1. Energy Effects in Binary Ionic Compounds and Lattice Energy

2. Consider Ionic Compounds

3. Consider Ionic Compounds
Not molecules

4. Consider Ionic Compounds
Not molecules
Have an arrangement of several ions all interacting with each other.

5. Consider Ionic Compounds
Not molecules
Have an arrangement of several ions all interacting with each other.
The solid is a regular arranged pattern of ions called a crystal lattice.

6. LiF…….An ionic compound between lithium and fluorine

7. LiF…….An ionic compound between lithium and fluorine
Li (s) ½ F2 (g)  LiF (s)
Li must be converted to a gas
1. Li (s)  Li (g) kJ/mol

8. LiF…….An ionic compound between lithium and fluorine
Li (s) ½ F2 (g)  LiF (s)
Li must be converted to a gas
Li must be ionized (ionization energy)
1. Li (s)  Li (g) kJ/mol
2. Li (g)  Li+ (g)+e kJ/mol

9. LiF…….An ionic compound between lithium and fluorine
Li (s) ½ F2 (g)  LiF (s)
Li must be converted to a gas
Li must be ionized (ionization energy)
F molecules need to be broken into atoms
1. Li (s)  Li (g) kJ/mol
2. Li (g)  Li+ (g)+e kJ/mol
3. 1/2F2 (g)  F (g) kJ/mol

10. LiF…….An ionic compound between lithium and fluorine
Li (s) ½ F2 (g)  LiF (s)
Li must be converted to a gas
Li must be ionized (ionization energy)
F molecules need to be broken into atoms
Form F ions
1. Li (s)  Li (g) kJ/mol
2. Li (g)  Li+ (g)+e kJ/mol
3. 1/2F2 (g)  F (g) kJ/mol
4. F (g) + e-  F- (g) kJ/mol

11. LiF…….An ionic compound between lithium and fluorine
Li (s) ½ F2 (g)  LiF (s)
Li must be converted to a gas
Li must be ionized (ionization energy)
F molecules need to be broken into atoms
Form F ions (electron affinity)
The ions are highly attracted to each other…Lattice energy
1. Li (s)  Li (g) kJ/mol
2. Li (g)  Li+ (g)+e kJ/mol
3. 1/2F2 (g)  F (g) kJ/mol
4. F (g) + e-  F- (g) kJ/mol
5. Li F-  LiF kJ/mol

12. An ionic compound forms between a metal and a nonmetal
Net change: kj/mol

13. An ionic compound forms between a metal and a nonmetal
Net change: kj/mol
- 617 kJ/mol of LiF
formed

14. An ionic compound forms between a metal and a nonmetal
The solid formed is a regular arranged pattern of ions called a crystal lattice
Net change: kj/mol
- 617 kJ/mol of LiF
formed

15. An ionic compound forms between a metal and a nonmetal
The solid formed is a regular arranged pattern of ions called a crystal lattice
Lattice energy of LiF is kJ/mol
Net change: kj/mol
- 617 kJ/mol of LiF
formed

16. As LiF (or another ionic compound dissolves in water………….

17. As LiF (or another ionic compound dissolves in water………….
Energy must be released to pull ions apart

18. As LiF (or another ionic compound dissolves in water………….
Energy must be released to pull ions apart
The quantity of energy released must be > or = to the lattice energy

19. As LiF (or another ionic compound dissolves in water………….
Energy must be released to pull ions apart
The quantity of energy released must be > or = to the lattice energy
Energy of hydration

20. Hydration of ion..dissolving
Requires and interaction with the polar water molecule
General rule of solubility…”Like dissolves like”
Nonpolar molecules require nonpolar solvents

21. Lattice Energies of Alkali Metals Halides (kJ/mol)
F Cl Br I-
Li Na
K Rb Cs

22. Lattice energy for LiF is -1047 kJ/mol….it dissolves in water
Lattice Energies of Alkali Metals Halides (kJ/mol)
F Cl Br I-
Li Na
K Rb Cs

23. Lattice energy for LiF is -1047 kJ/mol….it dissolves in water
Lattice energy for MgO is kJ/mol… it does not dissolve in water
Lattice Energies of Alkali Metals Halides (kJ/mol)
F Cl Br I-
Li Na
K Rb Cs

24. Lattice energy for LiF is -1047 kJ/mol….it dissolves in water
Lattice energy for MgO is kJ/mol… it does not dissolve in water
Lattice energy for NaF
-923kJ/mol
Lattice Energies of Alkali Metals Halides (kJ/mol)
F Cl Br I-
Li Na
K Rb Cs

25. Ion-Ion Interactions
Coulomb’s law states that the energy (E) of the interaction between two ions is directly proportional to the product of the charges of the two ions (Q1 and Q2) and inversely proportional to the distance (d) between them.

26. Predicting Forces of Attraction
Coulombs Law indicates the increases in the charges of ions will cause an increase in the force of attraction between a cation and an anion.
Increases in the distance between ions will decrease the force of attraction between them.

27. Size of Ions

28. Lattice Energy M+(g) + X-(g) ---> MX(s)
The lattice energy (U) of an ionic compound is the energy released when one mole of the ionic compound forms from its free ions in the gas phase.
M+(g) + X-(g) ---> MX(s)

29. Comparing Lattice Energies
Lattice Energies of Common Ionic Compounds
Compound
U(kJ/mol)
LiF
-1047
LiCl
-864
NaCl
-790
KCl
-720
KBr
-691
MgCl2
-2540
MgO
-3791

30. Practice Determine which salt has the greater lattice energy.
MgO and NaF
MgO and MgS

31. Lattice Energy Using Hess’s Law

32. Electron Affinity Cl(g) + e-(g) ---> Cl-(g)
Electron affinity is the energy change occurring when one mole of electrons combines with one mole of atoms or ion in the gas phase.
Step 4 in diagram on the last slide.
Cl(g) + e-(g) ---> Cl-(g)
ΔHEa = -349 kj/mole

33. Calculating U Na+(g) + e-(g) ---> Na(g) -HIE1
Na(g) ---> Na(s) -Hsub
Cl-(g) ---> Cl(g) + e-(g) -HEA
Cl(g) ---> 1/2Cl2(g) -1/2HBE
Na(s) + 1/2Cl2(g) ---> NaCl(s) Hf
Na+(g) + Cl-(g) ---> NaCl(s) U
U = Hf - 1/2HBE - HEA - Hsub - HIE1

34. Lattice energy for NaCl.