Ionic Bonding and Main-Group Elements Chapter 6

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

Metals versus Nonmetals Metals tend to form cations. Nonmetals tend to form anions.

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]

Cations and Anions Of Representative Elements

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

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

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

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

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+ ?

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. 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

Ionization Energy Abbreviation is E i, it has units of kJ/mol.

Ionization Energy02 Ionization Energy Trend: Ionization energies vary periodically; this is explained by the changes in Z eff.

Ionization Energy03

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.

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

Electron Affinity02 Electron Affinity Trend: Value of E ea results from interplay of nucleus-electron attraction, and electron–electron repulsion.

Electron Affinity03

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.

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 MgCl 2 :

Born-Haber Cycle for Determining Lattice Energy  H overall =  H 1 +  H 2 +  H 3 +  H 4 +  H 5 oooooo

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 kJ/mol Bond Dissociation Energy : +79 First Electron affinity -328 kJ/mol Lattice Energy-821 kJ/mol

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 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

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.

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

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)

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 +.

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)

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.

7A Halogens01 Halogens are reactive, & toxic nonmetals. Properties decrease down the group. HF is a weak acid, forms H–bonds.

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.”

Preparation of Chlorine 6HCl + 2KMnO4 + 2H+ --> 3Cl2 + 2MnO2 + 4H2O + 2K+