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Formation of Binary Ionic Compounds

Formation of Binary Ionic Compounds. Binary Ionic Compound. Binary- two Ionic- ions Compound- joined together. Binary Ionic Compound. Solid formed between a metal and a nonmetal The oppositely charged ions together have lower energy. Lattice Energy.

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Formation of Binary Ionic Compounds

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  1. Formation of Binary Ionic Compounds

  2. Binary Ionic Compound • Binary- two • Ionic- ions • Compound- joined together

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

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

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

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

  7. Li(s) + ½ F2(g) --> LiF(s) • Sublimation of solid Li • Li(s) --> Li(g) • Enthalpy for sublimation is 161 kJ/mol

  8. Li(s) + ½ F2(g) --> LiF(s) • Ionization of Li atoms to form Li+ • Li(g) --> Li+(g) + e- • Ionization is 520 kJ/mol

  9. Li(s) + ½ F2(g) --> LiF(s) • Dissociation of F2 molecules to F atoms • ½ F2(g) --> F(g) • 154 kJ/mol • Divide by two = 77 kJ/mol

  10. Li(s) + ½ F2(g) --> LiF(s) • Formation of F- ions • Electron affinity • F(g) + e- --> F-(g) • Electron affinity = - 328 kJ/mol

  11. Li(s) + ½ F2(g) --> LiF(s) • Formation of LiF(s) • Lattice energy • Li+(g) + F-(g) --> LiF(s) • Lattice energy = -1047 kJ/mol

  12. Li(s) + ½ F2(g) --> LiF(s) • Sum of these 5 processes yields the desired overall reaction • -617 kJ (per mole of LiF)

  13. Energy Diagram • Summarizes process • Notice that most of the processes are endothermic and unfavorable

  14. Energy Diagram • However, the large lattice energy makes the whole process worthwhile

  15. K(s) + ½ Cl2(g) --> KCl(s) • Sublimation of K = +64 kJ • Ionization of K = +419 kJ • Bond energy of Cl2 = +240 kJ • e- affinity of Cl = -349 kJ • Lattice energy = -690 kJ

  16. K(s) + ½ Cl2(g) --> KCl(s) • Net energy of formation equals the sum of the energy changes • Hfo = -436 kJ

  17. Lattice Energy Calculations • Lattice energy is important in contributing to the stability of the compounds

  18. Lattice Energy Calculations • Modified from Coulomb’s Law • Lattice energy = k(Q1Q2/r) • k = constant that depends on structure of solid

  19. Lattice energy = k(Q1Q2/r) • Q1 and Q2 = charges of ions • r = shortest distance between ions

  20. Lattice Energy • The higher the charge on each ion, the greater the lattice energy will be

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

  22. Li(s) + ½ Br2(g) --> LiBr(s) • Ionization of Li = +520 kJ/mol • e- affinity for Br = -324 kJ/mol • sublimation of Li = +161 kJ/mol • lattice energy = -787 kJ/mol • bond energy Br2 = +193 kJ/mol

  23. Li(s) + ½ Br2(g) --> LiBr(s) • Hfo = -334 kJ

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