1. 7 Ionic Bonding 7.1 Formation of Ionic Bonds: Donating and Accepting
Electrons
7.2 Energetics of Formation of Ionic Compounds
7.3 Stoichiometry of Ionic Compounds
7.4 Ionic Crystals
7.5 Ionic Radii

2. Sodium When sodium exposed in air, it becomes tarnished rapidly
Introduction (SB p.186)
Sodium
When sodium exposed in air, it becomes tarnished rapidly
 Reacts with oxygen in air
 Form a dull oxide layer on the metal surface

3. When sodium is placed in a bottle containing chlorine gas
Introduction (SB p.186)
When sodium is placed in a bottle containing chlorine gas
 Burns fiercely
 Gives a white coating of sodium chloride

4. Noble gases Very stable
Introduction (SB p.186)
Noble gases
Very stable
Rarely participate in chemical reactions and form bonds with other elements
 Octet configuration

5. Formation of compounds
Introduction (SB p.186)
Formation of compounds
Transfer or sharing of valence electron(s) takes place
Atoms achieve the electronic configuration of the nearest noble gas in the Periodic Table
Atoms are joined together by chemical bonds

6. Three types of chemical bonds
Introduction (SB p.186)
Three types of chemical bonds
1. Ionic bond
Electrostatic attraction between positively charged particles and negatively charged particles

7. Three types of chemical bonds
Introduction (SB p.186)
Three types of chemical bonds
2. Covalent bond
Electrostatic attraction between nuclei and shared electrons

8. Three types of chemical bonds
Introduction (SB p.186)
Three types of chemical bonds
3. Metallic bond
Electrostatic attraction between metallic cations and delocalized electrons (electrons that have no fixed positions)
Let's Think 1

9. Formation of Ionic Bonds: Donating and Accepting Electrons
7.1
Formation of Ionic Bonds: Donating and Accepting Electrons

10. 7. 1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p
Formed by a transfer of electrons from metallic atoms to non-metallic atoms
e.g. Formation of sodium chloride
Both the sodium ion and chloride ion attain the electronic configurations of noble gases which give rise to stability

11. Formation of ionic bond between sodium atom and chlorine atom
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
Formation of ionic bond between sodium atom and chlorine atom Cl Na
Sodium atom, Na
1s22s22p63s1
Chlorine atom, Cl
1s22s22p63s23p5

12. linked up together by ionic bond
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
Formation of ionic bond between sodium atom and chlorine atom + - Cl Na
linked up together by ionic bond
Sodium ion, Na+
1s22s22p6
Chloride ion, Cl-
1s22s22p63s23p6

13. Ionic Bonds: Donating and Accepting Electrons
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
Ionic Bonds: Donating and Accepting Electrons

14. Internuclear distance
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
Ionic Bonds: Donating and Accepting Electrons + –
Internuclear distance

15. Internuclear distance
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
Ionic Bonds: Donating and Accepting Electrons + – – +
Cationic radius (r+)
Anionic radius (r-)
Internuclear distance
Internuclear distance = r+ + r-

16. Ionic Bonds: Donating and Accepting Electrons
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
Ionic Bonds: Donating and Accepting Electrons
Ionic bonds are the strong non-directional electrostatic attraction between ions of opposite charges.

17. 7. 1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p
Electron transfer from a magnesium atom to two chlorine atoms
Electron transfer from two lithium atoms to an oxygen atom

18. Energetics of Formation of Ionic Compounds
7.2
Energetics of Formation of Ionic Compounds

19. Energetics of Formation of Ionic Compound
7.2 Energetics of Formation of Ionic Compounds (SB p.189)
Energetics of Formation of Ionic Compound
 Hf ø
macroscopic level
Na(s) Cl2(g)  NaCl(s)
Actually passing through many steps at the molecular level
microscopic level

20. Consider the formation of the ionic compound via a serious of steps:
7.2 Energetics of Formation of Ionic Compounds (SB p.189)
Consider the formation of the ionic compound via a serious of steps:
1. The conversion of the elements to the gaseous atoms (standard enthalpy change of atomization, )

21. Consider the formation of the ionic compound via a serious of steps:
7.2 Energetics of Formation of Ionic Compounds (SB p.189)
Consider the formation of the ionic compound via a serious of steps:
2. The conversion of the gaseous atoms to gaseous ions (ionization enthalpy, and electron affinity, )

