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Ionic Bonding and Main-Group Elements. Chapter 6. ns 2 np 6. ns 1. ns 2 np 1. ns 2 np 2. ns 2 np 3. ns 2 np 4. ns 2 np 5. ns 2. d 10. d 1. d 5. 4f. 5f. Ground State Electron Configurations of the Elements. 8.2. Metals versus Nonmetals. Metals tend to form cations.
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Ionic Bonding and Main-Group Elements Chapter 6
ns2np6 ns1 ns2np1 ns2np2 ns2np3 ns2np4 ns2np5 ns2 d10 d1 d5 4f 5f Ground State Electron Configurations of the Elements 8.2
Metals versus Nonmetals • Metals tend to form cations. • Nonmetals tend to form anions.
Ions and Ionic Radii 01 • Main-group metals donate electrons. from their valence shell. • Na: 1s2 2s2 2p6 3s1 = [Ne] 3s1 • Na+: 1s2 2s2 2p6 =[Ne] • Mg: 1s2 2s2 2p6 3s2 = [Ne] 3s2 • Mg2+: 1s2 2s2 2p6 =[Ne] • Al:1s2 2s2 2p6 3s2 3p1 = [Ne] 3s2 3p1 • Al3+ 1s2 2s2 2p6 = [Ne]
-1 -2 -3 +1 +2 +3 Cations and Anions Of Representative Elements
Ions and Ionic Radii 02 • Main-group nonmetals accept electrons into their valence shell. • N: 1s2 2s2 2p3 = [He] 2s2 2p3 • N3–: 1s2 2s2 2p6 = [He] 2s2 2p6 • O: 1s2 2s2 2p4 = [He] 2s2 2p4 • O2–: 1s2 2s2 2p6 = [He] 2s2 2p6 • F: 1s2 2s2 2p5 = [He] 2s2 2p5 • F–: 1s2 2s2 2p6 = [He] 2s2 2p6
Learning Check Predict the charge of the following elements Na [Ne]3s1 Na+ [Ne] Atoms lose electrons so that cation has a noble-gas outer electron configuration. Ca [Ar]4s2 Ca2+ [Ar] Al [Ne]3s23p1 Al3+ [Ne] H 1s1 H- 1s2 or [He] Atoms gain electrons so that anion has a noble-gas outer electron configuration. F 1s22s22p5 F- 1s22s22p6 or [Ne] O 1s22s22p4 O2- 1s22s22p6 or [Ne] N 1s22s22p3 N3- 1s22s22p6 or [Ne]
What neutral atom is isoelectronic with H- ? Na+: [Ne] Al3+: [Ne] F-: 1s22s22p6 or [Ne] O2-: 1s22s22p6 or [Ne] N3-: 1s22s22p6 or [Ne] Na+, Al3+, F-, O2-, and N3- are all isoelectronic with Ne H-: 1s2 same electron configuration as He
Ions and Ionic Radii 03 • Transition metals lose their valence-shell s-electrons before losing their d-electrons. Electrons with the highest n-quantum number are lost first. • Fe: 1s2 2s2 2p63s2 3p6 4s2 3d6 = [Ar] 4s2 3d6 • Fe2+: 1s2 2s2 3s2 3p6 3d6 = [Ar] 3d6 • Fe3+: 1s2 2s2 3s2 3p6 3d5 = [Ar] 3d5
Learning Check Predict the electronic structure of Mn2+ ? 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. Mn: [Ar]4s23d5 Mn2+: [Ar]4s03d5 or [Ar]3d5
Cation is always smaller than atom from which it is formed. Anion is always larger than atom from which it is formed. Size of Ions
Ionization energy is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state. Ei1 + X(g) X+(g) + e- Ei1 first ionization energy Ei2 + X+(g) X2(g) + e- Ei2 second ionization energy Ei3 + X2+(g) X3+ +e- Ei3 third ionization energy Ei1 < Ei2 < Ei3
Ionization Energy • Abbreviation is Ei, it has units of kJ/mol.
Ionization Energy 02 • Ionization Energy Trend: • Ionization energies vary periodically; this is explained by the changes in Zeff.
Electron Affinity • Energy change that occurs when an electron is added to an isolated atom in the gaseous state. • Abbreviation is Eea, it has units of kJ/mol.
