1
7
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

3
 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. noble gases)?
 Why are there compounds with chemical formulae H
NaCl
2
?
2
O and NaCl, but not H
3
O and
 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 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
Ion
Be 2+
O
2-
Sc
3+
Br -
Ba 2+
At s,p,d,f notation
Isoelectronic noble gas

7
Ion s,p,d,f notation
Be
2+
O
2-
1s
2
(2)
[He] 2s
2
,2p
6
(2,8)
Sc
3+
[Ne] 3s
2
,3p
6
(2,8,8)
Br
-
[Ar] 3d
10
,4s
2
,4p
6
(2,8,18,8)
Ba
2+
[Kr] 4d
10
,5s
2
,5p
6
(2,8,18,18,8)
At
-
[Xe] 4f
14
,5d
10
,6s
2
,6p
6
(2,8,18,32,18,8)
Isoelectronic noble gas
He
Ne
Ar
Kr
Xe
Rn

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

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
9
Sodium atom, Na
1s 2 2s 2 2p 6 3s 1
Chlorine atom, Cl
1s 2 2s 2 2p 6 3s 2 3p 5

Introduction (SB p.186)
Three types of chemical bonds
1. Ionic bond
Electrostatic attraction between positively charged particles and negatively charged particles
10

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

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

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)
13

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

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 electronattracting 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 electronattracting abilities.

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

19
Electronegativity and Types of Chemical Bonds
1. 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
1 st
I.E.
X + (g)

X 2+ (g) + e -

H
2 nd
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) + e -

X (g)

H
1 st
E.A.

X 2

(g)

H
2 nd
E.A.
Electron affinities can be positive or negative .

21
I.E. and E.A. only show e releasing/attracting power of
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
3. Electronegativity
The ability of an atom to attract electrons in a chemical bond .

23
Mulliken’s scale of electronegativity
Electronegativity(Mulliken) 
1
2
( IE

EA )
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

Pauling’s scale of electronegativity
26
IA IIA IIIA IVA VA VIA VIIA VIIIA
K
0.8
Rb
0.8
Cs
0.7
H
2.1
Li
1.0
Na
0.9
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
B
2.0
Al
1.5
Ga
1.6
In
1.7
Tl
1.8
C
2.5
Si
1.8
Ge
1.8
Sn
1.8
Pb
1.8
N
3.0
P
2.1
As
2.0
Sb
2.0
Bi
1.9
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
Kr
-
Xe
-
Rn
-
He
-
Ne
-
Ar
-

27
What trends do you notice about the EN values in the Periodic Table? Explain.
IA IIA IIIA IVA VA VIA VIIA VIIIA
K
0.8
Rb
0.8
Cs
0.7
H
2.1
Li
1.0
Na
0.9
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
B
2.0
Al
1.5
Ga
1.6
In
1.7
Tl
1.8
C
2.5
Si
1.8
Ge
1.8
Sn
1.8
Pb
1.8
N
3.0
P
2.1
As
2.0
Sb
2.0
Bi
1.9
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
Kr
-
Xe
-
Rn
-
He
-
Ne
-
Ar
-

28
The EN values increase from left to right across a Period.
IA IIA IIIA IVA VA VIA VIIA VIIIA
K
0.8
Rb
0.8
Cs
0.7
H
2.1
Li
1.0
Na
0.9
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
B
2.0
Al
1.5
Ga
1.6
In
1.7
Tl
1.8
C
2.5
Si
1.8
Ge
1.8
Sn
1.8
Pb
1.8
N
3.0
P
2.1
As
2.0
Sb
2.0
Bi
1.9
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
Kr
-
Xe
-
Rn
-
He
-
Ne
-
Ar
-

29
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.
IA IIA IIIA IVA VA VIA VIIA VIIIA
K
0.8
Rb
0.8
Cs
0.7
H
2.1
Li
1.0
Na
0.9
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
B
2.0
Al
1.5
Ga
1.6
In
1.7
Tl
1.8
C
2.5
Si
1.8
Ge
1.8
Sn
1.8
Pb
1.8
N
3.0
P
2.1
As
2.0
Sb
2.0
Bi
1.9
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
Kr
-
Xe
-
Rn
-
He
-
Ne
-
Ar
-

31
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 .

