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Homework Problems

Homework Problems Chapter 7 Homework Problems: 5, 6, 12, 18, 27, 28, 30, 34, 42, 46, 51, 58 a-e, 60 a-e, 65, 76, 82, 84, 96, 116. CHAPTER 7 Electron Configuration and the Periodic Table. Electron Configuration and the Periodic Table

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Homework Problems

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  1. Homework Problems Chapter 7 Homework Problems: 5, 6, 12, 18, 27, 28, 30, 34, 42, 46, 51, 58 a-e, 60 a-e, 65, 76, 82, 84, 96, 116

  2. CHAPTER 7 Electron Configuration and the Periodic Table

  3. Electron Configuration and the Periodic Table The organization of the periodic table reflects the electron configurations for the elements. Elements with the same (or similar) valence shell electron configurations have similar chemical and physical properties, and are grouped together. Example: Halogens F 1s2 2s2 2p5 = [He] 2s2 2p5 ( [He] 2s2 2p5 3s0 ) Cl 1s2 2s2 2p6 3s2 3p5 = [Ne] 3s2 3p5 Br 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 = [Ar] 4s2 3d10 4p5 Note that all of the halogens have the ns2 np5 configuration for their valence electrons, resulting in similarities in their properties.

  4. Organization of the Periodic Table

  5. Main Group Elements The valence electrons for the first 18 elements of the periodic table are given below. Notice how elements in the same group have the same number and type of valence electrons.

  6. Valence Electron Configuration of the Elements

  7. Free Elements in Chemical Equations Metals - These do not exist as individual “molecules” but (in the solid state) as crystals. Metals are therefore represented by the symbol for the element (Fe(s), Cu(s), Hg(l)) Metalloids - Like metals, these elements do not exist as individual molecules, and so they are represented in the same way as metals (B(s), Si(s), Ge(s)). Nonmetals - Representation can vary depending on the element Noble gases - He(g), Ne(g), Ar(g)… Diatomic molecules - H2(g), N2(g), O2(g), F2(g), Cl2(g), Br2(l), I2(s) Other nonmetals - C(s), P(s), S(s),Se(s). We assume the most stable solid crystal structure. Technically some of the above nonmetals exist as polyatomic molecules (P4, S8), but we generally ignore this.

  8. Effective Nuclear Charge (Zeff) It is useful in making predictions concerning the properties of atoms and ions (called periodic properties) to use a concept called the effective nuclear charge. effective nuclear charge - The nuclear charge seen by the valence electrons. This is different than the charge of the nucleus, because some of the nuclear charge is screened (blocked) by the core electrons.

  9. Approximate Value for Effective Nuclear Charge (Zeff) There is way of determining values for the effective nuclear charge experimentally. For our purposes, we will use an approximation method for finding Zeff Zeff (atomic number) - (number of core electrons) The above assumes that each core electron reduces the charge seen by the valence electrons by one unit. While crude, this approximation is sufficient for use in determining trends in periodic properties. Effective nuclear charge is most useful in discussions involving the main group elements.

  10. Finding Zeff If we know the electron configuration for an atom we can determine the number of core and valence electrons. From this, we can find the effective nuclear charge. Example: # core # valence O(8 e-) 1s2 2s2 2p4 2 6 Zeff = 8 - 2 = +6 Br(35 e-) [Ar] 4s2 3d10 4p5 28 7 Zeff = 35 - 28 = +7 Notice that the effective nuclear charge is just equal to the number of valence electrons.

  11. Zeff and Coulomb’s Law The attractive force acting between particles of opposite charge is described by Coulomb’s Law: F ~ Q1 Q2 Q1, Q2 = charge of particles d2 d = distance between particles When Q1 and Q2 are of opposite sign, the force is attractive. If we apply this to the attraction of a valence electron by the effective nuclear charge, then F ~ - (Zeff) (1) d2 The attractive force is stronger as Zeff increases in size and d decreases in size.

  12. Sizes of Atoms There are several ways in which values for size can be assigned for an atom. Nonbonding atomic radius (van der Waals radius) - Found from distance between nuclei of adjacent atoms in nonmetallic atomic solids. Bonding atomic radius - Found from distance between nuclei in diatomic molecules (nonmetals) or between adjacent atoms in metallic solids.

