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

5 4 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 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187) Three types of chemical bonds 1.Ionic bond (electrovalent bond)

6 5 Three types of chemical bonds 1.Ionic bond Introduction (SB p.186) Electrostatic attraction between positively charged particles and negatively charged particles

7 6 Three types of chemical bonds 2.Covalent bond Formed by sharing of electrons Introduction (SB p.186)

8 7 Three types of chemical bonds 2.Covalent bond Introduction (SB p.186) Electrostatic attraction between nuclei and shared electrons

9 8 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 9 The metallic bond strength increases with: 1. decreasing size of the metal atom (i.e. the metallic radius); 2. increasing number of valence electrons of the metal atom. 10.3 Factors affecting the strength of metallic bond (SB p.262)

11 10 Effect of number of valence electrons on metallic bond strength MetalNumber of valence electrons(s) Melting point ( o C) Sodium Magnesium Aluminium 123123 98 650 660 10.3 Factors affecting the strength of metallic bond (SB p.263)

12 11 Three types of chemical bonds 3. Metallic bond Introduction (SB p.186) Formed by sharing of a large number of delocalized electrons

13 12 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.

14 13 Ways to compare the electron-attracting ability of atoms 1.Ionization Enthalpy 2.Electron affinity 3.Electronegativity

15 14 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.

16 15 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.

17 16 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

18 17 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

19 18 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

20 19 Dots and Crosses Diagram

21 20 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.

22 21 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.

23 22 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.

24 23 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 

25 24 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.

26 25 N 3  O 2  F  P 3  S 2  Cl  Br  I  Ease of formation of anion 

27 26 (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

28 27 (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

29 28 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.

30 29 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)

31 30 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)

32 31 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.

33 32 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)

34 33 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)

35 34  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 = +108.3 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 = +121.1 kJ mol -1 4. First electron affinity of Cl(g) Cl(g) + e -  Cl - (g)  H = -349 kJ mol -1

36 35  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

37 36 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 The enthalpy of atomization of an element is the enthalpy change when 1mol of gaseous atoms is made from its elements in its standard state.

38 37 Q.7 Calculate the lattice enthalpy of NaCl  H f =  H 1 +  H 2 Lattice enthalpy =  H 2 =  H f -  H 1 = [(  411)  (+108.3 + 500 + 121.1  349)] kJ mol  1 =  791.4 kJ mol  1

39 38 -349 -791.4

40 39  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)] = (+150 + 736 + 121 - 364- 771) kJ mol -1 = -128 kJ mol -1

41 40  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)

42 41  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)] = [150+736+1450+2  121+2  (-364)+(-2602)] kJ mol -1 = -752 kJ mol -1

43 42 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

44 43  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)] = 150+736+1450+7740+3  121+3  (-364)+(-5440) = +3907 kJ mol -1

45 44 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)

46 45 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

47 46 Determination of Ionic Radii Pauling Scale Interionic distance (r + + r  ) can be determined by X-ray diffraction crystallography

48 47 7.5 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 

49 48 By additivity rule, Interionic distance = r + + r  For K + Cl , r + + r  = 0.314 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 + ) = 0.133 nm,r  (Cl  ) = 0.181 nm

50 49 For Na + Cl , r + + r  = 0.275 nm (determined by X-ray) Since r  (Cl  ) = 0.181 nm(calculated) r + (Na + ) = (0.275 - 0.181) nm = 0.094 nm

51 50 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

52 51 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

53 52 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

54 53 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

55 54 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.

56 55 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.  Ionic radius decreases as nuclear charge increases.

57 56 7.5 Ionic Radii (SB p.206) Isoelectronic to He(2) Isoelectronic to Ne(2,8) Isoelectronic to Ar(2,8,8)

58 57 7.5 Ionic Radii (SB p.206) H  is larger than most ions, why ?

59 58 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)


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