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. Bonding & Structure

3. Lewis Model
G.N. Lewis in 1916
Only the outermost (valence) electrons are involved significantly in bond formation
Successful in solving chemical problems

4. Why are some elements so reactive. (e. g. Na) and others inert (e. g
Why are some elements so reactive (e.g.Na) and others inert (e.g. noble gases)?
Why are there compounds with chemical formulae H2O and NaCl, but not H3O and NaCl2?
Why are helium and the other noble gases monatomic, when molecules of hydrogen and chlorine are diatomic?

5. Chemical bonds are strong. electrostatic forces holding atoms
Chemical bonds are strong electrostatic forces holding atoms or ions together, which are formed by the rearrangement (transfer or sharing) of outermost electrons
Atoms tend to form chemical bonds in such a way as to achieve the electronic configurations of the nearest noble gases (The Octet Rule )

6. Q.1 Write the s,p,d,f notation for the ions in the table below

8. Three types of chemical bonds
Introduction (SB p.186)
Three types of chemical bonds
1. Ionic bond (electrovalent bond)
Formed by transfer of electrons

9. Three types of chemical bonds
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
Three types of chemical bonds
1. Ionic bond (electrovalent bond) Cl Na
Sodium atom, Na
1s22s22p63s1
Chlorine atom, Cl
1s22s22p63s23p5

10. 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

11. Three types of chemical bonds
Introduction (SB p.186)
Three types of chemical bonds
Covalent bond
Formed by sharing of electrons

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

13. 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)

14. Three types of chemical bonds
Introduction (SB p.186)
Three types of chemical bonds
3. Metallic bond
Formed by sharing of a large number of delocalized electrons
Let's Think 1

15. Ionic bonds and Covalent bonds are only extreme cases of a continuum.
In real situation, most chemical bonds are intermediate between ionic and covalent

16. Ionic Bonds with incomplete transfer of electrons have covalent character.
Covalent Bonds with unequal sharing of electrons have ionic character.

17. Electronegativity and Types of Chemical Bonds
Ionic or covalent depends on the electron-attracting ability of bonding atoms in a chemical bond.
Ionic bonds are formed between atoms with great difference in their electron-attracting abilities
Covalent bonds are formed between atoms with small or no difference in their electron-attracting abilities.

18. Ways to compare the electron-attracting ability of atoms
Ionization Enthalpy
Electron affinity
Electronegativity

19. Electronegativity and Types of Chemical Bonds
Ionization enthalpy
The enthalpy change when one mole of electrons are removed from one mole of atoms or positive ions in gaseous state.
X(g)  X+(g) e H 1st I.E.
X+(g)  X2+(g) e H 2nd I.E.
Ionization enthalpies are always positive.

20. Electronegativity and Types of Chemical Bonds
2. Electron affinity
The enthalpy change when one mole of electrons are added to one mole of atoms or negative ions in gaseous state.
X(g) e-  X-(g) H 1st E.A.
X(g) e-  X2(g) H 2nd E.A.
Electron affinities can be positive or negative.

21. I.E. and E.A. only show e- releasing/attracting power of
free, isolated atoms
However, whether a bond is ionic or covalent depends on the ability of atoms to attract electrons in a chemical bond.

22. Electronegativity and Types of Chemical Bonds
The ability of an atom to attract electrons in a chemical bond.

23. Mulliken’s scale of electronegativity
Electronegativity(Mulliken) 
Nobel Laureate in Chemistry, 1966

24. Pauling’s scale of electronegativity
Nobel Laureate in Chemistry, 1954
Nobel Laureate in Peace, 1962

25. Pauling’s scale of electronegativity
calculated from bond enthalpies
cannot be measured directly
having no unit
always non-zero
the most electronegative element, F, is arbitrarily assigned a value of 4.00

26. Pauling’s scale of electronegativity

27. What trends do you notice about the EN values in the Periodic Table
What trends do you notice about the EN values in the Periodic Table? Explain.

28. The EN values increase from left to right across a Period.

29. What trends do you notice about the EN values in the Periodic Table
What trends do you notice about the EN values in the Periodic Table? Explain.
The EN values increase from left to right across a Period.
The atomic radius decreases from left to right across a Period.Thus,the nuclear attraction experienced by the bonding electrons increases accordingly.

30. The EN values decrease down a Group.

31. What trends do you notice about the EN values in the Periodic Table
What trends do you notice about the EN values in the Periodic Table? Explain.
The EN values decrease down a Group.
The atomic radius increases down a Group, thus weakening the forces of attraction between the nucleus and the bonding electrons.

32. Why are the E.A. of noble gases not shown ?

33. XeF2, XeF4 and XeF6 are present
Why are the E.A of noble gases not shown ?
EN is a measure of the ability of an atom to attract electrons in a chemical bond.
However, noble gases are so inert that they rarely form chemical bonds with other atoms.
XeF2, XeF4 and XeF6 are present

34. Formation of Ionic Bond
Atoms of Group IA and IIA elements ‘tend’ to achieve the noble gas structures in the previous Period by losing outermost electron(s).
In fact, formations of positive ions from metals are endothermic and not spontaneous.
I.E. values are always positive

35. Formation of Ionic Bond
Atoms of Group VIA and VIIA elements tend to achieve the noble gas structures in the same Periods by gaining electron(s)
First E.A. of Group VIA and VIIA elements are always negative.
 Spontaneous and exothermic processes

36. The oppositely charged ions are stabilized by coming close to each other to form the ionic bond.
Ionic bond is the result of electrostatic interaction between oppositely charged ions.
Interaction = attraction + repulsion

37. Dots and Crosses Diagram

38. Notes on Dots & Crosses representation
Electrons in different atoms are indistinguishable.
The dots and crosses do not indicate the exact positions of electrons.
Not all stable ions have the noble gas structures(Refer to pp.4-5).

