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
New Way Chemistry for Hong Kong A-Level Book 12
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
New Way Chemistry for Hong Kong A-Level Book 13
Introduction (SB p.186)
When sodium is placed in a bottle containing chlorine gas
Burns fiercely
Gives a white coating of sodium chloride
New Way Chemistry for Hong Kong A-Level Book 14
Noble gasesNoble gases
• Very stable
• Rarely participate in chemical reactions and form bonds with other elements
Octet configuration
Introduction (SB p.186)
New Way Chemistry for Hong Kong A-Level Book 15
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
Introduction (SB p.186)
New Way Chemistry for Hong Kong A-Level Book 16
Three types of chemical bondsThree types of chemical bonds
1. Ionic bond
Electrostatic attraction between positively charged particles and negatively charged particles
Introduction (SB p.186)
New Way Chemistry for Hong Kong A-Level Book 17
Three types of chemical bondsThree types of chemical bonds
2. Covalent bond
Electrostatic attraction between nuclei and shared electrons
Introduction (SB p.186)
New Way Chemistry for Hong Kong A-Level Book 18
Three types of chemical bondsThree types of chemical bonds
3. Metallic bond
Electrostatic attraction between metallic cations and delocalized electrons (electrons that have no fixed positions)
Introduction (SB p.186)
New Way Chemistry for Hong Kong A-Level Book 19
7.7.11Formation of Ionic Formation of Ionic
Bonds: Donating Bonds: Donating and Accepting and Accepting
ElectronsElectrons
New Way Chemistry for Hong Kong A-Level Book 110
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)
New Way Chemistry for Hong Kong A-Level Book 111
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
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
New Way Chemistry for Hong Kong A-Level Book 112
+ -
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
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
New Way Chemistry for Hong Kong A-Level Book 113
Ionic Bonds: Donating and Ionic Bonds: Donating and Accepting ElectronsAccepting Electrons
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
New Way Chemistry for Hong Kong A-Level Book 114
+ –
Internuclear distance
Ionic Bonds: Donating and Ionic Bonds: Donating and Accepting ElectronsAccepting Electrons
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
New Way Chemistry for Hong Kong A-Level Book 115
+ –
Internuclear distance
Cationic radius(r+)
+ –
Anionic radius(r-)
Internuclear distance = r+ + r-
Ionic Bonds: Donating and Ionic Bonds: Donating and Accepting ElectronsAccepting Electrons
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
New Way Chemistry for Hong Kong A-Level Book 116
Ionic Bonds: Donating and Ionic Bonds: Donating and Accepting ElectronsAccepting Electrons
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.
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
New Way Chemistry for Hong Kong A-Level Book 117
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
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.188)
New Way Chemistry for Hong Kong A-Level Book 118
7.7.22Energetics of FormatEnergetics of Format
ion of Ionic Compouion of Ionic Compoundsnds
New Way Chemistry for Hong Kong A-Level Book 119
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)
New Way Chemistry for Hong Kong A-Level Book 120
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, )
7.2 Energetics of Formation of Ionic Compounds (SB p.189)
New Way Chemistry for Hong Kong A-Level Book 121
Consider the formation of the ionic Consider the formation of the ionic compound via a serious of steps:compound via a serious of steps:
2. The conversion of the gaseous atoms to gaseous ions (ionization enthalpy, and electron affinity, )
7.2 Energetics of Formation of Ionic Compounds (SB p.189)
New Way Chemistry for Hong Kong A-Level Book 122
Consider the formation of the ionic Consider the formation of the ionic compound via a serious of steps:compound via a serious of steps:
3. The combination of the gaseous ions to form an ionic crystal (lattice enthalpy, )
7.2 Energetics of Formation of Ionic Compounds (SB p.189)
New Way Chemistry for Hong Kong A-Level Book 123
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
7.2 Energetics of Formation of Ionic Compounds (SB p.189)
New Way Chemistry for Hong Kong A-Level Book 124
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
7.2 Energetics of Formation of Ionic Compounds (SB p.190)
øø
New Way Chemistry for Hong Kong A-Level Book 125
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.
New Way Chemistry for Hong Kong A-Level Book 126
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
7.2 Energetics of Formation of Ionic Compounds (SB p.191)
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
New Way Chemistry for Hong Kong A-Level Book 127
7.2 Energetics of Formation of Ionic Compounds (SB p.192)
Electron affinities (in kJ mol–1) of some elements and ions
New Way Chemistry for Hong Kong A-Level Book 128
+ –
– +
øø
Na+ (g) + Cl-(g) NaCl(s) H lattice [Na+Cl-(s)]
ø
7.2 Energetics of Formation of Ionic Compounds (SB p.192)
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.
New Way Chemistry for Hong Kong A-Level Book 129
+ –
– +
+ –
– +
+ –
– +
Questions:
Why can’t L.E. be determined directly from experiments?
+ve or -ve?
7.2 Energetics of Formation of Ionic Compounds (SB p.192)
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
New Way Chemistry for Hong Kong A-Level Book 130
7.2 Energetics of Formation of Ionic Compounds (SB p.193)
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.
New Way Chemistry for Hong Kong A-Level Book 131
7.2 Energetics of Formation of Ionic Compounds (SB p.193)
Born-Haber Cycle for the formation of Born-Haber Cycle for the formation of sodium chloridesodium chloride
New Way Chemistry for Hong Kong A-Level Book 132
7.2 Energetics of Formation of Ionic Compounds (SB p.194)
• Or draw enthalpy level diagram to represent the enthalpy changes in the Born-Haber cycle
Example 7-2Example 7-2
New Way Chemistry for Hong Kong A-Level Book 133
7.2 Energetics of Formation of Ionic Compounds (SB p.196)
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.
