7 Ionic Bonding 7.1 Formation of 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

Sodium When sodium exposed in air, it becomes tarnished rapidly Introduction (SB p.186) Sodium When sodium exposed in air, it becomes tarnished rapidly  Reacts with oxygen in air  Form a dull oxide layer on the metal surface

When sodium is placed in a bottle containing chlorine gas Introduction (SB p.186) When sodium is placed in a bottle containing chlorine gas  Burns fiercely  Gives a white coating of sodium chloride

Noble gases Very stable Introduction (SB p.186) Noble gases Very stable Rarely participate in chemical reactions and form bonds with other elements  Octet configuration

Formation of compounds Introduction (SB p.186) Formation 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 bonds Introduction (SB p.186) Three types of chemical bonds 1. Ionic bond Electrostatic attraction between positively charged particles and negatively charged particles

Three types of chemical bonds Introduction (SB p.186) Three types of chemical bonds 2. Covalent bond Electrostatic attraction between nuclei and shared electrons

Three types of chemical bonds Introduction (SB p.186) Three types of chemical bonds 3. Metallic bond Electrostatic attraction between metallic cations and delocalized electrons (electrons that have no fixed positions) Let's Think 1

Formation of Ionic Bonds: Donating and Accepting Electrons 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons

7. 1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p 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

Formation of ionic bond between sodium atom and chlorine atom 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187) Formation of ionic bond between sodium atom and chlorine atom Cl Na Sodium atom, Na 1s22s22p63s1 Chlorine atom, Cl 1s22s22p63s23p5

linked up together by ionic bond 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187) Formation of ionic bond between sodium atom and chlorine atom + - Cl Na linked up together by ionic bond Sodium ion, Na+ 1s22s22p6 Chloride ion, Cl- 1s22s22p63s23p6

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

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

Internuclear distance 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187) Ionic Bonds: Donating and Accepting Electrons + – – + Cationic radius (r+) Anionic radius (r-) Internuclear distance Internuclear distance = r+ + r-

Ionic Bonds: Donating and Accepting Electrons 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187) Ionic Bonds: Donating and Accepting Electrons 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 Electron transfer from a magnesium atom to two chlorine atoms Electron transfer from two lithium atoms to an oxygen atom

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

Energetics of Formation of Ionic Compound 7.2 Energetics of Formation of Ionic Compounds (SB p.189) Energetics of Formation of Ionic Compound  Hf ø macroscopic level Na(s) + Cl2(g)  NaCl(s) Actually passing through many steps at the molecular level microscopic level

Consider the formation of the ionic compound via a serious of steps: 7.2 Energetics of Formation of Ionic Compounds (SB p.189) Consider the formation of the ionic compound via a serious of steps: 1. The conversion of the elements to the gaseous atoms (standard enthalpy change of atomization, )

Consider the formation of the ionic compound via a serious of steps: 7.2 Energetics of Formation of Ionic Compounds (SB p.189) Consider the formation of the ionic compound via a serious of steps: 2. The conversion of the gaseous atoms to gaseous ions (ionization enthalpy, and electron affinity, )

Consider the formation of the ionic compound via a serious of steps: 7.2 Energetics of Formation of Ionic Compounds (SB p.189) Consider the formation of the ionic compound via a serious of steps: 3. The combination of the gaseous ions to form an ionic crystal (lattice enthalpy, )

1. Standard Enthalpy Change of Formation (H f) ø 7.2 Energetics of Formation of Ionic Compounds (SB p.189) 1. Standard Enthalpy Change of Formation (H f) ø The enthalpy change when one mole of the ionic compound is formed from its constituent elements (in their standard states) under standard conditions. Na(s) + Cl2(g) NaCl(s) Hf = –411 kJ mol-1 ø

2. Standard Enthalpy Change of Atomization (H atom) ø 7.2 Energetics of Formation of Ionic Compounds (SB p.190) 2. Standard Enthalpy Change of Atomization (H atom) ø The enthalpy change when one mole of gaseous atoms is formed from an element in the standard state under standard conditions. Na(s) Na(g) H atom [Na(s)] = +109 kJ mol-1 ø Cl2(g) Cl(g) H atom [Cl2(g)] = +121 kJ mol-1 ø Questions: Why are the changes endothermic? What type of bond is broken in each case?

