Transcript Ionic Bonding

7
7.1
7.2
7.3
7.4
7.5
1
Ionic Bonding
Formation of Ionic Bonds: Donating and Accepting
Electrons
Energetics of Formation of Ionic Compounds
Stoichiometry of Ionic Compounds
Ionic Crystals
Ionic Radii
Bonding & Structure
2
Lewis Model

G.N. Lewis in 1916

Only the outermost (valence) electrons are
involved significantly in bond formation

Successful in solving chemical problems
3

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?
4

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 )
5
Q.1 Write the s,p,d,f notation for the ions in the table below
Ion
Be2+
O2Sc3+
BrBa2+
At6
s,p,d,f notation
Isoelectronic
noble gas
Ion
s,p,d,f notation
Isoelectronic
noble gas
Be2+
1s2 (2)
He
O2-
[He] 2s2,2p6 (2,8)
Ne
Sc3+
[Ne] 3s2,3p6 (2,8,8)
Ar
Br-
[Ar] 3d10,4s2,4p6 (2,8,18,8)
Kr
Ba2+
[Kr] 4d10,5s2,5p6 (2,8,18,18,8)
Xe
At-
[Xe] 4f14,5d10,6s2,6p6 (2,8,18,32,18,8)
Rn
7
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)
9
Na
Cl
Sodium atom, Na
1s22s22p63s1
Chlorine atom, Cl
1s22s22p63s23p5
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
Ionic bonds and Covalent bonds are only
extreme cases of a continuum.
In real situation, most chemical bonds
are intermediate between ionic and
covalent
15
Ionic Bonds with incomplete transfer
of electrons have covalent character.
Covalent Bonds with unequal sharing
of electrons have ionic character.
16
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.
17
Ways to compare the electron-attracting
ability of atoms
1. Ionization Enthalpy
2. Electron affinity
3. Electronegativity
18
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)
I.E.
X+(g)

X+(g)
+
e-
H 1st

X2+(g)
+
e-
H 2nd I.E.
Ionization enthalpies are always positive.
19
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)
X(g)
+
+
ee-


X-(g)
X2(g)
H 1st E.A.
H 2nd E.A.
Electron affinities can be positive or negative.
20
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.
21
Electronegativity and Types of Chemical Bonds
3. Electronegativity
The ability of an atom to attract electrons
in a chemical bond.
22
Mulliken’s scale of electronegativity
1
Electronegativity(Mulliken) 
( IE  EA )
2
Nobel Laureate in Chemistry,
1966
23
Pauling’s scale of electronegativity
Nobel Laureate in Chemistry, 1954
Nobel Laureate in Peace, 1962
24
Pauling’s scale of electronegativity
25

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
IA
H
2.1
Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7
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IIA
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
IIIA IVA
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
VA
N
3.0
P
2.1
As
2.0
Sb
2.0
Bi
1.9
VIA
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
VIIA VIIIA
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
He
Ne
Ar
Kr
Xe
Rn
-
What trends do you notice about the EN
values in the Periodic Table? Explain.
IA
H
2.1
Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7
27
IIA
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
IIIA IVA
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
VA
N
3.0
P
2.1
As
2.0
Sb
2.0
Bi
1.9
VIA
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
VIIA VIIIA
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
He
Ne
Ar
Kr
Xe
Rn
-
The EN values increase from left to
right across a Period.
IA
H
2.1
Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7
28
IIA
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
IIIA IVA
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
VA
N
3.0
P
2.1
As
2.0
Sb
2.0
Bi
1.9
VIA
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
VIIA VIIIA
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
He
Ne
Ar
Kr
Xe
Rn
-
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.
29
The EN values decrease down a Group.
IA
H
2.1
Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7
30
IIA
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
IIIA IVA
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
VA
N
3.0
P
2.1
As
2.0
Sb
2.0
Bi
1.9
VIA
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
VIIA VIIIA
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
He
Ne
Ar
Kr
Xe
Rn
-
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.
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Why are the E.A. of noble gases not shown ?
IA
H
2.1
Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7
32
IIA
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
IIIA IVA
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
VA
N
3.0
P
2.1
As
2.0
Sb
2.0
Bi
1.9
VIA
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
VIIA VIIIA
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
He
Ne
Ar
Kr
Xe
Rn
-
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
33
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
34
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
35
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
Dots and Crosses Diagram
37
Notes on Dots & Crosses representation



