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New Way Chemistry for Hong Kong A-Level Book 11 7.1Formation of Ionic Bonds: Donating and Accepting Electrons 7.2 Energetics of Formation of Ionic Compounds.

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Presentation on theme: "New Way Chemistry for Hong Kong A-Level Book 11 7.1Formation of Ionic Bonds: Donating and Accepting Electrons 7.2 Energetics of Formation of Ionic Compounds."— Presentation transcript:

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2 New Way Chemistry for Hong Kong A-Level Book Formation of Ionic Bonds: Donating and Accepting Electrons 7.2 Energetics of Formation of Ionic Compounds 7.3 Stoichiometry of Ionic Compounds 7.4Ionic Crystals 7.5Ionic Radii Ionic Bonding 7

3 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 Sodium

4 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

5 New Way Chemistry for Hong Kong A-Level Book 14 Noble gases Very stable Rarely participate in chemical reactions and form bonds with other elements  Octet configuration Introduction (SB p.186)

6 New Way Chemistry for Hong Kong A-Level Book 15 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 Introduction (SB p.186)

7 New Way Chemistry for Hong Kong A-Level Book 16 Three types of chemical bonds 1.Ionic bond Electrostatic attraction between positively charged particles and negatively charged particles Introduction (SB p.186)

8 New Way Chemistry for Hong Kong A-Level Book 17 Three types of chemical bonds 2.Covalent bond Electrostatic attraction between nuclei and shared electrons Introduction (SB p.186)

9 New Way Chemistry for Hong Kong A-Level Book 18 Three 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)

10 New Way Chemistry for Hong Kong A-Level Book Formation of Ionic Bonds: Donating and Accepting Electrons

11 New Way Chemistry for Hong Kong A-Level Book 110 Ionic 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)

12 New Way Chemistry for Hong Kong A-Level Book 111 Na Cl Sodium atom, Na 1s 2 2s 2 2p 6 3s 1 Chlorine atom, Cl 1s 2 2s 2 2p 6 3s 2 3p 5 Formation of ionic bond between sodium atom and chlorine atom 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)

13 New Way Chemistry for Hong Kong A-Level Book Sodium ion, Na + 1s 2 2s 2 2p 6 Chloride ion, Cl - 1s 2 2s 2 2p 6 3s 2 3p 6 linked up together by ionic bond Na Formation of ionic bond between sodium atom and chlorine atom Cl 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)

14 New Way Chemistry for Hong Kong A-Level Book 113 Ionic Bonds: Donating and Accepting Electrons 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)

15 New Way Chemistry for Hong Kong A-Level Book – Internuclear distance Ionic Bonds: Donating and Accepting Electrons 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)

16 New Way Chemistry for Hong Kong A-Level Book – Internuclear distance Cationic radius (r + ) + – Anionic radius (r - ) Internuclear distance = r + + r - Ionic Bonds: Donating and Accepting Electrons 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)

17 New Way Chemistry for Hong Kong A-Level Book 116 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.187)

18 New Way Chemistry for Hong Kong A-Level Book 117 Electron transfer from a magnesium atom to two chlorine atoms Electron transfer from two lithium atoms to an oxygen atom 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.188)

19 New Way Chemistry for Hong Kong A-Level Book Energetics of Formation of Ionic Compounds

20 New Way Chemistry for Hong Kong A-Level Book 119 Energetics of Formation of Ionic Compound Na(s) + Cl 2 (g)  NaCl(s) macroscopic level Actually passing through many steps at the molecular level microscopic level  Hf Hf ø 7.2 Energetics of Formation of Ionic Compounds (SB p.189)

21 New Way Chemistry for Hong Kong A-Level Book 120 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, ) 7.2 Energetics of Formation of Ionic Compounds (SB p.189)

22 New Way Chemistry for Hong Kong A-Level Book 121 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, ) 7.2 Energetics of Formation of Ionic Compounds (SB p.189)

23 New Way Chemistry for Hong Kong A-Level Book 122 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, ) 7.2 Energetics of Formation of Ionic Compounds (SB p.189)

24 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. 1. Standard Enthalpy Change of Formation (  H f ) ø Na(s) + Cl 2 (g) NaCl(s)  H f = –411 kJ mol -1 ø 7.2 Energetics of Formation of Ionic Compounds (SB p.189)

25 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 -1 ø 2. Standard Enthalpy Change of Atomization (  H atom ) The enthalpy change when one mole of gaseous atoms is formed from an element in the standard state under standard conditions. Cl 2 (g) Cl(g)  H atom [Cl 2 (g)] = +121 kJ mol -1 ø 7.2 Energetics of Formation of Ionic Compounds (SB p.190) ø

26 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) Mg 2+ (g) + e -  H I.E [Na(g)] = kJ mol Energetics of Formation of Ionic Compounds (SB p.190 – 191) 3. Ionization Enthalpy (  H I.E. ) The energy required to remove one mole of electrons from one mole of atoms or ions in the gaseous state.

