typical nonmetals. They have highly negative electron affinities, so they exist as anions in nature. They react directly with metals to form metal halides. of 56 the

58 Why are more molecules of I 2 seen in the molecular view relative to the number of Cl 2 molecules? Cl 2 is a gas at room temperature, whereas I 2 is a solid. 57 of the

59 Group 8A Noble Gases The noble gases have very large ionization energies. Their electron affinities are positive (can t form stable anions). Therefore, they are relatively unreactive. They are found as monatomic gases. of 58 the

60 of 59 the

61 The element bismuth (Bi, atomic number 83) is the heaviest member of group 5A. A salt of the element, bismuth subsalicylate, is the active ingredient in Pepto-Bismol, an over-the-counter medication for gastric distress. (a) Based on values presented in Figure 7.7 and Tables 7.5 and 7.6, what might you expect for the bonding atomic radius of bismuth? Bismuth is directly below antimony, Sb, in group 5A. Based on the observation that atomic radii increase as we go down a column, we would expect the radius of Bi to be greater than that of Sb, which is 1.39 Å. We also know that atomic radii generally decrease as we proceed from left to right in a period. Tables 7.5 and 7.6 each give an element in the same period, namely Ba and Po. We would therefore expect that the radius of Bi is smaller than that of Ba (2.15 Å) and larger than that of Po (1.40 Å). We also see that in other periods, the difference in radius between the neighboring group 5A and group 6A elements is relatively small. We might therefore expect that the radius of Bi is slightly larger than that of Po much closer to the radius of Po than to the radius of Ba. The tabulated value for the atomic radius on Bi is 1.48 Å, in accord with our expectations. 60 Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

62 (b) What accounts for the general increase in atomic radius going down the group 5A elements? The general increase in radius with increasing atomic number in the group 5A elements occurs because additional shells of electrons are being added, with corresponding increases in nuclear charge. The core electrons in each case largely screen the outermost electrons from the nucleus, so the effective nuclear charge does not vary greatly as we go to higher atomic numbers. However, the principal quantum number, n, of the outermost electrons steadily increases, with a corresponding increase in orbital radius. 61 Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

63 (c) Another major use of bismuth has been as an ingredient in lowmelting metal alloys, such as those used in fire sprinkler systems and in typesetting. The element itself is a brittle white crystalline solid. How do these characteristics fit with the fact that bismuth is in the same periodic group with such nonmetallic elements as nitrogen and phosphorus? The contrast between the properties of bismuth and those of nitrogen and phosphorus illustrates the general rule that there is a trend toward increased metallic character as we move down in a given group. Bismuth, in fact, is a metal. The increased metallic character occurs because the outermost electrons are more readily lost in bonding, a trend that is consistent with its lower ionization energy. 62 Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

64 (d) Bi 2 O 3 is a basic oxide. Write a balanced chemical equation for its reaction with dilute nitric acid. If 6.77 g of Bi 2 O 3 is dissolved in dilute acidic solution to make L of solution, what is the molarity of the solution of Bi 3+ ion? (d) Following the procedures described in Section 4.2 for writing molecular and net ionic equations, we have the following: Molecular equation: Bi 2 O 3 (s) + 6 HNO 3 (aq) 2 Bi(NO 3 )3(aq) + 3 H 2 O(l) Net ionic equation: Bi 2 O 3 (s) + 6 H + (aq) 2 Bi 3 + (aq) + 3 H 2 O(l) In the net ionic equation, nitric acid is a strong acid and Bi(NO 3 ) 3 is a soluble salt, so we need to show only the reaction of the solid with the hydrogen ion forming the Bi 3+ (aq) ion and water. To calculate the concentration of the solution, we proceed as follows (Section 4.5): 63 Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

65 (e) 209 Bi is the heaviest stable isotope of any element. How many protons and neutrons are present in this nucleus? (e) Recall that the atomic number of any element is the number of protons and electrons in a neutral atom of the element. (Section 2.3) Bismuth is element 83; there are therefore 83 protons in the nucleus. Because the atomic mass number is 209, there are = 126 neutrons in the nucleus. 64 Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

66 (f) The density of Bi at 25 is g cm 3. How many Bi atoms are present in a cube of the element that is 5.00 cm on each edge? How many moles of the element are present? (f) We can use the density and the atomic weight to determine the number of moles of Bi, and then use Avogadro s number to convert the result to the number of atoms. (Sections 1.4 and 3.4) The volume of the cube is 5.00) 3 cm 3 = 125 cm 3 (. Then we have 65 Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.