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).
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.
39
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.
40
Ease of formation of cation 
41
Ease of formation of cation 
Li+
Be2+
Na+
Mg2+
Al3+
K+
Ca2+
Sc3+
Rb+
Sr2+
Y3+
Zr4+
Cs+
Ba2+
La3+
Ce4+
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.
42
Ease of formation of anion 
43
Ease of formation of anion 
N3
O2
F
P3
S2
Cl
Br
I
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.
44
Q.2 Write the s,p,d,f notation for the following cations
Ion
Zn2+
Pb2+
V3+
Cr3+
Fe3+
Fe2+
Co2+
Ni2+
Cu2+
45
s,p,d,f notation
Q.2
Ion
Zn2+
Pb2+
V3+
Cr3+
Fe3+
Fe2+
Co2+
Ni2+
Cu2+
46
s,p,d,f notation
1s2,2s2,2p6,3s2,3p6,3d10
1s2,2s2,2p6,3s23p63d10,4s2,4p6,4d10,4f14,5s2,5p6,5d10,6s2
1s2,2s2,2p6,3s23p63d2
1s2,2s2,2p6,3s23p63d3
1s2,2s2,2p6,3s23p63d5
1s2,2s2,2p6,3s23p63d6
1s2,2s2,2p6,3s23p63d7
1s2,2s2,2p6,3s23p63d8
1s2,2s2,2p6,3s23p63d9
1.
18 - electrons group
e.g.
Zn2+
Fully-filled s, p & d-subshells
Ion
Zn2+
Pb2+
V3+
Cr3+
Fe3+
Fe2+
Co2+
Ni2+
Cu2+
47
s,p,d,f notation
1s2,2s2,2p6,3s2,3p6,3d10
1s2,2s2,2p6,3s23p63d10,4s2,4p6,4d10,4f14,5s2,5p6,5d10,6s2
1s2,2s2,2p6,3s23p63d2
1s2,2s2,2p6,3s23p63d3
1s2,2s2,2p6,3s23p63d5
1s2,2s2,2p6,3s23p63d6
1s2,2s2,2p6,3s23p63d7
1s2,2s2,2p6,3s23p63d8
1s2,2s2,2p6,3s23p63d9
2.
(18 + 2 ) - electron group
e.g.
Pb2+
Fully-filled s, p & d-subshells
Ion
Zn2+
Pb2+
V3+
Cr3+
Fe3+
Fe2+
Co2+
Ni2+
Cu2+
48
s,p,d,f notation
1s2,2s2,2p6,3s2,3p6,3d10
1s2,2s2,2p6,3s23p63d10,4s2,4p6,4d10,4f14,5s2,5p6,5d10,6s2
1s2,2s2,2p6,3s23p63d2
1s2,2s2,2p6,3s23p63d3
1s2,2s2,2p6,3s23p63d5
1s2,2s2,2p6,3s23p63d6
1s2,2s2,2p6,3s23p63d7
1s2,2s2,2p6,3s23p63d8
1s2,2s2,2p6,3s23p63d9
3.
Variable arrangements for ions of
transition metals.
ns2,np6, nd1 to ns2,np6, nd9
Ion
Zn2+
Pb2+
V3+
Cr3+
Fe3+
Fe2+
Co2+
Ni2+
Cu2+
s,p,d,f notation
1s2,2s2,2p6,3s2,3p6,3d10
1s2,2s2,2p6,3s23p63d10,4s2,4p6,4d10,4f14,5s2,5p6,5d10,6s2
1s2,2s2,2p6,3s23p63d2
1s2,2s2,2p6,3s23p63d3
1s2,2s2,2p6,3s23p63d5
1s2,2s2,2p6,3s23p63d6
1s2,2s2,2p6,3s23p63d7
1s2,2s2,2p6,3s23p63d8
1s2,2s2,2p6,3s23p63d9
Ti3+ [Ne] 3s23p63d1
49
Mn3+ [Ne] 3s23p63d4
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
50
Electronic arrangements of stable cations
Octet structure : cations of Group IA, IIA and IIIB
51
IA
IIA
IIIB
2,8
Na+
Mg2+
Al3+(IIIA)
2,8,8
K+
Ca2+
Sc3+
2,8,18,8
Rb+
Sr2+
Y3+
(a)
18 - electrons group
cations of Groups IB, IIB, IIIA and IVA
52
IB
IIB
IIIA
IVA
2,8,18
Cu+
Zn2+
Ga3+
--
2,8,18,18
Ag+
Cd2+
In3+
Sn4+
2,8,18,32,18
Au+
Hg2+
Tl3+
Pb4+
Less stable than ions with a noble gas structure.
53
IB
IIB
IIIA
IVA
2,8,18
Cu+
Zn2+
Ga3+
--
2,8,18,18
Ag+
Cd2+
In3+
Sn4+
2,8,18,32,18
Au+
Hg2+
Tl3+
Pb4+
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+
Zn2+
Ga3+
--
2,8,18,18
Ag+
Cd2+
In3+
Sn4+
2,8,18,32,18
Au+
Hg2+
Tl3+
Pb4+
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.
55
IB
IIB
IIIA
IVA
2,8,18
Cu+
Zn2+
Ga3+
--
2,8,18,18
Ag+
Cd2+
In3+
Sn4+
2,8,18,32,18
Au+
Hg2+
Tl3+
Pb4+
(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)
56
