(a) The ionization energy of potassium (K) is 4.34 eV and the electron affinity of chlorine (Cl) is -3.61 eV. The Madelung constant for the KCl structure is 1.748 and the distance between ions of opposite sign is 0.314 nm. On the basis of these data only, compute the cohesive energy of KCl per ion pair.(b) The observed cohesive energy of KCl is 6.42 eV per ion pair. On theassumption that the difference between this figure and that obtained in (a) is due to the exclusion principle repulsion, find the exponent n in the formula Br^-n for the potential energy arising from this source.

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Asked Mar 20, 2020
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(a) The ionization energy of potassium (K) is 4.34 eV and the electron affinity of chlorine (Cl) is -3.61 eV. The Madelung constant for the KCl structure is 1.748 and the distance between ions of opposite sign is 0.314 nm. On the basis of these data only, compute the cohesive energy of KCl per ion pair.


(b) The observed cohesive energy of KCl is 6.42 eV per ion pair. On the
assumption that the difference between this figure and that obtained in (a) is due to the exclusion principle repulsion, find the exponent n in the formula Br^-n for the potential energy arising from this source.

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Expert Answer

Step 1

a. Cohesive energy will be the negative of the Coulombic energy as given in Numerical, minus the difference between ionization energy of potassium and the electron affinity of chlorine:

Chemistry homework question answer, step 1, image 1

Hence Cohesive energy for KCL per ion =7.29eV.

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