27 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 Energetics of Formation of Ionic Compounds (SB p.191) 4. Electron affinity (ΔH E.A. ) 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 - O 2- (g)  H E.A [O(g)] = –844 kJ mol -1

28 New Way Chemistry for Hong Kong A-Level Book Energetics of Formation of Ionic Compounds (SB p.192) Electron affinities (in kJ mol –1 ) of some elements and ions

29 New Way Chemistry for Hong Kong A-Level Book – –+ø 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 ( ΔH lattice ) The enthalpy change when one mole of an ionic crystal is formed from its constituent ions in the gaseous state under standard conditions.

30 New Way Chemistry for Hong Kong A-Level Book – –+ + – – + + – – + 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)] ø 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

31 New Way Chemistry for Hong Kong A-Level Book Energetics of Formation of Ionic Compounds (SB p.193) Born-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.

32 New Way Chemistry for Hong Kong A-Level Book Energetics of Formation of Ionic Compounds (SB p.193) Born-Haber Cycle for the formation of sodium chloride

33 New Way Chemistry for Hong Kong A-Level Book 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 Example 7-2

34 New Way Chemistry for Hong Kong A-Level Book Energetics of Formation of Ionic Compounds (SB p.196) Lattice 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.

35 New Way Chemistry for Hong Kong A-Level Book Energetics of Formation of Ionic Compounds (SB p.196) Factors 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 enthalpy Check Point 7-2 Check Point 7-2

36 New Way Chemistry for Hong Kong A-Level Book Stoichiometry of Ionic Compounds

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

38 New Way Chemistry for Hong Kong A-Level Book 137 Mg (Group II) Cl (Group VII) Mg 2+ Cl - Elements involved Ions formed Ratio of ions Chemical formula Mg 2+ (Cl - ) 2 or MgCl Example magnesium chloride A. In Terms of Electronic Configuration 7.3 Stoichiometry of Ionic Compounds (SB p.197 – 198)

39 New Way Chemistry for Hong Kong A-Level Book 138 B. In Terms of Enthalpy Change of Formation 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 7-3 Check Point 7-3

40 New Way Chemistry for Hong Kong A-Level Book Ionic Crystals

41 New Way Chemistry for Hong Kong A-Level Book 140 Co-ordination number of Na + = 6 Co-ordination number of Cl - = 6 6 : 6 co-ordination Unit cell of NaCl Structure of Sodium Chloride 7.4 Ionic Crystals (SB p.201)

42 New Way Chemistry for Hong Kong A-Level Book Ionic Crystals (SB p.202) Face-centred cubic lattice

43 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. 7.4 Ionic Crystals (SB p.202)

44 New Way Chemistry for Hong Kong A-Level Book 143 Co-ordination number of Cs + = 8 Co-ordination number of Cl - = 8 8 : 8 co-ordination Structure of Caesium Chloride 7.4 Ionic Crystals (SB p.202) Simple cubic lattice

45 New Way Chemistry for Hong Kong A-Level Book 144 Co-ordination number of Ca + = 8 Co-ordination number of F - = 4 8 : 4 co-ordination Structure of Calcium Fluoride 7.4 Ionic Crystals (SB p.203) Face-centred cubic lattice

46 New Way Chemistry for Hong Kong A-Level Book 145 Type of structure Examples Radius Ratio (r + : r - )* Coordination Sodium chloride Na + Cl -, Na + Br -, K + Cl -, K + Br - < > : 6 Caesium chloride Cs + Cl -, Cs + Br -, Cs + I - > : 8 Calcium fluoride CaF 2, BaF 2, BaCl 2, SrF 2 > : Ionic Crystals (SB p.203) Some simple ionic structures Check Point 7-4 Check Point 7-4 Example 7-4 Example 7-4

47 New Way Chemistry for Hong Kong A-Level Book Ionic Radii

48 New Way Chemistry for Hong Kong A-Level Book 147 X-ray Photographic plate X-ray and electron diffraction technique 7.5 Ionic Radii (SB p.205)

49 New Way Chemistry for Hong Kong A-Level Book Ionic Radii (SB p.205) Electron density plot for sodium chloride crystal

50 New Way Chemistry for Hong Kong A-Level Book 149 A. 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

51 New Way Chemistry for Hong Kong A-Level Book 150 Size of ion vs size of atom Size 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