(b) (18+2) 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)
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], 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)
Inert pair effect increases down a Group.
Stability : Sn4+(18) > Sn2+(18+2)
Pb2+(18+2) > Pb4+(18)
58
moving down
Group IV
(b) (18+2) 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)
Inert pair effect increases down a Group.
Stability : Sb5+(18) > Sb3+(18+2)
Bi3+(18+2) > Bi5+(18)
59
moving down
Group V
(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
60
(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)
61
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.
62
First Electron Affinity (kJ mol1)
X(g) + e  X(g)
H
-73
Li
-60
Be
+18
B
C
-23 -122
O
F
-142 -322
Ne
+29
Na
-53
Mg
+21
Al
Si
P
S
Cl
-44 -134 -72 -200 -349
Ar
+35
Br
-325
Kr
+39
K
-48
63
He
+21
N
~0
E.A. becomes more –ve from Gp 1 to Gp 7
across a period
H
-73
Li
-60
Be
+18
B
C
-23 -122
O
F
-142 -328
Ne
+29
Na
-53
Mg
+21
Al
Si
P
S
Cl
-44 -134 -72 -200 -349
Ar
+35
Br
-324
Kr
+39
K
-48
64
He
+21
N
~0
The electrons added experience stronger
nuclear attraction when the atoms are getting
smaller across the period.
H
-73
Li
-60
Be
+18
B
C
-23 -122
O
F
-142 -328
Ne
+29
Na
-53
Mg
+21
Al
Si
P
S
Cl
-44 -134 -72 -200 -349
Ar
+35
Br
-324
Kr
+39
K
-48
65
He
+21
N
~0
+ve E.A. for Gp 2 and Gp 0 because the electron
added occupies a new shell / subshell
H
-73
Li
-60
Be
+18
B
C
-23 -122
O
F
-142 -328
Ne
+29
Na
-53
Mg
+21
Al
Si
P
S
Cl
-44 -134 -72 -200 -349
Ar
+35
Br
-324
Kr
+39
K
-48
66
He
+21
N
~0
Goes to a new subshell
Goes to a new shell
Be(2s2)  Be(2s22p1)
Ne(2p6)  Ne(2p63s1)
H
-73
Li
-60
Be
+18
B
C
-23 -122
O
F
-142 -328
Ne
+29
Na
-53
Mg
+21
Al
Si
P
S
Cl
-44 -134 -72 -200 -349
Ar
+35
Br
-324
Kr
+39
K
-48
67
He
+21
N
~0
More –ve E.A. for Gp 1 because the ions
formed have full-filled s-subshells.
E.g. Li(1s22s1)  Li(1s22s2)
H
-73
Li
-60
Be
+18
B
C
-23 -122
O
F
-142 -328
Ne
+29
Na
-53
Mg
+21
Al
Si
P
S
Cl
-44 -134 -72 -200 -349
Ar
+35
Br
-324
Kr
+39
K
-48
68
N
~0
He
+21
Less –ve E.A. for Gp 5 because the stable halffilled p-subshell no longer exists.
E.g. N(2s22p3)  N(2s22p4)
H
-73
Li
-60
Be
+18
B
C
-23 -122
O
F
-142 -328
Ne
+29
Na
-53
Mg
+21
Al
Si
P
S
Cl
-44 -134 -72 -200 -349
Ar
+35
Br
-324
Kr
+39
K
-48
69
He
+21
N
~0
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
Li
-60
Be
+18
B
C
N
O
F
-23 -122 ~0 -142 -328
Ne
+29
Na
-53
Mg
+21
Al
Si
P
S
Cl
-44 -134 -72 -200 -349
Ar
+35
Br
-324
Kr
+39
K
-48
70
He
+21
It is because the halide ions formed have fullfilled s/p subshells (octet structure).
H
-73
Li
-60
Be
+18
B
C
-23 -122
O
F
-142 -328
Ne
+29
Na
-53
Mg
+21
Al
Si
P
S
Cl
-44 -134 -72 -200 -349
Ar
+35
Br
-324
Kr
+39
K
-48
71
He
+21
N
~0
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.
72
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.
73
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)
K(g)
+
F(g)
IE1(K)
H1
EA1(F)
K+(g)
KF(g)
H2
+
F(g)
By Hess’s law, H1 = IE1(K) + EA1(F) + H2
74
Q.6
H1 = IE1(K) + EA1(F) + H2
(+)()
419.0103 J mol1  328.0103 J mol1 Q1Q2



