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1 7.1Formation of Ionic Bonds: Donating and Accepting Electrons 7.2 Energetics of Formation of Ionic Compounds 7.3 Stoichiometry of Ionic Compounds 7.4Ionic.

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Presentation on theme: "1 7.1Formation of Ionic Bonds: Donating and Accepting Electrons 7.2 Energetics of Formation of Ionic Compounds 7.3 Stoichiometry of Ionic Compounds 7.4Ionic."— Presentation transcript:

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2 1 7.1Formation 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 2 Bonding & Structure

4 3 Lewis Model  G.N. Lewis in 1916  Only the outermost (valence) electrons are involved significantly in bond formation  Successful in solving chemical problems

5 4  Why are some elements so reactive (e.g.Na) and others inert (e.g. noble gases)?  Why are there compounds with chemical formulae H 2 O and NaCl, but not H 3 O and NaCl 2 ?  Why are helium and the other noble gases monatomic, when molecules of hydrogen and chlorine are diatomic?

6 5  Chemical bonds are strong electrostatic forces holding atoms or ions together, which are formed by the rearrangement (transfer or sharing) of outermost electrons  Atoms tend to form chemical bonds in such a way as to achieve the electronic configurations of the nearest noble gases (The Octet Rule )

7 6 Q.1 Write the s,p,d,f notation for the ions in the table below

8 7

9 8 Three types of chemical bonds 1.Ionic bond (electrovalent bond) Formed by transfer of electrons Introduction (SB p.186)

10 9 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 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187) Three types of chemical bonds 1.Ionic bond (electrovalent bond)

11 10 Three types of chemical bonds 1.Ionic bond Introduction (SB p.186) Electrostatic attraction between positively charged particles and negatively charged particles

12 11 Three types of chemical bonds 2.Covalent bond Formed by sharing of electrons Introduction (SB p.186)

13 12 Three types of chemical bonds 2.Covalent bond Introduction (SB p.186) Electrostatic attraction between nuclei and shared electrons

14 13 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)

15 14 Three types of chemical bonds 3. Metallic bond Introduction (SB p.186) Formed by sharing of a large number of delocalized electrons

16 15 Ionic bonds and Covalent bonds are only extreme cases of a continuum. In real situation, most chemical bonds are intermediate between ionic and covalent

17 16 Ionic Bonds with incomplete transfer of electrons have covalent character. Covalent Bonds with unequal sharing of electrons have ionic character.

18 17 Electronegativity and Types of Chemical Bonds Ionic or covalent depends on the electron- attracting ability of bonding atoms in a chemical bond. Ionic bonds are formed between atoms with great difference in their electron-attracting abilities Covalent bonds are formed between atoms with small or no difference in their electron- attracting abilities.

19 18 Ways to compare the electron-attracting ability of atoms 1.Ionization Enthalpy 2.Electron affinity 3.Electronegativity

20 19 Electronegativity and Types of Chemical Bonds 1.Ionization enthalpy The enthalpy change when one mole of electrons are removed from one mole of atoms or positive ions in gaseous state. X(g)  X + (g) + e -  H 1 st I.E. X + (g)  X 2+ (g) + e -  H 2 nd I.E. Ionization enthalpies are always positive.

21 20 X(g) + e -  X - (g)  H 1 st E.A. X  (g) + e -  X 2  (g)  H 2 nd E.A. Electronegativity and Types of Chemical Bonds 2.Electron affinity The enthalpy change when one mole of electrons are added to one mole of atoms or negative ions in gaseous state. Electron affinities can be positive or negative.

22 21 I.E. and E.A. only show e - releasing/attracting power of free, isolated atoms However, whether a bond is ionic or covalent depends on the ability of atoms to attract electrons in a chemical bond.

23 22 Electronegativity and Types of Chemical Bonds 3.Electronegativity The ability of an atom to attract electrons in a chemical bond.

24 23 Mulliken’s scale of electronegativity Electronegativity(Mulliken)  Nobel Laureate in Chemistry, 1966

25 24 Pauling’s scale of electronegativity Nobel Laureate in Chemistry, 1954 Nobel Laureate in Peace, 1962

26 25 Pauling’s scale of electronegativity  calculated from bond enthalpies  cannot be measured directly  having no unit  always non-zero  the most electronegative element, F, is arbitrarily assigned a value of 4.00

27 26 Pauling’s scale of electronegativity

28 27 What trends do you notice about the EN values in the Periodic Table? Explain.

29 28 The EN values increase from left to right across a Period.

30 29 What trends do you notice about the EN values in the Periodic Table? Explain. The EN values increase from left to right across a Period. The atomic radius decreases from left to right across a Period.Thus,the nuclear attraction experienced by the bonding electrons increases accordingly.

31 30 The EN values decrease down a Group.

32 31 The EN values decrease down a Group. The atomic radius increases down a Group, thus weakening the forces of attraction between the nucleus and the bonding electrons. What trends do you notice about the EN values in the Periodic Table? Explain.