  13. Trends in Sizes of Atoms There are two general cases for which we can make predictions about the relative sizes of atoms. 1) Atoms in the same group of the periodic table. Since these atoms all have the same number of valence electrons (except for He in group 8A) the size depends on the value of n for the valence electrons. The general trend is that atomic size increases from top to bottom within a group. element* configuration radius (nm) O [He] 2s2 2p4 0.073 S [Ne] 3s2 3p4 0.103 Se [Ar] 4s2 3d10 4p4 0.117 Te [Kr] 5s2 4d10 5p4 0.143 *Note that for all of these atoms Zeff = +6.

  14. 2) Atoms in the same period (row) in the periodic table. In this case, the atoms all have the same value of n for the valence electrons. The determining factor is the value for Zeff, the effective nuclear charge. The larger the value for Zeff, the smaller the atom. Example: Consider the elements in the second period. element configuration Zeff radius (nm) Li 1s2 2s1 +1 0.152 Be 1s2 2s2 +2 0.112 B 1s2 2s2 2p1 +3 0.085 C 1s2 2s2 2p2 +4 0.077 N 1s2 2s2 2p3 +5 0.075 O 1s2 2s2 2p4 +6 0.073 F 1s2 2s2 2p5 +7 0.072 Ne 1s2 2s2 2p6 +8 0.070

  15. Sizes of Main Group Atoms

  16. Sizes of Ions There are three general cases we will consider. 1) Ions with the same charge within a group. In this case, Zeff will be the same for all the ions, and so the size of the ion will be controlled by the largest value of n for the electrons present. The general trend will be that size (radius) increases from top to bottom. Be2+ 0.031 nm 1s2 = [He] Mg2+ 0.065 nm 1s22s22p6 = [Ne] Ca2+ 0.099 nm 1s22s22p63s23p6 = [Ar] F- 0.136 nm 1s22s22p6 = [Ne] Cl- 0.181 nm 1s22s22p63s23p6 = [Ar] Br- 0.195 nm 1s22s22p63s23p64s23d104p6 = [Kr]

  17. 2) Different ions/atoms of the same element. In this case, the smaller the number of electrons (more positive the ion) the smaller the ion. Cu (29 e-) 0.128 nm 1s22s22p63s23p64s13d10 Cu+ (28 e-) 0.096 nm 1s22s22p63s23p63d10 Cu2+ (27 e-) 0.072 nm 1s22s22p63s23p63d9 Cl (17 e-) 0.099 nm 1s22s22p63s23p5 Cl- (18 e-) 0.181 nm 1s22s22p63s23p6

  18. 3) Ions/atoms with the same number of electrons. In this case, the larger the charge of the nucleus (atomic number) the smaller the ion. atom/ion Z radius (nm) N3- +7 0.171 O2- +8 0.140 F- +9 0.136 Ne +10 - * 1s22s22p6 Na+ +11 0.095 Mg2+ +12 0.065 Al3+ +13 0.050 In the above example all the atoms/ions have 10 electrons. Species with the same number and configuration of electrons are called isoelectronic. * Ion sizes are determined from studies of binary ionic compounds, and so cannot be directly compared to the size for Ne, which is determined in the gas phase or the condensed solid.

  19. Ionization Energy The ionization energy is defined as the energy required to remove one electron from an atom, ion, or molecule in the gas phase. X(g)  X+(g) + e- IE1 Since atoms in general have a large number of electrons, we can define the first ionization energy as the energy required to remove the first electron, the second ionization energy as the energy required to remove the second electron, and so forth. O(g)  O+(g) + e- 1st ionization enegy IE1 O+(g)  O2+(g) + e- 2nd ionization energy IE2 O2+(g)  O3+(g) + e- 3rd ionization energy IE3 : : : : : : : : O7+(g)  O8+(g) + e- 8th ionization energy IE8 Note that it will always be true that IE1 < IE2 < IE3 < ...

  20. Trends in 1st Ionization Energy 1) Atoms in the same group of the periodic table. Since all atoms in the same group have the same effective nuclear charge for the valence electrons, the determining factor for the ionization energy is the value of n for the valence electrons. The larger the value of n the further the valence electrons are from the nucleus, and so the easier to remove a valence electron. So 1st ionization energy decreases in going from top to bottom within a group. element* configuration IE1 (kJ/mol) Li [He] 2s1 520. Na [Ne] 3s1 496. K [Ar] 4s1 419. Rb [Kr] 5s1 403. * In all cases Zeff = +1.

