Solutions for Chapter 4 End-of-Chapter Problems Assignment Part 2

Problem 4.41. We use the de Broglie wavelength equation to calculate the wavelength of an electron of mass 9.1 10-31 kg traveling at 1.5 106 m·s–1: –34 6.626 10 J s = hm u = = 4.9 9.1 10 –31 kg 1.5 10 6 m s–1 10–10 m ≈ 490 pm Atoms have sizes that are measured in picometers (pm) so this wavelength is significant compared to the size of an atom. Problem 4.42. In the macroscopic world, particles are massive enough that their wave characteristics (wavelengths), although presumably present, are undetectable and hence do not significantly affect their properties. At the atomic level, particles have very small masses and both wave and particle properties can be observed (depending upon the observations that are made) and are important in understanding the interactions among matter and energy. Problem 4.43. (a) The de Broglie wavelength of an alpha particle, mass 6.6 of 1.5 107 m·s-1 is: –34 6.626 10 J s = hm u = = 6.7 6.6 10 –27 kg 1.5 107 m s–1 10–27 kg, traveling at a velocity 10–15 m = 6.7 fm (femtometer) (b) This very short wavelength is in the electromagnetic spectral region associated with highly energetic gamma rays. Problem 4.44. The standing waves on a guitar string are responsible for the sound produced by the guitar. The waves responsible for the sound producedby other musical instruments are described here. (a) The drumhead on drums is set in motion by the drumstick(s) and two-dimensional standing waves are set up on the head. (b) As air is blown across the mouthpiece by the flutist, standing waves are set up inside the tube. The wavelength of these waves is controlled by the length of the tube that is closed, which the player controls by opening and closing the valves on the instrument. You can get the same effect by blowing across the top of a bottle with a narrow opening, such as a soft-drink bottle. The frequency of the sound you produce can be changed by changing the amount of liquid in the bottle (thus changing the size of the air column you are setting in motion. (c) When a tuning fork is struck, its legs begin to vibrate. The standing waves are set up in the legs, very much like waves on a string, but a bit more complicated because the fork legs are thicker and only one end is fixed. The pitch of the fork depends upon its various dimensions (length, cross section of the legs, etc.) Problem 4.45. The contributions four scientists made to our understanding of atomic and electronic structure are briefly described here: (a) Bohr developed the theory of quantized electron energy levels. (b) Heisenberg developed the uncertainty principle, which states that, at very small (atomic) scale, we cannot accurately and simultaneously measure both the momentum and position of an electron. (c) Einstein showed that the quantized electromagnetic radiation proposed by Planck also quantitatively explained the photoelectric effect. (d) De Broglie developed the hypothesis that moving atomic-level objects behave like waves as well as particles and thus led to the dual wave/particle nature of matter as well as light. Problem 4.46. Tthe Heisenberg uncertainty principle tells us that the more we know about the location of an electron, the less we know about its momentum and vice versa. Problem 4.47. Bohr proposed that the electrons in an atom orbit the nucleus in a well-defined path, so it would be possible to determine precisely and simultaneously both the position of an electron (from the radius of the orbit) and its momentum (from its kinetic energy). However, the Heisenberg uncertainty principle states that we cannot determine precisely and simultaneously both the position and momentum of an electron in an atom. The de Broglie wave model states that the electron in the atom has properties of a wave as well as of a particle. Since a wave extends in space it is not possible to determine exactly where the electron is located at a specific time, which is consistent with Heisenberg uncertainty principle, and overcomes this limitation of the Bohr model. Problem 4.48. We can calculate exactly the position and the momentum of a rolling ball, because its wave properties are unobservable (too massive and too slow) and it behaves as a particle. We cannot calculate the same for a moving electron because it is very light and exhibits observable wave properties that make it impossible to simultaneously and precisely determine its position and momentum. Problem 4.49. The size of a hydrogen atom is a function of the kinetic and potential energy of the electron wave and the nucleus. As the radius of the electron cloud decreases the kinetic energy of the electron increases, raising the energy of the system. At the same time, the potential energy of the system decreases as the distance between the nucleus and the electron decreases, which lowers the energy of the system. These opposite effects result in a minimum energy of the system (electron and proton) at an electron wave radius of about 60 pm, the radius of the hydrogen atom. Problem 4.50. To compare the kinetic energy of particle A (mass 1.56 10-25 kg) and particle B (mass 4.25 10-24 kg), each traveling at a velocity of 3.15 105 m s-1, we need only compare their masses, because they are traveling at the same velocity. Particle B has a greater kinetic energy, m u2 2 , because it has more mass. Problem 4.51. The potential energy of an electron wave becomes lower as it gets closer to the nucleus, that is, as the wave becomes