Why are the E.A. of noble gases not shown ?
32
IA IIA IIIA IVA VA VIA VIIA VIIIA
H
2.1
Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
B
2.0
Al
1.5
Ga
1.6
In
1.7
Tl
1.8
C
2.5
Si
1.8
Ge
1.8
Sn
1.8
Pb
1.8
N
3.0
P
2.1
As
2.0
Sb
2.0
Bi
1.9
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
He
-
Ne
-
Ar
-
Kr
-
Xe
-
Rn
-

33
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.
2
4
6

34
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
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

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
36

37

38
 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
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
 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
+
+
+
+
+
Ease of formation of cation 
2+
2+
3+
2+
3+
2+
3+
2+
3+
4+
4+

42
 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
3 
3 
Ease of formation of anion 
2
2







44
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 s,p,d,f notation Ion
Zn 2+
Pb 2+
V 3+
Cr 3+
Fe 3+
Fe 2+
Co 2+
Ni
2+
Cu 2+

46
Ion
Zn 2+
Pb 2+
V 3+
Cr 3+
Fe 3+
Fe 2+
Co 2+
Ni 2+
Cu 2+ s,p,d,f notation
1s 2 ,2s 2 ,2p 6 , 3s 2 ,3p 6 ,3d 10
1s 2 ,2s 2 ,2p 6 ,3s 2 3p 6 3d 10 ,4s 2 ,4p 6 ,4d 10 ,4f 14 , 5s 2 ,5p 6 ,5d 10 ,6s 2
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 2
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 3
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 5
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 6
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 7
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 8
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 9

47
1. 18 - electrons group e.g. Zn 2+
Fully-filled s, p & d-subshells
Ion
Zn 2+
Pb 2+
V 3+
Cr 3+
Fe 3+
Fe 2+
Co 2+
Ni 2+
Cu 2+ s,p,d,f notation
1s 2 ,2s 2 ,2p 6 , 3s 2 ,3p 6 ,3d 10
1s 2 ,2s 2 ,2p 6 ,3s 2 3p 6 3d 10 ,4s 2 ,4p 6 ,4d 10 ,4f 14 , 5s 2 ,5p 6 ,5d 10 ,6s 2
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 2
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 3
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 5
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 6
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 7
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 8
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 9

48
2. (18 + 2 ) - electron group e.g. Pb 2+
Fully-filled s, p & d-subshells
Ion
Zn 2+
Pb 2+
V 3+
Cr 3+
Fe 3+
Fe 2+
Co 2+
Ni 2+
Cu 2+ s,p,d,f notation
1s 2 ,2s 2 ,2p 6 , 3s 2 ,3p 6 ,3d 10
1s 2 ,2s 2 ,2p 6 ,3s 2 3p 6 3d 10 ,4s 2 ,4p 6 ,4d 10 ,4f 14 , 5s 2 ,5p 6 ,5d 10 ,6s 2
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 2
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 3
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 5
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 6
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 7
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 8
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 9

3. Variable arrangements for ions of transition metals. ns 2 ,np 6 , n d 1 to ns 2 ,np 6 , n d 9
Ion
Zn 2+
Pb 2+
V 3+
Cr 3+
Fe 3+
Fe 2+
Co 2+
Ni 2+
Cu 2+
1s 2 ,2s 2 ,2p 6
Ti 3+ [Ne] 3s 2 3p 6 3d 1
,3s 2
1s 2 ,2s 2 ,2p 6 , 3s 2 ,3p 6 ,3d 10
3p s,p,d,f notation
6 3d 10 ,4s 2 ,4p 6 ,4d 10 ,4f 14 , 5s
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 2
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 3
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 5
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 6
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 7
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 8
1s 2 ,2s 2 ,2p 6 , 3s 2 3p 6 3d 9
Mn 3+ [Ne] 3s 2 3p 6 3d 4
2 ,5p 6 ,5d 10 ,6s 2
49

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

Electronic arrangements of stable cations
Octet structure : cations of Group IA, IIA and IIIB
IA IIA IIIB
2,8
2,8,8
2,8,18,8
Na +
K +
Rb +
Mg 2+ Al 3+ (IIIA)
Ca 2+ Sc 3+
Sr 2+ Y 3+
51

52
(a) 18 - electrons group cations of Groups IB, IIB, IIIA and IVA
IB IIB IIIA IVA
2,8,18 Cu +
2,8,18,18 Ag +
Zn 2+
Cd 2+
Ga 3+
In
--
3+ Sn 4+
2,8,18,32,18 Au + Hg 2+ Tl 3+ Pb 4+