33 32 Why are the E.A. of noble gases not shown ?

34 33 EN is a measure of the ability of an atom to attract electrons in a chemical bond. However, noble gases are so inert that they rarely form chemical bonds with other atoms. Why are the E.A of noble gases not shown ? XeF 2, XeF 4 and XeF 6 are present

35 34 Formation of Ionic Bond Atoms of Group IA and IIA elements ‘tend’ to achieve the noble gas structures in the previous Period by losing outermost electron(s). In fact, formations of positive ions from metals are endothermic and not spontaneous. I.E. values are always positive

36 35 Formation of Ionic Bond Atoms of Group VIA and VIIA elements tend to achieve the noble gas structures in the same Periods by gaining electron(s) First E.A. of Group VIA and VIIA elements are always negative.  Spontaneous and exothermic processes

37 36 The oppositely charged ions are stabilized by coming close to each other to form the ionic bond. Ionic bond is the result of electrostatic interaction between oppositely charged ions. Interaction = attraction + repulsion

38 37 Dots and Crosses Diagram

39 38 Notes on Dots & Crosses representation Electrons in different atoms are indistinguishable. The dots and crosses do not indicate the exact positions of electrons. Not all stable ions have the noble gas structures(Refer to pp.4-5).

40 39 Tendency for the Formation of Ions An ion will be formed easily if 1. The electronic structure of the ion is stable; 2. The charge on the ion is small; 3. The size of parent atom from which the ion is formed is small for an anion, or large for a cation.

41 40 For cation, Larger size of parent atom  less positive I.E. or less +ve sum of successive I.E.s,  easier formation of cation,  atoms of Group IA & Group IIA elements form cations easily.

42 41 Li + Be 2+ Na + Mg 2+ Al 3+ K + Ca 2+ Sc 3+ Rb + Sr 2+ Y 3+ Zr 4+ Cs + Ba 2+ La 3+ Ce 4+ Ease of formation of cation 

43 42 For anion, Smaller size of parent atom  more negative E.A. or more -ve sum of successive E.A.s,  easier formation of anion,  atoms of Group VIA & Group VIIA elements form anions easily.

44 43 N 3  O 2  F  P 3  S 2  Cl  Br  I  Ease of formation of anion 

45 44 Stable Ionic Structures Not all stable ions have the noble gas electronic structures. Q.2Write the s,p,d,f notation for each of the following cations. Use your answers to identify three types of commonly occurring arrangement of outershell electrons of cations other than the stable octet structure.

46 45 Q.2 Write the s,p,d,f notation for the following cations

47 46 Q.2

48 electrons group e.g. Zn 2+ Fully-filled s, p & d-subshells

49 48 Fully-filled s, p & d-subshells 2. ( ) - electron group e.g. Pb 2+

50 49 3. Variable arrangements for ions of transition metals. ns 2,np 6, nd 1 to ns 2,np 6, nd 9 Ti 3+ [Ne] 3s 2 3p 6 3d 1 Mn 3+ [Ne] 3s 2 3p 6 3d 4

51 50 Electronic arrangements of stable cations Ionization enthalpy determines the ease of formation of cations. Less positive I.E. or sum of I.E.s  easier formation of cations

52 51 Electronic arrangements of stable cations Octet structure : - cations of Group IA, IIA and IIIB

53 52 (a)18 - electrons group cations of Groups IB, IIB, IIIA and IVA

54 53 Less stable than ions with a noble gas structure.

55 54 Cu and Au form other ions because the nuclear charges are not high enough to hold the 18-electron group firmly.

56 55 Cu 2+ 2, 8, 17Au 3+ 2, 8, 18, 32, 16 The d-electrons are more diffused and thus less attracted by the less positive nuclei.

57 56 (b)(18+2) electrons Cations of heavier group members due to presence of 4f electrons. Tl[2,8,18,32], 5s 2, 5p 6, 5d 10, 6s 2, 6p 1 Tl 3+ [2,8,18,32], 5s 2, 5p 6, 5d 10 (18) Tl + [2,8,18,32], 5s 2, 5p 6, 5d 10, 6s 2 (18 + 2) Stability : Tl + > Tl 3+ due to extra stability of 6s electrons (inert pair effect)

58 57 (b)(18+2) electrons 6s electrons are poorly shielded by the inner 4f electrons. 6s electrons experience stronger nuclear attraction. Tl[2,8,18,32], 5s 2, 5p 6, 5d 10, 6s 2, 6p 1 Tl 3+ [2,8,18,32], 5s 2, 5p 6, 5d 10 (18) Tl + [2,8,18,32], 5s 2, 5p 6, 5d 10, 6s 2 (18 + 2)

