This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 7
7.1 (a) (b) (c) Periodic Properties of the Elements Visualizing Concepts
The light bulb itself represents the nucleus of the atom. The brighter the bulb, the more nuclear charge the electron “sees.” A frosted glass lampshade between the bulb and our eyes reduces the brightness of the bulb. The shade is analogous to core electrons in the atom shielding outer electrons (our eyes) from the full nuclear charge (the bare light bulb). Increasing the wattage of the light bulb mimics moving right along a row of the periodic table. The brighter bulb inside the same shade is analogous to having more protons in the nucleus while the core electron configuration doesn’t change. Moving down a family, both the nuclear charge and the core electron configuration changes. To simulate the addition of core electrons farther from the nucleus, we would add larger frosted glass shades as well as increase the wattage of the bulb to show the increase in Z. The effect of the shade should dominate the increase in wattage, so that the brightness of the light decreases moving down a column. 7.2 A billiard ball is an imperfect model for an atom. The ball has a definite “hard” boundary, while an atom has no definite edge and can be reshaped by interactions with other atoms. That said, the billiard ball is a more appropriate analogy for the nonbonding radius of a fluorine atom. The ball’s radius is spherical, and not deformed by interaction (bonding) with a second ball. If we use the billiard ball to represent the bonding atomic radius of a fluorine atom, we overestimate the bonding atomic radius. When atoms bond, attractive interactions cause their electron clouds to penetrate each other, bringing the nuclei closer together than during a nonbonding (billiard ball) collision. 7.3 (a) The bonding atomic radius of A, r A , is d 1 /2. The distance d 2 is the sum of the bonding atomic radii of A and X, r A + r X . Since we know that r A = d 1 /2, d 2 = r X + d 1 /2, r X = d 2 – d 1 /2. The length of the X-X bond is 2r X . 2r X = 2 (d 2 – d 1 /2) = 2d 2 – d 1 . (b) 164 7 Periodic Properties of the Elements
7.4 Solutions to Exercises Lines (a) and (b) coincide, but their directions are opposite. Line (a) goes from upper right to lower left, and line (b) from lower left to upper right. (c) From the diagram, we observe that the trends in bonding atomic radius (size) and ionization energy are opposite each other. As bonding atomic radius increases, ionization energy decreases, and vice versa. A(g) → A + (g) + e – A(g) + e – → A – (g) A(g) + A(g) → A + (g) + A – (g) ionization energy of A electron affinity of A ionization energy of A + electron affinity of A 7.5 The energy change for the reaction is the ionization energy of A plus the electron affinity of A. This process is endothermic for both nonmetals and metals. Considering data for Cl and Na from Figures 7.12 and 7.14, the endothermic ionization energy term dominates the exothermic electron affinity term, even for Cl which has the most exothermic electron affinity listed. 7.6 (a) (b) X + 2F 2 → XF 4 If X is a nonmetal, XF 4 is a molecular compound. If X is a metal, XF 4 is ionic. For an ionic compound with this formula, X would have a charge of 4+, and a much smaller bonding atomic radius than F – . X in the diagram has about the same bonding radius as F, so it is likely to be a nonmetal. Periodic Table; Effective Nuclear Charge
7.7 Mendeleev insisted that elements with similar chemical and physical properties be placed within a family or column of the table. Since many elements were as yet undiscovered, Mendeleev left blanks. He predicted properties for the “blanks” based on properties of other elements in the family and on either side. Assuming eka- means one place below or under, eka-manganese on Table 7.2 is technetium, Tc. According to Figure 7.2, many of the elements known since ancient times, Fe, Cu, Ag, Au, Hg, Sn, Pb, C and S, are present in nature in elemental form. These elements could be observed directly and their isolation did not require chemical processing. Even though Si is the second most abundant element in Earth’s crust, its abundant forms are compounds: silica, SiO2 (sand, amethyst, quartz, etc.) or silicates (clay, granite, mica, 7.8 7.9 165 7 Periodic Properties of the Elements Solutions to Exercises etc.). In order to be ‘discovered’, elemental Si had to be chemically removed from one of its compounds. Discovery of elemental silicon awaited the more sophisticated chemical techniques available in the nineteenth century. 7.10 (a) The verification of the existence of many new elements by accurately measuring their atomic weights spurred interest in a classification scheme. Mendeleev (and Meyer) noted that certain chemical and physical properties recur periodically when the elements are arranged by increasing atomic weight. The accurate atomic weights provided a common property on which to base a classification scheme of the elements. Moseley realized that the characteristic X-ray frequencies emitted by each element were related to a unique integer that he assigned to each element. We now know this integer as the atomic number, the number of protons in the nucleus of an atom. In general, atomic weight increases as atomic number increases, but there are a few exceptions. If elements are arranged by increasing atomic number, a few seeming contradictions in the Mendeleev table (the positions of Ar and K or Te and I) are eliminated. The main determining factor of physical and especially chemical properties is electron configuration. For electrically neutral elements, the number of electrons equals the number of protons, which in turn is the atomic number of an element. Atomic weight is related to mass number, protons plus neutrons. The number of neutrons in its nucleus does influence the mass of an atom, but mass is a minor or non-factor in determining properties. Effective nuclear charge, Z eff, is a representation of the average electrical field experienced by a single electron. It is the average environment created by the nucleus and the other electrons in the molecule, expressed as a net positive charge at the nucleus. It is approximately the nuclear charge, Z, minus the number of core electrons. Going from left to right across a period, nuclear charge increases while the number of electrons in the core is constant. This results in an increase in Z eff. Electrostatic attraction for the nucleus lowers the energy of an electron, while electron-electron repulsions increase this energy. The concept of effective nuclear charge allows us to model this increase in the energy of an electron as a smaller net attraction to a nucleus with a smaller positive charge, Z eff. In Be (or any element), the 1s electrons are not shielded by any core electrons, so they experience a much greater Z eff than the 2s electrons. Analyze/Plan. Zeff = Z−S. Find the atomic number, Z, of Na and K. Write their electron configurations and count the number of core electrons. Assume S = number of core electrons. Solve. Na: Z = 11; [Ne]3s1. In the Ne core there are 10 electrons. Zeff = 11−10 = 1. K: Z = 19; [Ar]4s1. In the Ar core there are 18 electrons. Zeff = 19−18 = 1. (b) (c) 7.11 (a) (b) 7.12 (a) (b) 7.13 (a) 166 7 Periodic Properties of the Elements
(b) Solutions to Exercises Analyze/Plan. Zeff = Z−S. Write the complete electron configuration for each element to show counting for Slater’s rules. S = 0.35 (# of electrons with same n) + 0.85 (# of electrons with (n−1) + 1 (# of electrons with (n−2). Solve. Na: 1s22s22p63s1. S = 0.35(0) + 0.85(8) + 1(2) = 8.8. Zeff = 11 – 8.8 = 2.2 K: 1s22s22p63s23p64s1. S = 0.35(0) + 0.85(8) + 1(10) = 16.8. Zeff = 19 – 16.8 = 2.2 (c) (d) For both Na and K, the two values of Zeff are 1.0 and 2.2. The Slater value of 2.2 is closer to the values of 2.51 (Na) and 3.49 (K) obtained from detailed calculations. Both approximations, ‘core electrons 100% effective’ and Slater, yield the same value of Zeff for Na and K. Neither approximation accounts for the gradual increase in Zeff moving down a group. 7.14 Follow the method in the preceding question to calculate Zeff values. (a) Si: Z = 14; [Ne]3s23p2. 10 electrons in the Ne core. Zeff = 14 – 10 = 4 Cl: Z = 17; [Ne]3s23p5. 10 electrons in the Ne core. Zeff = 17– 10 = 7 (b) (c) (d) Si: 1s22s22p63s23p2. S = 0.35(3) + 0.85(8) + 1(2) = 9.85. Zeff = 14 – 9.85 = 4.15 Cl: 1s22s22p63s23p5. S = 0.35(6) + 0.85(8) + 1(2) = 10.9. Zeff = 17 – 10.9 = 6.10 The Slater values of 4.15 (Si) and 6.10 (Cl) are closer to the results of detailed calculations, 4.29 (Si) and 6.12 (Cl). The Slater method of approximation more closely approximates the gradual increase in Zeff moving across a row. The ‘core 100%-effective’ approximation underestimates Zeff for Si but overestimates it for Cl. Slater values are closer to detailed calculations, and a better indication of the change in Zeff moving from Si to Cl. 7.15 Krypton has a larger nuclear charge (Z = 36) than argon (Z = 18). The shielding of electrons in the n = 3 shell by the 1s, 2s and 2p core electrons in the two atoms is approximately equal, so the n = 3 electrons in Kr experience a greater effective nuclear charge and are thus situated closer to the nucleus. Mg < P < K < Ti < Rh. The shielding of electrons in the n = 3 shell by 1s, 2s and 2p core electrons in these elements is approximately equal, so the effective nuclear charge increases as Z increases. 7.16 Atomic and Ionic Radii
