1. Chemistry: Atoms First Julia Burdge & Jason Overby Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 4 Periodic Trends of the Elements Kent L. McCorkle Cosumnes River College Sacramento, CA Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Periodic Trends of the Elements 4 4.1Development of the Periodic Table 4.2The Modern Periodic Table Classification of Elements 4.3Effective Nuclear Charge 4.4Periodic Trends in Properties of Elements Atomic Radius Ionization Energy Electron Affinity Metallic Character 4.5Electron Configuration of Ions Ions of Main Group Elements Ions of d-Block Elements 4.6Ionic Radius Comparing Ionic Radius with Atomic Radius Isoelectronic Series

3. Development of the Periodic Table In 1864, John Newlands noted that when the elements were arranged in order of atomic number that every eighth element had similar properties.  He referred to this as the law of octaves. In 1869, Dmitri Mendeleev and Lothar Meyer independently proposed the idea of periodicity. Mendeleev grouped elements (66) according to properties. Mendeleev predicted properties for elements not yet discovered, such as Ga. 4.1

4. Development of the Periodic Table However, Mendeleev could not explain inconsistencies such as argon coming before potassium in the periodic table, despite having a higher atomic mass. In 1913, Henry Moseley discovered the correlation between the number of protons (atomic number) and frequency of X-rays generated. Ordering the periodic table by atomic number instead of atomic mass enabled scientists to make sense of discrepancies. Entries today include atomic number and symbol; and are arranged according to electron configuration.

5. What elements would you expect to exhibit properties most similar to those of chlorine? Worked Example 4.1 Solution Fluorine, bromine, and iodine, the other nonmetals in Group 7A, should have properties most similar to those of chlorine. Strategy Because elements in the same group tend to have similar properties, you should identify elements in the same group (7A) as chlorine. Think About It Astatine (At) is also in Group 7A. Astatine, though, is classified as a metalloid, and we have to be careful comparing nonmetals to metalloids (or to metals). As a metalloid, the properties of astatine should be less similar to those of chlorine than the other members of Group 7A.

6. The Modern Periodic Table 4.2

7. Classification of Elements The main group elements (also called the representative elements) are the elements in Groups 1A through 7A.

8. Classification of Elements The noble gases are found in Group 8A and have completely filled p subshells.

9. The Modern Periodic Table The transition metals are found in Group 1B and 3B through 8B.  Group 2B have filled d subshells and are not transition metals.

10. The Modern Periodic Table The lanthanides and actinides make up the f-block transition elements.

11. The Modern Periodic Table There is a distinct pattern to the electron configurations of the elements in a particular group. For Group 1A: [noble gas]ns 1 For Group 2A: [noble gas]ns 2

12. The Modern Periodic Table The outermost electrons of an atom are called the valence electrons. Valence electrons are involved in the formation of chemical bonds. Similarity of valence electron configurations help predict chemical properties. For Group 1A: [noble gas]ns 1 valence core For Group 2A: [noble gas]ns 2 valencecore For Group 7A: [noble gas]ns 2 np 5 valencecore

13. Without using a periodic table, give the ground-state electron configuration and block designation (s-, p-, d-, or f-block) of an atom with (a) 17 electrons, (b) 37 electrons, and (c) 22 electrons. Classify each atom as a main group element or transition metal. Worked Example 4.2 Solution (a)1s 2 2s 2 2p 6 3s 2 3p 5, p-block, main group (b)1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1, s-block, main group (c)1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2, d-block, transition metal Strategy Use the figure at right to assign electrons to orbitals in the correct order. Recall that an s subshell has one orbital, a p subshell has three orbitals, and a d subshell has five orbitals. Remember, too, that each orbital can accommodate a maximum of two electrons. Think About It Consult the figure to confirm your answer.