Less stable than ions with a noble gas structure.
53
IB IIB IIIA IVA
2,8,18 Cu +
2,8,18,18 Ag +
Zn 2+
Cd 2+
Ga 3+
In
--
3+ Sn 4+
2,8,18,32,18 Au + Hg 2+ Tl 3+ Pb 4+

Cu and Au form other ions because the nuclear charges are not high enough to hold the 18-electron group firmly.
54
IB IIB IIIA IVA
2,8,18 Cu +
2,8,18,18 Ag +
Zn 2+
Cd 2+
Ga 3+
In
--
3+ Sn 4+
2,8,18,32,18 Au + Hg 2+ Tl 3+ Pb 4+

55
Cu 2+ 2, 8, 17 Au 3+ 2, 8, 18, 32, 16
The d-electrons are more diffused and thus less attracted by the less positive nuclei.
IB IIB IIIA IVA
2,8,18 Cu +
2,8,18,18 Ag +
Zn 2+
Cd 2+
Ga 3+
In
--
3+ Sn 4+
2,8,18,32,18 Au + Hg 2+ Tl 3+ Pb 4+

(b) (18+2) electrons
Cations of heavier group members due to presence of 4f electrons.
Tl [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 , 6s 2 , 6p 1
Tl 3+ [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 (18)
Tl + [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 , 6s 2 (18 + 2 )
Stability : Tl + > Tl 3+ due to extra stability of
6s electrons (inert pair effect)
56

(b) (18+2) electrons
Tl [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 , 6s 2 , 6p 1
Tl 3+ [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 (18)
Tl + [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 , 6s 2 (18 + 2 )
6s electrons are poorly shielded by the inner
4f electrons.
6s electrons experience stronger nuclear attraction.
57

(b) (18+2) electrons
Tl [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 , 6s 2 , 6p 1
Tl 3+ [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 (18)
Tl + [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 , 6s 2 (18 + 2 )
Inert pair effect increases down a Group.
Stability :Sn 4+ (18) > Sn 2+ (18+ 2 )
Pb 2+ (18+ 2 ) > Pb 4+ (18) moving down
Group IV
58

(b) (18+2) electrons
Tl [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 , 6s 2 , 6p 1
Tl 3+ [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 (18)
Tl + [2,8,18, 32 ], 5s 2 , 5p 6 , 5d 10 , 6s 2 (18 + 2 )
Inert pair effect increases down a Group.
Stability :Sb 5+ (18) > Sb 3+ (18+ 2 )
Bi 3+ (18+ 2 ) > Bi 5+ (18) moving down
Group V
59

60
(c) Cations of transition elements
-
variable oxidation numbers
Electronic configurations from ns 2 , np 6 , nd 1 to ns 2 , np 6 , nd 9
Fe [Ne] 3s 2 , 3p 6 , 3d 6 , 4s 2
Fe 2+ [Ne] 3s 2 , 3p 6 , 3d 6
Fe 3+ [Ne] 3s 2 , 3p 6 , 3d 5

61
(c) Cations of transition elements
Fe [Ne] 3s 2 , 3p 6 , 3d 6 , 4s 2
Fe 2+ [Ne] 3s 2 , 3p 6 , 3d 6
Fe 3+ [Ne] 3s 2 , 3p 6 , 3d 5
Which one is more stable, Fe 2+ (g) or Fe 3+ (g) ?
Fe 2+ (g) is more stable than Fe 3+ (g)
Energy is needed to remove electrons from
Fe 2+ (g) to give Fe 3+ (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)
Na
-53
K
-48
H
-73
Li
-60
Be
+18
Mg
+21
B
-23
Al
-44
C
-122
Si
-134
N
~0
P
-72
O
-142
S
-200
F
-322
Cl
-349
Br
-325
Ar
+35
Kr
+39
He
+21
Ne
+29

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

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

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

Goes to a new subshell Goes to a new shell
Be(2s 2 )  Be  (2s 2 2p 1 )
Ne(2p 6 )  Ne  (2p 6 3s 1 )
Na
-53
K
-48
H
-73
Li
-60
Be
+18
Mg
+21
B
-23
Al
-44
C
-122
Si
-134
N
~0
P
-72
O
-142
S
-200
F
-328
Cl
-349
Br
-324
Ar
+35
Kr
+39
He
+21
Ne
+29
67

68
More –ve E.A. for Gp 1 because the ions formed have full-filled s-subshells .
E.g. Li(1s 2 2s 1 )  Li  (1s 2 2s 2 )
H
-73
Li
-60
Na
-53
K
-48
Be
+18
Mg
+21
B
-23
Al
-44
C
-122
Si
-134
N
~0
P
-72
O
-142
S
-200
F
-328
Cl
-349
Br
-324
Ar
+35
Kr
+39
He
+21
Ne
+29