7.17 (a) (b) Atomic radii are determined by measuring distances between atoms (interatomic distances) in various situations. Bonding radii are calculated from the internuclear separation of two atoms joined by a chemical bond. Nonbonding radii are calculated from the internuclear separation between two gaseous atoms that collide and move apart, but do not bond. For a given element, the nonbonding radius is always larger than the bonding radius. When a chemical bond forms, electron clouds of the two atoms interpenetrate, bringing the two nuclei closer together and resulting in a smaller bonding atomic radius (Figure 7.6). (c) 167 7 Periodic Properties of the Elements
7.18 (a) Solutions to Exercises Since the quantum mechanical description of the atom does not specify the exact location of electrons, there is no specific distance from the nucleus where the last electron can be found. Rather, the electron density decreases gradually as the distance from the nucleus increases. There is no quantum mechanical “edge” of an atom. When nonbonded atoms touch, it is their electron clouds that interact. These interactions are primarily repulsive because of the negative charges of electrons. Thus, the size of the electron clouds determines the nuclear approach distance of nonbonded atoms. (b) 7.19 7.20 7.21 7.22 The atomic (metallic) radius of W is the interatomic W-W distance divided by 2, 2.74 Å/2= 1.37 Å. The distance between Si atoms in solid silicon is two times the bonding atomic radius from Figure 7.7. The Si−Si distance is 2 × 1.11 Å = 2.22 Å. From atomic radii, As–I = 1.19 Å + 1.33 Å = 2.52 Å. This is very close to the experimental value of 2.55 Å. Bi−I = 2.81 Å = r Bi + r I . From Figure 7.7, r I = 1.33 Å. r Bi = [Bi−I] – r I = 2.81 Å – 1.33 Å = 1.48 Å. 7.23 (a) (c) Atomic radii decrease moving from left to right across a row and (b) increase from top to bottom within a group. F < S < P <As. The order is unambiguous according to the trends of increasing atomic radius moving down a column and to the left in a row of the table. The vertical difference in radius is due to a change in principal quantum number of the outer electrons. The horizontal difference in radius is due to the change in electrostatic attraction between the outer electron and a nucleus with one more or one fewer proton. Adding or subtracting a proton has a much smaller radius effect than moving from one principal quantum level to the next. Si < Al < Ge < Ga. This order is predicted by the trends in increasing atomic radius moving to the left in a row and down a column of the periodic chart, assuming that changes moving down a column are larger [see part (a)]. That is, the order above assumes that the change from Si to Ge is larger than the change from Si to Al. This order is confirmed by the values in Figure 7.7. 7.24 (a) (b) 7.25 Plan. Locate each element on the periodic charge and use trends in radii to predict their order. Solve. (a) Be < Mg < Ca Na < Ca < Ba (b) Br < Ge < Ga (b) As < Sb < Sn (c) Si < Al < Tl 7.26 (a) (c) Be < Si < Al. This order assumes the increase in radius from the second to the third row is greater than the decrease moving right in the third row. Radii in Figure 7.7 confirm this assumption. Electrostatic repulsions are reduced by removing an electron from a neutral atom, Z eff increases, and the cation is smaller. 7.27 (a) 168 7 Periodic Properties of the Elements
(b) (c) Solutions to Exercises The additional electrostatic repulsion produced by adding an electron to a neutral atom causes the electron cloud to expand, so that the radius of the anion is larger than the radius of the neutral atom. Going down a column, the n value of the valence electrons increases and they are farther from the nucleus. Valence electrons also experience greater shielding by core electrons. The greater radial extent of the valence electrons outweighs the increase in Z, and the size of particles with like charge increases. As Z stays constant and the number of electrons increases, the electron-electron repulsions increase, the electrons spread apart, and the ions become larger. I– > I > I+ Going down a column, the increasing average distance of the outer electrons from the nucleus causes the size of particles with like charge to increase. Ca 2 + > Mg 2 + > Be 2 + Fe: [Ar]4s 2 3d 6 ; Fe 2 +: [Ar]3d 6 ; Fe 3 +: [Ar]3d 5 . The 4s valence electrons in Fe are on average farther from the nucleus than the 3d electrons, so Fe is larger than Fe 2 +. Since there are five 3d orbitals, in Fe 2 + at least one orbital must contain a pair of electrons. Removing one electron to form Fe 3 + significantly reduces repulsion, increasing the nuclear charge experienced by each of the other d electrons and decreasing the size of the ion. Fe > Fe 2 + > Fe 3 + 7.28 (a) (b) (c) 7.29 The size of the red sphere decreases on reaction, so it loses one or more electrons and becomes a cation. Metals lose electrons when reacting with nonmetals, so the red sphere represents a metal. The size of the blue sphere increases on reaction, so it gains one or more electrons and becomes an anion. Nonmetals gain electrons when reacting with metals, so the blue sphere represents a nonmetal. The order of radii is Ca > Ca 2 + > Mg 2 +, so the largest sphere is Ca, the intermediate one is Ca 2 +, and the smallest is Mg 2 +. (a) (b) An isoelectronic series is a group of atoms or ions that have the same number of electrons, and thus the same electron configuration. (i) Al 3 +: Ne (ii) Ti4 +: Ar (b) Sc3+: Ar (iii) Br–: Kr (iv) Sn2 +: Cd 7.30 7.31 7.32 (a) (c) Cl–: Ar Fe2+: [Ar]3d6. Fe2+ has 24 electrons. Neutral Cr has 24 electrons, [Ar]4s13d5. Because transition metals fill the s subshell first but also lose s electrons first when they form ions, many transition metal ions do not have isolectronic neutral atoms. Zn2+: [Ar]3d10; no isoelectronic neutral atom [same reason as (c)]. Sn4+: [Kr]4d10; no isoelectronic neutral atom [same reason as (c)], but Sn4+ is isoelectronic with Cd2+. Analyze/Plan. Follow the logic in Sample Exercise 7.4. Solve. Na+ is smaller. Since F− and Na+ are isoelectronic, the ion with the larger nuclear charge, Na+, has the smaller radius. (d) (e) 7.33 (a) 169 7 Periodic Properties of the Elements
(b) Solutions to Exercises Analyze/Plan. The electron configuration of the ions is [Ne] or [He]2s22p6. The ions have either 10 core electrons or 2 core electrons. Apply Equation 7.1 to both cases and check the result. Solve. F−: Z = 9. For 10 core electrons, Zeff = 9 – 10 = −1. While we might be able to interpret a negative value for Zeff, positive values will be easier to compare; we will assume a He core of 2 electrons. F−, Z = 9. Zeff = 9 − 2 = 7. Na+: Zeff = 11 – 2= 9 (c) Analyze/Plan. The electron of interest has n = 2. There are 7 other n = 2 electrons, and two n = 1 electrons. Solve. S = 0.35(7) + 0.85(2) + 1(0) = 4.15 F−: Zeff = 9 – 4.15 = 4.85. Na+: Zeff = 11 – 4.15 = 6.85 For isoelectronic ions, the electron configurations and therefore shielding values (S) are the same. Only the nuclear charge changes. So, as nuclear charge (Z) increases, effective nuclear charge (Zeff) increases and ionic radius decreases. K + (larger Z) is smaller. Cl – and K + : [Ne]3s23p6. 10 core electrons Cl – , Z = 17. Zeff = 17 – 10 = 7 K + , Z = 19. Zeff = 19 – 10 = 9 Valence electron, n = 3; 7 other n = 3 electrons; eight n = 2 electrons; two n = 1 electrons. S = 0.35(7) + 0.85(8) + 1(2) = 11.25 Cl – : Zeff = 17 – 11.25 = 5.75. K + : Zeff = 19 – 11.25 = 7.75 For isoelectronic ions, the electron configurations and therefore shielding values (S) are the same. Only the nuclear charge changes. So, as nuclear charge (Z) increases, effective nuclear charge (Zeff) increases and ionic radius decreases. (d) 7.34 (a) (b) (c) (d) 7.35 Analyze/Plan. Use relative location on periodic chart and trends in atomic and ionic radii to establish the order. (a) (c) Cl < S < K (b) K + < Cl – < S 2 – Even though K has the largest Z value, the n-value of the outer electron is larger than the n-value of valence electrons in S and Cl so K atoms are largest. When the 4s electron is removed, K + is isoelectronic with Cl – and S 2 –. The larger Z value causes the 3p electrons in K + to experience the largest effective nuclear charge and K + is the smallest ion. Se < Se 2 – < Te 2 – (b) Co 3 + < Fe 3 + < Fe 2 + (c) Ti 4 + < Sc 3 + < Ca (d) Be 2 +< Na + < Ne O2– is larger than O because the increase in electron repulsions that accompany addition of an electron causes the electron cloud to expand. S 2 – is larger than O 2 –, because for particles with like charges, size increases going down a family. S2 – is larger than K+ because the two ions are isoelectronic and K + has the larger Z and Zeff. 7.36 7.37 (a) (a) (b) (c) 170 7 Periodic Properties of the Elements
