The distance of balance between ions (Rb +) and (Cl-) in RbCl molecule is 0.279
ID: 3161310 • Letter: T
Question
The distance of balance between ions (Rb +) and (Cl-) in RbCl molecule is 0.279 nm. The ionization energy of rubidium is 4.18 eV and the affinity of chlorine is 3.62 eV.
A) Calculate the potential energy of attraction between ions in equilibrium, assuming that they behave as punctual charges. Determine the dissociation energy of RbCl, neglecting the energy of repulsion.
B) The experimental value of the dissociation energy is 4.37 eV. What is the energy of repulsion of the ions in the distance of equilibrium?
Explanation / Answer
A)
Potential energy = -k*q1*q2/r
= -9*10^9*(1.6*10^-19)^2/(0.279*10^-9)
= -8.26*10^-19 J
= -8.26*10^-19/(1.6*10^-19) eV
= -5.16 eV
B)
Energy of repulsion = 5.16 - 4.37 = 0.79 eV
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