Electron Configuration and Chemical Periodicity

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1 Electron Configuration and Chemical Periodicity 8-1

2 Electron Configuration and Chemical Periodicity 1 Development of the Periodic Table 2 Characteristics of Many-Electron Atoms 3 The Quantum-Mechanical Model and the Periodic Table 4 Trends in Three Key Atomic Properties 5 Atomic Structure and Chemical Reactivity 8-2

3 Table 1 Mendeleev s Predicted Properties of Germanium ( eka Silicon and Its Actual Properties 8-3 Property atomic mass appearance density molar volume specific heat capacity oxide formula oxide density sulfide formula and solubility chloride formula (boiling point) chloride density element preparation Predicted Properties of eka Silicon(E) 72amu gray metal 5.5g/cm 3 13cm 3 /mol 0.31J/g*K EO 2 4.7g/cm 3 ES 2 ; insoluble in H 2 O; soluble in aqueous (NH 4 ) 2 S ECl 4 ; (<100 0 C) 1.9g/cm 3 reduction of K 2 EF 6 with sodium Actual Properties of Germanium (Ge) 72.61amu gray metal 5.32g/cm cm 3 /mol 0.32J/g*K GeO g/cm 3 GeS 2 ; insoluble in H 2 O; soluble in aqueous (NH 4 ) 2 S GeCl 4 ; (84 0 C) 1.844g/cm 3 reduction of K 2 GeF 6 with sodium

4 Figure 1 Observing the Effect of Electron Spin The Stern-Gerlach experiment. 8-4

5 Table 2 Summary of Quantum Numbers of Electrons in Atoms Name Symbol Permitted Values Property principal n positive integers(1,2,3, ) orbital energy (size) angular momentum l integers from 0 to n-1 orbital shape (The l values 0, 1, 2, and 3 correspond to s, p, d, and f orbitals, respectively.) magnetic m l integers from -l to 0 to +l orbital orientation spin m s +1/2 or -1/2 direction of e - spin 8-5

6 Figure 2 Spectral evidence of energy-level splitting in many-electron systems. 8-6

7 Factors Affecting Atomic Orbital Energies The Effect of Nuclear Charge (Z effective ) Higher nuclear charge lowers orbital energy (stabilizes the system) by increasing nucleus-electron attractions. The Effect of Electron Repulsions (Shielding) Additional electron in the same orbital An additional electron raises the orbital energy through electron-electron repulsions. Additional electrons in inner orbitals Inner electrons shield outer electrons more effectively than do electrons in the same sublevel. 8-7

8 Figure 3 The effect of nuclear charge on orbital energy. 8-8

9 Shielding 8-9

10 The effect of orbital shape The two electron of Li have their highest probability in 1s region, but this region is penetrated by the 2s electron. The 2s electron experiences a nuclear charge which is more than +1 and less than +3, called Effective Nuclear Charge Z*. 8-10

11 Order for filling energy sublevels with electrons (n+l rule) Illustrating Orbital Occupancies The electron configuration n l # of electrons in the sublevel as s,p,d,f The orbital diagram (box or circle) 8-11

12 Pauli exclusion principle No atomic orbital can be assigned more than two electrons In an atom, two electrons having the same sequence of quantum numbers do not exist. Helium e a and e b e a : n = 1, l = 0 m l = 0 m s = 1/2 e b : n = 1, l = 0 m l = 0 m s = -1/2 8-12

13 A vertical orbital diagram for the Li ground state no color-empty light - half-filled dark - filled, spin-paired 8-13

14 SAMPLE PROBLEM 1 PROBLEM: Determining Quantum Numbers from Orbital Diagrams Write a set of quantum numbers for the third electron and a set for the eighth electron of the F atom. PLAN: 9F Use the orbital diagram to find the third and eighth electrons. 1s 2s 2p SOLUTION: The third electron is in the 2s orbital. Its quantum numbers are n = 2 l = 0 m l = 0 m s = + or -1/2 The eighth electron is in a 2p orbital. Its quantum numbers are n = 2 l = 1 m l = -1, 0, or +1 m s = + or -1/2 8-14

