New Way Chemistry for Hong Kong A-Level Book 11 7.1Formation of Ionic Bonds: Donating and Accepting...

New Way Chemistry for Hong Kong A-Level Book 11

7.17.1 Formation of Ionic Bonds: Donating and AcceptingFormation of Ionic Bonds: Donating and Accepting

ElectronsElectrons

7.2 7.2 Energetics of Formation of Ionic CompoundsEnergetics of Formation of Ionic Compounds

7.3 7.3 Stoichiometry of Ionic CompoundsStoichiometry of Ionic Compounds

7.47.4 Ionic CrystalsIonic Crystals

7.57.5 Ionic RadiiIonic Radii

Ionic BondingIonic Bonding77


Introduction (SB p.186)

When sodium exposed in air, it becomes tarnished rapidly

Reacts with oxygen in air

Form a dull oxide layer on the metal surface

SodiuSodiumm



When sodium is placed in a bottle containing chlorine gas

Burns fiercely

Gives a white coating of sodium chloride


Noble gasesNoble gases

• Very stable

• Rarely participate in chemical reactions and form bonds with other elements

Octet configuration



Formation of compoundsFormation of compounds

• Transfer or sharing of valence electron(s) takes place

• Atoms achieve the electronic configuration of the nearest noble gas in the Periodic Table

• Atoms are joined together by chemical bonds



Three types of chemical bondsThree types of chemical bonds

1. Ionic bond

Electrostatic attraction between positively charged particles and negatively charged particles




2. Covalent bond

Electrostatic attraction between nuclei and shared electrons




3. Metallic bond

Electrostatic attraction between metallic cations and delocalized electrons (electrons that have no fixed positions)



7.7.11Formation of Ionic Formation of Ionic

Bonds: Donating Bonds: Donating and Accepting and Accepting

ElectronsElectrons


Ionic BondsIonic Bonds

• Formed by a transfer of electrons from metallic atoms to non-metallic atoms

• e.g. Formation of sodium chloride

• Both the sodium ion and chloride ion attain the electronic configurations of noble gases which give rise to stability

7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)


Na Cl

Sodium atom, Na1s22s22p63s1

Chlorine atom, Cl1s22s22p63s23p5

Formation of ionic bond between Formation of ionic bond between sodium atom and chlorine atomsodium atom and chlorine atom



+ -

Sodium ion, Na+

1s22s22p6

Chloride ion, Cl-

1s22s22p63s23p6linked up together

by ionic bond

Na

Formation of ionic bond between Formation of ionic bond between sodium atom and chlorine atomsodium atom and chlorine atom

Cl



Ionic Bonds: Donating and Ionic Bonds: Donating and Accepting ElectronsAccepting Electrons



+ –

Internuclear distance




+ –

Internuclear distance

Cationic radius(r+)

+ –

Anionic radius(r-)

Internuclear distance = r+ + r-





Ionic bonds are the strong non-directional electrostatic attraction between ions of opposite charges.

Ionic bonds are the strong non-directional electrostatic attraction between ions of opposite charges.



Electron transfer from a magnesium atom to two chlorine atomsElectron transfer from a magnesium atom to two chlorine atoms

Electron transfer from two lithium atoms to an oxygen Electron transfer from two lithium atoms to an oxygen atomatom



7.7.22Energetics of FormatEnergetics of Format

ion of Ionic Compouion of Ionic Compoundsnds


Energetics of Formation of Ionic CompouEnergetics of Formation of Ionic Compoundnd

Na(s) + Cl2(g) NaCl(s)macroscopic level

Actually passing through many steps at the molecular level microscopic level

Hf

ø

7.2 Energetics of Formation of Ionic Compounds (SB p.189)


Consider the formation of the ionic Consider the formation of the ionic compound via a serious of steps:compound via a serious of steps:

1. The conversion of the elements to the gaseous atoms (standard enthalpy change of atomization, )




2. The conversion of the gaseous atoms to gaseous ions (ionization enthalpy, and electron affinity, )




3. The combination of the gaseous ions to form an ionic crystal (lattice enthalpy, )



The enthalpy change when one mole of the ionic compound is formed from its constituent elements (in their standard states) under standard conditions.

The enthalpy change when one mole of the ionic compound is formed from its constituent elements (in their standard states) under standard conditions.

