Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
As we move down in sulphates of group 2 their solubility decreases because there is a drastic decrease in hydration enthalpy as compared to decrease in lattice enthalpy.
Does the decrease in lattice enthalpy mean decrease in ionic character? But according to Fajans' rules the ionic character should increase or covalent character must decrease down the group.
Kindly explain this.
Lattice enthalpy is given by the Born-Lande equation: $${\displaystyle E=-{\frac {N_{A}Mz^{+}z^{-}e^{2}}{4\pi \epsilon _{0}r_{0}}}\left(1-{\frac {1}{n}}\right)}$$ This seems very complicated, but removing some constants, we can reduce it to say: $$E \propto \frac {z^+z^-}{r_0}$$ here, $z^+$ and $z^-$ are charge on cation and anion and $r_0$ is distance to closest ion.
Coming to your question, In the sulphate series of Alkaline earth metals, the charge remains the same for all members: what changes is the distance to closest ion, which is smaller for small ions such as $\ce{Be^2+}$, thereby causing their lattice energies to be high.
The decrease in lattice enthalpy may not neccessarily mean a decrease in ionic character. If two lattices have similar values of $r_0$, then a higher lattice enthalpy for one of them may imply larger ionic character due to large charges on either of the atoms.
Lattice energy is not at all related to ionic character. Ionic character is given by Fajans’ rule, whereas lattice enthalpy depends upon a whole set of reactions involved in the Born–Haber cycle. For instance:
