1. Periodic Relationships Among the Elements Chapter 8 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Acknowledgement Thanks to the McGraw-Hill Companies, Inc. for allowing classroom usage.

2. Electron Configurations of the Elements Valence Electrons the outer electrons of an atom, the ones involved in bonding (or chemical reaction) for the representative elements (main group elements), the number of valence electrons is equal to the A Group Number for the transition metal elements (the B group elements), the number of valence electrons is also equal to the B Group Number

3. Example: Which of the following electron configurations do represent similar chemical properties of their atoms? (i)1s 2 2s 2 2p 6 3s 2 (ii) 1s 2 2s 2 2p 3 (iii) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5 (iv) 1s 2 2s 1 (v) 1s 2 2s 2 2p 6 (vi) 1s 2 2s 2 2p 6 3s 2 3p 3 Hint: Two methods to solve this question. (1)Traditional method: Look for the valence electrons (shown in next slide). Atoms with same number of valence electrons are in the same group and thus will have similar chemical properties. 10th ed. Table 7.3, p. 308; 9th ed. Table 7.3, p. 300 and p. 318: Figure 8.2. (2)Easier method: Add total electrons. Since they are the atoms and thus the number of protons = number of electrons. Then find the atom symbol from periodic table to see which are in the same group (vertical column).

4. 8.2 ns 1 ns 2 ns 2 np 1 ns 2 np 2 ns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 d1d1 d5d5 d 10 4f 5f Ground State Electron Configurations of the Elements

5. Electron Configurations of Cations and Anions Na [Ne]3s 1 Na + [Ne] Ca [Ar]4s 2 Ca 2+ [Ar] Al [Ne]3s 2 3p 1 Al 3+ [Ne] Atoms lose electrons so that cation has a noble-gas outer electron configuration. H 1s 1 H - 1s 2 or [He] F 1s 2 2s 2 2p 5 F - 1s 2 2s 2 2p 6 or [Ne] O 1s 2 2s 2 2p 4 O 2- 1s 2 2s 2 2p 6 or [Ne] N 1s 2 2s 2 2p 3 N 3- 1s 2 2s 2 2p 6 or [Ne] Atoms gain electrons so that anion has a noble-gas outer electron configuration. Of Representative Elements 8.2 Remember: electrons are first removed from orbitals with the highest principal quantum number.

6. +1+2+3 -2-3 Cations and Anions Of Representative Elements 8.2

7. Na + : [Ne]Al 3+ : [Ne]F - : 1s 2 2s 2 2p 6 or [Ne] O 2- : 1s 2 2s 2 2p 6 or [Ne]N 3- : 1s 2 2s 2 2p 6 or [Ne] Na +, Mg 2+, Al 3+, F -, O 2-, and N 3- are all isoelectronic with Ne Term: Isoelectronic: have the same number of electrons Li +, H - : 1s 2- same electron configuration as He 8.2

8. Example: 1. Which of the following pairs are not isoelectronic? (a)K + and Cl - (b) Ca 2+ and S 2- (b) Al 3+ and N 3- (d) P 3- and Sc 3+ (c)Rh 3+ and Ir 3+ Answer is (c) because Rh and Ir are in the same group. Isoelectronic species can not be from the same group. 2. Which of the following pairs are not isoelectronic with Ar? Use the choices above. Hint: This question asks two concepts: (1) isoelectronic and (2) number of electrons in Ar. So the answers are (b) and (c).

9. Electron Configurations of Cations of Transition Metals 8.2 When a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n – 1)d orbitals. Fe: [Ar]4s 2 3d 6 Fe 2+ : [Ar]4s 0 3d 6 or [Ar]3d 6 Fe 3+ : [Ar]4s 0 3d 5 or [Ar]3d 5 Mn: [Ar]4s 2 3d 5 Mn 2+ : [Ar]4s 0 3d 5 or [Ar]3d 5

10. Example: What is the ion with +3 charges that has the electron configuration as [Ar]3d 3 ? (a) Sc 3+ (b) Cr 3+ (c) Co 3+ (d) Rh 3+ (e) Ir 3+ Hint: First, figure out the atom symbol for its atomic type. Since the cation has 18+3 = 21 electrons and thus there are 21+3 = 24 protons in this cation. Second, find the atom symbol from the periodic table, which indicates that the atom is Cr and thus the +3 cation is Cr 3+.

11. Example: A metal ion with a net +3 charge has five electrons in the 3d subshell. What is this metal? (a) Cr(b) Mn (c) Fe (d) Co(e) Ni Hint: Similar to the previous question. In order to have electrons in 3d, this ion must be from row number IV and is a transitional metal ion. This species has +3 charges, which indicates that it has three more protons than the electrons. According to the question that it has five electrons in the 3d subshell, and thus the total electrons in valence shells for its atomic type will be 8. Note that the transition metals (with d electrons) losing its s electrons prior to its d electrons and thus the valence electron configuration will be 4s 2 3d 6, which is Fe.

