1. Ionic Bonding
Edward Wen

2. Bonding Theories
Explain how and why atoms attach together to form molecules
Explain why some combinations of atoms are stable and others are not
why is water H2O, not HO or H3O
Can be used to predict the shapes of molecules
Can be used to predict the chemical and physical properties of compounds
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

3. Lewis Bonding Theory
One of the simplest bonding theories is called Lewis Theory
Lewis Theory emphasizes valence electrons to explain bonding
Using Lewis theory, we can draw models – called Lewis structures
aka Electron Dot Structures
Lewis structures allow us to predict many properties of molecules
such as molecular stability, shape, size, polarity
G.N. Lewis
( )
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

4. Why Do Atoms Bond?
Chemical bonds form because they lower the potential energy between the ions or atoms
The potential energy of the bonded atoms is less than the potential energy of the separate atoms
To calculate this potential energy, you need to consider the following interactions:
nucleus–to–nucleus repulsions
electron–to–electron repulsions
nucleus–to–electron attractions
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

5. Types of Bonds
We can classify bonds based on the kinds of atoms that are bonded together
Types of Atoms
Type of Bond
Bond Characteristic
metals to
nonmetals
Ionic
electrons
transferred
nonmetals to
Covalent
shared
metals
Metallic
pooled
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

6. Types of Bonding
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

7. Ionic Bonds When a metal atom loses electrons it becomes a cation
metals have low ionization energy (low Zeff), making it relatively easy to remove electrons from them
When a nonmetal atom gains electrons it becomes an anion
nonmetals have high electron affinities (high Zeff), making it advantageous to add electrons to these atoms
The oppositely charged ions are then attracted to each other, resulting in an ionic bond
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

8. Valence Electrons & Bonding
Because valence electrons are held most loosely, and
Because chemical bonding involves the transfer or sharing of electrons between two or more atoms,
Valence electrons are most important in bonding
Lewis theory focuses on the behavior of the valence electrons
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

9. Stable Electron Arrangements and Ion Charge
Metals form cations by losing enough electrons to get the same electron configuration as the previous noble gas
Nonmetals form anions by gaining enough electrons to get the same electron configuration as the next noble gas
The noble gas electron configuration must be very stable
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

10. Lewis Bonding Theory
Atoms bond because it results in a more stable electron configuration.
more stable = lower potential energy
Atoms bond together by either transferring or sharing electrons
Usually this results in all atoms obtaining an outer shell with eight electrons
octet rule
there are some exceptions to this rule—the key to remember is to try to get an electron configuration like a noble gas
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

11. Octet Rule
When atoms bond, they tend to gain, lose, or share electrons to result in eight valence electrons
ns2np6
noble gas configuration
Many exceptions
H, Li, Be, B attain an electron configuration like He
He = two valence electrons, a duet
Li loses its one valence electron
H shares or gains one electron
though it commonly loses its one electron to become H+
Be loses two electrons to become Be2+
commonly shares its two electrons in covalent bonds (4 valence electrons)
B loses three electrons to become B3+
commonly shares its three electrons in covalent bonds (6 valence electrons)
expanded octets for elements in Period 3 or below
using empty valence d orbitals
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

12. Lewis Theory Predictions for Ionic Bonding
The number of electrons a metal atom should lose or a nonmetal atom should gain in order to attain a stable electron arrangement
the octet rule
This allows us to predict the formulas of ionic compounds that result
also allows us to predict the relative strengths of the resulting ionic bonds from Coulomb’s Law
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

13. Predicting Ionic Formulas Using Lewis Symbols
Electrons are transferred until the metal loses all its valence electrons and the nonmetal has an octet
Numbers of atoms are adjusted so the electron transfer comes out even
Li2O
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

14. Energetics of Ionic Bond Formation
The ionization energy of the metal is endothermic
Na(s) → Na+(g) + 1 e ─ DH° = +496 kJ/mol
The electron affinity of the nonmetal is exothermic
½Cl2(g) + 1 e ─ → Cl─(g) DH° = −244 kJ/mol
Generally, the ionization energy of the metal is larger than the electron affinity of the nonmetal, therefore the formation of the ionic compound should be endothermic (DH > 0, unfavorable)
But the heat of formation of most ionic compounds is exothermic and generally large. Why?
Na(s) + ½Cl2(g) → NaCl(s) DH°f = −411 kJ/mol
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

15. Ionic Bonding & the Crystal Lattice
The extra energy is released from the formation of a structure in which every cation is surrounded by anions, and vice versa
Formed from the electrostatic attraction of the cations for all the surrounding anions
The crystal lattice maximizes the attractions between cations and anions, forming stable arrangement
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

16. Crystal Lattice Electrostatic attraction is nondirectional!!
no direct anion–cation pair
Ionic compound does not have molecule
the chemical formula is an empirical formula, simply giving the ratio of ions based on charge balance
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