(d) 7.38 Solutions to Exercises K+ is larger than Ca2+ because for isoelectronic ions, the ion with the smaller Z has the larger radius. Make a table of d(measured), d(ionic radii) and d(covalent radii). Use these values to make comparisons for (b) and (c). The estimated distances are just the sum of the various radii from Figure 7.8. (a) Li−F Na−Cl K−Br Rb−I (b) d(measured), Å 2.01 2.82 3.30 3.67 d(ionic radii), Å 2.09 2.83 3.34 3.72 d(covalent radii), Å 2.05 2.53 3.10 3.44 The agreement between measured distances in specific ionic compounds, and predicted distances based on ionic radii is not perfect. Ionic radii are averages compiled from distances in many ionic compounds containing the ion in question. The sum of these average radii may not give an exact match for the distance in any specific compound, but it will give good distance estimates for many ionic compounds. Also, there is uncertainty in all measured data. Note that all estimates from ionic radii are within 0.08 Å of the measured distances. Distance estimates from bonding atomic radii are not as accurate as those from ionic radii. This indicates that the bonding in these four compounds is more accurately described as ionic, rather than covalent. The details of these two models will be discussed in Chapter 8. (c) Ionization Energies; Electron Affinities
7.39 7.40 B(g) → B + (g) + 1e – ; (a) (b) 7.41 (a) B + (g) → B 2 +(g) + 1e – ; B 2 +(g) → B 3 +(g) + 1e – Sn(g) → Sn + (g) + 1e – ; Ti 3 +(g) → Ti 4 +(g) + 1e – According to Coulomb’s law, the energy of an electron in an atom is negative, because of the electrostatic attraction of the electron for the nucleus. In order to overcome this attraction, remove the electron and increase its energy; energy must be added to the atom. Ionization energy, ΔE for this process, is positive, regardless of the magnitude of Z or the quantum numbers of the electron. F has a greater first ionization energy than O, because F has a greater Z eff and the outer electrons in both elements are approximately the same distance from the nucleus. The second ionization energy of an element is greater than the first because Z eff is larger for the +1 cation than the neutral atom; more energy is required to overcome the larger Z eff. The effective nuclear charges of Li and Na are similar, but the outer electron in Li has a smaller n-value and is closer to the nucleus than the outer electron in Na. More energy is needed to overcome the greater attraction of the Li electron for the nucleus. Sn + (g) → Sn 2 +(g) + 1e – (b) (c) 7.42 (a) 171 7 Periodic Properties of the Elements
(b) Solutions to Exercises Sc: [Ar] 4s23d1; Ti: [Ar] 4s23d2. The fourth ionization of titanium involves removing a 3d valence electron, while the fourth ionization of Sc requires removing a 3p electron from the [Ar] core. The effective nuclear charges experienced by the two 3d electrons in Ti are much more similar than the effective nuclear charges of a 3d valence electron and a 3p core electron in Sc. Thus, the difference between the third and fourth ionization energies of Sc is much larger. The electron configuration of Li + is 1s 2 or [He] and that of Be + is [He]2s 1 . Be + has one more valence electron to lose while Li + has the stable noble gas configuration of He. It requires much more energy to remove a 1s core electron close to the nucleus of Li + than a 2s valence electron farther from the nucleus of Be + . In general, the smaller the atom, the larger its first ionization energy. According to Figure 7.12, He has the largest and Cs the smallest first ionization energy of the nonradioactive elements. Moving from F to I in group 7A, first ionization energies decrease and atomic radii increase. The greater the atomic radius, the smaller the electrostatic attraction of an outer electron for the nucleus and the smaller the ionization energy of the element. First ionization energies increase slightly going from K to Kr and atomic sizes decrease. As valence electrons are drawn closer to the nucleus (atom size decreases), it requires more energy to completely remove them from the atom (first ionization energy increases). Each trend has a discontinuity at Ga, owing to the increased shielding of the 4p electrons by the filled 3d subshell. Solve. (e) Te Ar (b) Be (c) Co (d) S (c) 7.43 (a) (b) 7.44 (a) (b) 7.45 7.46 Plan. Use periodic trends in first ionization energy. (a) (a) Ti. Effective nuclear charge increases moving both right across a row and up a family. The 4s valence electrons in Ti experience the greater Zeff and have greater first ionization energy than 6s electrons of Ba. Recall that transition metals like Ti lose ns electrons first when forming ions. Cu. The 4s electrons of Cu are closer to the nucleus and shielded mainly by an [Ar] core, while the 5s electrons of Ag are further from the nucleus and shielded by a [Kr] core. Recall that transition elements lose ns electrons first when forming ions. Cl. Effective nuclear charge increases moving both right across a row and up a family. Valence electrons in Cl, which is to the right and above Ge, experience the greater Z eff and have the larger first ionization energy. Sb. Zeff and first ionization energy increase moving up a family and right across a row. Even the excess nuclear charge (Z) associated with filling the 4f subshell between Sb and Pb does not totally offset these trends. Solve. (b) Sb3 +: [Kr]5s 2 4d1 0 (d) Te6+: [Kr]4d 1 0 (b) (c) (d) 7.47 Plan. Follow the logic of Sample Exercise 7.7. (a) (c) In3 +: [Kr]4d1 0 Te 2 –: [Kr]5s 2 4d 1 05p 6 or [Xe] 172 7 Periodic Properties of the Elements
(e) Hg 2 +: [Xe]4f 1 45d 1 0 7.48 (a) (b) (c) (e) 7.49 Cr3 +: [Ar]3d 3 Solutions to Exercises
(f) Rh3 +: [Kr]4d 6 [Note that In3+ and Te6+ (but not Sb3+ ) above are isoelectroic.] N3 –: [He]2s 2 2p 6 = [Ne], noble-gas configuration Sc 3 +: [Ar], noble-gas configuration Tl + : [Xe]6s 2 4f 1 45d 1 0 (d) Cu2 +: [Ar]3d 9 (f) Au + : [Xe]4f 1 45d 1 0 Plan. Follow the logic in Sample Exercise 7.7. Construct a mental box diagram for the outer electrons to determine how many are unpaired. Solve. (a) (b) Ni2 +: [Ar]3d 8 , 2 unpaired electrons. There are 5 degenerate d orbitals; 3 are filled, 2 contain single electrons. Sn 2 +: [Kr]5s 2 4d 1 0 , 0 unpaired electrons. The 5s and five 4d orbitals are all filled. Cu 2 +, 1 unpaired electron (b) Tl + , 0 unpaired electrons 7.50 7.51 (a) Analyze/Plan. Consider the definitions of ionization energy, electron affinity and the electron configuration of Ar. Solve. Argon is a noble gas, with a very stable core electron configuration. This causes the element to resist chemical change. Positive, endothermic, values for ionization energy and electron affinity mean that energy is required to either remove or add electrons. Valence electrons in Ar experience the largest Z eff of any element in the third row, because the nuclear buildup is not accompanied by an increase in screening. This results in a large, positive ionization energy. When an electron is added to Ar, the n = 3 electrons become core electrons which screen the extra electrons so effectively that Ar – has a higher energy than an Ar atom and a free electron. This results in a large positive electron affinity. 7.52 Li + 1e −
[He]2s 1 → Li −
[He]2s 2 ; Be
[He]2s 2 + 1e − → Be −
[He]2s 2 2 p1 Adding an electron to Li completes the 2s subshell. The added electron experiences essentially the same effective nuclear charge as the other valence electron, except for the repulsion of pairing electrons in an orbital. There is an overall stabilization; ΔE is negative. An extra electron in Be would occupy the higher energy 2p subshell. This electron is shielded from the full nuclear charge by the 2s electrons and does not experience a stabilization in energy; ΔE is positive. 7.53 Analyze/Plan. Consider the definitions of ionization energy and electron affinity, along with pertinent electron configurations. Solve.