smaller. Problem 4.52. In Figure 4.27, the potential energy is negative because it is proportional to the product of the charge on the electron (–1) and the charge on the nucleus (+1 for hydrogen). These particles with unlike charge attract one another, so the potential energy of the system is lowered by their m u2 interaction. Kinetic energy, 2 , is always positive, because mass is always positive and the velocity of the electron is squared, so, even if the velocity is negative (motion in a negative direction in the coordinate system used to describe the system), the function m·u2 is positive. Problem 4.53. The red circle in the Web Companion, Chapter 4, Section 4.7, page 1, represents the spherical electron wave that surrounds and interacts with the nucleus. The energy of the spherical electron wave is inversely proportional to the square of the radius of the wave. As the electron gets closer to the nucleus (the wave gets smaller) the kinetic energy, represented by the red curve on the graph, rises. If kinetic energy were the only energy in the system, the electron and proton would not come together. Problem 4.54. Electron spin is the property of the electron that makes it behave as though it is a tiny magnet Electron spin is responsible for electrons interacting with magnetic fields. Problem 4.55. In the experiment represented in Figure 4.28(a), the magnetic field does not affect the stream of helium atoms, because the spins of the two electrons in helium are opposite so their interactions with the field cancel each other out. The magnetic field does affect the stream of lithium atoms, Figure 4.28(b), because there are three electrons in the atom two of which are paired with one another; the third electron has no partner (is unpaired) and interacts with the magnetic field. Since half the atoms (on average) have the odd electron spin in one direction and the other half in the other direction, the interaction splits the stream of atoms into two streams, as shown. Problem 4.56. To tell whether atoms of an element have an odd number of electrons (or electrons that are not spin paired with one another), you might pass a stream of atoms of the element through a magnetic field like the one represented in Figure 4.28. Elements whose atoms have a single unpaired spin (an odd number of electrons) would split into two beams as the atoms with spin up and spin down electrons interact differently with the magnetic field. [Streams of atoms of elements whose atoms have more than one unpaired spin would be split into more than two streams by the magnetic field.] Problem 4.57. (a) A nitrogen atom has seven electrons. If three of the electrons have unpaired spins, that means the other four are spin-paired (two pairs of spin-paired electrons). (b) The electrons in a ground state atom are described by the lowest energy orbitals possible, so we might conclude that the energies of spatial orbitals that describe the three unpaired electrons are all the same. If one of the orbitals had a lower energy, then we would expect two electrons with opposite spins to be described by that orbital and to give an atom with only one unpaired spin. Problem 4.58. The Pauli exclusion principle states that only two electrons of opposite spin can occupy exactly the same region of space (orbital) in an atom. For lithium, the two lowest-energy electrons with paired spins occupy the orbital (shell) next to nucleus. The third electron, (which has the same spin as one or the other of the first two electrons), cannot share the same space with them. It has to be described by an orbital at a larger distance from the nucleus, and therefore has a lower ionization energy than the first two electrons. Problem 4.59. The second ionization energy for an atom is always greater than the first ionization energy, because, after the first electron is removed, a positive ion, more protons than electrons, has been formed. To remove an electron from this positive ion requires more energy (because of more electrostatic attraction) than to remove the first electron from a neutral atom. Another way to state this is to say that there is a greater effective nuclear attraction for the remaining electrons when one (or more) has been removed. Problem 4.60. (a) These ions and atom, Na+, Ne, F-, and O2-, all have ten electrons (two in the first shell and eight in the second), Atoms and ions of different elements that have the same number of electrons are called “isoelectronic.” (b) Of the four species in part (a), Na+ has the highest ionization energy. In all four species, the ten electrons have the same distribution about the nucleus, that is, two in the first shell and eight in the second. Thus, Na+, with the most protons, has the highest nuclear charge and holds its ten electrons with greater attraction than the other three ions and atom. Problem 4.61. Ionization energies, atomic radii, and electronegativities all depend on the nuclear charge, which increases across a row of the periodic table. Thus, attraction for electrons are more strongly attracted to the nucleus across a row of the table. (a) Ionization energy increases across a period, because the attraction for the electrons increases across a period and it becomes harder to remove electrons. (b) Atomic radii decrease across a row of the periodic table, because the attraction for the electrons increases across a period and they are held closer to the nucleus. (c) Electronegativity is a measure of the relative attraction of