Question

Use the ionization energies and electron affinities listed below to find the critical radius Rc inside 3. which the transfer of an electron is energetically favorable, for each of the ionic molecules KI, NaBr, LiF. Compare to the observed molecular separation, noting the dissociation energies, B IE (eV) B (eV) EA (eV) Ro (nm) K: 4.34 I: 3.06 KI: 0.305 3.31 Na: 5.14Br: 3.36NaBr: 0.250 3.74 5.91 F: 3.40 Li: 5.39 LiF: 0.156

Answer 1

All the given compounds KI NaBr and LiF are ionic compounds and in ionic compounds we have electrostatic force of attraction between anion and cation and the electrostatic force of attraction depends on the distance between two ions. Electrostatic force of attraction is directly proportional to the charge present on ions and its inversely proportional on the distance between the ions . The closer anion and cation higher will the attraction between these opposite charge species . As all the ionic compounds given have same charge of +1 on cation and -1 on anions the only parameter to differentiate their dissociation energy will the distance between ions which is critical radius as the distance decreases the force attraction as well as dissociation energy increase. LiF has the minimum separation between cation and anion so its dissociation energy is maximum.