Electron affinity : Br (g) [Ar ]4s 2 3d 10 4 p 5 + 1e − → Br − (g ) [Ar]4s 2 3d 10 4 p 6 When a Br atom gains an electron, the Br – ion adopts the stable electron configuration of Kr. Since the electron is added to the same 4p subshell as other outer electrons, it 173 7 Periodic Properties of the Elements
Electron affinity : Kr (g) [Ar]4s 2 3d 10 4 p 6 + 1e − → Solutions to Exercises experiences essentially the same attraction for the nucleus. Thus, the energy of the Br – ion is lower than the total energy of a Br atom and an isolated electron, and electron affinity is negative.
Kr − (g ) [Ar]4s 2 3d 10 4 p 6 5s 1 Energy is required to add an electron to a Kr atom; Kr – has a higher energy than the isolated Kr atom and free electron. In Kr – the added electron would have to occupy the higher energy 5s orbital; a 5s electron is farther from the nucleus and effectively shielded by the spherical Kr core and is not stabilized by the nucleus. 7.54 Ionization energy of F – : Electron affinity of F: F – (g) → F(g) + 1e – F(g) + 1e – → F – (g) The two processes are the reverse of each other. The energies are equal in magnitude but opposite in sign. I 1 (F – ) = –E (F) 7.55 Analyze/Plan. Consider the definitions of ionization energy and electron affinity, along with the appropriate electron configurations. Solve. (a) Ionization energy of Ne: Electron affinity of F: (b) (c) Ne(g) [He]2s22p6 F(g) + 1e− [He]2s22p5 → → Ne+(g) + [He]2s22p5 F−(g) [He]2s22p6 1e− The I1 of Ne is positive, while E1 of F is negative. All ionization energies are positive. One process is apparently the reverse of the other, with one important difference. The Z (and Zeff) for Ne is greater than Z (and Zeff) for F−. So we expect I1(Ne) to be somewhat greater in magnitude and opposite in sign to E1(F). [Repulsion effects approximately cancel; repulsion decrease upon I1 causes smaller positive value; repulsion increase upon E1 causes smaller negative value.]
+ 1e− → Mg(g) [Ne] 3s 2 7.56 Mg + (g) [Ne] 3s
1 This process is the reverse of the first ionization of Mg. The magnitude of the energy change for this process is the same as the magnitude of the first ionization energy of Mg, 738 kJ/mol. Properties of Metals and Nonmetals
7.57 7.58 The smaller the first ionization energy of an element, the greater the metallic character of that element. S < Si < Ge < Ca. S is a nonmetal, Si and Ge are metalloids, and Ca is a metal. We expect that electrical conductivity increases as metallic character increases. Since metallic character increases going down a column and to the left in a row, the order of increasing electrical conductivity is as shown above. 174 7 Periodic Properties of the Elements
7.59 Solutions to Exercises Analyze/Plan. Use first ionization energies for metals and metalloids from Figure 7.12 to inform our discussion. Solve. I1 for Al is 578 kJ/mol. This is similar to values for Ca and Sr group 2A, the alkaline earth metals. I1 values for the four metalloids in group 4A and 5A range from 762 kJ/mol for Ge to 947 kJ/mol for As. The low value of I1 for Al classifies it as a metal, rather than a metalloid. 7.60 Metallic character increases moving down a family and to the left in a period. Use these trends to select the element with greater metallic character. (a) Li (b) Na (c) Sn (d) Al 7.61 Analyze/Plan. Ionic compounds are formed by combining a metal and a nonmetal; molecular compounds are formed by two or more nonmetals. Solve. Ionic: MgO, Li 2 O, Y 2 O 3 ; molecular: SO 2 , P 2 O 5 , N 2 O, XeO 3 7.62 Follow the logic in Sample Exercise 7.8. Scandium is a metal, so we expect Sc2O3 to be ionic. Metal oxides are usually basic and react with acid to form a salt and water. We choose HNO3(aq) as the acid for our equation. Sc2O3(s) + 6HNO3(aq) → 2Sc(NO3)3(aq) + 3H2O(l). The net ionic equation is: Sc2O3(s) + 6H+(aq) → 2Sc3+(aq) + 3H2O(l) 7.63 (a) (b) When dissolved in water, an “acidic oxide” produces an acidic (pH < 7) solution. A “basic oxide” dissolved in water produces a basic (pH > 7) solution. Oxides of nonmetals are acidic. Example: SO 3 (g) + H 2 O(l) → H 2 SO 4 (aq). Oxides of metals are basic. Example: CaO (quick lime). CaO(s) + H 2 O(l) → Ca(OH) 2 (aq). 7.64 7.65 The more nonmetallic the central atom, the more acidic the oxide. In order of increasing acidity: CaO < Al 2 O 3 < SiO 2 < CO 2 < P 2 O 5 < SO 3 Analyze/Plan. Cl 2 O 7 is a molecular compound formed by two nonmetallic elements. More specifically, it is a nonmetallic oxide and acidic. Solve. (a) (b) (c) Dichlorineseptoxide Elemental chlorine and oxygen are diatomic gases. 2Cl 2 (g) + 7O 2 (g) → 2Cl 2 O 7 (l) Most nonmetallic oxides we have seen, such as CO 2 and SO 3 , are gases. However, oxides with more atoms, such as P 2 O 3 (l) and P 2 O 5 (s), exist in other states. A boiling point of 81°C is not totally unexpected for a large molecule like Cl 2 O 7 . Cl 2 O 7 is an acidic oxide, so it will be more reactive to base, OH – . Cl 2 O 7 (l) + 2OH – (aq) → 2ClO 4 – (aq) + H 2 O(l) (d) 7.66 (a) XCl 4 (l) + 2H 2 O(l) → XO 2 (s) + 4HCl(g) The second product is HCl(g). 175 7 Periodic Properties of the Elements
(b) Solutions to Exercises If X were a metal, both the oxide and the chloride would be high melting solids. If X were a nonmetal, XO 2 would be a nonmetallic, molecular oxide and probably gaseous, like CO 2 , NO 2 , and SO 2 . Neither of these statements describes the properties of XO 2 and XCl 4 , so X is probably a metalloid. Use the Handbook of Chemistry to find formulas and melting points of oxides, and formulas and boiling points of chlorides of selected metalloids. metalloid boron silicon germanium arsenic formula of oxide B2O3 SiO 2 GeO GeO 2 As 2 O 3 As 2 O 5 m.p. of oxide 460°C ~1700°C 710°C ~1100°C 315°C 315°C formula of chloride BCl 3 SiCl 4 GeCl 2 GeCl 4 AsCl 3 b.p. of chloride 12°C 58°C decomposes 84°C 132°C (c) Boron, arsenic, and, by analogy, antimony, do not fit the description of X, because the formulas of their oxides and chlorides are wrong. Silicon and germanium, in the same family, have oxides and chlorides with appropriate formulas. Both SiO 2 and GeO 2 melt above 1000°C, but the boiling point of SiCl 4 is much closer to that of XCl 4 . Element X is silicon. 7.67 (a) (b) (c) (d) 7.68 (a) (b) (c) (d) BaO(s) + H 2 O(l) → Ba(OH) 2 (aq) FeO(s) + 2HClO 4 (aq) → Fe(ClO 4 ) 2 (aq) + H 2 O(l) SO 3 (g) + H 2 O(l) → H 2 SO 4 (aq) CO 2 (g) + 2NaOH(aq) → Na 2 CO 3 (aq) + H 2 O(l) K 2 O(s) + H 2 O(l) → 2KOH(aq) P 2 O 3 (l) + 3H 2 O(l) → 2H 3 PO 3 (aq) Cr 2 O 3 (s) + 6HCl(aq) → 2CrCl 3 (aq) + 3H 2 O(l) SeO 2 (s) + 2KOH(aq) → K 2 SeO 3 (aq) + H 2 O(l) Group Trends in Metals and Nonmetals
7.69 (a) (b) (c) (d) (e) Na [Ne]3s 1 +1 +496 kJ/mol very reactive 1.54 Å Mg [Ne]3s 2 +2 +738 kJ/mol reacts with steam, but not H 2 O(l) 1.30 Å 176 7 Periodic Properties of the Elements
(b) Solutions to Exercises When forming ions, both adopt the stable configuration of Ne, but Na loses one electron and Mg two electrons to achieve this configuration. (c),(e) The nuclear charge of Mg (Z = 12) is greater than that of Na, so it requires more energy to remove a valence electron with the same n value from Mg than Na. It also means that the 2s electrons of Mg are held closer to the nucleus, so the atomic radius (e) is smaller than that of Na. (d) 7.70 (a) Mg is less reactive because it has a filled subshell and it has a higher ionization energy. Rb: [Kr]5s 1 , r = 2.11 Å Ag: [Kr]5s 1 4d 1 0, r = 1.53 Å The electron configurations both have a [Kr] core and a single 5s electron; Ag has a completed 4d subshell as well. The radii are very different because the 5s electron in Ag experiences a much greater effective nuclear charge. Ag has a much larger Z (47 vs. 37), and although the 4d electrons in Ag shield the 5s electron somewhat, the increased shielding does not compensate for the large increase in Z. (b) Ag is much less reactive (less likely to lose an electron) because its 5s electron experiences a much larger effective nuclear charge and is more difficult to remove. Ca and Mg are both metals; they tend to lose electrons and form cations when they react. Ca is more reactive because it has a lower ionization energy than Mg. The Ca valence electrons in the 4s orbital are less tightly held because they are farther from the nucleus than the 3s valence electrons of Mg. K and Ca are both metals; they tend to lose electrons and form cations when they react. K is more reactive because it has a lower ionization energy. The 4s valence electron in K is less tightly held because it experiences a smaller nuclear charge (Z = 19 for K versus Z = 20 for Ca) with similar shielding effects than the 4s valence electrons of Ca. Cs is much more reactive than Li toward H 2 O because its valence electron is less tightly held (greater n value), and Cs is more easily oxidized. The purple flame indicates that the metal is potassium (see Figure 7.26). 7.71 (a) (b) 7.72 (a) (b) (c) K 2 O 2 (s) + H 2 O(l) → H 2 O 2 (aq) + K 2 O (aq) potassium peroxide