14. Effective Nuclear Charge Effective nuclear charge (Z eff ) is the actual magnitude of positive charge that is “experienced” by an electron in the atom. In a multi-electron atom, electrons are simultaneously attracted to the nucleus and repelled by one another.  This results in shielding, where an electron is partially shielded from the positive charge of the nucleus by the other electrons.  Although all electrons shield one another to some extent, the most effective are the core electrons.  As a result, the value of Z eff increases steadily from left to right because the core electrons remain the same but Z increases. LiBeBCNOF Z3456789 Z eff 1.281.912.423.143.834.455.10 4.3

15. Effective Nuclear Charge In general, the effective nuclear charge is given by  Z is the nuclear charge or simply the number of protons in the nucleus.  σ is the shielding constant. Z eff increases from left to right across a period; changes very little down a column. Z eff = Z – σ LiBeBCNOF Z3456789 Z eff 1.281.912.423.143.834.455.10 4.3

16. Periodic Trends in Properties of Elements Atomic radius is the distance between the nucleus of an atom and its valence shell. 4.4 (a) Atomic radius in metals, or metallic radius, is half the distance between the nuclei of two adjacent, identical metal atoms. (b) Atomic radius in nonmetals, or covalent radius, is half the distance between adjacent, identical nuclei connected by a chemical bond.

17. Atomic Radius The atomic radius increases from top to bottom down a group.  Increasing n, so outermost shell lies farther from the nucleus Atomic radius decreases from left to right across a period.  Increasing Z eff which draws the valence shell closer to the nucleus Atomic radii (in picometers)

18. Atomic Radius Atomic radius decreases left to right across a period due to increased electrostatic attraction between the effective nuclear charge and the charge on the valence shell.

19. Referring only to a periodic table, arrange the elements P, S, and O in order of increasing atomic radius. Worked Example 4.3 Solution O < S < P Strategy Use the left-to-right (decreasing) and top-to-bottom (increasing) trends to compare the atomic radii of two of the three elements at a time. Think About It Consult the atomic radii figure to confirm the order. Note that there are circumstances under which the trends alone will be insufficient to compare the radii of two elements. Using only a periodic table, for example, it would not be possible to determine that bromine (r = 114 pm) has a smaller radius than silicon (r = 118 pm).

20. Ionization Energy Ionization energy (IE) is the minimum energy required to remove an electron from an atom in the gas phase. The result is an ion, a chemical species with a net charge. Sodium has an ionization energy of 495.8 kJ/mol. Specifically, 495.8 kJ/mol is the first ionization energy of sodium, IE 1 (Na), which corresponds to the removal of the most loosely held electron. Na(g) → Na + (g) + e −

21. Ionization Energy

22. In general, as Z eff increases, ionization energy also increases.  Thus, IE 1 increases from left to right across a period.

23. Ionization Energy Within a given shell, electrons with a higher value of l are higher in energy and thus, easier to remove.

24. Ionization Energy Removing a paired electron is easier because of the repulsive forces between two electrons in the same orbital.

25. Ionization Energy It is possible to remove additional electrons in subsequent ionizations, giving IE 1, IE 2, and so on. IE 1 (Na) = 496 kJ/mol IE 2 (Na) = 4562 kJ/mol Na(g) → Na + (g) + e − Na + (g) → Na 2+ (g) + e −

26. Ionization Energy It takes more energy to remove the 2nd, 3rd, 4th, etc. electrons because it is harder to remove an electron from a cation than an atom. It takes much more energy to remove core electrons than valence.  Core electrons are closer to nucleus.  Core electrons experience greater Z eff because of fewer filled shells shielding them from the nucleus.

27. Would you expect Na or Mg to have the greater first ionization energy (IE 1 )? Which should have the greater second ionization energy (IE 2 )? Worked Example 4.4 Solution IE 1 (Mg) > IE 1 (Na) because Mg is to the right of Na in the periodic table (i.e., Mg has the greater Z eff, so it is more difficult to remove its electron). IE 2 (Na) > IE 2 (Mg) because the second ionization of Mg removes a valence electron, whereas the second ionization of Na removes a core electron. Strategy Consider effective nuclear charge and electron configuration to compare the ionization energies. Na has one valence electron and Mg has two. Think About It The first ionization energies of Na and Mg are 496 and 738 kJ/mol, respectively. The second ionization energies of Na and Mg are 4562 and 1451 kJ/mol, respectively.

