The lattice energy of $NaCl,{\text{ NaF, KCl}}$ and $RbCl$follow the order:
A. $KCl < RbCl < NaCl < NaF$
B. $NaF < NaCl < KCl < RbCl$
C. $RbCl < KCl < NaCl < NaF$
D. $NaCl < RbCl < NaF < KCl$

Hint: Lattice energy: The amount of energy released when $1$mole of an ionic solid is formed from its gaseous ions.

Complete step by step answer:
Lattice energy depends upon charge and size of cation.
Lattice energy is directly proportional to charge on ion i.e. larger the magnitude of charge on ions, greater will be the attractive forces, higher is the value of lattice energy and inversely proportional to size of an ion i.e. smaller the size of ions, lesser is the internuclear distance and greater will be the interionic attraction. Hence, more will be the value of lattice energy. The given ionic solids are $NaCl,{\text{ NaF, KCl,}}$and$RbCl$.
In case of $NaF$and $NaCl$as ${F^ - }$is small in size than $C{l^ - }$then $NaF$has greater lattice energy than $NaCl$but in case of $KCl,{\text{ RbCl}}$as $N{a^ + }$is smaller than that of ${K^ + }$ion so $NaCl$has greater lattice energy than $KCl$ and $RbCl$.
$\therefore $The order of lattice energy is $NaF > NaCl > KCl > RbCl$.

Hence, the correct option is (C).

Note:
The important consequence of the lattice energies are:
(i) The greater the lattice enthalpy, the more the stability of ionic compounds.
(ii) The lattice energy is greater for small, highly charged ions.
(iii) The lattice energy affects the solubility of the ionic compounds.