new way chemistry for hong kong a-level book 1 1 chapter 7 ionic bonding 7.1 ionic bonds: donating...

New Way Chemistry for Hong Kong A-Level Book 1

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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


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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-


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Formation of ionic bond between sodium atom and chlorine atom

Na Cl

Sodium atom Na1s22s22p6

Chlorine atom Cl1s22s22p63s23p5



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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



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Ionic Bonds: Donating and Accepting Electrons

7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181)


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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.



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+ –

Internuclear distance



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+ –

Internuclear distance

Cationic radius(r+)

+ –

Anionic radius(r-)

Internuclear distance = r+ + r-




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Electron transfer from a magnesium atom to two chlorine atoms

Electron transfer from two lithium atoms to an oxygen atom.


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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


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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?


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


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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?


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


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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?


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


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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

+ –

– +


Lattice Enthalpy (H L.E.)

ø

Na+ (g) + Cl-(g) NaCl(s) H lattice [Na+Cl-(s)]ø


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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?


Na+ (g) + Cl-(g) NaCl(s) H lattice [Na+Cl-(s)]

ø


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Born-Haber Cycle for the formation of sodium chloride


Hatom[Na(s)]

HI.E.


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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


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)


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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)


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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)


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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.



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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.



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Diagram showing the two inter-penetrating face-centred cubic structure of Na+ and Cl- ions



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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?



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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



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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


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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



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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


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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.



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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?



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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+



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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

new way chemistry for hong kong a-level book 1 1 chapter 7 ionic bonding 7.1 ionic bonds: donating...

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