Ionic bonding

Chemistry B.Sc.-I Sem-Ist
Sant gadge Baba Amravati University, Amravati,
Late Ku. Durga K. Banmeru Science College, Lonar

1. B. Sc. I st year Sem-I st Chemistry Unit-1 : B) Ionic Bonding BY Dr. SURYAKANT B. BORUL M. Sc., M.Phil., Ph. D. Head & Assistant Professor, Department of Chemistry, Late Ku. Durga K. Banmeru Science College , Lonar. Dist . Buldana . 443302. LATE KU. DURGA K. BANMERU SCIENCE COLLEGE, LONAR, DIST. BULDANA. 443302 . (Affiliated to Sant Gadage Baba Amravati University Amravati; 2 (f) & 12 B; NAAC Accredited with ‘ C ’ grade)

2. Ionic Bonding- Defn - “ The chemical bond which is formed by transfer of one or more electrons from the valence shell of one atom to the valence shell of another atom is called as ionic bond.” Or “ The chemical bond which is formed due to electrostatics force of attraction in between opposite charge ions is called as ionic bond.” For example – Bond in between Na and Cl Na + Cl [Na + ] [ Cl - ] or Na + Cl - (2,8,1) (2,8,7) (2,8) (2,8,8)

3. Here, electrostatics force of attraction in between opposite charge ions Na + and Cl - in NaCl molecule.

4. The formation of an ionic bond is favoured when- 1) Metals has low ionization energy 2) Other elements has high electron affinity and 3) The resulting compound has lattice energy .

5. Types of Cations The formation of an ionic compound is due to atom attaining electronic configuration similar to that of inert gas element by an anion while the cation may achieve any one of the following configurations. 1. No valence electron (Ex.- H + ) 2. Ions with inert gas configuration– These ions have inert gas e. c. (ns 2 np 6 ) in their outermost shell. (Except foe n=1). Cations Inert Gas Li + He (1s 2 ) Na + , Mg 2+ , Al 3+ Ne (2s 2 2p 6 ) K + , Ca + , Sc 3+ , Ti 4+ Ar (3s 2 3p 6 ) Rb + , Sr 2+ , Y 3+ , Zr 4+ , Kr (4s 2 4p 6 ) Cs + , Ba 2+ , La 3+ , Ce 4+ Xe (5s 2 5p 6 )

6. 3. Ions with pseudo inert gas configuration (ns 2 np 6 nd 10 )– These ions have inert gas e.c . (ns 2 np 6 nd 10 ) in their outermost shell. Cations Electronic Configuration Ag + , Cd 2+ , In 3+ , Sn 4+ 4s 2 4p 6 4d 10 Au + , Hg 2+ ,Ti 3+ , Pb 4+ 5s 2 5p 6 5d 10 4. The inert s 2 pair configuration ( (n-1)s 2 p 6 d 10 ns 2 )– When the elements having valence shell configuration ns 2 np x (x=1,2,3) lose their p electrons only, cations with ns 2 configuration are formed.

7. Cations Electronic Configuration Ga + , Ge 2+ and As 3+ 4s 2 In + , Sn 2+ and Sb 3+ 5s 2 Ti + , Pb 2+ and Bi 3+ 6s 2 This is possible only when the energies of the ns & np electrons differ sufficiently so as to result in the stepwise ionization during the chemical bond formation. Therefore, only the post transition elements of group IIA, IV A and VA give such ions.

8. 5. The d and f ions:- The transition metal ions formed by the loss of the outer valence shell electrons without the ionization of the d electrons have the configuration of the outer shell as- ns 2 np 6 nd x (x= 1 to 9) and are classified as the d ions. Ex: Ti 2+ , V 2+ , Cr 2+ , Co 2+ etc. The ions are derived from the inner transition elements by the loss of the outer s and d electrons and have configuration. (n-1) s 2 ,(n-1)p 6 , (n-2)d 10 , (n-2)f 1-13 ex.- Lanthanide and actinide ions. 6. Ions with Irregular configurations:- These are certain ions that cannot be classified into any particular class Ex: Ga 4+

9. Energetics of Ionic Bond Formation Their are three types of energies are involved in the ionic bond formation. These are as follows Ionization energy Electron affinity or energy Lattice energy Ionization energy- “ The amount of energy required to remove the outer most electron from an isolated gaseous atom of an element in its ground state to form cation is called as ionization energy.”