7.73 (a) (b) (c) (d) 7.74 (a) (b) (c) (d) 2K(s) + Cl 2 (g) → 2KCl(s) SrO(s) + H 2 O(l) → Sr(OH) 2 (aq) 4Li(s) + O 2 (g) → 2Li 2 O(s) 2Na(s) + S(l) hydrogen peroxide → Na 2 S(s) 2Cs(s) + 2H 2 O(l) → 2CsOH(aq) + H 2 (g) Sr(s) + 2H 2 O(l) → Sr(OH) 2 (aq) + H 2 (g) 2Na(s) + O 2 (g) → Na2O2(s) (See Equation [7.21].) Ca(s) + I2(s) → CaI2(s) 177 7 Periodic Properties of the Elements
7.75 (a) Solutions to Exercises Analyze/Plan. Use first ionization energies from Figure 7.12 to inform our discussion. H (I1 = 1312 kJ/mol) fits between C (I1 = 1086 kJ/mol) and N (I1 = 1402 kJ/mol in the second period. (Ignore the value for O, which is anomalously low due to exaggerated repulsion effects in the small 2p orbitals.) Li (I1 = 520 kJ/mol) fits between Na (I1 = 496 kJ/mol) and Mg (I1 = 738 kJ/mol). H definitely fits among the nonmetals of the second row, and Li fits among the metals of the third row, so positions in these series are consistent with their assignment as nonmetal (H) and metal (Li). The reactions of the alkali metals with hydrogen and with a halogen are redox reactions. In both classes of reaction, the alkali metal loses electrons and is oxidized. Both hydrogen and the halogen gain electrons and are reduced. The product is an ionic solid, where either hydride ion, H – , or a halide ion, X – , is the anion and the alkali metal is the cation. Ca(s) + F 2 (g) → CaF 2 (s) Ca(s) + H 2 (g) → CaH 2 (s) (b) (c) 7.76 (a) (b) Both products are ionic solids containing Ca 2 + and the corresponding anion in a 1:2 ratio. 7.77 (a) (b) (c) (d) (e) (f) (b) F [He]2s 2 2p 5 –1 1681 kJ/mol reacts exothermically to form HF –328 kJ/mol 0.71 Å Cl [Ne]3s 2 3p 5 –1 1251 kJ/mol reacts slowly to form HCl –349 kJ/mol 0.99 Å F and Cl are in the same group, have the same valence electron configuration and common ionic charge. (c),(f) The n = 2 valence electrons in F are closer to the nucleus and more tightly held than the n = 3 valence electrons in Cl. Therefore, the ionization energy of F is greater, and the atomic radius is smaller. (d) In its reaction with H 2 O, F is reduced; it gains an electron. Although the electron affinity, a gas phase single atom property, of F is less negative than that of Cl, the tendency of F to hold its own electrons (high ionization energy) coupled with a relatively large exothermic electron affinity makes it extremely susceptible to reduction and chemical bond formation. Cl is unreactive to water because it is less susceptible to reduction. While F has approximately the same Z eff as Cl, its small atomic radius gives rise to large repulsions when an extra electron is added, so the overall electron affinity of F is smaller (less exothermic) than that of Cl. The n = 2 valence electrons in F are closer to the nucleus so the atomic radius is smaller than that of Cl. (e) (f) 178 7 Periodic Properties of the Elements
7.78 (a) Solutions to Exercises Plan. Predict the physical and chemical properties of At based on the trends in properties in the halogen (7A) family. Solve. F, at the top of the column, is a gas; I, immediately above At, is a solid; the melting points of the halogens increase going down the column. At is likely to be a solid at room temperature. All halogens form ionic compounds with Na; they have the generic formula NaX. The compound formed by At will have the formula NaAt. The term “inert” was dropped because it no longer described all the group 8A elements. In the 1960s, scientists discovered that Xe would react with substances such as F 2 and PtF 6 that have a strong tendency to remove electrons. Thus, Xe could not be categorized as an “inert” gas. The group is now called the noble gases. (b) 7.79 (a) (b) (c) 7.80 Xe has a lower ionization energy than Ne. The valence electrons in Xe are much farther from the nucleus than those of Ne (n = 5 vs n = 2) and much less tightly held by the nucleus; they are more “willing” to be shared than those in Ne. Also, Xe has empty 5d orbitals that can help to accommodate the bonding pairs of electrons, while Ne has all its valence orbitals filled. (a) (b) 2O 3 (g) → 3O 2 (g) Xe(g) + F 2 (g) → XeF 2 (g) Xe(g) + 2F 2 (g) → XeF 4 (s) Xe(g) + 3F 2 (g) → XeF 6 (s) (c) (d) S(s) + H 2 (g) → H 2 S(g) 2F 2 (g) + 2H 2 O(l) → 4HF(aq) + O 2 (g) Cl 2 (g) + H 2 O(l) → HCl(aq) + HOCl(aq) Ba(s) + H 2 (g) → BaH 2 (s) 2Li(s) + S(s) → Li 2 S(s) Mg(s) + F 2 (g) → MgF 2 (s) 7.81 7.82 (a) (b) (c) (d) Additional Exercises
7.83 Up to Z = 82, there are three instances where atomic weights are reversed relative to atomic numbers: Ar and K; Co and Ni; Te and I. In each case, the most abundant isotope of the element with the larger atomic number (Z) has one more proton, but fewer neutrons than the element with the smaller atomic number. The smaller number of neutrons causes the element with the larger Z to have a smaller than expected atomic weight. 7.84 (a) (b) 4s To a first approximation, s and p valence electrons do not shield each other, so we expect the 4s and 4p electrons in As to experience a similar Z eff. However, 179 7 Periodic Properties of the Elements Solutions to Exercises since s electrons have a finite probability of being very close to the nucleus (Figure 7.4), they experience less shielding than p electrons with the same nvalue. Since Z eff = Z – S and Z is the same for all electrons in As, if S is smaller for 4s than 4p, Z eff will be greater for 4s electrons and they will have a lower energy. 7.85 (a) (b) P: [Ne]3s23p3. Zeff = Z−S = 15 – 10 = 5. Four other n = 3 electrons, eight n = 2 electrons, two n = 1 electron. S = 0.35(4) + 0.85(8) + 1(2) = 10.2. Zeff = Z−S = 15 – 10.2 = 4.8. The approximation in (a) and Slater’s rules give very similar values for Zeff. The 3s electrons penetrate the [Ne] core electrons (by analogy to Figure 7.4) and experience less shielding than the 3p electrons. That is, S is greater for 3p electrons, owing to the penetration of the 3s electrons, so Z – S (3p) is less than Z – S (3s). The 3p electrons are the outermost electrons; they experience a smaller Z eff than 3s electrons and thus a smaller attraction for the nucleus, given equal n-values. The first electron lost is a 3p electron. Each 3p orbital holds one electron, so there is no preference as to which 3p electron will be lost. (c) (d) 7.86 Close approach by two positively charged nuclei is impossible because of the large electrostatic repulsion between like-charged particles at small distances. The additional space between the nuclei in a molecule like F 2 is occupied by bonding electrons, which are electrostatically stabilized by attraction to both nuclei. The electrons also provide a buffer between the two nuclei. (a) (b) Z eff = Z – S. According to our simple model, moving from C to N, causes Z to increase by 1 and S to remain the same, so Z eff is greater by 1 for N than for C. Using Slater’s rules, Z increases by 1 moving from C to N, but S also increases by 0.35, the value for an electron with the same n value as the one of interest. The increase in Zeff from C to N should be (1 − 0.35) = 0.65. In this case, Slater’s rules predict the change in Zeff more accurately than the simple model in (a). In O, [He]2s 2 2p 4 , one of the p orbitals is doubly occupied, which significantly increases electron-electron repulsion. This leads to an overall higher energy for the 2p electrons. In our model, this appears as a larger S and a smaller Z eff for O. The change from N to O is then smaller. 7.87 (c) (d) 7.88 Atomic size (bonding atomic radius) is strongly correlated to Z eff, which is determined by Z and S. Moving across the representative elements, electrons added to ns or np valence orbitals do not effectively screen each other. The increase in Z is not accompanied by a similar increase in S; Z eff increases and atomic size decreases. Moving across the transition elements, electrons are added to (n–1)d orbitals and become part of the core electrons, which do significantly screen the ns valence electrons. The increase in Z is accompanied by a larger increase in S for the ns valence electrons; Z eff increases more slowly and atomic size decreases more slowly. 180 7 Periodic Properties of the Elements