22. Consider the formation of the ionic compound via a serious of steps:
7.2 Energetics of Formation of Ionic Compounds (SB p.189)
Consider the formation of the ionic compound via a serious of steps:
3. The combination of the gaseous ions to form an ionic crystal (lattice enthalpy, )

23. 1. Standard Enthalpy Change of Formation (H f) ø
7.2 Energetics of Formation of Ionic Compounds (SB p.189)
1. Standard Enthalpy Change of Formation (H f) ø
The enthalpy change when one mole of the ionic compound is formed from its constituent elements (in their standard states) under standard conditions.
Na(s) Cl2(g) NaCl(s) Hf = –411 kJ mol-1 ø

24. 2. Standard Enthalpy Change of Atomization (H atom) ø
7.2 Energetics of Formation of Ionic Compounds (SB p.190)
2. Standard Enthalpy Change of Atomization (H atom) ø
The enthalpy change when one mole of gaseous atoms is formed from an element in the standard state under standard conditions.
Na(s) Na(g) H atom [Na(s)] = kJ mol-1 ø
Cl2(g) Cl(g) H atom [Cl2(g)] = +121 kJ mol-1 ø
Questions:
Why are the changes endothermic?
What type of bond is broken in each case?

25. 3. Ionization Enthalpy (HI.E.)
7.2 Energetics of Formation of Ionic Compounds (SB p.190 – 191)
3. Ionization Enthalpy (HI.E.)
The energy required to remove one mole of electrons from one mole of atoms or ions in the gaseous state.
Na(g) Na+(g) + e H I.E [Na(g)] = +494 kJ mol-1
Mg(g) Mg+(g) + e H I.E [Mg(g)] = +736 kJ mol-1
Mg+(g) Mg2+(g) + e- H I.E [Na(g)] = kJ mol-1
Questions:
Why are the changes endothermic?

26. 4. Electron affinity (ΔHE.A.)
7.2 Energetics of Formation of Ionic Compounds (SB p.191)
4. Electron affinity (ΔHE.A.)
The enthalpy change when one mole of electrons is added to one mole of atoms or ions in the gaseous state.
First electron affinity of O(g):
O(g) + e O-(g) H E.A [O(g)] = –142 kJ mol-1
Second electron affinity of O(g):
O-(g) + e O2-(g) H E.A [O(g)] = –844 kJ mol-1
Questions:
Why may E.A. have -ve or +ve values?

27. Electron affinities (in kJ mol–1) of some elements and ions
7.2 Energetics of Formation of Ionic Compounds (SB p.192)
Electron affinities (in kJ mol–1) of some elements and ions

28. 5. Lattice enthalpy ( ΔHlattice) ø
7.2 Energetics of Formation of Ionic Compounds (SB p.192)
5. Lattice enthalpy ( ΔHlattice) ø
The enthalpy change when one mole of an ionic crystal is formed from its constituent ions in the gaseous state under standard conditions.
Na+ (g) + Cl-(g) NaCl(s) H lattice [Na+Cl-(s)] ø + –

29. Why can’t L.E. be determined directly from experiments?
7.2 Energetics of Formation of Ionic Compounds (SB p.192)
Na+ (g) + Cl-(g) NaCl(s) H lattice [Na+Cl-(s)] ø
+ve or -ve?
L.E. can be calculated from the values of other experimentally determined enthalpy changes by constructing a Born-Haber cycle and applying Hess’s law + – + – + –
Questions:
Why can’t L.E. be determined directly from experiments?

30. 7.2 Energetics of Formation of Ionic Compounds (SB p.193)
Born-Haber Cycle
A simplified enthalpy level diagram used to calculate the lattice enthalpy of an ionic compound.
Two different routes to form an ionic compound
Route 1: Direct single-step reaction of the elements to form the ionic compound
Route 2: Consists of a number of steps. The enthalpy change of each step can be found from experiments, except the lattice enthalpy

31. Born-Haber Cycle for the formation of sodium chloride
7.2 Energetics of Formation of Ionic Compounds (SB p.193)
Born-Haber Cycle for the formation of sodium chloride

32. 7.2 Energetics of Formation of Ionic Compounds (SB p.194)
Or draw enthalpy level diagram to represent the enthalpy changes in the Born-Haber cycle
Example 7-2

33. 7.2 Energetics of Formation of Ionic Compounds (SB p.196)
Lattice enthalpy
A measure of ionic bond strength which in turn represents the strength of the ionic lattice.
The higher (more negative) the lattice enthalpy of an ionic lattice  The higher is the ionic bond strength  The more stable is the ionic lattice