69
Less –ve E.A. for Gp 5 because the stable halffilled p-subshell no longer exists.
E.g. N(2s 2 2p 3 )  N  (2s 2 2p 4 )
H
-73
He
+21
Li
-60
Na
-53
K
-48
Be
+18
Mg
+21
B
-23
Al
-44
C
-122
Si
-134
N
~0
P
-72
O
-142
S
-200
F
-328
Cl
-349
Br
-324
Ne
+29
Ar
+35
Kr
+39

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

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

Q.4 Why are the second E.A.s of O  (+844 kJmol  1 ) and S  (+532 kJmol  1 ) positive ?
O  (g) + e   O 2  (g)
S  (g) + e   S 2  (g)
The electrons added are repelled strongly by the negative ions.
72

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

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)
74
IE
1
(K)
EA
1
(F)
 H
2
K + (g) + F  (g)
By Hess’s law,  H
1
= IE
1
(K) + EA
1
(F) +  H
2

Q.6
 H
1
= IE
1
(K) + EA
1
(F) +  H
2

419.0

10
3
J mol

1
6.022

10
23 mol

1


328.0

10
3
J mol

1
6.022

10
23 mol

1
(+)(

)

Q
1
Q
4π

0
2
R

1.511

10

19 J

4π

8.854

(1.602
10

12 C 2

10
J

1 m

1

19 C) 2

2.170

10

10 m


1.511

10

19 J

1.063

10

18 J
 
9.119

10

19
J
75

76
 H
1
= IE
1
(K) + EA
1
(F) +  H
2
= 1.511
 10  19 J  1.063
 10  18 J =  9.119
 10  19 J
 H
2
=  1.063
 10  18 J
IE
1
(K) + EA
1
(F) = +1.511
 10  19 J
ΔH
1
 
9.119

10

19 J

When R  
Stability : K + (g) + F (g) < K(g) + F(g) by +1.511x10
-19 J
 H
2
=  1.063
 10  18 J
IE
1
(K) + EA
1
(F) = +1.511
 10  19 J
ΔH
1
 
9.119

10

19 J
77

I.E. and E.A. are NOT the driving forces for the formation of ionic bond.
 H
2
=  1.063
 10  18 J
IE
1
(K) + EA
1
(F) = +1.511
 10  19 J
ΔH
1
 
9.119

10

19 J
78

K + (g) & F  (g) tend to come close together in order to become stable.
 H
2
=  1.063
 10  18 J
IE
1
(K) + EA
1
(F) = +1.511
 10  19 J
ΔH
1
 
9.119

10

19 J
79

Coulomb stabilization is the driving force for the formation of ionic bond.
 H
2
=  1.063
 10  18 J
IE
1
(K) + EA
1
(F) = +1.511
 10  19 J
ΔH
1
 
9.119

10

19 J
80

When R = 2.170
 10  10 m
The most energetically stable state is reached.
 H
2
=  1.063
 10  18 J
IE
1
(K) + EA
1
(F) = +1.511
 10  19 J
ΔH
1
 
9.119

10

19 J
81

The ions cannot come any closer than R results in less stable states e as it will
 H
2
=  1.063
 10  18 J
IE
1
(K) + EA
1
(F) = +1.511
 10  19 J
ΔH
1
 
9.119

10

19 J
82

Repulsions between electron clouds and between nuclei > attraction between ions
 H
2
=  1.063
 10  18 J
IE
1
(K) + EA
1
(F) = +1.511
 10  19 J
ΔH
1
 
9.119

10

19 J
83

84
Repulsion :
V

1
12
R between opposite charges
Minimum V when R = 2.170  10  10 m
R
Attraction : V

1
R between opposite charges

What is the significance of the lowest energy state of the neutral state of K + F ?
 H
2
=  1.063
 10  18 J
IE
1
(K) + EA
1
(F) = +1.511
 10  19 J
ΔH
1
 
9.119

10

19 J
85

86
K F

K


F

Electrostatic attraction between positive nuclei and bond electron pair stabilizes the system

87
A. Formation of Ionic Crystals
Consider the formation of NaCl(s) from Na(s) & Cl
2
(g)
1
2
2
(g) NaCl(s)
 H
1  H
2
Na + (g) + Cl  (g)
By Hess’s law,  H f
=  H
1
+  H
2

 H
1 is the sum of four terms of enthalpy changes
1.
Standard enthalpy change of atomization of Na(s)
Na(s)  Na(g)  H = +108.3 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 Cl
2
(g)
1/2Cl
2
(g)  Cl(g)  H = +121.1 kJ mol -1
4.
First electron affinity of Cl(g)
Cl(g) + e  Cl (g)  H = -349 kJ mol -1
88

89
 H
2 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.
 H
L o
Na + (g) + Cl (g)  NaCl(s)

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.

