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Ionic Bonding and Main-Group Elements Chapter 6
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8.2 ns 1 ns 2 ns 2 np 1 ns 2 np 2 ns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 d1d1 d5d5 d 10 4f 5f Ground State Electron Configurations of the Elements
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Metals versus Nonmetals Metals tend to form cations. Nonmetals tend to form anions.
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Ions and Ionic Radii01 Main-group metals donate electrons. from their valence shell. –Na: 1s 2 2s 2 2p 6 3s 1 = [Ne] 3s 1 –Na + : 1s 2 2s 2 2p 6 = [Ne] –Mg: 1s 2 2s 2 2p 6 3s 2 = [Ne] 3s 2 –Mg 2+ : 1s 2 2s 2 2p 6 = [Ne] –Al:1s 2 2s 2 2p 6 3s 2 3p 1 = [Ne] 3s 2 3p 1 –Al 3+ 1s 2 2s 2 2p 6 = [Ne]
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+1+2+3 -2-3 Cations and Anions Of Representative Elements
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Ions and Ionic Radii02 Main-group nonmetals accept electrons into their valence shell. –N: 1s 2 2s 2 2p 3 = [He] 2s 2 2p 3 –N 3– : 1s 2 2s 2 2p 6 = [He] 2s 2 2p 6 –O: 1s 2 2s 2 2p 4 = [He] 2s 2 2p 4 –O 2– : 1s 2 2s 2 2p 6 = [He] 2s 2 2p 6 –F:1s 2 2s 2 2p 5 = [He] 2s 2 2p 5 –F – :1s 2 2s 2 2p 6 = [He] 2s 2 2p 6
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Na [Ne]3s 1 Na + [Ne] Ca [Ar]4s 2 Ca 2+ [Ar] Al [Ne]3s 2 3p 1 Al 3+ [Ne] Atoms lose electrons so that cation has a noble-gas outer electron configuration. H 1s 1 H - 1s 2 or [He] F 1s 2 2s 2 2p 5 F - 1s 2 2s 2 2p 6 or [Ne] O 1s 2 2s 2 2p 4 O 2- 1s 2 2s 2 2p 6 or [Ne] N 1s 2 2s 2 2p 3 N 3- 1s 2 2s 2 2p 6 or [Ne] Atoms gain electrons so that anion has a noble-gas outer electron configuration. Learning Check Predict the charge of the following elements
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Na + : [Ne]Al 3+ : [Ne]F - : 1s 2 2s 2 2p 6 or [Ne] O 2- : 1s 2 2s 2 2p 6 or [Ne]N 3- : 1s 2 2s 2 2p 6 or [Ne] Na +, Al 3+, F -, O 2-, and N 3- are all isoelectronic with Ne What neutral atom is isoelectronic with H - ? H - : 1s 2 same electron configuration as He
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Ions and Ionic Radii03 Transition metals lose their valence-shell s- electrons before losing their d-electrons. Electrons with the highest n-quantum number are lost first. –Fe:1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 = [Ar] 4s 2 3d 6 –Fe 2+ :1s 2 2s 2 3s 2 3p 6 3d 6 = [Ar] 3d 6 –Fe 3+ :1s 2 2s 2 3s 2 3p 6 3d 5 = [Ar] 3d 5
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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]4s 2 3d 5 Mn 2+ : [Ar]4s 0 3d 5 or [Ar]3d 5 Learning Check Predict the electronic structure of Mn 2+ ?
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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
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Ionization energy is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state. Ei 1 + X (g) X + (g) + e - Ei 2 + X + (g) X 2 (g) + e - Ei 3 + X 2+ (g) X 3+ +e - Ei 1 first ionization energy Ei 2 second ionization energy Ei 3 third ionization energy Ei 1 < Ei 2 < Ei 3
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Ionization Energy Abbreviation is E i, it has units of kJ/mol.
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Ionization Energy02 Ionization Energy Trend: Ionization energies vary periodically; this is explained by the changes in Z eff.
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Ionization Energy03
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Electron Affinity Energy change that occurs when an electron is added to an isolated atom in the gaseous state. Abbreviation is E ea, it has units of kJ/mol.
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X (g) + e - X - (g) F (g) + e - X - (g) O (g) + e - O - (g) H = -328 kJ/mol H = -141 kJ/mol ΔH < 0
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Electron Affinity02 Electron Affinity Trend: Value of E ea results from interplay of nucleus-electron attraction, and electron–electron repulsion.
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Electron Affinity03
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Ionic Bonds and Ionic Solids 01 Ionic bonds: Form when an element with a small E i value comes in contact with an element with a negative E ea value.
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Ionic Bonds and Ionic Solids Sodium Chloride (NaCl):
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Ionic Bonds and Ionic Solids 04 Born–Haber Cycle for NaCl:
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Ionic Bonds and Ionic Solids 05 Born–Haber Cycle for MgCl 2 :
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Born-Haber Cycle for Determining Lattice Energy H overall = H 1 + H 2 + H 3 + H 4 + H 5 oooooo
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Calculate the net energy change that takes place on formation of KF(s) from the elements : K(s) + ½ F 2 (g) KF(s) ΔH = ? kJ/mole Heat of sublimation:+89.2 kJ/mol First Ionization energy +418.8 kJ/mol Bond Dissociation Energy : +79 First Electron affinity -328 kJ/mol Lattice Energy-821 kJ/mol
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Calculate the net energy change that takes place on formation of KF(s) from the elements : K(s) + ½ F 2 (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 [F 2 (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 F 2 (g) KF(s)ΔH = -562 kJ/mole
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Alkali Metals All have one s electron outside noble gas core. All form M + ions.
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Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.
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Li & Na: Obtained by electrolysis. 2 LiCl(l) 2 Li(l) + Cl 2 (g) 2 NaCl(l)2 Na(l) + Cl 2 (g) K, Rb, & Cs: Obtained by reductive distillation. KCl(l) + Na(l) K(g) + NaCl(l) 2 RbCl(l) + Ca(l)2 Rb(g) + CaCl 2 (l) 2 CsCl(l) + Ca(l)2 Cs(g) + CaCl 2 (l) 1A Preparation 450°C in KCl 580°C in CaCl 2 850°C 750°C
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1A Compounds01 Reaction with Halogens 2 M(s) + X 2 2 MX(s) Reaction with Oxygen Forms oxide (Li 2 O), peroxide (Na 2 O 2 ), or superoxide (KO 2 ) Reaction with Hydrogen 2 M(s) + H 2 2 MH(s) Reaction with Nitrogen 6 Li(s) + N 2 2 Li 3 N(s)
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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. M 2+ ions are small than M +.
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2A Compounds01 Reaction with Halogens M(s) + X 2 MX 2 (s) Reaction with Oxygen 2M(s) + O 2 2 MO(s) Reaction with Hydrogen 2 Ca(s) + H 2 2 CaH 2 (s) Reaction with Water (only Ba and Radium are vigorous) Ba(s) + H 2 O Ba 2+ (aq) + 2 OH – (aq) + H 2 (g)
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3A Boron01 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.
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7A Halogens01 Halogens are reactive, & toxic nonmetals. Properties decrease down the group. HF is a weak acid, forms H–bonds.
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7A Halogens02 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.”
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Preparation of Chlorine 6HCl + 2KMnO4 + 2H+ --> 3Cl2 + 2MnO2 + 4H2O + 2K+
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