The equilibrium separation of the K + and F ions in KF is about 0.217 nm.

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The equilibrium separation of the K+ and F ions in KF is about 0.217 nm.

(a) Calculate the potential energy of attraction of the ions, assuming them to be point charges at this separation.

(b) The ionization energy of potassium is 4.34 eV and the electron affinity of F is 3.40 eV. Find the dissociation energy neglecting any energy of repulsion.

(c) The measured dissociation energy is 5.07 eV. Calculate the energy due to repulsion of the ions at the equilibrium separation.

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