Answer: Option (c) is the correct answer.
Explanation:
According to the Born-Lande equation,
E = [tex]\frac{N_{A}Mz^{+}z^{-}e^{2}}{4 \pi \epsilin_{o} r}[/tex]
where, [tex]N_{A}[/tex] = Avogadro's constant
r = distance between the anion and cation
M = Medlung constant
[tex]z^{+}[/tex] = charge on the cation
[tex]z^{-}[/tex] = charge on the anion
E = lattice energy
According to this expression, lattice energy is inversely proportional to the distance between the cation and anion. And, when we move down a group then there occurs an increase in the atomic radii of the atoms.
As a result, there will occur a decrease in their lattice energy. Since, the anionic part is same in all the given species but the cationic part is different.
In [tex]MgBr_{2}[/tex] the size of magnesium cation is smaller than calcium, barium and strontium ion. Hence, there will be least distance between Mg and Br ions.
Thus, we can conclude that the magnitude of lattice energy will be the highest in [tex]MgBr_{2}[/tex].