7.89 (a) bonded atoms P–H As – H Sb – H estimated distance 1.43 1.56 1.75 Solutions to Exercises The estimated distances in the table below are the sum of the radii of the group 5A elements and H from Figure 7.7. measured distance 1.419 1.519 1.707 In general, the estimated distances are a bit longer than the measured distances. This probably shows a systematic bias in either the estimated radii or in the method of obtaining the measured values. (b) The principal quantum number of the outer electrons and thus the average distance of these electrons from the nucleus increases from P (n = 3) to As (n = 4) to Sb (n = 5). This causes the systematic increase in M – H distance. 7.90 Ge – H distance = r Ge + r H = 1.22 + 0.37 = 1.59Å Ge – Cl distance = r Ge + r Cl = 1.22 + 0.99 = 2.21 Å 7.91 Y: [Kr]5s 2 4d 1 , Z = 39 La: [Xe]6s 2 5d 1 , Z = 57 Zr: [Kr] 5s 2 4d 2 , Z = 40 Hf: [Xe] 6s 2 4f 1 45d 2 , Z = 72 The completed 4f subshell in Hf leads to a much larger change in Z going from Zr to Hf (72 – 40 = 32) than in going from Y to La (57 – 39 = 18). The 4f electrons in Hf do not completely shield the valence electrons, so there is also a larger increase in Z eff. This larger increase in Z eff going from Zr to Hf leads to a smaller increase in atomic radius than in going from Y to La. 7.92 (a) (b) (c) Hg 2 + No. According to Figure 2.24, Hg is not essential for life, even in trace amounts. Zn 2 +: [Ar]3d 1 0, r = 0.88 Å; Cd 2 +: [Kr]4d 1 0, r = 1.09 Å; Hg 2 +: [Xe]4f 1 45d 1 0, r ≈ 1.19 Å. If there were no 4f electrons in Hg 2 +, we would expect an ionic radius of around 1.29 Å, an increase of ~0.20 Å due to the increase in principal quantum number of the valence electrons. However, the increase in Z due to filling of the 4f orbitals is not completely offset by shielding. The 5d valence electrons in Hg 2 + experience a greater than expected Z eff which largely offsets the increase in principal quantum number; ionic radius of Hg 2 + is smaller than expected. This phenomenon is known as the lanthanide contraction and affects the physical properties of elements in the sixth period and beyond. See also Solution 7.91. (d) (e) 7.93 (a) Since the ionic radius of Hg 2 + is similar to that of Cd 2 +, Hg 2 + will be physiologically more similar to Cd 2 +. By common knowledge, and verified with WebElements.com™, both Hg and Hg 2 + are extremely toxic to humans. 2Sr(s) + O 2 (g) → 2SrO(s) 181 7 Periodic Properties of the Elements
(b) Solutions to Exercises Assume that the corners of the cube are at the centers of the outermost O 2 – ions, and that the edges pass through the centers of perimeter Sr 2 + ions. The length of an edge is then r(O 2 –) + 2r(Sr 2 +) + r(O 2 –) = 2r(O 2 –) + 2r(Sr 2 +) = 2(1.32 Å) + 2(1.26 Å) = 5.16 Å. Density is the ratio of mass to volume.
d= mass SrO in cube vol cube = # SrO units × mass of SrO vol cube (c) Calculate the mass of 1 SrO unit in grams and the volume of the cube in cm 3 ; solve for number of SrO units. 103.62 g SrO mol × 1 mol SrO 6.022 × 10 23 SrO units
(1 × 10 −8 ) 3 cm 3 Å3 = 1.7207 × 10 − 22 = 1.721 × 10 − 22 g/SrO unit V = ( 5.16) 3 Å 3 × d= = 1.3739 × 10 − 22 = 1.37 × 10 − 22 cm 3 = 5.10 g/cm 3 = 4.07 units # of SrO units × 1.7207 × 10 −22 g/SrO unit 1.3739 × 10 − 22 cm 3 # of SrO units = 5.10 g/cm 3 × 1.3739 × 10 −22 cm 3 1.7207 × 10 − 22 g/SrO unit Since the number of formula units must be an integer, there are four SrO formula units in the cube. Using average values for ionic radii to estimate the edge length probably leads to the small discrepancy. 7.94 C: 1s 2 2s 2 2p 2 . I 1 through I 4 represent loss of the 2p and 2s electrons in the outer shell of the atom. The values of I 1 –I 4 increase as expected. The nuclear charge is constant, but removing each electron reduces repulsive interactions between the remaining electrons, so effective nuclear charge increases and ionization energy increases. I 5 and I 6 represent loss of the 1s core electrons. These 1s electrons are much closer to the nucleus and experience the full nuclear charge (they are not shielded), so the values of I 5 and I 6 are significantly greater than I 1 –I 4 . I 6 is larger than I 5 because all repulsive interactions have been eliminated. The statement is somewhat true, but more accurate if changed to read: “A negative value for the electron affinity of an atom occurs when the outermost electrons only incompletely shield the added electron from the nucleus.” This new statement totally explains the negative electron affinity of Br and the positive value for Kr. For Br – , the electron is added to the 4p subshell and is incompletely shielded by the “other” 4s and 4p electrons. For Kr – , the electron is added to the 5s subshell, which is effectively shielded by the spherical Kr core. O: [He]2s22p4 7.95 7.96 O2–: [He]2s22p6 = [Ne] 182 7 Periodic Properties of the Elements
O 3 –: [Ne]3s 1 Solutions to Exercises The third electron would be added to the 3s orbital, which is farther from the nucleus and more strongly shielded by the [Ne] core. The overall attraction of this 3s electron for the O nucleus is not large enough for O 3 – to be a stable particle. 7.97 (a) P: [Ne] 3s 2 3p 3 ; S: [Ne] 3s 2 3p 4 . In P, each 3p orbital contains a single electron, while in S one 3p orbital contains a pair of electrons. Removing an electron from S eliminates the need for electron pairing and reduces electrostatic repulsion, so the overall energy required to remove the electron is smaller than in P, even though Z is greater. C: [He] 2s 2 2p 2 ; N: [He] 2s 2 2p 3 ; O: [He] 2s 2 2p 4 . An electron added to a N atom must be paired in a relatively small 2p orbital, so the additional electron-electron repulsion more than compensates for the increase in Z and the electron affinity is smaller (less exothermic) than that of C. In an O atom, one 2p orbital already contains a pair of electrons, so the additional repulsion from an extra electron is offset by the increase in Z and the electron affinity is greater (more exothermic). Note from Figure 7.14 that the electron affinity of O is only slightly more exothermic than that of C, although the value of Z has increased by 2. O + : [He] 2s 2 2p 3 ; O 2 +: [He] 2s 2 2p 2 ; F : [He] 2s 2 2p 5 ; F + : [He] 2s 2 2p 4 . Both ‘coreonly’ [Zeff (F) = 7; Zeff (O+) = 6] and Slater [Zeff (F) = 5.2; Zeff (O+) = 4.9] predict that F has a greater Zeff than O+. Variation in Zeff does not offer a satisfactory explanation. The decrease in electron-electron repulsion going from F to F + energetically favors ionization and causes it to be less endothermic than the second ionization of O, where there is no significant decrease in repulsion. Mn 2 +: [Ar]3d 5 ; Mn 3 +; [Ar] 3d 4 ; Cr 2 +: [Ar] 3d 4 ; Cr 3 +: [Ar] 3d 3 ; Fe 2 +: [Ar] 3d 6 ; Fe 3 +: [Ar] 3d 5 . The third ionization energy of Mn is expected to be larger than that of Cr because of