52 New Way Chemistry for Hong Kong A-Level Book 151 B. 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

53 New Way Chemistry for Hong Kong A-Level Book 152 C. 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 + > Be 2+ > B 3+ (isoelectronic to He) 2.N 3– > O 2– > F – > Na + > Mg 2+ > Al 3+ (isoelectronic to Ne) 3. P 3– > S 2– > Cl – > K + > Ca 2+ (isoelectronic to Ar)

54 New Way Chemistry for Hong Kong A-Level Book 153 C. 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

55 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 7-5 Check Point 7-5 Example 7-5 Example 7-5

56 New Way Chemistry for Hong Kong A-Level Book 155 The END

57 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)

58 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 of aluminium+314 Standard enthalpy change of formation of aluminium oxide– Energetics of Formation of Ionic Compounds (SB p.195)

59 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

60 New Way Chemistry for Hong Kong A-Level Book 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.

61 New Way Chemistry for Hong Kong A-Level Book Energetics of Formation of Ionic Compounds (SB p.195) (b)ΔH f [Al 2 O 3 (s)] = 2 × ΔH atom [Al(s)] + 2 × (ΔH I.E.1 [Al(g)] + ΔH I.E.2 [Al(g)] + ΔH I.E.3 [Al(g)]) + 3 × ΔH atom [O 2 (g)] + 3 × (ΔH E.A.1 [O(g)] + ΔH E.A.2 [O(g)]) + ΔH lattice [Al 2 O 3 (s)] ΔH f [Al 2 O 3 (s)] =2 × (+314) + 2 × ( ) + 3 × (+248) + 3 × (– ) + ΔH lattice [Al 2 O 3 (s)] ΔH f [Al 2 O 3 (s)] = ΔH lattice [Al 2 O 3 (s)] ΔH lattice [Al 2 O 3 (s)]= ΔH f [Al 2 O 3 (s)] – ( ) = –1 669 – ( ) = – kJ mol –1 øø ø ø ø ø ø øø Back

62 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 particles Answer 7.2 Energetics of Formation of Ionic Compounds (SB p.196)

63 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)

64 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:ΔH atom [Mg(s)] = +150 kJ mol –1 ΔH I.E. [Mg(g)] = +736 kJ mol –1 ΔH I.E. [Mg + (g)] = kJ mol –1 ΔH atom [O 2 (g)] = +248 kJ mol –1 ΔH E.A. [O(g)] = –142 kJ mol –1 ΔH E.A. [O – (g)] = +844 kJ mol –1 ΔH f [MgO(s)] = –602 kJ mol –1 7.2 Energetics of Formation of Ionic Compounds (SB p.197) ø ø ø Answer

65 New Way Chemistry for Hong Kong A-Level Book 164 Back 7.2 Energetics of Formation of Ionic Compounds (SB p.197) (b)ΔH lattice [MgO(s)] = ΔH f [MgO(s)] – ΔH atom [Mg(s)] – ΔH I.E. [Mg(g)] – ΔH I.E. [Mg + (g)] – ΔH atom [O 2 (g)] – ΔH E.A. [O(g)] – ΔH E.A. [O – (g)] = [–602 – 150 – 736 – – 248 –(–142) – 844] kJ mol –1 = –3 888 kJ mol –1 ø ø ø ø

66 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)

67 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 AB 2 C. Answer Back 7.4 Ionic Crystals (SB p.204)

68 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)

69 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) ElementAtomic radius (nm)Ionic radius Be Mg Ca Sr Ba

70 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

71 New Way Chemistry for Hong Kong A-Level Book 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 Be 2+ ion, the electrons experience the greatest nuclear attraction. Therefore, the contraction in size of the electron cloud is the greatest when Be 2+ ion is formed, and the ratio of ionic radius to atomic radius of Be is the lowest.

72 New Way Chemistry for Hong Kong A-Level Book Ionic Radii (SB p.208) (c) The electronic configurations of both K + and Ca 2+ ions are 1s 2 2s 2 2p 6 3s 2 3p 6. Hence they have the same number and arrangement of electrons. However, Ca 2+ ion is doubly charged while K + ion is singly charged, so the outermost shell electrons of Ca 2+ ion experience a greater nuclear attraction. Hence, the ionic radius of Ca 2+ ion is smaller than that of K + ion. Back

73 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)Mg 2+, Na +, Al 3+, O 2–, F –, N 3– (a) Be < Mg < Ca < Sr < Ba < Ra (b) C < Si < Ge < Sn < Pb (c) F – < Cl – < Br – < I – (d) Al 3+ < Mg 2+ < Na + < F – < O 2– < N 3– Answer Back 7.5 Ionic Radii (SB p.208)


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