17. Lattice Energy NaCl(s) → Na+(g) + Cl-(g) DHlattice
Lattice energy: Energy required to dissociate the solid crystal into separate ions in the gas state:
NaCl(s) → Na+(g) + Cl-(g) DHlattice
always largely endothermic
Indicates the extra stability that accompanies the formation of the crystal lattice (the reverse of the above equation)
hard to measure directly, but can be calculated from knowledge of other processes
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

18. Determining Lattice Energy
The Born–Haber Cycle is devised to determine the lattice energy. Hess’ law!
Reactions involved:
Vaporization of metal: M(?) → M(g)
*Formation of nonmetal atom: X(?) → X(g)
**Ionization of metal atom: M(g) → M+(g) + ne-
**Electron affinity of nonmetal: X(g) + e-→ X-(g)
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

19. The Born-Haber cycle for lithium fluoride.
Edward Wen

20. Example: Use the Born-Haber cycle to find lattice energy for magnesium oxide
Given:
Electron Affinity (kJ/mol)
Ionization energy (kJ/mol)
Sublimation energy (kJ/mol)
Hf (kJ/mol)
O -141, 844
Mg 738, 1451
Mg 148
MgO (-601)
O(g) (249.4)
Find: ________ _____ + _____ Hlattice = ?
First write equations for the above processes:
_______________________ H1 = -141 kJ
_______________________ H2 = 844 kJ
_______________________ H3 = 738 kJ
_______________________ H4 = 1451 kJ
_______________________ H5 = 148 kJ
_______________________ H6 = -601 kJ
_______________________ H7 = kJ
Edward Wen

21. find: MgO(s)  O2-(g) + Mg2+(g) Hlattice = ?
O(g) + e-  O-(g) H1 = -141 kJ
O- (g) + e-  O2-(g) H2 = 844 kJ
Mg(g)  Mg+(g) + e- H3 = 738 kJ
Mg+(g)  Mg2+(g) + e- H4 = 1451 kJ
Mg(s)  Mg(g) H5 = 148 kJ
Mg(s) + ½O2 (g)  MgO(s) H6 = -601 kJ
½O2(g)  O(g) H7 = kJ
find: MgO(s)  O2-(g) + Mg2+(g) Hlattice = ?
Hlattice = 3890 kJ
Edward Wen

22. Trends in Lattice Energy Ion Size
The force of attraction between charged particles is inversely proportional to the distance between them
Larger ions mean the center of positive charge (nucleus of the cation) is farther away from the negative charge (electrons of the anion)
larger ion = weaker attraction
weaker attraction = smaller lattice energy
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

23. Lattice Energy vs. Ion Size
Metal chloride
Lattice energy (kJ/mol)
LiCl
834
NaCl
787
KCl
701
CsCl
657
Tro: Chemistry: A Molecular Approach, 2/e
Edward Wen

24. Trends in Lattice Energy Ion Charge
−910 kJ/mol
The force of attraction between oppositely charged particles is directly proportional to the product of the charges
Larger charge means the ions are more strongly attracted
larger charge = stronger attraction
stronger attraction = larger lattice energy
Of the two factors, ion charge is generally more important
Lattice Energy =
−3414 kJ/mol
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

25. Rank the following ionic compounds in order of increasing magnitude of lattice energy: CaO, KBr, KCl, SrO
First examine the ion charges and order by product of the charges
Ca2+& O2-, K+ & Br─, K+ & Cl─, Sr2+ & O2─
(KBr, KCl) ___ (CaO, SrO)
Then examine the ion sizes of each group and order by radius; larger < smaller
(KBr, KCl) same cation, Br─ > Cl─ (same Group)
(CaO, SrO) same anion, Sr2+ > Ca2+ (same Group)
KBr ___ KCl ___ (CaO, SrO)
KBr ___ KCl ___ SrO ___ CaO
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

26. (NaBr, LiBr) __ (MgS, SrS)
Rank the following ionic compounds in order of increasing magnitude of lattice energy: MgS, NaBr, LiBr, SrS
First examine the ion charges and order by product of the charges
Mg2+& S2-, Na+ & Br─, Li+ & Br─, Sr2+ & S2─
(NaBr, LiBr) __ (MgS, SrS)
Then examine the ion sizes of each group and order by radius; larger radius, ___ lattice energy
(NaBr, LiBr) same anion, Na+ > Li+ (same Group)
(MgS, SrS) same anion, Sr2+ > Mg2+ (same Group)
NaBr __ LiBr __ (MgS, SrS)
NaBr __ LiBr __ SrS __ MgS
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