the larger Z value of Mn. The third ionization energy of Fe is less than that of Mn because going from 3d 6 to 3d 5 reduces electron repulsions, making the process less endothermic than predicted by nuclear charge arguments. The group 2B metals have complete (n–1)d subshells. An additional electron would occupy an np subshell and be substantially shielded by both ns and (n–1)d electrons. Overall this is not a lower energy state than the neutral atom and a free electron. Valence electrons in Group 1B elements experience a relatively large effective nuclear charge due to the buildup in Z with the filling of the (n–1)d subshell. Thus, the electron affinities are large and negative. Group 1B elements are exceptions to the usual electron filling order and have the generic electron configuration ns 1 (n–1)d 1 0. The additional electron would complete the ns subshell and experience repulsion with the other ns electron. Going down the group, size of the ns subshell increases and repulsion effects decrease. That is, effective nuclear charge is greater going down the group because it is less diminished by repulsion, and electron affinities become more negative. (b) (c) (d) 7.98 (a) (b) 183 7 Periodic Properties of the Elements
7.99 (a) Solutions to Exercises For both H and the alkali metals, the added electron will complete an ns subshell (1s for H and ns for the alkali metals) so shielding and repulsion effects will be similar. For the halogens, the electron is added to an np subshell, so the energy change is likely to be quite different. True. Only He has a smaller estimated “bonding” atomic radius, and no known compounds of He exist. The electron configuration of H is 1s 1 . The single 1s electron experiences no repulsion from other electrons and feels the full unshielded nuclear charge. It is held very close to the nucleus. The outer electrons of all other elements that form compounds are shielded by a spherical inner core of electrons and are less strongly attracted to the nucleus, resulting in larger atomic radii. Ionization is the process of removing an electron from an atom. For the alkali metals, the ns electron being removed is effectively shielded by the core electrons, so ionization energies are low. For the halogens, a significant increase in nuclear charge occurs as the np orbitals fill, and this is not offset by an increase in shielding. The relatively large effective nuclear charge experienced by np electrons of the halogens is similar to the unshielded nuclear charge experienced by the H 1s electron. Both H and the halogens have large ionization energies. (b) (c) 7.100 Since Xe reacts with F 2 , and O 2 has approximately the same ionization energy as Xe, O 2 will probably react with F 2 . Possible products would be O 2 F 2 , analogous to XeF 2 , or OF 2 . O 2 (g) + F 2 (g) → O 2 F 2 (g) O 2 (g) + 2F 2 (g) → 2OF 2 (g) 7.101 O 2 < Br 2 < K < Mg. O 2 and Br 2 are (nonpolar) nonmetals. We expect O 2 , with the much lower molar mass, to have the lower melting point. This is confirmed by data in Tables 7.6 and 7.7. K and Mg are metallic solids (all metals are solids), with higher melting points than the two nonmetals. Since alkaline earth metals (Mg) are typically harder, more dense and higher melting than alkali metals (K), we expect Mg to have the highest melting point of the group. This is confirmed by data in Tables 7.4 and 7.5. Sr: [Kr]5s2, Z = 38. Ca: [Ar]4s2, Z = 20. Sr and Ca are in the same family. They have the same valence electron configuration so we expect their chemical properties to be similar. Both are metals that form cations with +2 charge. From trends, we expect Sr to have a larger covalent radius than Ca, and Sr2+ to have a larger ionic radius than Ca2+; Figure 7.8 corroborates this relationship. However, the size difference is not as great as we might expect. The [Kr] core of Sr includes a full 3d subshell, which increases the Z and Zeff for Sr and Sr2+ relative to Ca and Ca2+. Ca and Sr both react with water at room temperature to form Ca2+(aq) and Sr2+(aq). The predominant species in water supplies (very dilute aqueous solutions) and in the form for uptake by plants are Ca2+(aq) and Sr2+(aq). The main factors that determine transport properties are size and charge. Since Ca2+ and Sr2+ have the same charge and similar size, it is not surprising that they are transported together and assimilated by organisms similarly. Since they have similar properties, they are likely to interact with 7.102 184 7 Periodic Properties of the Elements Solutions to Exercises bio-molecules similarly. So, “normal” (nonradioactive) Sr acts like Ca, which is not only safe, but essential to organisms. This is why “normal” Sr is not very dangerous, but radioactive Sr is lethal. Radioactive Sr replaces Ca and is readily incorporated into organisms. 7.103 Moving one place to the right in a horizontal row of the table, for example, from Li to Be, there is an increase in ionization energy. Moving downward in a given family, for example from Be to Mg, there is usually a decrease in ionization energy. Similarly, atomic size decreases in moving one place to the right and increases in moving downward. Thus, two elements such as Li and Mg that are diagonally related tend to have similar ionization energies and atomic sizes. This in turn gives rise to some similarities in chemical behavior. Note, however, that the valences expected for the elements are not the same. That is, lithium still appears as Li + , magnesium as Mg 2 +. (a) Plan. Use qualitative physical (bulk) properties to narrow the range of choices, then match melting point and density to identify the specific element. Solve. Hardness varies widely in metals and nonmetals, so this information is not too useful. The relatively high density, appearance, and ductility indicate that the element is probably less metallic than copper. Focus on the block of nine main group elements centered around Sn. Pb is not a possibility because it was used as a comparison standard. The melting point of the five elements closest to Pb are: Tl, 303.5°C; In, 156.1°C; Sn, 232°C; Sb, 630.5°C; Bi, 271.3°C The best match is In. To confirm this identification, the density of In is 7.3 g/cm 3 , also a good match to properties of the unknown element. (b) In order to write the correct balanced equation, determine the formula of the oxide product from the mass data, assuming the unknown is In. 5.08 g oxide – 4.20 g In = 0.88 g O 4.20 g In/114.82 g/mol = 0.0366 mol In; 0.0366/0.0366 = 1 0.88 g O/16.00 g/mol = 0.0550 mol O; 0.0550/0.0366 = 1.5 Multiplying by 2 produces an integer ratio of 2 In: 3 O and a formula of In 2 O 3 . The balanced equation is: 4 In(s) + 3O 2 (g) → 2 In 2 O 3 (s) (c) According to Figure 7.2, the element In was discovered between 1843–1886. The investigator who first recorded this data in 1822 could have been the first to discover In. 7.104 Integrative Exercises
7.105 (a) ν = c/λ; 1 Hz = 1 s – 1
Ne: ν = Ca : ν = 2.998 × 10 8 m/s 14.610 Å 2.998 × 10 m/s 3.358 × 10 −10 m
8 × 1Å 1 × 10
−10 m = 2.052 × 10 17 s −1 = 2.052 × 10 17 Hz = 8.928 × 10 17 Hz 185 7 Periodic Properties of the Elements
Zn : ν = Zr : ν = Sn : ν = 2.998 × 10 8 m/s 1.435 × 10 −10 m 2.998 × 10 8 m/s 0.786 × 10 −10 m 2.998 × 10 8 m/s 0.491 × 10 −10 m = 20.89 × 10 17 Hz Solutions to Exercises = 38.14 × 10 17 = 38.1 × 10 17 Hz = 61.06 × 10 17 = 61.1 × 10 17 Hz (b) Element Ne Ca Zn Zr Sn Z 10 20 30 40 50 ν
2.052 × 10 1 7 8.928 × 10 1 7 20.89 × 10 1 7 38.14 × 10 1 7 61.06 × 10 1 7 ν 1 /2
4.530 × 10 8 9.449 × 10 8 14.45 × 10 8 19.5 × 10 8 24.7 × 10 8 (c) The plot in part (b) indicates that there is a linear relationship between atomic number and the square root of the frequency of the X-rays emitted by an element. Thus, elements with each integer atomic number should exist. This relationship allowed Moseley to predict the existence of elements that filled “holes” or gaps in the periodic table. For Fe, Z = 26. From the graph, ν 1 /2 = 12.5 × 10 8 , ν = 1.56 × 10 1 8 Hz.