Q.7 Calculate the lattice enthalpy of NaCl
 H f
=  H
1
+  H
2
Lattice enthalpy
=  H
2
=  H f
 H
1
= [(  411)  (+108.3 + 500 + 121.1  349)] kJ mol  1
=  791.4 kJ mol  1
91

92
-349
-791.4

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

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

• 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.
95

96
M is the Madelung constants that depends on the crystal structure

H lattice
 
MLQ

4

0
( r

Q

 r

)

97
L is the Avogadro’s constant

H lattice
 
MLQ

4

0
( r

Q

 r

)

98
Q
+ and Q
are the charges on the cation and the anion respectively

H lattice
 
MLQ

4

0
( r

Q

 r

)

99

0 is the permittivity of vacuum

H lattice
 
MLQ

4

0
( r

Q

 r

)

100
(r
+
+ r
-
) is the nearest distance between the nuclei of cation and anion r
+ r
is the ionic radius of the cation is the ionic radius of the anion

H lattice
 
MLQ

4

0
( r

Q

 r

)

101


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.

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
102

103
1. Considerations in terms of electronic configurations
Na + F  [Na] + [F] -
2,8,1 2,7 2,8 2,8  NaF
Mg + 2Cl  [Cl] [Mg] 2+ [Cl] -
2,8,2 2,8,7 2,8,8 2,8 2,8,8
 MgCl
2

2. Considerations in terms of enthalpy changes of formation
Based on the Born-Haber cycle & the
theoretically calculated lattice enthalpy, the values of  H
(e.g. MgCl , MgCl
2 f o of hypothetical compounds
, MgCl
3
) can be calculated.
The stoichiometry with the most negative
 H f o value is the most stable one .
104

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

106
Q.8 Three hypothetical formulae of magnesium chloride are proposed and their estimated lattice enthalpies are shown in the table below.
Hypothetical stoichiometry
MgCl MgCl
2
MgCl
3
Assumed structure
NaCl CaF
2
AlCl
3
 H
Estimated
L o (kJ mol -1 )
-771 -2602 -5440

107
 H f
[MgCl(s)] =  H at
[Mg(s)] + 1 st IE of Mg +  H at
[1/2Cl
2
(g)]
+ 1 st EA of Cl +  H
L
[MgCl(s)]
= (+150 + 736 + 121 - 364- 771) kJ mol -1
-1

108
Mg + (g)+Cl(g)

H at
[1/2Cl
2
]
Mg +
1
(g)+ Cl
2 2
(g) 1st EA[Cl]
Mg + (g)+Cl

(g)
1st IE[Mg]

H at
[Mg]
1
Mg(g)+ Cl
2 2
(g)
1
Mg(s)+ Cl
2 2
(g)

H f
[MgCl]<0

H
L
[MgCl]
MgCl(s)

109
 H f
[MgCl
2
(s)]
=  H at
[Mg(s)] + 1 st IE of Mg + 2 nd IE of Mg
+ 2  H at
[1/2Cl
2
(g)] + 2  1 st EA of Cl +  H
L
[MgCl
2
(s)]
= [150+736+1450+2  121+2  (-364)+(-2602)] kJ mol -1
-1

110
Mg 2+ (g)+2Cl
(g)
2

H at
[1/2Cl
2
]
Mg 2+ (g)+Cl
2
(g) 2

1st EA[Cl]
Mg 2+ (g)+2Cl

(g)
2nd IE[Mg]
Mg + (g)+Cl
2
(g)
1st IE[Mg]

H at
[Mg]
Mg(g)+Cl
2
(g)
Mg(s)+Cl
2
(g)

H f
[MgCl
2
] <0
MgCl
2
(s)

H
L
[MgCl
2
]

111
 H f
[MgCl
3
(s)]
=  H at
[Mg(s)] + 1 st IE of Mg + 2 nd IE of Mg + 3 rd IE of Mg
+ 3  H at
[1/2Cl
2
(g)] + 3  1 st EA of Cl +  H
L
[MgCl
3
(s)]
= 150+736+1450+7740+3  121+3  (-364)+(-5440)
-1