an atom for the shared electrons in a covalent bond with another atom. Electronegativity increases across a row because the nuclear charge increases and the atoms attract electrons more strongly, including those they share with other atoms. Problem 4.62. Ionization energies increase as you go from left to right in a row (period) on the periodic table. Across a period electrons are added to the same shell and are all about the same distance (on average) from the nucleus. Since the nuclear charge increases across a period, the electrons are all held more strongly and are closer to the nucleus (the shell gets smaller) across a period. Thus, the energy it takes to remove an electron (the ionization energy) increases across a period. Problem 4.63. Ionization energies decrease as you go from top to bottom in a group (family) on the periodic table. Down a family electrons in successive periods are added to shells that are successively larger and (on average) a greater distance (on average) from the nucleus. Although the nuclear charge increases across down a family, the “effective” nuclear charge remains about the same because the core charge on the elemental atoms in a family is the same. Since the electrons are held by about the same core charge but are a greater distance from the nucleus, the energy it takes to remove an electron (the ionization energy) decreases down a period. Problem 4.64. The number of valence electrons for the elements in the first three periods is given by the Roman numeral designation for its group (family). (a) Oxygen is in group VI and has six valence electrons. Oxygen is atomic number 8, so has a total of eight electrons, two of which are core electrons in the first shell. (b) Sodium is in group I and has one valence electron. Sodium is atomic number 11, so has a total of 11 electrons, 10 of which are core electrons in the first and second shells. (c) Chlorine is in group VII and has seven valence electrons. Chlorine is atomic number 17, so has a total of 17 electrons, 10 of which are core electrons in the first and second shells. (d) Argon is in group VIII and has eight valence electrons. Argon is atomic number 18, so has a total of 18 electrons, 10 of which are core electrons in the first and second shells. Problem 4.65. This table of the empirical formulas of the simplest hydrides of the second period elements is the background for this problem. element Li Be B C N O F Ne hydride LiH BeH2 BH3 CH4 NH3 OH2 FH none (a) As we go across the second period, we see that the number of hydrogen atoms in the empirical formulas of the simplest hydrides increases from one to four and then decreases to one (or none, if we count the noble gas, neon, which has not yet been found to form compounds). The ability to form bonds with hydrogen seems to be a maximum at Group 4, the carbon family, and to decrease as we go to higher or lower atomic numbers in the period. The contrasts between LiH and FH (or HF, as it is usually written) help to understand a bit about this series of compounds. LiH is an ionic salt of the Li+ cation and H– anion. HF is a weak acid that can react with water to form the H3O+ cation and F– anion. At the left end of the period, where electronegativity and ionization energy are low, the Li atom donates a good deal of its valence electron density to the H atom when they bond. At the right end of the period, where electronegativity and ionization energy are high, the F atom accepts a good deal of valence electron density from the H atom, making it possible for the proton to transfer to water to form H3O+. Thus, the H atom is an electron acceptor or electron donor, depending on the atom to which it bonds, so the “hydrides” are not strictly comparable with one another. They are quite polar (in opposite directions) at the ends of the period. In the middle of the period, CH4 is a hydrocarbon, which we have seen in previous chapters is nonpolar. Thus, the properties and formulas of the second period hydrides both follow trends that change toward the middle of the period. Note that the number of H atoms bonded to the atoms on the left of the period is equal to the number of valence electrons in the atom. The number of H atoms bonded to the atoms on the right of the period is equal to the number of electrons that are lacking to give the atom a completed shell of eight electrons. In the middle, carbon atoms match both these patterns. (b) The H atom has one electron in its valence shell. This is the same as the Li atom. Both atoms bond with one H atom to form, respectively, H2 and LiH. Thus, the H and Li atoms follow the same bonding pattern as we saw in the table and part (a). Since the electronegativities of the H atom and Li atom are more than one unit different, we do not expect the chemistry of the two hydrides to be similar. Indeed, H2 is nonpolar and a gas at room temperature, while LiH is an ionic solid at room temperature. We can also argue that an H atom lacks one electron of having a filled electron shell, which is similar to the F atom. Both atoms bond with one H atom to form, respectively, H2 and FH (HF). Thus, the H and F atoms follow the same bonding pattern as we saw in the table and part (a). Since the electronegativities of the H atom and F atom are more than one unit different, we do not expect the chemistry of the two hydrides to be similar. Indeed, H2 is a nonpolar gas at room temperature, while HF is a polar gas that dissolves in water to form an acidic solution [part (a)]. Another