  21. 2) Atoms in the same period (row) in the periodic table. The valence electrons all have the same value for n, and so it is Zeff that determines the ionization energy. The larger the value for Zeff the more difficult it is to remove an electron. So 1st ionization energy generally increases in going from left to right within a period. element configuration Zeff IE1 (kJ/mol) Li [He] 2s1 +1 520. Be [He] 2s2 +2 899. B [He] 2s2 2p1 +3 801. C [He] 2s2 2p2 +4 1086. N [He] 2s2 2p3 +5 1402. O [He] 2s2 2p4 +6 1314. F [He] 2s2 2p5 +7 1681. Ne [He] 2s2 2p6 +8 2081.

  22. Higher Ionization Energies As noted, ionization energies always increase as we go from removing the first electron to the second, third, and so forth. However, when we have removed all the valence electrons and begin removing core electrons there is a big jump in “Zeff”, and so usually a large jump in ionization energy, as can be seen below. atom/ion - configuration “Zeff” IE (kJ/mol) reaction Na 1s2 2s2 2p6 3s1 +1 496. Na(g)  Na+(g) + e- Na+ 1s2 2s2 2p6 +9 4560. Na+(g)  Na2+(g) + e- Na2+ 1s2 2s2 2p5 +9 6910. Na2+(g)  Na3+(g) + e- Mg 1s2 2s2 2p6 3s2 +2 738. Mg(g)  Mg+(g) + e- Mg+ 1s2 2s2 2p6 3s1 +2 1450. Mg+(g)  Mg2+(g) + e- Mg2+1s2 2s2 2p6 +10 7730. Mg2+(g)  Mg3+(g) + e-

  23. Ionization Energies For Elements 3-11 Notice that while the ionization always increases as more and more electrons are removed, the biggest jump occurs after all the valence electrons have been removed.

  24. Electron Affinity Electron affinity is defined as the energy change when an atom or molecule adds an electron in the gas phase. X(g) + e- X-(g) EA Unlike ionization energy, electron affinity can be either positive or negative. EA > 0 Must add energy to force the atom to add an electron EA < 0 Energy is released when atom adds an electron If EA is negative that means the atom wants to add an electron.

  25. Trends in Electron Affinity While there are no strong trends for electron affinity we can make the following observations. 1) EA is large and negative for the halogens. atom configuration EA (kJ/mol) F [He] 2s2 2p5 - 328. Cl [Ne] 3s2 3p5 - 349. Br [Ar] 4s2 3d10 4p5 - 325. I [Kr] 5s2 4d10 5p5 - 298. Explanation - There is space for one more electron in the valence shell. That electron will see a large effective nuclear charge (Zeff = +7) and so it is easy to add.

  26. Trends in Electron Affinity 2) EA is large and positive for the noble gases. Ne(g) + e- Ne-(g) Ne 1s2 2s2 2p6 Ne- 1s2 2s2 2p6 3s1 Explanation – The electron being added is going to a higher energy orbital. In fact, it will see an effective nuclear charge Zeff = 0, and so there is no strong attraction of the electron towards the atom. In fact, it takes energy to add the electron due to electron-electron repulsion.

  27. Examples of “Trend Questions” 1) Place the following in order from largest to smallest a) Ca, Cl, Mg b) P3-, S2-, Cl- 2) Which of the following has the largest 1st ionization energy? a) Cl or Br b) C or O

  28. Examples of “Trend Questions” 1) Place the following in order from largest to smallest a) Ca, Cl, Mg Ca > Mg ; Mg > Cl so Ca > Mg > Cl b) P3-, S2-, Cl- All have 18 e-, and so P3- > S2- > Cl- 2) Which of the following has the largest 1st ionization energy? a) Cl or Br Both are in the same group, so chlorine is larger. b) C or O Both are in the same row, so oxygen is larger.