39. Tendency for the Formation of Ions
An ion will be formed easily if
1. The electronic structure of the ion is stable;
2. The charge on the ion is small;
3. The size of parent atom from which the ion is formed is
small for an anion, or
large for a cation.

40. For cation, Larger size of parent atom
 less positive I.E. or less +ve sum of successive I.E.s,
 easier formation of cation,
 atoms of Group IA & Group IIA elements form cations easily.

41. Li+ Be2+ Na+ Mg2+ Al3+ K+ Ca2+ Sc3+ Rb+ Sr2+ Y3+ Zr4+
Ease of formation of cation 
Ease of formation of cation 
Li+ Be2+
Na+ Mg2+ Al3+
K+ Ca2+ Sc3+
Rb+ Sr2+ Y3+ Zr4+
Cs+ Ba2+ La3+ Ce4+

42. For anion, Smaller size of parent atom
 more negative E.A. or more -ve sum of
successive E.A.s,
 easier formation of anion,
atoms of Group VIA & Group VIIA elements form anions easily.

43. N3 O2 F P3 S2 Cl Br I Ease of formation of anion 

44. Stable Ionic Structures
Not all stable ions have the noble gas electronic structures.
Q.2 Write the s,p,d,f notation for each of the following cations.
Use your answers to identify three types of commonly occurring arrangement of outershell electrons of cations other than the stable octet structure.

45. Q.2 Write the s,p,d,f notation for the following cations

46. Q.2

47. 1. 18 - electrons group e.g. Zn2+
Fully-filled s, p & d-subshells

48. Fully-filled s, p & d-subshells
2. ( ) - electron group e.g. Pb2+
Fully-filled s, p & d-subshells

49. 3. Variable arrangements for ions of. transition. metals
3. Variable arrangements for ions of transition metals ns2,np6, nd1 to ns2,np6, nd9
Ti3+ [Ne] 3s23p63d1 Mn3+ [Ne] 3s23p63d4

50. Electronic arrangements of stable cations
Ionization enthalpy determines the ease of formation of cations.
Less positive I.E. or sum of I.E.s
 easier formation of cations

51. Electronic arrangements of stable cations
Octet structure : -
cations of Group IA, IIA and IIIB

52. (a) 18 - electrons group
cations of Groups IB, IIB, IIIA and IVA

53. Less stable than ions with a noble gas structure.

54. Cu and Au form other ions because the nuclear charges are not high enough to hold the 18-electron group firmly.

55. Cu2+ 2, 8, 17 Au3+ 2, 8, 18, 32, 16
The d-electrons are more diffused and thus less attracted by the less positive nuclei.

56. due to presence of 4f electrons.
(b) (18+2) electrons
Cations of heavier group members
due to presence of 4f electrons.
Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1
Tl3+ [2,8,18,32], 5s2, 5p6, 5d10 (18)
Tl+ [2,8,18,32], 5s2, 5p6, 5d10, 6s2 (18 + 2)
Stability : Tl+ > Tl3+ due to extra stability of
6s electrons (inert pair effect)

57. (b) (18+2) electrons Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1
6s electrons are poorly shielded by the inner 4f electrons.
6s electrons experience stronger nuclear attraction.

58. (b) (18+2) electrons Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1
Inert pair effect increases down a Group.
Stability : Sn4+(18) > Sn2+(18+2)
Pb2+(18+2) > Pb4+(18)
moving down Group IV

59. (b) (18+2) electrons Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1
Inert pair effect increases down a Group.
Stability : Sb5+(18) > Sb3+(18+2)
Bi3+(18+2) > Bi5+(18)
moving down Group V

60. (c) Cations of transition elements
- variable oxidation numbers
- Electronic configurations from
ns2, np6, nd1 to ns2, np6, nd9
Fe [Ne] 3s2, 3p6, 3d6, 4s2
Fe2+ [Ne] 3s2, 3p6, 3d6
Fe3+ [Ne] 3s2, 3p6, 3d5

61. (c) Cations of transition elements
Fe [Ne] 3s2, 3p6, 3d6, 4s2
Fe2+ [Ne] 3s2, 3p6, 3d6
Fe3+ [Ne] 3s2, 3p6, 3d5
Which one is more stable, Fe2+(g) or Fe3+(g) ?
Fe2+(g) is more stable than Fe3+(g)
Energy is needed to remove electrons from Fe2+(g) to give Fe3+(g)

62. B. Anions - with noble gas structures
Electron affinity determines the ease of formation of anions.
More -ve E.A. or sum of E.A.s
 more stable anion
Group VIA and Group VIIA elements form anions easily.