1. Standard Enthalpy Change 1. Standard Enthalpy Change of Formation (of Formation (H H ff))

øø

Na(s) + Cl2(g) NaCl(s) Hf = –411 kJ mol-1ø



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-1ø

2. Standard Enthalpy Change 2. Standard Enthalpy Change of Atomization (of Atomization (H H atomatom))

The enthalpy change when one mole of gaseous atoms is formed from an element in the standard state under standard conditions.

The enthalpy change when one mole of gaseous atoms is formed from an element in the standard state under standard conditions.

Cl2(g) Cl(g) H atom [Cl2(g)] = +121 kJ mol-1ø


øø


Questions: Why are the changes endothermic?

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

7.2 Energetics of Formation of Ionic Compounds (SB p.190 – 191)

3. Ionization Enthalpy (3. Ionization Enthalpy (HHI.E.I.E.))

The energy required to remove one mole of electrons from one mole of atoms or ions in the gaseous state.

The energy required to remove one mole of electrons from one mole of atoms or ions in the gaseous state.


Questions: Why may E.A. have -ve or +ve values?

First electron affinity of O(g):

O(g) + e- O-(g) H E.A [O(g)] = –142 kJ mol-1


4. Electron affinity (ΔH4. Electron affinity (ΔHE.A.E.A.))

The enthalpy change when one mole of electrons is added to one mole of atoms or ions in the gaseous state.

The enthalpy change when one mole of electrons is added to one mole of atoms or ions in the gaseous state.

Second electron affinity of O(g):

O-(g) + e- O2-(g) H E.A [O(g)] = –844 kJ mol-1



Electron affinities (in kJ mol–1) of some elements and ions


+ –

– +

øø

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

ø


5. Lattice enthalpy ( ΔH5. Lattice enthalpy ( ΔHlatticelattice))

The enthalpy change when one mole of an ionic crystal is formed from its constituent ions in the gaseous state under standard conditions.

The enthalpy change when one mole of an ionic crystal is formed from its constituent ions in the gaseous state under standard conditions.


+ –

– +

+ –

– +

+ –

– +

Questions:

Why can’t L.E. be determined directly from experiments?

+ve or -ve?


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

øL.E. can be calculated from the values of other experimentally determined enthalpy changes by constructing a Born-Haber cycle and applying Hess’s law



Born-Haber CycleBorn-Haber Cycle

• Two different routes to form an ionic compound

• Route 1: Direct single-step reaction of the elements to form the ionic compound

• Route 2: Consists of a number of steps. The enthalpy change of each step can be found from experiments, except the lattice enthalpy

A simplified enthalpy level diagram used to calculate the lattice enthalpy of an ionic compound.

A simplified enthalpy level diagram used to calculate the lattice enthalpy of an ionic compound.



Born-Haber Cycle for the formation of Born-Haber Cycle for the formation of sodium chloridesodium chloride



• Or draw enthalpy level diagram to represent the enthalpy changes in the Born-Haber cycle

Example 7-2Example 7-2



Lattice enthalpyLattice enthalpy

• The higher (more negative) the lattice enthalpy of an ionic lattice The higher is the ionic bond strength The more stable is the ionic lattice

A measure of ionic bond strength which in turn represents the strength of the ionic lattice.

A measure of ionic bond strength which in turn represents the strength of the ionic lattice.



Factors affect lattice enthalpyFactors affect lattice enthalpy

• Effect of ionic size:

The greater the ionic size The lower (or less negative) is the lattice

enthalpy

• Effect of ionic charge:

The greater the ionic charge The higher (or more negative) is the lattice

enthalpyCheck Point 7-2Check Point 7-2


7.7.33Stoichiometry of IonStoichiometry of Ion

ic Compoundsic Compounds


How can the stoichiometry of an ionic compound be determined?

7.3 Stoichiometry of Ionic Compounds (SB p.197)

StoichiometryStoichiometry

Stoichiometry of a compound is the simplest ratio of the atoms bonded to form the compound.

Stoichiometry of a compound is the simplest ratio of the atoms bonded to form the compound.


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

A. In Terms of Electronic A. In Terms of Electronic ConfigurationConfiguration

7.3 Stoichiometry of Ionic Compounds (SB p.197 – 198)


B. In Terms of Enthalpy Change of B. In Terms of Enthalpy Change of FormationFormation


• The more negative the enthalpy change of formation of an ionic compound

The greater is the driving force for its formation

The more stable the compound

Check Point 7-3Check Point 7-3


7.7.44Ionic CrystalsIonic Crystals


Co-ordination number of Na+ = 6

Co-ordination number of Cl- = 66 : 6 co-ordination

Unit cell of NaClStructure of Sodium ChlorideStructure of Sodium Chloride

7.4 Ionic Crystals (SB p.201)



Face-centred cubic lattice


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.