27. Ionic Bonding: Model vs. Reality
Lewis theory implies:
The attractions between ions are strong
Ionic compounds should have high melting points (m.p.) and boiling points (b.p.) because breaking down the crystal should require a lot of energy
the stronger the attraction (larger the lattice energy), the higher the m.p.
Ionic compounds do have high m.p. and b.p.
M.p. generally > 300 °C
Most ionic compounds are solids at room temperature (except ionic liquids, (C2H5)NH3+·NO3− m.p. 12°C)
Edward Wen

28. Properties of Ionic Compounds
Melting an ionic solid
Hard and brittle crystalline solids
all are solids at room temperature
Melting points generally > 300 C
The liquid state conducts electricity
the solid state does not conduct electricity
Many are soluble in water
the solution conducts electricity well
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

29. Practice – Which ionic compound below has the highest melting point?
KBr
CaCl2
MgF2
KBr (734 ºC)
CaCl2 (772 ºC)
MgF2 (1261 ºC)
More online resource
Tro: Chemistry: A Molecular Approach, 2/e
Edward Wen

30. Ionic Bonding Model vs. Reality
Lewis theory:
The positions of the ions in the crystal lattice are critical to the stability of the structure
moving ions out of position should therefore be difficult, and ionic solids should be hard
hardness is measured by rubbing two materials together and seeing which “streaks” or cuts the other
the harder material is the one that cuts or doesn’t streak
Ionic solids are relatively hard
compared to most molecular solids
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

31. Ionic Bonding Model vs. Reality
Lewis theory implies that if the ions are displaced from their position in the crystal lattice, that repulsive forces should occur
This predicts the crystal will become unstable and break apart. Lewis theory predicts ionic solids will be brittle.
Ionic solids are brittle. When struck they shatter. + - + - + -
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

32. Ionic Bonding Model vs. Reality
To conduct electricity, a material must have charged particles that are able to flow through the material
Lewis theory implies that, in the ionic solid, the ions are locked in position and cannot move around
Lewis theory predicts that ionic solids should not conduct electricity
Ionic solids do not conduct electricity
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

33. Ionic Bonding Model vs. Reality
Lewis theory implies
in the liquid state or when dissolved in water, the ions will have the ability to move around
Both a liquid ionic compound and an ionic compound dissolved in water should conduct electricity
Ionic compounds conduct electricity in the liquid state or when dissolved in water
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

34. Conductivity of NaCl: Solid, Solution, Molten
in NaCl(s), the ions are stuck in position and not allowed to move to the charged rods
in NaCl(aq), the ions are separated and allowed to move to the charged rods
in NaCl(l), the ions are mobile and allowed to move to the charged rods
Edward Wen
Tro: Chemistry: A Molecular Approach, 2/e

35. Example: Use the Born-Haber cycle to find lattice energy for magnesium oxide
Given:
Electron Affinity (kJ/mol)
Ionization energy (kJ/mol)
Sublimation energy (kJ/mol)
Hf (kJ/mol)
O -141, 844
Mg 738, 1451
Mg 148
MgO (-601)
O(g) (249.4)
Find: MgO(s)  O2-(g) + Mg2+(g) Hlattice = ?
First write equations for the above processes:
O(g) + e-  O-(g) H1 = -141 kJ
O- (g) + e-  O2-(g) H2 = 844 kJ
Mg(g)  Mg+(g) + e- H3 = 738 kJ
Mg+(g)  Mg2+(g) + e- H4 = 1451 kJ
Mg(s)  Mg(g) H5 = 148 kJ
Mg(s) + ½O2 (g)  MgO(s) H6 = -601 kJ
½O2 (g)  O(g) H7 = kJ
Edward Wen

36. How to find: MgO(s)  O2-(g) + Mg2+(g) Hlattice = ?
O(g) + e-  O-(g) H1 = -141 kJ
O- (g) + e-  O2-(g) H2 = 844 kJ
Mg(g)  Mg+(g) + e- H3 = 738 kJ
Mg+(g)  Mg2+(g) + e- H4 = 1451 kJ
Mg(s)  Mg(g) H5 = 148 kJ
Mg(s) + ½O2 (g)  MgO(s) H6 = -601 kJ
½O2(g)  O(g) H7 = kJ
How to find: MgO(s)  O2-(g) + Mg2+(g) Hlattice = ?
MgO(s)  Mg(s) + ½O2 (g) -H6 = 601 kJ
O- (g) + e-  O2-(g) H2 = 844 kJ
Mg+(g)  Mg2+(g) + e- H4 = 1451 kJ
O(g) + e-  O-(g) H1 = -141 kJ
Mg(g)  Mg+(g) + e- H3 = 738 kJ
Mg(s)  Mg(g) H5 = 148 kJ
½O2 (g)  O(g) H7 = kJ
Hlattice =-H6 + (H1 + H2 + H3 + H4 + H4 + H5 + H7) = 3890 kJ
Edward Wen