63. First Electron Affinity (kJ mol1) X(g) + e  X(g)H
-73 He
+21 Li
-60 Be
+18 B
-23 C
-122 N ~0 O
-142 F
-322 Ne
+29 Na
-53 Mg Al
-44 Si
-134 P
-72 S
-200 Cl
-349 Ar
+35 K
-48 Br
-325 Kr
+39

64. E.A. becomes more –ve from Gp 1 to Gp 7 across a periodH
-73 He
+21 Li
-60 Be
+18 B
-23 C
-122 N ~0 O
-142 F
-328 Ne
+29 Na
-53 Mg Al
-44 Si
-134 P
-72 S
-200 Cl
-349 Ar
+35 K
-48 Br
-324 Kr
+39

65. The electrons added experience stronger nuclear attraction when the atoms are getting smaller across the period. H
-73 He
+21 Li
-60 Be
+18 B
-23 C
-122 N ~0 O
-142 F
-328 Ne
+29 Na
-53 Mg Al
-44 Si
-134 P
-72 S
-200 Cl
-349 Ar
+35 K
-48 Br
-324 Kr
+39

66. +ve E.A. for Gp 2 and Gp 0 because the electron added occupies a new shell / subshell
-73 He
+21 Li
-60 Be
+18 B
-23 C
-122 N ~0 O
-142 F
-328 Ne
+29 Na
-53 Mg Al
-44 Si
-134 P
-72 S
-200 Cl
-349 Ar
+35 K
-48 Br
-324 Kr
+39

67. Goes to a new subshell Goes to a new shell H -73 He +21 Li -60 Be +18
Be(2s2)  Be(2s22p1)
Ne(2p6)  Ne(2p63s1) H
-73 He
+21 Li
-60 Be
+18 B
-23 C
-122 N ~0 O
-142 F
-328 Ne
+29 Na
-53 Mg Al
-44 Si
-134 P
-72 S
-200 Cl
-349 Ar
+35 K
-48 Br
-324 Kr
+39

68. More –ve E.A. for Gp 1 because the ions formed have full-filled s-subshells.
E.g. Li(1s22s1)  Li(1s22s2) H
-73 He
+21 Li
-60 Be
+18 B
-23 C
-122 N ~0 O
-142 F
-328 Ne
+29 Na
-53 Mg Al
-44 Si
-134 P
-72 S
-200 Cl
-349 Ar
+35 K
-48 Br
-324 Kr
+39

69. Less –ve E.A. for Gp 5 because the stable half-filled p-subshell no longer exists.
E.g. N(2s22p3)  N(2s22p4) H
-73 He
+21 Li
-60 Be
+18 B
-23 C
-122 N ~0 O
-142 F
-328 Ne
+29 Na
-53 Mg Al
-44 Si
-134 P
-72 S
-200 Cl
-349 Ar
+35 K
-48 Br
-324 Kr
+39

70. H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na
Q.3 Why are the first E.A.s of halogen atoms more negative than those of the O atom or the S atom ? H
-73 He
+21 Li
-60 Be
+18 B
-23 C
-122 N ~0 O
-142 F
-328 Ne
+29 Na
-53 Mg Al
-44 Si
-134 P
-72 S
-200 Cl
-349 Ar
+35 K
-48 Br
-324 Kr
+39

71. It is because the halide ions formed have full-filled s/p subshells (octet structure).
-73 He
+21 Li
-60 Be
+18 B
-23 C
-122 N ~0 O
-142 F
-328 Ne
+29 Na
-53 Mg Al
-44 Si
-134 P
-72 S
-200 Cl
-349 Ar
+35 K
-48 Br
-324 Kr
+39

72. The electrons added are repelled strongly by the negative ions.
Q.4 Why are the second E.A.s of O(+844 kJmol1) and S(+532 kJmol1) positive ?
O(g) + e  O2(g)
S(g) + e  S2(g)
The electrons added are repelled strongly by the negative ions.

73. Why is the E.A. of F less negative than that of Cl ?
Halogen F Cl Br I
E.A.
kJ mol1
328
349
325
295
The size of Fluorine atom is so small that the addition of an extra electron results in great repulsion among the electrons.
The 2nd electron shell is much smaller than the 3rd electron shell.

74. By Hess’s law, H1 = IE1(K) + EA1(F) + H2
Energetics of Formation of Ionic Compounds
A. Formation of an ion pair in gaseous state
Consider the formation of a KF(g) ion pair from K(g) & F(g)
H1
K(g) F(g) KF(g)
IE1(K)
K+(g)
H2
EA1(F)
+ F(g)
By Hess’s law, H1 = IE1(K) + EA1(F) + H2

75. Q.6
H1 = IE1(K) + EA1(F) + H2
(+)()

76. H1 = IE1(K) + EA1(F) + H2
= 1.5111019 J  1.0631018 J = 9.1191019 J
H2 = 1.0631018 J
IE1(K) + EA1(F) = 1019 J

77. Stability : K+(g) + F-(g) < K(g) + F(g) by +1.511x10-19J
When R 
Stability : K+(g) + F-(g) < K(g) + F(g) by x10-19J
H2 = 1.0631018 J
IE1(K) + EA1(F) = 1019 J