59 58 (b)(18+2) electrons Inert pair effect increases down a Group. Stability :Sn 4+ (18) > Sn 2+ (18+2) Pb 2+ (18+2) > Pb 4+ (18) moving down Group IV Tl[2,8,18,32], 5s 2, 5p 6, 5d 10, 6s 2, 6p 1 Tl 3+ [2,8,18,32], 5s 2, 5p 6, 5d 10 (18) Tl + [2,8,18,32], 5s 2, 5p 6, 5d 10, 6s 2 (18 + 2)

60 59 (b)(18+2) electrons Inert pair effect increases down a Group. Stability :Sb 5+ (18) > Sb 3+ (18+2) Bi 3+ (18+2) > Bi 5+ (18) moving down Group V Tl[2,8,18,32], 5s 2, 5p 6, 5d 10, 6s 2, 6p 1 Tl 3+ [2,8,18,32], 5s 2, 5p 6, 5d 10 (18) Tl + [2,8,18,32], 5s 2, 5p 6, 5d 10, 6s 2 (18 + 2)

61 60 (c) Cations of transition elements -variable oxidation numbers - Electronic configurations from ns 2, np 6, nd 1 to ns 2, np 6, nd 9 Fe[Ne] 3s 2, 3p 6, 3d 6, 4s 2 Fe 2+ [Ne] 3s 2, 3p 6, 3d 6 Fe 3+ [Ne] 3s 2, 3p 6, 3d 5

62 61 (c) Cations of transition elements Which one is more stable, Fe 2+ (g) or Fe 3+ (g) ? Fe 2+ (g) is more stable than Fe 3+ (g) Energy is needed to remove electrons from Fe 2+ (g) to give Fe 3+ (g) Fe[Ne] 3s 2, 3p 6, 3d 6, 4s 2 Fe 2+ [Ne] 3s 2, 3p 6, 3d 6 Fe 3+ [Ne] 3s 2, 3p 6, 3d 5

63 62 B. Anions - with noble gas structures Electron affinity determines the ease of formation of anions. More -ve E.A. or sum of E.A.s  more stable anion Group VIA and Group VIIA elements form anions easily.

64 63 H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -322 Ne +29 Na -53 Mg +21 Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -325 Kr +39 First Electron Affinity (kJ mol  1 ) X(g) + e   X  (g)

65 64 H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg +21 Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -324 Kr +39 E.A. becomes more –ve from Gp 1 to Gp 7 across a period

66 65 H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg +21 Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -324 Kr +39 The electrons added experience stronger nuclear attraction when the atoms are getting smaller across the period.

67 66 H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg +21 Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -324 Kr +39 +ve E.A. for Gp 2 and Gp 0 because the electron added occupies a new shell / subshell

68 67 H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg +21 Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -324 Kr +39 Goes to a new subshell Goes to a new shell Be(2s 2 )  Be  (2s 2 2p 1 ) Ne(2p 6 )  Ne  (2p 6 3s 1 )

69 68 H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg +21 Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -324 Kr +39 More –ve E.A. for Gp 1 because the ions formed have full-filled s-subshells. E.g. Li(1s 2 2s 1 )  Li  (1s 2 2s 2 )

70 69 H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg +21 Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -324 Kr +39 Less –ve E.A. for Gp 5 because the stable half- filled p-subshell no longer exists. E.g. N(2s 2 2p 3 )  N  (2s 2 2p 4 )

71 70 H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg +21 Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -324 Kr +39 Q.3 Why are the first E.A.s of halogen atoms more negative than those of the O atom or the S atom ?

72 71 H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg +21 Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -324 Kr +39 It is because the halide ions formed have full- filled s/p subshells (octet structure).

73 72 Q.4Why are the second E.A.s of O  (+844 kJmol  1 ) and S  (+532 kJmol  1 ) positive ? The electrons added are repelled strongly by the negative ions. O  (g) + e   O 2  (g) S  (g) + e   S 2  (g)

74 73 Why is the E.A. of F less negative than that of Cl ? The size of Fluorine atom is so small that the addition of an extra electron results in great repulsion among the electrons. HalogenFClBrI E.A. kJ mol  1  328  349  325  295 The 2 nd electron shell is much smaller than the 3 rd electron shell.

75 74 Energetics of Formation of Ionic Compounds A. Formation of an ion pair in gaseous state Consider the formation of a KF(g) ion pair from K(g) & F(g) H1H1 K(g) + F(g) KF(g) IE 1 (K) K + (g) H2H2 By Hess’s law,  H 1 = IE 1 (K) + EA 1 (F) +  H 2 EA 1 (F) + F  (g)

76 75  H 1 = IE 1 (K) + EA 1 (F) +  H 2 Q.6 (+)(  )

77 76  H 1 = IE 1 (K) + EA 1 (F) +  H 2 =  10  19 J   10  18 J =   10  19 J IE 1 (K) + EA 1 (F) =  10  19 J  H 2 =   10  18 J

78 77 When R   Stability : K + (g) + F - (g) < K(g) + F(g) by x J IE 1 (K) + EA 1 (F) =  10  19 J  H 2 =   10  18 J

79 78 IE 1 (K) + EA 1 (F) =  10  19 J  H 2 =   10  18 J I.E. and E.A. are NOT the driving forces for the formation of ionic bond.