10. M (g) M (g) + e - ; H = +I The energy required for this change is denoted by I. The energy is to be supplied in the process it is given a positive sign. The energy is measured in electron volts ( eV ) or Kcal / mole. The magnitude of ionization energy is a direct measure of ease of cation formation. If its value is low. Cation is readily formed. Alkali and alkaline earth metals have low values of ionization energy.

11. B) Electron affinity or energy- “ The amount of energy released when an electron is added to an isolated neutral gaseous atom in its ground state to produce an anion is called as electron affinity or energy.” X (g) + e - X - (g) ; H = -E It is denoted by E. It is the energy released, it is given a negative sign. The energy is measured in electron volts( eV ) or Kcal/ mole. Anion formation will be favoured if more energy is released in above process i.e. if electron affinity is high.

12. C) Lattice energy- It is related to the formation of an ionic solid from its ions. Lattice energy of an ionic crystal M + X - is defined in the following two ways- 1. The energy released when exact number of gaseous cations M + (g) and gaseous anions X - (g) come close together from infinity to from one mole of solid ionic crystal, M + X - (S) is called as lattice energy. M + (g) + X - (g) M + X - (S) + Energy released The energy required for removing ions of one mole of solid ionic crystal from their equilibrium positions in crystal to affinity is called as lattice energy. M + X - (S) + Energy supplied M + (g) + X - (g) It is denoted as U

13. Ionic Bonding- Defn - “ The chemical bond which is formed by transfer of one or more electrons from the valence shell of one atom to the valence shell of another atom is called as ionic bond.” Or “ The chemical bond which is formed due to electrostatics force of attraction in between opposite charge ions is called as ionic bond.” For example – Bond in between Na and Cl Na + Cl [Na + ] [ Cl - ] or Na + Cl - (2,8,1) (2,8,7) (2,8) (2,8,8)

14. C) Lattice energy- It is related to the formation of an ionic solid from its ions. Lattice energy of an ionic crystal M + X - is defined in the following two ways- 1. The energy released when exact number of gaseous cations M + (g) and gaseous anions X - (g) come close together from infinity to from one mole of solid ionic crystal, M + X - (S) is called as lattice energy. M + (g) + X - (g) M + X - (S) + Energy released The energy required for removing ions of one mole of solid ionic crystal from their equilibrium positions in crystal to affinity is called as lattice energy. M + X - (S) + Energy supplied M + (g) + X - (g) It is denoted as U

15. Factors favouring the Formation of Ionic Bond The formation of an ionic compound MX will be favoured if i ) The Ionization energy of element is low ii) Electron affinity or energy of X is high iii) Lattice energy of compound MX is high Calculation of lattice energy- Lattice energy may be calculated theoretically using Madelung constant or it may be determined experimentally using Born Haber cycle . Both the methods are discussed as-

16. Theoretical Calculation of Lattice energy using Madelung constant In 1918  Max Born  and  Alfred Landé  proposed that the lattice energy could be derived from the  electrostatic potential  of the ionic lattice and a repulsive potential energy term. Lattice energy can be theoretically calculated using the Born- Lande equation. Where e = Charge on electron (1.6022X 10 -19 C) Z + and Z - = Charge on cation and anion respectively N A = Avogadro number (6.023 x 10 23 ); n = Born Exponent r = Distance between nuclei of cation and anion in cm. U = Lattice energy of the ionic compound M = Madelung ( From name of Erwin Madelung , a German physicist)

