Question

Which set of compounds in the following pair of ionic compounds has a higher lattice energy:
I.$KCl$ or $MgO$
II.$LiF$ or $LiBr$
III.$Mg{F}_2$ or $NaCl$
A.$KCl$,$LiBr$,$Mg{F}_2$
B.$MgO$,$LiBr$,$Mg{F}_2$
C.$MgO$,$LiF$,$NaCl$
D.$MgO$,$LiF$,$Mg{F}_2$

Answer 1

Hint: To answer this question, you should recall the concept of Lattice energy. The lattice energy of any chemical entity is defined as the strength of the ionic bonds in an ionic compound. Smaller the size of the constituent ions and greater the charge, the stronger the force of attraction, the greater the lattice energy of the ionic solid.

Complete step by step answer:
Lattice energy depends on two factors:
Charge held by the Constituent Ions: Due to the presence of electrostatic forces between them, the individual ions in an ionic lattice are attracted to each other. The strength of the electrostatic force of attraction is directly proportional to the magnitude of the charge held by the constituent ions, i.e. the greater the charge, the stronger the force of attraction, the stronger the lattice.
Distance between the Ions: The electrostatic force and hence the lattice energy of an ionic compound is inversely proportional to the distance between the ions. The further the distance between the ions in a lattice, the weaker the electrostatic forces holding them together, the lower the lattice energy. Smaller atoms feature smaller interatomic distances in the ionic lattice and stronger binding forces. Therefore, the smaller the size of the constituent ions, the greater the lattice energy of the ionic solid.
Let us analyse the options systematically:
Among $$KCl$$ and $$MgO,K$$ has charge $$+1$$ and $$Mg$$ has charge $$+2.$$ Also $$,K+$$ have larger ionic radii than $$M{g}^{+2}$$. So, both the factors add up and the lattice energy of $$MgO$$ is higher than $$KCl.$$
$$LiF$$ and $$LiBr$$ have the same cation and same ionic charge. So lattice energy depends upon the size of the anion. $$F^{-}$$ is smaller than $$B{r}^{-}$$, therefore, lattice energy $$LiF$$ is higher than $$LiBr$$.
$$Mg{F}_2$$​ has a charge $$+2$$ while $$NaCl$$ has a charge $$+1$$ . Therefore, lattice energy $$Mg{F}_2$$ is higher than $$NaCl.$$

Hence, the correct option is option D.

Note:
Bond length in a molecule is defined as the distance between the centres of two covalently bonded atoms.
The length of the bond is determined by the number of bonded electrons. If bond order is high, there will be a stronger pull between two atoms and there will be shorter bond length. Therefore, bond length increases in the following order: triple bond > double bond > single bond.