Chapter 8 Electron Configuration and Chemical Periodicity 8
- Slides: 42
Chapter 8 Electron Configuration and Chemical Periodicity 8 -1 Dr. Wolf’s CHM 101
Electron Configuration and Chemical Periodicity 5. 1 Development of the Periodic Table 5. 2 Characteristics of Many-Electron Atoms 5. 3 The Quantum-Mechanical Model and the Periodic Table 5. 4 Trends in Some Key Periodic Atomic Properties 5. 5 The Connection Between Atomic Structure and Chemical Reactivity 8 -2 Dr. Wolf’s CHM 101
Mendeleev’s Periodic Law Arranged the 65 known elements by atomic mass and by recurrence of various physical and chemical properties. The Periodic Table today is very similar but arranged according to atomic number (number of protons). The arrangement led to families of elements with similar properties and at the time allowed for the prediction and properties of elements yet to be discovered. 8 -3 Dr. Wolf’s CHM 101
Table 8. 1 Mendeleev’s Predicted Properties of Germanium (“eka Silicon”) and Its Actual Properties 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 8 -4 Dr. Wolf’s CHM 101 Predicted Properties of eka Silicon(E) Actual Properties of Germanium (Ge) 72 amu gray metal 5. 5 g/cm 3 13 cm 3/mol 0. 31 J/g*K EO 2 4. 7 g/cm 3 ES 2; insoluble in H 2 O; soluble in aqueous (NH 4)2 S ECl 4; (<1000 C) 72. 61 amu gray metal 5. 32 g/cm 3 13. 65 cm 3/mol 0. 32 J/g*K Ge. O 2 4. 23 g/cm 3 Ge. S 2; insoluble in H 2 O; soluble in aqueous (NH 4)2 S Ge. Cl 4; (840 C) 1. 9 g/cm 3 reduction of K 2 EF 6 with sodium 1. 844 g/cm 3 reduction of K 2 Ge. F 6 with sodium
Remember from Chapter 4 Quantum Numbers and Atomic Orbitals An atomic orbital is specified by three quantum numbers. n the principal quantum number - a positive integer l the angular momentum quantum number - an integer from 0 to n-1 ml the magnetic moment quantum number - an integer from -l to +l The three quantum numbers are actually giving the energy of the electron in the orbital and a fourth q. n. is needed to describe a property of electrons called spin. The spin can be clockwise or counterclockwise. The spin q. n. , ms can be + ½ or - ½. The Pauli Exclusion Principle - No two electrons in the same atom can have the same four q. n. Since the first three q. n. define the orbital, this means only two electrons can be in the same orbital and they must have opposite spins. 8 -5 Dr. Wolf’s CHM 101
Table 8. 2 Summary of Quantum Numbers of Electrons in Atoms Name 8 -6 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 ml integers from -l to 0 to +l orbital orientation spin ms +1/2 or -1/2 direction of e- spin Dr. Wolf’s CHM 101
Factors Affecting Atomic Orbital Energies The Effect of Nuclear Charge (Zeffective) 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 (makes less stable) An additional electron raises the orbital energy through electron-electron repulsions. Additional electrons in inner orbitals (makes outer orbital less stable) Inner electrons shield outer electrons more effectively than do electrons in the same sublevel. 8 -7 Dr. Wolf’s CHM 101
The effect of another electron in the same orbital 8 -8 Dr. Wolf’s CHM 101
The effect of other electrons in inner orbitals 8 -9 Dr. Wolf’s CHM 101
The effect of orbital shape 8 -10 Dr. Wolf’s CHM 101
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 8 -11 Dr. Wolf’s CHM 101
empty A vertical orbital diagram for the Li ground state half-filled, spin-paired 8 -12 Dr. Wolf’s CHM 101