17. Madelung constant- It is a correction factor which takes into account the electrostatic forces exerted by neighboring ion pair. It entirely depends upon the arrangement of positive and negative ions in crystal, i.e. upon the geometry of the ionic crystal. It does not depends upon the nature of ions present in the crystal. It can be calculated by summing the mutual potential energies of all the ions a lattice. Values of Modelung constants for some common crystal are as- Crystal type Modelung constant NaCl 1.747558 CaCl 1.762670 CaF 2 5.03878 TiO 2 4.816

18. Born exponent- It is a repulsion exponent which allows for repulsive forced between the electron clouds of oppositely charged ions. It can be evaluated from the results of experimental measurements of the compressibility of the crystal. It is found that for all the crystals n lies in the neighbouring of 9. It depends upon configuration of ion. For the different configuration the values are as – He= 5, Ne=7, Ar =9, Kr=10, Xe =12 For a crystal having two ions of different electronic configuration average of the values given above is used. For ex.-in case of NaCl , Na + ion has configuration of neon (n=7) and Cl - has configuration of argon (n=9). Thus evaluation of lattice energy of NaCl the value of n to be used 8.

19. Experimental Determination of Lattice Energy using Born Haber cycle The lattice energy of an ionic solid like NaCl may be determined by using Born-Haber Cycle. It is a thermo-chemical cycle and was devised by Born and Haber in 1919. The cycle first relates the lattice energy of crystalline solid (unknown quantity) to other known thermo-chemical quantities. Then the use of Hess’s law to evaluate the unknown quantity.

20. Lattice energy of sodium chloride may be determined by using Born-Haber cycle as follows Sodium chloride may be considered to be formed solid sodium metal and gaseous chlorine by two different methods described below: Method 1: It is the direct combination of solid sodium and gaseous chlorine to give solid sodium chloride. The process may be represented by following equation. Na (s) + 1/2 Cl2 (g) → Na Cl (s) ;  H, = -414.2 kJ/mol

21. This equation tells us that when one mole of solid sodium combines with half mole of gaseous chlorine molecules, one mole of crystalline sodium chloride is formed. During this process 414.2 kJ mol of energy is also evolved. This energy is called heat of formation of sodium chloride and is represented by the symbol  H. Method 2 : It involves five different steps described below Step 1 : Sublimation of Sodium: In this process 1 mole of solid sodium Na (s) changes to gaseous sodium Na(s). The energy required for this process is S Na (Heat of sublimation of sodium) Its value is experimentally found out to be 108.7 kJ/mol. Na (s) → Na (g) ; S Na = 108.7 kJ/mol

22. Step 2: Dissociation of chlorine: In this process half mole of chlorine is dissociated into 1mole of chlorine atoms. The energy required for this process is 1/2 DCl 2 (where DCl 2 , is the heat of dissociation of one mole of chlorine). Experimental value of 1/2 DCl 2 is 112.95 kJ/mol Cl 2(g) Cl (g) ; 1/2 DCl 2 = 112.95 kJ/mol Step 3: Formation of sodium ions: 1 mole of gaseous sodium atoms are converted to sodium ions by removal of an electron from each of them. Energy required for this process is I Na .(Ionization energy of sodium) Its experimental value is 489.5 kJ/mol. Na (g) Na + (g) + e- I Na = 489.5 kJ/ mol

23. Step 4: Formation of chloride ions: One mole of chlorine atoms (formed in step 2) take up electrons given by sodium and are converted to negatively charged chloride ions. The process is accompanied by release of energy. By definition the energy released in this process is electron affinity of chlorine ( E Cl ). Its experimental value is -351.4 kJ/mol Cl (g) + e-→ Cl - (g) ; E Cl2 = -351 kJ/ mol Step 5: Formation of ionic crystal Na + Cl - (s) : Gaseous sodium and chloride ions formed instep (3) and (4) above combine to give solid sodium chloride crystal Na + Cl - (s) . Energy is related in this process also and by definition of lattice energy of NaCl .