80 79 IE 1 (K) + EA 1 (F) =  10  19 J  H 2 =   10  18 J K + (g) & F  (g) tend to come close together in order to become stable.

81 80 IE 1 (K) + EA 1 (F) =  10  19 J  H 2 =   10  18 J Coulomb stabilization is the driving force for the formation of ionic bond.

82 81 IE 1 (K) + EA 1 (F) =  10  19 J  H 2 =   10  18 J When R =  10  10 m The most energetically stable state is reached.

83 82 IE 1 (K) + EA 1 (F) =  10  19 J  H 2 =   10  18 J The ions cannot come any closer than R e as it will results in less stable states

84 83 IE 1 (K) + EA 1 (F) =  10  19 J  H 2 =   10  18 J Repulsions between electron clouds and between nuclei > attraction between ions

85 84 R Potential energy (V) Minimum V when R =  10  10 m Repulsion : between opposite charges Attraction : between opposite charges

86 85 IE 1 (K) + EA 1 (F) =  10  19 J  H 2 =   10  18 J What is the significance of the lowest energy state of the neutral state of K + F ?

87 86  KF  Electrostatic attraction between positive nuclei and bond electron pair stabilizes the system K F A covalent bond is formed

88 87 A. Formation of Ionic Crystals Consider the formation of NaCl(s) from Na(s) & Cl 2 (g) H2H2 By Hess’s law,  H f =  H 1 +  H 2 Na + (g) + Cl  (g) H1H1 HfHf Na(s) + Cl 2 (g) NaCl(s)

89 88  H 1 is the sum of four terms of enthalpy changes 1.Standard enthalpy change of atomization of Na(s) Na(s)  Na(g)  H = kJ mol -1 2.First ionization enthalpy of Na(g) Na(g)  Na + (g) + e -  H= +500 kJ mol -1 3.Standard enthalpy change of atomization of Cl 2 (g) 1/2Cl 2 (g)  Cl(g)  H = kJ mol First electron affinity of Cl(g) Cl(g) + e -  Cl - (g)  H = -349 kJ mol -1

90 89  H 2 is the lattice enthalpy of NaCl. It is the enthalpy change for the formation of 1 mole of NaCl(s) from its constituent ions in the gaseous state. Na + (g) + Cl - (g)  NaCl(s)  H L o

91 90 1.theoretical calculation using an ionic model, Or 2.experimental results indirectly with the use of a Born-Haber cycle. Direct determination of lattice enthalpy by experiment is very difficult, but it can be obtained from

92 91 Q.7 Calculate the lattice enthalpy of NaCl  H f =  H 1 +  H 2 Lattice enthalpy =  H 2 =  H f -  H 1 = [(  411)  (  349)] kJ mol  1 =  kJ mol  1

93

94 93 Lattice enthalpy is the dominant enthalpy term responsible for the -ve  H f of an ionic compound. More  ve  H L o  More stable ionic compound Lattice enthalpy is a measure of the strength of ionic bond.

95 94 Determination of Lattice Enthalpy From Born-Haber cycle (experimental method, refers to Q.7) From theoretical calculation based on the ionic model

96 95 Assumptions made in the calculation Ions are spherical and have no distortion of the charge cloud, I.e. 100% ionic. The crystal has certain assumed lattice structure. Repulsive forces between oppositely charged ions at short distances are ignored.

97 96 M is the Madelung constants that depends on the crystal structure

98 97 L is the Avogadro’s constant

99 98 Q + and Q - are the charges on the cation and the anion respectively

100 99  0 is the permittivity of vacuum

101 100 (r + + r - ) is the nearest distance between the nuclei of cation and anion r + is the ionic radius of the cation r - is the ionic radius of the anion

102 101 Stoichiometry of Ionic Compounds Stoichiometry of an ionic compound is the simplest whole number ratio of cations and anions involved in the formation of the compound. There are two ways to predict stoichiometry.

103 Considerations in terms of electronic configurations Atoms tend to attain noble gas electronic structures by losing or gaining electron(s). The ionic compound is electrically neutral. total charges on cation = total charges on anions

104 Considerations in terms of electronic configurations Na + F  [Na] + [F] - 2,8,1 2,7 2,8 2,8 Mg + 2Cl  [Cl] - [Mg] 2+ [Cl] - 2,8,2 2,8,7 2,8,8 2,8 2,8,8  NaF  MgCl 2

105 Considerations in terms of enthalpy changes of formation theoretically calculated lattice enthalpy Based on the Born-Haber cycle & the theoretically calculated lattice enthalpy, the values of  H f o of hypothetical compounds (e.g. MgCl, MgCl 2, MgCl 3 ) can be calculated. The stoichiometry with the most negative  H f o value is the most stable one The stoichiometry with the most negative  H f o value is the most stable one.

