Question

Why is the lattice energy of $MgO$ greater than $Mg{F}_2$ even though the size of oxygen is greater than fluorine?

Answer 1

Hint: Energy required to dissociate one mole of ionic solid to gaseous ionic constituents is the lattice energy. Lattice energy is directly proportional to the charge on the ion. Constituent ions with smaller size will have the greater lattice energy as smaller atoms have strong binding force and small interatomic distance in the ionic lattice.

Complete answer:
Lattice energy depends upon two factors i.e. charge on the ion and the size of the ion. The lattice energy is directly proportional to the charge on the ion which means that the greater the charge on an ion higher is the lattice energy while lattice energy is inversely proportional to the size of the ions which means that the smaller the size of the ion greater is the lattice energy.
But the lattice energy of $MgO$ and $Mg{F}_2$ can be calculated using the Kapustinskii equation which is as follows $E_L={\displaystyle \frac{121400n{z}_{+}{z}_{-}}{r_{+}+r_{-}}}(1-{\displaystyle \frac{34.5}{r_{+}+r_{-}}})$
Where,
 $n$ Is the total number of ions present in the empirical formula of the compound.
 $z_{+}$ is the total number of charge on the cation
 $z_{-}$ is the total number of charge on the anion
 $r_{+}$ is the radius of the cation
 $r_{-}$ Is the radius of the anion.
For $MgO$ the lattice energy can be calculated as follows:
 $E_L={\displaystyle \frac{121400\times 2\times 2\times 2}{72+140}}(1-{\displaystyle \frac{34.5}{72+140}})$
 $E_L=3840KJmo{l}^{-1}$
For $Mg{F}_2$ the lattice energy can be calculated as follows:
 $E_L={\displaystyle \frac{121400\times 3\times 2\times 1}{72+133}}(1-{\displaystyle \frac{34.5}{72+133}})$
 $E_L=3170KJmo{l}^{-1}$
Therefore, the lattice energy of $MgO$ is greater than that of $Mg{F}_2$ although the size of oxygen is greater than that of fluorine.

Note:
 Kapustinskii equation is used to calculate the lattice energy of ionic crystals. But the values calculated by Kapustinskii equation are five percent more than the real values in most of the cases. Kapustinskii equation also helps in the determination of the ionic radius if the lattice energy is known.