X (g) + e- X-(g) F (g) + e- X-(g) O (g) + e- O-(g) ΔH < 0 DH = -328 kJ/mol DH = -141 kJ/mol
Electron Affinity 02 • Electron Affinity Trend: • Value of Eea results from interplay of nucleus-electron attraction, and electron–electron repulsion.
Ionic Bonds and Ionic Solids 01 • Ionic bonds: • Form when an element with a small Eivalue comes in contact with an element with a negative Eeavalue.
Ionic Bonds and Ionic Solids Sodium Chloride (NaCl):
Ionic Bonds and Ionic Solids 04 Born–Haber Cycle for NaCl:
Ionic Bonds and Ionic Solids 05 Born–Haber Cycle for MgCl2:
o o o o o o DHoverall = DH1 + DH2 + DH3 + DH4 + DH5 Born-Haber Cycle for Determining Lattice Energy
Calculate the net energy change that takes place on formation of KF(s) from the elements:K(s) + ½ F2(g) KF(s) ΔH = ? kJ/mole
Calculate the net energy change that takes place on formation of KF(s) from the elements:K(s) + ½ F2(g) KF(s) ΔH = ? kJ/mole • K(s) K(g) Heat of sublimation +89.2 kJ/mol • K(g) K+(g) + e First Ionization energy +418.8 kJ/mol • ½ [F2(g) 2 F(g)] Bond Dissociation Energy +79 kJ/mol • F(g) + e F(g) First Electron affinity 328 kJ/mol • K+(g) + F(g) KF(s) Lattice Energy in KF 821 kJ/mol Sum = 562 kJ/mol K(s) + ½ F2(g) KF(s) ΔH = -562 kJ/mole
Alkali Metals • All have one s electron outside noble gas core. • All form M+ ions.
Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.
450°C in KCl 580°C in CaCl2 850°C 750°C 1A Preparation • Li & Na: Obtained by electrolysis. • 2 LiCl(l) 2 Li(l) + Cl2(g) • 2 NaCl(l) 2 Na(l) + Cl2(g) • K, Rb, & Cs: Obtained by reductive distillation. • KCl(l) + Na(l) K(g) + NaCl(l) • 2 RbCl(l) + Ca(l) 2 Rb(g) + CaCl2(l) • 2 CsCl(l) + Ca(l) 2 Cs(g) + CaCl2(l) 750°C
1A Compounds 01 • Reaction with Halogens • 2 M(s) + X2 2 MX(s) • Reaction with Oxygen • Forms oxide (Li2O), peroxide (Na2O2), or superoxide (KO2) • Reaction with Hydrogen • 2 M(s) + H2 2 MH(s) • Reaction with Nitrogen • 6 Li(s) + N2 2 Li3N(s)
Alkaline Earth Metals • Smaller atomic radii than 1A. • Higher melting and boiling points than 1A. • Less reactive than 1A. • Exhibit covalent and ionic bonding character. • All are highly electropositive metals. • M2+ ions are small than M+.
2A Compounds 01 • Reaction with Halogens • M(s) + X2 MX2(s) • Reaction with Oxygen • 2M(s) + O22 MO(s) • Reaction with Hydrogen • 2 Ca(s) + H2 2 CaH2(s) • Reaction with Water(only Ba and Radium are vigorous) • Ba(s) + H2O Ba2+(aq) + 2 OH–(aq) + H2(g)
3A Boron 01 • Elements in this group contain one semimetal, and four that are primarily metallic. • Boron is so different from the other elements in this group. • Boron is a semimetal, and forms covalent bonds. • It has many similarities to carbon and silicon.
7A Halogens 01 • Halogens are reactive, & toxic nonmetals. • Properties decrease down the group. • HF is a weak acid, forms H–bonds.
7A Halogens 02 • Fluorine and chlorine are strong oxidizing agents that are produced by electrolysis. • Fluorine is obtained from liquid HF. • Chlorine is obtained from molten NaCl or from the “chlor–alkali process.”
Preparation of Chlorine • 6HCl + 2KMnO4 + 2H+ --> 3Cl2 + 2MnO2 + 4H2O + 2K+