106 Considerations in terms of enthalpy changes of formation Or, we can determine the true stoichiometry by comparing the calculated  H f o values of hypothetical compounds with the experimentally determined  H f o value. The stoichiometry with  H f o value closest to the experimentally determined  H f o value is the answer.

107 106 Q.8 Three hypothetical formulae of magnesium chloride are proposed and their estimated lattice enthalpies are shown in the table below.

108 107  H f [MgCl(s)] =  H at [Mg(s)] + 1 st IE of Mg +  H at [1/2Cl 2 (g)] + 1 st EA of Cl +  H L [MgCl(s)] = ( ) kJ mol -1 = -128 kJ mol -1

109 108  H at [1/2Cl 2 ] Mg + (g)+Cl(g) Mg + (g)+Cl  (g) 1st EA[Cl] MgCl(s)  H L [MgCl]  H f [MgCl]<0 1st IE[Mg] Mg + (g)+ Cl 2 (g) Enthalpy (kJ mol  ) Mg(s)+ Cl 2 (g)  H at [Mg] Mg(g)+ Cl 2 (g)

110 109  H f [MgCl 2 (s)] =  H at [Mg(s)] + 1 st IE of Mg + 2 nd IE of Mg + 2  H at [1/2Cl 2 (g)] + 2  1 st EA of Cl +  H L [MgCl 2 (s)] = [   (-364)+(-2602)] kJ mol -1 = -752 kJ mol -1

111 110 Enthalpy (kJ mol  ) Mg(s)+Cl 2 (g) Mg(g)+Cl 2 (g)  H at [Mg] Mg + (g)+Cl 2 (g) 1st IE[Mg] Mg 2+ (g)+Cl 2 (g) 2nd IE[Mg] Mg 2+ (g)+2Cl (g) 2  H at [1/2Cl 2 ] Mg 2+ (g)+2Cl  (g) 2  1st EA[Cl] MgCl 2 (s)  H L [MgCl 2 ]  H f [MgCl 2 ] <0

112 111  H f [MgCl 3 (s)] =  H at [Mg(s)] + 1 st IE of Mg + 2 nd IE of Mg + 3 rd IE of Mg + 3  H at [1/2Cl 2 (g)] + 3  1 st EA of Cl +  H L [MgCl 3 (s)] =   (-364)+(-5440) = kJ mol -1

113 112 Enthalpy ( kJ mol  ) Mg(s)+3/2Cl 2 (g) Mg 3+ (g)+3Cl  (g) 3  1st EA[Cl] MgCl 3 (s)  H L [MgCl 3 ]  H f [MgCl 3 ] >0 Mg 3+ (g)+3Cl (g) 3  H at [1/2Cl 2 ] 3rd IE[Mg] Mg 3+ (g)+ Cl 2 (g)Mg 2+ (g)+ Cl 2 (g) 2nd IE[Mg] Mg + (g)+ Cl 2 (g) 1st IE[Mg]  H at [Mg] Mg(g)+ Cl 2 (g)

114 113 Since the hypothetical compound MgCl 2 has the most negative  H f value, and this value is closest to the experimentally determined one, the most probable formula of magnesium chloride is MgCl 2

115 114 Reasons for the discrepancy between calculated and experimental results of  H f  The ionic bond of MgCl 2 is not 100% pure. I,e. the ions are not perfectly spherical.  MgCl 2 has a different crystal structure from CaF 2  Short range repulsive forces between oppositely charged ions have not been considered. Check Point 7-2 Check Point 7-2

116 Energetics of Formation of Ionic Compounds (SB p.196) Factors affecting lattice enthalpy Effect of ionic size:  The greater the ionic size  The lower (or less negative) is the lattice enthalpy

117 Energetics of Formation of Ionic Compounds (SB p.196) Factors affect lattice enthalpy Effect of ionic charge:  The greater the ionic charge  The higher (or more negative) is the lattice enthalpy Check Point 7-3 Check Point 7-3

118 Ionic Crystals

119 Ionic Crystals (SB p.202) Crystal Lattice( 晶體格子 )

120 Ionic Crystals (SB p.202) In solid state, ions of ionic compounds are regularly packed to form 3-dimensional structures called crystal lattices( 晶體格子 )

121 Ionic Crystals (SB p.202) The coordination number (C.N.) of a given particle in a crystal lattice is the number of nearest neighbours of the particle. C.N. of Na + = 6

122 Ionic Crystals (SB p.202) The coordination number (C.N.) of a given particle in a crystal lattice is the number of nearest neighbours of the particle. C.N. of Cl  = 6

123 122 Identify the unit cell of NaCl crystal lattice Not a unit cell

124 123 Identify the unit cell of NaCl crystal lattice Not a unit cell

125 124 The unit cell of a crystal lattice is the simplest 3-D arrangement of particles which, when repeated 3-dimensionally in space, will generate the whole crystal lattice.

