Chapter 7
Ionic Bonding
7.1 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
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Chapter 7 Ionic Bonding (SB p.180)
Ionic Bonding
When a piece of sodium metal is allowed to react
with a jar of chlorine gas …...
+
ee
2
e
-
e1:1 ratio of
Na+ and ClNew Way Chemistry for Hong Kong A-Level Book 1
Chapter 7 Ionic Bonding (SB p.180)
Formation of ionic bond between sodium
atom and chlorine atom
Cl
Na
Sodium atom Na
1s22s22p6
3
Chlorine atom Cl
1s22s22p63s23p5
New Way Chemistry for Hong Kong A-Level Book 1
Chapter 7 Ionic Bonding (SB p.180)
Formation of ionic bond between sodium
atom and chlorine atom
+
Cl
Na
Na+
Sodium ion
1s22s22p6
4
-
Chloride
ion
Cl
linked up together 2 2 6 2 6
by ionic bond 1s 2s 2p 3s 3p
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7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181)
Ionic Bonds: Donating and Accepting Electrons
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7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181)
Donating and Accepting Electrons
Ionic bonds are the strong non-directional
electrostatic forces of attraction between oppositely
charged ions.
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7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181)
Donating and Accepting Electrons
+
–
Internuclear
distance
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7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181)
Donating and Accepting Electrons
+
–
+
Internuclear
distance
Cationic
radius
(r+)
Internuclear distance = r+ + r8
New Way Chemistry for Hong Kong A-Level Book 1
–
Anionic
radius
(r-)
7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.182)
Electron transfer from a magnesium atom to two chlorine atoms
Electron transfer from two lithium atoms to an oxygen atom.
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7.2 Energetics of Formation of Ionic Compounds (SB p. 183)
Energetics of Formation of Ionic Compound
 Hf
Na(s) + ½Cl2(g)

Actually passing
through many
steps at the
molecular level
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macroscopic level
NaCl(s)
microscopic level
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7.2 Energetics of Formation of Ionic Compounds (SB p. 184)
Standard Enthalpy Change of Atomization (H atom)
The enthalpy change when one mole of gaseous
atoms is formed from its elements in the defined
physical state under standard conditions.
Na(s)
1/2 Cl2(g)
Questions:
Na(g)
H atom [Na(s)] = +109 kJ mol-1
Cl(g) H atom [1/2Cl2(g)] = +121 kJ mol-1
Why are the changes endothermic?
What type of bond is broken in each case?
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7.2 Energetics of Formation of Ionic Compounds (SB p. 184)
Ionization Enthalpy (H I.E.)
The amount of energy required to remove one
mole of valence electrons from one mole of
atoms or ions in the gaseous state.
Na(g)
Na+(g) + e-
H I.E [Na(g)] = +494 kJ mol-1
Mg(g)
Mg+(g) + e-
H I.E [Mg(g)] = +736 kJ mol-1
Mg+(g)
Mg2+(g) + e-
H I.E [Na(g)] = +1 450 kJ mol-1
Questions:
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Why are the changes endothermic?
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7.2 Energetics of Formation of Ionic Compounds (SB p. 185)
Electron Affinity (H E.A.)
The energy change when one mole of electrons is
added to one mole of atoms or ions in the gaseous
state.
First electron affinity
O(g) + e-
O-(g)
H E.A [O(g)] = - 142 kJ mol-1
Second electron affinity
O-(g) + eQuestions:
13
O2-(g) H E.A [O(g)] = - 844 kJ mol-1
Why may E.A. have -ve or +ve values?
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7.2 Energetics of Formation of Ionic Compounds (SB p. 185)
Lattice Enthalpy (H L.E.)
The energy change when one mole of an ionic
crystal is formed from its constituent ions in the
gaseous state under standard conditions
Na+ (g) + Cl-(g)
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–
+
+
–
NaCl(s)
H lattice [Na+Cl-(s)]
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7.2 Energetics of Formation of Ionic Compounds (SB p. 185)
Na+ (g) + Cl-(g)
NaCl(s)
H lattice [Na+Cl-(s)]
+ve or -ve?
–
––
+++
+
+ +
–
– –
L.E. can be determined indirectly by
either:
(1) calculations basing on the knowledge
of electrostatics in Physics (assuming ions
are point charges); or
(2) calculations basing on Hess’s Law.
Questions Why can’t L.E. be determined
directly from experiments?
:
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7.2 Energetics of Formation of Ionic Compounds (SB p. 186)
Born-Haber Cycle for the formation of sodium chloride
Hatom[Na(s)]
HI.E.
