Chapter 8 Electron Configuration and Chemical Periodicity 8-1 Electron Configuration and Chemical Periodicity 8.1 Development of the Periodic Table 8.2 Characteristics of Many-Electron Atoms 8.3 The Quantum-Mechanical Model and the Periodic Table 8.4 Trends in Three Key Atomic Properties 8.5 Atomic Structure and Chemical Reactivity 8-2 1
8-3 The Fourth Quantum Number Three quantum numbers came out of Schroedinger s work, but he was dealing with single electron atoms and ions. Discrepancies became apparent upon examination of multielectron species Stern and Gerlach: fire a beam of H atoms thru a magnet. Unexpected result is that the incomingbeam is split into TWO separate beams, each containing HALF the H atoms. The orientation of the magnetic moment of the electron must also be quantized! Explain the results by introducing a fourth quantum number: the spin quantum number (m s ; either +1/2 (α) or -1/2 (β)) Observing the effect of electron spin. The Stern-Gerlach experiment. 8-4 2
8-5 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 electronelectron repulsions. Additional electrons in inner orbitals Inner electrons shield outer electrons more effectively than do electrons in the same sublevel. Shielding by inner electrons greatly lowers the Z eff felt by outer electrons. 8-6 3
Figure 8.2 Penetration and orbital energy. 8-7 Figure 8.3 Order for filling energy sublevels with electrons. Illustrating Orbital Occupancies The electron configuration n l # of electrons in the sublevel as s,p,d,f The orbital diagram (box or circle) 8-8 4
An orbital diagram for the Li ground state. no color-empty light - half-filled dark - filled, spin-paired 8-9 Determining Quantum Numbers from Orbital Diagrams PROBLEM: Write a set of quantum numbers for the third electron and a set for the eighth electron of the F atom. PLAN: Use the orbital diagram to find the third and eighth electrons. 9 F 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 - 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 1 2 8-10 5
8-11 Figure 8.4 Condensed ground-state electron configurations in the first three periods. 8-12 6
10/17/11 8-13 A periodic table of partial ground-state electron configurations. 8-14 7
Orbital filling and the periodic table. 8-15 PLAN: SOLUTION : (a) for K (Z = 19) Determining Electron Configuration PROBLEM: Using the periodic table on the inside cover of the text (not Figure 8.5 or Table 8.3), give the full and condensed electron 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) 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. 8-16 8
SAMPLE PROBLEM 8.2 continued Determining Electron Configuration (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. (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. 8-17 Figure 8.7 Defining metallic and covalent radii. 8-18 9
Figure 8.8 Atomic radii of the maingroup and transition elements. 8-19 Figure 8.9 Periodicity of atomic radius. 8-20 10
SAMPLE PROBLEM 8.3 Ranking Elements by Atomic Size PROBLEM: Using only the periodic table (not Figure 8.8), 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 as you go down; elements decrease in size as you go across a period. SOLUTION: (a) Sr > Ca > Mg (b) K > Ca > Ga (c) Rb > Br > Kr (d) Rb > Sr > Ca These elements are in Group 2A(2), and size decreases up the group. These elements are in Period 4, and size decreases across a period. Rb has a higher energy level and is far to the left. Br is to the left of Kr in Period 4. Ca is one energy level smaller than Rb and Sr. Rb is to the left of Sr in the same period. 8-21 Figure 8.10 Periodicity of first ionization energy (IE 1 ). 8-22 11
Figure 8.11 First ionization energies of the main-group elements. 8-23 Figure 8.12 The first three ionization energies of beryllium (in MJ/mol). For more data on sequential ionization energies of the elements, go to http://www.webelements.com. 8-24 12
SAMPLE PROBLEM 8.4 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 These three elements are all in Group 8A(18), IE decreases down a group. These are all in Period 5, 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. 8-25 8-26 13
SAMPLE PROBLEM 8.5 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 6 1012 1903 2910 4956 6278 22,230 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 3. 8-27 Figure 8.13 Electron affinities of the main-group elements. 8-28 14
Figure 8.14 Trends in three atomic properties. 8-29 Figure 8.15 Trends in metallic behavior. 8-30 15
Figure 8.16 The trend in acid-base behavior of element oxides. 8-31 Figure 8.17 Main-group ions and the noble gas electron configurations. 8-32 16
SAMPLE PROBLEM 8.6 Writing Electron Configurations of Main-Group Ions PROBLEM: 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) PLAN: 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) lose the np and ns or just the np electrons. 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 ) (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 - 8-33 Figure 8.18 The Period 4 crossover in sublevel energies. 8-34 17
Figure 8.19 Apparatus for measuring the magnetic behavior of a sample. 8-35 SAMPLE PROBLEM 8.7 Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions PROBLEM: Use condensed electron configurations to write the reaction for the formation of each transition metal ion, and predict whether the ion is paramagnetic. PLAN: SOLUTION: (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 1 3d 5 ) Cr 3+ ([Ar] 3d 3 ) + 3e - paramagnetic 8-36 (c) Hg 2+ (Z = 80) Hg ([Xe] 6s 2 4f 14 5d 10 ) Hg 2+ ([Xe] 4f 14 5d 10 ) + 2e - not paramagnetic (diamagnetic) 18
Figure 8.20 Depicting ionic radius. 8-37 Figure 8.21 Ionic vs. atomic radii. 8-38 19
SAMPLE PROBLEM 8.8 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 decrease in size going up the group. (b) S 2- > Cl - > K + (c) Au + > Au 3+ 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. The greater the + charge, the smaller the ion. 8-39 20