23
1
23
1
6.02210 mol
6.02210 mol
4π 0 R
19
2
(1.602

10
C)
 1.511 1019 J 
4π 8.854 1012 C2 J1m1 2.170 1010 m

 1.511 1019 J  1.063 1018 J
 9.11910 19J
75


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) = +1.5111019 J
ΔH1  9.119 1019 J
76
When R 
Stability : K+(g) + F-(g) < K(g) + F(g) by +1.511x10-19J
H2 = 1.0631018 J
IE1(K) + EA1(F) = +1.5111019 J
ΔH1  9.119 1019 J
77
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) = +1.5111019 J
ΔH1  9.119 1019 J
78
K+(g) & F(g) tend to come close together in order
to become stable.
H2 = 1.0631018 J
IE1(K) + EA1(F) = +1.5111019 J
ΔH1  9.119 1019 J
79
Coulomb stabilization is the driving force for the
formation of ionic bond.
H2 = 1.0631018 J
IE1(K) + EA1(F) = +1.5111019 J
ΔH1  9.119 1019 J
80
When R = 2.1701010 m
The most energetically stable state is reached.
H2 = 1.0631018 J
IE1(K) + EA1(F) = +1.5111019 J
ΔH1  9.119 1019 J
81
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) = +1.5111019 J
ΔH1  9.119 1019 J
82
Repulsions between electron clouds and between
nuclei > attraction between ions
H2 = 1.0631018 J
IE1(K) + EA1(F) = +1.5111019 J
ΔH1  9.119 1019 J
83
Potential energy (V)
1
Repulsion : V  12
R
between opposite charges
Minimum V when R = 2.170  1010 m
R
1
Attraction : V 
R
between opposite charges
84
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) = +1.5111019 J
ΔH1  9.119 1019 J
85
A covalent bond is formed
K
F