3

H at
[1/2Cl
2
]
Mg 3+ (g)+3Cl
(g)
Mg 3+
3
2
2
(g)
3

1st EA[Cl]
Mg 3+ (g)+3Cl

(g)
3rd IE[Mg]
112
Mg 2+
3
2
2
(g)
2nd IE[Mg]
1st IE[Mg]

H at
[Mg]
Mg +
3
2
2
(g)
3
2
2
(g)
Mg(s)+3/2Cl
2
(g)

H
L
[MgCl
3
]
MgCl
3
(s)

H f
[MgCl
3
] >0

113
Since the hypothetical compound MgCl
2 most negative

H f value, and has the this value is closest to the experimentally determined one, the most probable formula of magnesium chloride is
MgCl
2

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

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

H lattice
 
MLQ

4

0
( r

Q

 r

)
115

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

H lattice
 
MLQ

4

0
( r

Q

 r

)
Check Point 7-3
116

117
7.4

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

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
)
119

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
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 Cl 
= 6
121

122
Identify the unit cell of NaCl crystal lattice

123
Identify the unit cell of NaCl crystal lattice

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

128

129
• Simple cubic (primitive) structure
• Body-centred cubic (b.c.c.) structure
• Face-centred cubic (f.c.c.) structure

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

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

Face-centred cubic (f.c.c.) structure
132
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
- The empty spaces in a crystal lattice
Two types of interstitial sites in f.c.c. structure

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 45 o , 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.
45 o before behind

137
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 .

a c b a
139

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

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

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 .
142

143

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
144

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

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

7.4 Ionic Crystals (SB p.201)
Structure of Sodium Chloride
Unit cell of NaCl
147
Co-ordination number of Na + = 6
Co-ordination number of Cl = 6
6 : 6 co-ordination

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
with the neighboring unit cells of the crystal lattice.

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

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

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

152
Q.11 Calculate the net nos. of Na + and Cl ions in a unit cell of NaCl.
No. of Cl ions = 8

1

6

1

4
8 2
8 corners 6 faces
No. of Na + ions = 12




1




1

4
4
12 edges 1 body
Example 7-4

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
153

154

155

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

157
+

8

1
8

1

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.
158

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.
159

7.4 Ionic Crystals (SB p.203)
Structure of Calcium Fluoride  Link
Face-centred cubic lattice
160
Co-ordination number of Ca 2+ = 8
Co-ordination number of F = 4
8 : 4 co-ordination

7.4 Ionic Crystals (SB p.203)
CaF
2
Number of F  = 8
Number of Ca 2+ = 8

1
8

6

1
2

4
161

7.4 Ionic Crystals (SB p.203)
CaF
2
Only alternate cubical sites are occupied by
Ca 2+ in order to maintain electroneutrality .
162

7.4 Ionic Crystals (SB p.203)
Structure of Sodium oxide  Link
Antifluorite structure
Na
2
O vs CaF
2 fluorite structure
163

164

165
Number of Zn 2+ = 4
Number of S 2  =
8

1
8

6

1
2

4

166
Zinc sulphide, ZnS r r

 
0
0
.
.
074
184

0 .
402

 tetrahedral site
0 .
414
S 2  ions adopt f.c.c. structure
Alternate Td sites are occupied by Zn 2+ ions to ensure electroneutrality .

167

Ti 4+
O 2 
168
Number of Ti 4+ =
1

8

1
8

2
Number of O 2  =
2

4

1
2

4
2

169
4+
2 


170
Titanium(IV) oxide, r r

 
2 0 .
068
0 .
140

0 .
486

0 .
414
 octahedral site
O 2  ions adopt distorted h.c.p.(not f.c.c.) structure with alternate Oh sites occupied by Ti 4+ ions to ensure electroneutrality .

171 a b a h exagonal c losep acked

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.
172

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

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.
174

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 .
175

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 .
176

177
Factors governing the structures of ionic crystals
2. The relative size of cation and anion, r

In general, r

> r
+
, r

0
 r r

 
1 r
  r

C .
N .


178
Factors governing the structures of ionic crystals
If the cations are large, r
 
1 r
 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.

