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Hint: In order to determine whether a compound will be soluble in a solvent, always look into its dissolution enthalpy which is the sum of its lattice enthalpy and the solvation enthalpy(hydration enthalpy).
Complete step by step solution:
Lattice enthalpy or more specifically lattice dissociation enthalpy refers to the energy change when 1 mole of a solid ionic compound breaks into its constituent gaseous ions such that they are far apart from each other. It is always positive. It depends upon two factors namely (a) the charges on the constituent ions and (b) the ionic radii of the constituent ions. Greater the charge on the constituent ions, greater the lattice enthalpy and lesser the ionic radii of the constituent ions, more will be the lattice enthalpy.
Solvation enthalpy is the enthalpy change that occurs when 1 mole of gaseous ions are dissolved in a solvent to give an infinitely dilute solution. When the solvent is water, it is called hydration enthalpy and it is always negative. It depends upon the charge density of the ions. Higher the charge density, higher will be the solvation enthalpy.
BeO has a very high lattice enthalpy due to the high charge and small ionic radii of the ${ Be }^{ 2+ }$and ${ O }^{ 2- }$ ions. Its lattice enthalpy is even higher than its hydration enthalpy. Also due to the small size of ${ Be }^{ 2+ }$ ion it has high polarising power, so BeO has significant covalent nature (Fajan’s rules) and therefore it is insoluble in water. Whereas in BaO, the hydration enthalpy is able to overcome the lattice enthalpy and hence it dissolves in water.
Therefore the correct answer is (a) Lattice energy of BeO is higher than BaO due to the small size of ${ Be }^{ 2+ }$ ion and its covalent nature.
Note: Many ionic compounds have significant covalent nature especially compounds of ${ Li }^{ + }$, ${ Be }^{ 2+ }$and compounds with transition metal cations. In order to determine the covalency in an ionic compound qualitatively, we can use Fajan’s rules.
Complete step by step solution:
Lattice enthalpy or more specifically lattice dissociation enthalpy refers to the energy change when 1 mole of a solid ionic compound breaks into its constituent gaseous ions such that they are far apart from each other. It is always positive. It depends upon two factors namely (a) the charges on the constituent ions and (b) the ionic radii of the constituent ions. Greater the charge on the constituent ions, greater the lattice enthalpy and lesser the ionic radii of the constituent ions, more will be the lattice enthalpy.
Solvation enthalpy is the enthalpy change that occurs when 1 mole of gaseous ions are dissolved in a solvent to give an infinitely dilute solution. When the solvent is water, it is called hydration enthalpy and it is always negative. It depends upon the charge density of the ions. Higher the charge density, higher will be the solvation enthalpy.
BeO has a very high lattice enthalpy due to the high charge and small ionic radii of the ${ Be }^{ 2+ }$and ${ O }^{ 2- }$ ions. Its lattice enthalpy is even higher than its hydration enthalpy. Also due to the small size of ${ Be }^{ 2+ }$ ion it has high polarising power, so BeO has significant covalent nature (Fajan’s rules) and therefore it is insoluble in water. Whereas in BaO, the hydration enthalpy is able to overcome the lattice enthalpy and hence it dissolves in water.
Therefore the correct answer is (a) Lattice energy of BeO is higher than BaO due to the small size of ${ Be }^{ 2+ }$ ion and its covalent nature.
Note: Many ionic compounds have significant covalent nature especially compounds of ${ Li }^{ + }$, ${ Be }^{ 2+ }$and compounds with transition metal cations. In order to determine the covalency in an ionic compound qualitatively, we can use Fajan’s rules.
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