34. Factors affect lattice enthalpy
7.2 Energetics of Formation of Ionic Compounds (SB p.196)
Factors affect lattice enthalpy
Let's Think 2
Effect of ionic size:
 The greater the ionic size  The lower (or less negative) is the lattice enthalpy
Effect of ionic charge:
 The greater the ionic charge  The higher (or more negative) is the lattice enthalpy
Check Point 7-2

35. Stoichiometry of Ionic Compounds
7.3
Stoichiometry of Ionic Compounds

36. 7.3 Stoichiometry of Ionic Compounds (SB p.197)
Stoichiometry of a compound is the simplest ratio of the atoms bonded to form the compound.
How can the stoichiometry of an ionic compound be determined?

37. A. In Terms of Electronic Configuration
7.3 Stoichiometry of Ionic Compounds (SB p.197 – 198)
A. In Terms of Electronic Configuration
Example
magnesium chloride
Elements involved
Mg (Group II) Cl (Group VII)
Ions formed
Mg Cl- 2 1
Ratio of ions
Chemical formula
Mg2+(Cl-) or MgCl2

38. B. In Terms of Enthalpy Change of Formation
7.3 Stoichiometry of Ionic Compounds (SB p.198)
B. In Terms of Enthalpy Change of Formation
The more negative the enthalpy change of formation of an ionic compound
 The greater is the driving force for its formation  The more stable the compound
Check Point 7-3

39. 7.4
Ionic Crystals

40. Structure of Sodium Chloride
7.4 Ionic Crystals (SB p.201)
Structure of Sodium Chloride
Unit cell of NaCl
Co-ordination number of Na+ = 6
6 : 6 co-ordination
Co-ordination number of Cl- = 6

41. Face-centred cubic lattice
7.4 Ionic Crystals (SB p.202)
Face-centred cubic lattice

42. 7.4 Ionic Crystals (SB p.202)
A unit cell is the smallest basic portion of the crystal lattice that, when repeatedly stacked together at various directions, can reproduce the entire crystal structure.

43. Structure of Caesium Chloride
7.4 Ionic Crystals (SB p.202)
Structure of Caesium Chloride
Simple cubic lattice
Co-ordination number of Cs+ = 8
8 : 8 co-ordination
Co-ordination number of Cl- = 8

44. Face-centred cubic lattice
7.4 Ionic Crystals (SB p.203)
Structure of Calcium Fluoride
Face-centred cubic lattice
Co-ordination number of Ca+ = 8
8 : 4 co-ordination
Co-ordination number of F- = 4

45. Some simple ionic structures
7.4 Ionic Crystals (SB p.203)
Some simple ionic structures
Type of structure
Examples
Radius Ratio
(r+ : r-)*
Coordination
Sodium
chloride
Na+Cl-, Na+Br-,
K+Cl-, K+Br-
< 0.732
> 0.414
6 : 6
Caesium
Cs+Cl-, Cs+Br-,
Cs+I-
> 0.732
8 : 8
Calcium fluoride
CaF2, BaF2, BaCl2, SrF2
8 : 4
Example 7-4
Check Point 7-4

46. 7.5
Ionic Radii

47. X-ray and electron diffraction technique
7.5 Ionic Radii (SB p.205)
X-ray and electron diffraction technique
X-ray
Photographic plate

48. Electron density plot for sodium chloride crystal
7.5 Ionic Radii (SB p.205)
Electron density plot for sodium chloride crystal

49. A. Cations Smaller radius than the corresponding atom Reasons:
7.5 Ionic Radii (SB p.206)
A. Cations
Smaller radius than the corresponding atom
Reasons:
1. The number of electron shells decreases
2. No. of protons > No. of electrons (p/e ratio increases) The nuclear attraction is more effective to cause a contraction in the electron cloud

50. Size of ion vs size of atom
7.5 Ionic Radii (SB p.206)
Size of ion vs size of atom
Comparing relative atomic radii of some elements with the ionic radii of the corresponding ions

51. B. Anions Larger radius than the corresponding atom Reasons:
7.5 Ionic Radii (SB p.206)
B. Anions
Larger radius than the corresponding atom
Reasons:
1. Repulsion between newly added electron(s) with other electrons 2. No. of protons < No. of electrons (p/e ratio decreases) The nuclear attraction is less effective and there is an expansion of the electron cloud