78. I.E. and E.A. are NOT the driving forces for the formation of ionic bond.
H2 = 1.0631018 J
IE1(K) + EA1(F) = 1019 J

79. K+(g) & F(g) tend to come close together in order to become stable.
H2 = 1.0631018 J
IE1(K) + EA1(F) = 1019 J

80. Coulomb stabilization is the driving force for the formation of ionic bond.
H2 = 1.0631018 J
IE1(K) + EA1(F) = 1019 J

81. The most energetically stable state is reached.
When R = 2.1701010 m
The most energetically stable state is reached.
H2 = 1.0631018 J
IE1(K) + EA1(F) = 1019 J

82. The ions cannot come any closer than Re as it will results in less stable states
H2 = 1.0631018 J
IE1(K) + EA1(F) = 1019 J

83. Repulsions between electron clouds and between nuclei > attraction between ions
H2 = 1.0631018 J
IE1(K) + EA1(F) = 1019 J

84. Repulsion : Minimum V when R = 2.170  1010 m Attraction :
between opposite charges R
Potential energy (V)
Minimum V when R =  1010 m
Attraction :
between opposite charges

85. What is the significance of the lowest energy state of the neutral state of K + F ?
H2 = 1.0631018 J
IE1(K) + EA1(F) = 1019 J

86. K F A covalent bond is formed K F  K F 
Electrostatic attraction between positive nuclei and bond electron pair stabilizes the system

87. By Hess’s law, Hf = H1 + H2
A. Formation of Ionic Crystals
Consider the formation of NaCl(s) from Na(s) & Cl2(g)
Hf
Na(s) Cl2(g) NaCl(s)
Na+(g) Cl(g)
H1
H2
By Hess’s law, Hf = H1 + H2

88. H1 is the sum of four terms of enthalpy changes
1. Standard enthalpy change of atomization of Na(s)
  Na(s)  Na(g) H = kJ mol-1
2. First ionization enthalpy of Na(g)
  Na(g)  Na+(g) + e- H= +500 kJ mol-1
3. Standard enthalpy change of atomization of Cl2(g)
  1/2Cl2(g)  Cl(g) H = kJ mol-1
4.  First electron affinity of Cl(g)
  Cl(g) + e-  Cl-(g) H = -349 kJ mol-1

89. H2 is the lattice enthalpy of NaCl.
It is the enthalpy change for the formation of 1 mole of NaCl(s) from its constituent ions in the gaseous state.
Na+(g) + Cl-(g)  NaCl(s) HLo

90. Direct determination of lattice enthalpy by experiment is very difficult, but it can be obtained from
1. theoretical calculation using an ionic model, Or
2. experimental results indirectly with the use of a Born-Haber cycle.

91. Q.7 Calculate the lattice enthalpy of NaCl
Hf = H1 +H2
Lattice enthalpy
= H2
= Hf - H1
= [(411)  (  349)] kJ mol1
= 791.4 kJ mol1

92. -349
-791.4

93. Lattice enthalpy is a measure of the strength of ionic bond.
Lattice enthalpy is the dominant enthalpy term responsible for the -ve Hf of an ionic compound.
More ve HLo  More stable ionic compound
Lattice enthalpy is a measure of the strength of ionic bond.

94. Determination of Lattice Enthalpy
From Born-Haber cycle (experimental method, refers to Q.7)
From theoretical calculation based on the ionic model

95. Assumptions made in the calculation
Ions are spherical and have no distortion of the charge cloud, I.e. 100% ionic.
The crystal has certain assumed lattice structure.
Repulsive forces between oppositely charged ions at short distances are ignored.

96. M is the Madelung constants that depends on the crystal structure

97. L is the Avogadro’s constant

98. Q+ and Q- are the charges on the cation and the anion respectively

99. 0 is the permittivity of vacuum

100. (r+ + r-) is the nearest distance between the nuclei of cation and anion
r+ is the ionic radius of the cation
r- is the ionic radius of the anion

101. Stoichiometry of Ionic Compounds
Stoichiometry of an ionic compound is the
simplest whole number ratio of cations
and anions involved in the formation of
the compound.
There are two ways to predict stoichiometry.

102. 1. Considerations in terms of electronic configurations
Atoms tend to attain noble gas electronic structures by losing or gaining electron(s).
The ionic compound is electrically neutral.
total charges on cation = total charges on anions

103.  NaF  MgCl2 1. Considerations in terms of electronic configurations
Na F  [Na]+ [F]-
2,8,1 2, ,8 2,8
 NaF
Mg Cl  [Cl]- [Mg]2+ [Cl]-
2,8,2 2,8, ,8,8 2,8 2,8,8
 MgCl2

104. 2. Considerations in terms of enthalpy changes of formation
Based on the Born-Haber cycle & the theoretically calculated lattice enthalpy, the values of Hfo of hypothetical compounds (e.g. MgCl , MgCl2 , MgCl3) can be calculated.
The stoichiometry with the most negative Hfo value is the most stable one.

105. 2. Considerations in terms of enthalpy changes of formation
Or, we can determine the true stoichiometry by comparing the calculated Hfo values of hypothetical compounds with the experimentally determined Hfo value.
The stoichiometry with Hfo value closest to the experimentally determined Hfo value is the answer.

106. Q. 8. Three hypothetical formulae of magnesium chloride
Q.8 Three hypothetical formulae of magnesium chloride are proposed and their estimated lattice enthalpies are shown in the table below.

