Formation of Binary Ionic Compounds

Binary Ionic Compound n Binary- two n Ionic- ions n Compound- joined together

Binary Ionic Compound n Solid formed between a metal and a nonmetal n The oppositely charged ions together have lower energy

Lattice Energy n The change in energy that takes place when separated gaseous ions are packed together to form an ionic solid

Lattice Energy n The energy released when an ionic solid is formed n M + (g) + X - (g) --> MX (s) n Sign will be negative b/c process is exothermic

Energy Changes n Look at the formation of an ionic solid from its elements n Keep in mind that energy is a STATE FUNCTION!!

Li (s) + ½ F 2(g) --> LiF (s) n Sublimation of solid Li n Li (s) --> Li (g) n Enthalpy for sublimation is 161 kJ/mol

Li (s) + ½ F 2(g) --> LiF (s) n Ionization of Li atoms to form Li + n Li (g) --> Li + (g) + e - n Ionization is 520 kJ/mol

Li (s) + ½ F 2(g) --> LiF (s) n Dissociation of F 2 molecules to F atoms n ½ F 2(g) --> F (g) n 154 kJ/mol n Divide by two = 77 kJ/mol

Li (s) + ½ F 2(g) --> LiF (s) n Formation of F - ions n Electron affinity n F (g) + e - --> F - (g) n Electron affinity = kJ/mol

Li (s) + ½ F 2(g) --> LiF (s) n Formation of LiF (s) n Lattice energy n Li + (g) + F - (g) --> LiF (s) n Lattice energy = kJ/mol

Li (s) + ½ F 2(g) --> LiF (s) n Sum of these 5 processes yields the desired overall reaction n -617 kJ (per mole of LiF)

Energy Diagram n Summarizes process n Notice that most of the processes are endothermic and unfavorable

Energy Diagram n However, the large lattice energy makes the whole process worthwhile

K (s) + ½ Cl 2(g) --> KCl (s) n Sublimation of K = +64 kJ n Ionization of K = +419 kJ n Bond energy of Cl 2 = +240 kJ n e - affinity of Cl = -349 kJ n Lattice energy = -690 kJ

K (s) + ½ Cl 2(g) --> KCl (s) n Net energy of formation equals the sum of the energy changes n  H f o = -436 kJ

Lattice Energy Calculations n Lattice energy is important in contributing to the stability of the compounds

Lattice Energy Calculations n Modified from Coulomb’s Law n Lattice energy = k(Q 1 Q 2 /r) n k = constant that depends on structure of solid

Lattice energy = k(Q 1 Q 2 /r) n Q 1 and Q 2 = charges of ions n r = shortest distance between ions

Lattice Energy n The higher the charge on each ion, the greater the lattice energy will be

Lattice Energy n This value counteracts the higher endothermic ionization energies, thus resulting in a more stable energetically stable crystal

Li (s) + ½ Br 2(g) --> LiBr (s) n Ionization of Li = +520 kJ/mol n e - affinity for Br = -324 kJ/mol n sublimation of Li = +161 kJ/mol n lattice energy = -787 kJ/mol n bond energy Br 2 = +193 kJ/mol

Li (s) + ½ Br 2(g) --> LiBr (s) n  H f o = -334 kJ