7.4 Ionic Crystals (SB p.203)
Summary : -
Interstitial site
Tetrahedral Octahedral Cubical
179
Coordination r
 r

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, NH
4
Cl
NH
4
+ is large

7.4 Ionic Crystals (SB p.203)
Summary : -
Interstitial site
Coordination r
 r

Examples
Cubical
8 : 4
0.732 – 1.000
*CaF
2
, BaF
2
, BaCl
2
, SrF
2
180

181
Q.17
cos

ABC

BA
BC
 r
 r

 r

 cos 45

A
C r
+ r

45
° r

B r
 r


0 .
414

182 r
 r


0 .
414
0 .
732
 r
 
0 .
414 r


183 r
 r


0 .
732

184
Silver chloride, AgCl r
 
0
0
.
126
.
181

0 .
696

0 .
414 r

 octahedral site
Cl  ions adopt f.c.c. structure
All Oh sites are occupied by Ag + ions to ensure electroneutrality.

185
Silver chloride, AgCl r
 
0
0
.
126
.
181

0 .
696

0 .
414 r

 octahedral site
Ag + ions adopt an open f.c.c. structure
All Oh sites are occupied by Cl  ions to ensure electroneutrality.

186
Copper(I) bromide, CuBr r r

 
0
0
.
.
074
195

0 .
379

 tetrahedral site
0 .
414
Br  ions adopt f.c.c.
structure
Only alternate Td sites are occupied by
Cu + ions to ensure electroneutrality.

Copper(II) bromide, CuBr
2 r r

 
0 .
071

0 .
364

0 .
414
0 .
195
 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
187
 Not favourable .

Copper(II) bromide, CuBr
2
Or, Cu 2+ 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.
188

Cu 2+
Br 
4 : 2 coordination
189

190
Copper(II) chloride, CuCl
2
Layer structure
0.295nm
0.230 nm

191
Copper(II) chloride, CuCl
2
6 : 3 coordination

192
7.5

193

194
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
spherical Symmetrical distribution along x, y and z axes
 almost spherical
Li + 1s 2
Na + [He] 2s 2 , 2p x
2 , 2p y
2 , 2p z
2

2
3p x
2 , 3p y
2 , 3p z
2
Zn 2+ [Ar] 3d xy
2 , 3d xz
2 , 3d yz
2 3d x
 y
3d z
2
2
Symmetrical distribution along xy, xz, yz planes and along x, y and z axes  almost spherical

196
+


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

198
By additivity rule ,
Interionic distance = r
+
For K + Cl  ,
+ r
 r
+
+ r

= 0.314 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 + ) = 0.133 nm, r

(Cl  ) = 0.181 nm

199
For Na + Cl  , r
+
+ r

= 0.275 nm (determined by X-ray)
Since r

(Cl  ) = 0.181 nm(calculated) r
+
(Na + ) = (0.275 - 0.181) nm = 0.094 nm

200
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


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

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).

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
203

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
204

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

207
1. 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
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.
 Z eff
 nuclear charge (Z)
 Ionic radius decreases as nuclear charge increases.

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

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

211

3 


3 

212
Example 7-5
Check Point 7-5

Introduction (SB p.186)
Why do two atoms bond together?
Answer configuration which brings stability.
213

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

7.2 Energetics of Formation of Ionic Compounds (SB p.195)
Given the following data:
First electron affinity of oxygen
ΔH (kJ mol –1 )
–142
Second electron affinity of oxygen +844
Standard enthalpy change of atomization of oxygen +248
Standard enthalpy change of atomization of aluminium
Standard enthalpy change of formation of aluminium oxide
+314
–1669
215

7.2 Energetics of Formation of Ionic Compounds (SB p.195)
Answer
ΔH (kJ mol –1 )
First ionization enthalpy of aluminium
Second ionization enthalpy of aluminium
+577
+1820
Third ionization enthalpy of aluminium
(a) (i) Construct a labelled Born-Haber cycle for the formation of aluminium oxide.
+2740
(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.
216

7.2 Energetics of Formation of Ionic Compounds (SB p.195)
(a) (i)
217
(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.