107. Hf[MgCl(s)] = Hat[Mg(s)] + 1st IE of Mg + Hat[1/2Cl2(g)]
+ 1st EA of Cl + HL[MgCl(s)]
= ( ) kJ mol-1
= kJ mol-1

108. Enthalpy (kJ mol) HL[MgCl] Hf[MgCl]<0 Mg(s)+ Cl2(g)
Hat[1/2Cl2]
Mg+(g)+Cl(g)
Mg+(g)+Cl(g)
1st EA[Cl]
1st IE[Mg]
Mg+(g)+ Cl2(g)
MgCl(s)
HL[MgCl]
Hat[Mg]
Mg(g)+ Cl2(g)
Hf[MgCl]<0

109. = -752 kJ mol-1 Hf[MgCl2(s)]
= Hat[Mg(s)] + 1st IE of Mg + 2nd IE of Mg
+ 2Hat[1/2Cl2(g)] + 21st EA of Cl + HL[MgCl2(s)]
= [ 121+2(-364)+(-2602)] kJ mol-1
= kJ mol-1

110. Enthalpy (kJ mol) HL[MgCl2] Hf[MgCl2] <0 Mg(s)+Cl2(g)
Mg2+(g)+2Cl(g)
2Hat[1/2Cl2]
Mg2+(g)+2Cl(g)
21st EA[Cl]
Mg2+(g)+Cl2(g)
2nd IE[Mg]
MgCl2(s)
HL[MgCl2]
Mg+(g)+Cl2(g)
1st IE[Mg]
Mg(g)+Cl2(g)
Hat[Mg]
Hf[MgCl2] <0

111. = +3907 kJ mol-1 Hf[MgCl3(s)]
= Hat[Mg(s)] + 1st IE of Mg + 2nd IE of Mg + 3rd IE of Mg
+ 3Hat[1/2Cl2(g)] + 31st EA of Cl + HL[MgCl3(s)]
= 121+3(-364)+(-5440)
= kJ mol-1

112. Enthalpy ( kJ mol) Hf[MgCl3] >0 Mg3+(g)+3Cl(g) 3Hat[1/2Cl2]
3rd IE[Mg]
Mg3+(g)+ Cl2(g)
Mg3+(g)+3Cl(g)
31st EA[Cl]
MgCl3(s)
HL[MgCl3]
Enthalpy ( kJ mol)
Mg2+(g)+ Cl2(g)
2nd IE[Mg]
Hf[MgCl3] >0
Mg+(g)+ Cl2(g)
1st IE[Mg]
Hat[Mg]
Mg(g)+ Cl2(g)
Mg(s)+3/2Cl2(g)

113. Since the hypothetical compound MgCl2 has the most negative Hf value, and
this value is closest to the experimentally determined one,
the most probable formula of magnesium chloride is
MgCl2

114. Reasons for the discrepancy between calculated and experimental results of Hf
The ionic bond of MgCl2 is not 100% pure I,e. the ions are not perfectly spherical.
MgCl2 has a different crystal structure from CaF2
Short range repulsive forces between oppositely charged ions have not been considered.
Check Point 7-2

115. Factors affecting lattice enthalpy
7.2 Energetics of Formation of Ionic Compounds (SB p.196)
Factors affecting lattice enthalpy
Effect of ionic size:
 The greater the ionic size  The lower (or less negative) is the lattice enthalpy

116. Factors affect lattice enthalpy
7.2 Energetics of Formation of Ionic Compounds (SB p.196)
Factors affect lattice enthalpy
Effect of ionic charge:
 The greater the ionic charge  The higher (or more negative) is the lattice enthalpy
Check Point 7-3

117. 7.4
Ionic Crystals

118. Crystal Lattice(晶體格子)
7.4 Ionic Crystals (SB p.202)
Crystal Lattice(晶體格子)

119. 7.4 Ionic Crystals (SB p.202)
In solid state, ions of ionic compounds are regularly packed to form 3-dimensional structures called crystal lattices(晶體格子)

120. 7.4 Ionic Crystals (SB p.202)
The coordination number (C.N.) of a given particle in a crystal lattice is the number of nearest neighbours of the particle.
C.N. of Na+
= 6

121. 7.4 Ionic Crystals (SB p.202)
The coordination number (C.N.) of a given particle in a crystal lattice is the number of nearest neighbours of the particle.
C.N. of Cl
= 6

122. Identify the unit cell of NaCl crystal lattice
Not a unit cell

123. Identify the unit cell of NaCl crystal lattice
Not a unit cell

124. The unit cell of a crystal lattice is the simplest 3-D arrangement of particles which, when repeated 3-dimensionally in space, will generate the whole crystal lattice.

125. The unit cell of a crystal lattice is the simplest 3-D arrangement of particles which, when repeated 3-dimensionally in space, will generate the whole crystal lattice.

126. The unit cell of a crystal lattice is the simplest 3-D arrangement of particles which, when repeated 3-dimensionally in space, will generate the whole crystal lattice.