F

K


Electrostatic attraction between positive
nuclei and bond electron pair stabilizes
the system
86
A. Formation of Ionic Crystals
Consider the formation of NaCl(s) from Na(s) & Cl2(g)
1
Na(s) +
Cl2(g)
2
Hf
H1
NaCl(s)
H2
Na+(g)
+
Cl(g)
By Hess’s law, Hf = H1 + H2
87
H1 is the sum of four terms of enthalpy changes
1.
Standard enthalpy change of atomization of Na(s)
Na(s)  Na(g)
2.
H = +108.3 kJ mol-1
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)
4.
First electron affinity of Cl(g)
Cl(g) + e-  Cl-(g)
88
H = +121.1 kJ mol-1
H = -349 kJ mol-1
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.
HLo
89
Na+(g) + Cl-(g)  NaCl(s)
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.
90
Q.7 Calculate the lattice enthalpy of NaCl
Hf = H1 +H2
Lattice enthalpy
= H2
= Hf - H1
= [(411)  (+108.3 + 500 + 121.1  349)] kJ mol1
= 791.4 kJ mol1
91
-349
-791.4
92
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.
93
Determination of Lattice Enthalpy
• From Born-Haber cycle (experimental
method, refers to Q.7)
• From theoretical calculation based on
the ionic model
94
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.
95
M is the Madelung constants that
depends on the crystal structure
MLQ Q
H lattice  
4 0 (r  r )
96
L is the Avogadro’s constant
MLQ Q
H lattice  
4 0 (r  r )
97
Q+ and Q- are the charges on the
cation and the anion respectively
MLQ Q
H lattice  
4 0 (r  r )
98
0 is the permittivity of vacuum
MLQ Q
H lattice  
4 0 (r  r )
99
(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
MLQ Q
H lattice  
4 0 (r  r )
100
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.

101
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
1.
Considerations in terms of electronic
configurations
Na
2,8,1
Mg
2,8,2
+
F
2,7
+ 2Cl
2,8,7
 MgCl2
103