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7.2 Energetics of Formation of Ionic Compounds (SB p. 187)
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New Way Chemistry for Hong Kong A-Level Book 1
7.2 Energetics of Formation of Ionic Compounds (SB p. 187)
By Hess’s law,
ΔHf [NaCl(s)]
= ΔHatom[Na(s)] + ΔHI.E.[Na(g)] + ΔHatom[Cl2(g)]
+ ΔHE.A.[Cl(g)] + ΔHlattice [NaCl(s)]
i.e. ΔHf [NaCl(s)]
= 109 + 494 + 121 + (-364) +ΔHlattice [NaCl(s)]
ΔHlattice [NaCl(s)]
= ΔHf [NaCl(s)] +[109 + 494 + 121 + (-364)]
= -411 - [109 + 494 + 121 + (-364)] = -711 kJ mol-1
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7.3 Stoichiometry of Ionic Compounds (SB p. 189)
Stoichiometry of Ionic Compounds
Stoichiometry is the simplest ratio of the atoms
bonded together in a compound.
How can the stoichiometry of an ionic compound
be determined?
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7.3 Stoichiometry of Ionic Compounds (SB p. 189)
In Terms of Electronic Configuration
magnesium chloride
Example
Elements involved
Ions formed
Mg2+
Ratio of ions
1
Chemical formula
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Mg (Group II)
Cl (Group VII)
Cl-
2
Mg2+(Cl-)2
or
New Way Chemistry for Hong Kong A-Level Book 1
MgCl2
7.4 Ionic Crystals (SB p. 193)
Ionic Crystals
Unit cell of NaCl
Structure of Sodium Chloride
Co-ordination number of Na+ = 6
21
Co-ordination number of
Cl-
=6
6:6 co-ordination
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7.4 Ionic Crystals (SB p. 193)
A unit cell is the smallest basic portion of the
crystal lattice that, when repeatedly stacked
together at various directions, can reproduce
the entire crystal structure.
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7.4 Ionic Crystals (SB p. 193)
Ionic Crystals
corner(Cl-)
edge(Na+)
face(Cl-)
Question
Determine the number of Na+ and Cl- in a unit cell
of sodium chloride respectively.
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7.4 Ionic Crystals (SB p. 193)
Diagram showing the two inter-penetrating
face-centred cubic structure of Na+ and Cl- ions
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7.4 Ionic Crystals (SB p. 194)
Ionic Crystals
How to describe
the structure?
Structure of Caesium Chloride
(CsCl)
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Co-ordination number of Cs+ = 8
Co-ordination number of Cl- = 8
8:8 co-ordination
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7.4 Ionic Crystals (SB p. 194)
The structure is actually two
inter-penetrating simple
cubic structure of Cs+ and Clions
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7.4 Ionic Crystals (SB p. 195)
Some simple ionic structures
Type of
structure
Examples
Sodium
chloride
Na+Cl-,
Na+Br-,
K+Cl-,
K+BrCs+Cl-,
Cs+Br-,
Cs+I-
Caesium
chloride
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Radius Ratio Coordination
(r+ : r-)*
< 0.732
6:6
> 0.414
> 0.732
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8:8
7.5 Ionic Radii (SB p. 196)
Ionic Radii
Photographic
plate
X-ray
The technique of X-ray diffraction
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7.5 Ionic Radii (SB p. 196)
Electron density map
Electron density map found by X-ray diffraction
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7.5 Ionic Radii (SB p. 197)
Comparing relative atomic radii of some elements with
the ionic radii of the corresponding ions.
Size of ion vs size of atom
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7.5 Ionic Radii (SB p. 197)
Size of cation < size of atom
Reasons:
(1) The number of electron shell decreases
(2) No. of protons > no. of electrons (p/e ratio increases).
The nuclear attraction is more effective to cause a
contraction in the electron cloud.
Size of anion > size of atom
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Reasons:
(1) Repulsion between newly added electron(s) with other
electrons
(2) No. of protons < no. of electrons (p/e ratio decreases).
The nuclear attraction is less effective and there is an
expansion of the electron cloud.
New Way Chemistry for Hong Kong A-Level Book 1
7.5 Ionic Radii (SB p. 198)
Variation of ionic radii of the first 20 elements
in the Periodic Table
isoelectronic ions
Why ionic radius
decreases along
the isoelectronic
series?
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7.5 Ionic Radii (SB p. 198)
Isoelectronic ions are ions with the same
number of electrons.
The following are examples of isoelectronic series:
1. H-, Li+, Be2+, B3+
2. N3-, O2-, F-, Na+, Mg2+, Al3+
3. P3-, S2-, Cl-, K+, Ca2+
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Reason
Isoelectronic ions have the same number of
electrons. An increase in the number of protons
implies an increase in the p/e ratio which leads
to a contraction of the electron cloud.
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The END
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