New Way Chemistry for Hong Kong A-Level Book 134
7.2 Energetics of Formation of Ionic Compounds (SB p.196)
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
New Way Chemistry for Hong Kong A-Level Book 135
7.7.33Stoichiometry of IonStoichiometry of Ion
ic Compoundsic Compounds
New Way Chemistry for Hong Kong A-Level Book 136
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.
New Way Chemistry for Hong Kong A-Level Book 137
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)
New Way Chemistry for Hong Kong A-Level Book 138
B. In Terms of Enthalpy Change of B. In Terms of Enthalpy Change of FormationFormation
7.3 Stoichiometry of Ionic Compounds (SB p.198)
• 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
New Way Chemistry for Hong Kong A-Level Book 140
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)
New Way Chemistry for Hong Kong A-Level Book 141
7.4 Ionic Crystals (SB p.202)
Face-centred cubic lattice
New Way Chemistry for Hong Kong A-Level Book 142
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.202)
New Way Chemistry for Hong Kong A-Level Book 143
Co-ordination number of Cs+ = 8
Co-ordination number of Cl- = 88 : 8 co-ordination
Structure of Caesium ChlorideStructure of Caesium Chloride
7.4 Ionic Crystals (SB p.202)
Simple cubic lattice
New Way Chemistry for Hong Kong A-Level Book 144
Co-ordination number of Ca+ = 8
Co-ordination number of F- = 48 : 4 co-ordination
Structure of Calcium FluorideStructure of Calcium Fluoride
7.4 Ionic Crystals (SB p.203)
Face-centred cubic lattice
New Way Chemistry for Hong Kong A-Level Book 145
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
7.4 Ionic Crystals (SB p.203)
Some simple ionic structuresSome simple ionic structures
Check Point 7-4Check Point 7-4Example 7-4Example 7-4
New Way Chemistry for Hong Kong A-Level Book 147
X-ray
Photographic plate
X-ray and electron diffraction X-ray and electron diffraction techniquetechnique
7.5 Ionic Radii (SB p.205)
New Way Chemistry for Hong Kong A-Level Book 148
7.5 Ionic Radii (SB p.205)
Electron density plot for sodium chloride crystal
New Way Chemistry for Hong Kong A-Level Book 149
A. CationsA. Cations
7.5 Ionic Radii (SB p.206)
• 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
New Way Chemistry for Hong Kong A-Level Book 150
Size of ion vs size of atomSize of ion vs size of atom
7.5 Ionic Radii (SB p.206)
Comparing relative atomic radii of some elements with the ionic radii of the
corresponding ions
New Way Chemistry for Hong Kong A-Level Book 151
B. AnionsB. Anions
7.5 Ionic Radii (SB p.206)
• 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
New Way Chemistry for Hong Kong A-Level Book 152
C. Isoelectronic IonsC. Isoelectronic Ions
7.5 Ionic Radii (SB p.206)
• 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)
New Way Chemistry for Hong Kong A-Level Book 153
C. Isoelectronic IonsC. Isoelectronic Ions
7.5 Ionic Radii (SB p.206)
• 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
New Way Chemistry for Hong Kong A-Level Book 154
isoelectronic ions Why ionic radius decreases along the isoelectronic series?
7.5 Ionic Radii (SB p.206)
Check Point 7-5Check Point 7-5Example 7-5Example 7-5
New Way Chemistry for Hong Kong A-Level Book 156
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
Introduction (SB p.186)
New Way Chemistry for Hong Kong A-Level Book 157
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
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
New Way Chemistry for Hong Kong A-Level Book 158
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.
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
Answer
New Way Chemistry for Hong Kong A-Level Book 159
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.
New Way Chemistry for Hong Kong A-Level Book 160
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
(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
New Way Chemistry for Hong Kong A-Level Book 161
What are the forces that hold atoms together in molecules and ions in ionic compounds?
Back
Electrostatic attractions between oppositely charged particlesAnswer
7.2 Energetics of Formation of Ionic Compounds (SB p.196)
New Way Chemistry for Hong Kong A-Level Book 162
(a) Draw a Born-Haber cycle for the formation of magnesium oxide.
(a) The Born-Haber cycle for the formation of MgO:
Answer
7.2 Energetics of Formation of Ionic Compounds (SB p.197)
New Way Chemistry for Hong Kong A-Level Book 163
(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
7.2 Energetics of Formation of Ionic Compounds (SB p.197)
øø
ø
Answer
New Way Chemistry for Hong Kong A-Level Book 164
Back
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
ø
ø
ø
ø
New Way Chemistry for Hong Kong A-Level Book 165
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
7.3 Stoichiometry of Ionic Compounds (SB p.201)
New Way Chemistry for Hong Kong A-Level Book 166
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)
New Way Chemistry for Hong Kong A-Level Book 167
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
7.4 Ionic Crystals (SB p.205)
New Way Chemistry for Hong Kong A-Level Book 168
The following table gives the atomic and ionic radii of some Group IIA elements.
7.5 Ionic Radii (SB p.208)
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
New Way Chemistry for Hong Kong A-Level Book 169
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).
7.5 Ionic Radii (SB p.208)
Answer
New Way Chemistry for Hong Kong A-Level Book 170
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.
New Way Chemistry for Hong Kong A-Level Book 171
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
New Way Chemistry for Hong Kong A-Level Book 172
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
7.5 Ionic Radii (SB p.208)
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