Co-ordination number of Cs+ = 8

Co-ordination number of Cl- = 88 : 8 co-ordination

Structure of Caesium ChlorideStructure of Caesium Chloride


Simple cubic lattice


Co-ordination number of Ca+ = 8

Co-ordination number of F- = 48 : 4 co-ordination

Structure of Calcium FluorideStructure of Calcium Fluoride


Face-centred cubic lattice


Type of structure

ExamplesRadius Ratio

(r+ : r-)*Coordination

Sodium

chloride

Na+Cl-, Na+Br-,

K+Cl-, K+Br-

< 0.732

> 0.4146 : 6

Caesium

chloride

Cs+Cl-, Cs+Br-,

Cs+I-> 0.732 8 : 8

Calcium fluoride

CaF2, BaF2, BaCl2, SrF2

> 0.732 8 : 4


Some simple ionic structuresSome simple ionic structures

Check Point 7-4Check Point 7-4Example 7-4Example 7-4


7.7.55 Ionic RadiiIonic Radii


X-ray

Photographic plate

X-ray and electron diffraction X-ray and electron diffraction techniquetechnique

7.5 Ionic Radii (SB p.205)



Electron density plot for sodium chloride crystal


A. CationsA. Cations


• Smaller radius than the corresponding atom

• Reasons:

1. The number of electron shells 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 ion vs size of atomSize of ion vs size of atom


Comparing relative atomic radii of some elements with the ionic radii of the

corresponding ions


B. AnionsB. Anions


• Larger radius than the corresponding 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


C. Isoelectronic IonsC. Isoelectronic Ions


• They have the same number of electrons

• Sizes decrease along the isoelectronic series:

1. H– > Li+ > Be2+ > B3+

(isoelectronic to He)

2. N3– > O2– > F– > Na+ > Mg2+ > Al3+ (isoelectronic to Ne)

3. P3– > S2– > Cl– > K+ > Ca2+

(isoelectronic to Ar)


C. Isoelectronic IonsC. Isoelectronic Ions


• Reason:

They 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


isoelectronic ions Why ionic radius decreases along the isoelectronic series?


Check Point 7-5Check Point 7-5Example 7-5Example 7-5


The END


Why do two atoms bond together? How does covalent bond strength compare with ionic bond

strength?

Back

The two atoms tend to achieve an octet configuration which brings stability. Answer



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 ofaluminium +314

Standard enthalpy change of formation ofaluminium oxide –1669



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.


Answer



(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.



(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

ø øø

ø

ø

ø

øø ø

Back


What are the forces that hold atoms together in molecules and ions in ionic compounds?

Back

Electrostatic attractions between oppositely charged particlesAnswer



(a) Draw a Born-Haber cycle for the formation of magnesium oxide.

(a) The Born-Haber cycle for the formation of MgO:

Answer



(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


øø

ø

Answer


Back


(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

ø

ø

ø

ø


Give two properties of ions that will affect the value of the lattice enthalpy of an ionic compound.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.

Answer

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

(1 at the centre of the unit cell)

Number of atom B = 8 × = 2

(shared along each edge)

Number of atom C = 8 × = 1

(shared at each corner)

∴ The formula of the compound is AB2C.

41

41

81 8

1

Answer

Back7.4 Ionic Crystals (SB p.204)


The diagram on the right shows a unit cell of titanium oxide. What is the coordination number of

(a) titanium; and

(b) oxygen?(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.

Answer

Back



The following table gives the atomic and ionic radii of some Group IIA elements.


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


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



(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.



(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


Arrange the following atoms or ions in an ascending order of their sizes:

(a) Be, Ca, Sr, Ba, Ra, Mg

(b) Si, Ge, Sn, Pb, C

(c) F–, Cl–, Br–, I–

(d) Mg2+, Na+, Al3+, O2–, F–, N3–

(a) Be < Mg < Ca < Sr < Ba < Ra

(b) C < Si < Ge < Sn < Pb

(c) F– < Cl– < Br– < I–

(d) Al3+ < Mg2+ < Na+ < F– < O2– < N3–

Answer

Back


New Way Chemistry for Hong Kong A-Level Book 11 7.1Formation of Ionic Bonds: Donating and Accepting...

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Transcript of New Way Chemistry for Hong Kong A-Level Book 11 7.1Formation of Ionic Bonds: Donating and Accepting...