15 Figure 8 Orbital occupancy for the first 10 elements, H through Ne. 8-15

16 8-16 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

17 Figure 9 Condensed ground-state electron configurations in the first three periods. 8-17

18 Figure 10 Similar reactivites within a group 8-18

19 8-19 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

20 8-20 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

21 Figure 11 A periodic table of partial ground-state electron configurations 8-21

22 Figure 12 The relation between orbital filling and the periodic table 8-22

23 SAMPLE PROBLEM 2 Determining Electron Configuration PROBLEM: Using the periodic table on the inside cover of the text (not Figure 8.12 or Table 8.4), give the full and condensed electrons configurations, partial orbital diagrams showing valence electrons, and number of inner electrons for the following elements: (a) potassium (K: Z = 19) (b) molybdenum (Mo: Z = 42) (c) lead (Pb: Z = 82) PLAN: SOLUTION: (a) for K (Z = 19) Use the atomic number for the number of electrons and the periodic table for the order of filling for electron orbitals. Condensed configurations consist of the preceding noble gas and outer electrons. full configuration 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 condensed configuration partial orbital diagram [Ar] 4s 1 There are 18 inner electrons s 1 3d 4p

24 SAMPLE PROBLEM 2 continued (b) for Mo (Z = 42) full configuration condensed configuration partial orbital diagram 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 5 [Kr] 5s 1 4d 5 There are 36 inner electrons and 6 valence electrons. 5s 1 4d 5 5p (c) for Pb (Z = 82) full configuration 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 2 condensed configuration partial orbital diagram [Xe] 6s 2 4f 14 5d 10 6p 2 There are 78 inner electrons and 4 valence electrons s 2 6p 2

25 Defining metallic and covalent radii 8-25

26 Atomic radii of the maingroup and transition elements. 8-26

27 Periodicity of atomic radius 8-27

28 SAMPLE PROBLEM 3 Ranking Elements by Atomic Size PROBLEM: Using only the periodic table (not Figure 8.15)m rank each set of main group elements in order of decreasing atomic size: (a) Ca, Mg, Sr (b) K, Ga, Ca (c) Br, Rb, Kr (d) Sr, Ca, Rb PLAN: Elements in the same group increase in size and you go down; elements decrease in size as you go across a period. SOLUTION: (a) Sr > Ca > Mg These elements are in Group 2A(2). (b) K > Ca > Ga These elements are in Period 4. (c) Rb > Br > Kr (d) Rb > Sr > Ca Rb has a higher energy level and is far to the left. Br is to the left of Kr. Ca is one energy level smaller than Rb and Sr. Rb is to the left of Sr. 8-28

29 Periodicity of first ionization energy (IE 1 ) 8-29

30 First ionization energies of the main-group elements. 8-30

31 The first three ionization energies of beryllium (in MJ/mol). For more data on sequential ionization energies of the elements, go to or click on the button below. 8-31

32 SAMPLE PROBLEM 4 Ranking Elements by First Ionization Energy PROBLEM: Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE 1 : (a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs PLAN: IE decreases as you proceed down in a group; IE increases as you go across a period. SOLUTION: (a) He > Ar > Kr (b) Te > Sb > Sn (c) Ca > K > Rb (d) Xe > I > Cs Group 8A(18) - IE decreases down a group. Period 5 elements - IE increases across a period. Ca is to the right of K; Rb is below K. I is to the left of Xe; Cs is furtther to the left and down one period. 8-32

33 8-33 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

34 SAMPLE PROBLEM 5 Identifying an Element from Successive Ionization Energies PROBLEM: Name the Period 3 element with the following ionization energies (in kj/mol) and write its electron configuration: IE 1 IE 2 IE 3 IE 4 IE 5 IE ,230 PLAN: Look for a large increase in energy which indicates that all of the valence electrons have been removed. SOLUTION: The largest increase occurs after IE 5, that is, after the 5th valence electron has been removed. Five electrons would mean that the valence configuration is 3s 2 3p 3 and the element must be phosphorous, P (Z = 15). The complete electron configuration is 1s 2 2s 2 2p 6 3s 2 3p

