A) \[~\text{LiF }>\text{ NaF }>\text{ KF }>\text{ RbF }>\text{ CsF}\]
B) \[\text{CsF }>\text{ RbF }>\text{ KF }>\text{ NaF }>\text{ LiF}\]
C) \[\text{LiF }>\text{ KF }>\text{ NaF }>\text{ CsF }>\text{ RbF}\]
D) \[\text{CsF }>\text{ KF }>\text{ NaF }>\text{ RbF }>\text{ LiF}\]
Correct Answer: B
Solution :
\[\text{LiI NaI KI RbI CsI (solubility)}\] \[\text{Lattice energy}\frac{\text{1}}{{{\text{r}}_{\text{+}}}\text{+ }{{\text{r}}_{\text{-}}}}\] Since, \[{{\text{r}}_{-}}\text{ }{{\text{r}}_{+}}\], the sum will not change too much as \[{{\text{r}}_{+}}\] increases. As a result the lattice energy will not change significantly as \[{{\text{r}}_{+}}\] increases. Thus decreases in lattice energy is not as fast as decreases in hydration energy. Thus, more decrease in hydration energy results in decrease in solubility. Thus, solubility decreases as \[\text{LiI NaI KI RbI CsI}\] With a small anion as \[({{F}^{-}})\], the lattice energy decreases more rapidly along a series of salts with increasing cation size and thus solubility increases. \[~~\text{LiF}<\text{NaF}<\text{KF}<\text{RbF}<\text{CsF}\]
You need to login to perform this action.
You will be redirected in
3 sec