SAMPLE PROBLEM 8. 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: Use the orbital diagram to find the third and eighth electrons. 9 F 1 s 2 s 2 p SOLUTION: The third electron is in the 2 s orbital. Its quantum numbers are n= 2 l= 0 ml = 0 ms= +1/2 The eighth electron is in a 2 p orbital. Its quantum numbers are n= 8 -13 2 Dr. Wolf’s CHM 101 l= 1 ml = -1 ms = -1/2
Orbital occupancy for the first 10 elements, H through Ne. 8 -14 Dr. Wolf’s CHM 101
Hund’s rule 8 -15 Dr. Wolf’s CHM 101
Condensed ground-state electron configurations in the first three periods. 8 -16 Dr. Wolf’s CHM 101
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A periodic table of partial ground-state electron configurations 8 -19 Dr. Wolf’s CHM 101
The relation between orbital filling and the periodic table 8 -20 Dr. Wolf’s CHM 101
General pattern for filling the sublevels 8 -21 Dr. Wolf’s CHM 101
SAMPLE PROBLEM 8. 2 Determining Electron Configuration 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) PROBLEM: PLAN: 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. SOLUTION: (a) for K (Z = 19) full configuration 1 s 22 p 63 s 23 p 64 s 1 condensed configuration [Ar] 4 s 1 There are 18 inner electrons. partial orbital diagram 4 s 1 8 -22 Dr. Wolf’s CHM 101
SAMPLE PROBLEM 8. 2 continued (b) for Mo (Z = 42) full configuration 1 s 22 p 63 s 23 p 64 s 23 d 104 p 65 s 14 d 5 condensed configuration [Kr] 5 s 14 d 5 partial orbital diagram There are 36 inner electrons and 6 valence electrons. 5 s 1 4 d 5 (c) for Pb (Z = 82) full configuration 1 s 22 p 63 s 23 p 64 s 23 d 104 p 65 s 24 d 105 p 66 s 24 f 145 d 106 p 2 condensed configuration partial orbital diagram 8 -23 6 s 2 Dr. Wolf’s CHM 101 6 p 2 [Xe] 6 s 24 f 145 d 106 p 2 There are 78 inner electrons and 4 valence electrons.
Defining metallic and covalent radii Knowing the Cl radius and the C-Cl bond length, the C radius can be determined. 8 -24 Dr. Wolf’s CHM 101
Trends in the Periodic Table Atomic radii of the maingroup and transition elements. 8 -25 Dr. Wolf’s CHM 101
Trends in the Periodic Table Periodicity of atomic radius 8 -26 Dr. Wolf’s CHM 101
SAMPLE PROBLEM 8. 3 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 PLAN: Ranking Elements by Atomic Size (b) K, Ga, Ca (c) Br, Rb, Kr (d) Sr, Ca, Rb Elements in the same group decrease in size as you go up; elements decrease in size as you go across a period. SOLUTION: (a) Sr > Ca > Mg These elements are in Group 2 A(2). (b) K > Ca > Ga These elements are in Period 4. (c) Rb > Br > Kr Rb has a higher energy level and is far to the left. Br is to the left of Kr. (d) Rb > Sr > Ca Ca is one energy level smaller than Rb and Sr. Rb is to the left of Sr. 8 -27 Dr. Wolf’s CHM 101
Trends in the Periodic Table Periodicity of first ionization energy (IE 1) Energy required to remove one outermost electron. 8 -28 Dr. Wolf’s CHM 101
First ionization energies of the main-group elements Trends in the Periodic Table 8 -29 Dr. Wolf’s CHM 101
SAMPLE PROBLEM 8. 4 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 PLAN: Ranking Elements by First Ionization Energy (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs IE increases as you proceed up in a group; IE increases as you go across a period. SOLUTION: 8 -30 (a) He > Ar > Kr Group 8 A(18) - IE decreases down a group. (b) Te > Sb > Sn Period 5 elements - IE increases across a period. (c) Ca > K > Rb Ca is to the right of K; Rb is below K. (d) Xe > I > Cs I is to the left of Xe; Cs is further to the left and down one period. Dr. Wolf’s CHM 101
