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New Way Chemistry for Hong Kong A-Level Book 1
1
Chapter 7Chapter 7Ionic BondingIonic Bonding
7.17.1 Ionic Bonds: Donating and Accepting Electrons Ionic Bonds: Donating and Accepting Electrons
7.2 7.2 Energetics of Formation of Ionic Compounds Energetics of Formation of Ionic Compounds
7.3 7.3 Stoichiometry of Ionic Compounds Stoichiometry of Ionic Compounds
7.47.4 Ionic Crystals Ionic Crystals
7.57.5 Ionic Radii Ionic Radii
New Way Chemistry for Hong Kong A-Level Book 1
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Ionic Bonding
When a piece of sodium metal is allowed to react with a jar of chlorine gas …...
+
e-
e-
e-
e-
Chapter 7 Ionic Bonding (SB p.180)
1:1 ratio of Na+ and Cl-
New Way Chemistry for Hong Kong A-Level Book 1
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Formation of ionic bond between sodium atom and chlorine atom
Na Cl
Sodium atom Na1s22s22p6
Chlorine atom Cl1s22s22p63s23p5
Chapter 7 Ionic Bonding (SB p.180)
New Way Chemistry for Hong Kong A-Level Book 1
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Na Cl
+ -
Sodium ion Na+
1s22s22p6
Chloride ion Cl-
1s22s22p63s23p6linked up together
by ionic bond
Formation of ionic bond between sodium atom and chlorine atom
Chapter 7 Ionic Bonding (SB p.180)
New Way Chemistry for Hong Kong A-Level Book 1
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Ionic Bonds: Donating and Accepting Electrons
7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181)
New Way Chemistry for Hong Kong A-Level Book 1
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Donating and Accepting Electrons
Ionic bonds are the strong non-directional electrostatic forces of attraction between oppositely charged ions.
Ionic bonds are the strong non-directional electrostatic forces of attraction between oppositely charged ions.
7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181)
New Way Chemistry for Hong Kong A-Level Book 1
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Donating and Accepting Electrons
+ –
Internuclear distance
7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181)
New Way Chemistry for Hong Kong A-Level Book 1
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+ –
Internuclear distance
Cationic radius(r+)
+ –
Anionic radius(r-)
Internuclear distance = r+ + r-
7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181)
Donating and Accepting Electrons
New Way Chemistry for Hong Kong A-Level Book 1
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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.
New Way Chemistry for Hong Kong A-Level Book 1
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Energetics of Formation of Ionic Compound
Na(s) + ½Cl2(g) NaCl(s)macroscopic level
Actually passing through many steps at the molecular level
microscopic level
7.2 Energetics of Formation of Ionic Compounds (SB p. 183)
Hf
ø
New Way Chemistry for Hong Kong A-Level Book 1
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The enthalpy change when one mole of gaseous atoms is formed from its elements in the defined physical state under standard conditions.
Questions: Why are the changes endothermic?
What type of bond is broken in each case?
7.2 Energetics of Formation of Ionic Compounds (SB p. 184)
Na(s) Na(g) H atom [Na(s)] = +109 kJ mol-1
ø1/2 Cl2(g) Cl(g) H atom [1/2Cl2(g)] = +121 kJ mol-1ø
Standard Enthalpy Change of Atomization (H atom)ø
New Way Chemistry for Hong Kong A-Level Book 1
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The amount of energy required to remove one mole of valence electrons from one mole of atoms or ions in the gaseous state.
Questions: Why are the changes endothermic?
7.2 Energetics of Formation of Ionic Compounds (SB p. 184)
Ionization Enthalpy (H I.E.)
ø
Na(g) Na+(g) + e- H I.E [Na(g)] = +494 kJ mol-1øMg(g) Mg+(g) + e- H I.E [Mg(g)] = +736 kJ mol-1
ø
Mg+(g) Mg2+(g) + e- H I.E [Na(g)] = +1 450 kJ mol-1
ø
New Way Chemistry for Hong Kong A-Level Book 1
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The energy change when one mole of electrons is added to one mole of atoms or ions in the gaseous state.
Questions: Why may E.A. have -ve or +ve values?
7.2 Energetics of Formation of Ionic Compounds (SB p. 185)
Electron Affinity (H E.A.)
ø
First electron affinity
O(g) + e- O-(g) H E.A [O(g)] = - 142 kJ mol-1
Second electron affinity
O-(g) + e- O2-(g) H E.A [O(g)] = - 844 kJ mol-1
øø
New Way Chemistry for Hong Kong A-Level Book 1
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The energy change when one mole of an ionic crystal is formed from its constituent ions in the gaseous state under standard conditions
+ –
– +
7.2 Energetics of Formation of Ionic Compounds (SB p. 185)
Lattice Enthalpy (H L.E.)