24. It is represented by U NaCl . Its value is to be determine fro other values. Na + (g) + Cl - (g) → Na + Cl - ; U NaCl = ? According to Hess’s law the energy change in method (1) must be equal to total of energy changes of all steps in method (2) i.e.  H f = S Na + ½ D Cl 2 + I Na + E Cl2 + U NaCl Putting the actual values we gets. -414.2 = +108.7 +1/2*225.9 + 489.5 - 351.4 + U NaCl Therefore U NaCl = -414.2 -108.7 – 112.95- 489.5 + 351.4 U NaCl = -773.95KJ/Mol

25. Representation of Born –Haber Cycle for the formation of NaCl ionic crystal

26. Experimental Determination of Lattice Energy using Born Haber cycle The lattice energy of an ionic solid like NaCl may be determined by using Born-Haber Cycle. It is a thermo-chemical cycle and was devised by Born and Haber in 1919. The cycle first relates the lattice energy of crystalline solid (unknown quantity) to other known thermo-chemical quantities. Then the use of Hess’s law to evaluate the unknown quantity.

27. Lattice energy of sodium chloride may be determined by using Born-Haber cycle as follows Sodium chloride may be considered to be formed solid sodium metal and gaseous chlorine by two different methods described below: Method 1: It is the direct combination of solid sodium and gaseous chlorine to give solid sodium chloride. The process may be represented by following equation. Na (s) + 1/2 Cl2 (g) → Na Cl (s) ;  H, = -414.2 kJ/mol

28. This equation tells us that when one mole of solid sodium combines with half mole of gaseous chlorine molecules, one mole of crystalline sodium chloride is formed. During this process 414.2 kJ mol of energy is also evolved. This energy is called heat of formation of sodium chloride and is represented by the symbol  H. Method 2 : It involves five different steps described below Step 1 : Sublimation of Sodium: In this process 1 mole of solid sodium Na (s) changes to gaseous sodium Na(s). The energy required for this process is S Na (Heat of sublimation of sodium) Its value is experimentally found out to be 108.7 kJ/mol. Na (s) → Na (g) ; S Na = 108.7 kJ/mol

29. Step 2: Dissociation of chlorine: In this process half mole of chlorine is dissociated into 1mole of chlorine atoms. The energy required for this process is 1/2 DCl 2 (where DCl 2 , is the heat of dissociation of one mole of chlorine). Experimental value of 1/2 DCl 2 is 112.95 kJ/mol Cl 2(g) Cl (g) ; 1/2 DCl 2 = 112.95 kJ/mol Step 3: Formation of sodium ions: 1 mole of gaseous sodium atoms are converted to sodium ions by removal of an electron from each of them. Energy required for this process is I Na .(Ionization energy of sodium) Its experimental value is 489.5 kJ/mol. Na (g) Na + (g) + e- I Na = 489.5 kJ/ mol

30. Step 4: Formation of chloride ions: One mole of chlorine atoms (formed in step 2) take up electrons given by sodium and are converted to negatively charged chloride ions. The process is accompanied by release of energy. By definition the energy released in this process is electron affinity of chlorine ( E Cl ). Its experimental value is -351.4 kJ/mol Cl (g) + e-→ Cl - (g) ; E Cl2 = -351 kJ/ mol Step 5: Formation of ionic crystal Na + Cl - (s) : Gaseous sodium and chloride ions formed instep (3) and (4) above combine to give solid sodium chloride crystal Na + Cl - (s) . Energy is related in this process also and by definition of lattice energy of NaCl .