7.2 Energetics of Formation of Ionic Compounds (SB p.195)
Back
218
(b) ΔH f
[Al
2
O
3
(s)] = 2 × ΔH atom
[Al(s)] + 2 × (ΔH
I.E.1
[Al(g)]
+ ΔH
I.E.2
[Al(g)] + ΔH
I.E.3
[Al(g)])
+ 3 × ΔH atom
[O
2
(g)] + 3 × (ΔH
E.A.1
[O(g)]
+ ΔH
E.A.2
[O(g)]) + ΔH lattice
[Al
2
O
3
(s)]
ΔH f
[Al
2
O
3
(s)] = 2 × (+314) + 2 × (+577 + 1 820 + 2 740)
+ 3 × (+248) + 3 × ( –142 + 844)
+ ΔH lattice
[Al
2
O
3
(s)]
ΔH f
[Al
2
O
3
(s)] = + 628 + 10 274 + 744 + 2 106 + ΔH lattice
[Al
2
O
3
(s)]
ΔH lattice
[Al
2
O
3
(s)] = ΔH f
[Al
2
O
3
(s)] – (628 + 10 274 + 744 + 2 106)
= –1 669 – (628 + 10 274 + 744 + 2 106)
= –15 421 kJ mol –1

7.2 Energetics of Formation of Ionic Compounds (SB p.197)
(a) Draw a Born-Haber cycle for the formation of magnesium oxide.
(a) The Born-Haber cycle for the formation of MgO:
Answer
219

7.2 Energetics of Formation of Ionic Compounds (SB p.197)
(b) Calculate the lattice enthalpy of magnesium oxide by means of the Born-Haber cycle drawn in (a).
Given: ΔH atom
[Mg(s)] = +150 kJ mol –1
ΔH
I.E.
[Mg(g)] = +736 kJ mol –1
ΔH
I.E.
[Mg + (g)] = +1 450 kJ mol –1
ΔH atom
[O
2
(g)] = +248 kJ mol –1
ΔH
E.A.
[O(g)] = –142 kJ mol –1
ΔH
E.A.
[O – (g)] = +844 kJ mol –1
ΔH f
[MgO(s)] = –602 kJ mol –1
Answer
220

7.2 Energetics of Formation of Ionic Compounds (SB p.197)
221
(b) ΔH lattice
[MgO(s)]
= ΔH f
[MgO(s)] – ΔH atom
[Mg(s)]
– ΔH
I.E.
[Mg(g)] – ΔH
I.E.
[Mg + (g)] – ΔH atom
[O
2
(g)]
– ΔH
E.A.
[O(g)] – ΔH
E.A.
[O – (g)]
= [ –602 – 150 – 736 – 1 450 – 248 –(–142) – 844] kJ mol –1
= –3 888 kJ mol –1
Back

7.3 Stoichiometry of Ionic Compounds (SB p.201)
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
222

7.4 Ionic Crystals (SB p.204)
Back
Write down the formula of the compound that possesses the lattice structure shown on the right:
223
To calculate the number of each type of particle present in the unit cell:
Number of atom A = 1
Answer
(1 at the centre of the unit cell)
Number of atom B = 8 ×
1
4
= 2
1
(shared along each edge)
4
Number of atom C = 8 ×
1
8
= 1
1
(shared at each corner)
8
∴ The formula of the compound is AB
2
C.

7.4 Ionic Crystals (SB p.205)
Back
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.
224

7.5 Ionic Radii (SB p.208)
The following table gives the atomic and ionic radii of some
Group IIA elements.
Element
Be
Mg
Ca
Sr
Ba
Atomic radius (nm)
0.112
0.160
0.190
0.215
0.217
Ionic radius
0.031
0.065
0.099
0.133
0.135
225

7.5 Ionic Radii (SB p.208)
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
226

7.5 Ionic Radii (SB p.208)
227
(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 Be 2+ ion, the electrons experience the greatest nuclear attraction. Therefore, the contraction in size of the electron cloud is the greatest when Be 2+ ion is formed, and the ratio of ionic radius to atomic radius of Be is the lowest.

7.5 Ionic Radii (SB p.208)
(c) The electronic configurations of both K + and Ca 2+ ions are
1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 . Hence they have the same number and arrangement of electrons. However, Ca 2+
K + ion is doubly charged while ion is singly charged, so the outermost shell electrons of Ca 2+ ion experience a greater nuclear attraction. Hence, the ionic radius of
Ca 2+ ion is smaller than that of K + ion.
Back
228

7.5 Ionic Radii (SB p.208)
Arrange the following atoms or ions in an ascending order of their sizes:
(a) Be < Mg < Ca < Sr < Ba < Ra
(a) Be, Ca, Sr, Ba, Ra, Mg
(b) C < Si < Ge < Sn < Pb
(b) Si, Ge, Sn, Pb, C
(c) F – < Cl – < Br – < I –
(c) F – , Cl – , Br – , I –
(d) Al 3+ < Mg 2+ < Na + < F – < O 2 – < N 3 –
(d) Mg 2+ , Na + , Al 3+ , O 2– , F – , N 3–
Answer
Back
229