Question

Which set of compounds in the following pair of ionic compounds has the 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

The correct option is D $MgO,LiF,$ $Mg{F}_2$

Lattice energy depends upon two factors:

• charge on ion (greater the charge higher is the lattice energy)
• size of ion (smaller the size of ion greater is the lattice energy)

1. Among $KCl$ and $MgO$, $K$ has charge $+1$ and $Mg$ has charge $+2$. Also, $K^{+}$ has larger ionic radii than $M{g}^{+2}$. So, both the factors add up and the lattice energy of $MgO$ is higher than $KCl$.
   
2. $LiF$ and $LiBr$ have the same cation and same ionic charge. So lattice energy depends upon the size of anion. $F^{-}$ is smaller than $B{r}^{-}$, therefore, lattice energy of $LiF$ is higher than $LiBr$.
3. $Mg{F}_2$ has a charge $+2$ while $NaCl$ has a charge $+1$. Therefore, lattice energy of $Mg{F}_2$ is higher than $NaCl$.