Question
The ionic substances $\mathrm{KF}, \mathrm{CaO}$, and $\mathrm{Sc} \mathrm{N}$ are isoelectronic (they have the same number of electrons). Examine the lattice energies for these substances in Table $8.2$, and account for the trends you observe.
Step 1
First, let's recall the definition of lattice energy: Lattice energy is the energy required to separate one mole of an ionic solid into its gaseous ions. It is a measure of the strength of the electrostatic forces between the ions in the solid. Show more…
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The substances $\mathrm{NaF}$ and $\mathrm{CaO}$ are isoelectronic (have the same number of valence electrons). (a) What are the charges on each of the cations in each compound? (b) What are the charges of each of the anions in each compound? (c) Without looking up lattice energies, which compound is predicted to have the larger lattice energy? (d) Using the lattice energies in Table $8.1,$ predict the lattice energy of ScN.
NaCl and KF have the same crystal structure. The only difference between the two is the distance that separates cations and anions. Based on the lattice energies, would you expect the Na-Cl or the K-F distance to be longer? Explain your choice. NaCl = 788 kJ/mol KF = 808 kJ/mol
Consider the isoelectronic ions: $K^{+}, S^{2}, C 1$ and $C a^{2 t}$. The radii of these ionic species follow the order (a) $\mathrm{Ca}^{2 *}>\mathrm{K}+>\mathrm{Cl}>\mathrm{S}^{2-}$ (b) $\mathrm{Cl}>\mathrm{S}^{2-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}$ (c) $\mathrm{S}^{2-}>\mathrm{Cl}>\mathrm{K}^{*}>\mathrm{Ca}^{2+}$ (d) $\mathrm{K}^{+}>\mathrm{Ca}^{2+}>\mathrm{S}^{2-}>\mathrm{Cl}^{-}$
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