way to look at the H atom is to consider that is has a half-filled electron shell, one electron in a shell that can accommodate two. This is quite like carbon with four electrons in a shell that can accommodate eight. Since the electronegativities of C and H are quite close, they form a nonpolar gaseous compound, CH4, that is more similar to H2 than either LiH or HF. These differences and similarities are the reason we have chosen to locate the element hydrogen in the group IV family of the periodic table, instead of the more conventional location in group I (or occasionally in group VII). (c) The third period elements have electron shell structures similar to those in the second period (Figure 4.31), except that the electrons are not as tightly held (lower ionization energies) and the electronegativities are somewhat lower for atoms in the same families. Based on the similarity in electronic structure, we would predict that the empirical formulas of the simplest hydrides of Al, Si, and P would be AlH3, SiH4, and PH3. We would also predict that SiH4 would be relatively nonpolar. These predictions are borne out by experimental observations. Problem 4.66. (a) The plot of first ionization energies from the interactive periodic table in the Web Companion is shown here. The saw tooth shape with peaks at noble gases and minima at the alkali metals are like the plots in Figures 4.4 and 4.30. The smaller features, for example, the two, three, three steps between the minima and the peaks are evident here as well. By extending the plot beyond element 20, we see other features, the 10-element plateaus evident in periods four (19-36) and five (37-54) and following a 14-element lower plateau in period six (55-86). At the end of the plateau, there is a drop in ionization energy at the first of the last six elements in the period (which also show the smaller stair step feature). The 14-element low plateau is also evident in period seven, but the rest of the period is incomplete. (b) The plot of second-ionization energies is shown here. The saw tooth features and the smaller intermediate features noted above in part (a) are also evident here, except, as we found in the solution for Problem 4.5, moved one atomic number (element) higher. The shell structure of a monopositive elemental ion appears to be the same and behave the same as that of the neutral atom one atomic number lower (which has the same number of electrons). Since the nuclear charge of the ion is one higher, the actual ionization energies are higher. [Note the difference in the energy scales here and in part (a).] (c) The plot of third ionization energies is shown here. Although the smaller features are becoming blurred, they are still present and we see that dipositive elemental ions have ionization properties similar to the monopositive ions and neutral atoms with the same number of electrons. Since the nuclear charge on the dipositive ions are higher than on the corresponding monopositive ions or neutral atoms, the ionization energies are higher still, than those in parts (a) and (b). Problem 4.67. Was your classmate able to solve your periodicity problem based on the interactive periodic table in the Web Companion? Explain why or why not. Problem 4.68. [NOTE: This problem is probably misplaced and should go in Section 4.11, Wave Equations and Atomic Orbitals, or in the section General Problems. It is helpful to discuss its structure and problems with the s, p, d, f nomenclature for atomic orbitals.] In this three-dimensional periodic table model of eight stacked, round wooden discs with the symbols for the elements on the discs, each of the second tier discs represents parts of two periods or rows in the electron shell model. For example, the top disc shown here has the last six elements from period two on the outer ring and the first two elements of period three on the inner ring. Similarly, the lower disc represents the rest of period three and beginning of period four. [The two small discs at the top of the stack contain the two elements of the first period (on the top disc) and the first two elements of the second period (on the lower disc).] The problem with this representation is that periods are not kept together and can be confusing, if you do not realize that different periods are represented on the same disc. What is required is to read from the outside toward the center to get the ordering of shells and subshells correct on each disc. In terms of the structure of the shell and subshell structure of the periodic table, the inner ring is the two-electron subshell in each shell (s-block elements), the second ring is the six-electron subshell (p-block elements), the third ring is the 10 electron subshell (d-block elements), and the fourth ring is the 14 electron subshell (f-block elements). (See Section 4.11 and the electronic configurations there for an explanation of this nomenclature.) [NOTE: The "Periodic Roundtable" is an "invention" of Gary Katz (PhD biochemist), P.O. Box 156, Cabot, VT 05647. He sells it in a hardwood version for about $50. It has been reviewed in the Journal of Chemical Education, 2000, 77(2), 164, by Glen Rogers, from Allegheny College.] Problem 4.69. [NOTE: This problem is probably misplaced and should go in Section 4.11, Wave Equations and Atomic Orbitals, or in the section General Problems. It is helpful to discuss its structure and problems with the s, p, d, f nomenclature for atomic orbitals.] (a) The central cylinder of