12. Periodic Variation in Physical Properties

13. 8.3

14. Sizes of atoms: Trends size increases while going down a Group Why? Because orbital size increases with increasing n number, the size of atom increases. The highest energy electrons can be farther away from the nucleus. size decreases going across Period Why? effective nuclear charge increases. The larger the effective nuclear charge, the stronger hold of the nucleus on electrons. The electron clouds shrink. Remember n does NOT increases--e - cannot be farther from nucleus.

15. Cation is always smaller than atom from which it is formed. This is because the nuclear charge remains the same but the reduced electron repulsion resulting from removal of electrons make the electron clouds shrink. Anion is always larger than atom from which it is formed. This is because the nuclear charge remains the same but electron repulsion resulting from the additional electron enlarges the electron clouds. 8.3 Ionic Radius

16. From top to bottom, both the atomic and ionic radius increase within group. Across a period the anions are usually larger than cations. For ions derived from different groups, size comparison is meaningful only if the ions are isoelectronic. E.g. Na + (Z=11) is smaller than F - (Z=9). **For isoelectronic series (or species), the larger the effective nuclear charge results in a smaller radius. Simply, the more positively charged ion, smaller the size; the more negatively charged ion, the larger the size. Radius of tripositive ions < dipositive ions<unipositive ions Al 3+ <Mg 2+ <Na + Radius of uninegative ions < dinegative ions O 2- > F - (oxide ion is larger than fluoride ion)

18. Example: Which of the following is a correct order of atomic radii? (a) Ar < P < Na < Ca < Cs(b) Cs < Rb < K < Na < Li (c) Na < Mg < Al < Si < P(d) Rb < Mg < C < F < He (e) Li < H < Al < K < Ar Answer: (a) Example: Which of the following is a correct order of ionic radii? (a)Na + < Mg 2+ < Al 3+ < O 2- < F - < N 3- (b) Al 3+ < Mg 2+ < Na + < F - < O 2- < N 3- (c) F - < O 2- < N 3- < Na + < Mg 2+ < Al 3+ (d) Al 3+ < Mg 2+ < Na + < N 3- < O 2- < F - (e) Na + < Mg 2+ < Al 3+ < O 2- < N 3- < F - Answer: (b)

19. Ionization energy (I.E.) is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state. I 1 + X (g) X + (g) + e - I 2 + X + (g) X 2 + (g) + e - I 3 + X 2+ (g) X 3 + (g) + e - I 1 first ionization energy, remove 1 st electron I 2 second ionization energy, remove 2 nd electron I 3 third ionization energy, remove 3 rd electron 8.4 I 1 < I 2 < I 3

20. General Trend in First Ionization Energies 8.4 Increasing First Ionization Energy

21. First IE: Trends down a Group remember that Cs is more reactive than Na Easier to remove an e- from Cs than from Na As size increases, I.E. decreases Trends across a Period remember that Cl gains rather than loses an e- Easier to remove an e- from Al than from Cl As size decreases, I.E. increases

22. Exceptions: Ionization Energy 1.Occur between Group 2A and 3A Group 3A: ns 2 np 1. one single electron in p orbital. removing the first p electron is less than expected Why? This p electron is shielded by inner ns 2 electrons. Less energy is needed to remove a single p electron than to remove a pair of s electron of the same n level. 2. Occur Group 5A and 6A 5A: ns 2 np 3 6A: ns 2 np 4 removing the fourth p electron is less than expected Why? 2nd electron in a p orbital increases the electron-electron repulsion, which makes it easier to ionize an atom of Group 6A.

23. Example: Arrange the following in order of the increasing first ionization energy: Na, Cl, Al, S, Cs Hint: Ionization energy increases across a row of the periodic table and decreases down a column or group. Cs < Na < Al < S < Cl

24. Electron affinity (EA) is the negative of the energy change that occurs when an electron is accepted by an atom in the gaseous state to form an anion. X (g) + e - X - (g) 8.5 F (g) + e - X - (g) O (g) + e - O - (g)  H = -328 kJ/mol EA = +328 kJ/mol  H = -141 kJ/mol EA = +141 kJ/mol

25. 8.5 Variation of Electron Affinity With Atomic Number (H – Ba) Overall trend: Becomes more negative from left to right across the period, and from bottom to top in the group. The values varied little in the group. The EA of metals are generally lower than those of nonmetals.

26. Properties of Oxides Across a Period basicacidic 8.6 Across a period, oxides change from basic to amphoteric to acidic. Going down a group, the oxides become more basic.

27. Generally speaking, metal oxides and hydroxides are basic; while nonmetal oxides are acidic.