127. Unit cell
Unit cell

128. A crystal lattice with a cubic unit cell is known as a cubic structure.

129. Three Kinds of Cubic Structure
Simple cubic (primitive) structure
Body-centred cubic (b.c.c.) structure
Face-centred cubic (f.c.c.) structure

130. Simple cubic structure
Space filling
Space lattice
Unit cell 
View this Chemscape 3D structure

131. Body-centred cubic (b.c.c.) structure
Space lattice
Space filling
Unit cell 
View this Chemscape 3D structure

132. Face-centred cubic (f.c.c.) structure View this Chemscape 3D structure
Unit cell
View this Chemscape 3D structure

133. The f.c.c. structure can be obtained by stacking the layers of particles in the pattern abcabc...

134. Interstitial sites - The empty spaces in a crystal lattice
Two types of interstitial sites in f.c.c. structure
Octahedral site
Tetrahedral site

135. Octahedral site : -
It is the space between the 3 spheres in one layer and 3 other spheres in the adjacent layer.

136. When the octahedron is rotated by 45o , the octahedral site can also be viewed as the space confined by 4 spheres in one layer and 1 other sphere each in the upper and lower layers respectively.
before
behind
45o

137. Tetrahedral site : -
It is the space between the 3 spheres of one layer and a fourth sphere on the upper layer.

138. In an f.c.c. unit cell, the tetrahedral site is the space bounded by a corner atom and the three face-centred atoms nearest to it.

139. c b a

140. Top layer Q.9 Label a, b, and c as Oh sites or Td sites c is Oh site
a and b are Td sites
c is Oh site
Top layer

141. 13 Oh sites 4 Th sites in the front 4 Th sites at the back
Q.10 Identify the Oh sites and Td sites in the f.c.c. unit cell shown below. 2 1
13 Oh sites 3 4 5 6
4 Th sites in the front 1 3 5 6 9 8 7
4 Th sites at the back 7 8 2 4 11 10 12 13
View Octahedral sites and tetrahedral sites

142. Since ionic crystal lattice is made of two kinds of particles, cations and anions,
the crystal structure of an ionic compound can be considered as two lattices of cations and anions interpenetrating with each other.

143. The crystal structures of
sodium chloride,
caesium chloride and
calcium fluoride are discussed.

144. Sodium Chloride (The Rock Salt Structure)
f.c.c. unit cell of larger Cl- ions, with all Oh sites occupied by the smaller Na+ ions
Oh sites of f.c.c. lattice of Cl- ions

145. A more open f.c.c. unit cell of smaller Na+ ions with all Oh sites occupied by the larger Cl- ions.
Oh sites of f.c.c. lattice of Na+ ions

146. F.C.C. Structure of Sodium Chloride
View this Chemscape 3D structure

147. 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

148. Only the particles in the centre (or body) of the unit cell belong to the unit cell entirely. Particles locating on the faces, along the edges or at the corner of a unit cell are shared with the neighboring unit cells of the crystal lattice.

149. Fraction of particles occupied by a unit cell
1/2
Body
Face
Edge
Corner 1
1/2
Fraction of particles occupied by a unit cell

150. Fraction of particles occupied by a unit cell
1/4
Body
Face
Edge
Corner 1
1/2
1/4
Fraction of particles occupied by a unit cell

151. Fraction of particles occupied by a unit cell
1/8
Body
Face
Edge
Corner 1
1/2
1/4
1/8
Fraction of particles occupied by a unit cell

152. Q. 11. Calculate the net nos. of Na+ and Cl-
Q.11 Calculate the net nos. of Na+ and Cl- ions in a unit cell of NaCl.
No. of Cl- ions =
8 corners
6 faces
No. of Na+ ions =
12 edges
1 body
Example 7-4

153. Structure of Caesium Chloride  Link
7.4 Ionic Crystals (SB p.202)
Structure of Caesium Chloride  Link
Cs+ ions are too large to fit in the octahedral sites.
Thus Cl ions adopt the more open simple cubic structure with the cubical sites occupied by Cs+ ions.
Simple cubic lattice

154. A cubical site is the space confined by 4 spheres in one layer and 4 other spheres in the adjacent layer.

155. Size of interstitial sites : -
Cubical > octahedral > tetrahedral
In f.c.c. structure
In simple cubic structure

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

157. Q.12
Number of Cs+ = 1
Number of Cl =

158. Structure of Calcium Fluoride  Link
7.4 Ionic Crystals (SB p.203)
Structure of Calcium Fluoride  Link
Fluorite structure
It can be viewed as a simple cubic structure of larger fluoride ions with alternate cubical sites occupied by smaller calcium ions.

159. Structure of Calcium Fluoride  Link
7.4 Ionic Crystals (SB p.203)
Structure of Calcium Fluoride  Link
Alternately, it can be viewed as an expanded f.c.c. structure of smaller calcium ions with all tetrahedral sites occupied by larger fluoride ions.

160. Face-centred cubic lattice
7.4 Ionic Crystals (SB p.203)
Structure of Calcium Fluoride  Link
Face-centred cubic lattice
Co-ordination number of Ca2+ = 8
8 : 4 co-ordination
Co-ordination number of F- = 4

161. CaF2 Q.13 (a) Number of F = 8 Number of Ca2+ =
7.4 Ionic Crystals (SB p.203)
Q.13 (a)
CaF2
Number of F = 8
Number of Ca2+ =

162. 7.4 Ionic Crystals (SB p.203)
Q.13 (b)
CaF2
Only alternate cubical sites are occupied by Ca2+ in order to maintain electroneutrality.