31. It is represented by U NaCl . Its value is to be determine fro other values. Na + (g) + Cl - (g) → Na + Cl - ; U NaCl = ? According to Hess’s law the energy change in method (1) must be equal to total of energy changes of all steps in method (2) i.e.  H f = S Na + ½ D Cl 2 + I Na + E Cl2 + U NaCl Putting the actual values we gets. -414.2 = +108.7 +1/2*225.9 + 489.5 - 351.4 + U NaCl Therefore U NaCl = -414.2 -108.7 – 112.95- 489.5 + 351.4 U NaCl = -773.95KJ/Mol

32. Representation of Born –Haber Cycle for the formation of NaCl ionic crystal

33. Problem based on Lattice Energy To calculate the lattice energy of NaCl crystal the data is- Sublimation energy of Na ( S Na )=108.710 Kj / Mol Dissociation energy for Cl 2 ( D Cl2 )=225.9 Kj / Mol Ionization energy for Na (g) (E Cl2 ) =489.5 Kj / Mol Electron affinity for Cl (g) ( I Na ) =-351.4 Kj / Mol Heat of formation of NaCl (  H f ) = -414.2 Kj / Mol Ans - We know that,  H f = S Na + ½D Cl2 + I Na + E Cl2 + U NaCl Na + + Cl - → Na + Cl - ; U NaCl = ? U NaCl = -414.2 -108.7 – 112.95- 489.5 + 351.4 U NaCl = -773.95KJ/Mol

34. 2) To calculate the heat of reaction of KF from its elements from the following data by use of Born Haber cycle. Sublimation energy of K ( S K )=87.8 Kj / Mol Dissociation energy for F 2 ( D F2 )=158.9 Kj / Mol Ionization energy for K (g) ( I K )= 414.2 Kj / Mol Electron affinity for F (g) ( E F2 )= -334.7 Kj / Mol Lattice energy for KF( U KF )= -807.5 Kj / Mol Heat of formation of KF (  H f ) = ? Ans - We know that,  H f = S K + ½D F2 + I K + E F2 + U KF K (S) + ½ F 2(g) → KF (S) ;  H f =?  H f = 87.8 + ½(158.9) + 414.2 + (-334.7) + (-807.5)  H f = 87.8 + 79.45 + 414.2 + (-1142.2)= 581.45-1142.2=-560.75 Heat of formation of KF (  H f ) = -560.75KJ/Mol

35. 3) To calculate the heat of reaction of MgF 2 from its elements by using of Born Haber cycle. Thermochemical data as- Sublimation energy of Mg ( S Mg )=146.4 Kj / Mol Dissociation energy for F 2 ( D F2 )=158.9 Kj / Mol Ionization energy for Mg (g) ( I Mg )= 2184.0 Kj / Mol Electron affinity for F (g) ( E F2 )= -334.7 Kj / Mol Lattice energy for MgF 2 ( U )= -2922.5 Kj / Mol Heat of formation of MgF 2 (  H f ) = ? Ans - We know that,  H f = S Mg + D F2 + I Mg + E F2 + U MgF2 Mg (S) + F 2(g) → MgF 2(S) ;  H f =?  H f = 146.4 + 158.9 + 2184 + 2(-334.7) +(-2922.5)  H f = 146.4 + 158.9 + 2184 -669.4 -2922.5=2489.3-3591.9 Heat of formation of MgF2 (  H f ) = -1102.6KJ/Mol

36. Solvation of Ions and Solvation Energy Q.-Explain the term Solvation energy. Ans - “ The interaction that takes place when a substance is introduced in a solvent is called as solvation and the energy associated with this is called as solvation energy . ” Water is called a polar solvent because in its molecule the oxygen atom is partly negatively charged and each hydrogen atom is partly positively changed as -

37. When sodium chloride is introduced in such a solvent, the negative end of water molecule attract the positive ions, and the positive end attract the negative ions of the crystal. These attraction forces exerted by the water molecules weaken the attractions existing among the ions in the crystal. Hence some of the ions in the crystal are pulled away from their positions in crystal lattice as -

38. Once the Na + and Cl - ions are broken away from the ionic lattice, following two processes occur same time. 1. Each sodium ion is surrounded by a definite but unknown number of water molecules, say ‘x’, with their negative ends (oxygen ends) pointing towards it, as shown in figure. Na + + xH 2 O = [Na(H 2 O) x ] + This process is called solvation of sodium ion and the energy change associated with it is called as solvation energy of sodium ion, (  H s ) Na+. The chemical species [Na(H 2 O) x ] + is called solvated or acquated sodium ion and may also be represented as [Na ( aq ) ] + .