Trends in the Periodic Table 8 -31 Dr. Wolf’s CHM 101 The first three ionization energies of beryllium (in MJ/mol)
SAMPLE PROBLEM 8. 5 PROBLEM: PLAN: Identifying an Element from Successive Ionization Energies Name the Period 3 element with the following ionization energies (in k. J/mol) and write its electron configuration: IE 1 IE 2 IE 3 IE 4 IE 5 1012 1903 2910 4956 6278 IE 6 22, 230 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 5 th valence electron has been removed. Five electrons would mean that the valence configuration is 3 s 23 p 3 and the element must be phosphorous, P (Z = 15). The complete electron configuration is 1 s 22 p 63 s 23 p 3. 8 -32 Dr. Wolf’s CHM 101
Trends in the Periodic Table Electron Affinity: Energy change to add one electron. In most cases, EA negative (energy released because electron attracted to nucleus 8 -33 Dr. Wolf’s CHM 101 Electron affinities of the main-group elements
Trends in three atomic properties 8 -34 Dr. Wolf’s CHM 101
Trends in metallic behavior 8 -35 Dr. Wolf’s CHM 101
Trends in the Periodic Table 8 -36 Dr. Wolf’s CHM 101 Properties of Monatomic Ions Main-group ions and the noble gas configurations
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 1 A(1), 2 A(2), 6 A(16), and 7 A(17) are usually isoelectronic with the nearest noble gas. Metals in Groups 3 A(13) to 5 A(15) can lose their np or ns and np electrons. SOLUTION: (a) Iodine (Z = 53) is in Group 7 A(17) and will gain one electron to be isoelectronic with Xe: I ([Kr]5 s 24 d 105 p 5) + e. I- ([Kr]5 s 24 d 105 p 6) (b) Potassium (Z = 19) is in Group 1 A(1) and will lose one electron to be isoelectronic with Ar: K ([Ar]4 s 1) K+ ([Ar]) + e(c) Indium (Z = 49) is in Group 3 A(13) and can lose either one electron or three electrons: In ([Kr]5 s 24 d 105 p 1) In+ ([Kr]5 s 24 d 10) + e. In ([Kr]5 s 24 d 105 p 1) In 3+([Kr] 4 d 10) + 3 e- 8 -37 Dr. Wolf’s CHM 101
Magnetic Properties of Transition Metal Ions A species with unpaired electrons exhibits paramagnetism. It is attracted by an external magnetic field. Species with all paired e’s, not attracted. . . . diamagnetic 8 -38 Dr. Wolf’s CHM 101
SAMPLE PROBLEM 8. 7 PROBLEM: 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) PLAN: Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions (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. SOLUTION: (a) Mn 2+(Z = 25) Mn([Ar]4 s 23 d 5) (b) Cr 3+(Z = 24) Cr([Ar]4 s 23 d 6) (c) Hg 2+(Z = 80) Hg([Xe]6 s 24 f 145 d 10) Mn 2+ ([Ar] 3 d 5) + 2 e. Cr 3+ ([Ar] 3 d 5) + 3 e- paramagnetic Hg 2+ ([Xe] 4 f 145 d 10) + 2 enot paramagnetic (is diamagnetic) 8 -39 Dr. Wolf’s CHM 101
Ionic vs. atomic radius 8 -40 Dr. Wolf’s CHM 101
SAMPLE PROBLEM 8. 8 PROBLEM: Ranking Ions by Size Rank each set of ions in order of decreasing size, and explain your ranking: (a) Ca 2+, Sr 2+, Mg 2+ PLAN: (b) K+, S 2 -, Cl - (c) Au+, Au 3+ 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+ (b) S 2 - > Cl - > K+ (c) Au+ > Au 3+ 8 -41 These are members of the same Group (2 A/2) and therefore decrease in size going up the group. The ions are isoelectronic; S 2 - has the smallest Zeff and therefore is the largest while K+ is a cation with a large Zeff and is the smallest. The higher the + charge, the smaller the ion. Dr. Wolf’s CHM 101
End of Chapter 8 8 -42 Dr. Wolf’s CHM 101
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