λ = c/ν = 2.998 × 10 8 m/s 1.56 × 10
18 −1 (d) s × 1Å 1 × 10 −10 m = 1.92 Å (e) λ = 0.980 Å = 0.980 × 10 – 10 m
ν = c/λ = 2.998 × 10 8 m/s 0.980 × 10 −10 m = 30.6 × 10 17 Hz; ν 1 / 2 = 17.5 × 10 8 From the graph, ν 1 /2 = 17.5 × 10 8 , Z = 36. The element is krypton, Kr. 186 7 Periodic Properties of the Elements
7.106 (a) (b) Solutions to Exercises Li: [He]2s 1 . Assume that the [He] core is 100% effective at shielding the 2s valence electron Z eff = Z – S ≈ 3 – 2 = 1+. The first ionization energy represents loss of the 2s electron. ΔE = energy of free electron (n = ∞) – energy of electron in ground state (n = 2) ΔE = I 1 = [–2.18 × 10 – 18 J (Z 2 /∞2)] – [–2/18 × 10 – 18 J (Z 2 /2 2 ) ] ΔE = I 1 = 0 + 2.18 × 10 – 18 J (Z 2 /2 2 )
For Li, which is not a one-electron particle, let Z = Z eff. ΔE ≈ 2.18 × 10 – 18 J (+1 2 /4) ≈ 5.45 × 10 – 19 J/atom
(c) Change the result from part (b) to kJ/mol so it can be compared to the value in
Table 7.4. 5.45 × 10 −19 6.022 × 10 23 atom 1 kJ J × × = 328 kJ/mol atom mol 1000 J The value in Table 7.4 is 520 kJ/mol. This means that our estimate for Z eff was a lower limit, that the [He] core electrons do not perfectly shield the 2s electron from the nuclear charge. (d) From Table 7.4, I 1 = 520 kJ/mol.
520 kJ mol × 1000 J kJ × 1 mol 6.022 × 10
23 atoms = 8.6350 × 10 −19 J/atom Use the relationship for I 1 and Z eff developed in part (b).
2 Z eff = 4(8.6350 × 10 −19 J) 2.18 × 10 −18 J = 1.5844 = 1.58; Z eff = 1.26 This value, Zeff = 1.26, based on the experimental ionization energy, is greater than our estimate from part (a), which is consistent with the explanation in part (c). 7.107 (a) E = hc/λ; 1 nm = 1 × 10 – 9 m; 58.4 nm = 58.4 × 10 – 9 m; 1 eV = 96.485 kJ/mol, 1 eV - mol = 96.485 kJ
E= 6.626 × 10 −34 J - s × 2.998 × 10 8 m/s 58.4 × 10 −9 m = 3.4015 × 10 −18 = 3.40 × 10 −18 J/photon 3.4015 × 10 −18 J 1 kJ 6.022 × 10 23 photons 1 eV - mol × × × = 21.230 = 21.2 eV photon 1000 J mol 96.485 kJ (b) (c) Hg(g) → Hg + (g) + 1e – I 1 = E 58.4 – E K = 21.23 eV 10.75 eV = 10.48 = 10.5 eV
10.48 eV × 96.485 kJ = 1.01 × 10 3 kJ/mol 1 eV - mol (d) From Figure 7.12, iodine (I) appears to have the ionization energy closest to that of Hg, approximately 1000 kJ/mol. 187 7 Periodic Properties of the Elements
7.108 (a) Na(g) → Na + (g) + 1e – Cl(g) + 1e – → Cl – (g) Na(g) + C1(g) → Na + (g) + Cl – (g) (b) (c) Solutions to Exercises (ionization energy of Na) (electron affinity of Cl) ΔH = I 1 (Na) + E 1 (Cl) = +496 kJ – 349 kJ = +147 kJ, endothermic
The reaction 2Na(s) + Cl 2 (g) → 2NaCl(s) involves many more steps than the reaction in part (a). One important difference is the production of NaCl(s) versus NaCl(g). The condensation NaCl(g) → NaCl(s) is very exothermic and is the step that causes the reaction of the elements in their standard states to be exothermic, while the gas phase reaction is endothermic. Mg 3 N 2 Mg 3 N 2 (s) + 3H 2 O(l) → 3MgO(s) + 2NH 3 (g) The driving force is the production of NH 3 (g). 7.109 (a) (b) (c) After the second heating, all the Mg is converted to MgO. Calculate the initial mass Mg.
0.486 g MgO × 24.305 g Mg 40.305 g MgO = 0.293 g Mg x = g Mg converted to MgO; y = g Mg converted to Mg 3 N 2 ; x = 0.293 – y
⎛ 40.305 g MgO ⎞ ⎛ 100.929 g Mg 3 N 2 ⎟ ⎜ g MgO = x ⎜ ⎜ 24.305 g Mg ⎟ ; g Mg 3 N 2 = y ⎜ 72.915 g Mg ⎝ ⎠ ⎝ ⎞ ⎟ ⎟ ⎠ g MgO + g Mg 3 N 2 = 0.470
⎛ 40.305 ⎞ ⎛ 100.929 ⎞ (0.293 − y) ⎜ ⎟ + y⎜ ⎟ = 0.470 ⎝ 24.305 ⎠ ⎝ 72.915 ⎠ (0.293 – y)(1.6583) + y(1.3842) = 0.470 –1.6583 y + 1.3842 y = 0.470 – 0.48588 –0.2741 y = –0.016 y = 0.05794 = 0.058 g Mg in Mg 3 N 2
g Mg 3 N 2 = 0.05794 g Mg × mass % Mg 3 N 2 = 100.929 g Mg 3 N 2 72.915 g Mg = 0.0802 = 0.080 g Mg 3 N 2 0.0802 g Mg 3 N 2 0.470 g (MgO + Mg 3 N 2 ) × 100 = 17% (The final mass % has two sig figs because the mass of Mg obtained from solving simultaneous equations has two sig figs.) 188 7 Periodic Properties of the Elements
(d) 3Mg(s) + 2NH 3 (g) → Mg 3 N 2 (s) + 3H 2 (g)
6.3 g Mg × 1 mol Mg = 0.2592 = 0.26 mol Mg 24.305 g Mg Solutions to Exercises 2.57 g NH 3 × 1 mol NH 3 = 0.1509 = 0.16 mol NH 3 17.031 g NH 3 2 mol NH 3 = 0.1728 = 0.17 mol NH 3 3 mol Mg 0.2592 mol Mg × 0.26 mol Mg requires more than the available NH 3 so NH 3 is the limiting reactant.
0.1509 mol NH 3 × 3 mol H 2 2 mol NH 3 × 2.016 g H 2 mol H 2 = 0.4563 = 0.46 g H 2 (e) ° ΔH rxn = ΔH ° Mg 3 N 2 (s) + 3 ΔH ° H 2 (g ) − 3ΔH ° Mg(s) − 2ΔH ° NH 3 (g ) f f f f = –461.08 kJ + 0 – 3(0) – 2(–46.19) = –368.70 kJ 7.110 (a) (b) (c) r Bi = r BiBr3 − r Br = 2.63 Å − 1.14 Å = 1.49 Å
Bi 2 O 3 (s) + 6HBr(aq) → 2BiBr 3 (aq) + 3H 2 O(l) Bi 2 O 3 is soluble in acid solutions because it act as a base and undergoes acidbase reactions like the one in part (b). It is insoluble in base because it cannot acts as an acid. Thus, Bi 2 O 3 is a basic oxide, the oxide of a metal. Based on the properties of its oxide, Bi is characterized as a metal. Bi: [Xe]6s 2 4f 1 45d 1 06p 3 . Bi has five outer electrons in the 6p and 6s subshells. If all five electrons participate in bonding, compounds such as BiF 5 are possible. Also, Bi has a large enough atomic radius (1.49 Å) and low-energy orbitals available to accommodate more than four pairs of bonding electrons. The high ionization energy and relatively large negative electron affinity of F, coupled with its small atomic radius, make it the most electron withdrawing of the halogens. BiF 5 forms because F has the greatest tendency to attract electrons from Bi. Also, the small atomic radius of F reduces repulsions between neighboring bonded F atoms. The strong electron withdrawing properties of F are also the reason that only F compounds of Xe are known. 4KO 2 (s) + 2CO 2 (g) → 2K 2 CO 3 (s) + 3O 2 (g) K, +1; O, –1/2 (O 2 – is superoxide ion); C, +4; O, –2 → K, +1; C, +4; O, –2; O, 0
18.0 g CO 2 × 18.0 g CO 2 × 1 mol CO 2 44.01 g CO 2 1 mol CO 2 44.01 g CO 2 × × 4 mol KO 2 2 mol CO 2 3 mol O 2 2 mol CO 2 × × 71.10 g KO 2 1 mol KO 2 32.00 g O 2 1 mol O 2 = 58.2 g KO 2 (d) (e) 7.111 (a) (b) (c) = 19.6 g O 2 189 ...
View Full Document