163. Na2O vs CaF2 Structure of Sodium oxide  Link Antifluorite structure
7.4 Ionic Crystals (SB p.203)
Structure of Sodium oxide  Link
Antifluorite structure
Na2O vs CaF2
fluorite structure

164. Zinc blende structure - Link
Q.14(a)
Zinc blende structure Link

165. Q.14(a)
ZnS
Number of Zn2+ = 4
Number of S2 =

166. Q.14(b) Zinc sulphide, ZnS  tetrahedral site
S2 ions adopt f.c.c. structure
Alternate Td sites are occupied by Zn2+ ions to ensure electroneutrality.

167. The unit cell is not a cube
Q.15
Rutile structure - Link
The unit cell is not a cube

168. Ti4+
Q.15(a)
O2
Number of Ti4+ =
TiO2
Number of O2 =

169. Q.15(a)
C.N. of Ti4+ = 6
C.N. of O2 = 3
 6 : 3 coordination

170. Q.15(b) Titanium(IV) oxide, TiO2  octahedral site
O2 ions adopt distorted h.c.p.(not f.c.c.) structure with alternate Oh sites occupied by Ti4+ ions to ensure electroneutrality.

171. hexagonal close-packedb a

172. 1. Close Packing Considerations
Factors governing the structures of ionic crystals
1. Close Packing Considerations
Ions in ionic crystals tend to pack as closely as possible. (Why ?)
To strengthen the ionic bonds formed
To maximize the no. of ionic bonds formed.

173. 1. Close Packing Considerations
Factors governing the structures of ionic crystals
1. Close Packing Considerations
The larger anions tend to adopt face-centred cubic structure (cubic closest packed, c.c.p.)
The smaller cations tend to fill the interstitial sites as efficiently as possible

174. Q.16 Is there no limit for the C.N. ? Explain your answer.
Factors governing the structures of ionic crystals
Q.16 Is there no limit for the C.N. ? Explain your answer.
The anions tend to repel one another when they approach a given cation.
The balance between attractive forces and repulsive forces among ions limits the C.N. of the system.

175. Q.16 Is there no limit for the C.N. ? Explain your answer.
Factors governing the structures of ionic crystals
Q.16 Is there no limit for the C.N. ? Explain your answer.
If the cations are small, less anions can approach them without significant repulsions.
Or, if the cations are small, they choose to fit in the smaller tetrahedral sites with smaller C.N. to strengthen the ionic bonds formed.

176. Q.16 Is there no limit for the C.N. ? Explain your answer.
Factors governing the structures of ionic crystals
Q.16 Is there no limit for the C.N. ? Explain your answer.
If the cations are large, they choose to fit in the larger octahedral sites or even cubical sites with greater C.N. to maximize the no. of ionic bonds formed.

177. 2. The relative size of cation and anion,
Factors governing the structures of ionic crystals
2. The relative size of cation and anion,
In general, r > r+,

178. If the cations are large,
Factors governing the structures of ionic crystals
If the cations are large,
more anions can approach the cations without significant repulsions.
Or, the large cations can fit in the larger interstitial sites with greater C.N. to maximize the no. of bonds formed.

179. Summary : - NH4+ is large Interstitial site Tetrahedral Octahedral
7.4 Ionic Crystals (SB p.203)
Summary : -
Interstitial site
Tetrahedral
Octahedral
Cubical
Coordination
4 : 4
6 : 6
8 : 8
0.225 – 0.414
0.414 – 0.732
0.732 – 1.000
Examples
ZnS, most copper(I) halides
Alkali metal halides except CsCl
CsCl, CsBr
CsI, NH4Cl
NH4+ is large

180. Summary : - *CaF2, BaF2, BaCl2, SrF2 Examples 0.732 – 1.000 8 : 4
7.4 Ionic Crystals (SB p.203)
Summary : -
*CaF2, BaF2, BaCl2, SrF2
Examples
0.732 – 1.000
8 : 4
Coordination
Cubical
Interstitial site

181. Q.17 r A B
45° r r+ C

182. the range of ratio that favours octahedral arrangement.

183. the optimal ratio for cations and anions to be in direct contact with each other in the cubical sites.
Q.18

184. Q.18(a) Silver chloride, AgCl  octahedral site
Cl ions adopt f.c.c. structure
All Oh sites are occupied by Ag+ ions to ensure electroneutrality.

185. Q.18(a) Silver chloride, AgCl  octahedral site
Ag+ ions adopt an open f.c.c. structure
All Oh sites are occupied by Cl ions to ensure electroneutrality.

186. Q.18(b) Copper(I) bromide, CuBr  tetrahedral site
Br ions adopt f.c.c. structure
Only alternate Td sites are occupied by Cu+ ions to ensure electroneutrality.

187. Q.18(c) Copper(II) bromide, CuBr2  tetrahedral site
Br ions adopt f.c.c. structure
Only 1/4 Td sites are occupied by Cu+ ions to ensure electroneutrality.
Some Br may form less bonds than others
 Not favourable.
Q.18(c)

188. Q.18(c) Copper(II) bromide, CuBr2
Or, Cu2+ ions adopt a very open f.c.c. structure
All the Td sites are occupied by Br ions to ensure electroneutrality.
However, this structure is too open to form strong ionic bonds.
 non-cubic structure is preferred.