the threedimensional periodic table shown here (and with other views in the chapter opening illustration) corresponds to the elements in the two far left and six far right groups of the conventional periodic table. These elements are often called the representative elements. Often, the two columns of elements at the right are called the s-block elements and the six on the right, the pblock elements. (See Section 4.11 and the electronic configurations there for an explanation of this nomenclature.) The broader loop is attached to the central cylinder between the second and third columns of the representative elements, beginning at period four, that is, between the alkaline earth metals and the boron-family elements. This is where the interrupting block of elements (10 per period) is found in the conventional periodic table. This block of elements is often called the transition metals or the d-block transition metals. The thinner loop is attached to the broader loop between the first and second transition metal elements, beginning at period six. This loop of 14 elements is found at the bottom of the conventional table and often linked by thin lines to the place where this block of elements interrupts the table in the transition series. Often, each of the periods in this block of elements is named separately as the lanthanide and actinide series, as a pointer to where they come in the sequence of elements. Together this block of elements is also known as the f-block transition metals. (b) The representative elements are shown in the “circular” section of the spiral periodic table and on the cylindrical section of the three-dimensional table. The d-block transition metals are shown in the “squarish” extension toward the lower right of the spiral table and on the broader loop of the three-dimensional table. The f-block transition metals are shown in the “rectangular” extension toward the upper right of the spiral table and on the thinner loop of the threedimensional table. Note that both the spiral and three-dimensional tables represent the elements in a continuous series. In the conventional table (even the long form shown in the Web Companion interactive periodic table under “Electron Configuration”), there is a “break” at the end of each period which signals the completion of one of our shells, but does not signal how the elements simply continue in the same pattern in the next period. Problem 4.70. (a) The Schrödinger wave equation relates the wave properties of the electron to its kinetic energy. FALSE. The Schrödinger wave equation relates the wave properties of the electron to its total energy, which is a sum of kinetic and potential energy. (b) The wave function ( ) is a mathematical description of an allowed energy state (orbital) for an electron. TRUE. (c) The wave function ( ) is proportional to the probability of finding the electron in a given location near the nucleus. FALSE. The square of wave function ( 2) is proportional to probability of finding the electron in given location around the nucleus . (d) Orbitals with the same energy are called degenerate. TRUE. (e) Orbitals can be thought of as a boxes into which electrons can be placed. FALSE. Orbitals describe calculated energy states of an electron and its corresponding spatial distribution. Orbitals do not exist (except as mathematical constructs), if there is no electron to be described. (f) Each calculated energy level corresponds to a spatial probability distribution for the electron. TRUE. Keep in mind that each of the energy levels represented by a horizontal line on an energy diagram for an atomic or molecular system describes a different spatial probability distribution for the electron wave. For a degenerate set of orbitals, each of the individual energy levels represents a different orbital shape and/or orientation, as in Figure 4.36 for the n = 2 orbitals. Problem 4.71. (a) This electron configuration for Al, 1s2 2s2 2p43s23p3, has the correct number of electrons, 13, but the lower energy 2p orbital is not completely filled, as it should be for the lowest energy configuration. The ground-state configuration for Al is 1s2 2s2 2p63s23p1. (b) This electron configuration for P, 1s22s22p63s23p2, represents only 14 electrons, but P has 15 electrons. The ground-state configuration for P is 1s22s22p63s23p3. (c) This electron configuration for B, 1s2 2s0 2p3, has the correct number of electrons, 5, but the lower energy 2s orbital is empty, so this is not the lowest energy configuration. The groundstate configuration for B is 1s22s22p1. Problem 4.72. The number of valence electrons located in an s orbital in each of these elements is (remember that the number of electrons in any s orbital is limited to two): (a) potassium 1 (b) fluorine 2 (c) magnesium 2 (d) boron 2 Problem 4.73. The number of valence electrons located in a p orbital in each of these elements is (remember that the number of electrons in any set of three degenerate p orbitals is limited to six): (a) phosphorus 3 (b) aluminum 1 (c) lithium 0 (d) bromine 5 (e) sulfur 4 Problem 4.74. Consider atoms A, B, and C with these electron configurations: C = 1s22s22p63s23p6 A = 1s22s22p63s2 B = 1s22s22p63s23p4 Atom A will have the largest third ionization energy. In all three atoms, the first two electrons will come from the third shell and the first and second ionization energies will be in the order: C > B > A, because ionization energies increase across a period. However, the third electron in A