35 Electron affinities of the main-group elements. 8-35

36 Trends in three atomic properties. 8-36

37 Trends in metallic behavior. 8-37

38 The trend in acid-base behavior of element oxides. 8-38

39 Main-group ions and the noble gas configurations. 8-39

40 SAMPLE PROBLEM 6 Writing Electron Configurations of Main-Group Ions PROBLEM: PLAN: SOLUTION: (a) Iodine (Z = 53) is in Group 7A(17) and will gain one electron to be isoelectronic with Xe: I ([Kr]5s 2 4d 10 5p 5 ) + e - I - ([Kr]5s 2 4d 10 5p 6 ) 8-40 Using condensed electron configurations, write reactions for the formation of the common ions of the following elements: (a) Iodine (Z = 53) (b) Potassium (Z = 19) (c) Indium (Z = 49) Ions of elements in Groups 1A(1), 2A(2), 6A(16), and 7A(17) are usually isoelectronic with the nearest noble gas. Metals in Groups 3A(13) to 5A(15) can lose their np or ns and np electrons. (b) Potassium (Z = 19) is in Group 1A(1) and will lose one electron to be isoelectronic with Ar: K ([Ar]4s 1 ) K + ([Ar]) + e - (c) Indium (Z = 49) is in Group 3A(13) and can lose either one electron or three electrons: In ([Kr]5s 2 4d 10 5p 1 ) In + ([Kr]5s 2 4d 10 ) + e + In ([Kr]5s 2 4d 10 5p 1 ) In 3+ ([Kr] 4d 10 ) + 3e -

41 The Period 4 crossover in sublevel energies. 8-41

42 Apparatus for measuring the magnetic behavior of a sample. 8-42

43 SAMPLE PROBLEM 7 Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions PROBLEM: PLAN: SOLUTION: Use condensed electron configurations to write the reaction for the formation of each transition metal ion, and predict whether the ion is paramagnetic. (a) Mn 2+ (Z = 25) (b) Cr 3+ (Z = 24) (c) Hg 2+ (Z = 80) Write the electron configuration and remove electrons starting with ns to match the charge on the ion. If the remaining configuration has unpaired electrons, it is paramagnetic. (a) Mn 2+ (Z = 25) Mn([Ar]4s 2 3d 5 ) Mn 2+ ([Ar] 3d 5 ) + 2e - paramagnetic (b) Cr 3+ (Z = 24) Cr([Ar]4s 2 3d 6 ) Cr 3+ ([Ar] 3d 5 ) + 3e - paramagnetic 8-43 (c) Hg 2+ (Z = 80) Hg([Xe]6s 2 4f 14 5d 10 ) Hg 2+ ([Xe] 4f 14 5d 10 ) + 2e - not paramagnetic (is diamagnetic)

44 Depicting ionic radius. 8-44

45 Ionic vs. atomic radius. 8-45

46 SAMPLE PROBLEM 8 Ranking Ions by Size PROBLEM: Rank each set of ions in order of decreasing size, and explain your ranking: (a) Ca 2+, Sr 2+, Mg 2+ (b) K +, S 2-, Cl - (c) Au +, Au 3+ PLAN: Compare positions in the periodic table, formation of positive and negative ions and changes in size due to gain or loss of electrons. SOLUTION: (a) Sr 2+ > Ca 2+ > Mg 2+ These are members of the same Group (2A/2) and therefore decrease in size going up the group. (b) S 2- > Cl - > K + The ions are isoelectronic; S 2- has the smallest Z eff and therefore is the largest while K + is a cation with a large Z eff and is the smallest. (c) Au + > Au 3+ The higher the + charge, the smaller the ion. 8-46

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