189. Q.18(c)
4 : 2 coordination
Cu2+
Br

190. Copper(II) chloride, CuCl2
Layer structure
0.295nm
0.230 nm

191. Copper(II) chloride, CuCl2
6 : 3 coordination

192. 7.5
Ionic Radii

193. Ionic Radii
Ionic radius is the approximate radius of the spherical space occupied by the electron cloud of an ion in all directions in the ionic crystal.

194. Q.19 Why is the electron cloud of an ion always spherical in shape ?
Stable ions always have fully-filled electron shells or subshells. The symmetrical distribution of electrons accounts for the spherical shapes of ions.

195. Q.19 Cl [Ne] 3s2, 3px2, 3py2, 3pz2 Li+ 1s2
Symmetrical distribution along x, y and z axes   almost spherical
spherical
Li+ 1s2
Na+ [He] 2s2, 2px2, 2py2, 2pz2
Cl [Ne] 3s2, 3px2, 3py2, 3pz2
Zn2+ [Ar] 3dxy2, 3dxz2, 3dyz2, ,
Symmetrical distribution along xy, xz, yz planes and along x, y and z axes  almost spherical

196. Determination of Ionic Radii Pauling Scale
Interionic distance (r+ + r) can be determined by
X-ray diffraction crystallography

197. Electron density plot for sodium chloride crystal
7.5 Ionic Radii (SB p.205)
Cl
Na+
Electron density map for NaCl
Contour having same electron density
r++ r
r++ r
Electron density plot for sodium chloride crystal

198. By additivity rule,
Interionic distance = r+ + r
For K+Cl,
r+ + r = nm (determined by X-ray)
Since K+(2,8,8) and Cl(2,8,8) are isoelectronic, their ionic radii can be calculated.
r+(K+) = nm, r(Cl) = nm

199. For Na+Cl,
r+ + r = nm (determined by X-ray)
Since r(Cl) = nm(calculated)
r+(Na+) = ( ) nm = nm

200. Limitation of Additivity rule
In vacuum, the size of a single ion has no limit according to quantum mechanics.
The size of the ion is restricted by other ions in the crystal lattice.
Interionic distance < r+ + r

201. Evidence :
Electron distribution is not perfectly spherical at the boundary due to repulsion between electron clouds of neighbouring ions.

202. Ionic radius depends on the bonding environment.
For example, the ionic radius of Cl ions in NaCl (6 : 6 coordination) is different from that in CsCl (8 : 8 coordination).

203. Ionic radius vs atomic radius
7.5 Ionic Radii (SB p.206)
Radii of cations < Radii of corresponding parent atoms
cations have one less electron shell than the parent atoms
Ionic radius vs atomic radius

204. Ionic radius vs atomic radius
7.5 Ionic Radii (SB p.206)
p/e of cation > p/e of parent atom  Less shielding effect
stronger nuclear attraction for outermost electrons
 smaller size
Ionic radius vs atomic radius

205. Ionic radius vs atomic radius
7.5 Ionic Radii (SB p.206)
Radii of anions > Radii of corresponding parent atoms
Ionic radius vs atomic radius

206. Ionic radius vs atomic radius
7.5 Ionic Radii (SB p.206)
p/e of anion < p/e of parent atom  more shielding effect
weaker nuclear attraction for outermost electrons
 larger size
Ionic radius vs atomic radius

207. Periodic trends of ionic radius
Ionic radius increases down a Group
Ionic radius depends on the size of the outermost electron cloud.
On moving down a Group, the size of the outermost electron cloud increases as the number of occupied electron shells increases.

208. Periodic trends of ionic radius
2. Ionic radius decreases along a series of isoelectronic ions of increasing nuclear charge
The total shielding effects of isoelectronic ions are approximately the same.
 Zeff  nuclear charge (Z)
 Ionic radius decreases as nuclear charge increases.

209. Isoelectronic to He(2) Isoelectronic to Ar(2,8,8)
7.5 Ionic Radii (SB p.206)
Isoelectronic to He(2)
Isoelectronic to Ar(2,8,8)
Isoelectronic to Ne(2,8)

210. H is larger than most ions, why ?
7.5 Ionic Radii (SB p.206)
H is larger than most ions, why ?

211. Q.20
H > N3
The nuclear charge (+1) of H is too small to hold the two electrons which repel each other strongly within the small 1s orbital.
Or,
p/e of H (1/2) < p/e of N3 (7/10)

212. The END
Example 7-5
Check Point 7-5

213. Why do two atoms bond together?
Introduction (SB p.186)
Let's Think 1
Why do two atoms bond together?
Answer
The two atoms tend to achieve an octet configuration which brings stability.

214. How does covalent bond strength compare with ionic bond strength?
Introduction (SB p.186)
Let's Think 1
How does covalent bond strength compare with ionic bond strength?
They are similar in strength.
Both are electrostatic attractions between charged particles.
Answer
Back

215. 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

216. 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.

217. 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.

218. 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

219. 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

220. 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

221. 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

222. 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

223. 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

224. 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.

225. 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

226. 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

227. 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.

228. 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

229. 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