28. Electron Affinity Electron affinity (EA) is the energy released when an atom in the gas phase accepts an electron. Cl(g) + e − → Cl − (g)

29. Electron Affinity Like ionization energy, electron affinity increases from left to right across a period as Z eff increases.  Easier to add an electron as the positive charge of the nucleus increases.

30. Electron Affinity It is easier to add an electron to an s orbital than to add one to a p orbital with the same principal quantum number.

31. Electron Affinity Within a p subshell, it is easier to add an electron to an empty orbital than to add one to an orbital that already contains an electron.

32. ProcessElectron Affinity More than one electron may be added to an atom. While many first electron affinities are positive, subsequent electron affinities are always negative.  Considerable energy is required to overcome the repulsive forces between the electron and the negatively charged ion. O(g) + e − → O − (g) O − (g) + e − → O 2− (g) EA 1 = 141 kJ/mol EA 2 = −741 kJ/mol

33. For each pair of elements, indicate which one you would expect to have the greater first electron affinity, EA 1 : (a) Al or Si, (b) Si or P. Worked Example 4.5 Solution EA 1 (Si) > EA 1 (Al) because Si is to the right of Al and therefore has a greater Z eff. EA 1 (Si) > EA 1 (P) because although P is to the right of Si, adding an electron to a P atom requires placing it in a partially occupied 3p orbital. The energy cost of pairing electrons outweighs the energy advantage of adding an electron to an atom with a larger Z eff. Strategy Consider effective nuclear charge and electron configuration to compare the ionization energies. (a) Al is in Group 3A and Si is in Group 4A. Al has three valence electrons ([Ne]3s 2 3p 1 ), and Si has four valence electrons ([Ne]3s 2 3p 2 ). (b) P is in Group 5A, so it has five valence electrons ([Ne]3s 2 3p 3 ). Think About It The first electron affinities of Al, Si, and P are 42.5, 134, and 72.0 kJ/mol, respectively.

34. Metallic Character Metals tend to  Be shiny, lustrous, malleable, and ductile  Be good conductors of both heat and electricity  Have low ionization energies (commonly form cations)

35. Metallic Character Nonmetals tend to  Vary in color and are not shiny  Be brittle, rather than malleable  Be poor conductors of both heat and electricity  Have high electron affinities (commonly form anions)

36. Metallic Character Metalloids are elements with properties intermediate between those of metals and nonmetals.

37. Metallic Character Many of the periodic trends of the elements can be explained using Coulomb’s law, which states that the force (F) between two charged objects (Q 1 and Q 2 ) is directly proportional to the product of the two charges and inversely proportional to the distance (d) between the objects squared. F α Q1×Q2d2Q1×Q2d2

38. For carbon and nitrogen, use their effective nuclear charges and atomic radii to compare the attractive force between the nucleus in each atom and the valence electron that would be removed by the first ionization. Worked Example 4.6 Solution For C: F α For N: F α Note that it doesn’t matter what units we use for the distance between the charges. We are not trying to calculate a particular attractive force, only compare the magnitude of these two attractive forces. Strategy Z eff for C and N are 3.14 and 3.83, respectively; the radii of C and N are 77 pm and 75 pm, respectively. IE 1 are 1086 kJ/mol (C) and 1402 kJ/mol (N). The charge on the valence electron in each case is -1. (3.14)×(-1) (77 pm) 2 = -5.3×10 -4 (3.83)×(-1) (75 pm) 2 = -6.8×10 -4 Think About It The negative sign indicates that the force is attractive rather than repulsive. The calculated number for nitrogen is about 28 percent larger than for carbon.