126 125 The unit cell of a crystal lattice is the simplest 3-D arrangement of particles which, when repeated 3-dimensionally in space, will generate the whole crystal lattice.

127 126 The unit cell of a crystal lattice is the simplest 3-D arrangement of particles which, when repeated 3-dimensionally in space, will generate the whole crystal lattice.

128 127 Unit cell

129 128 A crystal lattice with a cubic unit cell is known as a cubic structure.

130 129 Three Kinds of Cubic Structure Simple cubic (primitive) structure Body-centred cubic (b.c.c.) structure Face-centred cubic (f.c.c.) structure

131 130 Simple cubic structure View this Chemscape 3D structure Unit cell  Space filling Space lattice

132 131 Body-centred cubic (b.c.c.) structure View this Chemscape 3D structure Unit cell  Space filling Space lattice

133 132 Face-centred cubic (f.c.c.) structure View this Chemscape 3D structure Unit cell

134 133 The f.c.c. structure can be obtained by stacking the layers of particles in the pattern abcabc...

135 134 Interstitial sites - The empty spaces in a crystal lattice Two types of interstitial sites in f.c.c. structure Octahedral site Tetrahedral site

136 135 Octahedral site : - It is the space between the 3 spheres in one layer and 3 other spheres in the adjacent layer.

137 136 When the octahedron is rotated by 45 o, the octahedral site can also be viewed as the space confined by 4 spheres in one layer and 1 other sphere each in the upper and lower layers respectively. 45 o before behind

138 137 Tetrahedral site : - It is the space between the 3 spheres of one layer and a fourth sphere on the upper layer.

139 138 In an f.c.c. unit cell, the tetrahedral site is the space bounded by a corner atom and the three face-centred atoms nearest to it.

140 139 c b a a

141 140 Q.9 Label a, b, and c as Oh sites or Td sites a and b are Td sites c is Oh site Top layer

142 141 Q.10 Identify the Oh sites and Td sites in the f.c.c. unit cell shown below. 13 Oh sites Th sites in the front Th sites at the back View Octahedral sites and tetrahedral sites

143 142 Since ionic crystal lattice is made of two kinds of particles, cations and anions, the crystal structure of an ionic compound can be considered as two lattices of cations and anions interpenetrating with each other.

144 143 The crystal structures of sodium chloride, caesium chloride and calcium fluoride are discussed.

145 144 Sodium Chloride (The Rock Salt Structure) f.c.c. unit cell of larger Cl - ions, with all Oh sites occupied by the smaller Na + ions Oh sites of f.c.c. lattice of Cl - ions

146 145 A more open f.c.c. unit cell of smaller Na + ions with all Oh sites occupied by the larger Cl - ions. Oh sites of f.c.c. lattice of Na + ions

147 146 F.C.C. Structure of Sodium Chloride View this Chemscape 3D structure

148 147 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)

149 148 Only the particles in the centre (or body) of the unit cell belong to the unit cell entirely. Particles locating on the faces, along the edges or at the corner of a unit cell are shared with the neighboring unit cells of the crystal lattice.

150 149 Fraction of particles occupied by a unit cell 1/21 CornerEdgeFaceBody 1/2

151 150 Fraction of particles occupied by a unit cell 1/41/21 CornerEdgeFaceBody 1/4

152 151 Fraction of particles occupied by a unit cell 1/81/41/21 CornerEdgeFaceBody 1/8

153 152 Q.11 Calculate the net nos. of Na + and Cl - ions in a unit cell of NaCl. No. of Cl - ions =No. of Na + ions = 8 corners 6 faces 12 edges 1 body Example 7-4 Example 7-4

154 153 Structure of Caesium Chloride  Link Structure of Caesium Chloride  Link 7.4 Ionic Crystals (SB p.202) Simple cubic lattice Cs + ions are too large to fit in the octahedral sites. Thus Cl  ions adopt the more open simple cubic structure with the cubical sites occupied by Cs + ions.

155 154 A cubical site is the space confined by 4 spheres in one layer and 4 other spheres in the adjacent layer.

156 155 Size of interstitial sites : - Cubical > octahedral > tetrahedral In f.c.c. structure In simple cubic structure

157 156 Co-ordination number of Cs + = 8 Co-ordination number of Cl - = 8 8 : 8 co-ordination 7.4 Ionic Crystals (SB p.202) Simple cubic lattice Structure of Caesium Chloride  Link Structure of Caesium Chloride  Link

158 157 Q.12 Number of Cs + = 1 Number of Cl  =

159 158 Structure of Calcium Fluoride  Link Structure of Calcium Fluoride  Link 7.4 Ionic Crystals (SB p.203) It can be viewed as a simple cubic structure of larger fluoride ions with alternate cubical sites occupied by smaller calcium ions. Fluorite structure

160 Ionic Crystals (SB p.203) Alternately, it can be viewed as an expanded f.c.c. structure of smaller calcium ions with all tetrahedral sites occupied by larger fluoride ions. Structure of Calcium Fluoride  Link Structure of Calcium Fluoride  Link

161 160 Co-ordination number of Ca 2+ = 8 Co-ordination number of F - = 4 8 : 4 co-ordination 7.4 Ionic Crystals (SB p.203) Face-centred cubic lattice Structure of Calcium Fluoride  Link Structure of Calcium Fluoride  Link

162 Ionic Crystals (SB p.203) Q.13 (a) Number of F  = 8 Number of Ca 2+ = CaF 2

163 Ionic Crystals (SB p.203) Q.13 (b) CaF 2 Only alternate cubical sites are occupied by Ca 2+ in order to maintain electroneutrality.