3. Ionization Enthalpy (HI.E.) 7.2 Energetics of Formation of Ionic Compounds (SB p.190 – 191) 3. Ionization Enthalpy (HI.E.) The energy required to remove one mole of 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: Why are the changes endothermic?

4. Electron affinity (ΔHE.A.) 7.2 Energetics of Formation of Ionic Compounds (SB p.191) 4. Electron affinity (ΔHE.A.) The enthalpy change when one mole of electrons is added to one mole of atoms or ions in the gaseous state. First electron affinity of O(g): O(g) + e- O-(g) H E.A [O(g)] = –142 kJ mol-1 Second electron affinity of O(g): O-(g) + e- O2-(g) H E.A [O(g)] = –844 kJ mol-1 Questions: Why may E.A. have -ve or +ve values?

Electron affinities (in kJ mol–1) of some elements and ions 7.2 Energetics of Formation of Ionic Compounds (SB p.192) Electron affinities (in kJ mol–1) of some elements and ions

5. Lattice enthalpy ( ΔHlattice) ø 7.2 Energetics of Formation of Ionic Compounds (SB p.192) 5. Lattice enthalpy ( ΔHlattice) ø The enthalpy 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) NaCl(s) H lattice [Na+Cl-(s)] ø + –

Why can’t L.E. be determined directly from experiments? 7.2 Energetics of Formation of Ionic Compounds (SB p.192) Na+ (g) + Cl-(g) NaCl(s) H lattice [Na+Cl-(s)] ø +ve or -ve? L.E. can be calculated from the values of other experimentally determined enthalpy changes by constructing a Born-Haber cycle and applying Hess’s law + – + – + – Questions: Why can’t L.E. be determined directly from experiments?

7.2 Energetics of Formation of Ionic Compounds (SB p.193) Born-Haber Cycle A simplified enthalpy level diagram used to calculate the lattice enthalpy of an ionic compound. 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

Born-Haber Cycle for the formation of sodium chloride 7.2 Energetics of Formation of Ionic Compounds (SB p.193) Born-Haber Cycle for the formation of sodium chloride

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 7-2

7.2 Energetics of Formation of Ionic Compounds (SB p.196) Lattice enthalpy A measure of ionic bond strength which in turn represents the strength of the ionic lattice. 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

Factors affect lattice enthalpy 7.2 Energetics of Formation of Ionic Compounds (SB p.196) Factors affect lattice enthalpy Let's Think 2 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 enthalpy Check Point 7-2

Stoichiometry of Ionic Compounds 7.3 Stoichiometry of Ionic Compounds

7.3 Stoichiometry of Ionic Compounds (SB p.197) Stoichiometry of a compound is the simplest ratio of the atoms bonded to form the compound. How can the stoichiometry of an ionic compound be determined?

A. In Terms of Electronic Configuration 7.3 Stoichiometry of Ionic Compounds (SB p.197 – 198) A. In Terms of Electronic Configuration Example magnesium chloride Elements involved Mg (Group II) Cl (Group VII) Ions formed Mg2+ Cl- 2 1 Ratio of ions Chemical formula Mg2+(Cl-)2 or MgCl2

B. In Terms of Enthalpy Change of Formation 7.3 Stoichiometry of Ionic Compounds (SB p.198) B. In Terms of Enthalpy Change of Formation 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 7-3

7.4 Ionic Crystals

Structure of Sodium Chloride 7.4 Ionic Crystals (SB p.201) Structure of Sodium Chloride Unit cell of NaCl Co-ordination number of Na+ = 6 6 : 6 co-ordination Co-ordination number of Cl- = 6

Face-centred cubic lattice 7.4 Ionic Crystals (SB p.202) Face-centred cubic lattice