[Na]+ [F]2,8 2,8

 NaF
[Cl]- [Mg]2+ [Cl]2,8,8 2,8
2,8,8
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.
104
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.
105
Q.8
Three hypothetical formulae of magnesium chloride
are proposed and their estimated lattice enthalpies
are shown in the table below.
Hypothetical
stoichiometry
MgCl
MgCl2
MgCl3
Assumed
structure
NaCl
CaF2
AlCl3
Estimated
-771 -2602 -5440
o
-1
HL (kJ mol )
106
Hf[MgCl(s)] = Hat[Mg(s)] + 1st IE of Mg + Hat[1/2Cl2(g)]
+ 1st EA of Cl + HL[MgCl(s)]
= (+150 + 736 + 121 - 364- 771) kJ mol-1
= -128 kJ mol-1
107
Mg+(g)+Cl(g)
Enthalpy (kJ mol)
Hat[1/2Cl2]
Mg+(g)+
1
Cl (g)
2 2
Mg+(g)+Cl(g)
1st IE[Mg]
1
Mg(g)+ Cl2(g)
2
Hat[Mg]
HL[MgCl]
1
2
Mg(s)+ Cl2(g)
Hf[MgCl]<0
108
1st EA[Cl]
MgCl(s)
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)]
= [150+736+1450+2121+2(-364)+(-2602)] kJ mol-1
= -752 kJ mol-1
109
Mg2+(g)+2Cl(g)
2Hat[1/2Cl2]
Mg2+(g)+Cl2(g)
21st EA[Cl]
Enthalpy (kJ mol)
Mg2+(g)+2Cl(g)
2nd IE[Mg]
HL[MgCl2]
Mg+(g)+Cl2(g)
1st IE[Mg]
Mg(g)+Cl2(g)
Hat[Mg]
Mg(s)+Cl2(g)
Hf[MgCl2] <0
110
MgCl2(s)
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)]
= 150+736+1450+7740+3121+3(-364)+(-5440)
= +3907 kJ mol-1
111
Mg3+(g)+3Cl(g)
3Hat[1/2Cl2]
Mg3+(g)+ 3 Cl2(g)
2
Enthalpy ( kJ mol)
3rd IE[Mg]
3 Cl (g)
2
2
Mg+(g)+ 3 Cl2(g)
2
1st IE[Mg]
Hat[Mg]
112
Mg3+(g)+3Cl(g)
HL[MgCl3]
Mg2+(g)+
2nd IE[Mg]
31st EA[Cl]
Mg(g)+
3Cl (g)
2
2
Mg(s)+3/2Cl2(g)
MgCl3(s)
Hf[MgCl3] >0
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
113
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
114
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
MLQ Q
H lattice  
4 0 (r  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
MLQ Q
H lattice  
4 0 (r  r )
Check Point 7-3
116
7.4
Ionic Crystals
117
7.4 Ionic Crystals (SB p.202)
Crystal Lattice(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
Identify the unit cell of NaCl crystal lattice
Not a unit cell
122
Identify the unit cell of NaCl crystal lattice
Not a unit cell
123
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.
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
Unit cell
Unit cell
127
A crystal lattice with a cubic unit cell is
known as a cubic structure.
128
Three Kinds of Cubic Structure
• Simple cubic (primitive) structure
• Body-centred cubic (b.c.c.) structure
• Face-centred cubic (f.c.c.) structure
129
Simple cubic 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
Unit cell
View this Chemscape 3D structure
132
The f.c.c. structure can be obtained by stacking the
layers of particles in the pattern abcabc...
133
Interstitial sites
- The empty spaces in a crystal lattice
Two types of interstitial sites in f.c.c. structure
Octahedral site
Tetrahedral site
134
Octahedral site : It is the space between the 3 spheres in one
layer and 3 other spheres in the adjacent
layer.
135
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.
45o
before
behind
136
Tetrahedral site : It is the space between the 3 spheres of one
layer and a fourth sphere on the upper layer.
137
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.
138
b
c
a
139
a
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
140
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
6
5
1
5
3
6
9
8
7
8
2
4 Th sites at
the back
7
4
11
10
4 Th sites in
the front
12
13
View Octahedral sites and tetrahedral sites
141
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
The crystal structures of
sodium chloride,
caesium chloride and
calcium fluoride are discussed.
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
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
145
F.C.C. Structure of Sodium Chloride
View this Chemscape 3D structure
146
7.4 Ionic Crystals (SB p.201)
Structure of Sodium Chloride
Co-ordination number of Na+ = 6
Co-ordination number of
147
Cl-
=6
Unit cell of NaCl
6 : 6 co-ordination
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.
148
1/2
Body
Face
1
1/2
Edge
Corner
Fraction of particles occupied by a unit cell
149
1/4
Body
Face
Edge
1
1/2
1/4
Corner
Fraction of particles occupied by a unit cell
150
1/8
Body
Face
Edge
Corner
1
1/2
1/4
1/8
Fraction of particles occupied by a unit cell
151
Q.11 Calculate the net nos. of Na+ and Clions in a unit cell of NaCl.
No. of
Cl-
ions =
1
1
8    6    4
8
 2
8 corners 6 faces
No. of
Na+
ions =
1
12     1  4
4
12 edges
Example 7-4
152
1 body
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.
153
Simple cubic lattice
A cubical site is the space confined
by 4 spheres in one layer and 4 other
spheres in the adjacent layer.
154
Size of interstitial sites : Cubical > octahedral > tetrahedral
In f.c.c. structure
In simple cubic structure
155
7.4 Ionic Crystals (SB p.202)
Structure of Caesium Chloride  Link
Simple cubic lattice
Co-ordination number of Cs+ = 8
Co-ordination number of Cl- = 8
156
8 : 8 co-ordination
Q.12
Number of Cs+ = 1
Number of
157
Cl
1
= 8  1
8
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
Co-ordination number of Ca2+ = 8
Co-ordination number of F- = 4
160
8 : 4 co-ordination
7.4 Ionic Crystals (SB p.203)
Q.13 (a)
CaF2
Number of F = 8
1
1
Number of Ca2+ = 8   6   4
8
2
161
7.4 Ionic Crystals (SB p.203)
Q.13 (b) CaF2
Only alternate cubical
sites are occupied by
Ca2+ in order to maintain
electroneutrality.
162
7.4 Ionic Crystals (SB p.203)
Structure of Sodium oxide  Link
Antifluorite structure
Na2O
vs
CaF2
fluorite structure
163
Q.14(a)
Zinc blende
structure
- Link
164
Q.14(a)
ZnS
Number of Zn2+ = 4
1
1
Number of S2 = 8   6   4
8
2
165
Q.14(b)
Zinc sulphide, ZnS
r 0.074