39. Electron Configurations of Ions To write the electron configuration of an ion formed by a main group element: 1) Write the configuration for the atom. 2) Add or remove the appropriate number of electrons. Na: 1s 2 2s 2 2p 6 3s 1 Na + : 1s 2 2s 2 2p 6 Cl: 1s 2 2s 2 2p 6 3s 2 3p 5 Cl − : 1s 2 2s 2 2p 6 3s 2 3p 6 10 electrons total, isoelectronic with Ne 18 electrons total, isoelectronic with Ar 4.5

40. Write electron configurations for the following ions of main group elements: (a) N 3-, (b) Ba 2+, and (c) Be 2+. Worked Example 4.7 Solution (a)[He]2s 2 2p 6 or [Ne] (b)[Kr]5s 2 4d 10 5p 6 or [Xe] (c)1s 2 or [He] Strategy First write electron configurations for the atoms. Then add electrons (for anions) and remove electrons (for cations) to account for the charge. Think About It Be sure to add electrons to form an anion, and remove electrons to form a cation.

41. Ions of d-Block Elements Ions of d-block elements are formed by removing electrons first from the shell with the highest value of n. For Fe to form Fe 2+, two electrons are lost from the 4s subshell not the 3d. Fe can also form Fe 3+, in which case the third electron is removed from the 3d subshell. Fe: [Ar]4s 2 3d 6 Fe 2+ : [Ar]3d 6 Fe: [Ar]4s 2 3d 6 Fe 3+ : [Ar]3d 5

42. Write electron configurations for the following ions of d-block elements: (a) Zn 2+, (b) Mn 2+, and (c) Cr 3+. Worked Example 4.8 Solution (a)[Ar]3d 10 (b)[Ar]3d 5 (c)[Ar]3d 3 Strategy First write electron configurations for the atoms. Then add electrons (for anions) and remove electrons (for cations) to account for the charge. The electrons removed from a d-block element must come first from the outermost s subshell, not the partially filled d subshell. Think About It Be sure to add electrons to form an anion, and remove electrons to form a cation. Also, double-check to make sure that electrons removed from a d-block elment come first from the ns subshell and then, if necessary, from the (n – 1)d subshell.

43. Ionic Radius The ionic radius is the radius of a cation or an anion. When an atom loses an electron to become a cation, its radius decreases due in part to a reduction in electron-electron repulsions in the valence shell. A significant decrease in radius occurs when all of an atom’s valence electrons are removed. 4.6

44. Comparing Ionic Radius with Atomic Radius When an atom gains one or more electrons and becomes an anion, its radius increases due to increased electron- electron repulsions.

45. Isoelectronic Series An isoelectronic series is a series of two or more species that have identical electron configurations, but different nuclear charges. O 2 − : 1s 2 2s 2 2p 6 F − : 1s 2 2s 2 2p 6 Ne: 1s 2 2s 2 2p 6 isoelectronic

46. Identify the isoelectronic series in the following group of species, and arrange them in order of increasing radius: K +, Ne, Ar, Kr, P 3-, S 2-, and Cl -. Worked Example 4.9 Solution The isoelectronic series includes K +, Ar, P 3-, and Cl - (18 electrons each). In order of increasing radius: K + < Ar < Cl - < S 2- < P 3-. Strategy Isoelectronic series are species with identical electron configurations but different nuclear charges. Determine the number of electrons in each species. The radii of isoelectronic series members decreases with increasing nuclear charge. Think About It With identical electron configurations, the attractive force between the valence electrons and the nucleus will be strongest for the largest nuclear charge. Thus, the larger the nuclear charge, the closer in the valence electrons will be pulled and the smaller the radius will be.

47. Chapter Summary: Key Points 4 Development of Periodic Table The Modern Periodic Table Classification of Elements Effective Nuclear Charge Periodic Trends in Properties of Elements Atomic Radius Ionization Energy Electron Affinity Metallic Character Electron Configuration of Ions Ions of Main Group Elements Ions of d-Block Elements Ionic Radius Comparing Ionic Radius with Atomic Radius Isoelectronic Series

48. Write electron configurations for the following atoms/ions. K (long-hand) S 2- (long-hand) Cu (short-hand) Al 3+ (long-hand) Fe 2+ (short-hand) Group Quiz #4 48 48

49. For each periodic trend indicated, identify the atom or ion with a larger value: Radius: Cl or Cl 1- Radius: O 2- or Na 1+ Ionization Eng: Na or K Ionization Eng: CaorSe Electron Affinity: Cor F Group Quiz #5 49