7.4 Ionic Crystals (SB p.202) 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.

Structure of Caesium Chloride 7.4 Ionic Crystals (SB p.202) Structure of Caesium Chloride Simple cubic lattice Co-ordination number of Cs+ = 8 8 : 8 co-ordination Co-ordination number of Cl- = 8

Face-centred cubic lattice 7.4 Ionic Crystals (SB p.203) Structure of Calcium Fluoride Face-centred cubic lattice Co-ordination number of Ca+ = 8 8 : 4 co-ordination Co-ordination number of F- = 4

Some simple ionic structures 7.4 Ionic Crystals (SB p.203) Some simple ionic structures 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 Cs+Cl-, Cs+Br-, Cs+I- > 0.732 8 : 8 Calcium fluoride CaF2, BaF2, BaCl2, SrF2 8 : 4 Example 7-4 Check Point 7-4

7.5 Ionic Radii

X-ray and electron diffraction technique 7.5 Ionic Radii (SB p.205) X-ray and electron diffraction technique X-ray Photographic plate

Electron density plot for sodium chloride crystal 7.5 Ionic Radii (SB p.205) Electron density plot for sodium chloride crystal

A. Cations Smaller radius than the corresponding atom Reasons: 7.5 Ionic Radii (SB p.206) A. 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 atom 7.5 Ionic Radii (SB p.206) Size of ion vs size of atom Comparing relative atomic radii of some elements with the ionic radii of the corresponding ions

B. Anions Larger radius than the corresponding atom Reasons: 7.5 Ionic Radii (SB p.206) B. 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 Ions They have the same number of electrons 7.5 Ionic Radii (SB p.206) C. 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 Ions Reason: 7.5 Ionic Radii (SB p.206) C. 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

7.5 Ionic Radii (SB p.206) isoelectronic ions Why ionic radius decreases along the isoelectronic series? Example 7-5 Check Point 7-5

The END

Introduction (SB p.186) Let's Think 1 Why do two atoms bond together? How does covalent bond strength compare with ionic bond strength? The two atoms tend to achieve an octet configuration which brings stability. Answer Back

Example 7-2 Given the following data: ΔH (kJ mol–1) 7.2 Energetics of Formation of Ionic Compounds (SB p.195) Example 7-2 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 of aluminium +314 Standard enthalpy change of formation of aluminium oxide –1669

Example 7-2 Answer ΔH (kJ mol–1) 7.2 Energetics of Formation of Ionic Compounds (SB p.195) Answer Example 7-2 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.

Example 7-2 7.2 Energetics of Formation of Ionic Compounds (SB p.195) (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.

7.2 Energetics of Formation of Ionic Compounds (SB p.195) Back Example 7-2 (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

7.2 Energetics of Formation of Ionic Compounds (SB p.196) Let's Think 2 What are the forces that hold atoms together in molecules and ions in ionic compounds? Electrostatic attractions between oppositely charged particles Answer Back

7.2 Energetics of Formation of Ionic Compounds (SB p.197) Check Point 7-2 (a) Draw a Born-Haber cycle for the formation of magnesium oxide. The Born-Haber cycle for the formation of MgO: Answer

7.2 Energetics of Formation of Ionic Compounds (SB p.197) Check Point 7-2 (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 ø ø Answer ø

7.2 Energetics of Formation of Ionic Compounds (SB p.197) Check Point 7-2 (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 Back

7.3 Stoichiometry of Ionic Compounds (SB p.201) Check Point 7-3 Give two properties of ions that will affect the value of the lattice enthalpy of an ionic compound. Answer 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. Back

7.4 Ionic Crystals (SB p.204) Back Example 7-4 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. Answer

Check Point 7-4 Answer Back 7.4 Ionic Crystals (SB p.205) Back Check Point 7-4 The diagram on the right shows a unit cell of titanium oxide. What is the coordination number of (a) titanium; and (b) oxygen? Answer (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.

7.5 Ionic Radii (SB p.208) Example 7-5 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

Example 7-5 Answer Explain briefly the following: 7.5 Ionic Radii (SB p.208) Example 7-5 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

Example 7-5 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.

Example 7-5 Back 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

7.5 Ionic Radii (SB p.208) Check Point 7-5 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