 0.402  0.414
r 0.184
 tetrahedral site
S2 ions adopt f.c.c. structure
Alternate Td sites are occupied by Zn2+
ions to ensure electroneutrality.
166
Q.15
Rutile structure - Link
The unit cell is not a cube
167
Q.15(a)
Ti4+
O2
1
Number of
= 1 8  2
8
1
2
Number of O = 2  4   4
2
Ti4+
168
TiO2
Q.15(a)
C.N. of Ti4+ = 6
C.N. of O2 = 3
169
 6 : 3 coordination
Q.15(b)
Titanium(IV) oxide,
TiO
r 2 0.068

r

0.140
 0.486  0.414
 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.
170
hexagonal close-packed
a
b
a
171
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 (cubic closest packed, 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
Factors governing the structures of ionic crystals
2. The relative size of cation and anion,
In general, r > r+,
r
0  1
r
r
 C.N . 
r
177
r
r
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.
178
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
ZnS, most
copper(I)
halides
Alkali metal
halides except
CsCl
CsCl, CsBr
CsI, NH4Cl
r
r
Examples
NH4+ is large
179
7.4 Ionic Crystals (SB p.203)
Summary : Interstitial
site
Cubical
Coordination
8:4
r
r
Examples
180
0.732 – 1.000
*CaF2, BaF2, BaCl2, SrF2
Q.17
BA
r

cosABC 

 cos 45
BC r  r
A
r
45°
r
C
181
r+
B
r
 0.414
r
r
 0.414
r
r
0.732   0.414
r
the range of ratio that favours
octahedral arrangement.
182
r
 0.732
r
the optimal ratio for cations and
anions to be in direct contact with
each other in the cubical sites.
Q.18
183
Q.18(a)
Silver chloride, AgCl
r 0.126

 0.696  0.414
r 0.181
 octahedral site
Cl ions adopt f.c.c. structure
All Oh sites are occupied by Ag+
ions to ensure electroneutrality.
184
Q.18(a)
Silver chloride, AgCl
r 0.126

 0.696  0.414
r 0.181
 octahedral site
Ag+ ions adopt an open f.c.c. structure
All Oh sites are occupied by Cl ions to
ensure electroneutrality.
185
Q.18(b)
Copper(I) bromide, CuBr
r 0.074

 0.379  0.414
r 0.195
 tetrahedral site
Br ions adopt f.c.c. structure
Only alternate Td sites are occupied by
Cu+ ions to ensure electroneutrality.
186
Copper(II) bromide, CuBr2
r 0.071

