(a) Based on the lattice energies of MgCl2 and SrCl2 given in Table 8.2, what is the range of values that you would expect for the lattice energy of CaCl2? (b) Using data from Appendix C, Figure 7.9, and Figure 7.11 and the value of the second ionization energy for Ca, 1145 kJ/mol, calculate the lattice energy of CaCl2.
Problem 30E(a) Based on the lattice energies of MgCl and SrCl2given in Tab2e 8.2, what is the range ofvalues that you would expect for the lattice energy of CaCl (b) Using d2a from AppendixC, Figure 7.9, and Figure 7.11 and the value of the second ionization energy for Ca, , calculate the lattice energy of CaCl . 2Solution 30EStep-1(a)Here we have to find the range of lattice energy of CaCl . 2In this question it has been given that, the lattice energy of MgCl and SrCl2 from the t2ble 8.2 isfound to be 2326 kJ/mol and 2127 kJ/mol.Mg, Ca and Sr all are present in a same group i.e group IIA. It is known that as we go down agroup, the atomic size increases. So the order of increasing atomic size of Mg, Ca and Sr areMg < Ca < Sr .Here all the atoms are having same charge and forming compounds with chlorine. As the size ofCa is lying in between Mg and Sr, the lattice energy of Ca must be expected to lie in between therange of Mg to Sr.Thus the range of lattice energy of CaCl is in be2een 2127 kJ/mol to 2326 kJ/mol.Step-2(b) Here we have to calculate the lattice energy of CaCl . 2Given:Sublimation energy of calcium = 178.20 kJ/mol1st ionization energy of calcium = 589.50 kJ/mol2nd ionization energy of Ca =1145 kJ/molBond dissociation of Cl = 22 kJ/mol -Electron affinity of Cl = -349 kJ/molEnthalpy of reaction = -326 kJ\nIn order to calculate the lattice energy 1st we have to write the chemical equation for thedissociation of CaCl . 2Over all reaction : Ca 2+ (g) + 2Cl (g) CaCl 2 Ca (s) Ca (g) 178.20 kJ/mol Ca (g) Ca (g)+ e 589.50 kJ/mol Ca (g) Ca (g) + e 1145 kJ/mol- Cl 2) 2Cl (g) 244 kJ/mol - 2Cl (g) 2Cl (g) 2(-349 kJ/mol) Ca (g) + 2Cl (g) CaCl (s) Lattice energy 2 Ca (s) + Cl (g) 2CaCl (s) -326 kJ2ol178.20 + 589.50 +1145 +244 + 2(-349) + Lattice energy = -326 kJ/mol 1458.7+ Lattice energy = -326 kJ/mmol Lattice energy = -1784.7 kJ/molThus the lattice energy of CaCl is found to be =21784.7 kJ/mol.