52. C. Isoelectronic Ions They have the same number of electrons
7.5 Ionic Radii (SB p.206)
C. Isoelectronic Ions
They have the same number of electrons
Sizes decrease along the isoelectronic series:
1. H– > Li+ > Be2+ > B3+ (isoelectronic to He)
2. N3– > O2– > F– > Na+ > Mg2+ > Al (isoelectronic to Ne)
3. P3– > S2– > Cl– > K+ > Ca2+ (isoelectronic to Ar)

53. C. Isoelectronic Ions Reason:
7.5 Ionic Radii (SB p.206)
C. Isoelectronic Ions
Reason:
They have the same number of electrons. An increase in the number of protons implies an increase in the p/e ratio which leads to a contraction of the electron cloud

54. 7.5 Ionic Radii (SB p.206)
isoelectronic ions
Why ionic radius decreases along the isoelectronic series?
Example 7-5
Check Point 7-5

55. The END

56. Introduction (SB p.186)
Let's Think 1
Why do two atoms bond together? How does covalent bond strength compare with ionic bond strength?
The two atoms tend to achieve an octet configuration which brings stability.
Answer
Back

57. Example 7-2 Given the following data: ΔH (kJ mol–1)
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
Example 7-2
Given the following data:
ΔH (kJ mol–1)
First electron affinity of oxygen –142
Second electron affinity of oxygen +844
Standard enthalpy change of atomization of oxygen +248
Standard enthalpy change of atomization of aluminium +314
Standard enthalpy change of formation of aluminium oxide –1669

58. Example 7-2 Answer ΔH (kJ mol–1)
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
Answer
Example 7-2
ΔH (kJ mol–1)
First ionization enthalpy of aluminium +577
Second ionization enthalpy of aluminium
Third ionization enthalpy of aluminium
(a) (i) Construct a labelled Born-Haber cycle for the formation of aluminium oxide.
(Hint: Assume that aluminium oxide is a purely ionic compound.)
(ii) State the law in which the enthalpy cycle in (i) is based on.
(b) Calculate the lattice enthalpy of aluminium oxide.

59. Example 7-2 7.2 Energetics of Formation of Ionic Compounds (SB p.195)
(ii) The enthalpy cycle in (i) is based on Hess’s law which states that the total enthalpy change accompanying a chemical reaction is independent of the route by means of which the chemical reaction is brought about.

60. 7.2 Energetics of Formation of Ionic Compounds (SB p.195)
Back
Example 7-2
(b) ΔHf [Al2O3(s)] = 2 × ΔHatom[Al(s)] + 2 × (ΔHI.E.1 [Al(g)]
+ ΔHI.E.2 [Al(g)] + ΔHI.E.3 [Al(g)])
+ 3 × ΔHatom [O2(g)] + 3 × (ΔHE.A.1 [O(g)]
+ ΔHE.A.2 [O(g)]) + ΔHlattice[Al2O3(s)]
ΔHf [Al2O3(s)] = 2 × (+314) + 2 × ( )
+ 3 × (+248) + 3 × (– )
+ ΔHlattice [Al2O3(s)]
ΔHf [Al2O3(s)] = ΔHlattice[Al2O3(s)]
ΔHlattice[Al2O3(s)] = ΔHf [Al2O3(s)] – ( )
= –1 669 – ( )
= – kJ mol–1 ø

61. 7.2 Energetics of Formation of Ionic Compounds (SB p.196)
Let's Think 2
What are the forces that hold atoms together in molecules and ions in ionic compounds?
Electrostatic attractions between oppositely charged particles
Answer
Back

62. 7.2 Energetics of Formation of Ionic Compounds (SB p.197)
Check Point 7-2
(a) Draw a Born-Haber cycle for the formation of magnesium oxide.
The Born-Haber cycle for the formation of MgO:
Answer

63. 7.2 Energetics of Formation of Ionic Compounds (SB p.197)
Check Point 7-2
(b) Calculate the lattice enthalpy of magnesium oxide by means of the Born-Haber cycle drawn in (a).
Given: ΔHatom [Mg(s)] = +150 kJ mol–1
ΔHI.E. [Mg(g)] = +736 kJ mol–1
ΔHI.E. [Mg+(g)] = kJ mol–1
ΔHatom [O2(g)] = +248 kJ mol–1
ΔHE.A. [O(g)] = –142 kJ mol–1
ΔHE.A. [O–(g)] = +844 kJ mol–1
ΔHf [MgO(s)] = –602 kJ mol–1 ø ø
Answer ø