  29. Ion Formation By knowing the electron configuration of an atom we can often predict the ion that will form from that atom. That is particularly true for main group elements. The general tendency for the main group elements is to either add electrons to have completely filled outer s and p orbitals, or to lose electrons to completely empty the outermost s and p orbitals. Which of the above will occur depends on which process involves the fewest electrons. In terms of a general statement we can say the following: Main group (representative) elements tend to add or lose electrons to achieve an electron configuration that is the same as that of the nearest noble gas.

  30. Examples Group 2A elements – 2 valence electron Be 1s2 2s2 = [He]2s2 Mg 1s2 2s22p63s2 = [Ne]3s2 Ca 1s2 2s2 2p6 3s2 3p6 4s2 = [Ar]4s2 In all cases, the outermost s and p orbitals can be emptied by removing two electron (the valence electrons). So the ions expected to form are Be2+, Mg2+, and Ca2+. Group 6A elements – 6 valence electrons O 1s2 2s2 2p4 = [He]2s2 2p4 S 1s2 2s2 2p6 3s2 3p4 = [Ne]3s2 3p4 Se 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4 = [Ar]4s2 3d10 4p4 In this case we can fill the outermost (valence) s and p orbitals by adding two electrons. So the ions expected to form are O2-, S2-, and Se2-.

  31. Common Ions For Main Group Elements The common ions formed from the main group elements are indicated below. In a few cases main group elements can forms ions with different charges. For example, thallium (Tl), a group 3A element, forms ions with a +1 and +3 charge (Tl+, Tl3+). Lead (Pb), a group 4A element, can for ions wit a +2 and +4 charge.

  32. Electron Configurations For Metal Cations We may write electron configurations for ions just as we can for atoms. For anions the same rules used for atoms apply. For cations we must use a slightly different approach. 1) Write the electron configuration for the atom. 2) Remove electrons to form the cation. Start by removing electrons from orbitals with the largest value of n. If there are two orbitals with the same value of n, begin with the orbital with the largest value of . Example: What are the electron configurations for Fe, Fe2+ and Fe3+?

  33. Example: What are the electron configurations for Fe2+ and Fe3+? Fe(26 e-) [Ar] 4s2 3d6 = 1s2 2s2 2p6 3s2 3p6 4s2 3d6 Fe2+(24 e-) [Ar] 3d6 Fe3+(23 e-) [Ar] 3d5 Transition metal ions often form by creating empty s-orbitals, empty d-orbitals, or half filled d-orbitals, but there are many exceptions to this general tendency.

  34. Properties of Metals and Nonmetals Metals Are usually solids at room temperature. Are shiny, lustrous, malleable, and ductile. Are good conductors of heat and electricity. Tend to form cations in ionic compounds. Do not usually form molecular compounds. Form oxides that are ionic and usually are bases. Nonmetals Exhibit a variety of colors, and states (s, l, g). Are not shiny, and are brittle rather than malleable. Are poor conductors of heat and electricity. Tend to form anions in ionic compounds. Form molecular compounds with other nonmetals. Form oxides that are molecular compounds and usually acids.

  35. Examples of Metal and Nonmetal Oxides Metals CaO(s) + H2O(l)  Ca(OH)2(aq) Li2O(s) + H2O(l)  2 LiOH(s) Nonmetals SO3(g) + H2O(l)  H2SO4(aq) N2O5(g) + H2O(l)  2 HNO3(aq)

  36. Metallic Character of Elements Metals tend to lose electrons to form cations, while nonmetals tend to gain electrons to form anions. Based on the periodic trends in ionization energy (which measures how easy it is to remove an electron from an atom) and electron affinity (which measures how easy it is to add an electron to an atom) we can make the following prediction about the “metallic character” of an element. Metallic character decreases in going from bottom to top and from left to right in the periodic table. Metallic character can be related to electronegativity, a concept we will introduce in the next chapter.

  37. Trends in Metallic Character

  38. End of Chapter 7 “When I first saw the periodic table, it hit me with the force of revelation - it embodied, I was convinced, eternal truths...I thought of Mendeleev as a sort of Moses, bearing the tablets of the God-given periodic law.” - Oliver Sachs Two ions are talking to each other in a solution. One says: "Are you a cation or an anion?" The other replies, "Oh, I'm a cation." The first asks, "Are you sure?" The reply - "I'm positive." - anonymous

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