164 163 Structure of Sodium oxide  Link Structure of Sodium oxide  Link 7.4 Ionic Crystals (SB p.203) Na 2 O vs CaF 2 Antifluorite structure fluorite structure

165 164 Q.14(a) Zinc blende structure - Link

166 165 Q.14(a) Number of Zn 2+ = 4 Number of S 2  = ZnS

167 166  tetrahedral site S 2  ions adopt f.c.c. structure Alternate Td sites are occupied by Zn 2+ ions to ensure electroneutrality. Zinc sulphide, ZnS Q.14(b)

168 167 Q.15 The unit cell is not a cube Rutile structure - Link

169 168 Q.15(a) Number of O 2  = TiO 2 Ti 4+ O2O2 Number of Ti 4+ =

170 169 Q.15(a) C.N. of Ti 4+ = 6 C.N. of O 2  = 3  6 : 3 coordination

171 170  octahedral site O 2  ions adopt distorted h.c.p.(not f.c.c.) structure with alternate Oh sites occupied by Ti 4+ ions to ensure electroneutrality. Titanium(IV) oxide, TiO 2 Q.15(b)

172 171 a b a hexagonal close-packed

173 172 Factors governing the structures of ionic crystals 1.Close Packing Considerations Ions in ionic crystals tend to pack as closely as possible. (Why ?) To strengthen the ionic bonds formed To maximize the no. of ionic bonds formed.

174 173 Factors governing the structures of ionic crystals 1.Close Packing Considerations The larger anions tend to adopt face-centred cubic structure (cubic closest packed, c.c.p.) The smaller cations tend to fill the interstitial sites as efficiently as possible

175 174 Factors governing the structures of ionic crystals Q.16Is there no limit for the C.N. ? Explain your answer. The anions tend to repel one another when they approach a given cation. The balance between attractive forces and repulsive forces among ions limits the C.N. of the system.

176 175 Factors governing the structures of ionic crystals Q.16Is there no limit for the C.N. ? Explain your answer. If the cations are small, less anions can approach them without significant repulsions. Or, if the cations are small, they choose to fit in the smaller tetrahedral sites with smaller C.N. to strengthen the ionic bonds formed.

177 176 Factors governing the structures of ionic crystals Q.16Is there no limit for the C.N. ? Explain your answer. If the cations are large, they choose to fit in the larger octahedral sites or even cubical sites with greater C.N. to maximize the no. of ionic bonds formed.

178 177 Factors governing the structures of ionic crystals 2. The relative size of cation and anion, In general, r  > r +,

179 178 Factors governing the structures of ionic crystals If the cations are large, more anions can approach the cations without significant repulsions. Or, the large cations can fit in the larger interstitial sites with greater C.N. to maximize the no. of bonds formed.

180 179 Interstitial site TetrahedralOctahedralCubical Coordination4 : 46 : 68 : – – – Examples ZnS, most copper(I) halides Alkali metal halides except CsCl CsCl, CsBr CsI, NH 4 Cl 7.4 Ionic Crystals (SB p.203) Summary : - NH 4 + is large

181 Ionic Crystals (SB p.203) Summary : - *CaF 2, BaF 2, BaCl 2, SrF 2 Examples – : 4Coordination Cubical Interstitial site

182 181 Q.17 AB C rr r+r+ rr 45 °

183 182 the range of ratio that favours octahedral arrangement.

184 183 the optimal ratio for cations and anions to be in direct contact with each other in the cubical sites. Q.18

185 184  octahedral site Cl  ions adopt f.c.c. structure All Oh sites are occupied by Ag + ions to ensure electroneutrality. Silver chloride, AgCl Q.18(a)

186 185  octahedral site Ag + ions adopt an open f.c.c. structure All Oh sites are occupied by Cl  ions to ensure electroneutrality. Silver chloride, AgCl Q.18(a)

187 186 Copper(I) bromide, CuBr  tetrahedral site Br  ions adopt f.c.c. structure Only alternate Td sites are occupied by Cu + ions to ensure electroneutrality. Q.18(b)

188 187 Copper(II) bromide, CuBr 2  tetrahedral site Br  ions adopt f.c.c. structure Only 1/4 Td sites are occupied by Cu + ions to ensure electroneutrality. Some Br  may form less bonds than others  Not favourable. Q.18(c)

189 188 Copper(II) bromide, CuBr 2 All the Td sites are occupied by Br  ions to ensure electroneutrality. Q.18(c) Or, Cu 2+ ions adopt a very open f.c.c. structure However, this structure is too open to form strong ionic bonds.  non-cubic structure is preferred.