64. 7.2 Energetics of Formation of Ionic Compounds (SB p.197)
Check Point 7-2
(b) ΔHlattice [MgO(s)]
= ΔHf [MgO(s)] – ΔHatom [Mg(s)]
– ΔHI.E. [Mg(g)] – ΔHI.E. [Mg+(g)] – ΔHatom [O2(g)]
– ΔHE.A. [O(g)] – ΔHE.A. [O–(g)]
= [–602 – 150 – 736 – – 248 –(–142) – 844] kJ mol–1
= –3 888 kJ mol–1 ø
Back

65. 7.3 Stoichiometry of Ionic Compounds (SB p.201)
Check Point 7-3
Give two properties of ions that will affect the value of the lattice enthalpy of an ionic compound.
Answer
The charges and sizes of ions will affect the value of the lattice enthalpy.
The smaller the sizes and the higher the charges of ions, the higher (i.e. more negative) is the lattice enthalpy.
Back

66. 7.4 Ionic Crystals (SB p.204)
Back
Example 7-4
Write down the formula of the compound that possesses the lattice structure shown on the right:
To calculate the number of each type of particle present in the unit cell:
Number of atom A = 1
(1 at the centre of the unit cell)
Number of atom B = 8 × = 2
(shared along each edge)
Number of atom C = 8 × = 1
(shared at each corner)
∴ The formula of the compound is AB2C.
Answer

67. Check Point 7-4 Answer Back
7.4 Ionic Crystals (SB p.205)
Back
Check Point 7-4
The diagram on the right shows a unit cell of titanium oxide. What is the coordination number of
(a) titanium; and
(b) oxygen?
Answer
(a) The coordination number of titanium is 6 as there are six oxide ions surrounding each titanium ion.
(b) The coordination number of oxygen is 3.

68. 7.5 Ionic Radii (SB p.208)
Example 7-5
The following table gives the atomic and ionic radii of some Group IIA elements.
Element
Atomic radius (nm)
Ionic radius Be
0.112
0.031 Mg
0.160
0.065 Ca
0.190
0.099 Sr
0.215
0.133 Ba
0.217
0.135

69. Example 7-5 Answer Explain briefly the following:
7.5 Ionic Radii (SB p.208)
Example 7-5
Explain briefly the following:
(a) The ionic radius is smaller than the atomic radius in each element.
(b) The ratio of ionic radius to atomic radius of Be is the lowest.
(c) The ionic radius of Ca is smaller than that of K (0.133 nm).
Answer

70. Example 7-5 7.5 Ionic Radii (SB p.208)
(a) One reason is that the cation has one electron shell less than the corresponding atom. Another reason is that in the cation, the number of protons is greater than the number of electrons. The electron cloud of the cation therefore experiences a greater nuclear attraction. Hence, the ionic radius is smaller than the atomic radius in each element.
(b) In the other cations, although there are more protons in the nucleus, the outer most shell electrons are further away from the nucleus, and electrons in the inner shells exhibit a screening effect. Be has the smallest atomic size. In Be2+ ion, the electrons experience the greatest nuclear attraction. Therefore, the contraction in size of the electron cloud is the greatest when Be2+ ion is formed, and the ratio of ionic radius to atomic radius of Be is the lowest.

71. Example 7-5 Back 7.5 Ionic Radii (SB p.208)
(c) The electronic configurations of both K+ and Ca2+ ions are 1s22s22p63s23p6. Hence they have the same number and arrangement of electrons. However, Ca2+ ion is doubly charged while K+ ion is singly charged, so the outermost shell electrons of Ca2+ ion experience a greater nuclear attraction. Hence, the ionic radius of Ca2+ ion is smaller than that of K+ ion.
Back

72. 7.5 Ionic Radii (SB p.208)
Check Point 7-5
Arrange the following atoms or ions in an ascending order of their sizes:
(a) Be, Ca, Sr, Ba, Ra, Mg
(b) Si, Ge, Sn, Pb, C
(c) F–, Cl–, Br–, I–
(d) Mg2+, Na+, Al3+, O2–, F–, N3–
(a) Be < Mg < Ca < Sr < Ba < Ra
(b) C < Si < Ge < Sn < Pb
(c) F– < Cl– < Br– < I–
(d) Al3+ < Mg2+ < Na+ < F– < O2– < N3–
Answer
Back