39. 2) Each chloride ion is surrounded by definite but unknown number of water molecules, say ‘y’ with their positive ends (hydrogen ends) pointing towards it, as shown Cl - + yH 2 O = [ Cl (H 2 O) y ] - The process is called solvation of chloride ion and the energy change associated with it is called solvation energy of chloride ion, (  H s ) Cl - . The chemical species [ Cl (H 2 O) y ] - is called solvated or aquated chloride ion and may also be represented as [ Cl ( aq ) ] -

40. The total process may be written as: NaCl (s) + ( x+y )H 2 O [Na (H 2 O) x ] + + [ Cl (H 2 O)y] - or NaCl (s) + aq Na + ( aq ) + Cl - aq ) fig- Born-Haber cycle for determination of salvation energy.

41. Calculation of Solvation Energy The energy changes during solvation of sodium and chloride ions may be calculated using a Born-Haber type cycle as given in figure 1.6 Here L is the heat of solution of NaCl at infinite dilution (i.e. the total amount of heat evolved or absorbed when one mole of sodium chloride dissolved in such a large excess of water, that further addition of water does not produce any heat change). U NaCl is the lattice energy of NaCl . ( HS) Na+ and (  H s ) Cl - are the solvation energies of sodium and chloride ions.

42. Since heat of solution of NaCl at infinite dilution (L) and lattice energy of NaCl ( U NaCl ) are experimentally known, the solvation energies Na + and Cl - ions can be calculated from following relation. L = U NaCl + (  H S ) Na+ + (  H S ) cı - It gives us the sum of solvation energies of sodium and chloride ions. (There is no purely thermochemical way to separate this sum into two parts corresponding to sodium and chloride ions.

43. Factors affecting solvation and solvation energy Solvation energy and lattice energy: The dissolution of an ionic compound in polar solvent is favoured if the attraction between solvent molecules and ions, exceeds the attraction among the ions in a crystal lattice or in other words if the energy of solvation of ions exceeds the lattice energy of the crystal. Dielectric constant and solvation energy: For a given ion and the solvent the dielectric constant and the solvent energy are related by following equation, called Born equation.

44. Here, H = Solvation energy of gaseous ion, r = ionic radius and or C = charge on the ion, D = dielectric constant of the solvent. From this equation it is evident that increase in the magnitude of dielectric constant increases the solvation energy. iii) Ionic size: Both solvation energy and lattice energy are increased by decreases in cation and anion size. It is therefore difficult to relate solubility to size of ion.

45. iv) However the two opposite charges are not of the same magnitude and in general other factors being equal solubility increases with increase in cation or anion size. v) Ionic charge: With increasing cation or anion charge, the lattice energy increases much more rapidly than the solvation energy. This results in decrease of solubility. vi) Electronic configuration of cations and their polarising effect: If the anion is more readily polarized by the cation, than is the solvent, the lattice energy will increase more than solvation energy and the solubility will decrease. If the solvent is more readily polarized by the cation, the solubility will increase.

46. The ions having pseudo inert gas configuration Ag + , Pb ++ , Hg ++ , etc. have high anion polarizing effect, hence their salts ( AgCl , PbCl , HgCl ) have lower solubility in water, As compared to these, the alkaline earth cations (Ca + , Ba ++ etc.) having inert gas type configuration, have low anion polarizing effect, hence their halides CaCl 2 , BaCl 2 , are readily soluble in water.

47. Thank you