190 189 Q.18(c) 4 : 2 coordination Cu 2+ Br 

191 190 Copper(II) chloride, CuCl 2 Layer structure 0.295nm nm

192 191 Copper(II) chloride, CuCl 2 6 : 3 coordination

193 Ionic Radii

194 193 Ionic Radii Ionic radius is the approximate radius of the spherical space occupied by the electron cloud of an ion in all directions in the ionic crystal.

195 194 Q.19 Why is the electron cloud of an ion always spherical in shape ? Stable ions always have fully-filled electron shells or subshells. The symmetrical distribution of electrons accounts for the spherical shapes of ions.

196 195 Q.19 Li + 1s 2 Na + [He] 2s 2, 2p x 2, 2p y 2, 2p z 2 Cl  [Ne] 3s 2, 3p x 2, 3p y 2, 3p z 2 Zn 2+ [Ar] 3d xy 2, 3d xz 2, 3d yz 2,, spherical Symmetrical distribution along x, y and z axes   almost spherical Symmetrical distribution along xy, xz, yz planes and along x, y and z axes  almost spherical

197 196 Determination of Ionic Radii Pauling Scale Interionic distance (r + + r  ) can be determined by X-ray diffraction crystallography

198 Ionic Radii (SB p.205) Electron density plot for sodium chloride crystal Cl  Contour having same electron density Na + r + + r  Electron density map for NaCl r + + r 

199 198 By additivity rule, Interionic distance = r + + r  For K + Cl , r + + r  = nm (determined by X-ray) Since K + (2,8,8) and Cl  (2,8,8) are isoelectronic, their ionic radii can be calculated. r + (K + ) = nm,r  (Cl  ) = nm

200 199 For Na + Cl , r + + r  = nm (determined by X-ray) Since r  (Cl  ) = nm(calculated) r + (Na + ) = ( ) nm = nm

201 200 Limitation of Additivity rule In vacuum, the size of a single ion has no limit according to quantum mechanics. The size of the ion is restricted by other ions in the crystal lattice. Interionic distance < r + + r 

202 201 Evidence : Electron distribution is not perfectly spherical at the boundary due to repulsion between electron clouds of neighbouring ions.

203 202 Ionic radius depends on the bonding environment. For example, the ionic radius of Cl  ions in NaCl (6 : 6 coordination) is different from that in CsCl (8 : 8 coordination).

204 203 Ionic radius vs atomic radius Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206) Radii of cations < Radii of corresponding parent atoms cations have one less electron shell than the parent atoms

205 204 Ionic radius vs atomic radius Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206) p/e of cation > p/e of parent atom  Less shielding effect  stronger nuclear attraction for outermost electrons  smaller size

206 205 Ionic radius vs atomic radius Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206) Radii of anions > Radii of corresponding parent atoms

207 206 Ionic radius vs atomic radius Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206) p/e of anion < p/e of parent atom  more shielding effect  weaker nuclear attraction for outermost electrons  larger size

208 207 Periodic trends of ionic radius 1.Ionic radius increases down a Group Ionic radius depends on the size of the outermost electron cloud. On moving down a Group, the size of the outermost electron cloud increases as the number of occupied electron shells increases.

209 208 Periodic trends of ionic radius 2.Ionic radius decreases along a series of isoelectronic ions of increasing nuclear charge The total shielding effects of isoelectronic ions are approximately the same.  Z eff  nuclear charge (Z)  Ionic radius decreases as nuclear charge increases.

210 Ionic Radii (SB p.206) Isoelectronic to He(2) Isoelectronic to Ne(2,8) Isoelectronic to Ar(2,8,8)

211 Ionic Radii (SB p.206) H  is larger than most ions, why ?

212 211 Q.20H  > N 3  The nuclear charge (+1) of H  is too small to hold the two electrons which repel each other strongly within the small 1s orbital. Or, p/e of H  (1/2) < p/e of N 3  (7/10)

213 212 The END Check Point 7-5 Check Point 7-5 Example 7-5 Example 7-5

214 213 Why do two atoms bond together? The two atoms tend to achieve an octet configuration which brings stability. Answer Introduction (SB p.186)

215 214 How does covalent bond strength compare with ionic bond strength? Back They are similar in strength. Both are electrostatic attractions between charged particles. Answer Introduction (SB p.186)

216 215 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)

217 216 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

218 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.

219 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

220 219 (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)

221 220 (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

222 221 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

223 222 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)

224 223 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)

225 224 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)

226 225 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

227 226 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

228 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.

229 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

230 229 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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