must come from the second electron shell, which is closer to the nucleus than the third, so its electrons are much more tightly held. Thus, a good deal more energy will be required to remove the third electron from A than from either B or C, in which the third electron still comes from the third shell. Problem 4.75. Electron energy diagrams for the sulfur atom (16 electrons) and the sulfide ion (18 electrons), with Figure 4.37 as a template and Worked Example 4.60 as a guide, are: sulfur atom sulfide ion Problem 4.76. Electron energy diagrams for the lithium (3 electrons) and sodium (11 electrons) atoms, with Figure 4.37 as a template and Worked Example 4.60 as a guide, are: lithium atom sodium atom Problem 4.77. (a) The commonality of the series of atoms from H through Ar with the series of monocations He+ through K+ and the series of dications Li2+ through Ca2+ is that each series has the same pattern of electron configurations. That is, the electron configuration for H, He+, and Li2+ is 1s1, for He, Li+, and Be2+, it is 1s2, and so on up through Ar, K+, and Ca2+ with an electron configuration 1s22s22p63s23p6. (b) We found that the patterns of first, second, and third ionization energies of elemental gaseous atoms as a function of atomic number (solutions to Problems 4.5 and 4.66) are very similar, except displaced to one higher atomic number as we go from first to second to third ionization energy. If the electron configurations of the atom and ions losing electrons are the same, then it makes sense that the patterns of next electron loss should also be the same. Problem 4.78. (a) Figure 4.37 showed the relative energy levels only up through the 4s orbital energy. The energy level scheme in the Web Companion interactive periodic table for “Electron Configuration” (reproduced here) carries the energy levels up through 7p. Energy increases from the bottom to the top of the scheme. The d (5-fold degenerate) and f (7-fold degenerate) orbital energies (see Consider This 4.59) are included and can help you see how the periodic table is built up with its s, p, d, and f blocks. (b) The electron configuration for Li that you get from the Web Companion (shown above) is exactly what we wrote in Worked Example 4.60, with two 1s spin-paired electrons and one 2s electron. (c) Our results in Consider This 4.62(a) give electron configurations for the elements Na through Ar that are exactly analogous to those for Li through Ne in Table 4.4 with the electron energies corresponding to 3s and 3p orbitals instead of 2s and 2p. These are also the configurations you get from the Web Companion, as shown here for the Na and Ar configurations. Na Ar Problem 4.79. (a) As you use the Web Companion interactive periodic table to examine the electron configurations of the first series of transition metals, Sc through Zn, you find that the pattern is not quite regular. The electron configurations for V, Cr, and Mn exemplify this irregularity: Given the patterns for Sc, Ti, and V, our expectation would be that the electron configuration for Cr would be [Ar]184s23d4, instead of what is observed, [Ar]184s13d5. If the 4s energy levels were much lower in the energy than the 3d, we would not expect this sort of irregularity. However, if the levels are rather close in energy, then the Cr atom might be lower in energy if the six highest energy electrons are all described by different orbitals. This arrangement spreads the electrons out a bit more so that electron-electron repulsion is a little less. The evidence suggests that the 4s and 3d electrons are of about the same energy. (b) Note, as you found for simpler atoms in Consider This 4.62(c) and (d), that the electrons in the singly-occupied orbitals in these transition metals, all have the same spin. This arrangement of spins also serves to keep the electrons a bit farther apart on the average, further helps to reduce electron-electron repulsion, and hence helps stabilize the atom. It appears that, when electrons in an atom are in singly-occupied orbitals, the lowest energy state is when all the unpaired electrons have the same spin. This observation is known as Hund’s rule. (c) The electron configurations of the elements from Y through Cd do not follow exactly the same pattern as those from Sc through Zn. It appears that the 5s and 4d energy levels must be even closer together than the 4s and 3d levels, since electrons in this series often go preferentially to the 4d rather than 5s levels, as in Nb, Ru, Rh, and Pd. Problem 4.80. (a) The inset in Figure 4.27 shows that E is a minimum at R = 2. (b) Using the value of R you from part (a) in equations (4.29) and (4.30) and assuming that the proportionality constant is the same in both cases, H, we find these potential and kinetic energies: H H ; when E is a minimum, PE = PE = R 2 H H KE = 2 ; when E is a minimum, KE = 4 R We see that the relationship of the kinetic energy to the potential energy of the proton-electron wave system is: KE = – 12 PE. (c) The total energy at its minimum for the proton-electron wave system as a function of H from part (b) is: H H H – = 4 2 4 The ionization energy for the hydrogen atom, IE = 1312 kJ·mol–1, is the energy required to overcome the overall binding energy, E, of the proton-electron system. Using this information, we can find the value of the proportionality constant, H: H IE = 1312 kJ·mol–1 = –E = ; H = 5248 kJ·mol–1 4 E = KE + PE = Problem 4.81. (a) For an electron (charge –1) attracted by a nucleus of charge +Z, the potential energy is: Z Z –1 PE = r r (b) By analogy with the transformation of equation (4.16) to equation (4.30), we can write the potential energy for a spherical electron wave attracted to the nucleus in part (a), using the proportionality constant, H, from Problem 4.80(b): ZH PE = R (c) If the kinetic energy of the electron wave is independent of the nuclear charge, the total energy of a one-electron ion with a nuclear charge +Z, as a function of R, H, and Z. is: ZH H E = KE + PE = 2 – R R (d) Three methods for finding the value of R that makes the total energy in part (c) a minimum are: • Method (i), using calculus, gives this result: dE 2H ZH 2H ZHR =0= 3 + 2 = 3 dR R R R 2 2H = ZHR R= Z • Method (ii), for Z = 1, 2, 3, and 5, gives R = 2 (from Figure 4.27), 1, 0.67, and 0.40 from the plot below. (Substituting these values of Z into the result from method (i) gives the same R values we get from the plot, so the methods produce the same results.) -0.500 -1.000 -1.500 -2.000 E -2.500 -3.000 -3.500 -4.000 -4.500 Z=2 Z=3 -5.000 Z=5 -5.500 -6.000 2.000 1.800 1.600 1.400 1.200 1.000 0.800 0.600 0.400 0.200 0.000 -6.500 R • Method (iii), assuming KE = –PE/2 gives: ZH H 2 = 2R R Solving for R gives the same result as in method (i). (e) Substitute the result from part (d) into the expression for E from part (c) to get the minimum for the total energy of a one-electron ion as a function of H and Z: E= ZH H Z2 H Z2 H Z2 H – = – = – 2 Z 4 2 4 2 Z2 For He+, Li2+, and B4+, the Z values are 2, 3, and 5, so the total energies are H, (9/4)H, and (25/4)H. From Problem 4.80(c), we know that H = 5248 kJ·mol–1, so the ionization energies (– E) are 5248, 11,808, and 32,800 kJ·mol–1, respectively. As you see in Table 4.2, these are exactly the ionization energies for the final electron in these three atoms. You can calculate the last ionization energy for any atom. You can check your results for the final ionization energy for any of the atoms through Ne in the table. (The last few values in the table are only given to three significant figures.) Problem 4.82. [NOTE: The sources for this response are the Department of Physics and Astronomy, Arizona State University, Tempe, AZ (acept.la.asu.edu/PiN/rdg/elmicr/versus.shtml) and Allen R. Sampson, Advanced Research Systems. The on-line materials have been extensively edited to include only what is necessary to see how the information in Chapter 4 is used to respond to Problem 4.82.] The attractive feature of optical microscopy is that it is so easy: samples can be analyzed in air or water, the images are in natural color with magnifications of up to one hundred to one thousand times, and modern semiconductor electronics with charge-coupled devices allow image processing. The optical microscope should dominate the field of microscopy, but it doesn't. The scanning electron microscope (SEM) is the microscope of choice because of its depth of focus and resolving capability. Examination of this figure shows a striking contrast between an optical and SEM image of a radiolarian (tiny marine creature) at the same magnification. [Figure taken from J.I. Goldstein et al., eds., Scanning Electron Microscopy and X-Ray Microanalysis, (Plenum Press, New York, 1980).] optical SEM In the optical micrograph taken at high resolution only a section of the radiolarian is in sharp focus. In the SEM image the whole specimen is in focus. The three-dimensional appearance of this image is a direct result of the large depth of field of the SEM. It is this large depth of field in the SEM images that is one of its most attractive features. The other important feature of the SEM is resolving power, which is the smallest detail that a microscope can resolve, or "see". Electron microscopy takes advantage of the wave nature of rapidly moving electrons. The resolving power of electron microscopes is orders of magnitude better than that of an optical microscope because the wavelength of the probing beam is orders of magnitude smaller. The limit on what size can be resolved – regardless of instrument type -is set by the wavelength. The wavelength of the visible light used in optical microscopes is between 400 and 700 nm. The resolving power of high-quality light microscopes is limited by diffraction of the imaging light to about 200 nm (0.2 m). Scanning electron microscopy uses electrons with energies of a few thousand electron volts (eV), energies a thousand times greater than that of visible light (2 to 3 eV). The wavelength of the electron is given by h/m·u. For 3600 eV electrons, the wavelength is 0.02 nm (20 pm). The resolving power of ordinary electron microscopes is about 1 nm (not as good as the theoretical value because construction details limit resolving power), a value that can be pushed to 0.1 nanometer. Problem 4.83. Einstein’s relativity theory shows that the effective mass of a moving particle, mmoving, is related to its rest mass, mrest, its velocity, u, and the speed of light, c, by this equation: mrest mmoving = 2 1- uc (a) The “relativistic mass” of a 145 g baseball thrown at 44 m·s–1 is: 145 g 145 g 145 g mmoving = = = = 145 g 2 –14 –1 1 1 2.2 10 44m s 1 3.0 108 m s –1 At such a slow rate of travel, the baseball is not affected by relativity and batters do not have to be concerned about learning relativity theory (at least in order to play baseball). (b) For the alpha particle in Problem 4.3(b) moving at 1.5 107 m·s–1 we have an effective mass of: 6.6 10 –24 g mmoving = 1.5 10 7 m s–1 1 2 8 3.0 