ø
Na+ (g) + Cl-(g) NaCl(s) H lattice [Na+Cl-(s)]ø
New Way Chemistry for Hong Kong A-Level Book 1
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+ –
– +
+ –
– +
+ –
– +
Questions:
Why can’t L.E. be determined directly from experiments?
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.
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.
+ve or -ve?
7.2 Energetics of Formation of Ionic Compounds (SB p. 185)
Na+ (g) + Cl-(g) NaCl(s) H lattice [Na+Cl-(s)]
ø
New Way Chemistry for Hong Kong A-Level Book 1
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Born-Haber Cycle for the formation of sodium chloride
7.2 Energetics of Formation of Ionic Compounds (SB p. 186)
Hatom[Na(s)]
HI.E.
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7.2 Energetics of Formation of Ionic Compounds (SB p. 187)
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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
7.2 Energetics of Formation of Ionic Compounds (SB p. 187)
ø
ø ø
ø
ø
ø
øø
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Stoichiometry of Ionic Compounds
Stoichiometry is the simplest ratio of the atoms bonded together in a compound.Stoichiometry is the simplest ratio of the atoms bonded together in a compound.
How can the stoichiometry of an ionic compound be determined?
7.3 Stoichiometry of Ionic Compounds (SB p. 189)
New Way Chemistry for Hong Kong A-Level Book 1
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Mg (Group II) Cl (Group VII)
Mg2+ Cl-
Elements involved
Ions formed
Ratio of ions
Chemical formula Mg2+(Cl-)2 or MgCl2
1 2
Example magnesium chloride
In Terms of Electronic Configuration
7.3 Stoichiometry of Ionic Compounds (SB p. 189)
New Way Chemistry for Hong Kong A-Level Book 1
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Ionic Crystals
Structure of Sodium Chloride
Co-ordination number of Na+ = 6
Co-ordination number of Cl- = 66:6 co-ordination
Unit cell of NaCl
7.4 Ionic Crystals (SB p. 193)
New Way Chemistry for Hong Kong A-Level Book 1
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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.
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.
7.4 Ionic Crystals (SB p. 193)
New Way Chemistry for Hong Kong A-Level Book 1
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Ionic Crystals
corner(Cl-)
face(Cl-)edge(Na+)
Question
Determine the number of Na+ and Cl- in a unit cell of sodium chloride respectively.
7.4 Ionic Crystals (SB p. 193)
New Way Chemistry for Hong Kong A-Level Book 1
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Diagram showing the two inter-penetrating face-centred cubic structure of Na+ and Cl- ions
7.4 Ionic Crystals (SB p. 193)
New Way Chemistry for Hong Kong A-Level Book 1
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Ionic Crystals
Structure of Caesium Chloride (CsCl)
Co-ordination number of Cs+ = 8Co-ordination number of Cl- = 8 8:8 co-ordination
How to describe the structure?
How to describe the structure?
7.4 Ionic Crystals (SB p. 194)
New Way Chemistry for Hong Kong A-Level Book 1
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The structure is actually two inter-penetrating simple cubic structure of Cs+ and Cl- ions
The structure is actually two inter-penetrating simple cubic structure of Cs+ and Cl- ions
7.4 Ionic Crystals (SB p. 194)
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Some simple ionic structures7.4 Ionic Crystals (SB p. 195)
Type of structure
Examples Radius Ratio
(r+ : r-)*
Coordination
Sodium
chloride
Na+Cl-, Na+
Br-,
K+Cl-, K+Br-
< 0.732
> 0.414
6:6
Caesium
chloride
Cs+Cl-,
Cs+Br-,
Cs+I-
> 0.732 8:8
New Way Chemistry for Hong Kong A-Level Book 1
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Ionic Radii
X-rayPhotographic plate
The technique of X-ray diffraction
7.5 Ionic Radii (SB p. 196)
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Electron density map
Electron density map found by X-ray diffraction
7.5 Ionic Radii (SB p. 196)
New Way Chemistry for Hong Kong A-Level Book 1
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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 atomSize of ion vs size of atom
New Way Chemistry for Hong Kong A-Level Book 1
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Size of cation < size of atomSize of cation < size of atomReasons:(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 atomSize of anion > size of atom
Reasons:(1) Repulsion between newly added electron(s) with other electrons(2) No. of protons < no. of electrons (p/e ratio decreases). The nuclear attraction is less effective and there is an expansion of the electron cloud.
7.5 Ionic Radii (SB p. 197)
New Way Chemistry for Hong Kong A-Level Book 1
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Variation of ionic radii of the first 20 elements in the Periodic Table
isoelectronic ions
Why ionic radius decreases along the isoelectronic series?
7.5 Ionic Radii (SB p. 198)
New Way Chemistry for Hong Kong A-Level Book 1
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Isoelectronic ions are ions with the same number of electrons.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+
7.5 Ionic Radii (SB p. 198)
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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.