Electron Configuration and Chapter 8 Electron Configuration and Chemical Periodicity

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 Some Key Periodic Atomic Properties 8.5 The Connection Between Atomic Structure and Chemical Reactivity

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.

Mendeleev’s Predicted Properties of Germanium (“eka Silicon”) and Its Actual Properties Table 8.1 Predicted Properties of eka Silicon(E) Actual Properties of Germanium (Ge) Property atomic mass 72amu 72.61amu appearance gray metal gray metal density 5.5g/cm3 5.32g/cm3 molar volume 13cm3/mol 13.65cm3/mol specific heat capacity 0.31J/g*K 0.32J/g*K oxide formula EO2 GeO2 oxide density 4.7g/cm3 4.23g/cm3 sulfide formula and solubility ES2; insoluble in H2O; soluble in aqueous (NH4)2S GeS2; insoluble in H2O; soluble in aqueous (NH4)2S chloride formula (boiling point) ECl4; (<1000C) GeCl4; (840C) chloride density 1.9g/cm3 1.844g/cm3 element preparation reduction of K2EF6 with sodium reduction of K2GeF6 with sodium

Remember from Chapter 7 - 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.

Table 8.2 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 ml integers from -l to 0 to +l orbital orientation spin ms +1/2 or -1/2 direction of e- spin

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.

The effect of another electron in the same orbital

The effect of other electrons in inner orbitals

The effect of orbital shape

Order for filling energy sublevels with electrons Illustrating Orbital Occupancies The electron configuration # of electrons in the sublevel n l as s,p,d,f The orbital diagram

A vertical orbital diagram for the Li ground state empty A vertical orbital diagram for the Li ground state half-filled filled, spin-paired

SAMPLE PROBLEM 8.1 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. 9F 1s 2s 2p SOLUTION: The third electron is in the 2s orbital. Its quantum numbers are n = l = ml = ms= 2 +1/2 The eighth electron is in a 2p orbital. Its quantum numbers are n = l = ml = ms= 2 1 -1 -1/2

Orbital occupancy for the first 10 elements, H through Ne.

Hund’s rule

Condensed ground-state electron configurations in the first three periods.

A periodic table of partial ground-state electron configurations

The relation between orbital filling and the periodic table

General pattern for filling the sublevels

SAMPLE PROBLEM 8.2 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: 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) condensed configuration partial orbital diagram full configuration 1s22s22p63s23p64s1 [Ar] 4s1 There are 18 inner electrons. 4s1

SAMPLE PROBLEM 8.2 continued (b) for Mo (Z = 42) condensed configuration partial orbital diagram full configuration 1s22s22p63s23p64s23d104p65s14d5 [Kr] 5s14d5 There are 36 inner electrons and 6 valence electrons. 5s1 4d5 (c) for Pb (Z = 82) 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2 condensed configuration partial orbital diagram full configuration [Xe] 6s24f145d106p2 There are 78 inner electrons and 4 valence electrons. 6s2 6p2

Defining metallic and covalent radii Knowing the Cl radius and the C-Cl bond length, the C radius can be determined.

Atomic radii of the main-group and transition elements. Trends in the Periodic Table Atomic radii of the main-group and transition elements.

Trends in the Periodic Table Periodicity of atomic radius

SAMPLE PROBLEM 8.3 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 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 2A(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.

Periodicity of first ionization energy (IE1) Trends in the Periodic Table Periodicity of first ionization energy (IE1) Energy required to remove one outermost electron.

First ionization energies of the main-group elements Trends in the Periodic Table

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 IE1: (a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs PLAN: IE increases as you proceed up in a group; IE increases as you go across a period. SOLUTION: (a) He > Ar > Kr Group 8A(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.

The first three ionization energies of beryllium (in MJ/mol) Trends in the Periodic Table The first three ionization energies of beryllium (in MJ/mol)

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: IE1 IE2 IE3 IE4 IE5 IE6 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 IE5, that is, after the 5th valence electron has been removed. Five electrons would mean that the valence configuration is 3s23p3 and the element must be phosphorous, P (Z = 15). The complete electron configuration is 1s22s22p63s23p3.

Electron affinities of the main-group elements Trends in the Periodic Table Electron affinities of the main-group elements Electron Affinity: Energy change to add one electron. In most cases, EA negative (energy released because electron attracted to nucleus

Trends in three atomic properties

Trends in metallic behavior

Main-group ions and the noble gas configurations Trends in the Periodic Table Main-group ions and the noble gas configurations Properties of Monatomic Ions

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) can lose their np or ns and np electrons. SOLUTION: (a) Iodine (Z = 53) is in Group 7A(17) and will gain one electron to be isoelectronic with Xe: I ([Kr]5s24d105p5) + e- I- ([Kr]5s24d105p6) (b) Potassium (Z = 19) is in Group 1A(1) and will lose one electron to be isoelectronic with Ar: K ([Ar]4s1) K+ ([Ar]) + e- (c) Indium (Z = 49) is in Group 3A(13) and can lose either one electron or three electrons: In ([Kr]5s24d105p1) In+ ([Kr]5s24d10) + e- In ([Kr]5s24d105p1) In3+([Kr] 4d10) + 3e-

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

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. (a) Mn2+(Z = 25) (b) Cr3+(Z = 24) (c) Hg2+(Z = 80) PLAN: 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) Mn2+(Z = 25) Mn([Ar]4s23d5) Mn2+ ([Ar] 3d5) + 2e- paramagnetic (b) Cr3+(Z = 24) Cr([Ar]4s23d6) Cr3+ ([Ar] 3d5) + 3e- paramagnetic (c) Hg2+(Z = 80) Hg([Xe]6s24f145d10) Hg2+ ([Xe] 4f145d10) + 2e- not paramagnetic (is diamagnetic)

Ionic vs. atomic radius

SAMPLE PROBLEM 8.8 Ranking Ions by Size PROBLEM: Rank each set of ions in order of decreasing size, and explain your ranking: (a) Ca2+, Sr2+, Mg2+ (b) K+, S2-, Cl - (c) Au+, Au3+ 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) Sr2+ > Ca2+ > Mg2+ These are members of the same Group (2A/2) and therefore decrease in size going up the group. The ions are isoelectronic; S2- has the smallest Zeff and therefore is the largest while K+ is a cation with a large Zeff and is the smallest. (b) S2- > Cl - > K+ (c) Au+ > Au3+ The higher the + charge, the smaller the ion.

End of Chapter 8