10 m s –1 = 6.6 10 –24 g 1 2.5 10 –3 6.6 10–24 g 6.6 10–24 g = = 6.6 10–24 g 0.999 0.998 Within the number of significant digits in the data, this velocity does not affect the mass of the alpha particle and the calculations in Problem 4.3(b) do not need correction. (c) For an electron moving at 50% the speed of light, the u/c ratio is 0.50, so we have for the effective mass of the electron: 9.1 10–31 kg 9.1 10–31 kg 9.1 10–31 kg mmoving = = = = 10.5 10–31 kg 2 0.75 0.87 1 0.50 = The effective mass of the moving electron is about 14% higher than the rest mass. Substitution into the de Broglie wavelength equation gives: h 6.63 10 –34 J s = = 10–12 m = 4.2 pm –31 8 –1 = 4.2 m u 10.5 10 kg 1.50 10 m s This electron has a wavelength comparable to gamma radiation. Problem 4.84. (a) This problem, finding the wavelength of a proton that has been accelerated to 50% the speed of light, is identical to Problem 4.83(c), except that the rest mass of the particle in this case is 1.67 10-27 kg: 1.67 10 –27 kg mmoving = = 1.92 10–27 kg 0.87 = 6.63 10–34 J s –27 8 –1 = 2.3 1.92 10 kg 1.50 10 m s 10–15 m = 2.3 fm (femtometer) (b) Let’s do this problem in reverse. First we will calculate the kinetic energy of one electron moving at 90% the speed of light (u/c = 0.90) and then multiply by Avogadro’s number to get the energy of a mole of these electrons. 9.1 10–31 kg 9.1 10–31 kg 9.1 10–31 kg mmoving = = = = 20.9 10–31 kg 2 0.19 0.44 1 0.90 The kinetic energy of this electron is: mmoving u2 20.9 10 –31 kg 0.90 3.00 108 m s–1 2 = 2 2 = 7.6 10–14 J = 7.6 10–11 kJ A mole of these electrons has: KE (mole) = (7.6 10–11 kJ)(6.02 1023 mol–1) = 4.6 1010 kJ·mol–1 This is enough energy to raise the temperature of 11 million liters of water from 0 C to 100 C. KE (one electron) = Problem 4.85. (a) The representation of the periodic table shown in the “Electron Configuration” module in the Web Companion interactive periodic table looks like this (without the block designations): The version of the periodic table on the inside front cover of this book is the one you will probably see most often, because it conserves space and can easily be printed large enough to be easily readable on a standard size page. Compared to the form from the Companion shown here, we see that the f-block elements (often called the inner transition series or the lanthanides and actinides on the book cover) have been moved below the rest of the table and the s- and d-blocks moved together to make the table less long (wide). Using the form from the book cover, we have to be careful to take note of the jump in atomic number (and hence properties) between elements 56 (barium) and 71 (lutetium) and between elements 88 (radium) and 103 (lawrencium) where the inner transition series elements have been shifted out. (b) In the spiral periodic table, Figure 4.34, the s- and p-block elements, usually called the representative elements. together form the “circular” (actually a spiral) portion on the left, The d-block elements are represented by the broader rectangular extension of the spiral on the right and the f-block elements by the narrower rectangular extension. The most interesting difference between the spiral table and the one above is that the spiral table is continuous, that is, the noble gas elements are flanked by a halogen and an alkali metal, rather than ending a period. The families of elements (periodicity) are represented by elements going out on a radius (or stacked in successive layers), but the periods themselves are a little more difficult to discern. (c) The three-dimensional table shown in Problem 4.69 and the chapter opening illustration is related to the one above much as described for the spiral table in part (b). The cylinder that is the tallest part of the three-dimensional table is a helix (spiral), much like the circular spiral in the spiral table and also consists of the s- and p-block (representative) elements. Each cycle around the cylinder represents a period of the table and the families are in the columns down the surface of cylinder. The loop extending directly from the cylinder are the d-block transition elements and “interrupt” the spiral, effectively making each new loop of spiral longer, as they make the table above longer (wider). Another loop, the f-block elements, extends from the d-block element loop, which, again, effectively lengthens the spiral. [NOTE: There is a somewhat difference between the ways the “linear” tables inside the cover of the text and in part (a) and the spiral tables described here and in part (b) treat the initial elements of the f- and d-blocks. Note that the linear tables place Lu and Lr in the family that includes the first elements in the d-block, whereas, in the spiral tables, La and Ac are in this position. The argument for putting La and Ac in this position is that these elements have no f electrons in their valence shells and their electron configurations are (n-1)d1ns2, exactly the same as the first elements in the d-block (periods 4 and 5). The electron configurations of Lu and Lr are (n-2)f14(n-1)d1ns2, which are also like the electron configurations of the first elements in the d-block, if the filled inner shell of f electrons are assumed to be core electrons not valence electrons. The predominant chemistry for all four elements, formation of 3+ metal cations, is the same as that for the 4th and 5th period elements in this family, so it is difficult to differentiate between these conventions on that basis.]