 0.364  0.414
r 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
 Not favourable.
187
Q.18(c)
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.
188
Q.18(c)
4 : 2 coordination
Cu2+
Br
189
Copper(II) chloride, CuCl2
Layer structure
0.230 nm
190
0.295nm
Copper(II) chloride, CuCl2
6 : 3 coordination
191
7.5
192
Ionic Radii
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.
193
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.
194
Q.19
Li+
spherical
1s2
Symmetrical distribution
along x, y and z axes
 almost spherical
Na+ [He] 2s2, 2px2, 2py2, 2pz2
Cl [Ne] 3s2, 3px2, 3py2, 3pz2
Zn2+
2,
2,
[Ar] 3dxy 3dxz 3dyz
2,
3dx2  y2 ,3dz2
2
Symmetrical distribution along xy, xz, yz planes
and along x, y and z axes  almost spherical
195
2
Determination of Ionic Radii
Pauling Scale
Interionic distance (r+ + r) can be
determined by
X-ray diffraction crystallography
196
7.5 Ionic Radii (SB p.205)
Cl
Na+
Electron density
map for NaCl
Contour having
same electron
density
r++ r
Cl
Na+
r++ r
Electron density plot for sodium chloride crystal
197
By additivity rule,
Interionic distance = r+ + r
For K+Cl,
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
198
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
199
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
200
Evidence :
Electron distribution is not perfectly spherical at
the boundary due to repulsion between electron
clouds of neighbouring ions.
201
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).
202
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
Ionic radius vs atomic radius
205
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
Periodic trends of ionic radius
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.
207
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.
208
7.5 Ionic Radii (SB p.206)
Isoelectronic
to He(2)
Isoelectronic
to Ne(2,8)
209
Isoelectronic
to Ar(2,8,8)
7.5 Ionic Radii (SB p.206)
H is larger than most ions, why ?
210
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)
211
The END
Example 7-5
212
Check Point 7-5
Introduction (SB p.186)
Why do two atoms bond together?
Answer
The two atoms tend to achieve an octet
configuration which brings stability.
213
Introduction (SB p.186)
How does covalent bond strength
compare with ionic bond strength?
They are similar in strength.
Both are electrostatic attractions between
charged particles.
Answer
214
Back
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
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
215
–1669
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
Answer
ΔH (kJ mol–1)
First ionization enthalpy of aluminium
+577
Second ionization enthalpy of aluminium
+1820
Third ionization enthalpy of aluminium
+2740
(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.
216
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
(a) (i)
(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.
217
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
Back
(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 × (+577 + 1 820 + 2 740)
+ 3 × (+248) + 3 × (–142 + 844)
+ ΔHlattice [Al2O3(s)]
ΔHf [Al2O3(s)] = + 628 + 10 274 + 744 + 2 106 + ΔHlattice[Al2O3(s)]
ΔHlattice[Al2O3(s)] = ΔHf [Al2O3(s)] – (628 + 10 274 + 744 + 2 106)
= –1 669 – (628 + 10 274 + 744 + 2 106)
= –15 421 kJ mol–1
218
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: ΔHatom [Mg(s)] = +150 kJ mol–1
ΔHI.E. [Mg(g)] = +736 kJ mol–1
ΔHI.E. [Mg+(g)] = +1 450 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
220
Answer
7.2 Energetics of Formation of Ionic Compounds (SB p.197)
(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 – 1 450 – 248 –(–142) – 844] kJ mol–1
= –3 888 kJ mol–1
Back
221
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:
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)
1
Number of atom B = 8 ×
=2
4
1
(shared along each edge)
4
1
Number of atom C = 8 ×
=1
8
1
(shared at each corner)
8
∴ The formula of the compound is AB2C.
223
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.
225
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
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)
(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.
227
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
228
7.5 Ionic Radii (SB p.208)
Arrange the following atoms or ions in an ascending order of
their sizes:
(a) Be, Ca, Sr, Ba, Ra, Mg
(a) Be < Mg < Ca < Sr < Ba < Ra
(b) C < Si < Ge < Sn < Pb
(b) Si, Ge, Sn, Pb, C
(c) F– < Cl– < Br– < I–
(c) F–, Cl–, Br–, I–
(d) Al3+ < Mg2+ < Na+ < F– < O2– < N3–
(d) Mg2+, Na+, Al3+, O2–, F–, N3–
Answer
Back
229